Objective Chemistry for NEET 2020 [1 ed.] 9788126598335, 9788126588756

Citation preview

Chapter at a Glance with flowcharts and tables for quick revision of all important concepts.

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Solved Examples placed in the same order as the topics in the feature Chapter at a Glance.

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Practice Exercises based on latest NEET pattern in big number, arranged topic-wise.

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Previous Years' NEET Questions (2007-2017) with solutions covered chapter-wise.

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Hints and Explanations for tricky and difficult questions.

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Three Mock Tests to develop examination temperament.

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Solved NEET 2018 Chemistry paper

Chapter at a Glance, an attractive feature with concepts summarized in a systematic flow, has been incorporated to enhance the quick-learning. The enormous number of practice exercises based on latest NEET pattern has been extensively developed based on NCERT Chemistry books of Class 11 and 12. These are arranged topic-wise under two levels of difficulty and include Previous Years’ NEET Questions. The designing of questions is strictly in accordance with the topic that aids students in approaching the corresponding problems of the topic under study. Apart from the Answer Key, the distinctive feature, Hints and Explanations of tricky and difficult questions have also been included to simplify learning. At the end of the book, three Mock tests have been provided to give the students an experience of attempting the actual examination. About Maestro Series l Idea: World-class content developed by “Master teachers” adapted to the needs of medical aspirants. l

Process: Developed by best-in class teachers in the field after in depth study of syllabus and relative weightage of topics.

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Approach: Designing to meet the objective of: Knowledge: Conceptual strength provided by authoritative yet precise content as per syllabus requirement. v

Comprehension: Supported with illustrations and effective pedagogy. v

Application: Assessment as per medical entrance through thought-provoking exercises as per the pattern of NEET (previously AIPMT) papers. v

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ISBN 978-81-265-9833-5

Objective

CHEMISTRY

NEET

SINGH MEHTA ASOKAN

Features of the book include: Focused on NEET examination Chapters summarized in a systematic flow for quick revision Selected topic-wise Practice Questions to cover all important concepts Previous Years' NEET Questions (2007-2017) with solutions covered chapter-wise Three NEET Mock Tests for self-evaluation Includes solved NEET 2018 Chemistry paper

9 788126 598335

2020

FOR

NEET

Visit us at https://www.wileyindia.com/resources/

Objective CHEMISTRY

l

ABOUT THE BOOK The book Objective Chemistry for NEET is an endeavour to help students to prepare for NEET (National Eligibility cum Entrance Test) and other medical entrance examinations. NEET-UG has replaced All India Pre-Medical Test (AIPMT), and all individual MBBS and BDS exams conducted by states or colleges themselves for admissions into medical and dental courses.

FOR

HIGHLIGHTS OF THE BOOK

K. Singh Vipul Mehta Asokan K. K.

2020

Objective

CHEMISTRY FOR

NEET

K. Singh Vipul Mehta Asokan K. K.

Objective Chemistry for NEET Copyright © 2019 by Wiley India Pvt. Ltd., 4436/7, Ansari Road, Daryaganj, New Delhi-110002. Cover Image: Fabrizio Denna/Getty Images All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or scanning without the written permission of the publisher. Limits of Liability: While the publisher and the author have used their best efforts in preparing this book, Wiley and the author make no representation or warranties with respect to the accuracy or completeness of the contents of this book, and specifically disclaim any implied warranties of merchantability or fitness for any particular purpose. There are no warranties which extend beyond the descriptions contained in this paragraph. No warranty may be created or extended by sales representatives or written sales materials. Disclaimer: The contents of this book have been checked for accuracy. Since deviations cannot be precluded entirely, Wiley or its author cannot guarantee full agreement. As the book is intended for educational purpose, Wiley or its author shall not be responsible for any errors, omissions or damages arising out of the use of the information contained in the book. This publication is designed to provide accurate and authoritative information with regard to the subject matter covered. It is sold on the understanding that the Publisher is not engaged in rendering professional services. Other Wiley Editorial Offices: John Wiley & Sons, Inc. 111 River Street, Hoboken, NJ 07030, USA Wiley-VCH Verlag GmbH, Pappellaee 3, D-69469 Weinheim, Germany John Wiley & Sons Australia Ltd, 42 McDougall Street, Milton, Queensland 4064, Australia John Wiley & Sons (Asia) Pte Ltd, 1 Fusionpolis Walk #07-01 Solaris, South Tower Singapore 138628 John Wiley & Sons Canada Ltd, 22 Worcester Road, Etobicoke, Ontario, Canada, M9W 1L1 First Edition: 2018 Reprint: 2019 ISBN: 978-81-265-9833-5 ISBN: 978-81-265-8875-6 (ebk) www.wileyindia.com Printed at:

Preface

Chemistry plays a central role among all sciences as well as in our day-to-day life in general. The general chemistry course of CBSE for classes 11 and 12 is designed to give the student an appreciation of this fact and provide a sound foundation of basic facts and concepts. The chemistry syllabus for NEET (earlier AIPMT) is closely aligned with this and the entrance examination aims to test the competency of the medical aspirants in all three branches of Chemistry – Physical, Inorganic and Organic. The current pattern of examination aims to foster problem-solving skills in students based on the strength or core concepts and their application. The book Objective Chemistry for NEET is designed to prepare the medical aspirants for NEET and other medical entrance examination along these lines. The book is aimed at students who want to ace their subjects by rigorous practice. The coverage of the book is in accordance with the latest NEET syllabus and covers all question-types asked in the target examination. It is the endeavor of the authors of the book to offer it as a must-have resource for these students. It is our advice to all the students going through this book is to solve the problems in a timed manner. Here is a brief overview of this book: 1.  Chapter at a Glance: A summarized, relevant, lucid and up to date coverage of the concepts at the beginning of every chapter as per the NEET curriculum. These can be used as the eleventh hour revision notes. 2.  Solved Examples: Ample in number, these are placed in the same order as the topics in the feature Chapter at a Glance and give students first-hand experience of how to apply concepts to problem-solving. 3.  Practice Exercises: This section examines the skills of the students to attempt objective problems as per the latest NEET pattern. They are further divided into two levels. LEVEL I is the basic level, and LEVEL II is the advanced level. The problems are arranged section-wise to build proper context with the concept. 4.  Previous Years’ NEET Questions: These are solved objective questions from previous years’ NEET (AIPMT) papers. This section will help the student to get acquainted with the actual NEET questions and to test the skills learnt in chapter. 5.  Hints and Explanations: These are provided for most of the questions in Practice Exercises and Previous Years’ NEET Questions. Special care has been taken to answer questions in the simplest and shortest possible way so that the NEET aspirants develop an attitude for solving problems in the least time. 6.  Mock Tests: This section, is provided at the end of this book, helps in a holistic revision of all chapters of Chemistry after the completion of the syllabus. These tests are designed after the analysis of the past NEET papers, and the difficulty level of the questions in these tests will be very close to the actual NEET test.  We hope that the students would benefit from the experience of using the book. It would be the privilege and reward for the authors if the book will help any student improve his/ her score even by a small percentage. We will be grateful for all the readers (particularly students) for their evaluation, of this book. Their comments, criticism and suggestions will be incorporated in the subsequent editions. Dr. K. Singh [email protected] Vipul Mehta [email protected] Asokan K.K. [email protected]

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About the Authors

Dr. K. Singh Dr. K. Singh has done his post-graduation in Chemistry with special paper in Organic Chemistry and his Ph.D in Water Pollution from Veer Kunwar Singh University, Ara, Bihar. He is an accomplished faculty in Chemistry, imparting quality education to mainly engineering and also medical aspirants across north India. He has a vast teaching experience of more than 21 years and has taught more than sixty thousand students. He has churned out thousands of students successful in competitive examinations, many of whom attained top hundred ranks and studied in institutes of their choice. Many of these students are doing exceedingly well in the field of research and in corporate sector. Dr. Singh is one of the very few faculties from Bihar who had the privilege of teaching at Kota in early days. He then went on to establish the first Coaching Institute at Patna, Bihar, which is based on Kota pattern. As a leading educationist of Patna, he very successfully stopped the migration of the students from Bihar to Kota. Vipul Mehta With over 10 years of experience in the education industry, Vipul Mehta, an IIT alumnus has excelled in many roles in the past decade, from being a teacher, a mentor to an entrepreneur. He has taught over 20,000 students for various Indian competitive examinations and international Olympiads and has produced numerous all-India toppers and Olympiad medalists in Chemistry. Vipul has addressed over 1000 educational seminars and has also written a series of articles for leading newspapers to help students excel in competitive examination. In his current role, he is working on his venture Edhola (www.edhola.com), an initiative that provides a one-stop solution to students from class 8 to 12 for education abroad. Asokan K. K. Mr Asokan K. K. started his career as a PGT for Chemistry in Kendriya Vidyalaya and has served a long tenure in different capacities in Jawahar Navodaya Vidyalaya across Kerala. For the past 15 years teaching, he has concentrated on teaching Inorganic Chemistry for Engineering and Medical entrance examinations in Brilliant, Pala. He is also the Academic Facilitator for NIOS (National Institute of Open Schooling) in Kochi Centre. His earlier publications include Chemistry books for Vidyarthimithram Publications, Kottayam.

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Directions to Use the Book

 1. To start with, read the chapter thoroughly from your text books, notes and NCERT books. You must have a comprehensive knowledge of the entire topic whose questions you are going to crack.  2. Then, go to the corresponding chapters in this book and quickly go through Chapter at a Glance provided at the beginning of each chapter.  3. Now approach the exhaustive topic-wise question bank, to self asses your understanding of the topic. Before you start practicing the questions, set an alarm of 40 minutes in your watch.  4. It is recommended that you to aim at attempting at least 45 questions in a single sitting.  5. As the alarm rings, stop solving the paper.  6. Check answers with those given in the Answer Key.  7. If you have doubts, then check the explanations given at the end of the chapter.  8. If you are not satisfied, then ask the doubts from your respective teachers at school or institute.  9. If you do not have access to a trainer, you can email the undersigned. 10. After completing your complete syllabus, you can start solving previous years’ papers to understand the weightage of each chapter from the examination point of view. 11. As a last step in your preparation, start with Mock Tests based on the pattern of NEET and check your answer from the provided Answer Key list. ALL THE BEST and Enjoy Solving the MCQs!

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CONTENTS

Prefaceiii About the Authors v Directions to Use the Book vii

PHYSICAL CHEMISTRY 1.  Some Basic Concepts of Chemistry Chapter at a Glance Solved Examples Practice Exercises Level I Laws of Chemical Combinations Calculations of Moles, Empirical and Molecular Formula  Stoichiometry, Limiting Reagent, and POAC Concentration Terms  Equivalent Weight Concept  Level II Calculations of Moles, Empirical and Molecular Formula  Stoichiometry, Limiting Reagent, and POAC Concentration Terms  Equivalent Weight Concept  Previous Years’ NEET Questions Answer Key Hints and Explanations

2.  Structure of Atom Chapter at a Glance Solved Examples Practice Exercises Level I Early Models, Electromagnetic Waves and Planck’s Theory Bohr’s Model, Calculation of Radius, Velocity and Energy  Hydrogen Spectrum and Rydberg Equation Heisenberg Uncertainty Principle and de Broglie Equation Photoelectric Effect Schrödinger’s Wave Equation Aufbau Principle, Hund’s Rule and Pauli’s Exclusion Principle Level II Early Models, Electromagnetic Waves, Planck’s Theory Bohr’s Model, Calculation of Radius, Velocity and Energy  Hydrogen Spectrum and Rydberg Equation

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3 3 8 13 13 13 13 13 14 15 15 15 16 16 17 17 19 19

29 29 34 36 36 36 37 37 38 38 39 39 40 40 40 41

Heisenberg Uncertainty Principle and de Broglie Equation Photoelectric Effect Schrödinger’s Wave Equation Aufbau Principle, Hund’s Rule and Pauli’s Exclusion Principle Previous Years’ NEET Questions Answer Key Hints and Explanations

3.  States of Matter Chapter at a Glance   Gaseous State   Liquid State Solved Examples Practice Exercises Level I Gas Laws Ideal Gas Law Dalton’s Law Graham’s Law of Effusion and Diffusion Kinetic Theory of Gases and Maxwell Boltzmann Distribution Curve Compressibility Factor (Z), Liquefaction of Gas, van der Waals Equation  Liquid State Level II  Gas Laws Ideal Gas Law Dalton’s Law Graham’s Law of Effusion and Diffusion Kinetic Theory of Gases and Maxwell Boltzmann Distribution Curve Compressibility Factor (Z), Liquefaction of Gas, van der Waals Equation  Liquid State Previous Years’ NEET Questions Answer Key Hints and Explanations

4. Thermodynamics Chapter at a Glance Solved Examples Practice Exercises Level I Heat, Work, Internal Energy, Enthalpy, Heat Capacity and First Law of Thermodynamics Entropy and Second Law of Thermodynamics Gibbs Free Energy and Spontaneity of a Reaction

42 42 42 43 43 45 46

55 55 55 60 61 66 66 66 66 67 68 68 69 70 70 70 70 71 71 72 73 74 74 75 76

83 83 88 91 91 91 92 92

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CONTENTS

Thermochemistry, Hess’ Law and Kirchoff ’s Equation 92 Level II  94 Heat, Work, Internal Energy, Enthalpy, Heat Capacity and First Law of Thermodynamics 94 Entropy and Second Law of Thermodynamics 95 Gibbs Free Energy and Spontaneity of a Reaction 96 Thermochemistry, Hess’ Law and Kirchoff ’s Equation 97 Previous Years’ NEET Questions 98 Answer Key 101 Hints and Explanations 101

5.  Chemical Equilibrium Chapter at a Glance Solved Examples Practice Exercises Level I Law of Mass Action  Relation between Kp and KC, Characteristics of Equilibrium Constant K Reaction Quotient (Q), Extent of Reaction Calculation of K in Different Reactions and Degree of Dissociation Thermodynamics of Equilibrium Le Chatelier’s Principle Level II  Law of Mass Action  Relation between Kp and KC and Characteristics of Equilibrium Constant K Reaction Quotient (Q) and Extent of Reaction Calculation of K in Different Reactions and Degree of Dissociation Thermodynamics of Equilibrium Le Chatelier’s Principle Previous Years’ NEET Questions Answer Key  Hints and Explanations

6.  Ionic Equilibrium Chapter at a Glance Solved Examples Practice Exercises  Level I Strength of Acids/Bases and Conjugate Acid-Base Pairs pH for Weak/Strong (Monobasic/Monoacidic) Acids/Bases, Kw , Common Ion Effect and Mixtures of Acids and Bases Salt Hydrolysis  Buffer Solutions  Solubility and Solubility Product Level II Strength of Acids/Bases and Conjugate Acid-Base Pairs

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111 111 115 118 118 118 119 120 120 121 121 122 122 122 123 123 125 125 127 128 129

137 137 140 143 143 143

143 144 145 146 146 146

pH for Weak/Strong (Monobasic/Monoacidic) Acids/Bases, Kw , Common Ion Effect and Mixtures of Acids and Bases Salt Hydrolysis  Buffer Solutions  Solubility and Solubility Product Previous Years’ NEET Questions Answer Key Hints and Explanations

7.  Redox Reactions Chapter at a Glance Solved Examples Practice Exercises Level I Oxidation Number Concept, Oxidizing and Reducing Agents Applications of Activity Series Balancing of Redox Reactions  n-Factor and Equivalent Weight Redox Titrations  Level II Oxidation Number Concept, Oxidizing and Reducing Agents Applications of Activity Series Balancing of Redox Reactions n-Factor and Equivalent Weight  Redox Titrations  Previous Years’ NEET Questions Answer Key Hints and Explanations

8.  Solid State Chapter at a Glance Solved Examples Practice Exercises  Level I Types of Unit Cell, Simplest Formula, Coordination Number, Packing Fraction Density of Unit Cell Close-Packed Structures Tetrahedral and Octahedral Voids Types of Ionic Crystals, Radius Ratio Imperfections in Solids Level II Types of Unit Cell, Simplest Formula, Coordination Number, Packing Fraction Density of Unit Cell Close-Packed Structures Tetrahedral and Octahedral Voids Types of Ionic Crystals, Radius Ratio Bragg’s Law Imperfections in Solids Electrical Properties Previous Years’ NEET Questions

146 147 147 148 148 150 151

159 159 165 169 169 169 170 170 170 171 172 172 173 173 173 174 174 175 175

183 183 190 192 192 192 192 193 193 193 194 195 195 195 195 195 195 196 196 197 197

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CONTENTS Answer Key  Hints and Explanations

9. Solutions Chapter at a Glance Solved Examples Practice Exercises Level I Concentration Terms Vapor Pressure, Raoult’s Law and Relative Lowering of Vapor Pressure Elevation in Boiling Point and Depression in Freezing Point Osmosis, Reverse Osmosis and Osmotic Pressure van’t Hoff Factor and Abnormal Colligative Properties Level II Concentration Terms Vapor Pressure, Raoult’s Law and Relative Lowering of Vapor Pressure Elevation in Boiling Point and Depression in Freezing Point Osmosis, Reverse Osmosis and Osmotic Pressure van’t Hoff Factor and Abnormal Colligative Properties Previous Years’ NEET Questions Answer Key  Hints and Explanations

10. Electrochemistry Chapter at a Glance Solved Examples Practice Exercises Level I Construction of Cells and Reference Electrodes Electrochemical Series, Calculation of Eo for the Cell and Feasibility of Cell Nernst Equation and Concentration Cells Electrolysis and Faraday’s Laws of Electrolysis Conductance, Kohlrausch’s Law, Batteries and Corrosion Level II  Construction of Cells and Reference Electrodes Electrochemical Series, Calculation of Eo for the Cell and Feasibility of Cell Nernst Equation and Concentration Cells Electrolysis and Faraday’s Laws of Electrolysis Conductance, Kohlrausch’s Law, Batteries and Corrosion Previous Years’ NEET Questions Answer Key Hints and Explanations

11.  Chemical Kinetics Chapter at a Glance Solved Examples

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198 199

205 205 212 216 216 216 217 218 218 218 219 219 219 220 220 221 221 223 223

231 231 236 240 240 240 241 241 242 243 244 244 244 245 246 247 249 252 252

261 261 265

Practice Exercises Level I Rate of the Reaction Order and Molecularity Integrated Rate Law and Half-Life (t1/2) Methods to Find Order of the Reaction Replacement of Concentration Terms by Other Variables Arrhenius Equation and Collision Theory Mechanism of Complex Reactions Level II Rate of the Reaction Order and Molecularity Integrated Rate Law and Half-Life (t1/2) Methods to Find Order of the Reaction Replacement of Concentration Terms by Other Variables Arrhenius Equation and Collision Theory Mechanism of Complex Reactions Previous Years’ NEET Questions Answer Key Hints and Explanations

12.  Surface Chemistry Chapter at a Glance Solved Examples Practice Exercises Level I Adsorption Adsorption Isotherms Catalyst Colloids, Coagulation, Flocculation and Emulsion Level II Adsorption Adsorption Isotherms Catalyst Colloids, Coagulation, Flocculation and Emulsion Previous Years’ NEET Questions Answer Key Hints and Explanations

xi 268 268 268 269 269 270 271 271 271 272 272 272 272 273 273 273 274 274 277 278

287 287 293 295 295 295 296 296 297 299 299 300 300 301 303 303 304

INORGANIC CHEMISTRY 13. Classification of Elements and Periodicity in Properties Chapter at a Glance Solved Examples Practice Exercises Level I Development of Periodic Table and Modern Periodic Table Shielding Effect and Effective Nuclear Charge Atomic and Ionic Radius Ionization Enthalpy

309 309 311 315 315 315 316 316 316

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xii

CONTENTS

Electron Gain Enthalpy and Electron Affinity Electronegativity Level II Development of Periodic Table and Modern Periodic Table Shielding Effect and Effective Nuclear Charge Atomic and Ionic Radius Ionization Enthalpy Electron Gain Enthalpy and Electron Affinity Electronegativity Previous Years’ NEET Questions Answer Key Hints and Explanations 

14. Chemical Bonding and Molecular Structure Chapter at a Glance Solved Examples Practice Exercises  Level I Kossel–Lewis Approach to Chemical Bonding Ionic or Electrovalent Bond Bond Parameters The Valence Shell Electron Pair Repulsion (VSEPR) Theory Valence Bond Theory Hybridization Molecular Orbital Theory Bonding in Some Homonuclear Diatomic Molecules Hydrogen Bonding Level II Kossel–Lewis Approach to Chemical Bonding Ionic or Electrovalent Bond Bond Parameters The Valence Shell Electron Pair Repulsion (VSEPR) Theory Valence Bond Theory Hybridization Molecular Orbital Theory Hydrogen Bonding Previous Years’ NEET Questions Answer Key Hints and Explanations

15. Hydrogen Chapter at a Glance Solved Examples Practice Exercises Level I Dihydrogen Hydride Water Hydrogen Peroxide Heavy Water and Hard Water Level II Dihydrogen

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317 317 318 318 318 318 319 319 320 320 321 322

325 325 332 337 337 337 337 338 338 339 339 340 340 340 340 340 340 341 341 342 342 343 343 343 346 347

357 357 362 366 366 366 367 367 368 369 369 369

Hydride Water Hydrogen Peroxide Heavy Water and Hard Water Previous Years’ NEET Questions Answer Key Hints and Explanations 

16. The s-Block Elements Chapter at a Glance   Group I (Alkali Metals)   Group II (Alkaline Earth Metals) Solved Examples Practice Exercises  Level I General Trends in Physical and Chemical Properties of Elements Anomalous Properties of the First Element of Each Group and Diagonal Relationships Preparation and Properties of Some Important Compounds of Sodium Compounds of Calcium  Biological Significance of Na, K, Mg and Ca Level II General Trends in Physical and Chemical Properties of Elements Anomalous Properties of the First Element of Each Group and Diagonal Relationships Preparation and Properties of Some Important Compounds of Sodium Compounds of Calcium  Biological Significance of Na, K, Mg and Ca Previous Years’ NEET Questions Answer Key Hints and Explanations

17. The p-Block Elements Chapter at a Glance   Group 13 Elements    Group 14 Elements   Group 15 Elements   Group 16 Elements – Chalcogens   Group 17 Elements – Halogens   Group 18 Elements – Noble Gases Solved Examples Practice Exercises Level I Group 13 Group 14 Group 15 Group 16 Group 17 Group 18 Level II Group 13 Group 14

370 371 371 371 372 373 373

377 377 377 379 382 385 385 385 386 386 387 387 387 387 388 388 388 389 389 390 391

395 395 395 400 404 410 416 420 422 428 428 428 430 431 433 433 434 435 435 435

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CONTENTS Group 15 Group 16 Group 17 Group 18 Previous Years’ NEET Questions  Answer Key Hints and Explanations

18. General Principles and Processes of Isolation of Elements Chapter at a Glance Solved Examples Practice Exercises Level I Occurrence of Metals Concentration of Ore Extraction of Crude Metal from Concentrated Ore Thermodynamic Principles of Metallurgy Electrochemical Principles of Metallurgy Refining of Metals Alloys and Amalgams Level II Occurrence of Metals Concentration of Ore Extraction of Crude Metal from Concentrated Ore Thermodynamic Principles of Metallurgy Electrochemical Principles of Metallurgy Refining of Metals Alloys and Amalgams Previous Years’ NEET Questions Answer Key Hints and Explanations

19. The d- and f-Block Elements Chapter at a Glance Solved Examples Practice Exercises Level I General Properties of the Transition Elements (d-Block) Some Important Compounds of Transition Elements General Properties of the Inner Transition Elements (f-Block) Some Applications of d- and f-Block Elements Level II General Properties of the Transition Elements (d-Block) Some Important Compounds of Transition Elements General Properties of the Inner Transition Elements (f-Block) Some Applications of d- and f-Block Elements Previous Years’ NEET Questions Answer Key Hints and Explanations

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436 437 438 438 439 441 442

453 453 461 464 464 464 464 465 465 466 466 466 467 467 467 467 468 468 469 469 469 471 471

475 475 481 486 486 486 487 487 488 488 488 489 490 490 490 492 492

20.  Coordination Compounds Chapter at a Glance Solved Examples Practice Exercises Level I Werner’s Theory of Coordination Compounds Definition of Some Important Terms Pertaining to Coordination Compounds Nomenclature of Coordination Compounds Isomerism in Coordination Compounds Bonding in Coordination Compounds Bonding in Metal Carbonyls Importance and Applications of Coordination Compounds Level II Werner’s Theory of Coordination Compounds Definition of Some Important Terms Pertaining to Coordination Compounds Nomenclature of Coordination Compounds Isomerism in Coordination Compounds Bonding in Coordination Compounds Bonding in Metal Carbonyls Stability of Coordination Compounds Previous Years’ NEET Questions Answer Key Hints and Explanations 

21.  Environmental Chemistry Chapter at a Glance Solved Examples Practice Exercises Level I Atmospheric Pollution Water Pollution Soil Pollution Strategies to Control Environmental Pollution and Green Chemistry Previous Years’ NEET Questions Answer Key Hints and Explanations

xiii

497 497 507 513 513 513 513 514 515 515 516 517 517 517 517 518 518 518 519 520 520 522 523

533 533 538 541 541 541 541 542 542 542 543 543

ORGANIC CHEMISTRY 22. Organic Chemistry – Some Basic Principles and Techniques Chapter at a Glance   General Concepts of Organic Chemistry  Purification and Characterization of Organic Compounds   Qualitative Analysis of Organic Compounds   Quantitative Estimation of Elements Solved Examples Practice Exercises Level I Hybridization, Shapes of Simple Molecules and Nomenclature

547 547 547 556 557 559 560 567 567 567

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xiv

CONTENTS

Isomerism and Dipole Moment Reaction Intermediates Electrophiles, Nucleophiles, Inductive Effect, Resonance and Hyperconjugation Purification and Characterization of Organic Compounds Qualitative Analysis of Organic Compounds Quantitative Estimation of Elements Level II  Reaction Intermediates Isomerism and Dipole Moment Electrophiles, Nucleophiles, Inductive Effect, Resonance and Hyperconjugation Purification and Characterization of Organic Compounds Qualitative Analysis of Organic Compounds  Quantitative Estimation of Elements Previous Years’ NEET Questions Answer Key Hints and Explanations

23.  Aliphatic Hydrocarbons

568 569 569 571 571 571 571 571 572 572 572 573 573 573 576 576

585

Chapter at a Glance  Alkanes  Alkenes  Alkynes Solved Examples Practice Exercises Level I Alkanes Nomenclature and Isomerism Methods of Preparation Conformations Physical Properties Chemical Reactions Alkenes Nomenclature and Structure Physical Properties Methods of Preparation Chemical Reactions Alkynes

585 585 588 590 592 597 597

Chemical Reactions Level II Alkanes Chemical Reactions Alkenes Methods of Preparation Chemical Reactions Alkynes Methods of Preparation Chemical Reactions Previous Years’ NEET Questions Answer Key Hints and Explanations

601 601 601 602 602 602 603 603 603 604 607 607

24.  Aromatic Hydrocarbons

615

Chapter at a Glance Solved Examples

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597 597 598 598 598 598 599 599 599

615 620

Practice Exercises Level I Nomenclature  Structure and Physical Properties Chemical Reactions Level II  Structure and Physical Properties Chemical Reactions Previous Years’ NEET Questions Answer Key Hints and Explanations

25. Organic Compounds Containing Halogens Chapter at a Glance   Alkyl Halides   Aryl Halides Solved Examples Practice Exercises Level I Alkyl Halides Aryl Halides Level II  Alkyl Halides Aryl Halides Previous Years’ NEET Questions Answer Key Hints and Explanations

26.  Alcohols, Phenols and Ethers

625 625 625 625 626 627 627 628 630 631 632

637 637 637 642 646 650 650 650 652 653 653 655 656 658 659

665

Chapter at a Glance  Alcohols  Phenols  Ethers Solved Examples Practice Exercises Level I Alcohols Phenols Ethers Level II Alcohols Phenols Ethers Previous Years’ NEET Questions Answer Key Hints and Explanations

665 665 670 675 676 682 682 682 684 686 686 686 688 689 690 691 692

27.  Aldehydes and Ketones

699

Chapter at a Glance Solved Examples Practice Exercises Level I Nomenclature Structure and Isomerism Methods of Preparation

699 703 708 708 708 708 709

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CONTENTS Physical Properties Chemical Reactions Tests to Distinguish Between Aldehydes and Ketones Level II Methods of Preparation Chemical Reactions Previous Years’ NEET Questions Answer Key Hints and Explanations

709 709 710 711 711 711 713 716 716

28.  Carboxylic Acids and its Derivatives

725

Chapter at a Glance Solved Examples Practice Exercises Level I Nomenclature Structure and Physical Properties of Carboxylic Acids  Acidic Character of Carboxylic Acids  Methods of Preparation of Carboxylic Acids Chemical Reactions of Carboxylic Acids  Chemical Reactions of Carboxylic Acid Derivatives Level II Methods of Preparation of Carboxylic Acids Chemical Reactions of Carboxylic Acids  Physical Properties and Chemical Reactions of Carboxylic Acids Derivatives Previous Years’ NEET Questions Answer Key Hints and Explanations

725 730 733 733 733

29. Organic Compounds Containing Nitrogen Chapter at a Glance Solved Examples Practice Exercises Level I Structure Physical Properties  Methods of Preparation  Basicity  Chemical Reactions  Level II Methods of Preparation Chemical Reactions  Previous Years’ NEET Questions Answer Key Hints and Explanations

30. Polymers Chapter at a Glance Solved Examples Practice Exercises Level I

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734 734 735 736 736 737 737 737 738 739 741 741

749 749 755 763 763 763 764 764 765 766 766 766 767 768 771 772

779 779 786 788 788

Classification of Polymers  Types of Polymerization Reactions  Rubbers  Molecular Mass of Polymers Biodegradable Polymers Uses of Polymers Level II  Classification of Polymers Types of Polymerization Reactions  Rubbers  Molecular Mass of Polymers Biodegradable Polymers Uses of Polymers Previous Years’ NEET Questions Answer Key Hints and Explanations

31. Biomolecules Chapter at a Glance  Carbohydrates  Proteins   Nucleic Acids  Vitamins Solved Examples Practice Exercises Level I Carbohydrates Proteins Nucleic Acids Vitamins Level II  Carbohydrates Proteins Nucleic Acids Vitamins Previous Years’ NEET Questions Answer Key Hints and Explanations

32.  Chemistry in Everyday Life Chapter at a Glance Solved Examples Practice Exercises Level I Medicines Chemicals in Food Cleansing Agents Previous Years’ NEET Questions Answer Key Hints and Explanations

xv 788 788 789 789 789 789 789 789 790 790 791 791 791 791 793 793

799 799 799 803 806 806 807 809 809 809 810 811 811 811 811 812 813 813 814 815 815

821 821 824 826 826 826 827 828 828 829 829

Mock Tests Mock Tests I Mock Test II Mock Test III Answer Key NEET 2018 Paper

831 835 839 843 P1

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PHYSICAL CHEMISTRY

Chapter 1_Some Basic Concepts of Chemistry.indd 1

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3

1

Some Basic Concepts of Chemistry

Chapter at a Glance 1. Matter: It is anything that occupies space. (a) (b)

It exists in three physical states viz. (i) Gas (ii) Liquid, and (iii) Solid. Chemically it can be classified into: (i) Elements, (ii) Compounds, and (iii) Mixture Matter

Physical classification

Gas

Liquid

Chemical classification

Solid

Pure substance

Elements

Mixture

Compounds

Organic

Homogenous

Heterogenous

Inorganic

2. Elements These cannot be decomposed further into simpler products by any physical and chemical means. Majority of them are metals, few are metalloids and non-metals. Oxygen is the most abundant element, whereas Si, Al, Fe are the second most, third most and fourth most abundant element of earth. They can be further classified into metals and non-metals. 3. Compounds Two or more different elements in a fixed ratio or proportion of their mass are known as compounds. 4. Mixture It consists of two or more substances in any proportion in which compounds do not lose their identity. They may be homogeneous or heterogeneous. Majority of heterogeneous mixture can be easily separated by physical methods. Homogeneous composition mixtures are difficult to separate by physical methods, (e.g., alloys). 5. Precision and Accuracy (a) Precision refers to closeness of various measurements for the same quantity to each other and to the average. (b) Accuracy refers to an agreement of a particular value to the true value of the result.

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OBJECTIVE CHEMISTRY FOR NEET

6. Significant Figures It is the total number of digits in a number that results from a measurement which is known with certainty. (a) The rule for determining the number of significant figures is as follows: (i) All non-zero digits are significant. (ii) Zero proceeding to first non-zero digit is not significant. (iii) Zero between two non-zero digits is significant. (iv) Zero at the end or right of a number is significant provided it is on the right side at the decimal point. (v)  Some numbers are called exact numbers and have an infinite number of significant figures. For example, 12 inches in a foot. (vi)  If number is written in scientific notation, all digits are significant. For example 5.026 × 1023 has 4 significant figures. (b) Addition and subtraction: The result cannot have more digits to the right of the decimal point than either of the original numbers. For example, 16.24 + 13.0 + 1.221 = 30.461. But 13.0 has only one digit after decimal point hence result will be 30.4 or around to 30.5. (c) Multiplication and division: The number of significant figures in the answer should not be greater than the number of significant figures in the least precise measurement. For example, 2.4 × 1.21 = 2.904, as 2.4 has two significant figures, therefore, the result should be 2.9. 7. Measurement of Properties Properties of matter like length, area, volume, etc., are quantitative in nature and they must be suffixed by certain units. (a) The international system of units (SI) is a well-accepted system to express. Physical quantity

Symbol

Length Mass Time Electric current Temperature Amount of substance Luminous intensity

l m t I T n lv

Name of unit

Symbol

Meter Kilogram Second Ampere Kelvin Mole Candela

m kg s A K mol cd

(b) The SI system allows the use of prefix to indicate the multiples or sub-multiples of a unit. Multiple 10-24

Prefix yocto

Symbol y

10-21

zepta

z

10-18

atto

a

10-15

femto

F

10-12

pico

P

-9

nano

n

-6

micro

-3

milli

µ m

-2

centi

c

-1

deci

d

1

deca

da

10 10 10 10 10 10

(Continued)

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Some Basic Concepts of Chemistry

Multiple 10

Prefix hecto

Symbol h

103

2

kilo

k

6

mega

M

9

giga

G

12

tera

T

15

peta

P

18

exa

E

21

zeta

Z

24

yotta

Y

10 10 10 10 10 10 10

8. Laws of Chemical Combination These were basic laws based on the experimental observations of many scientists in the 18th and early 19th centuries that governed the combination of atoms in the formation of compounds. (a) Law of conservation of mass: Given by Antoine Lavoisier who conducted several experiments on combustion. It states that no detectable gain or loss of mass occurs in chemical reactions, that is, mass is conserved. (b) Law of definite proportions: Given by Joseph Proust, it states that in a given chemical compound, the elements are always combined in the same proportions by mass. It is also known as law of definite composition. (c) Law of multiple proportions: Given by Dalton, it states that whenever two elements form more than one compound, the different masses of one element that combine with the same mass of the other element are in the ratio of small whole numbers. (d) Gay–Lussac’s law of gaseous volumes: It states that when measured at the same temperature and pressure, the ratios of the volumes of reacting gases are small whole numbers. (e) Avogadro’s law: It states that equal volumes of different gases at the same temperature and pressure contain the same number of molecules. This constant number of molecules was later determined as Avogardro’s constant and defined as mole. The law could be stated as: one mole of any gas contains 6.023 × 1023 molecules and it occupies a volume of 22.4 L at STP. 9. Dalton’s Atomic Theory (a) (b) (c) (d)

Every element is composed of small indivisible particles called atoms. Atoms of the same element are alike, but different elements have different types of atoms. Atoms of different elements can combine in simple ratio to form compounds. The relative number and kind of atoms are always the same in a given compound.

10. Atomic and Molecular Masses (a) Atomic masses: It is the ratio of mass of one atom of an element to 1/12th part mass of carbon = 12 (12C), which is an isotope of carbon. The unit in which atomic mass measurements are reported is the atomic mass unit (amu). One atomic mass unit is equal to 1/12th of the mass of one 12C atom. Thus, Atomic mass=

Mass of one atom of an element Mass of one atom of 12 C

´ 12

By definition, the mass of a single atom of the 12C isotope is exactly 12 atomic mass units, or 12 amu. 1 amu = 1.66 × 10−24 g



So, the actual mass of an element is (amu) × 1.66 × 10−24 g. Nowadays, amu has been replaced by u which stands for unified mass.

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OBJECTIVE CHEMISTRY FOR NEET

(b) Average atomic mass: Most of the elements exist as isotopes and since there is a large difference in the natural abundance of these isotopes, the average mass of an element is a weighted average of the masses of the different 35 35 isotopes. For example, chlorine has two isotopes, 17 Cl(75%) and 17 Cl(25%) Average mass = 35 ×



75 25 + 37 × = 35.5 100 100

(c) Atomic weight: The weighted average of all the naturally occurring isotopes of an atom is known as the atomic weight of the element. (i) Dulong-Petit law: It is applicable to metals and states that Atomic weight × specific heat = 6.4 where specific heat is in g cal−1. (d) Molecular mass: It is defined as the sum of the atomic masses of the elements in the molecule. For example, the molecular mass of water (H2O) is determined from the atomic weights of hydrogen and oxygen as 18.02 amu. Molecular mass = 2 × vapor density (e) Formula mass: It is the sum of the atomic masses of the elements in the formula. It is used for substances which do not contain discrete molecules as their constituent units. For example, in case of sodium chloride, formula mass is used instead of molecular mass as in the solid state it does not exist as a single entity; instead positive and negative entities are arranged in three dimensional structures. So, it has a formula (NaCl) mass of 58.44 u, determined by adding the atomic weights of sodium and chlorine. For more complex ionic compounds, the formula mass is calculated the same way. For example, the formula mass of Cu(NO3)2, is 187.52 u. 11. Mole Concept and Molar Masses (a) Mole: A mole of any substance contains the same number of particles as the number of atoms in exactly 12 g of the 12C isotope of carbon. It is the SI unit for the amount of substance. The number of particles in a mole is known as Avogadro’s constant (previously known as Avogadro’s number) and is equal to 6.023 × 1023. (b) Gram atomic mass: A mole of atoms of any element has a mass in grams equal to the atomic weight of the element. It is the absolute mass in grams of 6.023 × 1023 atoms of an element. (c) Gram molecular mass: The mass of a mole of a substance is often called the molar mass. It is the absolute mass expressed in grams of 6.023 × 1023 molecules of a substance. Thus, Number of moles=



Given mass of substance in grams Molar Mass

(d) Molecular weight: The molecular weight of a compound is the sum of the atomic weights of the atoms in the formula of the compound. (e) Gram molar volume: It is the volume occupied by one gram molecular mass of any gas at STP. 12. Percentage Composition: In general, the percentage composition of an element in a compound can be calculated using the relation

Mass % of an element =

Mass of the element in the compound he compound Molar mass of th

´ 100

13. Empirical Mass and Molecular Mass (a) Empirical formula: It represents the simplest relative whole number of atoms of each element present in the molecule of a substance. (b) Molecular formula: It represents the actual number of atoms of each element present in one molecule of a substance.

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Some Basic Concepts of Chemistry

Molecular formula = n × Empirical formula Molecular mass n= Empricial formula mass 14. Equivalent Weight (Eq. wt.) (a) Eq.wt.of an element =

Atomic weight of an element

(b) Eq.wt.of a compound =

Valence of an elementt Molecular mass of compound Total charge on ionicc part

(c) Eq.wt.of an element in a redox reaction=

Molecular mass Number of eleectrons participating in the reaction

15. Stoichiometry and Stoichiometric Calculations (a)  The quantitative relationships among the reacting materials and the products formed are known as stoichiometry. (b) Stoichiometric calculations: These are based on the use of balanced chemical equation for the reaction, in which each kind of atom in the reactant and product side is balanced. When reactants are mixed in exactly the mass ratio as determined from the balanced equation, the mixture is said to be stoichiometric. The coefficients used with reactants and products to balance these are called stoichiometric coefficients. A balanced equation also mentions the physical state of reactants and products. The three main steps in stoichiometric calculations for any chemical reaction are: (i) Determine the number of moles of starting substance. (ii) Determine the mole ratio of the desired substance to the starting substance. (iii)  Calculate the desired substance in the units specified in the problem (i.e., in terms of mass, mole, volume or density). 16. Limiting Reagent (a)  In the case of a chemical reaction, if specific amounts of each reactant are mixed, the reactant that produces the least amount of product is called the limiting reagent. In other words, the amount of product formed is limited by the reactant that is completely consumed. (b)  The following sequence of steps is helpful in working limiting reagent problems. (i) Calculate the amount of product (moles or grams, as needed) formed from each reactant. (ii) Determine which reactant is limiting. (iii)  Once the limiting reactant is identified, the amount of product formed can be determined. It is the amount determined by the limiting reactant. (iv) To determine how much of the other reactant remains unreacted, we calculate the amount of the other reactant required to react with the limiting reactant, and then subtract this amount from the starting quantity of the reactant. 17. Principle of Atom Conservation (POAC) The principle states that moles of atoms of an element are conserved throughout the reaction (provided the reactants are converted to products completely). Advantages of using POAC method over stoichiometry: (a) No need to balance the equation. (b) Complete reactions are not required. 18. Methods for Expressing Concentration in Solutions Mass of A (a) Mass fraction of A = Mass of A + Mass of B

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OBJECTIVE CHEMISTRY FOR NEET

(b) Mass percent: This expresses the mass of solute per 100 g of solution. Mass of solute

% by mass(solute) =

Mass of solute + Mass of solvent Mass of solute = × 100 Mass of solvent

    

(c) Mole fraction of A (xA): =

nA nA + nB

× 100

x A + xB = 1

nA = number of moles of A = (d) Molarity (M) Molarity(M ) =

W A (given mass ) M A (molar mass )

Number of moles of solute A

Volume of solution(liter) W A × 100 M= M A × V (mL )

(e) Molality (m) =

Number of moles of solute A Mass of solvent(kg) W A × 100 M A × W (g )

m= (f ) Normality (N) =

Number of gram equivalents of solute A Volume of solution(V in L) Molar mass n W A × 100 N= Eq.wt. × V (mL )

Eq.wt.=

W A × 100 × n M A × V (mL )

N=

 where n is the number of electron participating or acidity or basicity. (g) Parts per million(ppm)=

Mass of solute A Mass of solution

× 106

ppm is used whenever concentration is very low.

(h) Percentage of an element in a compound =

Mass of an element Molecular maass of compound

× 100

Solved Examples 1. A gas container contains 35 g nitrogen gas, 64 g O2 gas and 0.112 L of CH4 at STP. What is average molecular weight of the gas mixture? (1) 40.32 g mol-1 (2) 30.44 g mol-1 (3) 38.29 g mol-1 (4) 25.06 g mol-1

Chapter 1_Some Basic Concepts of Chemistry.indd 8

Solution (2) We have

M Avg = M1x1 + M 2 x 2 + M 3 x 3 

(1)

where M1, M2, M3 is molecular weight, x1, x2, x3 is the mole fraction of components of gas mixture and MAvg is the average molecular weight.

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Some Basic Concepts of Chemistry

Number of moles can be calculated as nN 2 =



Weight of substance Mol. wt. Weight of substance 0.205 = 197 Weight of substance = 0.205 ´ 197 = 40.385 g

Moles fractions can be calculated as

xCH4

1.25 2 , x O2 = , (1.25 + 2 + 0.005) (1.25 + 2 + 0.005)

0.005 = (1.25 + 2 + 0.005)

Therefore, M Avg =

(1) 50 mol of O2 (2) 100 mol of O2 (3) 125 mol of O2 (4) 200 mol of O2 Solution (3) The reaction involved is C8H18(g ) +

25 O2(g ) → 8 CO2(g ) + 9H 2O(g ) 2



Weight of C8H18 = 0.8 × 1.425 × 103 = 1140 g



Number of moles of C8H18 =





1140 = 10 mol 114 No. of moles of O2 = ( 25/2)

No. of moles of C8H18 1 ⇒ Number of moles of O2 = 125 mol

3. 149 g of potassium chloride (KCl) is dissolved in 10 L of an aqueous solution. Determine the molarity of the solution (K = 39, Cl = 35.5) (1) 0.1 M (2) 0.05 M (3) 0.2 M (4) 0.4 M Solution (3) We know

5. There are 10 g of mixture of NaCl and NaBr. If the amount of sodium is 25% of the weight of total mixture, calculate the amount of NaCl and NaBr present in the mixture. (Given, atomic weights of Na, Cl and Br are 23, 35.5 and 80, respectively).

28 ´ 1.25 + 32 ´ 2 + 16 ´ 0.005 = 30.44 g mol -1 3.255

2. Assuming that petrol is octane (C8H18) and has density 0.8 g mL−1, 1.425 L of petrol on complete combustion will consume



We know that       Number of moles(n ) =

35 64 0.112 = 1.25; nO2 = = 2; nCH4 = = 0.005 28 32 22.4

xN2 =





Molarity =

Number of moles of KCl Volume of solution (L )

149/74.5 = = 0.2 M   10

4. The mass of BaCO3 produced when excess of CO2 is bubbled through a solution of 0.205 mol Ba(OH)2 is (1) 81 g (2) 40.5 g (3) 20.25 g (4) 162 g Solution (2) The reaction involved is Ba(OH )2 + CO 2 → BaCO3 + H 2O Mol. wt. of BaCO3 = 137 + 12 + 16 × 3 = 197 g

Chapter 1_Some Basic Concepts of Chemistry.indd 9

9

(1) (2) (3) (4)

NaCl = 1.5734 g, NaBr = 1.5734 g NaCl = 8.4266 g, NaBr = 1.5734 g NaCl = 8.4266 g, NaBr = 8.4266 g NaCl = 1.5734 g, NaBr = 8.4266 g

Solution (4) Let the weight of NaCl be a.

Therefore, weight of NaBr = (10 – a)



Number of moles of NaCl = Na in NaCl

a = Number of moles of 58.5



Therefore, weight of Na in NaCl =



Number of moles of NaBr =

a × 23 (1) 58.5

(10 - a ) = Number of moles 103

of Na in NaBr

Therefore, weight of Na in NaBr =



From Eq. (1) and Eq. (2), we get

(10 - a ) × 23 (2) 103

a (10 - a ) ´ 23 + ´ 23 = 0.25 ´ 10 Þ a = 1.5734 g 58.5 103

Weight of NaBr (10 – a) = 8.4266 g

6. The number of water molecules present in a drop of water (volume = 0.0018 mL) at room temperature is (1) 6.023 ´ 1019 (2) 1.084 × 1018 (3) 4.84 × 1017 (4) 5.023 × 1023 Solution (1) We know that Density =

Mass Volume

Weight of 0.0018 mL = 0.0018 g (as rH2O = 1 g mL-1) Weight Molecular weight 0.0018 = 1 × 10 –4 = 18

Number of moles =

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OBJECTIVE CHEMISTRY FOR NEET Solution

Therefore, number of water molecules =   6.023 ´ 10

23

´ 1 ´ 10

-4

(2) Considering the given reactions, we have

= 6.023 × 10

19

7. A sample of coal gas contains 50% H2, 30% CH4, 14% CO and 6% C2H4. 100 mL of coal gas is mixed with 150 mL of O2 and the mixture is exploded. What will be the volume and composition of mixture when cooled to original conditions? (1) (2) (3) (4)

4KClO 3 → 3KClO 4 + KCl



12KClO → 8KCl + 4KClO3 12Cl 2 + 24KOH → 12KClO + 12H 2O + 12KCl



Adding these equations, we get 12Cl 2 + 24KOH →  3KClO4 + 21KCl + 12H 2O

% by vol CO2 = 29.167%, O2 = 41.666% % by vol CO2 = 58.334%, O2 = 41.666% % by vol CO2 = 58.334%%, O2 = 83.332% % by vol CO2 = 29.167%, O2 = 83.332%

Solution



Mol. wt. of KClO 4 = 138.5 g mol-1



As 3 × 138.5 g KClO 4 is formed by 12 × 71 g of Cl 2 ,



Therefore, 100 g KClO 4 will be formed from

(2) We have, volume of H2 = 50 mL, volume of CH4 = 30 mL, volume of CO = 14 mL and volume of C2H4 = 6 mL

The reactions involved are

Initial volume Final volume Initial volume Final volume

1 H 2 (g ) + O 2 (g ) 2 50 mL 25 mL 0 0

C H 4 + 2O2 → 30 mL 60 mL 0 0 CO

+

14 mL 0

Initial volume Final volume

C 2H 4 + 3O 2 → 6 mL 18 mL 0 0

H 2O( l ) 0 25 mL

2CO2 + 2H 2O( l ) 0 0 12 mL 12 mL



Since H2O(l) can be neglected. Final mixture consists of



   O2 = [150 − (25 + 60 + 7 + 18)] = 40 mL



   CO2 = (30 + 14 + 12) = 56 mL 56 % of CO2 = × 100 = 58.334% 96 40    % of O 2 = × 100 = 41.666% 96

8. From the following reaction sequence Cl 2 + 2KOH → KCl + KClO + H 2O 3KClO → 2KCl + KClO3 4KClO 3 → 3KClO 4 + KCl

Calculate the mass of chlorine needed to produce 100 g of KClO 4. (1) 410.1 g (2) 205.05 g (3) 102.525 g (4) 307.575 g

Chapter 1_Some Basic Concepts of Chemistry.indd 10

K 2Cr2O7 + 14HCl → 2KCl + 2CrCl 3 + 7H 2O + 3Cl 2

If 68 g sample that is 96% K2Cr2O7 is allowed to react with 320 mL of HCl solution having density of 1.15 g mL−1 and containing 30% by weight HCl, what mass of Cl2 is generated? (Given atomic mass of K and Cr are 39 and 52, respectively). (1) 43.52 g (2) 39.62 g (3) 46.015 g (4) 35.73 g

CO2(g ) 0 14 mL

12 ´ 71 ´ 100 = 205.05 g of Cl 2 3 ´ 138.5

9. Chlorine gas can be produced in the laboratory by the reaction

CO2(g ) + 2H 2O( l ) 0 0 30 mL 60 mL

1 O2 → 2 7 mL 0

Initial volume Final volume

→

=

Solution (3) The reaction involved is K 2Cr2O7 + 14HCl → 2KCl + 2CrCl 3 + 7H 2O + 3Cl 2

Number of moles of K2Cr2O7 =



Number of moles of HCl =



68 × 0.96 = 0.222 mol 294

1.15 × 320 × 0.3 = 3.024 mol 36.5 Since, HCl is the limiting reagent, therefore Number of moles of HCl Number of moles of Cl 2 = 14 3 3.024 × 3 = 0.648 mol 14



Number of moles of Cl 2 =



Weight of Cl2 = 0.648 × 71 = 46.015 g

10. A polystyrene, having formula Br3C6 H 3 (C8H8 )n was prepared by heating styrene with tribromobenzoyl peroxide in the absence of air. If it was found to contain 10.46% bromine by weight, find the value of n. (1) 16 (2) 17 (3) 18 (4) 19

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Some Basic Concepts of Chemistry Solution (4) Let the weight of polystyrene prepared be 100 g. Therefore, number of moles of Br in 100 g of polystyrene 10.46 = = 0.1308 mol 80 From the formula of polystyrene, we have, No. of moles of Br = 3 × mol of Br3C6H3(C8H8)n Weight 3 × 100 0.1308 = 3 × = Mol.wt. 315 + 104n



On solving we get n = 19.

11. 5000 mL of 0.6 M H2SO4 acid is mixed with 2 L of 98% (w/V ) of H2SO4 (specific gravity 1.84). If the specific gravity of the resulting mixture is 1.4, what is the molarity of the mixture? (Given that the specific gravity of 0.6 M H2SO4 is 1.02 and also consider there is no loss of mass due to mixing.) (1) 1.833 M (2) 3.6674 M (3) 7.3348 M (4) 4.584 M Solution (2) Total volume of H2SO4 = 5 × 1.02 + 2 × 1.84 = 8.78 L or 8780 mL

8.78 = 6.27143 L or 6271.43 mL 1.4



We know that



Total number of moles of H 2SO 4 Molarity = Total volume of solution(L )



13. The vapor density of a mixture containing NO2 and N2O4 is 38.3 at 27°C. Calculate the moles of NO2 in 100 mol mixture. (1) 33.48 (2) 44.66 (3) 76.46 (4) 91.22 Solution (1) Molecular weight of mixture of NO2 and N2O4 = 38.3 × 2 = 76.6

Let x mole of NO2 be present in the mixture.



Wt. of NO2 + Wt. of N2O4 = Total wt. of mixture x ´ 46 + (100 - x ) ´ 92 = 100 ´ 76.6 x = 33.48 mol

14. In a victor Meyer determination of the relative molecular mass of benzene, the heating vessel was maintained at 120°C. A mass of 0.1528 g of benzene was used and the volume of displaced air collected over water at 15°C, was 48 cm3. The barometric pressure was 743 mm Hg. Calculate the relative molecular mass of benzene. The vapor pressure of water at 15°C is 13 mm Hg. (1) 39.13 (2) 78.26 (3) 117.39 (4) 156.52

Therefore, total volume of solution     =

=



0.98 × 2 × 103 88 = 3.6674 M 6.27143

5 × 0.6 +

12. The density of 3 M solution of Na 2S 2O3 is 1.25 g mL-1. Calculate, the mole fraction of Na2SO3.

Solution (2) Actual pressure of displaced air = 743 – 13 = 730 mm Hg = 730/760 atm

We have, 15°C = 15 + 273 = 288 K, V = 48 cm3 = 48 × 10– 3 L and w = 0.1528 g



Therefore, M =

(1) (2) (3) (4)

Solution



Therefore, number of moles of Na 2S 2O3 = 3



Weight of Na 2S 2O 3 = 3 × 158 = 474 g



We know, volume of solution = 1 L = 1000 mL



Therefore, weight of solution = 1000 × 1.25 = 1250 g



Thus, weight of water = 1250 – 474 = 776 g



Mole fraction of Na2SO3 Na   2    S 2O 3 =

No. of moles of Na 2S 2O3 oles H 2O No.of moles Na 2S 2O3 + No.of mo

3 = 0.065    = 3 + 776 /18

Chapter 1_Some Basic Concepts of Chemistry.indd 11

wRT 0.1528 ´ 0.82 ´ 288 = 78.26 g mol-1 = 730 pV ´ 0.048 760

15. A mixture of FeO and Fe3O4 when heated in air to constant weight, gains 5% in its weight. Find out composition of mixture.

(1) 0.065 (2) 0.13 (3) 0.26 (4) 0.39

(1) Given that molarity of Na 2S 2O3 is 3 mol L−1.

11

FeO = 20.25%, Fe3O4 = 79.25% FeO = 79.75%, FeO = 20.25% FeO = 21%, Fe3O4 = 79% FeO = 79%, Fe3O4 = 21%

Solution (1) The reactions involved are 1 O2 → Fe2O3 2 1 + O2 → 3Fe2O3 2

2 FeO + 2 Fe3O4

Let the weight of FeO be a g and weight of Fe3O4 be b g.



   a + b = 100



Now, 2 ´ 72 g FeO gives 160 g Fe2O3



Therefore, a g FeO gives Fe2O3 =

(1)

160 ´ a g 144

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OBJECTIVE CHEMISTRY FOR NEET



Also, 2 ´ 232 g of Fe3O4 gives Fe2O3 = 160 ´ 3 g



160 ´ 3 ´ b g Therefore, b g Fe3O4 gives Fe2O3 = 464







From Eq. (1) and Eq. (2), we get



a = 20.25 g; FeO = 20.25%



b = 79.75 g; Fe3O4 = 79.75%

160 a 160 ´ 3 ´ b + = 105  144 464

         nO2 = 5 ´ 0.8923

(2)

16. A sample of NaCl contains NaNO3 as an impurity. 250 mL of its solution was prepared by dissolving 1.25 g of the sample. 25 mL of this solution required 17.75 mL of M/10 AgNO3 solution. Calculate the composition of the solution in g L−1. (1) (2) (3) (4)

−1

5 × 0.8923 =

NaCl = 4.1536 g L , NaNO3 = 0.4232 g L NaCl = 4.1536 g L−1, NaNO3 = 0.8464 g L−1 NaCl = 2.0768 g L−1, NaNO3 = 0.4232 g L−1 NaCl = 2.0768 g L−1, NaNO3 = 0.8464 g L−1

(2) In 250 mL of solution, let the weight of NaCl be a. Therefore, weight of NaNO3 = (1.25 – a) g

In 25 mL of solution, a (1.25 - a ) Weight of NaCl = and weight of NaNO3 = 10 10

VO 2 22.4

VO 2 22.4

⇒ VO2 = 99.93 L

18. On heating 60 mL of a mixture of equal volumes of chlorine and its gaseous oxide and cooling to atmospheric temperature, the resulting gas measured is 75 mL. Treatment of this resulting gas mixture with caustic soda (absorbs chlorine) solution resulted in contraction to 15 mL. Assuming that all measurements were made at the same temperature and pressure, deduce the formula of oxide. (Consider that due to heating entire chlorine oxide is decomposed). (1) Cl2O7 (2) Cl2O (3) ClO2 (4) Cl2O4

−1

Solution

At STP    n =



Solution (2) Let the formula of chlorine oxide be ClaOb

Given: Volume of ClaOb = 30 mL and Volume of Cl2 = 30 mL a b The reaction is Cl a Ob → Cl 2 + O2 2 2



 



After cooling,

30   15 a 15 b

VTotal = VCl 2 (originally present ) + VCl 2 (formed ) + VO2 (formed )



The reaction is NaCl + AgNO3 → AgCl + NaNO3







Number of moles of NaCl = Number of moles of AgNO3 = 1 17.75 ´ ´ 10 -3 10



or     a + b = 3



Given: NaOH absorbs 60 mL of Cl2 as 15 mL is left after contraction out of 75 mL.



So,



From Eq. (1), b = 1



Therefore, formula of oxide is Cl2O.

a 1 17.75 ´ = ´ 10 -3 Þ a = 1.04 g 10 58.8 10

1.0384 = 4.16 g L-1  Concentration of NaCl = ( 250 /1000)



Concentration of NaNO3 =

0.2116 = 0.8464 g L-1 ( 250 /1000)

17. What volume of oxygen gas at NTP is necessary for complete combustion of 20 L of propane measured at 0°C and 760 mm pressure? (1) 24.98 L (2) 33.31 L (3) 99.93 L (4) 133.24 L



(1) C10H8 (2) C14H10 (3) C18H14 (4) None of these Solution (1) Calculation of empirical formula

(760 / 760) ´ 20 =

0.8923 mol 0.0821 ´ 273 [Using Ideal gas equation]

The reaction is C 3H8(g ) + 5 O2(g ) ® 3 CO2(g ) + 4 H 2O(g )

Chapter 1_Some Basic Concepts of Chemistry.indd 12

60 = 30 + 15a ⇒ a = 2

Element %

Number of moles of C 3H8 Number of moles of O2 = 1 5

(1)

19. Naphthalene (compound of C and H) contains 93.71% carbon. If its molar mass is 128 g mol–1, calculate its molecular formula.

Solution (3) Number of moles of C 3H8 =

75 = 30 + 15a + 15b

Atomic Relative Simple mass number ratio of of moles moles

Simplest whole number ratio

C

93.71 12

7.809

1.3 × 3

4

H

6.29 1

6.29

1×3

3



Thus, empirical formula is C4H3.



Calculation of molecular formula

1/4/2018 5:08:35 PM

Some Basic Concepts of Chemistry Emprical formula mass = 12 ´ 4 + 3 ´ 1 = 51 g 128 Molecular mass = = 2.51 n= Emprical formula mass 51





13

Molecular formula = Empirical formula × 2.51 = C 4H 3 × 2.51 = C10H8

Practice Exercises Level I

8. The number of atoms in 52 amu of He is (1) 13 × 1023 (2) 1.3 × 1023 (3) 13 (4) 103

Laws of Chemical Combinations 1. Potassium combines with two isotopes of chlorine (35Cl and 37Cl) respectively to form two samples of KCl. Their formation follows the law of (1) (2) (3) (4)

constant proportions. multiple proportions. reciprocal proportions. None of these.

2. Irrespective of the source, pure sample of water always yields 88.89% mass of oxygen and 11.11% mass of hydrogen. This is explained by the law of (1) (2) (3) (4)

conservation of mass. constant composition. multiple proportion. constant volume.

Calculations of Moles, Empirical and Molecular Formula 3. A sample contains 200 atoms of hydrogen, 0.05 g atom of nitrogen, 10−20 g atom of oxygen. What is the approximate number of total atoms? (1) 200 (2) 6223 (3) 3 × 1022 (4) 5 × 1022 4. The number of gram molecules of oxygen in 6.022 × 1024 molecules of CO is (1) 10 g molecules (2) 5 mg molecules (3) 1 g molecules (4) 0.5 g molecules 5. The element A (At. wt. = 75) and B (At. wt. = 32) combine to form a compound X. If 3 mol of B combine with 2 mol of A to give 1 mol of X, the weight of 5 mol of X is (1) 246 g (2) 1230 amu (3) 1.23 kg (4) None of these 6. Total number of protons in 36 mL of water at 4°C (where r of water = 1 g mL−1) is (1) 20 (2) 16 (3) 20 NA (4) 16 NA 7. Under similar conditions, oxygen and nitrogen are taken in the same mass. The ratio of their volume will be (1) 7:8 (2) 3:5 (3) 6:5 (4) 9:2

Chapter 1_Some Basic Concepts of Chemistry.indd 13

9. The number of electrons in 2 g ion of nitrate ion ( NO3- ) is (1) 64 (2) 64 NA (3) 32 (4) 32 NA 10. The molecular mass of CO2 is 44 g and Avogadro’s number is 6.02 × 1023. Calculate the mass of one molecule (in g) of CO2.

(1) 7.31 × 10–23 g (2) 3.65 × 10–23 g (3) 1.01 × 10–23 g (4) 2.01 × 10–23 g 11. The number of silver atoms present in a 90% pure silver wire weighing 10 g is (At. wt. of Ag = 108)

(1) 8 × 1022 (2) 0.62 × 1023 (3) 5 × 1022 (4) 6.2 × 1029 12. An organic compound on analysis was found to contain 0.014% of nitrogen. If its molecule contains two N atoms, then the molecular mass of the compound is

(1) 200 (2) 2000 (3) 20,000 (4) 200000 13. An element, X, has three isotopes X20, X21 and X22. The percentage abundance of X20 is 90% and its average atomic mass of the element is 20.11. The percentage abundance of X21 should be (1) 9% (2) 8% (3) 10% (4) 0% 14. The chloride of a metal contains 71% chlorine by weight and the vapor density of it is 50. The atomic weight of the metal will be (1) 29 (2) 58 (3) 35.5 (4) 71

Stoichiometry, Limiting Reagent, and POAC 15. A mixture contains FeSO 4 and Fe2(SO 4 )3 . If both FeSO 4 and Fe2(SO 4 )3 provide equal number of sulphate ions then, the ratio of Fe2 + and Fe3+ ions in mixture is (1) 1:2 (2) 2:3 (3) 2:1 (4) 3:2 16. A gas mixture of 3 L of propane and butane on complete combustion at 25°C produced 10 L of CO2. The propane and butane are respectively

1/4/2018 5:08:35 PM

14

OBJECTIVE CHEMISTRY FOR NEET (1) 1, 2 L (2) 2, 1 L (3) 1.5 L each (4) None of these

17. A mixture contains n mol of H2 and 2n mol of CH4. The ratio of number of C:H atoms in the mixture is (1) 1/5 (2) 2/3 (3) 4/5 (4) 1/3 18. One mole of potassium chlorate is thermally decomposed and excess of aluminium is burnt in the gaseous product. How many moles of aluminium oxide are formed? (1) 1 (2) 1.5 (3) 2 (4) 3 19. 2 mol of H2S and 11.2 L SO2 at NTP reacts to form x mol of sulphur; x is



SO2 + 2H2S → 3S + 2H2O

(1) 37.3 g (2) 3.73 g (3) 56 g (4) 5.6 g 26. 2.4 kg of carbon is made to react with 1.35 kg of aluminium to form Al4C3. The maximum amount (in kg) of aluminium carbide formed is (1) 5.4 (2) 3.75 (3) 1.05 (4) 1.8 27. Hydrogen evolved at NTP on complete reaction of 27 g of Al with excess of aqueous NaOH would be (Chemical reaction: 2Al + 2NaOH + 2H2O → 2NaAlO2 + 3H2) (1) 22.4 L (2) 44.8 L (3) 67.2 L (4) 33.6 L 28. On reduction with hydrogen, 3.6 g of an oxide of metal left 3.2 g of the metal. If the atomic weight of the metal is 64, the simplest formula of the oxide would be

(1) 1.5 (2) 3 (3) 11.2 (4) 6 20. A mixture of CO and CO2 having a volume of 20 mL is mixed with x mL of oxygen and electrically sparked. The volume after explosion is (16 + x) mL under the same conditions. What would be the residual volume if 20 mL of the original mixture is treated with aqueous NaOH ?

(1) M2O3 (2) M2O (3) MO (4) MO2 29. The reaction is Fe2(SO 4 )3 + BaCl 2 → BaSO 4 + FeCl 3

5 1 mol (2) mol 2 2 3 (3) mol (4) 2mol 4 (1)

(1) 12 mL (2) 10 mL (3) 9 mL (4) 8 mL 21. Consider the reactions 2A → 2B, B → 2C, 3C → 4D. The number of moles of D formed starting 4 mol of A, are (1) 8 (2) 16 (3) 4 (4) 10.67 22. A mixture of CuSO4 × 5H 2O and MgSO 4 × 7H 2O is heated until all the water is driven off. If 5.0 g of a mixture gives 3 g of anhydrous salts, what is the percentage by mass of CuSO4 × 5H 2O in the original mixture? (1) 44% (2) 64% (3) 74% (4) 94% 23. Element X reacts with oxygen to produce a pure sample of X 2O3 . In an experiment it is found that 1 g of X produces 1.16 g of X 2O3 . Calculate the atomic weight of X. (Given atomic weight of oxygen, 16.0 g mol -1.) (1) 67 (2) 100.2 (3) 125 (4) 150 24. How many grams of phosphoric acid (H3PO4) would be needed to neutralize 100 g of magnesium hydroxide Mg(OH)2 ? (1) 66.7 g (2) 252 (3) 112.6 g (4) 168 g 25. Calculate the weight of iron which will be converted into its oxide by the action of 18 g of steam on it from the reaction 2Fe + 3H 2O → Fe2O3 + 3H 2.

Chapter 1_Some Basic Concepts of Chemistry.indd 14

How many moles of BaCl 2 is needed to produce 1/2 mol of FeCl 3 ?

30. 5 mol of CH4 is burned with 8 mol of O2, then calculate the moles of CO2 formed and remaining moles of excess reagent. (1) 4, 1 (2) 1, 4 (3) 0, 5 (4) 5, 0 31. Calcium carbonate reacts with aqueous HCl to give CaCl 2 according to the reaction, CaCO3(s) + 2HCl(aq) → CaCl 2(aq) + CO2(g) + H 2O(l). The mass of CaCO3 required to react completely with 25 mL of 0.75 M HCl is (1) 0.1 g (2) 0.84 g (3) 8.4 g (4) 0.94 g 32. The ratio of the molar amounts of H2S needed to precipitate the metal ions from 20 mL each of 1 M Cd(NO3)2 and 0.5 M CuSO4 is



Cd(NO3)2 + H2S → CdS↓ + 2HNO3

(1) 1:1 (2) 2:1 (3) 1:2 (4) Indefinite

Concentration Terms 33. The molarity of pure water is almost (1) 5.55 M (2) 55.55 M (3) 2 M (4) 1/18 M

1/4/2018 5:08:37 PM

Some Basic Concepts of Chemistry 34. 3.0 molal NaOH solution has a density of 1.11 g mL−1 . The molarity of the solution is (1) 2.97 M (2) 3.05 M (3) 3.64 M (4) 3.050 M 35. 50 mL of 0.01 M FeSO4 will react with what volume of 0.01 M KMnO4 solution in acid medium? (1 mol KMnO4 requires 5 mol of FeSO4 for complete reaction) (1) 50 mL (2) 25 mL (3) 100 mL (4) 10 mL 36. The number of H+ ions present in 100 mL of 0.001 M H2SO4 solution will be (1) 120.4 × 10 (2) 1.20 × 10 (3) 6.023 × 1020 (4) 6.023 × 1021 19

(1) 18 (2) 29.9 (3) 24 (4) 108 2. The molecular weight of green vitriol is M0. The weight of 10−3 NA molecules of it is (1) M0 g (2) M0 mg (3) 103 M0 g (4) 10−3 M0 mg 3. Cortisone is a molecular substance containing 21 atoms of carbon per molecule. The weight percentage of carbon in cortisone is 69.98%. What is the molecular weight of cortisone? (1) 176.5 (2) 252.2 (3) 287.6 (4) 360.1

20

37. Molarity of 1 g H2SO4 solution in 1 L water is nearly (1) 0.1 (2) 0.20 (3) 0.05 (4) 0.01

4. A crystalline hydrated salt on being rendered anhydrous, looses 45.6% of its weight. The percentage composition of anhydrous salt is Al = 10.5%, K = 15.1%, S = 24.8% and O = 49.6%. The empirical formula of the crystalline salt is (1) KAlS 2O8 ×12H 2O (2) K 2 Al 2S 2O8 ×12H 2O

Equivalent Weight Concept 38. 0.5 g of metal on oxidation gave 0.79 g of its oxide. The equivalent weight of the metal is (1) 10 (2) 14 (3) 20 (4) 40 39. 74.5 g of the metallic chloride contains 35.5 g of chlorine. The equivalent weight of the metal is (1) 19.5 (2) 35.5 (3) 39.0 (4) 78.0 40. The sulphate of an element contains 42.2% element. The equivalent weight of the metal would be

(3) KAl 2S 2O8 ×12H 2O (4) None of these 5. The atomic weight of a triatomic gas is a. The correct formula for the number of moles of gas in its w g is 3w w (2) a 3a a (3) 3wa (4) 3w 6. If NA is Avogadro’s number, then the number of valence electrons in 4.2 g of nitride ion ( N 3- ) is (Given one atom of N has 5 valence electrons.) (1)

(1) 2.4 NA (2) 4.2 NA (3) 1.6 NA (4) 3.2 NA

(1) 17.0 (2) 35.0 (3) 51.0 (4) 68.0 41. 0.1 g of a metal gave on reaction with dilute acid at STP 34.2 mL hydrogen gas. The equivalent weight of the metal is (1) 32.7 (2) 48.6 (3) 64.2 (4) 16.3 42. The equivalent weight of a metal is 4.5 and the molecular weight of its chloride is 80. The atomic weight of the metal is (1) 18 (2) 9 (3) 4.5 (4) 36

Level II Calculations of Moles, Empirical and Molecular Formula 1. One atom of an element weighs 3.98 × 10−23 g. Its atomic mass is

Chapter 1_Some Basic Concepts of Chemistry.indd 15

15



7. Chlorophyll contains 2.68% of magnesium by mass. Calculate the number of magnesium atoms in 3 g of chlorophyll. (1) 2.01 × 1021 atoms (2) 6.023 × 1023 atoms (3) 1.7 × 1020 atoms (4) 2.8 × 1022 atoms

8. Two flasks of equal volumes are evacuated, then one is filled with gas A and other with gas B at the same temperature and pressure. The weight of B was found to be 0.80 g while the weight of gas A is found to be 1.40 g. What is the weight of one molecule of B in comparison to one molecule of A? (1) (2) (3) (4)

1.40 times as heavy as A. 0.40 times as heavy as A. 0.57 times as heavy as A. 0.80 times as heavy as A.

9. The mass of carbon present in 0.5 mol of K4[Fe(CN)6] is (1) 1.8 g (2) 18 g (3) 3.6 g (4) 36 g

1/4/2018 5:08:37 PM

16

OBJECTIVE CHEMISTRY FOR NEET

10. A spherical ball of radius 7 cm contains 56% iron. If density is 1.4 g cm−3, number of moles of Fe present approximately is (1) 10 (2) 15 (3) 20 (4) 25 11. The total number of protons in 8.4 g of MgCO3 is (NA = 6.02 × 1023) (1) 2.52 × 1022 (2) 2.52 × 1024 (3) 3.01 × 1024 (4) 3.01 × 1022 12. The specific heat of a metal is 0.16. Its approximate atomic mass would be (1) 32 (2) 16 (3) 40 (4) 64 13. Hemoglobin contains 0.33% of iron by weight. The molecular mass of hemoglobin is approximately 67200. The number of iron atoms (At. mass of Fe = 56) present in one molecule of hemoglobin is (1) 6 (2) 1 (3) 4 (4) 2 14. The charge on 1 g ion of Al3 + is (e represents magnitude of charge on 1 electron) (1) 1/27 NAe C (2) 1/3 NAe C (3) 1/9 NAe C (4) 3 NAe C 15. A partially dried clay mineral contains 8% water. The original sample contained 12% water and 45% silica. Percentage of silica in the partially dried sample is nearly (1) 50% (2) 49% (3) 55% (4) 47%

Stoichiometry, Limiting Reagent, and POAC 16. 510 mg of liquid on vaporization in Victor Meyer’s apparatus displaces 67.2 cm3 of air at (STP). The molecular weight of the liquid is (1) 130 (2) 17 (3) 170 (4) 1700 17. When 100 g of ethylene polymerizes to polyethylene according to equation of nCH 2 = CH 2 ® ( CH 2 CH 2) n . The weight of polyethylene produced will be n (1) g (2) 100 g 2 100 (3) g (4) 100 n g n 18. A hydrocarbon C10Hx requires 32.5 mol of O2 for combustion of 2.5 mol. Calculate value of x? (1) 24 (2) 32 (3) 12 (4) 22 19. An ore contains 1.34% of the mineral argentite, Ag2S, by weight. How many grams of this ore would have

Chapter 1_Some Basic Concepts of Chemistry.indd 16

to be processed in order to obtain 1 g of pure solid silver Ag? (1) 74.6 g (2) 85.7 g (3) 134.0 g (4) 171.4 g 20. 8 g of sulphur is burnt to form SO2 which is oxidized by Cl2 water. The solution is treated with BaCl2 solution. The amount of BaSO4 precipitated is (1) 1 mol (2) 0.5 mol (3) 0.24 mol (4) 0.25 mol 21. 100 mL of CH4 and C2H2 were exploded with excess of O2. After explosion and cooling, the mixture was treated with KOH, where a reduction of 165 mL was observed. Therefore, the composition of the mixture is (1) CH4 = 35 mL ; C2H2 = 65 mL (2) CH4 = 65 mL ; C2H2 = 35 mL (3) CH4 = 75 mL; C2H2 = 25 mL (4) CH4 = 25 mL; C2H2 = 75 mL 22. An element (X) reacts with hydrogen leading to formation of a class of compounds that is analogous to hydrocarbons. 5 g of X forms 5.628 g of a mixture of two compounds of X ( XH 4 and X 2H6 ) in the molar ratio of 2:1. Determine the molar mass of X. (1) 28 (2) 58 (3) 72 (4) 83

Concentration Terms 23. A copper sulphate solution contains 1.595 % of CuSO4 by weight. Its density is 1.2 g mL−1, Its molarity will be (1) 0.12 (2) 0.06 (3) 1.20 (4) 1.595 24. What volume of a 1.36 M HCl solution should be added to a 200 mL 2.4 M HCl solution and finally diluted to 500 mL so that molarity of final HCl solution becomes 1.24 M? (1) 29.2 mL (2) 102.94 mL (3) 46.34 mL (4) 9.4 mL 25. What volume of 0.010 M NaOH (aq) is required to react completely with 30 g of an aqueous acetic acid solution in which mole fraction of acetic acid is 0.15? (1) 108.55 L (2) 18.55 L (3) 34.66 L (4) 42 L 26. An aqueous solution of urea containing 18 g urea in 1500 cm3 of solution has a density of 1.052 g cm−3. If the molecular weight of urea is 60, then the molality of solution is (1) 0.2 (2) 0.192 (3) 0.064 (4) 1.2 27. N2 and H2 are mixed in 14:3 mass ratio. After certain time ammonia was found to be 40% by mol. The mole

1/4/2018 5:08:38 PM

Some Basic Concepts of Chemistry (1) 12 (2) 24 (3) 48 (4) 6

fraction of N2 at that time in the mixture of N2, H2 and NH3 is (1) 0.15 (2) 0.3 (3) 0.45 (4) none of these 28. 25.4 g of iodine and 14.2 g of chlorine are made to react completely to yield a mixture of ICl and ICl3. Calculate the ratio of moles of ICl and ICl3. (1) 1:1 (2) 1:2 (3) 1:3 (4) 2:3 29. 25.0 mL of HCl solution gave, on reaction with excess AgNO3 solution 2.125 g of AgCl. The molarity of HCl solution is (1) 0.25 (2) 0.6 (3) 1.0 (4) 0.75

Equivalent Weight Concept 30. Equivalent weight of H3PO4 in the following reaction is

37. A metallic oxide contains 60% of the metal. The equivalent weight of the metal is (1) 12 (2) 24 (3) 40 (4) 48 38. The weight of a metal of equivalent weight 12, which will give 0.475 g of its chloride, is (1) 0.12 g (2) 0.16 g (3) 0.18 g (4) 0.24 g 39. 0.84 g of a metal hydride contains 0.042 g of hydrogen. Its equivalent weight is (1) 80 (2) 40 (3) 60 (4) 20 40. A bivalent metal has the equivalent weight of 12. The molecular weight of its oxide will be (1) 24 (2) 34 (3) 36 (4) 40

2 H3PO4 + 3Ca(OH)2 → Ca3(PO4)2 + 6H2O (1) 98 (2) 49 (3) 32.66 (4) 24.5

31. 0.534 g Mg displaces 1.415 g Cu from the salt solution of Cu. Equivalent weight of Mg is 12. The equivalent weight of Cu would be

Previous Years’ NEET Questions 1. An element, X, has the following isotopic composition: X : 90% X : 8.0% 202 X : 2.0% 200

199

(1) 15.9 (2) 47.7 (3) 31.8 (4) 8.0 32. The oxide of an element possesses the formula M2O3. If the equivalent weight of the metal is 9, then the atomic weight of the metal will be (1) 9 (2) 18 (3) 27 (4) None of these. 33. In m1 g of a metal A displaces m2 g of another metal B from its salt solutions and if the equivalent weights are E1 and E2, respectively, then the equivalent weight of A can be expressed by m ´ E2 m (1) E1 = 1 ´ E 2 (2) E1 = 2 m1 m2

m1 m ´ m2 ´ E2 (3) E1 = 1 (4) E1 = E2 m2 34. The equivalent weight of iron in Fe2O3 would be (1) 18.6 (2) 28 (3) 56 (4) 112.0 35. When a metal is burnt, its weight is increased by 24%. The equivalent weight of the metal will be (1) 2 (2) 24 (3) 33.3 (4) 76 36. 1.5 g of a divalent metal displaced 4 g of copper (At. wt. = 64) from a solution of copper sulphate. The atomic weight of the metal is

Chapter 1_Some Basic Concepts of Chemistry.indd 17

17



The weighted average atomic mass of the naturally occurring element X is closest to (1) 199 amu (2) 200 amu (3) 201 amu (4) 202 amu (AIPMT 2007)

2. Concentrated aqueous sulphuric acid is 98% H2SO4 by mass and has a density of 1.80 g mL−1. The volume of acid required to make 1 L of 0.1 M H2SO4 solution is (1) 5.55 mL (2) 11.10 mL (3) 16.65 mL (4) 22.20 mL (AIPMT 2007) 3. How many moles of lead (II) chloride will be formed from a reaction between 6.5 g of PbO and 3.2 g of HCl? (1) 0.029 (2) 0.044 (3) 0.333 (4) 0.011 (AIPMT 2008) 4. An organic compound contains carbon, hydrogen and oxygen. Its elemental analysis gave C, 38.71% and H, 9.67%. The empirical formula of the compound would be (1) CH4O (2) CH3O (3) CH2O (4) CHO (AIPMT 2008)

1/4/2018 5:08:39 PM

18

OBJECTIVE CHEMISTRY FOR NEET

5. 10 g of hydrogen and 64 g of oxygen were filled in a steel vessel and exploded. Amount of water produced in this reaction will be

at STP, the moles of HCl(g) formed is equal to (1) 1 mol of HCl(g). (2) 2 mol of HCl(g). (3) 0.5 mol of HCl(g). (4) 1.5 mol of HCl(g).

(1) 1 mol (2) 2 mol (3) 3 mol (4) 4 mol

(AIPMT 2014) (AIPMT 2009)

6. 25.3 g of sodium carbonate, Na 2CO3 is dissolved in enough water to make 250 mL of solution. If sodium carbonate dissociates completely, molar concentration of sodium ion, Na + and carbonate ions, CO23 are respectively (Molar mass of Na 2CO3 = 106 g mol -1 ) (1) 1.910 M and 0.955 M (2) 1.90 M and 1.910 M (3) 0.477 M and 0.477 M (4) 0.955 M and 1.910 M (AIPMT PRE 2010) 7. The number of atoms in 0.1 mol of a triatomic gas is ( N A = 6.02 ´ 1023 mol -1 ) (1) 1.806 ´ 1023 (2) 3.600 ´ 1023 (3) 1.800 ´ 1022 (4) 6.026 ´ 1022 (AIPMT PRE 2010) 8. Mole fraction of the solute in a 1.00 molal aqueous solution is (1) 1.7700 (2) 0.1770 (3) 0.0177 (4) 0.0344 (AIPMT PRE 2011) 9. How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0 M HNO3? The concentrated acid is 70% HNO3. (1) 45.0 g conc. HNO3 (3) 70.0 g conc. HNO3

(2) 90.0 g conc. HNO3 (4) 54.0 g conc. HNO3 (NEET 2013)

10. An excess of AgNO3 is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromium(III) chloride. The number of moles of AgCl precipitated would be (1) 0.001 (2) 0.002 (3) 0.003 (4) 0.01

(1) Mg, 0.16 g (2) O2, 0.16 g (3) Mg, 0.44 g (4) O2, 0.28 g (AIPMT 2014) 14. A mixture of gases contains H2 and O2 gases in the ratio of 1:4 (w/w). What is the molar ratio of the two gases in the mixture? (1) 4:1 (2) 16:1 (3) 2:1 (4) 1:4 (AIPMT 2015) 15. The number of water molecules is maximum in (1) 18 g of water. (2) 18 mol of water. (3) 18 molecules of water. (4) 1.8 g of water. (RE AIPMT 2015) 16. If Avogadro number NA, is changed from 6.022 × 1023 mol−1 to 6.022 × 1020 mol−1, this would change (1) the ratio of chemical species to each other in a balanced equation. (2) the ratio of elements to each other in a compound. (3) the definition of mass in units of grams. (4) the mass of one mole of carbon. (RE AIPMT 2015) 17. 20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample? (1) 60% (2) 84% (3) 75% (4) 96% (RE AIPMT 2015)

(NEET 2013) 11. 6.02 × 1020 molecules of urea are present in 100 mL of its solution. The concentration of solution is (1) 0.02 M (2) 0.01 M (3) 0.001 M (4) 0.1 M (NEET 2013) 12. When 22.4 L of H2(g) is mixed with 11.2 L of Cl2(g), each

Chapter 1_Some Basic Concepts of Chemistry.indd 18

13. 1.0 g of magnesium is burnt with 0.56 g O2 in a closed vessel. Which reactant is left in excess and how much? (At. wt. Mg = 24; O = 16)

18. What is the mole fraction of the solute in a 1.00 m aqueous solution? (1) 0.0354 (2) 0.0177 (3) 0.177 (4) 1.770 (RE AIPMT 2015) 19. What is the mass of the precipitate formed when 50 mL of 16.9% solution of AgNO3 is mixed with 50 mL of 5.8%

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Some Basic Concepts of Chemistry

19

(1) 60, 40 (2) 20, 30 (3) 30, 20 (4) 40, 30

NaCl solution? (At. wt. Ag = 107.8, N = 14, O = 16, Na = 23, Cl = 35.5) (1) 7 g (2) 14 g (3) 28 g (4) 3.5 g

(NEET-II 2016)

(RE AIPMT 2015)

21. Which of the following is dependent on temperature? (1) Molarity (2) Mole fraction (3) Weight percentage (4) Molality

20. Suppose the elements X and Y combine to form two compounds XY2 and X3Y2. When 0.1 mol of XY2 weighs 10 g and 0.05 mol of X3Y2 weighs 9 g, the atomic weights of X and Y are

(NEET 2017)

Answer Key Level I  1. (4)

 2. (2)

 3. (3)

 4. (1)

 5. (3)

 6. (3)

 7. (1)

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10. (1)

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13. (1)

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17. (1)

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20. (1)

21. (4)

22. (3)

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26. (4)

27. (4)

28. (2)

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35. (4)

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37. (4)

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Level II  1. (3)

 2. (2)

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 4. (1)

 5. (2)

 6. (1)

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 9. (4)

10. (3)

11. (2)

12. (3)

13. (3)

14. (4)

15. (4)

16. (3)

17. (2)

18. (3)

19. (2)

20. (4)

21. (1)

22. (1)

23. (1)

24. (2)

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27. (1)

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29. (2)

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33. (1)

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40. (4)

Previous Years’ NEET Questions  1. (1)

 2. (1)

 3. (1)

 4. (2)

 5. (4)

 6. (1)

 7. (1)

 8. (3)

 9. (1)

10. (1)

11. (2)

12. (1)

13. (1)

14. (1)

15. (2)

16. (4)

17. (2)

18. (2)

19. (1)

20. (4)

21. (1)

Hints and Explanations Level I

6. (3) Number of protons in 1 molecule of H2O = (2 + 8) = 10

3.  (3) Total number of atoms = 200 + (0.05 × NA) + (10-20 ×



Number of molecules of H2O in 36 mL =



Therefore, number of protons =

NA) ≅ 3 × 1022 4. (1) Number of gram molecules =

6.022 ´ 1024 = 10 6.022 ´ 1023

5. (3) Formula of X = A 2B3

36 ´ N A ´ 10 = 20 NA 18

7. (1) Ratio of volume of O2:N2 = Ratio of no. of moles O2:N2



Weight of 1 molecule of X = (75 × 2 + 32 × 3) amu





Weight of 1 mol of X = 246 g





Therefore, weight of 5 mol of X = 246 × 5 g = 1.23 kg

Chapter 1_Some Basic Concepts of Chemistry.indd 19

36 ´ NA 18

Let the mass taken be x x Therefore, no. of moles of O2 = and no. of moles of N2 32 x = 28

1/4/2018 5:08:40 PM

20



OBJECTIVE CHEMISTRY FOR NEET

Hence, ratio of volume of O2:N2 =

8. (3) We know 1 amu =

1 g NA

x / 32 28 7 = = x / 28 32 8



(52 / NA )

Number of moles =



Therefore, number of atoms =

4

aL

(52 / NA ) ´ N

A

4

3a L

(3 - a )L

NO3- = (7 + 8 ´ 3 + 1) = 32



Number of NO3- ions in 2 g = 2 ´ NA



Therefore, total number of electrons = 32 × (2 × NA) = 64 NA

4( 3 - a ) L



Total volume of CO2 produced = 3a + 4( 3 - a ) = 10 Þ a = 2



Therefore, volume of propane = 2 L and volume of butane =1L

17. (1) We know 1 mol = 6.023 × 1023 or NA atoms n mol of H2 contains = 2 × n × NA hydrogen atoms

10. (1) We have Molecular Mass NA 44 = 6.02 × 1023

Mass of 1 molecule =

       = 7.308 × 10 1. (3) Number of Ag atoms 1

-23

g

10 ´ (0.9) ´ (6.22 ´ 1023 ) = 5 ´ 1022 108



2n mol of CH4 contains = 2 × n × NA carbon atoms



+ 2 × n × 4 × NA hydrogen atoms



Total number of hydrogen atoms = (8n + 2n) NA



Total number of carbon atoms = 2n NA



Therefore, ratio of number of C:H =

0.014 =

KClO 3 → KCl +

( 2 ´ 14) ´ 100 Þ M = 200, 000 M

No. of moles



Let the percentage of X21 be x. Then

No. of moles

x (10 - x ) + 22 ´ 100 100

x =get 9%x = 9% Solving, we

We know Molecular mass of gas = 2 × vapor density



M + 35.5 × x = 50 × 2 

(1)



35.5 ´ x ´ 100 (2) Also,   71 = (M + 35.5x )



On solving Eq. (1) and Eq. (2), we get x = 2, M = 29

15. (4) Let the number of moles of FeSO4 and Fe2(SO 4 )3 be a and b respectively.

Number of moles of SO24 ion in FeSO4 = a



Number of moles of SO24 ion in Fe2SO4 = 3b



Given that



Number of moles of Fe2+ = a and number of moles of Fe2+ = 2b

Chapter 1_Some Basic Concepts of Chemistry.indd 20

3 2

3 2 × = 1 mol 2 3

19. (1)

14. (1) Let the formula be MClx and atomic weight be M

1mol

3 O2 2 3 2

3 2 Al + O2 → Al 2O3 2

13. (1) We have M avg = M1 X 1 + M 2 X 2 + M 3 X 3

20.11 = 20 ´ 0.90 + 21 ´

2n NA 1 = 10n NA 5

18. (1) The reactions involved are as follows:

12. (4) Let the molecular mass of the compound be M. Therefore,



(From Eq. (1))

C 4H10 + 3O2 → 4CO2 + 5H 2O

= 13

9. (2) Number of electrons in one ion of NO3-

=

a 3 =  2b 2

16. (2) The reactions involved are as follows: C 3H 4 + 4O2 ® 3CO2 + 2H 2O





Therefore, ratio of Fe2+ to Fe3+ is



SO2 + 2 H 2S → 11.2 At t = 0 2 mol 22.4 (excess) = 0.5 mol (Limiting reagent) 0 2 - 0.5 × 2 At t = t end = 1 mol



3S + 2 H 2O 0

3 × 0.5

0

2 × 0.5

= 1.5 mol = 1 mol

Therefore, x = 1.5 mol

20. (1) Let the volume of CO be a.

Therefore, volume of CO2 = 20 – a



The reaction is CO +

a = 3b(1)

At t = 0

a

At t = t end 0

1 O2 → CO2 2 x 0 1 x- a a 2

1/4/2018 5:08:42 PM

Some Basic Concepts of Chemistry

Final volume can be calculated as (CO 2 )initially present + (CO 2 )formed + (O 2 )left



1ö æ 16 + x = ( 20 - a ) + a + ç x - ÷ ´ a è 2ø



a 16 + x = 20 + x - Þ a = 8 2

Thus, volume of CO = 8 mL and volume of CO2 = 12 mL

21. (4) We have

2A 2 A®®2B 2B No.of No.ofmoles moles 4 4 4 4







25. (1) The reaction is 2Fe + 3H 2O → Fe2O3 + 3H 2 From the reaction,



Number of moles of Fe =



2 18 2 × = 3 18 3 2 18 Therefore, weight of Fe = ´ ´ 56 = 37.33 g 3 18

26. (4)

nfinal

Therefore, weight of Al4C3 (in kg) =

( 5- x ) moles 249.5

(5 - x ) × 120 g 246

(5 - x ) x ´ 159.5 + ´ 120 = 3 249.5 246 0.639x + 2.439 - 0.487 x = 3 0.561 x= Þ x = 3.69 g 0.152 % of CuSO4 × 5H 2O=

At t = 0 At t = t final

( Excess )

0 3 27 mol ´ 2 27

0

Therefore, volume of H2 evolved at STP =

3 27 × × 22.4 = 33.6 L 2 27

28. (2) Let the formula of oxide be M2Ox. M 2O x → M

Applying POAC on M, we have 2 ´ (No. of moles of M 2O x ) = 1 ´ (No. of moles of M) 2´

3.69 ´ 100 = 73.81% or » 74% 5

3.6 3.2 = 1´ 64 ( 2 ´ 64 + 16 x ) x =1





Applying POAC on X

29. (3) The balanced reaction is



1 × (No. of moles of X) = 2 × (No. of moles of X2O3)



Let the atomic weight of X be a 1 1.6 = 2× ⇒ a = 150 a ( 2a + 48)

Therefore, formula of oxide is M2O.

Fe2(SO 4 )3 + 3BaCl 2 → 3BaSO 4 + 2FeCl 3 nBaCl 2 3

At t =0

3 Mg(OH )2 + 2 H 3PO4 ® Mg 3(PO4 )2 + 6 H 2O

Chapter 1_Some Basic Concepts of Chemistry.indd 21

=

nFeCl 3 2

⇒ nBaCl 2 =

3 1 3 × = mol 2 2 4

30. (1) The reaction is

24. (3) The reaction is

No.of moles of Mg(OH )2 No.of moles of H 3PO 4 = 3 2 (100 / 58) ( x / 98) = Þ x = 112.63 g 3 2

50 144 × = 1.8 kg. 4 1000

27 = 1mol 27

23. (4) The reaction is X → X 2O3

1×

50 mol 4

2 Al + 2NaOH + 2H 2O ® 2NaAlO2 + 3H 2

( Limiting )

x mol 249.5

MgSO 4 ⋅ 7H 2O → MgSO 4 + 7H 2O

0

27. (4)

x × 159.5 g Weight of CuSO4 = 249.5

Weight of MgSO 4 =

3

50 ´ 3 ö æ çè 200 ÷ mol 4 ø

0

CuSO4 ⋅ 5H 2O → CuSO4 + 5H 2O



Al 4C 3

13.5 ´ 10 2.4 ´ 10 27 12 = 50 mol = 200 mol (Limiting reagent) (excess)

3C ® 4D 32 8 ´ 4= = 10.67 No.of moles 8 3 3

( 5- x ) moles 246

®

3C

ninitial

22. (3) Let the weight of CuSO4 × 5H 2O and MgSO 4 × 7H 2O be x g and (5 − x) g respectively.



+

4 Al 3

BB ® ® 22C C No.of No.of moles moles 44 88

x mol 249.5

2 No. of moles of Fe = No. of moles of H 2O 3



 



21

At t =t end



CH 4 + 2O2 ® CO2 + 5 8 (excess) (Limiting reagent) 8 8 0 52 2 = 1 mol = 4 mol

2H 2 O

8 ´2 2 = 8 mol

1/4/2018 5:08:45 PM

22

OBJECTIVE CHEMISTRY FOR NEET



Therefore, number of moles of CO2 = 4 mol



Number of moles of excess reagent = 1 mol

38. (2) We know Eq. wt. of Metal = Eq. wt. of oxide 0.5 0.79 = (Eq. wt.)metal (Eq. wt.)metal + (Eq. wt.)oxide

31. (4) From the reaction, we have

0.79 0.5 = Þ (Eq wt.)metal = 14 (Eq wt.)metal (Eq wt.)metal + 8

nCaCO3

n = HCl 1 2 Mass of CaCO3 25 ´ 0.75 ´ 10 -3 = 100 2

39. (3) Eq. wt. of metal chloride = Eq. wt. of Cl

Therefore, mass of CaCO3 = 0.9375 g or 0.94 g.

41. (1) Eq. wt. of Metal = Eq. wt. of H2

32. (2) We have +

Cd(NO 3 )2 No. of millimoles

+

No. of millimoles 20 ´ 0.5 = 10 millimoles Therefore, ratio of H2S =

0.1 34.2 = ×2 (Eq wt. of metal ) 22400

H 2S ® CdS + 2HNO 3

20 ´ 1 = 20 millimoles CuSO4



74.5 35.5 = ⇒ (Eq. wt.)M = 39 (Eq. wt.)M + 35.5 35.5

20 millimoles



H 2S ® CdS + 2HNO 3

Level II

10 millimoles

Therefore, Eq. wt. of metal = 32.7

1. (3) We know Atomic mass = (mass of 1 atom) × N A     = ( 3.98 × 10 -23 ) × (6.022 × 1023 ) = 24

20 2 = 20 × 0.5 1

3. (4) Let the molecular weight of cortisone be M.

33. (2) Density of pure H2O = 1 g mL−1

Mass of 21 carbon atoms = 12 × 21 = 252 g



(1/18) 1000 Molarity = = = 55.55 M (1/1000) 18 34. (1) Let the volume of solution be V mL

69.98 252 = ⇒ M = 360.1 100 M 4. (1)

No.of molesof NaOH We know Molarity = Weight of solvent(kg)

Element

No. of moles of NaOH Wt. of solvent (kg) aOH × 1000 No.of moles of Na 3= 1.11V - No. of moles of NaOH × 40 (No. of moles of NaOH/V ) × 1000 3= 1.11 - (No. of moles of NaOH/V ) × 40 No. of moles of NaOH Molarity = × 1000 V = 0.00297 × 100 = 2.97 M 3=

35. (4) 1 mol KMnO4 requires 5 mol of FeSO4.

1 Therefore, 50 × 0.01 M FeSO4 requires × 50 × 0.01 mol 5 KMnO4.



1 Also, ´ 50 ´ 0.01 = 0.01 ´ V Þ V = 10 mL 5 -4



= 2 × 10 × 6.023 × 10

           

= 1.20 × 1020 (1/ 98) 1 7. (4) Molarity = 3 = ≈ 0.01M 1 98

Chapter 1_Some Basic Concepts of Chemistry.indd 22

23

No. of moles

Ratio

Al

10.5

10.5 = 0.38 27

0.38 =1 0.38

K

15.1

15.1 = 0.38 39

0.38 =1 0.38

S

24.8

24.8 = 0.775 32

0.775 =2 0.38

O

49.6

49.6 = 3 -1 16

3.1 =8 0.38

Therefore, empirical formula of anhydrous salt is KAlS2O8.

36. (2) Number of H+ ions = (100 × 0.001 × 10 -3 ) × 2 × N A

Weight (%)

5. (2) As gas is triatomic, therefore, mol. wt. = 3a Thus, number of moles =

w 3a

6. (1) No. of valence electrons in N 3- = (5 + 3) = 8

No. of valence electrons in 4.2 g of N3N 3- =

4.2 × N A × 8 = 2.4 N A 14

1/4/2018 5:08:50 PM

Some Basic Concepts of Chemistry

7. (1) Number of Mg atoms =



2.68 3 × × NA 100 24

We know that Mass of liquid = Number of moles × Molecular weight

    = 2.01 ´ 1021 atoms

8. (3) Given, Number of moles of A = Number of moles of B

1.4 0.8 1.4 = ⇒ MA = M B or M B = 0.57 M A M A MB 0.8



23

9. (4) Mass of carbon = (0.5 × 6 ) × 12 g = 36 g



17. (2) The reaction involved is nCH 2

Number of moles of ethylene =



From the reaction, we have

56  4  ×  p (7 )3 × 1.4  100  3 10. (3) Number of moles of iron = = 20.12 56 11. (2) Total number of protons can be calculated as 8.4 × (6.02 × 1023 ) × (12 × 1 + 6 × 1 + 3 × 8) 84 = 2.52 × 1024 =

510 ´ 10 -3 = 3 ´ 10 -3 Þ Molecular weight = 170 g Molecular weight CH 2 → (CH 2 CH 2 )n

100 mol 28

No.of moles of C 2H 4 No.of moles of (CH 2 CH 2 )n = n 1 100 28n



Number of moles of (CH 2 CH 2 )n =



Therefore,



Hence, weight of polyethylene = 100 g

Weight of polyethylene 100 = 28n 28n

12. (3) We know that

Specific heat × atomic mass = 6.4



Atomic mass =

18. (3) The reaction involved is x x  C10H x +  10 +  O2 → 10CO2 + H 2O  4 2

6.4 = 40 0.16

(No.of moles of C10H x ) No.of moles of O 2 = 1 (10 + x/4) 2.5 32.5 ⇒ x = 12 ⇒ = 1 (10 + x/4)

13. (3) Let the number of iron atoms be x. Therefore, 0.33 x × 56 = ⇒x=4 100 67200 14. (4) Charge on one Al3 + ion = 3 (+e)

19. (2) Applying POAC on Ag



Charge on 1 g ion = (1 × NA)(3e)

1 ´ nAg = 2 ´ nAg 2 S



   = 3 NAe C

1´

15. (4)

SiO2 + H2O + Impurity



Therefore, weight of ore =



“100 g”

1 Wt.of Ag 2S = 2´ Þ WAg 2 S = 1.148 g 108 248

“y” g 45 12

D

43

SiO2 + H2O + Impurity

a% = a 100 8 ×y 100

Chapter 1_Some Basic Concepts of Chemistry.indd 23

20. (4) We have S ¾Several ¾¾ ® BaSO4 steps

Applying POAC on S 1 × nS = 1 × nBaSO4

(92 - a) ×y 100

(92 - a ) ´ y = 43 (1) 100 (92 -aa ) ´ y = 45 ((21)) 100 ´ y = 43 100 45 aa = 43a = 92 45 - 45a Þ a = 47% ´ay) (2) = 45 ( 2) Dividing byÞEq. (1), we´get (100 92 -Eq. 43 45 a = Þ 43a = 92 ´ 45 - 45a Þ a = 47% (92 - a ) 43

16. (3) Number of moles of vapor =

y

67.2 × 10 -3 = 3 × 10 -3 22.4

1.1481 = 85.68 g (1.34 /100)

1×

8 = 1 × nBaSO4 ⇒ nBaSO4 = 0.25 32

21. (1) Let x mL of CH4 is reacted with oxygen. The reactions are CH 4 + 2O 2 → CO2 + 2H 2O x mL

CH +

2 2 (100- x )mL

2 x mL

x mL

5 O 2 → 2 CO 2 + H 2O 2 2(100- x )mL

5 (100- x )mL 2



Reduction when treated with KOH = Volume of CO2







Therefore, CH4 = 35 mL and C2H2 = 65 mL

165 = x + 2(100 – x) ⇒ x = 35 mL

1/4/2018 5:08:54 PM

24

OBJECTIVE CHEMISTRY FOR NEET

22. (1) The reaction is X → X 2H6 + XH 4

      =

Applying POAC on X, we get 1 ´ nX = 2 ´ nX 2 H6 + 1 ´ nXH4 5 = 2 ´ n X 2 H6 + 1 ´ n X 2 H6 (Mol. wt.)X





n X 2 H6 =

5 (1) 3(Mol. wt.)X

nX 2 H6 ´ [(Mol. wt.)X ´ 2 + 6] + nXH4 ´ [(Mol. wt.)X + 4] = 5.628

27. (1) Let the weight of N2 be 14x g.

Therefore, number of moles of N2 =



Let the weight of H2 be 3x g.



Therefore, number of moles of H2 =



The reaction involved is

nX 2 H6 ´ [(Mol. wt.)X ´ 2 + 6] + 2nX 2 H6 ´ [(Mol. wt.)X + 4] = 5.628  (2)



At t = 0

From Eq. (1) and Eq. (2), we get

At t = t

23. (1) Let the volume be V mL. Then Weight of CuSO4 =

1.595 × 1.2V 100

æ 1.595 ö ´ 1.2V ÷ çè ø 100 Molarity = = 0.12 M 159.5 ´ (V /1000)



We know that





Therefore, total number of HCl moles = (1.36V + 200 × 2.4) × 10-3







On solving, we get V = 102.94 mL

Final molarity = 1.24 =



From the reaction, we have nNaOH = nCH3 COOH



Given, mole fraction of acetic acid = 0.15, therefore, mole fraction of H2O = 0.85



Let the number of moles of CH3COOH be a



Therefore, number of moles of H2O =

Therefore, I2 in the reaction giving ICl3 will be 25.4 – a g



The reactions are I2

Moles of solute Volume of solution

0.185 ⇒ V = 18.5 L V

Moles of urea We know Molality = weight of solvent(kg)

Chapter 1_Some Basic Concepts of Chemistry.indd 24

+

Cl 2

a mol 254

→

2ICl 2a 254

a mol 254 +

I2

3Cl 2

( 25.4 - a ) mol 254

a × 0.85 0.15

26. (2) Weight of solution = 1.052 × 1500 = 1578 g

nNH3 nN 2 + nH2 + nNH3



→

3( 25.4 - a ) mol 254

2ICl 3 2( 25.4 - a ) 254

3( 25.4 - a )  a Total weight of Cl2 = 14.2 =  +  × 71 254  254 25.4 a= 2

 a  a × 60 +  × 0.85 × 18 = 30 ⇒ a = 0.185  0.15 

0.01 =

3x 0 2 3x - 3a 2a 2

28. (1) Let the I2 in the reaction giving ICl be a.

NaOH + CH 3COOH → CH 3COONa + H 2O

We know Molarity =

3x mol 2

( 2a ) × 100 = 40  x    3x   2 - a  +  2 - 3a  + 2a    a x 0.4 2 = 0.4 ⇒ = = a-x a 0.6 3 x   - a  2 xN2 = = 0.15 ( 2 x - 2a )

(1.36V + 200 × 2.4) (500 /1000)

25. (2) Let the volume be V L. The reaction is



x 2 x -a 2

Mole fraction of NH 3 =

24. (2) Let the volume be V mL.

14 x x = mol 28 2

N 2 + 3H 2 → 2NH 3

(Mol. wt.)X = 28



(18/60) = 0.192 (1578 - 18)/1000



Therefore, ratio of moles of ICl to ICl3 is =

( 2a / 254)

 25.4 - a  2  254 

a 25.4 / 2 = ( 25.4 - a ) 25.4 - 25.4 / 2 = 1:1

=

1/4/2018 5:08:57 PM

Some Basic Concepts of Chemistry M1V1 = M 2V2 18 × V1 = 0.1 × 1 0.1 × 1 V1 = = 0.00555 L or 5.55mL 18

29. (2) The reaction is HCl + AgNO3 → AgCl ↓ + HNO 3

25

From the reaction, we have nHCl = nAgCl

2.125 ⇒ M = 0.6 3. (1) The reaction involved is 143.5 98 98   PbO   +  2HCl ® PbCl2 + H2O 30. (3) Equivalent wt. of H 3PO 4 = = = 32 .66 (No. of replaceble H ) 3 Amount given  6.5 g    3.2 g 98 98 H 3PO 4 = = = 32.66 No. of moles   0.02 mol  0.087 mol (No. of replaceble H ) 3

Therefore, 25 × M × 10 -3 =

31. (3) We have Eq. wt. of Mg = Eq. wt. of Cu 0.534 1.415 = (Eq. wt.)Mg (Eq. wt.)Cu 0.534 1.415 = ⇒ (Eq. wt.)Cu = 31.8 12 (Eq. wt.)Cu



As, PbO is a limiting reagent, 0.029 mol of PbCl2 will be formed.

4. (2) The empirical formula of the compound is calculated as %

% At. wt.

Simplest ratio

C

38.71

38.71 12

1

H

9.67

9.67 1

3

O

51.62

51.62 16

1

32. (3) Atomic weight = 9 × 3 = 27 33. (1) Equivalent wt. of A = Equivalent wt. of B m1 m2 m = ⇒ E1 = 1 × E 2 E1 E 2 m2 56 = 18.6 3 9. (4) Equivalent wt. of metal hydride = Equivalent wt. of H2 3 34. (1) Equivalent wt. =

0.84 0.042 = Þ (Eq. wt.)Metal hydride = 20 (Eq. wt.)Metal hydride 2/ 2



Thus, the empirical formula is CH3O.

5. (4) The reaction involved is 1 H 2 + O 2 → H 2O 2

40. (4) Atomic wt. of metal = 12 × 2 = 24

Molecular wt. of oxide (MO) = 24 + 16 = 40

Previous Years’ NEET Questions 1. (1) Average atomic mass of element X can be calculated as follows.



10 g of hydrogen contains 5 mol.



64 g of oxygen contains 2 mol.



1 mol of oxygen gives 2 mol of water.



Therefore, 2 mol of oxygen will produce 4 mol of water.

w Na 2 CO3 % of isotope-1 × atomicmass-1 + % of isotope-2 × atomicmass-2 25.3 g 6. (1) We have nNa 2 CO3 = = = 0.238 mol + % of isotope-3 × atomicmass-3 M Na 2 CO3 106 g mol -1 90 8 2 ´ 200 + ´ 199 + ´ 202 100 100 100 = 180 + 15.92 + 4.04 = 199.6 amu =



Na 2CO3(aq)  2Na + (aq) + CO23- (aq)

2. (1) The molarity of H2SO4 is calculated as Molarity = =

r(solution) × w(H 2SO 4 ) M (H 2SO4 ) (1800 g L-1 ) × 0.98 = 18 mol L-1 98 g mol -1

Using the relation, M1V1 = M 2V2, the volume of acid can be calculated

Chapter 1_Some Basic Concepts of Chemistry.indd 25

The reaction involved is

The number of moles of nCO2- = 0.238

nNa + = 0.238 × 2 = 0.476 and

3



Therefore, Molarity =

Number of molesof solute Volumeof solution

Molarity of Na + =

0.476 ´ 1000 = 1.91 M 250

Molarity of CO23- =

0.238 ´ 1000 = 0.95 M 250

1/4/2018 5:09:01 PM

26

OBJECTIVE CHEMISTRY FOR NEET

7. (1) We know 1 mol = Avogadro number of molecules





13. (1) The reaction involved is

Therefore, 0.1 mol contains = 0.1 ´ N A molecules = 0.1 ´ N A ´ 3(as triatomic) = 1.806 ´ 1023 atoms

Molality × Mol.wt. of H 2O w H2 O

1 × 18 1000 = 0.018

=



Initial moles 0.0416 Final moles (0.0416 - 2 ´ 0.0175) = 0.0066 mol

Since, amount of O2 available is 0.0175 mol. Therefore, O2 is the limiting reagent and Mg is the excess reagent.



Excess of Mg left = 0.0066 × 24 = 0.158 g

14. (1) We know that Number of moles (n ) =

nMoles of HNO3 = M ´ V =

250 ´ 2 = 0.5 mol 1000



It is given that 70% HNO3, that means, 70 g of HNO3 is present in 100 g solution.



Therefore, 0.5 × 63 g of HNO3 is present in 100 0.5 × 63 × = 45 g of conc. HNO3 70

10. (1) The reaction involved is [Cr(H 2O)4 Cl 2 ]Cl + AgNO3 → [Cr(H 2O)4 Cl 2 ]NO3 + AgCl ↓ n(Complex solution) = M × V

The complex [Cr(H2O)4Cl2]Cl has one ionizable Cl–, therefore, 0.001 mol will precipitate the same amount of AgCl.

11. (2) We know Number of moles=

number of molecules NA

=

6.02 × 10 = 10 -3 mol 6.023 × 1023

  



20

n ×1000 Molar concentration (M) = Vsolution =

-3

10 × 1000 = 0.01M 100

12. (1) 1 mol of H2 needs 1 mol of Cl2, therefore,

22.4 L needs 22.4 L of Cl2



Available volume of Cl2 = 11.2 L



So, Cl2 is the limiting reagent here.

Chapter 1_Some Basic Concepts of Chemistry.indd 26

nH2 =

Given that

mH2 mO2

nH2 nO2

=

; nO2 =

mO2 M O2

1 . Therefore, 4

mH2

=

mH2 M H2

Given mass (m ) Molar mass (M )

mO2

×

M O2 M H2

=

1 32 4 × = = 4 :1 4 2 1

15. (2) 1 mol of water contains 6.023 × 1023 molecules. Therefore, 18 mol water contains = 18 × 6.023 × 1023 molecules. 16. (4) 12 g of carbon = 6.022 × 1023 atoms (mass of 1 mol)

= 0.01 × (100/1000) = 0.001mol

1 O2 ® MgO 2 0.0175



9. (1) We know



+

Mg

8. (3) Mole fraction can be calculated as xsolute =

Thus, 1 mol of Cl2 will give 1 mol of HCl.



Mass of 1 mol of carbon =

12 × 6.022 × 1020 = 12 × 10−3g 6.022 × 1023

17. (2) The decomposition reaction is MgCO3 ∆ → MgO + CO 2 20 = 0.238 mol 84



Number of moles of MgCO3 =



1 mol of MgCO3 gives 1 mol of MgO.



Therefore, 0.238 mol of MgCO3 will give 0.238 mol of MgO.



The amount of MgO obtained in gram is 0.238 × 40 g = 9.52 g.



Amount of magnesium oxide obtained = 8 g



Therefore, Experimental yield 8 % purity = ×100 = × 100 = 84% Theoretical yield 9.523

18. (2) Number of moles of water = xsolute =

1000 = 55.5 mol 18

nsolute 1 = = 0.0177 nsolute + nH2 O 1 + 55.5

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Some Basic Concepts of Chemistry 19. (1) 16.9% solution of AgNO3 implies that 8.45 g of AgNO3 is present in 50 mL solution.

Similarly, 5.8% of NaCl indicates 2.9 g of NaCl is present in 50 mL solution.



Number of moles of AgNO3 =



8.45g = 0.049 mol 170 g mol -1 2.9 g = 0.049 mol Number of moles of NaCl = 58.5 g mol -1 The reaction AgNO3 Initial amount Final amount



+ NaCl

0.049 0

→ AgCl

0.049 0

0 0.049

+ NaNO3 0 0.049

Then number of moles of XY2 is

Chapter 1_Some Basic Concepts of Chemistry.indd 27

10 = 0.1 AX + 2 AY

AX + 2 AY = 100 (1)



 



Then number of moles of X3Y2 is nX 3 Y2 =



9 = 0.05 3 AX + 2 AY

3 AX + 2 AY = 180 (2)







On solving Eq. (1) and Eq. (2), we get AX = 40 and AY = 30.

21. (1) Molarity of a solution is defined as the number of moles of solute per liter of solution.

Therefore, mass of AgCl precipitated = 0.049 mol × 143.3 g mol−1 = 7 g

20. (4) Let us assume that atomic weight of X is AX and Y is AY.

nXY2 =

27

Molarity =

Molesof solute Liter of solution

Since, volume is a function of temperature, therefore, molarity changes with temperature.

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Chapter 1_Some Basic Concepts of Chemistry.indd 28

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2

Structure of Atom

Chapter at a Glance 1. (a)   The smallest indivisible particle is known as an atoms. All atoms have three fundamental s­ ubatomic particles: Mass

Particle

Electric Charge (C)

Kilograms (kg)

Electron

-1.6 ´ 10-19

9.109382 ´ 10-31 5.485799 ´ 10-4

e-

J.J. Thomson

Proton

+1.6 ´ 10-19

1.672622 ´ 10-27

1.007276

p

Goldstein

0

1.674927 ´ 10-27

1.008665

n

Chadwick

Neutron

Atomic mass units (u)

Symbol Discoverer

Location Outside the nucleus Inside the nucleus Inside the nucleus

(b) Quantization of charge was given by Millikan oil drop experiment as q = ne where q = charge on any particle, n is any integer (……., -3, -2, -1, 0, 1, 2, 3…) and e is electronic charge (1.6 ´ 10-19 C) 2. Atomic Models (a) Thomson model: It is also known as plum pudding model. Thomson proposed that positive charge is spread over a sphere in which electrons are embedded to make the atom neutral as a whole. (b) Rutherford experiment: It leads to the discovery of the nucleus. A beam of a-particles was directed on a thin gold foil of about 0.0004 cm thickness. The following observations were made: (i) Most of the a-particles were able to pass through the foil straight. (ii) A small fraction of them got deflected through small angles. (iii) A very few did not pass through the foil and were deflected back through an angle of 90° or more. (iv) Observations from Rutherford’s experiment indicated that a positively charged heavy mass occupies only a small volume in an atom. The term nucleus was coined by Rutherford for this positively charged mass in the atom. (v) Radius of nucleus = 10-15 m and density = 108 tonnes cm-3. 3. Drawbacks of Rutherford Model



(a) Rutherford’s atomic model could not explain the stability of the atom. (b) It was not able to explain the line spectra for various elements. (c) It was unable to explain the energies of electrons and their distribution around the nucleus.   Distance of closest approach 4kZe 2 r0 = mv 2  where Z is atomic number, e is charge of electron, m is mass of a-particle and v is the velocity of a-particle.

4. Moseley’s Experiment u µ ( Z - 1)2 or u = a( Z - b )



 Here, u = frequency of X-rays, Z = atomic number of the element, a and b are constants.

Chapter 2_Structure of Atom.indd 29

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30

OBJECTIVE CHEMISTRY FOR NEET

5. Representation of Atom with Electrons and Neutrons (a) Atomic number (Z ) = Number of protons in the nucleus of an atom = number of electrons in a neutral atom.   Protons and neutrons present in the nucleus are collectively known as nucleons. (b) Mass number: The total number of nucleons is termed as mass number (A) of the atom.   Mass number (A) = number of protons (Z ) + number of neutrons (n)  An atom can be represented as ZA X where X is the symbol of the element, and A and Z represent the mass number and atomic number, respectively. (c) Isotopes: The element species having the same atomic numbers, but different atomic mass, for example, for 1 2 example, 1 H, 1 H and 13 H . (d) Isobars: The atoms of different elements having the same mass number but different atomic number, for 40 40 40 example, 18 Ar , 19 K and 20 Ca . (e) Isotones: The atoms of different elements having the same number of neutrons, for example, 14 15 16 6 C, 7 N and 8 O. (All have eight neutrons.) (f ) Isodiaphers: Two or more element species having the same difference between numbers of neutrons (n) and 23 19 protons (p) or having the same value of (A–Z). For example, 11 Na and 9 F have n - p = 1. (g) Isoesters: Two compounds which have the same number of total electrons and same number of atoms. For example N2O and CO2 has total electron = 22 and number of atoms in a molecule is three. 6. Planck’s Quantum Theory According to Planck’s quantum theory, electromagnetic radiation can be viewed as a stream of discrete small packets or quanta of energy. Energy is emitted in the form of small packets called quanta or photon. E = hu =h

c l

c  u =  l

where h is Planck’s constant and is equal to 6.626 ´ 10-34 Js, l = wavelength, u is frequency and c = velocity of light. 1 1 and T = l n where n is wave number and T is time period. Also, n =

7. Photoelectric Effect It is the phenomenon of emission of electrons from a metallic surface when exposed to light radiation. The electrons which are emitted are known as photo electrons. Einstein explained this effect based on Planck’s theory. (a) Maximum kinetic energy of photoelectrons = Absorbed energy − Work function   Work function (W0 or f) is the minimum energy (hu0) required by an electron to escape from the metal surface. K.E. = hu - W 0 1 1 1 2 mv = hu - hu 0 = hc  -  2  l l0 



 where n0 and l 0 are threshold frequency and wavelength respectively. (b) The minimum potential at which the photoelectric current reaches zero, is called the stopping potential, V0. eV0 = h(u - u0 )



(c) R  ate of emission of photoelectrons from a metallic surface is directly proportional to the intensity of light falling on it. (d) Kinetic energy of photoelectrons is directly proportional to the frequency of incident radiation, but independent of intensity of light. (e) For emission of photoelectrons the incident light frequency must be greater than threshold frequency.

Chapter 2_Structure of Atom.indd 30

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Structure of Atom

31

8. Blackbody Radiation A body which can emit or absorb radiations of all frequencies is known as blackbody and radiation emitted by it is known as blackbody radiation. 9. Line Spectrum of Hydrogen (a) The wave number of any line of atomic spectra of hydrogen atom in visible region is given by 1 1 1 = n = RH  2 - 2  l  n1 n2      where, RH is Rydberg constant for hydrogen and is equal to 109678 cm−1. (b) The series are named as: Lyman series Balmer series Paschen series Brackett series Pfund series Humphrey series

n1

n2

1 2 3 4 5 6

2, 3, 4, 5 3, 4, 5, 6 4, 5, 6, 7 5, 6, 7 6, 7 7

Observed in Ultraviolet region Visible region Infrared red region Infrared red region Infrared red region Infrared red region

10. Bohr’s Atomic Model It is applicable to the hydrogen or hydrogen like atom such as He+, Li2+, Be3+, B4+, etc. According to Bohr’s model electrons are moving in stationary orbits around the nucleus. (a) Angular momentum of the electron is given by nh mvr = 2p (b) The radii of the stationary states is known as Bohr radius and can be expressed as rn = n 2 a0, where a0 = 52.9 pm. n2 The radius of nth shell is given by rn = a0 Z  where n number of the shell and Z is atomic number. Energy of electron in the nth shell Z2 ; RH = - 13.6 eV n2  where RH is Rydberg constant for hydrogen and its value is 2.18 × 10−18 J.   Velocity of electron in the nth shell Z vn = 2.188 × 108 × cm s -1 n En = RH

11. Dual Nature of Matter (a) de Broglie equation: According to de Broglie, matter also exhibits wave like behavior. The wavelength of a matter wave, l is given by the equation h h =       l = mv p  where h is Planck’s constant, m is the particle’s mass, v is its velocity and p is the momentum.  or           l =

h = 2 Em

h 2mqV

 where E is kinetic energy of particle; q is the charge of particle and V is the potential difference.

Chapter 2_Structure of Atom.indd 31

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32



OBJECTIVE CHEMISTRY FOR NEET

(b) H  eisenberg Uncertainty Principle: It states that it is impossible to measure simultaneously both the position and momentum of any microscopic particle with accuracy. h Δx ⋅ Δ p ≥ 4p  where, Δ x is the uncertainty in position and Dp is the uncertainty in momentum.

12. Quantum Mechanical Model of Atom: The fundamental equation of wave mechanics is the Schrödinger wave equation which provides a satisfactory description of an atom in these terms of duality of wave and particle. d 2y d 2y d 2y 8p 2m + 2 + 2 + 2 ( E - V )y = 0 dx 2 dy dz h where y is the amplitude of the wave at a point with coordinates x, y and z. E is the total energy. 13. Quantum Numbers: Set of four integers required to define an electron completely in an atom. (a) Principal quantum number (n): It describes the size and energy of the shell. n = 1, 2, 3, 4…. represents K, L, M, N shells and so on. (b) Azimuthal quantum number (l ): It describes the shape of the electron cloud and number of subshells in a shell. For a given value of n, l can have values 0 to n - 1. l = 0 → for s subshell l = 1 → for p subshell h Angular momentum = l (l + 1) 2p (c) Magnetic quantum number (ml ): It describes the orientations of the subshell. Its values varies from −l to +l. (d) Spin quantum number (ms ): It describes the spin of the electron. It has value +1/2 or −1/2. h Spin angular momentum = s ( s + 1) 2p 14. Some Important Points to Remember (a) (b) (c) (d) (e) (f ) (g)

Number of subshells in a shell = n Maximum number of orbitals in a shell = n2 Maximum number of electrons in a subshell = 2(2l + 1) Maximum number of electrons in a shell = 2n2 Maximum number of subshells in a shell = 2l + 1 Number of spherical nodes (radial node) = n − l − l Number of angular nodes/nodal plane = l

15. Shapes of Subshells (a) s-Subshell

(b) p-Subshells x

x

x y

y z px

Chapter 2_Structure of Atom.indd 32

y z

z py

pz

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Structure of Atom

(c) d-Subshells

x

33

x y

z y

z dxz

dxy x

y

x z

y z

dyz

dx2-z2

dz2

(d) f-subshells: These subshells have complicated shapes with seven lobes. 16. Nodal Plane: It is the plane at which the probability of finding electrons present in that orbital is zero. The number of nodal planes is ‘l  ’. (a) (b) (c) (d) (e) (f )

The number of nodal planes in s-orbital is zero. The number of nodal planes in p-orbitals is one. The number of nodal planes in d-orbitals is two. Total number of radial nodes = n - l - 1. Total number of angular nodes = l. Total number of nodes = n - 1.

17. Rules for filling of orbitals in atom (a) Aufbau principle (i) In the case of atoms, electrons occupy the available orbitals in the subshells of lowest energy. This is known as the Aufbau principle which determines the assignment of all the electrons in an atom into specific shells and subshells is known as the element’s electronic configuration. 1s 2s

2p

3s

3p

3d

4s

4p

4d

5s

5p

5d

6s

6p

6d

4f

7s

(ii) T  he energy of the orbital is given by (n + l ) rule. Lower the (n + l ) value for an orbital, lower is the energy. If the value of (n + l ) for two orbital is same, then the orbital having higher value of n has higher energy.

Chapter 2_Structure of Atom.indd 33

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34

OBJECTIVE CHEMISTRY FOR NEET

(b) Pauli’s exclusion principle (i) This states that no two electrons in one atom can have all four quantum numbers the same. (ii) By permutating the quantum numbers, the maximum number of electrons which can be contained in each main energy level can be calculated. Subshell s p d f

Number of orbitals 1 3 5 7

Maximum number of electrons 2 6 10 14

(c) Hund’s rule of maximum multiplicity  It states that pairing of electrons in orbitals of p, d and f subshells does not take place till each orbital belonging to that subshell has got one electron each.

Solved Examples 1. The neutron is attracted towards

Solution

(1) positive charged particles. (2) negative charged particles. (3) not attracted by any charge. (4) none of these.

(4) Energy of electron in first Bohr’s orbit of hydrogen atom 2.18 × 10 -18 is E = J n2 E2 = -

Solution (3) Neutron is an uncharged particle. Hence (3) is correct. 2. For which of the following species, Bohr theory does not apply? (1) H (2) He+ (3) Li2+ (4) Na+

(4) Bohr theory is not applicable to multi-electron species. 3. If the radius of second Bohr orbit of hydrogen atom is r2. The radius of third Bohr orbit will be 4 (1) r2 (2) 4r2 9 9 (3) r2 (4) 9r2 4 Solution (3) Radius of Bohr orbit in hydrogen can be expressed as n 2h 2 r= 2 4p mZe 2 r2 22 9 Therefore, = ⇒ r3 = r2 r3 32 4 4. The ionization energy of the ground state hydrogen atom is 2.18 ´ 10–18 J. The energy of an electron in its second orbit would be (1) –1.09 ´ 10 J (2) –2.18 ´ 10 J (3) –4.36 ´ 10–18 J (4) –5.45 ´ 10–19 J

Chapter 2_Structure of Atom.indd 34

5. The velocity of electron in the ground state hydrogen atom is 2.18 ´ 106 ms–1. Its velocity in the second orbit would be (1) 1.09 ´ 106 ms–1 (2) 4.38 ´ 106 ms–1 (3) 5.5 ´ 105 ms–1 (4) 8.76 ´ 106 ms–1 Solution

Solution

–18

2.18 × 10 -18 J = - 5.45 × 10 -19 J 22

–18

(1) We know that velocity of electron in nth Bohr’s orbit is Z given by v = 2.18 × 106 × ms -1 n

For hydrogen, Z = 1, therefore, 2.18 × 106 ms -1 1 2.18 × 106 v2 = ms -1 = 1.09 × 106 ms -1 2 v1 =



6. The speed of the electron in the first orbit of the hydrogen atom in the ground state is (c is the velocity of light) c c (2) 1.37 1370 c c (3) (4) 13.7 137 (1)

Solution (4) Velocity of electron is given by v=

h = 2.189 × 108 cms -1 2p mr

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35

Structure of Atom

We know   c = 3 ´ 1010 cm s-1 c 3 ´ 1010 137 c = = Þv = Therefore 8 v 2.189 ´ 10 1 137

Solution (3) We know

7. The wave number of first line of Balmer series of hydrogen atom is 15200 cm-1. What is the wave number of first line of Balmer series of Li2+ ion? (1) 15200 cm-1 (2) 6080 cm-1 (3) 76000 cm-1 (4) 1,36,800 cm-1 Solution (4) For Li2+, we have n = n Hydrogen × Z 2 = 15200 × 9



8. The degeneracy of the level of hydrogen atom that has R energy - H is 16 (1) 16 (2) 4 (3) 2 (4) 1 R (1) We have     E n = - H2 n RH R - 2 = - H ⇒n = 4 Therefore, n 16 R Hence, 4th subshell has energy - H . For n = 4, the value 16 of l is 0, 1, 2 and 3. n

l

m

Total orbitals

4

0

0

1

1

-1, 0, 1

3

2

-2, -1, 0, +1, +2

5

3

-3, 2, 1, 0, +1, +2, +3

Degeneracy

7 16

9. The wave number of the shortest wave length transition in Balmer series of atomic hydrogen will be (1) 4215Å (2) 1437 Å (3) 3942 Å (4) 3647 Å  1 1 1 = RZ 2  2 - 2  l  n1 n2 

1  1 = 109678 × 1 ×  2 - 2      2 ∞  2

-5

l = 3.647 × 10 cm or 3647 Å 10. Magnetic moments of V (Z = 23), Cr (Z = 24) and Mn (Z = 25) are x, y, z. Hence (1) z < y < x (2) x = y = z (3) x < z < y (4) x < y < z

Chapter 2_Structure of Atom.indd 35

Electronic configuration

Unpaired electrons

Magnetic moment

V ( Z = 23)

[Ar] 3d 34s2

3

15 BM

Cr (Z = 24)

[Ar] 3d 54s1

6

48 BM

Mn (Z = 25)

[Ar] 3d 54s2

5

35 BM

(1) 3.31 ´ 10–3 Å (2) 1.33 ´ 10–3 Å (3) 3.13 ´ 10–2 Å (4) 1.31 ´ 10–2 Å Solution (2) According to de Broglie’s relation, we have l =

Solution

(4) We know

Element

11. What is the de Broglie wavelength? Assume that one mole of protons has a mass equal to 1 g? (h = 6.626 ´ 10–27 erg s)

= 1, 36, 800 cm -1

Solution

n(n + 2) BM

Magnetic moment =

where n is the number of unpaired electron.



We know 1 mol = 6.023 ´ 1023 protons = 1 g 1 g Therefore, mass of one proton = 6.023 ´ 1023 From Eq. (1), we get

h   (1) mv

h mv 6.626 ´ 10 -27 = ´ 6.023 ´ 1023 1 ´ 3 ´ 108

l=

= 13.302 ´ 10 -12 cm or

1.33 ´ 10 -3 ´ 10 -8 cm or 1.33 ´ 10 -3 Å

12. An electron is moving with a kinetic energy of 4.55 ´ 10–25 J. What will be de Broglie wavelength for this electron? (1) 5.28 ´ 10–7 m (2) 7.28 ´ 10–7 m (3) 2 ´ 10–10 m (4) 3 ´ 10–5 m Solution (2) We know    K.E. = 2          v =



1 mv 2 = 4.55 × 10 -25 2 2 × 4.55 × 10 -25 = 106 ⇒ v = 103 ms -1 9.1 × 10 -31

From de Broglie’s equation, l=

l=

h mv

6.626 ´ 10 -34 = 7.28 ´ 10 -7 m 9.1 ´ 10 -31 ´ 103

13. The quantum number not obtained from the Schrödinger’s wave equation is (1) n (2) l (3) m (4) s

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36

OBJECTIVE CHEMISTRY FOR NEET

Solution

Solution

(4) Quantum numbers n, l and m can be obtained from Schröding­er equation while s is obtained from spectral evidence.

(1) For n = 3, the set of quantum numbers possible are: n

l

3

0

0

1 2

-1, 0, +1 -2, -1, 0, +1, +2

14. Number of waves made by an electron in one complete revolution in third Bohr orbit is (1) 2 (2) 3 (3) 4 (4) 1

16. The two electrons present in an orbital are distinguished by

Solution

(1) (2) (3) (4)

(2) Circumference of third orbit is 2pr3.

According to Bohr’s, angular momentum of electron in third orbit is



mvr3 = 3

h 2p

or

h 2p r3 = (1) mv 3

From de Broglie equation, we have l =

h (2) mv

From Eq. (1) and Eq. (2), we get

principal quantum number. azimuthal quantum number. magnetic quantum number. spin quantum number.

Solution (4) The two electrons present in an orbital are distinguished by their spin quantum number. It refers to the two possible orientations of the spin axis of an electron and can take the values +1/2 and -1/2 for a given value of m. 17. Assuming that a 25 W bulb emits monochromatic yellow light of wave length 0.57 mm. The rate of emission of quanta per second will be

2p r3 l= Þ 2p r3 = 3l 3

m

Circumference of third orbit is three times the wavelength of electron or number of waves made by Bohr electron in one complete revolution in third orbit is three.

(1) 5.89 ´ 1015 s–1 (2) 7.28 ´ 1017 s–1 (3) 5 ´ 1010 s–1 (4) 7.18 ´ 1019 s–1 Solution (4) Let n be the quanta evolved per second. Then  hc  n   = 25 Js -1  l

15. Which of the following sets of quantum numbers is not allowed?

n × 6.626 × 10 -34 × 3 × 108 = 25 0.57 × 10 -6

(1) n = 3, l = 1, m = +2 (2) n = 3, l = 1, m = +1 (3) n = 3, l = 0, m = 0 (4) n = 3, l = 2, m = ± 2

n = 7.18 × 1019 s -1

Practice Exercises Level I 4. X2- has 56 electrons, the atomic number X is

Early Models, Electromagnetic Waves and Planck’s Theory

(1) 56 (2) 58 (3) 28 (4) 54

1. 11Na23 and 12Mg24 are

5. The frequency in Hertz of a radiation of wavelength 500 nm is

(1) isotopes. (2) isobars. (3) isodiaphers (4) isotones. 2. In the following reaction 3Li + ? ¾¾® 2He + 1H , the missing particles is 6

4

3

(1) neutron. (2) proton. (3) electron. (4) deuterium. 3. The increasing order (lowest first) for the magnitude of e/m (charge/mass) for electron (e), proton (p), neutron (n) and alpha particle (a) is (1) e, p, n, a (2) n, p, e, a (3) n, p, a, e (4) n, a, p, e

Chapter 2_Structure of Atom.indd 36

(1) 6 Hz (2) 1014 Hz (3) 6 × 1014 Hz (4) 0.6 × 1014 Hz 6. Particle in cathode ray has the same charge/mass ratio as (1) a -particle. (2) b -particle. (3) g -particle. (4) proton. 7. During Muliken’s oil drop experiment, out of the following, which is not a possible charge on oil droplet? (1) 1.6 × 10−19 C (2) 2.4 × 10−19 C (3) 3.2 × 10−19 C (4) 4.8 × 10−19 C

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Structure of Atom 8. The wave number of radiation of wavelength 500 nm is (1) 5 ´ 10 m (2) 2 ´ 10  m (3) 2 ´ 106 m-1 (4) 500 ´ 10–9 m-1 –7

-1

7

-1

9. The ratio of the energy of a photon of 2000 Å wavelength radiation to that of 4000 Å radiation is (1) 1/4 (2) 1/2 (3) 2 (4) 4 10. Radio city broadcasts on a frequency of 5,090 kHz. What is the wavelength of electromagnetic radiation emitted by the transmitter? (1) 10.3 m (2) 58.9 m (3) 60.5 m (4) 75.5 m

Bohr’s Model, Calculation of Radius, Velocity and Energy 11. The kinetic and potential energy (in eV) of an electron present in third Bohr’s orbit of hydrogen atom are respectively (1) –1.51, –3.02 (2) 1.51, –3.02 (3) –3.02, 1.51 (4) 1.51, –1.51 12. The ratio of velocity of the electron in the third and fifth orbit of Li2+ would be (1) 3:5 (2) 5:3 (3) 25:9 (4) 9:25 13. The ratio of energy of the electron in ground state of the hydrogen to electron in first excited state of He+ is (1) 1:4 (2) 1:1 (3) 1:8 (4) 1:16 14. In which of the following one electron system, the radius of first Bohr orbit is minimum? (1) H atom (2) Deuterium ion (3) He+ ion (4) Li2+ ion 15. If radius of second stationary orbit (in Bohr’s atom) is R, then the radius of third orbit will be (1) R/3 (3) R/9

(2) 9 R (4) 2.25 R

16. Which state of Be has the same orbit radius as that of the ground state of hydrogen atom?

18. The spacing between the orbits in terms of distance is maximum in the case of (1) first and second. (2) second and third. (3) third and fourth. (4) fourth and fifth. 19. The ratio of potential energy and total energy of an electron in a Bohr orbit of hydrogen-like species (1) 2 (2) −2 (3) 1 (4) −1 20. The spacing between the orbits in terms of energy is maximum in the case of (1) first and second. (2) second and third. (3) third and fourth. (4) fourth and fifth. 21. Splitting of spectral lines when atoms are subjected to strong electric field is called (1) Zeeman effect. (2) Stark effect. (3) decay. (4) disintegration.

Hydrogen Spectrum and Rydberg Equation 22. Transition of an electron from n = 2 to n = 1 level results (for a H atom) in (1) IR spectrum. (2) UV spectrum. (3) visible spectrum. (4) X-ray spectrum. 23. Hydrogen atom consists of a single electron but so many lines appear in the spectrum of atomic hydrogen because (1) sample contains some impurity. (2) experiment is done on collection of atoms. (3) hydrogen atom splits to form more than one different species. (4) some different isotope of hydrogen atom may be present. 24. Rydberg has given the equation for all visible radiation kn 2 in the hydrogen spectrum as l = 2 . The value of k n -4 in terms of Rydberg constant is (1) 4R (2)

3+

(1) 3 (2) 2 (3) 4 (4) 5

(3)

the hydrogen atom only. all elements. any atomic or ionic species having one electron only. the hydrogen molecule.

Chapter 2_Structure of Atom.indd 37

R 4

4 (4) R R

25. The first emission line of hydrogen atomic spectrum in Balmer series appears at (R = Rydberg constant)

17. Bohr model can explain spectrum of (1) (2) (3) (4)

37

(1)

5R 3R cm -1 (2) cm -1 4 36

(3)

9R 7R cm -1 cm -1 (4) 144 400

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38

OBJECTIVE CHEMISTRY FOR NEET

26. The ionization energy of hydrogen atom in the ground state is x kJ. The energy required for an electron to jump for second orbit to third orbit is 5x (1) (2) 5 x 36 x (3) 7.2 x (4) 6 27. The emission spectrum of He+ ion is the consequence of transition of electron from orbit n2 to orbit n1. Given that 2n2 + 3n1 = 18 and 2n2 – 3n1 = 6, then what will be the maximum number of spectral lines in atomic spectrum when electron transits from n2 to orbit n1? (1) 10 (2) 15 (3) 20 (4) 21 28. If the electron of hydrogen atom is excited to the fourth excited state, then the total spectral lines falling in Paschen series are equal to (1) 2 (2) 10 (3) 3 (4) 4 29. 242 nm is just sufficient to ionize sodium atom. Ionization energy of sodium atom will be -1

(1) 494 kJ mol (2) 24.7 kJ mol

-1

(3) 988 kJ mol -1 (4) 247 kJ mol -1 30. What according to the Bohr model would be the radius of the electron orbit in the first excited state of the Li2+ ions? (1) 0.751 Å (2) 0.705 Å (3) 0.925 Å (4) 0.952 Å 31. Which of the following transitions of an electron in a hydrogen atom would emit a photon of minimum wavelength? (1) 8 to 3 (2) 6 to 1 (3) 2 to 1 (4) 7 to 5 32. Ratio between longest wavelengths of hydrogen atom in Lyman series to the shortest wavelength in Balmer series of He+ is 4 36 (2) 3 5 5 1 (3) (4) 4 9 (1)

33. Which of the following transition between energy levels in the hydrogen atom produces the fourth line of the second series in the hydrogen spectrum? (1) n2 = 5 ® n1 = 1 (2) n2 = 4 ® n1 = 1 (3) n2 = 6 ® n1 = 2 (4) n2 = 5 ® n1 = 2 34. Three atomic states of a hydrogen-like atom are shown in the figure. The transition from C to B yields a photon of wavelength 364.6 nm and the transition from B to A yields a photon of wavelength 121.5 nm, then the transition from C to A will yield a photon of wavelength

Chapter 2_Structure of Atom.indd 38

C 364.6 nm

B

121.5 nm A

(1) 91.2 nm (2) 243.1 nm (3) 486.1 nm (4) None of these

Heisenberg Uncertainty Principle and de Broglie Equation 35. Wavelength associated with Virar-F local train having mass 100 × 103 kg moving with the speed of 23.76 km h-1 is (Planck’s constant = 6.6 × 10-34 Js) (1) 10−31 Å (2) 10−35 Å (3) 10−29 Å (4) 10−40 Å 36. Calculate the wavelength of a track runner who completes the run of 150 m in 12.1 s, if its weight is 50 kg. (1) 9.11 ´ 10–34 m (2) 8.92 ´ 10–37 m (3) 1.12 ´ 10–45 m (4) none of these 37. When applied, the uncertainty principle has significance in case of (1) moving train. (2) spinning cricket ball. (3) moving a-particle. (4) All the above. 38. The momentum of a particle which has a de Broglie wavelength of 0.1 nm is (in kg ms–1) (1) 7.08 × 10–21 (2) 6.62 × 10–24 (3) 6.58 × 10–14 (4) 5.89 × 10–9 39. A ball of mass 200 g is moving with a velocity of 10 ms–1. If the error in measurement of velocity is 0.1%, the uncertainty in its position is (1) 3.3 ´ 10–3 m (2) 3.3 ´ 10–27 m (3) 5.3 ´ 10–25 m (4) 2.64 ´ 10–31 m

Photoelectric Effect 40. The kinetic energy of the electron emitted when light of frequency 3.5 ´ 1015 Hz falls on a metal surface having threshold frequency, 1.5 ´ 1015 Hz is (h = 6.6 ´ 10-34 Js) (1) 1.32 ´ 10-18 J (2) 3.3 ´ 10-18 J (3) 6.6 ´ 10-19 J (4) 1.98 ´ 10-19 J 41. Photoelectrons are ejected from a metal surface using photons of energy 4 × 10-20 J. The de Broglie wavelength of the electron emitted with maximum K.E. = 59 Å. What is the photoelectric threshold in joules? (1) 3.313 × 10-20 (2) 1.131 × 10-20 (3) 1.331 × 10-20 (4) 1.673 × 10-20

1/4/2018 5:09:00 PM

Structure of Atom 42. Photoelectric emission is observed from a surface for frequencies u1 and u2 of the incident radiation (where u1>u2). If the maximum K.E. of the photoelectrons in two cases are in the ratio of 1:k, then the threshold frequency v0 is given by

(1) (u1 − u2)/k - 1 (2) (ku1 − u2)/k - 1 (3) (ku2 − u1)/k - 1 (4) (u1 − u2)/k

Schrödinger’s Wave Equation 43. The number of nodal planes in a px orbital is (1) one. (2) two. (3) three. (4) zero. 44. Non-directional orbital is (1) 3s (2) 4f (3) 4d (4) 4p 45. Which of the following is the correct order of probability of being found close to the nucleus? (1) s > p > d > f (2) f > d > p > s (3) p > d > f > s (4) d > f > p > s 46. A 3p orbital has (1) (2) (3) (4)

two non-spherical nodes. two spherical nodes. one spherical and one non-spherical node. one spherical and two non-spherical nodes.

47. If travelling at equal speeds, the longest wavelength of the following matter waves is that of (1) electron. (2) proton. (3) neutron. (4) alpha particle. 48. Orbital angular momentum for 3f orbital electron is h h (1) 3 (2) 2 2p 2p h (3) 2 3 (4) 0 2p 49. Which of the following subshells does not exist? (1) 7s (2) 3d (3) 3f (4) 5d 50. The difference in angular momentum associated with the electron in two successive orbits of hydrogen atom is (1) h/p (2) h/2p (3) h/2 (4) (n - 1) h/2p 51. The quantum number not obtained from the Schrödinger’s wave equation is (1) n (2) l (3) m (4) s 52. Which one of the orbitals has zero probability of finding electrons in xz plane? (1) px (2) py (3) pz (4) dxz

Chapter 2_Structure of Atom.indd 39

39

53. Which quantum number determines shape of the ­orbital? (1) Principal (2) Angular (3) Magnetic (4) Spin 54. Radial nodes are maximum in (1) 4s (2) 4p (3) 3d (4) 5f 55. Number of nodal planes (planes of zero electron density) in the dxy orbital is (1) 1 (2) 2 (3) 0 (4) 0 56. The number of orbitals in a subshell is equal to (1) n2 (2) 2l (3) 2l + 1 (4) m 57. Which of the following relates to photons both as wave motion and as a stream of particles? (1) Interference (2) E = mc2 (3) Diffraction (4) E = hu 58. Probability of finding the electron in the orbital is (1) 100% (2) 5-10% (3) 90-95% (4) 50-60% 59. The wave mechanical model of atom is based upon (1) de Broglie concept of dual nature. (2) Heisenberg’s uncertainty principle. (3) Schrödinger wave equation. (4) All of these.

Aufbau Principle, Hund’s Rule and Pauli’s Exclusion Principle 60. Which of the following has the maximum number of unpaired electrons? (1) Mn (2) Ti (3) V (4) Al 61. Total spin resulting from a d 5 configuration is (1) 1 (2) 1/2 (3) 5/2 (4) 3/2 62. Which set of quantum numbers (n, l, m, s) represents the outermost electron in a gaseous aluminium atom? 1 1 (2) 2, 1, -1, + 2 2 1 1 (3) 3, 0, 0, + (4) 3, 1, -1, + 2 2 (1) 2, 1, 0, +

63. Electronic configuration of Ni is [Ar]3d8,4s2. The electronic configuration of next element is (1) [Ar]3d104s1 (2) [Ar]3d94s2 (3) 3d84s24p1 (4) None 64. The electronic configuration of the element which is just above the element with atomic number 43 in the same periodic group is

1/4/2018 5:09:01 PM

40

OBJECTIVE CHEMISTRY FOR NEET (1) 1s22s22p63s23p63d54s2 (2) 1s22s22p63s23p63d104s24p5 (3) 1s22s22p63s23p63d64s1 (4) 1s22s22p63s23p63d104s14p6

65. Which of the following is having the maximum number of unpaired electrons?

(1) Mg2+ (2) Ti3+ (3) V3+ (4) Fe2+

Level II Early Models, Electromagnetic Waves, Planck’s Theory 1. A 1000 W radio transmitter operates at a frequency of 880 kilocycle s-1. How many photons per second does it emit? (1) 2.01 × 1029 (2) 1.72 × 1030 (3) 1.51 × 1029 (4) 1.77 × 1031 2. The wave number of a spectral line is 5 × 105 m–1. The energy corresponding to this line will be (1) 3.39 ´ 10–23 kJ (2) 9.93 ´ 10–23 kJ (3) 3.45 ´ 10–24 J (4) 9.93 ´ 10–23 J 3. The eyes of certain members of reptile family pass a visual signal to the brain when the visual receptors are struck by photons of wavelength 890 nm. If a total energy of 3.15 × 10-14 J is required to trip signal, what is the ­minimum number of photons that must strike the receptor? (1) 3.05 × 1019 (2) 1.72 × 109 (3) 1.41 × 105 (4) 2.75 × 1010 4. One molecule of a substance absorbs one quantum of energy. The energy involved when 1.5 mol of the ­substance absorbs red light of frequency 7.5 × 1014 s-1 will be (1) 2.99 ´ 105 J (2) 3.23 ´ 105 J (3) 4.48 ´ 105 J (4) 2.99 ´ 106 J 5. Assuming that a 25 W bulb emits monochromatic yellow light of wavelength 0.57 mm. The rate of emission of quanta per second will be (1) 5.89 ´ 1015 s–1 (2) 7.28 ´ 1017 s–1 (3) 5 ´ 1010 s–1 (4) 7.18 ´ 1019 s–1

6. A certain dye absorbs light of wavelength 4500 Å and then fluorescence light of 5000 Å. Assuming that, under given conditions 50% of the absorbed energy is re-emitted out as fluorescence, calculate the ratio of quanta emitted to the number of quanta absorbed. (1) 0.55 (2) 2.1 (3) 1.8 (4) 0.75

Chapter 2_Structure of Atom.indd 40

Bohr’s Model, Calculation of Radius, Velocity and Energy 7. The potential energy of an electron in the hydrogen atom is –6.8 eV. Indicate in which excited state, the electron is present? (1) First (2) Second (3) Third (4) Fourth 8. Which of the following statements is true? (1) Spacing between energy levels n = 1 and n = 2 in hydrogen atom is greater than that of n = 2 and n = 3. (2) Spacing between energy levels n = 1 and n = 2 in hydrogen atom is equal to that n = 2 and n = 3. (3) Spacing between energy levels n = 1 and n = 3 in hydrogen atom is less than that of n = 2 and n = 3. (4) None of these. 9. What is the potential energy of an electron present in N shell of the Be3+ ion? (1) –3.4 eV (2) –6.8 eV (3) –13.6 eV (4) –27.2 eV 10. What is the atomic number (Z) corresponding to which fourth orbit would fit inside the first Bohr orbit of hydrogen atom? (1) 3 (2) 4 (3) 16 (4) 25 11. If velocity of an electron in first Bohr orbit of hydrogen atom is x, its velocity in third orbit will be x (1) (2) 3 x 3 x (3) 9 x (4) 9 12. The ionization potential for an electron in ground state of the hydrogen atom is 13.6 eV. What would be the ionization potential for the electron in the first excited state of hydrogen atom? (1) 13.6 eV (2) 6.8 eV (3) 3.4eV (4) 27.2 eV 13. The potential energy of the electron in an orbit of ­hydrogen atom would be e2 (1) −mv2 (2) r (3) -

mv 2 e2 (4) 2 2r

14. If the revolutions per second by the electron in third orbit of H is d, then the revolutions per second by the electron in second orbit of He+ is (1) d (2) 13.5 d (3) 1.5 d (4) 0.07 d

1/4/2018 5:09:02 PM

Structure of Atom 15. The binding energy of an electron in the ground state of the He atom is equal to 24 eV. The energy required to remove both the electrons from the atom will be (1) 59 eV (2) 81 eV (3) 79 eV (4) None of these 16. The difference of nth and (n + 1)th Bohr’s radius of hydrogen atom is equal to its (n - 1)th Bohr’s radius. The value of n is (1) 1 (2) 2 (3) 3 (4) 4 17. The time taken by an electron for one complete revolution in the nth Bohr orbit of the hydrogen atom is (1) (2) (3) (4)

inversely proportional to n2. directly proportional to n3. directly proportional to n. directly proportional to (n/h)2.

18. If the kinetic energy of electron moving in fourth orbit of hydrogen is Î, then the total energy in first orbit of Li2+ is (1) –144 Î (2) –0.0069 Î (3) –(27/9) Î (4) –Î 19. The electron in a hydrogen atom makes transition from M shell to L, the ratio of magnitude of initial to final acceleration is (1) 9:4 (2) 81:16 (3) 4:9 (4) 16:81 20. The ionization potential difference is 2.55 eV for two of the Bohr orbits of the atomic hydrogen of quantum numbers n1 and n2, (n1 < n2). What are the values of n1 and n2? (1) 2, 3 (2) 3, 4 (3) 2, 4 (4) 3, 5 21. If the same energy is supplied to electron in ground state of hydrogen as well as He+, electron jumps to fifth main shell in hydrogen, then final orbit of electron in He+ is (1) second. (2) first. (3) third. (4) fourth. 22. The ionization energy of He+ is 19.6 × 10-18 J per atom. The energy of the first excited state of Li2+ is (1) 19.6 × 10-18 J (2) 5 × 10-18 J (3) 4.41 × 10-17 J (4) 4.41 × 10-18 J

Hydrogen Spectrum and Rydberg Equation 23. Which electronic transition in a hydrogen atom, starting from the orbit n = 7, will produce infrared light of ­wavelength 2170 nm? (1) n = 7 to n = 6 (2) n = 7 to n = 5 (3) n = 7 to n = 4 (4) n = 7 to n = 3 24. An electron in a hydrogen atom in its ground state absorbs 1.5 times as much energy as the minimum

Chapter 2_Structure of Atom.indd 41

41

required escape from the atom. What is the wavelength of the emitted electron? (1) 4.7 Å (2) 4.7 nm (3) 9.4 Å (4) 9.4 nm 25. The angular momentum of an electron in a Bohr’s orbit 2h of hydrogen atom is . The wavelength of spectral line p emitted when the electron falls from this level to next lower level is 9 RH Å (2) 16 × 25 Å 16 × 25 9 RH 7 RH 144 (3) Å (4) Å 144 7 RH (1)

26. The energy of I, II and III energy levels of a certain atom 4E are E , and 2E respectively. A photon of wavelength l 3 is emitted during a transition from III to I. What will be the wavelength of emission for transition II to I? l (2) l 2 (3) 2l (4) 3l (1)

27. What is the series limit for Paschen series of the He+ ion in Å? (1) 2502 (2) 2520 (3) 2250 (4) 2052 28. Which element has hydrogen-like spectrum whose lines have wavelengths one-fourth of atomic hydrogen? (1) He+ (2) Li2+ (3) Be3+ (4) B4+ 29. The emission spectra are observed as a consequence of transition of electron from higher energy state to ground state in Li2+ ion. Six different types of photons are observed during the emission spectra; then what is the excitation state of Li2+ ion? (1) Third (2) Fourth (3) Second (4) Fifth 30. If l1, l2, and l3, are wavelengths of photon corresponding to first, second Lyman series and first Balmer series respectively, which of the following statement is correct? (1) l2 = l1 + l3 (2) l2 = l1l3/(l1 + l3) (3) l1 + l2 + l3 = 0 (4) l 22 = l 12 + l 23 31. When electrons are de-exciting from nth orbit of hydrogen atoms, 15 spectral lines are formed. The shortest wavelength among these will be 11 900 R (2) 11R 900 35 36 (3) (4) 36R 35R (1)

1/4/2018 5:09:03 PM

42

OBJECTIVE CHEMISTRY FOR NEET

Heisenberg Uncertainty Principle and de Broglie Equation

Schrödinger’s Wave Equation 39. Select the correct statements from the following:

32. If the uncertainty Dx in the position is along x-axis, then uncertainty in the momentum is along (1) x-axis. (2) y-axis. (3) z-axis. (4) any axis. 33. If the radius of first Bohr orbit is a1, then de Broglie wavelength of electron in third orbit is nearly. (1) 2pa1 (2) 6pa1 (3) 9pa1 (4) 16pa1 34. If E1, E2 and E3 are kinetic energy of electron, alpha ­particle and proton having same de Broglie wave length, then (1) E1 > E2 >E3 (2) E1< E2< E3 (3) E2< E3 < E1 (4) E1 = E2 = E3

(3)

h (4) p

h pm

36. An electron is moving with a K.E. of 4.55 × 10 J. What will be the de Broglie wavelength for this electron? (1) 5.28 × 10−7 m (2) 7.27 × 10−7 m (3) 2 × 10−10 m (4) 3 × 10−5 m

Photoelectric Effect 37. l0 is the threshold wavelength of a metal for photoelectron emission. If the metal is exposed to the light of wavelength l, then the velocity of ejected electron will 2h ( l 0 - l ) K . The value of K is m (1) speed of light (2) 1 be

c 1 (4) l0 l ll0

38. When photons of energy 4.25 eV strike the surface of a metal A. The ejected photoelectrons have maximum kinetic energy TA (expressed in eV) and de Broglie wavelength l(1). The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy 4.2 eV is TB where TB = (TA - 1.5). If de Broglie wave-length of these photoelectrons lB = 2lA, then which of the following is not correct? (1) (2) (3) (4)

The work function of A is 2.25 eV. The work function of B is 3.7 eV. TA = 2.0 eV TB = 0.75 eV

Chapter 2_Structure of Atom.indd 42

41. The magnetic quantum number signifies (1) (2) (3) (4)

h 2p

−25

(3)

40. In the absence of external magnetic field, f sub-shell is

size of the orbital. shape of the orbital. orientation of orbital in space. nuclear stability.

42. Wave function versus distance from nucleus graph of an orbital is shown below. Wave function

1 h (2) 2m p

(1) (I), (II), (IV) (2) (I), (II), (III) (3) (II), (III), (IV) (4) (I), (III), (IV)

(1) 5-fold degenerate. (2) 3-fold degenerate. (3) 7-fold degenerate. (4) non-degenerate.

35. If uncertainty in position and momentum are equal, the uncertainty in velocity would be (1)

(I)  Splitting of spectra line occurs when placed in a magnetic field or in an electric field. (II) In case of 1s-orbital, the density of the electron cloud is the greatest near the nucleus and falls off with the distance. (III) Electron density is concentrated along a particular direction in case of 2p-orbital. (IV) A p-orbital can take maximum of six electrons.

r

The number of nodal spheres of this orbital is (1) 1 (2) 2 (3) 3 (4) 4 43. Number of waves in third Bohr orbit of hydrogen will be (1) 3 (2) 6 (3) 9 (4) 12 44. Which of the following statements is correct? (1) (n - 1) d subshell has lower energy than ns subshell. (2) (n - 1) d subshell has higher energy than ns subshell. (3) (n + 1) d subshell has lower energy than nf subshell (4) nf subshell has lower energy than (n + 2)s subshell. 45. If the number of orbitals of a particular type are (3l + 1), but spin quantum numbers only + 1/2 and -1/2, then d-type orbital will contain a maximum of ________ ­electrons. (1) 10 (2) 14 (3) 2 (4) 5

1/4/2018 5:09:05 PM

Structure of Atom 46. Which set of quantum numbers is not possible for electron in third shell? (1) (2) (3) (4)

(AIPMT 2007)

47. Out of the following which subshell has maximum energy? (1) 3d (2) 4s (3) 5s (4) 4p 48. Which series of subshells is arranged in the order of increasing energy for multi-electron atoms? (1) 6s, 4f, 5d, 6p (2) 4f, 6s, 5d, 6p (3) 5d, 4f, 6s, 6p (4) 4f, 5d, 6s, 6p 49. The subshell that arises after f is called the g subshell. How many electrons may occupy the g subshell? (1) 9 (2) 7 (3) 5 (4) 18

Aufbau Principle, Hund’s Rule and Pauli’s Exclusion Principle 50. Which of the following violates the Pauli exclusion principle? (2)

(3)

(4)

2. If uncertainty in position and momentum are equal, then uncertainty in velocity is (1)

1 h h (2) m p p

(3)

1 h h (4) 2m p 2p (AIPMT 2008)

3. The measurement of the electron position is associated with an uncertainty in momentum, which is equal to 1 × 10-18 g cm s-1. The uncertainty in electron velocity is (mass of an electron is 9 × 10-28 g) (1) 1 × 1011 cm s-1 (3) 1 × 106 cm s-1

(2) 1 × 109 cm s-1 (4) 1 × 105 cm s-1 (AIPMT 2008)

4. Maximum number of electron in a subshell of an atom is determined by the following (1) 2n2 (2) 4l + 2 (3) 2l + 1 (4) 4l - 2 (AIPMT 2009) 5. Which of the following is not permissible arrangement of electrons in an atom?

51. Which of the following is a colored ion? (1) Cu+(aq) (2) Na+(aq) (3) Cu2+ (aq) (4) K+ (aq) 52. The value of the magnetic moment of a particular ion is 2.83 B.M. The ion is (1) Fe2+ (2) Ni2+ (3) Mn2+ (4) Co3+ 53. If an ion of 25Mn has a magnetic moment of 3.873 B.M, then, Mn is in which state (1) + 2 (2) + 3 (3) + 4 (4) + 5

(1) (2) (3) (4)

n = 3, l = 2, m = -2, s = -1/2 n = 4, l = 0, m = 0, s = -1/2 n = 5, l = 3, m = 0, s = -1/2 n = 3, l = 2, m = -3, s = -1/2

(AIPMT 2009)

6. A 0.66 kg ball is moving with a speed of 100 ms-1. The associated wavelength will be (h = 6.6 × 10−34 Js) (1) 6.6 × 10−34 m (2) 1.0 × 10−35 m (3) 1.0 × 10−32 m (4) 6.6 × 10−32 m (AIPMT MAINS 2010) 7. The total number of atomic orbitals in fourth energy level of an atom is

Previous Years’ NEET Questions 1. Consider the following sets of quantum number: n

l

m

s

(I)

3

0

0

+½

(II)

2

2

1

+½

(III)

4

3

−2

−½

(IV)

1

0

−1

−½

(V)

3

2

3

+½

Chapter 2_Structure of Atom.indd 43

Which of the following sets of quantum number is not possible? (1) (I) and (III) (2) (II), (III) and (IV) (3) (I), (II), (III) and (IV) (4) (II), (IV), and (V)

n = 3, l = 2, m = – 1, s = +1/2 n = 3, l = 2, m = –1, s = –1/2 n = 3, l = 2, m = 0, s = +1/2 n = 3, l = 3, m = 0, s = –1/2

(1)

43

(1) 4 (2) 8 (3) 16 (4) 32 (AIPMT PRE 2011) 8. The energies E1 and E2 of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths, that is, l1 and l2 will be (1) l 1 =

1 l2 2

(2) l 1= l 2

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44

OBJECTIVE CHEMISTRY FOR NEET (3) l 1= 2l 2

(1) 10 (2) 6 (3) 4 (4) 2

(4) l 1= 4l 2 (AIPMT PRE 2011)

9. If n = 6, the correct sequence of filling of electrons will be (1) (2) (3) (4)

ns → np → (n − 1)d → (n − 2)f ns → (n − 2)f → (n − 1)d → np ns → n(n − 1)d → (n − 2)f → np ns → (n − 2)f → np → (n − 1)d

(NEET 2013)  Z2 16. Based on equation E = -2.178 × 10 -18 J  2  certain conn  clusions are written. Which of them is not correct? (1) The negative sign in equation simply means that the energy of electron bound to the nucleus is lower than it would be if the electrons were at the infinite distance from the nucleus. (2) Larger the value of n, the larger is the orbit radius. (3) Equation can be used to calculate the change in energy when the electron changes orbit. (4) For n = 1, the electron has a more negative energy than it does for n = 6 which means that the electron is more loosely bound in the smallest allowed orbit.

(AIPMT PRE 2011) 10. According to the Bohr theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon? (1) n = 5 to n = 3 (2) n = 6 to n = 1 (3) n = 5 to n = 4 (4) n = 6 to n = 5 (AIPMT MAINS 2011) 11. Maximum number of electrons in a subshell with l = 3 and n = 4 is (1) 14 (2) 16 (3) 10 (4) 12

17. What is the maximum number of orbitals that can be identified with the following quantum numbers?

(AIPMT PRE 2012) 12. The correct set of four quantum numbers for the valence electron of rubidium atom (Z = 37) is (1) 5, 1, +1/2 (2) 6, 0, 0, +1/2 (3) 5, 0, 0, +1/2 (4) 5, 1, 0, +1/2 (AIPMT PRE 2012) 13. The orbital angular momentum of a p electron is given as (1)

h (2) 2p

h 3 2p

(3)

3h (4) 2p

6

14. The value of Planck’s constant is 6.63 × 10–34 Js. The speed of light is 3 × 1017 nm s–1. Which value is closest to the wavelength in nanometer of a quantum of light with frequency of 6 × 1015 s–1? (1) 10 (2) 25 (3) 50 (4) 75 (NEET 2013) 15. What is the maximum numbers of electrons that can be associated with the following set of quantum numbers?

Chapter 2_Structure of Atom.indd 44



n = 3, l = l, m = 0 (1) 1 (2) 2 (3) 3 (4) 4 (AIPMT 2014)

18. Calculate the energy corresponding to light of ­wave­le ­ ngth 45 nm (Planck’s constant h = 6.63 ´ 10-34 Js: speed of light c = 3 ´ 108 m s-1) (1) 6.67 ´ 1015 J (2) 6.67 ´ 1011 J (3) 4.42 ´ 10-15 J (4) 4.42 ´ 10-18 J (AIPMT 2014)

h 2p

(AIPMT MAINS 2012)

n = 3, l = 1 and m = –1

(NEET 2013)

19. The angular momentum of electron in d orbital is equal to (1)

2h

(3) 0 h





(2) 2 3 h (4)

6h (AIPMT 2015)

20. The number of d-electrons in Fe2+ (Z = 26) is not equal to the number of electrons in which one of the following? (1) p-electrons in Cl (Z = 17) (2) d-electrons in Fe (Z = 26) (3) p-electrons in Ne (Z = 10) (4) s-electrons in Mg (Z = 12) (AIPMT 2015)

1/4/2018 5:09:07 PM

Structure of Atom 21. Two electrons occupying the same orbital are distinguished by (1) (2) (3) (4)

principal quantum number. magnetic quantum number. azimuthal quantum number. spin quantum number.

45

(1) 6 (2) 10 (3) 14 (4) 2 (NEET-II 2016) 24. Which one is the wrong statement?

(NEET-I 2016) 22. Which of the following pairs of d-orbitals will have ­electron density along the axes? (1) dxz, dyz (2) dz2, dx2-y2 (3) dxy, dx2-y2 (4) dz2, dxz (NEET-II 2016)

23. How many electrons can fit in the orbital for which n = 3

(1) The uncertainty principle is ΔE × Δt ≥ h/4p. (2) Half-filled and fully filled orbitals have greater stability due to greater exchange energy, greater ­ symmetry and more balanced arrangement. (3) The energy of 2s orbitals is less than energy of 2p orbitals in case of hydrogen-like atoms. h (4) de-Broglie’s wavelength is given by l = , where mv m = mass of the particle, v = group velocity of the ­particle.

and l = 1?

(NEET 2017)

Answer Key Level I  1. (4)

 2. (1)

 3. (4)

 4. (4)

 5. (3)

 6. (2)

 7 (2)

 8. (3)

 9. (3)

10.  (2)

11.  (2)

12.  (2)

13.  (2)

14.  (4)

15.  (4)

16.  (2)

17.  (3)

18.  (4)

19.  (1)

20.  (1)

21.  (2)

22.  (2)

23.  (2)

24.  (3)

25.  (1)

26.  (1)

27.  (1)

28.  (1)

29.  (1)

30.  (2)

31.  (2)

32.  (1)

33.  (3)

34.  (1)

35.  (3)

36.  (2)

37.  (3)

38.  (2)

39.  (4)

40.  (1)

41.  (1)

42.  (2)

43.  (1)

44.  (1)

45.  (1)

46.  (3)

47.  (1)

48.  (3)

49.  (3)

50.  (2)

51.  (4)

52.  (2)

53.  (2)

54.  (1)

55.  (2)

56.  (3)

57.  (4)

58.  (3)

59.  (3)

60.  (1)

61.  (3)

62.  (4)

63.  (1)

64.  (1)

65.  (4)

Level II  1. (2)

 2. (2)

 3. (3)

 4. (3)

 5. (4)

 6. (1)

 7. (1)

 8. (1)

 9. (4)

10.  (4)

11.  (1)

12.  (3)

13.  (1)

14.  (2)

15.  (3)

16.  (4)

17.  (2)

18.  (1)

19.  (4)

20.  (3)

21.  (2)

22.  (3)

23.  (3)

24.  (1)

25.  (4)

26.  (4)

27.  (4)

28.  (1)

29.  (1)

30.  (2)

31.  (4)

32.  (4)

33.  (2)

34.  (3)

35.  (1)

36.  (2)

37.  (3)

38.  (4)

39.  (2)

40.  (3)

41.  (3)

42.  (1)

43.  (1)

44.  (2)

45.  (2)

46.  (4)

47.  (3)

48.  (1)

49.  (4)

50.  (3)

51.  (3)

52.  (2)

53.  (3)

Chapter 2_Structure of Atom.indd 45

1/4/2018 5:09:07 PM

46

OBJECTIVE CHEMISTRY FOR NEET

Previous Years’ NEET Questions  1. (4)

 2. (3)

 3. (2)

 4. (1)

 5. (4)

 6. (2)

 7. (3)

 8. (3)

 9. (2)

10.  (4)

11.  (1)

12.  (3)

13.  (1)

14.  (3)

15.  (4)

16.  (4)

17.  (1)

18.  (4)

19.  (4)

20.  (1)

21.  (4)

22.  (2)

23.  (4)

24.  (3)

Hints and Explanations Level I 2. (1) The reaction is 3 Li 6 +

Z

X A ® 2He4 + 1H 3





6+A=4+3⇒A=1





3+Z=2+1⇒Z=0



Therefore, the missing particle is 0n1 or neutron. e e 0 = for electron; = for proton; = me mp mn

3. (4) We know

for neutron and

2e = for a-particle 4mp

Therefore, increasing order of

e is n < a < p < e m

5. (3) Frequency can be calculated as u=

c 3 ´ 108 = = 6 ´ 1014 Hz l 500 ´ 10 -9

7. (2) Charge of any particle can be expressed as integral multiple of electronic charge (e).

q = n(e) where n ∈ I, e = 1.6 × 10−19 C



Therefore, 2.4 × 10−19 ≠ n (1.6 × 10−19)

8. (3) Wave number can be expressed as n =

1 1 = = 2 ´ 106 m -1 l 500 ´ 10-9

9. (3) The ratio of energy of photon is E1 hc ´ l 2 l 2 4000 2 = = = = E 2 l 1 ´ hc l 1 2000 1



10. (2) Wavelength can be expressed as l=

c 3 ´ 108 = = 58.9 m n 5090 ´ 103

11. (2) The energy of electron can be expressed as





Chapter 2_Structure of Atom.indd 46

æ Z2 ö E = -13.6 ´ ç 2 ÷ èn ø (1)2 E = -13.6 ´ 2 = -1.51 eV ( 3)



Therefore, P.E. = 2E = 2 × (−1.51) = −3.02 eV



K.E. = −E = −(−1.51 eV) = 1.51 eV Z n

12. (2) We have v µ

Therefore,

v Li 2+ ,3 v Li 2+ ,5

=

5 3

æ Z2 ö 13. (2) The energy of the electron is  E = -13.6 ´ ç 2 ÷ èn ø (1)2 ( -13.6 )´ 2 E H,1 (1) = 1 Therefore, the ratio is      = ( 2)2 1 E He+ ,2 ( -13.6 )´ 2 ( 2) n2 14. (4) We know r µ Z 1 Given n = 1, therefore, r µ Z

Therefore, radius for Li2+ is minimum. n2 Z r2 ( 2)2 9 Therefore, = 2 ⇒ r3 = R = 2.25 R r3 ( 3) 4

15. (4) We know r µ

16. (2) Given that  rBe3+ , n = rH ,1 (0.529)´

n2 (1)2 = (0.529)´ Þn = 2 4 1

17. (3) Any atomic or ionic species having one electron only. 18. (4) We have r µ

n2 Z

Since, Z is the same, therefore, r µ n 2. Hence, (r5 – r4) is maximum. Z2 20. (1) The energy of electron is E µ 2 n 1 Since Z is the same, therefore, E µ 2 . Hence, (E1 – E2) is n maximum.

24. (3) We know

æ 1 1 1 ö = RZ 2 ç 2 - 2 ÷ l è nf ni ø

1/4/2018 5:09:10 PM

Structure of Atom

   

l=



1 1 ö R(n 2 - 4) æ 1 = R(1)2 ç 2 - 2 ÷ = l 4n 2 è ( 2) n ø

Therefore, k =

4 n R (n 2 - 4)



4 R



é1 1ù 5 Therefore, E 3 - E 2 = E1 ê 2 - 2 ú = ´x ë 2 3 û 36

27. (1) Given

Since, E2 – E1 is maximum, therefore, ΔE corresponding to the transition 6 → 1 is maximum. Hence l is minimum for this transition.

l1 = 364.1 nm

E1 n2

Ionization energy E n =

DE = E n2 - E n1

34. (1)

1 5R æ 1 1ö æ 5 ö = R(1)2 ç 2 - 2 ÷ = ç cm -1 ÷R = l 2 3 4 9 36 ´ è ø è ø 6. (1) Given I.E. = E1 = x 2 25. (1) n =



ΔEmax will correspond to lmax



2

2n2 + 3n1 = 18

(1)



  2n2 – 3n1 = 6

(2)



From Eq. (1) and Eq. (2), we get







Maximum number of spectral lines =

( Dn )( Dn + 1) 2 4 ´ ( 5) = = 10 2

    

hc hc hc + = l1 l2 l



Therefore,

1 1 1 + = l1 l2 l



Also, l =





l=

n=3

29. (1) I.E. of Na atom =

1240 eV/atom 242

1240 ´ 1.6 ´ 10         = 242

-19

J/atom

1240 × 1.6 × 10 -19 × 10 -3 × (6.023 × 1023 ) kJ mol -1 242        = 493.78 kJ mol−1

   =

30. (2) Radius of electron can be expressed as

r = (0.529Å )´

n2 Z



Radius of electron in first excited orbit



rLi 2+ , 2 = (0.529Å ) ´

( 2)2 = 0.705 Å 3

31. (2) According to Planck’s quantum theory, we have



Chapter 2_Structure of Atom.indd 47

36. (2) l =

n=4

Paschen series

DE = hn =

hc 1 or ΔE ∝ l l

l1l2 ( 364.6 ) ´ (121.5) = ⇒ l = 91.2 nm l1 + l2 ( 364.6 + 121.5)

35. (3) From de Broglie equation, we have

28. (1) Number of spectral lines from n = 5 to n = 3 falling in Paschen series n=5

l

l2 = 121.5 nm



n2 = 6 and n1 = 2

47

38. (2) l =

h = mn

6.62 ´ 10-34 m ⇒ l = 10−29 Å 5 ö æ (100 ´ 103 )ç 23.76 ´ ÷ 18 ø è

h 6.62 ´ 10 -34 = =8.92 ´ 10 -37 m mn æ 150 ö (50) ´ ç è 12.1÷ø h h 6.62 ´ 10 -34 ⇒ mn = = mn l 0.1 ´ 10 -9

Therefore, p = mn = 6.62 × 10−24 kg ms−1

39. (4) According to Heisenberg’s uncertainty principle, we have h         Δx × Δv ≥ 4pm

        Dx ³

6.62 ´ 10 -34 (0.01)( 4p ´ 200 ´ 10 -31 )

        Dx ³ 2.64 ´ 10 -31 m 40. (1) According to photoelectric experiment, we have hu = hu 0 + (K.E.)max

h(u - u 0 ) = (K.E.)max (K.E.)max = 6.62 ´ 10 -34( 3.5 - 1.5)´ 1015 J =1.32 ´ 10 -18 J

41. (1) We know      E = f + (K.E )max 4 × 10-20 = f +

p2 2m

where p is the momentum.

1/4/2018 5:09:15 PM

48



OBJECTIVE CHEMISTRY FOR NEET

From de Broglie’s equation, we have p = Therefore, 4 ´ 10 4 × 10-20 = f +

-20

h2 =f + 2ml 2

h l

-34 2

(6.62 × 10 ) ⇒ f = 3.31 ´ 10-20 J 2 × (9.1 × 10-31 )(59 × 10-10 )2

42. (2) According to photoelectric experiment   hu = hu 0 + (K.E.)1   hu 2 = hu 0 +(K.E.)2



(K.E.)1 1 h(u1 - u 0 ) = = (K.E.)2 k h(u 2 - u 0 ) Therefore, u 2 - u 0 = k(u1 - u 0 ) Þ u 0 =

ku1 - u 2 k -1

46. (3) Number of radial/spherical nodes = n – l – 1

Number of angular nodes = n

For 3p orbital, we know n = 3, l = 1

Therefore, number of spherical or radial nodes = 3 – 1−1 = 1



Number of non-spherical or angular nodes = 1

48. (3) For 3f orbital, we have l = 3 h Therefore, orbital angular Momentum = l(l + 1) 2p h h h = 12 =2 3 = 3( 3 + 1) 2p 2p 2p 49. (3) The value of l can go from 0 to n – 1.

If n = 3, l = 0, 1, 2, therefore, 3f is not possible.

50. (2) Angular momentum =

nh 2p

Let the two successive orbits be n and (n + 1)



Therefore, the difference in angular momentum h h nh = = (n + 1) 2p 2p 2p 54. (1) Number of radial nodes = n – l – 1

Radial nodes in 4s = 4 – 0 – 1 = 3



Radial nodes in 4p = 4 – 1 – 1 = 2



Radial nodes in 3d = 3 – 2 – 1 = 0



Radial nodes in 5f = 5 – 3 – 1 = 1

55. (2) There are two nodal planes in dxy orbital, that is, xz and yz. 61. (3) Total spin from d 5 = 5 ´



Mg: [Ne]3s2 ⇒ Mg2+: [Ne] → 0 unpaired electron



Ti: [Ar] 3d24s2 ⇒ Ti3+: [Ar]3d1 → 1 unpaired electron



V: [Ar] 3d34s2 ⇒ V3+: [Ar]3d2 → 2 unpaired electron



Fe: [Ar] 3d64s2 ⇒ Fe2+: [Ar]3d6 → 4 unpaired electron

Level II 1. (2) Let the number of photons per second be n





65. (4) The electronic configurations are:

1 5 = 2 2

Therefore, n(hu) = 1000 J s−1



n(6.62 × 10−34 × 880 × 103) = 1000 ⇒ n = 1.72 × 1030 2. (2) From Planck’s quantum theory, we have



æ1ö E = hc ç ÷ = hc(5 ´ 105 ) èl ø



E = 6.62 × 10−34 × 3 × 108 × 5 × 105 J = 9.93 × 10−23 kJ

3. (3) Let the number of photons be n We know

E=

nhc l

hc æ ö Therefore, n ç = 3.15 ´ 10-14 Þ n = 1.41 ´ 105 -9 ÷ ´ 890 10 è ø 4. (3) Energy can be expressed as



E = nN Ahn



  = (1.5 × 6.023 × 1023) × (6.62 × 10−34 × 7.5 × 1014) J



  = 4.48 × 105 J

5. (4) Let the quanta emitted per second be n We know





E=

Therefore, n ´

nhc l

hc = 25 ⇒ n = 7.18 × 1019 s−1 (0.57 ´ 10-6 )

6. (1) Given 0.5 Eabsorbed = Eemitted  

0.5 ´

nabsorbed (hc ) nemitted (hc ) = labsorbed lemitted



0.5 ´

0.5 ´ 5000 nabsorbed nemitted n = 0.55 = Þ emitted = 4500 5000 4500 nabsorbed

7. (1) We know  P.E. = 2E Therefore,   E =

-6.8 = -3.4 2

Hence, the electron is present in first excited state.

62. (4) The electronic configuration of Al is [Ne] 3s2 3p1

9. (4) N-shell of the Be3+ ion implies n = 4





3p orbital contain outermost electron. Therefore, for 3p 1 1 orbital n = 3, l = 1, m = −1, 0, 1, s = + , 2 2

Chapter 2_Structure of Atom.indd 48



E Be3+ ,4 = -13.6 ´

42 = -13.6 eV 42

P.E. = 2E = −27.2 eV

1/4/2018 5:09:17 PM

Structure of Atom  n2  10. (4) Radius of electron, r = (0.529 Å ) ×   (1) Z Given rZ ,4 < rH , 1

Therefore, from Eq. (1), we get







Hence, Z = 25 is the correct option.

0.529 ´

16 12 < 0.529 ´ ⇒ Z > 16 Z 1

vµ

11. (1) We know

19. (4) Acceleration is given by a =

-13.6 eV = −3.4 eV 4



(E2) =



Therefore, energy required to free electron, if it is present in first excited state = 0 – E2 = 3.4 eV





æ 1 ö     P.E. = 2 E = 2 ç - mv 2 ÷ = -mv 2 è 2 ø v 14. (2) Number of revolutions of electrons per second = 2pr Therefore, number of revolutions per second

( Z /n ) Z 2 µ 3 n 2 /Z n (No. of rev/s)H ,3 2 (1)2 ´ ( 2)3 = 3 = (No. of rev/s)He+, 2 ( 3) ´ ( 2)2 27

µ





27 d = 13.5 d 2



    

Therefore, time taken µ



Given Z = 1, therefore, time taken µ n 3.

Chapter 2_Structure of Atom.indd 49

n 2 /Z n 3 µ Z /n Z 2

1 24 1 1 76 ´ = 1- 2 ⇒ 2 = 25 4 n n 25 ´ 4

Since, no integral value of n is possible; therefore, electron will remain ground state of He+.

é1 1 ù 2 2 IE He+ = 13.6 Z He + êë12 - ¥ 2 úû = 13.6 Z He+ ( where Z He+ = 2)



Now, for Li2+, we have

2pr n



2 -18 13.6 ´ Z He J atom -1 (1) + = 19.6 ´ 10

2 Multiplying and dividing by Z He +, we get

é Z Li2 2+ 2 (E1 )Li 2+ = -13.6 Z He + ê 2 êë Z He+

23. (3) We have

ù 9 2 ú = -13.6 Z He+ ´ 4 úû

From Eq. (1), we have (E1 )Li 2+ = -19.6 × 10 -18 ×

Z n2 (constant ) and r = (constant ) n Z





So,



(n + 1) – (n) = (n – 1) ⇒ n = 4

We know v =

æ1 1ö æ1 1 ö 13.6 ç 2 - 2 ÷ = 13.6 ´ ( 2)2 ç 2 - 2 ÷ è1 5 ø è1 n ø



2





Given that     IE He+ = 19.6 ´ 10-18 J



17. (2) Time taken to complete one revolution =

Given E H ,5 - E H ,1 = E He+ ,n - E He+ ,1

é1 1 ù (E1 )Li 2+ = -13.6 Z Li2 2+ ê 2 - 2 ú = -13.6 Z Li2 2+ ë1 ¥ û

16. (4) Given: r(n + 1) – r(n) = r(n – 1) 2

Z2 n2



( 2)2 = 24 + 54.4 (1)2

= 78.4 eV

2

(a )n=3 ( 2)4 16 = = (a )n=2 ( 3)4 81

22. (3) The ionization enthalpy (I.E.) is expressed as

15. (3) Energy required to remove both electron = Binding energy (He → He+) + first ionization energy = 24 + 13.6 ´



v2 r

Z3 ( Z 2 /n 2 ) ⇒ aµ 4 2 n (n /Z )

21. (2) We have E = -13.6 ´

1         E = - K.E. = - mv 2 2

Therefore, ( No. of rev/s)He+, 2 =

Therefore,    a µ



1 13. (1) We know  K.E. = mv 2 2



Therefore, (K.E.)Li 2+ ,1 = 144(Î)

   (E )Li 2+ ,1 = -(K.E.)Li 2+ ,1 = -144 Î

Z n

12. (3) Energy of electron in the first excited state



Z2 and K.E. = −E n2 Z2 Therefore, K.E. µ 2 n (K.E.)H ,4 (1)2 (1)2 1 = ´ = (K.E.)Li 2+ ,1 ( 4)2 ( 3)2 144 18. (1) We know E µ



v3 1 1 1 = Þ v 3 = v1 = x v1 3 3 3



49

9 = -4.41 × 10 -17 Jatom -1 4

é1 1 1ù = RH ê 2 - 2 ú l n n 2 û ë 1

Given l = 2170 nm = 2170 × 10−9 m, RH = 1.09677 × 107 m−1

1/4/2018 5:09:21 PM

50

OBJECTIVE CHEMISTRY FOR NEET 109 é1 1ù = 1.09677 ´ 107 ê 2 - 2 ú 2170 ëx 7 û é1 1ù êë x 2 - 7 2 úû = 0.0420 1 1 = 0.0420 + Þx=4 x2 49



3( 3 + 1) = 6 photons 2

=







Thus, excitation state of Li2+ ion will be third.

30. (2) l2

24. (1) Given 1.5 E = E + (K.E.) (where E = 13.6 eV)

E2

l1

E1



K.E. = 0.5 × 13.6 × 1.6 × 10−19 J





h h Therefore, l = = mv 2m(K.E.)



From the above figure, we have





     l =

6.62 ´ 10-34 2 ´ 9.1 ´ 10-31 ´ (0.5 ´ 13.6 ´ 1.6 ´ 10-19 )

     l = 4.7 Å

25. (4) Angular momentum =

Therefore,





26. (4) III

2E

II



hc hc or E = l l 4E hc E hc -E = or = 3 x 3 x

2E - E =











On dividing Eq. (1) by Eq. (2), we get







(1) (2)

E x = Þ x = 3l ( E /3) l

1 1 ö æ 1 = (10973731.6 )´ ( 2)2 ç 2 - 2 ÷ m -1 l è3 ¥ ø or     l = 2052 Å

28. (1) As Given



Therefore, Z = 2 as E He+ = ( 2)2 E H

Number of different photons emitted, when returning to Dn( Dn + 1) ground state = 2 (where Δn = 4 - 1)

Chapter 2_Structure of Atom.indd 50

36 35R

l =

Therefore, mv =



Also,

l=

           l=



We know





nh 2p

nh 2pr h h × 2p r 2p r = = mv nh n 2p (a ) ´ 32 = 6p a 1 3

34. (3) From de Broglie, we have l =

h 2m(K.E.)

me < mp < ma

Therefore,

E1 > E 3 > E 2

35. (1) Given Δx = Δp Therefore, Δx = mΔv

EZ = 4E4

29. (1) In the third excited state, n = 4



Shortest wavelength corresponds to maximum energy or E6 – E1 1 é 1 1 ù R( 35) = Rê 2 - 2 ú= Therefore, for hydrogen atom l 36 ë1 6 û



1 E∝ l





hc hc hc l1l3 + = ⇒ l2 = ( l1 + l3 ) l1 l3 l2

33. (2) Angular momentum is given by mvr =

æ 1 1 ö 27. (4) For Paschen series, we have n = RH Z 2 ç 2 - 2 ÷ m -1 è3 n ø

n=1

4E/3 E

I

n=2

Dn( Dn + 1) 31. (4) Number of spectral lines = = 15, where 2 Δn = n − 1 (n – 1)(n) = 30 ⇒ n = 6

nh 2p

nh 2h = ⇒n = 4 2p p



n=3

E1 + E3 = E2



1 144 é 1 1 ù R (7 ) = RH (1)2 ê 2 - 2 ú = H Þ l = Å l 7 RH ë 3 4 û 144



l3 E3



h 4p m h 1 h m( Δv )2 = ⇒ Δv = 4p m 2m p Δx × Δv =

36. (2) From de Broglie, we have l =

l=

6.62 ´ 10-34 2(1.6 ´ 10-19 )( 4.55 ´ 10-25 )

h 2m(K.E.) = 7.27 ´ 10-7 m

37. (3) From photoelectric effect, we have

1/4/2018 5:09:24 PM

Structure of Atom hc hc = + (K.E.) l l0 hc 1 ( l0 - l ) = mv 2 ll0 2

44. (2) Using (n + l) rule



hc 1 2h ( l0 - l ) = m ´ ( l0 - l )K ll0 2 m

nf → n + l = n + 4 = n + 4



c K= ll0



38. (4)     4.25 = fA + TA (1)



(n – 1)d → n + l = n – 1 + 3 = n + 2



ns → n + l = n + 0 = n



(n + 2)s → n + l = n + 2 + 0 = n + 2



(n + 1)d → n + l = n + 1 + 3 = n + 4

45. (2) Number of orbital = 3l + 1





4.2 = fB + TB (2)



Number of orbitals in d = 3( 2) + 1 = 7



Given

lB = 2l A



Therefore, total number of electrons = 7 × 2 = 14



Therefore,



        TA = 4TB(3)



        TB = TA – 1.5



From Eq. (3) and Eq. (4), we get



Therefore,







Hence, the ion is Ni2+: [Ar] 3d8



Using TA and TB values in Eq. (1) and Eq. (2), we get





h 2h = 2m(TB ) 2m(TA )

51

51. (3) Ions that contain unpaired electrons are known as paramagnetic and impart colour to solution. (4)

TB = 0.5 eV and TA = 2 eV

fA = 2.25 eV and fB = 3.7 eV

43. (1) Number of waves = n



Cu2+: [Ar]3d9 contain one unpaired electron.

52. (2) Magnetic moment can be calculated as m = n(n + 2) n(n + 2) = 2.83 ⇒ n = 2

53. (3) Magnetic moment can be calculated as m = n(n + 2) n(n + 2) = 3.873 ⇒n = 3







The electronic configuration of Mn is [Ar] 3d54s2. Since, the ion contains three unpaired electrons, therefore, it must be in +4 oxidation state, that is, Mn4+: [Ar] 3d3.

Therefore, number of waves = 3

Previous Years’ NEET Questions 1. (4) Principal quantum number (n)

Subsidiary quantum number (l)

Magnetic quantum number (me)

Spin quantum number (ms)

1

0

0

+1/2, −1/2

2

0

0

+1/2, −1/2

2

1

−1, 0, +1

+1/2, −1/2

3

0

0

+1/2, −1/2

3

1

−1, 0, +1

+1/2, −1/2

3

2

−2, −1, 0, +1,+2

+1/2, −1/2

4

0

0

+1/2, −1/2

4

1

−1, 0, +1

+1/2, −1/2

4

2

−2, −1, 0, +1,+2

+1/2, −1/2

4

3

−3,−2, −1, 0, +1,+2,+3

+1/2, −1/2

2. (3) Heisenberg’s uncertainty principle is Dx × Dpx ³



h 4p



where ∆x is the uncertainty in position and ∆px is uncertainty in momentum.



Given ∆x = ∆px. Therefore,

Chapter 2_Structure of Atom.indd 51

h 4p h 2 (mDv ) ³ 4p 1 h Dv ³ 2m p

( Dp )



2

³

1/4/2018 5:09:25 PM

52

OBJECTIVE CHEMISTRY FOR NEET

3. (2) Given Dp = m × Dv = 1 ´ 10-18 g cm s-1 and m = 9 × 10-28 g

Therefore,

Dv =

      =

Dp m 10-18 g cm s-1 = 0.1 ´ 1010 cm s-1 9 ´ 10-28 g

æ h ö 13. (1) Orbital angular momentum is given by ç ÷ (l(l + 1) è 2p ø For p orbital, l = 1, thus orbital angular momentum h æ h ö ç ÷ 1(1 + 1) = p 2 2p è ø



4. (2) Maximum number of electrons in a subshell is determined by 2(2l + 1) or 4l + 2.

14 (3) We have

5. (4) The possible values of m = −l to +l including zero. Therefore, for l = 2, m = 2 to + 2 (range) .Thus, for l = 2, m = −3 is not possible.



6. (2) From de Broglie equation, we have

l=



h 6.6 ´ 10-34 = = 10−35 m mv 0.66 ´ 100

7. (3) For n = 4, the possible values of l is 0, 1, 2, 3.

l = 0 4s ® 1orbital ; l = 1 4p ® 3 orbitals

l = 2 4d ® 5 orbital ; l = 3 4 f ® 7 orbitals Therefore the number of orbitals is 1 + 3 + 5 + 7=16 8. (3) We have    E1 =

Therefore,

hc hc and E 2 = l1 l2

E1 l2 25 l2 = Þ = E 2 l1 50 l1 l2 1 = Þ l1 = 2l2 l1 2



9. (2) Filling is done in accordance with Aufbau principle, that is, (n + l) rule, lower (n + l) being preferred. If the value of (n +l) is the same, then lower value of n is preferred. Therefore, the order of filling is



ns → (n − 2)f → (n −1)d → np.

10. (4) The various lines obtained in the spectrum can be related to transition between different energy states. When an electron undergoes transition from initial state ni to final state nf, the energy change is given by æ 1 1 ö DE = 2.18 ´ 10-18 ç ÷ ç (n ) 2 (n ) 2 ÷ i è f ø



Therefore, for transition ni = 6 to nf = 5 will give rise to the least energetic photon.

11. (1) Given that n = 4 and l = 3, so it is 4f subshell. The total number of electrons = 2(2l + 1) = 2(2 × 3 + 1) = 14 electrons. 12. (3) The electronic configuration of Rb (Z = 37) is [Kr] 5s1. Now, n = 5, l = 0, ml = 0 and ms = +1/2.

Chapter 2_Structure of Atom.indd 52

c l 3 ´ 1017 nms-1 = 50 nm l= 6 ´ 1015 s-1

v=

15. (4) The given set of quantum number represents the 3pz orbital which can accommodate maximum two electrons. 16. (4) Free electron corresponds to being present in orbit n = ¥ and is infinitely away from the nucleus, and thus, its energy is taken as zero. When the electron is present in any other stationary state (orbit) n, it is attracted to the nucleus and its energy is lowered. The energy of electron in orbit n is thus larger in absolute value than zero, but is denoted with a negative sign, since it is being lowered. The negative sign for the energy thus represents the relative stability of the state with reference to zero energy state n = ¥. 17. (1) 3p has m = 0, so only one orbital is possible, that is, 3pz. 18. (4) Using Planck’s quantum theory E=



hc 6.63 ´ 10-34 ´ 3 ´ 108 = = 4.4 ´ 10-18 J l 45 ´ 10-9

19. (4) Angular momentum is given by l(l +1)h

For d-orbital, l = 2. Therefore, l(l +1)h = 2( 2 + 1)h = 6 h

20. (1) The electronic configuration of the given elements is as follows:



17





26





10





12



The number of d electrons in Fe2+ is 6, which is not equal to the number of p-electrons in 17Cl.

Cl = 1s22s22p63s23p5 ; total number of p-electrons are 11.

Fe = 1s22s22p63s23p64s23d 6; total number of d-electrons are 6. Ne = 1s22s22p6 ; total number of p-electrons are 6.

Mg = 1s22s22p63s2; total number of s-electrons are 6.

21. (4) Electrons present in the same orbital differ in their spin quantum number value. 22. (2) From the shapes of the d-orbitals, it can be seen that the lobes of dz 2 and dx 2 – y 2 orbitals are along the axes.

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Structure of Atom Hence, the electron density is concentrated along the axes and these are called axial orbitals. The lobes of dxy, dyz and dxz orbitals are not along the axes and these are called non-axial orbitals. y

z +

+ +



3dx2- y2

Chapter 2_Structure of Atom.indd 53

x

-

-

53

23. (4) For n = 3 and l = 1, the orbital is 3p and it can hold two electrons as per Pauli’s exclusion principle. 24. (3) In case of hydrogen or hydrogen-like species, the energy of electron is determined only by the principal quantum number. The energy of all the orbitals in a given shell is the same, irrespective of the shape of the orbital. Thus, the order of energy levels is 1s < 2s = 2p < 3s = 3p = 3d

x

+

3dz2

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Chapter 2_Structure of Atom.indd 54

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3

States of Matter

Chapter at a Glance   Gaseous State 1. The Gas Laws (a) Boyle’s law: At isothermal (constant temperature) condition, for a definite amount of gas 1 V pV  k1 p1V1  p2V2 p

 Here, p1 is pressure of gas 1, V1 is volume of gas 1, p2 is pressure of gas 2 and V2 is volume of gas 2. T1

pV

T1

p

T2

T2

T3

T3 V

p T1 T2 p

T3

T1 > T2 > T3

1/V

Force Area 1 N 1 kg 1Pa = 2  m m  s2

Pressure 

  1 atm = 101325 Pa = 76 cm of Hg = 760 mm of Hg   1 mm of Hg = 1 torr; 1 bar = 105 Pa (b) Charles’ law: At isobaric (constant pressure) condition, for a definite amount of gas V T V  k2 T V1 T1  V2 T2

Chapter 3_States of matter.indd 55

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56

OBJECTIVE CHEMISTRY FOR NEET

 Here, V1 is volume of gas 1, T1 is temperature of gas 1; V2 is volume of gas 2 and T2 is temperature of gas 2. (c) Gay–Lussac’s law: It states that at constant volume, pressure of a fixed amount of gas varies directly with the temperature. p T p  constant T

V1 TN 2

3 × n1R × 300 = 2 x J 2



Solution

2

= 7×

(1) density. (2) pressure. (3) K.E. per mol (4) urms Solution

7urms( N 2 )

3RTN 2 28

⇒ TN 2 = 2TH2 or TN 2 > TH2

20. For two gases A and B with molecular weights MA and MB, it is observed that at a certain temperature T1 the mean velocity of A is equal to the root mean square velocity of B. Thus the mean velocity of A can be made equal to the mean velocity of B if (1) (2) (3) (4)

A is at temperature T and B at T ′ , T > T ′. A is lowered to a temperature T2 , T2 < T while B is at T. Both A and B are raised to a higher temperature. Both A and B are placed at lower temperature.

(2) We have (u ) = 8RT1 and (u ) = 3RT1 avg A rms B pM A MB Given 

(uavg )A = (urms )B. Therefore, (uavg )A =

Chapter 3_States of matter.indd 64

r=

pM RT



Option (1): Since M B = M A , therefore, at the same presTB TA sure rA = rB. But if pressure is different then rA ≠ rB.



Option (2): Pressure of the gases would be equal if their densities are equal otherwise not.



3 Option (3): K.E per mol = RT . It will be different for the 2 two gases.

Option (4): urms =

3RT TA T , since = B , therefore, M M A MB

urms of A = urms of B

Solution



(2)

From Eq. (1) and Eq. (2), we get

(4) We know 3RTH2

(1)

22. If for two gases of molecular weights MA and MB at temperature TA and TB, TAMB = TBMA, then which property has the same magnitude for both the gases

(4) TH2 > 7TN 2

(3) We have urms (H2) =

3 nRT 2

n = n1

19. The rms speed of hydrogen is 7 times the rms speed of nitrogen. If T is the temperature of the gas, then (1) TH2 = TN 2 (3) TH2 < TN 2

(2) 2N molecules of O2 (4) N/4 molecule of O2

(1) Total kinetic energy is given by

Solution (2) We know

21. The kinetic energy of N molecules of O2 is x J at –123°C. Another sample of O2 at 27°C has a kinetic energy of 2x J. The latter sample contains

8 = 3p

8RT2 8RT and (uavg )B = pM A pM B

MA MB

23. Helium atom is two times heavier than a hydrogen molecule. At 298 K, the average kinetic energy of a Helium atom is (1) (2) (3) (4)

two times that of hydrogen molecule. same as that of a hydrogen molecule. four times that of a hydrogen molecule. half that of a hydrogen molecule.

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States of Matter Solution

3 kT . As it does 2 not depend on mass of the atom, therefore, it would be same for both hydrogen and helium.

(2) The average kinetic energy of an atom is

24. The ratio between the rms velocity of H2 at 50 K and that of O2 at 800 K is (1) 4 (2) 2 (3) 1 (4) 1/4



(1) (2) (3) (4)



urms(O2 at 800 K) =



3RT M

3R ´ 50 urms(H 2 at 50 K) = 2 ´ 10 -3

the constant a is negligible and not b. the constant b is negligible and not a. both constants a and b are negligible. both the constants a and b are not negligible.

Solution



(1)

(1) van der Waals equation is given by  n 2a   p + V 2  (V − nb ) = nRT

3R ´ 800 (2) 32 ´ 10 -3

From Eq. (1) and Eq. (2) we get urms(H 2 ) = urms(O2 )

3R ´ 50 3 2 ´ 10 -3 = 25 ´ 10 = 1 3R ´ 800 25 ´ 103 -3 32 ´ 10



At high pressures, b cannot be ignored as the volume of the gas is very low. At high temperatures a can be ignored



Therefore,   p(V − b ) = RT pV = RT + pb pV RT pb = + RT RT RT pb Z = 1+ RT

25. The temperature of an ideal gas is increased from 140 K to 560 K. If at 140 K the root mean square velocity of the gas molecule is V, at 560 K it becomes (1) 5u (2) 2u (3) u/2 (4) u/4

28. The compressibility factor for a given gas is 0.927 at 273 K and 100 atm. Calculate the amount of gas required to fill a gas cylinder of 100 L capacity under given conditions. (Mol. wt. of gas is 30.)

Solution (2) We know urms =

3RT M



At 140 K, urms can be expressed as u =



At 540 K, urms can be expressed as u¢ =

3R ´ 140 M

3R ´ 560 3R ´ 140 ´ 4 3R ´ 140 = =2 = 2u M M M

26. At 100°C and 1 atm, if the density of liquid water is 1.0 g cm-3 and that of water vapor is 0.0006 g cm-3, then the volume occupied by water molecule in one liter of steam at that temperature is (1) 6 cm3 (2) 60 cm3 (3) 0.6 cm3 (4) 0.06 cm3 Solution (3) Mass can be expressed as V × r

Mass of 1 L of water vapor = V × r = 1000 × 0.0006 = 0.6 g

Chapter 3_States of matter.indd 65

0.6 = 0.6 cm 3 1

27. The behavior of a real gas is usually depicted by plotting compressibility factor Z versus p at a constant temperature. At high temperature and high pressure, Z is usually more than one. This fact can be explained by van der Waals equation when

Solution (3) We know urms =

Therefore, volume of liquid water =

65

(1) 16.4 kg (2) 14.44 kg (3) 4 kg (4) 10.5 kg Solution (2) For real gas, we have pV = ZnRT W ´ 0.0821 ´ 273 30 100 ´ 100 ´ 30 W= Þ W = 14.439 kg 0.927 ´ 0.0821 ´ 273

100 ´ 100 = 0.927 ´

29. Using van der Waals equation, calculate the constant, a (atm L2 mol-2) when two moles of a gas confined in a four liter flask exerts a pressure of 11 atm at a temperature of 300 K. The value of b is 0.05 L mol-1. (1) 6.5 (2) 2.23 (3) 23.2 (4) 0.85

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66

OBJECTIVE CHEMISTRY FOR NEET

Solution



(1) van der Waals equation is given by

Substituting V = 4 L, p = 11 atm, T = 300 K , b = 0.05 L mol-1 and n = 2 in Eq. (1), we get æ ( 2)2 ´ a ö 11 + ( 4 - 2 ´ 0.05) = 2 ´ 0.082 ´ 300 çè ( 4)2 ÷ø

2        p + n a  (V − nb ) = nRT (1) V 2  

Þ a = 6.5 atm L2 mol -2

Practice Exercises Level I

7. Absolute zero is defined as the temperature

Gas Laws 1. At a constant temperature, what should be the percentage increase in pressure for a 5% decrease in the volume of gas? (1) 5% (2) 10% (3) 5.26% (4) 4.26% 2. At constant temperature, in a given mass of an ideal gas (1) the ratio of pressure and volume always remains constant. (2) volume always remains constant. (3) pressure always remains constant. (4) the product of pressure and volume always remains constant. 3. Air at sea level is dense. This can be explained by (1) Boyle’s law. (2) Charle’s law. (3) Avogadro’s law. (4) Dalton’s law. 4. When the product of pressure and volume is plotted against pressure for a given amount of a gas, the line obtained is (1) (2) (3) (4)

parallel to x-axis. parallel to y-axis. linear with positive slope. linear with negative slope.

(1) (2) (3) (4)

at which all molecular motion ceases. at which liquid helium boils. at which ether boils. All of the above.

8. At a constant pressure, what should be the percentage increase in the temperature in Kelvin for a 10% increase in volume? (1) 10% (2) 20% (3) 5% (4) 50% 9. If p, V, T represent pressure, volume and temperature of a gas, the correct representation of Boyle’s law is (1) V ∝

1 (at constant p ) (2) pV = RT T

(3) V ∝ 1 (at constant T ) (4) pV = nRT p

Ideal Gas Law 10. 6 g each of the following gases at 87°C and 750 mm pressure are taken. Which of them will have the least volume? (1) HF (2) HCl (3) HBr (4) HI 11. In the equation pV = nRT, which one cannot be the numerical value of R?

5. The volume of gas is 100 mL at 100°C. If pressure remains same then at what temperature will the volume be 200 mL? (1) 200°C (2) 473°C (3) 746°C (4) 50°C 6. I, II, III are three isotherms respectively at temperatures T1, T2 and T3. The temperatures will be in the order

(1) 8.31 × 107 erg K-1 mol-1 (2) 8.31 × 107 dyne cm K-1 mol-1 (3) 8.31 J K–1 mol–1 (4) 8.31 atm K-1 mol–1 12. Two flasks A and B have capacity 1 L and 2 L and each of them contains 1 mol of a gas. The temperatures of the flasks are adjusted such that average speed of molecules in A is twice as those in B. The pressure in flask A would be (1) same as that in B. (2) half of that in B. (3) twice of that in B. (4) 8 times of that in B.

p III

II

I

V

(1) T1 = T2 = T3 (3) T1 > T2 > T3

Chapter 3_States of matter.indd 66

(2) T1 < T2 < T3 (4) T1 > T2 = T3

13. Two containers A and B contain the same gas. If the pressure, volume and absolute temperature of the gas in A are twice as compared to that of gas in B, and if the mass of the gas B is x g, the mass of gas in A is (1) x g (2) 4x g (3) 2/x g (4) 2x g

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States of Matter 14. To expel half the mass of air from a large flask at 27°C, it must be heated to (1) 54°C (2) 177°C (3) 277°C (4) 327°C 15. Temperature and pressure in Delhi is 47°C and 800 mm Hg whereas in Mumbai is 27°C and 750 mmHg. What is ratio of densities of air in Delhi and Mumbai? (1) 1:1.63 (2) 1:0.957 (3) 1:1.12 (4) 1:1 16. Two separate bulbs contain ideal gas A and B. The density of gas A is twice that of B. The molecular mass of A is half that of B. The two gases are at the same temperature. The ratio of pressure A to that of gas B is 1 2 1 (3) 4 (4) 4 (1) 2 (2)

17. Slope of plot between pV and p at constant temperature is (1) 0 (2) 1 (3) 1/2 (4) 1/3 18. Flasks A and B of equal size contain 2 g of H2 and 2 g of N2 respectively at the same temperature. The number of molecules in flask A is (1) (2) (3) (4)

same as those in flask B. less than that in flask B. greater than those in flask B. exactly half of those in flask B.

19. For an ideal gas, the number of moles per liter in terms of its pressure p, gas constant R and temperature T is (1) pT (2) pRT R (3)

p (4) RT p RT

20. An LPG cylinder can withstand pressure difference of 15 atm across its boundaries. If at room temperature (27°C) it is filled with 2 atm pressure, determine the temperature at which it will explode? (1) 216°C (2) 2127°C (3) 2400°C (4) 27°C 21. What is the final temperature if a sample of ammonia gas, initially at a pressure of 3 atm, a temperature of 500 K, and a volume of 275 L is changed to a volume of 200 L and a pressure of 2.50 atm? (1) 303 K (2) 436 K (3) 573 K (4) 825 K

Chapter 3_States of matter.indd 67

67

22. If the pressure of a given mass of gas is reduced to half and temperature is doubled simultaneously then the volume will be (1) same as before. (2) twice as before. (3)

1 th as before. (4) None of the above. 4

23. The density of neon gas is highest at (1) STP. (2) 0°C, 2 atm (3) 273°C, 1 atm (4) 273°C, 2 atm 24. Which of the following gases would have the largest density at 25°C and 1 atm pressure? (1) Methane, CH4 (2) Acetylene, C2H2 (3) Ethylene, C2H4 (4) Propane, C3H8 25. The density of a gas is 16 at STP. At what temperature will its density be 14 if pressure remains constant? (1) 50°C (2) 39°C (3) 57°C (4) 43°C 26. The density of O2 gas at 25°C is 1.458 mg L-1 at 1 atm pressure. At what pressure will O2 have the density twice the value? (1) 0.5 atm/25°C (2) 2 atm/25°C (3) 4 atm/25°C (4) none

Dalton’s Law 27. A gaseous mixture contains 1 g of H2, 4 g of He, 7 g of N2 and 8 g of O2. The gas having the highest partial pressure is (1) H2 (2) O2 (3) He (4) N2 28. A closed vessel contains equal volume of N2 and O2 at a pressure of p. If N2 is removed from the system then the pressure will be (1) p (2) 2p (3) p/2 (4) p2 29. Steam distillation is based on (1) Boyle’s law. (2) Charle’s law. (3) Avogadro’s law. (4) Dalton’s law. 30. A gaseous mixture contains 56 g of N2, 44 g of CO2 and 16 g of CH4. The total pressure of the mixture is 720 mm Hg. The partial pressure of CH4 is (1) 180 mm (2) 360 mm (3) 540 mm (4) 720 mm 31. Air contains 79% N2 and 21% O2 by volume. If the barometric pressure is 750 mm Hg. The partial pressure of oxygen is

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68

OBJECTIVE CHEMISTRY FOR NEET (1) 157.5 mm Hg (2) 175.5 mm Hg (3) 315.0 mm Hg (4) none

32. Same masses of CH4 and H2 is taken in a container. The partial pressure caused by H2 is (1)

Kinetic Theory of Gases and Maxwell Boltzmann Distribution Curve 41. A sample of gas is at 0°C. The temperature at which rms speed of the molecule will be doubled is

8 1 (2) 9 9

(1) 103°C (2) 273°C (3) 723°C (4) 819°C

1 (4) 1 2 3. Aqueous tension of water depends upon 3

42. According to kinetic theory of gases

(3)

(1) the pressure exerted by given mass of a gas is proportional to mean square velocity of the molecules at constant volume. (2) the pressure exerted by the gas is proportional to the root mean square velocity of the molecules. (3) the root mean square velocity is inversely proportional to the temperature. (4) the mean K.E. of the molecule is directly proportional to the temperature in degree centigrade.

(1) the amount of water taken. (2) volume of container in which water is present. (3) temperature. (4) some external other factors.

Graham’s Law of Effusion and Diffusion 34. Which of the following pairs will diffuse at the same rate? (1) CO2 and N2O (2) CO2 and NO (3) CO2 and CO (4) N2O and NO 35. According to Graham’s law at a given temperature, the ratio of the rates of diffusion rA/rB of gases A and B is given by p M  (1)  A   A  p M 

1/2

(3)  p A   M B   p   M  B A

1/2

B



B



 M A   pA  (2)   M   p 

1/2

(4)  M A   pB   M   p  B A

1/2

B

43. At the same temperature and pressure, which of the ­follow­ing gases will have the highest kinetic energy per mole? (1) Hydrogen (2) Oxygen (3) Methane (4) All are the same 44. The kinetic energy of a mole of ideal gas in calories is approximately equal to (1) (2) (3) (4)

B

36. A balloon having 8 g Ne gas and radius 20 cm is pricked and 7 g of Ne gas effused from it. What will be the radius of balloon under similar conditions?

3 times its absolute temperature. 2 times its absolute temperature. 4 times its absolute temperature. 2/3 times its absolute temperature.

45.

(1) 5 cm (2) 10 cm (3) 15 cm (4) 20 cm 37. Molecular weight of a gas that diffuses twice as rapidly as the gas with molecular weight 64 is

Fraction of molecules

(1) 16 (2) 8 (3) 64 (4) 6.4 38. Helium gas diffuses four times faster than gas X. Molar mass of gas X is (1) 8 (2) 72 (3) 16 (4) 64 39. The ratio of rates diffusion of SO2, O2 and CH4 is (1) 1: 2:2 (2) 1:2:4 (3) 2: 2:1 (4) 1:2: 2 40. The pair of gases which can be most easily separated from effusion technique is (1) D2 and H2 (2) CH4 and CD4 (3) C12H4 and C14H4 (4) U235F6 and U238F6

Chapter 3_States of matter.indd 68

C1

Velocity

The velocity C1 in the above figure is given by the relation (1) C1 = 3RT (2) C1 = 2RT M M (3) C1 = 8RT (4) C1 = 3RT pM M 46. In a closed vessel, a gas is heated from 300 K to 600 K, the kinetic energy becomes/remains (1) half. (2) double. (3) same. (4) four times.

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States of Matter 47. A closed flask contains water in all its three states at 0°C. In this situation average kinetic energy of water molecule will be (1) (2) (3) (4)

same in all the three states. the greatest in vapor state. the greatest in liquid state. the greatest in solid state.

48. If p is the pressure of gas, then the kinetic energy per unit volume of the gas is (1) p/2 (2) p (3) 3p/2 (4) 2p 49. Which of the following gases would have the highest rms speed at 0°C?

56. The root mean square speed of CH4 molecules at 25°C is about 0.56 km s-1. What is the root mean square speed of a H2 molecule at 25°C? (1) 0.070 km s-1 (2) 0.20 km s-1 (3) 1.1 km s-1 (4) 1.6 km s-1

Compressibility Factor (Z), Liquefaction of Gas, van der Waals Equation 57. The van der Waals equation explains the behavior of (1) ideal gases. (2) real gases. (3) vapor. (4) non-real gases. 58. The compressibility factor for an ideal gas is (1) 1.5 (2) 1 (3) 2 (4) ¥

(1) O2 (2) CO2 (3) SO3 (4) CO 50. Assume that the container is filled with the mixture of SO3 and Ne. The molecular weight of SO3 is 80 g mol-1 and the atomic weight of Ne is 20 g mol-1. The average velocity of SO3 molecules is (1) (2) (3) (4)

one fourth that of a Ne atom. one half that of a Ne atom. the same as a Ne atom. two times that of a Ne atom.

51. The ratio of rms velocity to average velocity of gas molecules at a particular temperature is (1) 1.086:1 (2) 1:1.086 (3) 2:1.086 (4) 1.086:2 52. The temperature at which H2 has the same rms speed (at 1 atm) as that of O2 at STP is (1) 37K (2) 17K (3) 512K (4) 27K 53. Identify a postulate of kinetic theory among the following. (1) (2) (3) (4)

An atom is indivisible. Gases combine in simple ratio. There is no influence of gravity on gas molecules. None of the above.

54. An ideal gas molecule is present at 27°C. By how many degree centigrade its temperature should be raised so that its urms, ump and uav all may double? (1) 900°C (2) 108°C (3) 927°C (4) 81°C 55. A certain gas is at a temperature of 350 K. If the temperature is raised to 700 K, the average translational kinetic energy of the gas will (1) (2) (3) (4)

remain constant. increase by a factor of 2. increase by a factor of square root of 2. decrease by a factor of square root of 2.

Chapter 3_States of matter.indd 69

69

59. The value of van der Waals constant a is maximum for (1) He (2) N2 (3) CH4 (4) NH3 60. Table gives values of a for different gases



O2

H2

NH3

CH4

1.310

1.390

4.17

2.253

Therefore which can most easily liquefied is (1) O2 (2) NH3 (3) H2 (4) CH4

61. The compressibility factor for gas obeying van der Waals’ equation of state is given by (where V is molar volume) (1)

V a (2) a V − − V − b RTV RTV V − b

(3) V − b − RTV (4) RTV − V − b V V a a 62. Out of the following gases, which one has least value of van der Waals constant a? (1) CO2 (2) NH3 (3) CH4 (4) H2 63. A given gas cannot be liquefied if its temperature is (1) (2) (3) (4)

equal to its critical temperature. greater than its critical temperature. smaller than its critical temperature. equal to its inversion temperature.

64. Which of the following is incorrect statement? (1) van der Waals’ constant a is a measure of attractive force. (2) van der Waals’ constant b is also called co-volume or excluded volume per mole. (3) b is expressed in L mol–1. (4) b is one-fourth of critical volume.

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OBJECTIVE CHEMISTRY FOR NEET

65. a /V 2 given in van der Waals equation is for (1) (2) (3) (4)

internal pressure. intermolecular attraction. both (1) and (2). temperature correction.

66. The temperature at which real gases obey the ideal gas laws over a wide range of pressure is called (1) (2) (3) (4)

critical temperature. Boyle temperature. inversion temperature. reduced temperature.

67. Which of the following statement is not true? (1) The pressure of the real gas is equal to the pressure calculated for an ideal gas. (2) The van der Waals’ equation helps to calculate the pressure and volume of real gases. (3) Real gas molecules do occupy a finite, but small, volume. (4) None of these. 68. Which of the following gases is the least likely to behave ideally? (1) He (2) N2 (3) HCl (4) H2

Liquid State 69. Which of the following is not true? (1) Density of solids is greater than gases. (2) Molecules of solid possess vibratory motion. (3) Molecules of gases possess greater kinetic energy than molecules of solid and liquid. (4) Gases like liquids possess definite volume. 70. With the increasing molecular mass of a liquid, the viscosity (1) decreases. (2) increases. (3) no effect. (4) all wrong. 71. Which of the following expression regarding the unit of coefficient of viscosity is not true? (1) dyne cm−2 s (2) dyne cm−2 s−1 (3) Nm−2 s (4) 1 poise = 10−1 Nm−2 s

Level II Gas Laws 1. The volume of a large irregularly shaped tank is determined as follows. The tank is first evacuated, and then it is connected to a 50 L cylinder of compressed helium gas. The gas pressure in the cylinder, originally at 21 atm, falls to 7.0 atm without a change in temperature. What is the volume of the tank?

Chapter 3_States of matter.indd 70

(1) 100 L (2) 150 L (3) 200 L (4) 300 L 2. A spherical air bubble is rising from the depth of a lake where pressure is p atm and temperature is T Kelvin. The percentage increase in its radius when it comes to the free surface of lake will be (Assume temperature and pressure at the surface be respectively p/4 and 2T Kelvin.) (1) 100% (2) 50% (3) 40% (4) 200% 3. Which of the following graphs does not represent isobar given by Charles’ law?

V

V

(1)

V

V

(2) T(ºC) T(ºC)

T(K) T(K)

(3) Both (1) and (2) (4) None of these 4. A gas cylinder containing cooking gas can withstand a pressure of 14.9 atm. The pressure gauge of cylinder indicates 12 atm at 27°C. Due to sudden fire in building the temperature starts rising. The temperature at which the cylinder will explode is (1) 42.5°C (2) 67.8°C (3) 99.5°C (4) 25.7°C

Ideal Gas Law 5. 3.2 g S on heating occupies a volume of 780 mL at 450°C and 723 mm Hg pressure. Formula of sulphur is (1) S2 (2) S (3) S4 (4) S8 6. 1 mol of an ideal gas A with 300 mm of Hg is separated from 2 mol of another ideal gas B with 300 mm of Hg in a closed container at the same temperature. If the separation is removed then the total pressure is (1) 200 mm of Hg (2) 300 mm of Hg (3) 500 mm of Hg (4) 600 mm of Hg 7. The volume of balloon filled with 4 g He at 22°C and 720 mm of Hg is (1) 25.565 L (2) 20 L (3) 15 L (4) 30 L 8. 0.2 mol sample of hydrocarbon CxHy yields after complete combustion with excess O2 gas, 0.8 mol of CO2, 1.0 mol of H2O. Hence hydrocarbon is (1) C4H10 (2) C4H8 (3) C4H5 (4) C8H16 9. Excess F2(g) reacts at 150°C and 1.0 atm pressure with Br2(g) to give a compound BrFn. If 423 mL of Br2(g) at the

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States of Matter

71

same temperature and pressure produced 4.2 g of BrFn. what is n ? (At.wt.Br = 80, F = 19) (1) 3 (2) 1 (3) 5 (4) 7 CO2 2.13 atm 1.5 L

10. Assume center of Sun to consist of gases whose average molecular weight is 2. The density and pressure of the gas are 1.3 g cm-3 and 1.12 × 109 atm respectively. The temperature of Sun is (1) 2 × 103 K (2) 2 × 105 K (3) 2 × 107 K (4) 2 × 109 K 11. 5.40 g of an unknown gas at 27°C occupies the same volume as 0.14 g of hydrogen at 17°C and same pressure. The molecular weight of unknown gas is (1) 79.8 (2) 81 (3) 79.2 (4) 83 12. In the figure below mercury columns of 10 cm each are trapped between gas column of 10 cm each. If p atm = 75 cm of Hg then the gas pressure in the topmost column will be p gas = ? Hg

Ar 1.15 atm 2.0 L

H2 1.0 L 0.861 atm

(1) 1.41 atm (2) 2.41 atm (3) 3.41 atm (4) 1.12 atm 16. When 2 g gas A is introduced into an evacuated flask kept at 25°C, the pressure is found to be 1 atm. If 3 g of another gas B is further added to same flask, the total pressure becomes 1.5 atm. The ratio of molecular weights is (1) 1:1 (2) 1:2 (3) 1:3 (4) 1:4

Graham’s Law of Effusion and Diffusion 17. The valves X and Y in the below figure are opened simultaneously. The white fumes of NH4Cl will first formed at

10 cm 10 cm

Hg

NH3

10 cm 10 cm

X Hg

(1) 55 cm of Hg (2) 35 cm of Hg (3) 65 cm of Hg (4) 45 cm of Hg 13. An open vessel containing air is heated from 300 K to 400 K. The fraction of air originally present which goes out of it is (1) 3/4 (2) 1/4 (3) 2/3 (4) 1/8

Dalton’s Law 14. A mixture consisting of 0.10 mol of N2, 0.05 mol of O2 and 0.20 mol of CH4 and an unknown amount of CO2 occupied a volume of 9.6 L at 27°C and 1.0 atm pressure. How many moles of CO2 are there in this sample? (1) 0.04 mol (2) 0.39 mol (3) 0.05 mol (4) 0.10 mol 15. In the following figure, when the two stop corks are opened, the total pressure inside the flask will be

Chapter 3_States of matter.indd 71

HCl B

A

C

Y

(1) A (2) B (3) C (4) A, B and C simultaneously 18. A balloon filled with moist air has developed a pinhole. It is quickly plunged into a tank of dry air at the same pressure. In a short while (1) it will collapse. (2) it will enlarge. (3) no change will take place. (4) can’t be predicted. 19. The mole fraction He is 0.4 in gaseous mixture He and CH4. If both the gaseous are effusing through the constant area of the orifice of the container, then what will be % composition by volume of CH4 gas effusing out initially? (1) 50% (2) 40% (3) 43% (4) 75% 20. A mixture of H2 and O2 in 2:1 volume is allowed to diffuse through a porous partition what is the composition of gas coming out initially (1) 1:2 (2) 4:1 (3) 8:1 (4) 1:4

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72

OBJECTIVE CHEMISTRY FOR NEET

21. The rate of diffusion of methane at a given temperature is twice that of gas X. The molecular weight of X is (1) 64.0 (2) 32.0 (3) 4.0 (4) 8.0

28. According to the kinetic theory of gases

22. Vegetables are canned, while they are steaming hot because (1) (2) (3) (4)

the heat inside will seal the jars. the heat increases the atmospheric pressure. the heat creates more pressure inside the jars. when the jars cool, a vacuum inside will help to seal the jars.

23. Which of the following mixture of gases cannot be separated by diffusion method? (1) NO + C2H6 (2) NO + NO2 (3) CO + CO2 (4) C2H4 + C2H6 24. Some moles of SO2 diffuse through a small opening in 20 s. Same number of moles of an unknown gas diffuses through the same opening in 60 s. Molecular mass of the unknown gas is 60 (1) (64)2 ×    20 

 20  (2) (64)2 ×    60 



2

(3) (64) ×  60   20 

(4) (64) ×  20   60 

2

Kinetic Theory of Gases and Maxwell Boltzmann Distribution Curve 25. If EK is the kinetic energy per mole of a gas, then (1) pV = 3 E K (2) p = 3 VE K 2 2 2 (3) pV = E K (4) 3pV = E K 3 26. On increasing temperature, the fraction of total gas molecule which has acquired most probable velocity will (1) increase. (2) decrease. (3) remains constant. (4) cannot say without knowing pressure. 27. Distribution of molecules with velocity is represented by the following curve. Point A in the curve shifts to the higher value of velocity if

A

Number of molecules u

Chapter 3_States of matter.indd 72

(1) T is increased. (2) V is increased. (3) p is increased. (4) All of the above.

(1) (2) (3) (4)

there are intermolecular attractions. molecules have considerable volume. no intermolecular attractions. the velocity of molecules decreases after each collision.

29. Let the most probable velocity of hydrogen molecules at a temperature T °C is V0. Suppose all the molecules dissociate into atoms when temperature is raised to (2 T + 273)°C then the new rms velocity is (1)

2 (2) V0 3

(3) 2 3 V0 (4)

3( 2 + 273) V0 T 6 V0

30. Which of the following statements concerning the kinetic theory of gases is (are) correct? (I) Molecules make elastic collisions with each other and with the walls of their container. (II) The average kinetic energy of a large number of molecules of mass, M, is proportional to M1/2 at a given temperature. (III) The molecules of a gas are in constant random motion. (IV) All the molecules of a gas have the same kinetic energy at a given temperature. (1) I, II, III, IV (2) I, II, III (3) I, III, IV (4) I, III 31. Compare the root mean square speed of an O2 molecule with that of CH4 molecule at the same temperature and pressure. (1) The speed is the same, since the weight of O2 and of CH4 are both 16 g mol-1. (2) The speed is the same because at the same temperature all gas molecules have the same mean square speed. (3) CH4 is 2 times faster, because the molecular weight of O2 is 2 times greater than the molecular weight of CH4. (4) CH4 is 1.41 times faster, since at equal temperature of all gas molecules have the same kinetic energy. 32. Consider two 1 L flasks, one containing O2, the other containing He, each at STP. Which of the following statement is NOT true regarding these gases? (1) Each flask contains the same number of atoms or molecules. (2) The gases in each flask have the same average kinetic energy. (3) The gases in each flask have the same density. (4) The pressure in each flask is the same.

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States of Matter 33. Molecular velocities of two gases at the same temperature are u1 and u2 and their molecular masses are m1 and m2 respectively. Which of the following expression is correct? (1) m1 = m2 u12 u22

(2) m1u1 = m2u2

(3) m1 = m2 (4) m1u12 = m2u22 u1 u2 34. The kinetic molecular theory of gases predicts pressure to rise as the temperature of a gas increases because (1) the average kinetic energy of the gas molecules decreases. (2) gas molecules collide more frequently with the container walls. (3) gas molecules collide less frequently with the container walls. (4) gas molecules collide less energetically with the container walls. 35. If the absolute temperature of a sample of gas in a fixed volume container is quadrupled, then the root mean square speed in the initial state ui and that in the final stage uf would be related as (1) uf = ui (2) uf = ui 4 2 (3) uf = 2ui (4) uf = 4ui

Compressibility Factor (Z), Liquefaction of Gas, van der Waals Equation 36. The compressibility factor of He as a real gas at room temperature is (1) unity (2) 1− a RTV RTV (3) 1+ pb (4) 1− a RT 37. At the critical temperature (1) (2) (3) (4)

liquid and vapor exist in equilibrium. the meniscus between liquid and vapor disappears. vapor state does not exist at all. the vapor condense into solid.

38. The compressibility factor Z = pV nRT a will be T= Rb

of a gas above

(1) always less than unity. (2) always equal to unity. (3) always greater than unity. (4) depends on pressure. 39. A gas deviates from ideal behavior at a high pressure because its molecules

Chapter 3_States of matter.indd 73

(1) (2) (3) (4)

73

have kinetic energy. are bound by covalent bonds. have different shapes. attract one another.

40. At critical temperature, which kind of forces is dominating? (1) Attractive forces (2) Repulsive forces (3) Both are the same (4) Cannot be determined 41. For a non-zero volume of molecules having no force of attraction, the variation of compressibility factor, Z vs p is best represent by

Z

II

1.0 I

III

p (atm)

(1) I (2) II (3) III (4) All of the above 42. A gas is easily liquefied (1) above critical temperature and below critical pressure. (2) below critical temperature and above critical pressure. (3) below critical temperature and critical pressure. (4) above critical temperature and critical pressure. 43. van der Waals constant a and b are related with (1) attractive force and bond energy of molecules respectively. (2) volume and repulsive forces of molecules respectively. (3) shape and repulsive forces of molecules respectively. (4) attractive force and volume of the molecules respectively. 44. For non-zero value of force of attraction between gas molecules, gas equation will be 2

(1) pV = nRT - n a (2) pV = nRT + nbp V nRT (3) pV = nRT (4) p = V −b 45. Which of the following is an incorrect statement? (1) Ideal gases have Z = 1 and cannot be liquefied. (2) When Z > 1, real gases are difficult to compress. (3) When Z > 1, real gases are easier to compress. (4) When Z < 1, real gases are easier to compress.

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74

OBJECTIVE CHEMISTRY FOR NEET

46. One mole of SO2 gas occupies a volume of 900 mL at 24 atm pressure at 27°C. Then the gas will show (1) (2) (3) (4)

negative deviation from ideal behavior. positive deviation from ideal behavior. no deviation from ideal behavior. Cannot be determined.

47. At low pressure van der Waals equation for 4 mol of a real gas will have its simplified form as pV (1) =4 RT + pb (3)

pV



pV =1 RT + 4pb



=4 (2) RT − 4a V pV =4 (4) RT − a V

48. Which of the following statement is true for van der Waals gas constant a and b? (1) a depends on size and shape, b depends only on size of molecule. (2) b depends on size and shape, a depends only on size of molecule. (3) Both a and b depends on shape and size of molecule. (4) Both a and b depends only on size of molecule. 49. Dominance of strong repulsive forces among the molecules of real gas (Z = compressibility factor) (1) (2) (3) (4)

depends on Z and indicated by Z = 1 depends on Z and indicated by Z < 1 depends on Z and indicated by Z > 1 is independent of Z.

50. If for a gas, critical parameters are pC = 2 atm, TC = 47°C, then approximate value of VC will be (1) 10 L (2) 5 L (3) 20 L (4) 2 L 51. When do you expect a real gas to behave like an ideal gas? (1) (2) (3) (4)

When both the temperature and pressure are low. When both the temperature and pressure are high. When the temperature is high and pressure is low. When the temperature is low and pressure is high.

Liquid State 52. Viscosity of a liquid is increased by (1) (2) (3) (4)

increase in temperature. decrease in molecular size. increase in molecular size. none of the above.

53. If n1 and n2 are the coefficients of viscosity of two liquids, r1 and r2 their densities and t1 and t2 the flow times in Ostwald viscometer, then

Chapter 3_States of matter.indd 74

(1)

n1 r1t 2 = n2 r2t1

n rt (3) 1 = 1 1 n2 r2t 2

(2)

n1 r2t 2 = n2 r1t1

(4)

n1 r2t1 = n2 r1t 2



Previous Years’ NEET Questions 1. If an ideal gas expands at a constant temperature, it indicates that (1) (2) (3) (4)

the number of the molecules of gas increases. kinetic energy of molecule decreases. pressure of the gas increases. kinetic energy of molecules remains the same. (AIPMT 2008)

2. What volume of oxygen gas (O2) measured at 0°C and 1 atm, is needed to burn completely 1 L of propane gas (C3H8) measured under the same conditions? (1) 10 L (2) 7 L (3) 6 L (4) 5 L (AIPMT 2008) 3. The pressure exerted by 6.0 g of methane gas in a 0.03 m3 vessel at 129°C is (atomic masses: C = 12.01, H = 1.01 and R = 8.314 JK−1 mol−1) (1) 13409 Pa (2) 41648 Pa (3) 31684 Pa (4) 215216 Pa (AIPMT MAINS 2010) 4. By what factor does the average velocity of a gaseous molecule increase when the temperature (in Kelvin) is doubled? (1) 1.4 (2) 2.0 (3) 2.8 (4) 4.0 (AIPMT PRE 2011) 5. Two gases A and B having the same volume diffuse through a porous partition in 20 and 10 seconds respectively. The molecular mass of A is 49 u. Molecular mass of B will be (1) 25.00 u (2) 50.00 u (3) 12.25 u (4) 6.50 u (AIPMT PRE 2011) 6. A gaseous mixture was prepared by taking equal moles of CO and N2. If the total pressure of the mixture was found 1 atm, the partial pressure of the nitrogen (N2) in the mixture is (1) 1 atm (2) 0.5 atm (3) 0.8 atm (4) 0.9 atm (AIPMT PRE 2011)

1/4/2018 5:10:35 PM

States of Matter 7. A bubble of air is underwater at temperature 15°C and the pressure 1.5 bar. If the bubble rises to the surface where the temperature is 25°C and the pressure is 1.0 bar, what will happen to the volume of the bubble? (1) (2) (3) (4)

Volume will become greater by a factor of 2.5. Volume will become greater by a factor of 1.6. Volume will become greater by a factor of 1.1. Volume will become greater by a factor of 0.70. (AIPMT MAINS 2011)

8. 50 mL of each gas A and of gas B takes 150 and 200 s respectively for effusing through a pin-hole under the similar condition. If molecular mass of gas is 36, the molecular mass of gas A will be (1) 96 (2) 128 (3) 21 (4) 64

11. Maximum deviation from ideal gas is expected from (1) H2(g) (2) N2(g) (3) CH4(g) (4) NH3(g) (NEET 2013) 12. Equal masses of H2, O2 and methane have been taken in a container of volume V at temperature 27°C in identical conditions. The ratio of the volumes of gases H2:O2: methane would be (1) 8:16:1 (2) 16:8:1 (3) 16:1:2 (4) 8:1:2 (AIPMT 2014) 13. A gas such as carbon monoxide would be most likely to obey the ideal gas law at (1) (2) (3) (4)

(AIPMT PRE 2012) 9. A certain gas takes three times as long to effuse out as helium. Its molecular mass will be (1) 27 u (2) 36 u (3) 64 u (4) 9 u (AIPMT MAINS 2012) 10. For real gases, van der Waals equation is written as  an 2  p + (V − nb ) = nRT  V 2 

where a and b are van der Waals constants.



Two sets of gases are:

(I) He < H2 < CO2 < O2 (I) O2 < He < H2 < CO2 (I) H2 < He < O2 < CO2 (I) H2 < O2 < He < CO2

(RE-AIPMT 2015) 14. Equal moles of hydrogen and oxygen gases are placed in a container with a pin-hole through which both can escape. What fraction of the oxygen escapes in the time required for one-half of the hydrogen to escape?

(NEET-I 2016)

The gases given in Set I in increasing order of b and gases given in Set II in decreasing order of a, are arranged below. Select the correct order from the following: (1) (2) (3) (4)

high temperatures and high pressures. low temperatures and low pressures. high temperatures and low pressures. low temperatures and high pressures

(1) 1/8 (2) 1/4 (3) 3/8 (4) 1/2

(I) O2, CO2, H2 and He (II) CH4, O2 and H2

75

(II) CH4 > H2 > O2 (II) H2 > O2 > CH4 (II) CH4 > O2 > H2 (II) O2 > CH4 > H2

15. A 20 L container at 400 K contains CO2(g) at pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO). The volume of the container is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when pressure of CO2 attains its maximum value, will be

(Given that: SrCO3(s)  SrO(s) + CO2(g ), Kp = 1.6 atm) (1) 10 L (2) 4 L (3) 2 L (4) 5 L (NEET 2017)

(AIPMT MAINS 2012)

Answer Key Level I  1. (3)

 2. (4)

 3. (1)

 4. (1)

 5. (2)

 6. (3)

 7. (1)

 8. (1)

 9. (3)

10.  (4)

11.  (4)

12.  (4)

13.  (4)

14.  (4)

15.  (4)

16.  (3)

17.  (1)

18.  (3)

19.  (3)

20.  (2)

21.  (1)

22.  (4)

23.  (2)

24.  (4)

25.  (2)

26.  (2)

27.  (3)

28.  (3)

Chapter 3_States of matter.indd 75

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76

OBJECTIVE CHEMISTRY FOR NEET

29.  (4)

30.  (1)

31.  (1)

32.  (1)

33.  (3)

34.  (1)

35.  (3)

36.  (2)

37.  (1)

38.  (4)

39.  (1)

40.  (1)

41.  (4)

42.  (1)

43.  (4)

44.  (1)

45.  (2)

46.  (2)

47.  (2)

48.  (3)

49.  (4)

50.  (2)

51.  (1)

52. (2)

53.  (3)

54.  (1)

55.  (2)

56.  (4)

57.  (2)

58.  (2)

59.  (4)

60.  (2)

61.  (1)

62.  (4)

63.  (2)

64.  (4)

65.  (2)

66.  (2)

67. (1)

68.  (3)

69. (4)

70.  (2)

 1. (1)

 2. (1)

 3. (4)

 4. (3)

 5. (4)

 6. (2)

 7. (1)

 8. (1)

9.  (3)

10.  (3)

11.  (1)

12.  (1)

13.  (2)

14.  (1)

15.  (1)

16.  (3)

17.  (3)

18.  (1)

19.  (3)

20.  (3)

21.  (1)

22.  (4)

23.  (1)

24. (3)

25.  (3)

26  (2)

27.  (1)

28.  (3)

29.  (4)

30.  (4)

31.  (4)

32.  (3)

33.  (4)

34.  (2)

35.  (3)

36.  (3)

37.  (2)

38.  (3)

39.  (4)

40.  (1)

41.  (2)

42.  (2)

43.  (4)

44.  (1)

45.  (3)

46.  (1)

47.  (2)

48.  (1)

49.  (3)

50.  (2)

51.  (3)

52.  (3)

53.  (3)

71.  (3)

Level II

Previous Years’ NEET Questions  1. (4)

 2. (4)

 3. (2)

 4. (1)

 5. (3)

 6. (2)

 7. (2)

 8. (3)

 9. (2)

10.  (4)

11.  (4)

12.  (3)

13. (3)

14.  (1)

15.  (4)

Hints and Explanations Level I

10. (4) From ideal gas equation, we know V =

1. (3) We know p1V1 = p2V2 p 0.95



Therefore, pV = p2(0.95V ) Þ p2 =



Percentage increase in pressure can be calculated as æ p ö - p ÷ ´ 100 ç p2 - p 0.95 è ø = ´ 100 = = 5.26 % p p

2. (4) pV = constant 5. (2) According to Charle’s law

V1 V2 = T1 T2

100 200 = Þ T2 = 746 K or 473°C 373 T2 8. (1) We have

V1 V2 V 1.1 V = Þ = Þ T2 = 1.1T T1 T2 T T2

Hence, percentage of increase in temperature = 10%

Chapter 3_States of matter.indd 76

nRT p



Therefore, V µ n



Since, nHI is minimum, therefore, VHI is minimum.

12. (4) The average speed is given by 8RT 8pV = pM pM

uavg =

Given (uavg )A = 2(uavg )B





8p A ´ 1 8 pB ´ 2 = Þ p A = 8 pB pM pM

13. (4) From ideal gas equation pV = nRT

       pV = nB RT (1)







On dividing Eq. (2) by Eq. (1), we get

2p × 2V = nA × 2T (2)

4=

nA ´ 2 nB

1/4/2018 5:10:37 PM

States of Matter

or    2nB = nA



x (Mass of A ) or   2 × Þ Mass of A = 2x g = M M

r1 T2 = r2 T1 16 T2 = Þ T2 = 312 K or 39°C 14 273

14. (4) We know n1T1 = n2T2 n ´ 300 = 0.5n ´ T2 Þ T2 = 600 K or 327°C 15. (4) Density is given by r =

26. (2) We have r =

pM RT

pM rRT or p = RT M p A rA ´ M B pA 2 ´ 2 4 Þ = = = 1 1 pB rB ´ M A pB

18. (3) nH2 =

2 = 1, therefore, number of molecules of H2 = NA 2

nN 2 =



Hence, number of molecules in flask A is greater than that in flask B.

pV = n (1) RT p If V = 1 L, then Eq. (1) becomes n = RT p p 20. (2) We know 1 = 2 T1 T2 19. (3) From ideal gas equation



Therefore,

21. (1)    

2 17 = Þ T2 = 150 ´ 17 = 2550 K or 2127°C 300 T2  



22. (4) From combined gas law, we have

p1V1 p2V2 = T1 T2

where p is the partial pressure and x is the mole fraction. pµx



Number of moles can be calculated as nH2 =



1 4 7 8 = 0.5 ; nHe = = 2; nN2 = = 0.25; nO2 = = 0.25 2 2 28 32

As mole fraction of helium would be highest, therefore, it will also have highest partial pressure.

28. (3) Given that pN 2 + pO2 = p p 2



Also, pN 2 = pO2 =



If N2 is removed, net pressure = pO2 =

p 2

30. (1) Mole fraction of CH4 can be calculated as xCH4 =

p1V1 p2V2 = T1 T2

3 ´ 275 2.5 ´ 200 = Þ T2 = 303 K T2 500

p1 p2 1 Þ p2 = 2 atm p2

27. (3) Using Dalton’s law p = ptotal x

N 2 1 = , therefore, number of molecules of N2 = A 14 28 14



pM or r µ p RT r1 = r2 1 = 2

800 ´ M Air ´ R ´ 300 1 ( r )Delhi ( r )Delhi Þ = = ( r )Mumbai R ´ 320 ´ 750 ´ M Air ( r )Mumbai 1 16. (3) We know r =

16 /16 1 = = 0.25 æ 56 44 16 ö ( 2 + 1 + 1) + + ç ÷ è 28 44 16 ø



Partial pressure is given by pCH4 = xCH4 × pTotal



= 0.25 × 720 = 180 mm

31. (1) We have xO2 = 0.21 pO2 = ptotal × xO2

pV ( p / 2 )V2 = ⇒ V2 = 4V ( 2T ) T

= 0.21 × 750 = 157.5 mm Hg 32. (1) Mole fraction of H2 can be calculated as



pM p or r µ RT T 1 2 Option (1): r µ ; Option (2): r µ 273 273 1 2 Option (3): r µ ; Option (4): r µ 546 546



Therefore, density of neon is highest at 0°C and 2 atm.



23. (2) Density is given by r =

24. (4) We know r =

pM or r µ M RT

Therefore, density of propane (C3H8) is the maximum.

25. (2) We have r =

Chapter 3_States of matter.indd 77

pM 1 or r µ RT T

77

x H2

æxö ç ÷ 8 è2ø = = æ x ö æxö 9 ç ÷+ç ÷ è 16 ø è 2 ø

8 Therefore, partial pressure = xH2 ´ pTotal = ´ pTotal 9

34. (1) The rate of diffusion is directly proportional to its molar mass. Since, CO2 and N2O have same molar mass, therefore, they will diffuse at the same rate. n n 36. (2) We have      1 = 2 V1 V2

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78

OBJECTIVE CHEMISTRY FOR NEET 49. (4) Root mean square speed is given by

(8/M ) (1/M ) = 4 4 3 3 p ( 20) pr 3 3 20 r = 1/3 = 10 cm (8 ) 37. (1) From Graham’s law of diffusion

urms =

r1 M2 = r2 M1

(uavg )SO3 (uavg )Ne

39. (1) We have rSO2 : rO2 = M O2 : M SO2

(uavg )SO3





rSO2 : rO2 = 1 : 2 (1)





rO2 : rCH4 = M CH4 : M O2 





rO2 : rCH4 = 1 : 2 (2)



Therefore, from (1) and (2), we get rSO2 : rO2 : rCH4 = 1 : 2 : 2

(uavg )Ne 51. (1) We have  

41. (4) From root mean square speed, we have

(urms )1 T = 1 (urms )2 T2

1 273 = Þ T2 = 1092 K or 819°C T2 2 43. (4) The average kinetic energy per mole (Ek) is equal to kinetic energy per molecule (E K) multiplied by the Avoga-

dro’s number (NA).

3 RT 2

Therefore,

(K.E.)1 300 = Þ (K.E.)2 = 2(K.E.)1 (K.E.)2 600

3 8. (3) Kinetic energy is given by K.E. = nRT 4 2

Chapter 3_States of matter.indd 78

1 20 Þ (uavg )SO3 = (uavg )Ne 2 80

3RT 3R ´ 273 = Þ T @ 17 K 2 32

54. (1) We know urms , ump , uav g ∝ T



Therefore,

(urms )2 T = 2 (urms )1 T1

Hence, temperature should be raised by 900°C.

55. (2) Average translational kinetic energy =

3 RT ´ n 2

K.E. µT (K.E.)2 T2 = (K.E.)1 T1 (K.E.)2 700 = Þ (K.E.)2 = 2(K.E.)2 (K.E.)1 350

2RT M

K.E. 3 nRT 3 = = p Volume 2 V 2

=

3RT pM urms = ´ 8RT u Avg M

   

3 ´ 2 ´ T (cal) 2

46. (2) We know K.E. µT

M Ne M SO3

52. (2) Given that urms, H2 = urms, O2

K.E./ mol = 3T 45. (2) C1 = ump =

=

2 T = T2          1 T2 = 4T1 ⇒ T2 = 4 × 300 = 1200 K or 927°C

As temperature is constant, therefore, kinetic energy per mole is same for all. K.E. =

8RT 1 or uavg µ (As rest are pM M

3p urms = = 1.086 : 1 8 u Avg

3 3 E k = N A ´ kT = RT 2 2

44. (1) Kinetic energy per mole is given by

Molecular weight of CO is least, therefore, (urms)CO is maximum.

50. (2) We know uavg = constant)

2 64 = Þ M gas = 16 1 M gas



3RT 1 or urms µ M M

56. (4) Root mean square speed is given by urms =

Therefore, urms, CH4 urms, H2

=

3RT M

3RT 2 ´ 16 3RT

0.56 1 = Þ urms, H2 = 1.6 km s -1 urms, H2 8

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States of Matter 60. (2) Higher the value of a, higher is the force of attraction; easier it is to liquefy the gas. a ö æ 61. (1) We know    ç p + 2 ÷(V - b ) = RT V ø è

Therefore,       p +

V=











 CxH y +  x + 

a RT = V 2 (V - b )

V a (V - b ) (VRT )

62. (4) a is a measure of attractive forces and is proportional to molar mass.



p1V1 p2V2 = T1 T2

p1 p2 = T1 T2

12 14.9 = Þ T2 = 99.5°C 300 T2

Let the formula of sulphur be Sx.



Therefore, n =



From Eq. (1), we get

3.2 32 x

3.2 (723 /760) × (780 × 10 −3 ) = ⇒x=8 32 x 0.0821 × 723

Hence, the formula of sulphur is S8.

n 1 Br2 + F2 → BrFn 2 2 nBr2 nBrFn = 1/ 2 1 1 × 423 4.2 ⇒n = 5 = 0.5 × 1000 × 0.0821 × 423 (80 + 19n )

10. (3) r =

pM 1.12 ´ 109 ´ 2 ÞT = = 2 ´ 107 K RT 0.0821 ´ 1.3

æ 0.14 R ´ 290 ö 5.40 pç ´ ´ R ´ 300 ÷= p ø M è 2 M = 79.8 g 12. (1) We have pgas + 10 + 10 = patm

n1 ´ 300 = n2 ´ 400 3 n2 = n1 4

nRT 7. (1) Volume can be calculated as V = p

Therefore, fraction of air left =

n1 - n2 1 = n1 4

14. (1) pTotal V = nTotal RT



1 ´ 9.6 = (0.1 + 0.05 + 0.2 + x) ´ (0.0821) ´ (300)

   x = 0.04 mol 15. (1) pTotal V = nTotal RT é 2.13 ´ 1.5 1 ´ 0.861 1.15 ´ 2 ù pTotal (1.5 + 1 + 2) = ê + + RT RT RT úû ë RT pTotal = 1.41 atm

pTotal (V A + VB ) = (1 + 2)RT æ 1 ´ RT 2RT ö pTotal ç + ÷ = 3RT Þ pTotal = 300 mm Hg è 300 300 ø

pgas = 75 − 20 = 55 cm of Hg

13. (2) We have        n1T1 = n2T2

6. (2) We have  pTotal VTotal = nTotal RT

Chapter 3_States of matter.indd 79

nH2 O

9. (3) The reaction is

pV 5. (4) According to ideal gas equation n = (1) RT

=

11. (1) From ideal gas equation, pV = nRT

4 p 4 3 p ´ p r13 ´ p r2 3 = 4 3 T 2T 8r13 = r23 r2 = 2r1 or 100% increase 4. (3) From combined gas law, we have

nCO2

Hence, the hydrocarbon is C4H10.

1. (1) Let the volume of the tank be V L. We know

2. (1) We know

=

( y/ 2) x 1 0.2 0.8 = ⇒x=4 x 1 0.2 1 = ⇒ y = 10 1 ( y/ 2)

Level II

p1V1 = p2V2 21 × 50 = 7(V + 50) ⇒ V = 100 L

y y  O2 → xCO2 + H 2O 4 2 nCx H y

RT a (V - b ) V 2 pV V a = RT (V - b ) VRT Z=

( 4 /4) ´ (0.0821) ´ ( 295) = 25.625 L (720/760 )

8. (1) The combustion reaction is

p=



79

16. (3) We know    pV = nRT

    1 ´ V =

2 ´ R ´ 298 (1) MA

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80

OBJECTIVE CHEMISTRY FOR NEET æ 2 3 ö 1.5 ´ V = ç + ÷ R ´ 298 (2) M M B ø è A







On dividing Eq. (2) by Eq. (1), we get æ 2 3 ö + ç ÷ M MB ø 1 M 1.5 = è A Þ A= MB 3 æ 2 ö ç ÷ è MA ø

rms velocity is given by urms =



 



On dividing Eq. (2) by Eq. (1), we get

Vnew = 6 V0 Vrms, O2

31. (4)

19. (3) From Graham’s law of diffusion rHe p = He rCH4 pCH4

M CH4

Vrms, CH4

rH2 rO2 rH2 rO2



=

=

pH 2

M O2

pO2

M H2

35. (3) rms velocity is given by urms =

rCH4 rgas X

=

2 32 8 = 1 2 1

1 ui = Þ uf = 2ui 4 uf 36. (3) We have

MX 16



runknown

2

=

M unknown 60 æ 60 ö = Þ M unknown = ç ÷ ´ 64 M SO2 20 è 20 ø

25. (3) Average kinetic energy per mole is given by 3 RT (1) 2





EK =



Also, for 1 mol

pV = RT(2)



From Eq. (1) and Eq. (2), we get EK =

3 2 pV Þ pV = E K 2 3

2RT 9. (4) Most probable velocity is given by ump = 2 M



Chapter 3_States of matter.indd 80

 

æ a ö ç p + 2 ÷(Vm - b ) = RT V m ø è

For He, a can be neglected. Therefore, p(Vm - b ) = RT

23. (1) The molecular weights of NO and C2H6 are the same, thus, they cannot be separated by diffusion. rSO 2

or urms µ T

ui T = 1 uf T2

MX 2= Þ M X = 64 16

24. (3)

3RT M

Therefore,

Therefore, composition of gas coming out initially is 8:1.

21. (1) We have

16 32

1 1 m1u12 = m2u22 Þ m1u12 = m2u22 2 2

3 Percentage of CH4 coming out = ´ 100 = 43% 7

20. (3)   

=

33. (4) For the same temperature, kinetic energy remains same. Therefore,

M He

rHe 0.4 16 4 = = rCH4 0.6 4 3

3R( 2t + 546 ) (2) 1

Vnew =

Vnew 3R ´ 2(t + 273) = = 6 V0 R ´ (t + 273)

18. (1) As pmoist gas > pdry gas, therefore, gas will flow from higher pressure to lower pressure and the balloon will collapse.

 

3RT M



2R(t + 273) V0 = (1) 2



Therefore, Z =

pVm pb = 1+ RT RT

44. (1) van der Waals equation is given by



æ a ö ç p + 2 ÷(Vm - b ) = RT V m ø è Neglecting b in the above equation, we get æ an 2 ö ç p + 2 ÷ ´ V = nRT V ø è pV = nRT -

an 2 V

46. (1) Compressibility factor can be calculated as Z=

pVm 24 × (900 /1000 ) = 0.87 = RT 0.0821 × 300

As Z < 1, the gas will show negative deviation from ideal behavior.

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States of Matter 47. (2) van der Waals equation is given by æ an 2 ö ç p + 2 ÷(V - nb ) = nRT V ø è



Substituting n = 4 in the above equation, we get pV 4a ö æ pV = 4 ç RT =4 ÷Þ 4a ö V ø æ è RT ç ÷ V ø è



Dividing Eq. (2) by Eq. (1), we get TC 8 a æ 27b 2 ö = ç ÷ = 160 pC 27 bR è a ø



   (uavg)2 = 1.44 (uavg)1

5. (3) The rate of diffusion is related to the molecular mass by the relation rA = rB





where r =



Substituting the values in Eq. (1), we get

Volume diffused (V ) Time interval (t )

V/20 = V/10

MB 49 MB 49 Þ MB = = 12.25 u 49 4

1 = 2

6. (2) We know pN 2 = xN 2 ´ pTotal

As  nN = nCO Þ xN = 0.5 2



2. (4) The reaction involved is

Substituting the values, we get = 0.5 atm

7. (2) According to combined gas law, we have p1V1 p2V2 = T1 T2 1.5 ´ V1 1 ´ V2 = Þ V2 = 1.55 V1 » 1.6 V1 288 298 8. (3) Using the relationship

5 Vol



Given that the volume of C3H8 at STP = 1 L, therefore the volume of O2 at STP = 5 L.

3. (2) From ideal gas equation, we have pV = nRT

   pV = w RT (1) M



Substituting the values in Eq. (1), we get 6 8.314 ´ 402 ´ 16.05 0.03 = 41647.7 Pa » 41648 Pa

p=

Chapter 3_States of matter.indd 81

V A / tA  M B  = VB / t B  M A 

1/2



 



Given that VA = VB = 50 mL; tA = 150 s; tB = 200 s; MB = 36 u. Substituting in Eq. (1), we get

C 3H8(g) + 5 O2(g) ® 4 H 2O(g) + 3 CO2(g) 1 Vol

2

pN 2 = 0.5 ´ 1 atm

3 K.E. = kT 2

where k is the Boltzmann’s constant. Kinetic energy of gaseous molecules depends on temperature only.

(1)



8b = 160 Þ b = 1.642 R VC = 3b = 4.926 L or 5 L approx

1. (4) We know

MB  MA



2

Previous Years’ NEET Questions

T1 T2

As T2 = 2T1, therefore

VC = 3b

=



a pC = = 2 (1) 27b 2 8a TC = = 320 (2) 27bR

50. (2) We have

(uavg )1

At low pressure, b can be neglected æ an 2 ö ç p + 2 ÷ V = nRT V ø è nRT an 2 p= - 2 V V an 2 pV = nRT V

8RT pM

4. (1) We have    uavg =

(uavg )2

81

50 /150  36  = 50 /200  M A 

1/2

⇒



200  36  = 150  M A 

(1)

1/2

⇒ M A = 21 u

9. (2) As per kinetic theory of gases, r µ (1/M)1/2 r2  M 2  = r1  M1 

1/2

 Vgas   t   4  ⇒ × =   t   VHe   M 

1/2

 t   2  = ⇒ M 1/2 = 6 ⇒ M = 36 u  3t   M 1/2 

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OBJECTIVE CHEMISTRY FOR NEET

10. (4) In the van der Waals equation, constant a is the measure of magnitude of intermolecular forces. It can be seen that as the molar mass increases, a increases and also constant b increases with the size of the molecule. 11. (4) A real gas, deviates from ideal behavior for two important reasons: (a) The model of an ideal gas assumes that gas molecules are infinitesimally small, but real molecules do take up some space. (b) In an ideal gas, there would be no attractions between molecules, but in a real gas molecules experience weak attractions towards each other.

In case of NH3, the force of attraction between the molecules is highest due to hydrogen bonding, hence, it deviates maximum from the ideal gas.

(b) At high temperature, the molecules are moving so rapidly and are so far apart that the attraction between them is hardly felt. 14. (1) Given

Let us assume that rate of escape of O2 molecule = rO2



Rate of escape of H2 molecule = rH2



According to Graham’s law rO2



V∝n 





Let the mass of H2, O2 and methane be x.



The ratio of the number of moles of gases x/2 : x/32 : x/16 = 16:1:2

(a) At low temperature, the space between the molecules is so large that the volume occupied by the molecules is insignificant.

Chapter 3_States of matter.indd 82

=

M H2 M O2

=

1 1 =  16 4

Fraction of H2 molecules escaped in time t = 1/2



Fraction of O2 molecules escaped in time t = x



Substituting in Eq. (1) we get rO2 rH2

=

(1)

1/2t 1 = x /t 4

1 1 = x 8

(at constant T and p)

13. (3) The law pV = nRT is called ideal because it follows from the assumptions of kinetic theory, which describes an ideal gas. The molecules of an ideal gas have negligible volume and no attraction or repulsion for each other. At low temperature and high pressure, Carbon monoxide will show ideal gas behavior due to following two reasons:

rH2



12. (3) According to Avogadro’s principle, the volume of a gas is directly proportional to its number of moles, n.

nO2 = nH2

  15. (4) We have SrCO3 (s)    SrO(s) + CO2(g)

Also,



From combined gas law p1V1 = p2V2(1)



Substituting p1 = 0.4 atm, V1 = 20 L and p2 = 1.6 atm in Eq. (1), we get

K p = pCO2 = 1.6 atm

0.4 ´ 20 = 1.6 ´ V2 0.4 ´ 20 V2 = =5L 1.6

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4

Thermodynamics

Chapter at a Glance 1. Thermodynamics: It is a branch of chemistry that deals with exchange in heat energy during chemical and physical transformations. 2. System: It is part of universe which is under study, research and question. Depending upon the variations in energy and matter, we can classify systems as: (a) Open system: In this system, continuous exchange of heat energy and matter takes place together, for example, hot cup of coffee at room temperature. (b) Closed system: In this system, there is continuous exchange of energy, but matter remains constant, for example, human body. (c) Isolated system: In this system, there is neither exchange of energy nor matter, for example, thermoflask, ice box. (This is a near ideal system and no perfectly isolated system is possible). 3. Surroundings: It constitutes biotic and abiotic factors that affect the physical and chemical properties of the system. 4. Thermodynamic Properties System has two types of properties (a) Intensive properties: These are properties which do not depend upon the quantity of matter present in the system, such as pressure, temperature, specific heat, refractive index, surface tension, viscosity, boiling and melting point. (b) Extensive properties: These are properties which depend upon the quantity of matter present in the system, such as volume, mass, energy, enthalpy and entropy, etc. 5. State Functions and Variables: Any property of the system that defines the state of the system, but is independent of the path that brings about that change in state is known as state function, for example, pressure (p), temperature (T ), volume (V ), composition or amount (n), etc. These variables are called state variables. 6. Thermodynamic Processes (a) Adiabatic: Processes where there is no exchange of heat, that is, Dq = 0 (b) Isothermal: Processes in which the temperature of the system remains fixed throughout, that is, DT = 0. (c) Isochoric: Processes in which the volume of the system remains constant throughout, that is, DV = 0. (d) Isobaric: Processes in which the pressure of the system remain constant throughout, that is, Dp = 0. (e) Cyclic: In these processes, the system undergoes a number of different processes and finally returns to the initial state, that is DU = 0 and DH = 0. The initial and final states are identical. (f ) Reversible: Processes in which energy change in each step can be reversed in direction by making an infinitesimally small change in any property of the system. (g) Irreversible: Processes in which the system or surroundings are not restored to their initial state at the end of process. All processes occurring spontaneously in nature are irreversible. 7. Zeroth Law of Thermodynamics: This law states that if two bodies are in thermal equilibrium with some third body, then they may also be in equilibrium with each other.

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OBJECTIVE CHEMISTRY FOR NEET

8. System Characteristics (a) Internal change: The total energy contained in a thermodynamic system is called internal energy (U). It is represented by U. The SI unit of internal entry is joule (J ). U = Kinetic energy + Potential energy + Rotational energy + Electronic energy + … (any kind of energy)    (i) It is one of the state properties.    (ii) Absolute value of internal energy of the system cannot be calculated. Change of internal energy (DU ) can be calculated using bomb calorimeter.  (iii) For an isothermal process, DU = 0 (b) Work: It is defined as the action displacement of the system against some force originating from surrounding and acting on the boundary of the system. If work (w) is done by the system, its energy decreases and if work is done on the system, its energy increases. Change in internal energy of the system is equal to amount of adiabatic work done on the system. DU = U 2 - U1 = wadiabatic (c) Heat: It is the energy transferred from a hot object to a cold object.    (i) The amount of heat transferred (q) is proportional to the difference in temperatures (q ∝T ). (ii) Change in internal energy of the system is equal to heat supplied to the system.        DU = U 2 - U1 = q (iii) When heat is absorbed by the system from the surroundings, q is positive; and when heat is given out by the system, q is negative. 9. First Law of Thermodynamics: It states that the energy can neither be created nor destroyed but can be transformed from one form to the other. If heat q is supplied to a system with internal energy U1, then a part of it is used by the system to do work (−w) and remaining is used to raise the internal energy to U2. Then by first law of thermodynamics,            DU = U 2 - U1 = q + w q = DU - ( -w ) 10. Application of First Law (a) (b)

Expansion takes place w = Negative; work is done by system V2 > V1 where V2 is final volume and V1 is initial volume. Compression takes place w = Positive; work is done on the system V2 < V1 where V2 is final volume and V1 is initial volume. So, we can say that DU = q + p DV

(c) If a system is doing work isothermally and reversibly, then the amount of work done is w = -2.303 nRT log

V2 p1 or w = -2.303 nRT log V1 p2

where n = Number of moles, R = Gas constant, T = Temperature in Kelvin, V1 = Initial volume, V2 = Final volume, p1 = Initial pressure, and p2= Final pressure. 11. Enthalpy The enthalpy of a system may be defined as the sum of the internal energy and the product of its pressure and volume. It is denoted by the symbol H. DH = DU + w DH = DU + Dng RT where ∆ng = nP - nR is the difference between stoichiometric coefficients of all gaseous products and all gaseous reactants. It is a number and does not represent moles.

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Thermodynamics

85

12. Kirchhoff’s Equation The variation of heat of reaction with temperature can be ascertained by Kirchhoff ’s equation. DH T 2 - DH T 1 T2 - T1

= Cp

Kirchhoff ’s equation at constant volume can be written as: UT 2 - UT 1 T2 - T1

= CV

13. Heat Capacity It is the quantity of heat required to raise the temperature of the system by one degree. Heat capacity =

Dq DT

(a) Heat capacity is of two types:

æ DH ö (i) Heat capacity at constant pressure C p = ç è DT ÷ø p

æ DU ö (ii) Heat capacity at constant volume CV = ç è DT ÷ø V (b) Relation between Cp and CV for an ideal gas is Cp – CV = R

For 1 g of substance        C p - CV =

R M

where M is the molar mass of the substance. (c) Adiabatic expansion of an ideal gas: Cp   pV g = constant  where g =       CV   g

 T1   p1   T  =  p  2 2

g -1

p  = 2  p1 

1-g

(d) T  he molar heat capacity (Cm) is the amount of heat required to raise the temperature of one mole of the substance by 1°C (or K). (e) Specific heat or specific heat capacity can be defined as the heat required to raise the temperature of a unit mass by 1°C (or K), at a specified temperature. The amount of heat required to bring about a temperature change (∆T ) in mass m of a substance with specific heat c, can be obtained by the relation: q = c ´ m ´ DT 14. Thermochemistry The study of energy or heat changes accompanying a chemical reaction or a physical change is called thermochemistry. Reactions which involve absorption of heat are called endothermic reactions, whereas reactions involving release of heat are called exothermic reactions. (a) Enthalpy of reaction: It is the amount of heat absorbed or evolved in a reaction, when the number of moles of reactants (as represented by the balanced chemical equation change) completely converted into the products. It is the difference between the heat content or enthalpies of the products and the reactants and represented as ∆H = Heat of products - Heat of reactants ∆H r = Enthalpies of products - Enthalpies of reactants

     ∆H r = ∑ ai H products - ∑ bi H reactants i

Chapter 4_Thermodynamics.indd 85

i

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OBJECTIVE CHEMISTRY FOR NEET

 The summation sign indicates that the enthalpies for all reactants and products are summed and ai and bi indicate the stoichiometric coefficients of the reactants and products in a balanced chemical equation. (b) Standard enthalpy of reactions: The heat change for a reaction taking place at standard conditions of temperature and pressure, that is, 298 K and 1 bar, is called standard heat of reaction or standard enthalpy change and is denoted by ∆H °. All the participating substances (reactants and products) are in their standard states. (c) Enthalpy of formation: The amount of heat evolved or absorbed when one mole of substance is formed from its constituent elements. It is represented by ∆Hf . If all the substances are in their standard state, that is, at 298 K and 1 atmospheric pressure, the enthalpy of reaction is known as standard enthalpy of formation (∆H °) (can be exothermic or endothermic). (d) Enthalpy of combustion: It is the enthalpy change when one mole of the substance is completely burnt in excess of air (always exothermic). The standard heat of combustion, ∆cH °, of a substance is the amount of heat released when 1 mol of a fuel substance is completely burned in pure oxygen gas, with all reactants and products brought to 25°C and 1 bar of pressure. (e) Enthalpy of neutralization: The amount of heat evolved when one gram equivalent of an acid is neutralized by one gram equivalent base. (f ) Enthalpy of solution: The amount of heat evolved or absorbed when given quantity of solute is dissolved in large quantity of solvent at constant pressure (can be endothermic or exothermic). (g) Enthalpy of atomization: It is the heat absorbed by one mole of substance to get converted into individual atoms in gaseous state by breaking all chemical bonds (only endothermic). (h) Enthalpy of hydration: The amount of heat evolved when a given quantity of anhydrous salt absorb sufficient amount of solvent to form hydrated salt (only exothermic). (i) Enthalpy of vaporization: The amount of heat absorbed when one mole of liquid substance gets converted into vapor state at its boiling points (only endothermic). (j) Enthalpy of sublimation: The amount of heat absorbed when one mole of solid substance gets converted into vapor state at temperature below its melting point without undergoing into liquid state (only endothermic). (k) Enthalpy of condensation: Amount of heat evolved when one mole of gaseous substance gets condensed into the liquid state (only exothermic). (l) Enthalpy of fusion: Amount of heat absorbed in converting one mole of liquid into solid at its melting point (only endothermic). 15. Bond Energy (B.E): The energy required to break a specific kind of bond in one mole of a molecule is known as bond dissociation energy.

Enthalpy of reaction= å Bond energy of reactants - å Bond energy of products

16. Thermochemical Equations and Rules for their Manipulation A balanced chemical equation that indicates the enthalpy change accompanying the reaction is called thermochemical equation. It gives the physical states of the reactants and products. (a) Hess’s law of heat summation: The sum of heat changes during a chemical reaction is constant, irrespective of route or path it follows. In other words, the total amount of heat changes is the sum of heat released or gained during the reaction. For example, in the reaction A ® B DH = x kJ 1 2 A ¾x¾ ® C ¾x¾ ® B DH = x = x1 + x2 kJ

+x

-x

x6 x4 A ¾¾3 ® P ¾+¾ ® D ¾¾5 ® Q ¾+¾ ®B DH = x = - x3 + x 4 + x5 + x6

17. Lattice Enthalpy and Born–Haber Cycle (a) Lattice enthalpy: It is the energy required to completely separate the ions in one mole of a solid compound from each other to form a cloud of gaseous ions. NaCl(s) ® Na + (g ) + Cl - (g )

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Thermodynamics

87

(b) B  orn–Haber cycle: Lattice enthalpy cannot be measured directly, but we can use Hess’ law and some other experimental data and calculate it by using a set of alternate paths from the free elements to the solid ionic compound. This is called a Born–Haber cycle and is used to calculate lattice energies, as shown below for the formation of sodium chloride. Na(g)

Ionization enthalpy Electron gain enthalpy

Na(s) + 1 Cl2(g) 2

Cl-(g) + Na+(g) Lattice enthalpy

Enthalpy of atomization

Bond dissoclation energy

Cl(g)

Enthalpy of formation

NaCl(s)

18. Entropy It is the disorderness/randomness of the substance or system. (a) The absolute value of entropy cannot be determined. (b) It is represented by S and unit is J mol−1 K−1. The change of entropy ∆S for physical changes is given by DH fusion DSfusion = T q (c) Entropy for chemical reactions DSrev = rev T 19. Gibbs Energy (a) G  ibbs energy is represented by G. It is the minimum amount of energy available with the system to do the useful work at a given time. (b) Standard Gibb’s free energy can be calculated as: o o DG o = å Gproduct - å Greactant

Change in Gibbs’s energy is given by DG = DH - T DS (c) wnon-expansion = −∆G or wUseful = −∆G



Thus decrease in Gibb’s free energy = Maximum useful work

20. Gibbs Energy and Spontaneity (a) T  he sole criterion to decide spontaneity of the reaction is decrease in free energy (i.e., ∆G = −ve). ∆U, ∆H and ∆S is not the sole criteria to decide the spontaneous reaction. (b) Hence, the following relationships can help us determine if a reaction is spontaneous. ∆H +ve

∆S +ve

Relationship between ∆H and T∆S

∆G

Reaction

T∆S > ∆H

Spontaneous

+ve

+ve

−ve +ve

+ve

T∆S < ∆H All temperatures

−ve +ve

Spontaneous

−ve

All temperatures

−ve +ve

Non-spontaneous Non-spontaneous

21. Gibb’s Energy Change and Equilibrium (a) Calculation of Equilibrium constant

Chapter 4_Thermodynamics.indd 87

DG = DG o + RT log Q A +BC + D [C][D] Q= [ A ][B]

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88

OBJECTIVE CHEMISTRY FOR NEET

  At equilibrium, DG = 0 and Q = K eq , Thus, we get 0 = ∆G ° + RT ln K eq ∆G ° = - RT ln K eq ∆G ° = -2.303RT log K eq   This is known as van’t Hoff ’s equation. 22. Second Law of Thermodynamics: It states that the entropy of the universe keeps on increasing or it is possible to construct a system which is able to convert heat by a cyclic process from a low temperature to a high temperature. T 23. Efficiency of the Carnot’s engine is given by h = 1 - 2 T1 24. Third law of thermodynamics states that at absolute zero, the entropy of a substance may become zero.

Solved Examples 1. A monoatomic gas X and a diatomic gas Y, both initially at the same temperature and pressure are compressed adiabatically from a volume V to V/2, which gas will be at higher temperature? (1) X (2) Y (3) Both are same (4) Can’t say

(1) 59.54 J mol−1 (2) 5954 J mol−1 (3) 594.5 J mol−1 (4) 320.6 J mol−1 Solution (2) We know

Solution (1) We have



T1V1g -1 = T2V2g -1

   

T2  V1  = T1  V2 

= 5954.13 J mol -1 = 2g -1

4. The molar heat capacity of water in equilibrium with ice at constant pressure is

Since g is more for gas X, therefore, the temperature will also be more for it.

(Specific heat capacity of water = 4.18 JK−1 g−1) (1) 292.0 K (2) 290.8 K (3) 298.0 K (4) 293.7 K

Solution (2) Work done in expansion = p × ∆V





We have 1 L atm = 101.3 J



   = 3 × (5 − 3) = 6 L atm

(1) zero (2) infinity (∞) (3) 40.45 kJK−1mol−1 (4) 75.48 JK−1mol−1 Solution (2) We know that heat capacity at constant pressure is given by  ∆H  Cp =   ∆T  p

Since the phase transformation, that is, ice  water, takes place at a fixed temperature so ∆T = 0.



Therefore,

Let ∆T be the change in temperature of water.



Therefore, p∆V = m × C × ∆T



     607.8 = 180 × 4.184 × ∆T ⇒ ∆T = 0.81 K



      Tf = T1 + ∆T = 290.8 K

3. One mole of ice is converted into water at 273 K. The entropies of H2O(s) and H2O(l) are 38.20 and 60.01 J mol−1 K−1 respectively. The enthalpy change for the conversion is

Cp =

∆H =∞ 0

5. For A → B, ∆H = 4 k cal mol -1, DS = 10 cal mol -1 K -1. Reaction is spontaneous when temperature will be (1) 400 K (2) 300 K (3) 500 K (4) 200 K

w = 6 × 1.013 J = 607.8 J



Chapter 4_Thermodynamics.indd 88

At equilibrium ∆G = 0, therefore, Eq. (1) becomes ∆H = T ∆S = 273 × (60.01 - 38.20)

g -1

2. A sample of oxygen gas expands its volume from 3 L to 5 L against a constant pressure of 3 atm. If the work done during expansion is used to heat 10 mol of water initially present at 290 K, its final temperature will be

∆G = ∆H - T∆S(1)

Solution (3) Gibbs free energy can be expressed as ∆G = ∆H - T ∆S (1)



  



When ∆G = 0, Eq. (1) becomes T=

∆H ∆S

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Thermodynamics

T=

4 = 0.4 × 103 = 400 K 10 × 10 -3







When T > 400 K, then, ∆G will be negative and reaction will be spontaneous.

1 9. H 2(g ) + O2(g ) → H 2O(l) 2

B.E. (H   H) = x1; B.E. (O   O) = x2; B.E. (O   H) = x3



Latent heat of vaporization of water liquid into water vapor = x4, then ∆Hf (heat of formation of liquid water) is

6. For a certain reaction the change in enthalpy and change in entropy are 40.63 kJ mol−1 and 100 JK−1 respectively. What is the value of DG at 27°C? (1) 10630 J (2) −10630 J (3) −10630 KJ (4) 10630 KJ Solution ∆G = ∆H - T∆S(1)

(1) We know

Substituting ∆H = 40.63 × 103 J mol−1 = 40630 J mol−1, ∆S = 100 JK−1 and T = 27 + 273 = 300 K in Eq. (1), we get DG = DH - T DS = 40630 - 300 ´ 100 = 10630 J



Positive value of ∆G indicates that the reaction is not possible.

7. The difference between the heats of reaction at constant pressure and a constant volume for the reaction 2C6 H6 ( l ) + 15 O2(g ) → 12CO2(g ) + 6H 2O(l) at 25°C in kJ is

x2 x - x 3 + x 4 (2) 2 x 3 - x1 - 2 - x 4 2 2 x2 x2 (3) x1 + - 2 x 3 - x 4 (4) x1 + - 2 x 3 + x 4 2 2 (1) x1 +

Solution

1 O (g) → H2O(g) 2 2 ∆H = –2 × bond formation energy of (O   H) + bond 1 breaking energy of H2 + bond breaking energy of O2 2 x2 ∆H = x1 + - 2 x 3 2 (3) The reaction involved is H2(g) +





H2O(g) → H2O(l);



Therefore,

H2(g) +





∆H 2 = x1 +

∆H1 = – x4

1 O (g) → H2O(l); ∆H2 2 2 x2 - 2x3 - x4 2

10. Heat of neutralization of oxalic acid is −53.35 kJ mol−1 using NaOH. Hence ∆H of H 2C 2O4  C 2O24- + 2H + is

(1) − 7.43 (2) 3.72 (3) − 3.72 (4) 7.43 Solution

89

(1) 5.88 kJ (2) −5.88 kJ (3) −13.7 kcal (4) 7.9 kJ Solution

DH = DU + Dng RT (1)

(1) We know

Here         ∆ng = np - nr = 12 - 15 = -3



From Eq. (1), we get ∆H - ∆E = ∆ng RT = -3 × 8.314 × 298 = -7.43 kJ

(4) By the definition of heat of neutralization, we have 1 1 H 2C 2O4 + NaOH → Na 2C 2O4 + H 2O; ∆H = -53.35 kJ 2 2 1 1 H 2C 2O4 + OH - → C 2O 24- + H 2O; ∆H = -53.35 kJ (1) 2 2



Subtracting Eq. (2) from Eq. (1) we get 1 1 H 2C 2O4 → C 2O24- + H+ ; ∆H = 3.95 kJ 2 2

(1) H + + OH - → H 2O

H 2C 2O4 → C 2O24- + 2H+ ; ∆H = 7.9 kJ

(2) H 2O + H + → H 3O+

11. Find out the calorific value of glucose

(3) 2H 2 + O2  2H 2O (4) CH 3COOH+ NaOH  CH 3COONa +H 2O



(1) Since heat of neutralization of strong acid and strong base is equal to the heat of formation of water, that is, NaOH + HCl → NaCl + H 2O + Q where Q is heat of neutralization. +

-

+

-

Na + OH + H + Cl → Na + Cl + H 2O + Q -

H + OH → H 2O + Q +

Chapter 4_Thermodynamics.indd 89

C6 H12O6 + 6O2 ® 6CO2 + 6H 2O; DH = - 2900 kJ mol -1 (1) 15.34 kJ g −1 (2) 13.92 kJ g −1 (3) 18.50 kJ g −1 (4) 16.11 kJ g −1

Solution

-

∆H = -57.3 kJ (2)

 

8. Heat of neutralization of a strong acid by a strong base is equal to ∆H of

+

H+ + OH - → H 2O;



Solution (4) Heat evolved from 1 mol glucose = 2900 kJ

or heat evolved from 180 g glucose = 2900 kJ



Therefore, heat evolved from 1 g glucose 2900 = = 16.11 kJg -1 180

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90

OBJECTIVE CHEMISTRY FOR NEET

12. For the reaction,

15. The heat of transition for carbon from the following is

C7H8(l) + 9O2(g) → 7CO2(g) + 4H2O(l), the calculated heat of reaction is 232 kJ mol−1 and observed heat of reaction is 50.4 kJ mol−1, then the resonance energy is



C(Diamond ) + O2(g ) → CO2(g ) ; ∆H = -94.3 k cal    C( Amorphous) + O2(g ) → CO2(g ) ; ∆H = -97.6 k cal (1) 3.3 kJ mol−1 (2) 3.3 kcal mol−1 (3) –3.3 kJ mol−1 (4) –3.3 kcal mol−1

(1) –182.2 kJ mol−1 (2) +182.2 kJ mol−1 (3) 172 kJ mol−1 (4) −172 kJ mol−1

Solution

Solution

(2) C(Diamond ) + O2(g ) ® CO2(g ); DH = -94.3 k cal (1)

(1) As we know that, Resonance energy = ∆H °(observed ) - ∆H °(calculated ) = (50.4 - 232.6 ) kJmol -1





C( Amorphous) + O2(g ) ® CO2(g ); DH = -97.6 k cal (2)



Subtracting Eq. (2) from Eq. (1), we get C(Diamond ) ® C( Amorphous); DH = +3.3 kcal mol -1

= -182.2 kJmol -1 13. Ionization energy of Al = 5137 kJ mol–1, (DH) hydration of Al3+ = – 4665 kJ mol–1. (DH)hydration for Cl− = – 381 kJ mol–1. Which of the following statement is correct? (1) AlCl3 would remain covalent in aqueous solution. (2) Only at infinite dilution AlCl3 undergoes ionization. (3) In aqueous solution AlCl3 becomes ionic. (4) None of these.

16. The ∆H fo for CO2(g), CO(g) and H2O(g) are −393.5, −110.5 and −241.8 kJ mol−1 respectively. The standard enthalpy change in kJ for the reaction CO2(g) + H2(g) → CO(g) + H2O(g) is (1) +524.1 (2) +41.2 (3) −262.5 (4) −41.2 Solution

Solution

(2)       C(s) + O2(g) → CO2(g ); ∆H = -393.5 kJ mol -1  (1)

(3) If AlCl3 is present in ionic state in aqueous solution, it will have Al3+ and 3Cl– ions.



AlCl 3(s)  Al 3+(aq ) + 3Cl - (aq )





Standard heat of hydration of Al3+ and 3Cl− ions is –4665 + 3 × (–381) kJ mol–1 = −5808 kJ mol−1



We know, ionization energy of Al = 5137 kJ mol–1



Since, hydration energy overcomes ionization energy, therefore, AlCl3 would be ionic in aqueous solution.

14. The standard heat of formation values of SF6(g), S(g) and F(g) are −1100, 275 and 80 kJ mol−1 respectively. Then the average S   F bond energy in SF6 is (1) 301 kJ mol−1 (2) 320 kJ mol−1 (3) 309 kJ mol−1 (4) 280 kJ mol−1 Solution (3) S(s) + 3F2(g ) → SF6 (g); ∆H f = -1100 kJ mol -1 S(s) → S(g); ∆H = +275 kJ mol -1 1 F2(g) → F(g); ∆H = 80 kJ mol -1 2

We know that,



Heat of formation = Bond energy of reactants − Bond energy of products -1100 = ( 275 + 6 ´ 80) - 6 ´ B.E (S - F )

     B.E (S - F ) = 309 kJ mol -1

Chapter 4_Thermodynamics.indd 90



1    C(s) + O 2(g) ® CO(g ); DH = -110.5 kJ mol -1 (2) 2 1 H 2(g) + O 2(g) → H 2O(g ); ∆H = -241.8 kJ mol -1 (3) 2 From Eq. (2) + Eq. (3) − Eq. (1), we get ∆H = -110.5 - 241.8 + 393.5 = 41.2 kJ mol -1

17. Given that 2H2(g) + O2(g) → H2O(g), ∆H = –115.4 kcal the bond energy of H   H and O   O bond respectively is 104 kcal and 119 kcal, then the O   H bond energy of water vapor is (1) 110.6 kcal mol−1 (2) –110.6 kcal mol−1 (3) 105 kcal mol−1 (4) −105 kcal mol−1 Solution (1) We know Enthalpy of reaction = ∑ B.E. (reactants) - ∑ B.E. ( products)

For the reaction 2H   H(g) + O O(g) ® 2H   O   H(g), ∆H = –115.4 kcal, B.E (H   H) = 104 kcal and B.E (O   O) = 119 kcal.



Since, one H2O molecule contains two O   H bonds, therefore, -115.4 = ( 2 × 104) + 119 - 4 × B.E(O - H) 4 × B.E (O - H) = ( 2 × 104) + 119 + 115.4 ( 2 × 104) + 119 + 115.4 = 110.6 kcal mol -1 B.E (O - H) = 4

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Thermodynamics 18. At a particular temperature

Solution 0.2 × 2 × 400 = 0.16 1000 600 × 0.1 Number of equivalent of KOH added = = 0.06 1000 Number of equivalents of acid and bases which neutralized each other is 0.06



H+(aq) + OH−(aq) → H2O(l); ∆H = −57.1 kJ

(1) Number of equivalent of H2SO4 taken =



The approximate heat evolved when 400 mL of 0.2 M H2SO4 is mixed with 600 mL of 0.1 M KOH solution will be



(1) 3.426 kJ (2) 13.7 kJ (3) 5.2 kJ (4) 55 kJ

91



Therefore, heat evolved = 0.06 × 57.1 kJ = 3.426 kJ

Practice Exercises Level I Heat, Work, Internal Energy, Enthalpy, Heat Capacity and First Law of Thermodynamics 1. For an isochoric process, the change in (1) volume is zero. (2) pressure is zero. (3) internal energy is zero. (4) temperature is zero. 2. In thermodynamics, a process is called reversible when (1) surroundings and system change into each other. (2) there is no boundary between system and surroundings. (3) the surroundings are always in equilibrium with the system. (4) the system changes into the surroundings spontaneously. 3. The work done during the process when 1 mol of gas is allowed to expand freely into vacuum is (1) zero. (2) positive. (3) negative. (4) either of these. 4. An intensive property is that property which depends upon (1) (2) (3) (4)

the nature of the substance. the amount of the substance. both the amount and nature of the substance. neither the nature nor the amount of the substance.

5. Evaporation of water is (1) a process in which neither heat is evolved nor absorbed. (2) a process accompanied by chemical reaction. (3) an exothermic change. (4) an endothermic change. 6. When a certain mass of an ideal gas is adiabatically compressed so that its volume is reduced to 1/32 times, its absolute temperature is increased 4 times. The number of atoms in a molecule (atomicity) of the gas is (1) 3 (2) 4 (3) 2 (4) 1

Chapter 4_Thermodynamics.indd 91

7. The heat content of the system is called (1) internal energy. (2) enthalpy. (3) free energy. (4) entropy. 8. For a fixed amount of perfect gas, which of these statements must be true? (1) (2) (3) (4)

U and H each depend only on T. Cp is constant. pdV = nRdT for every infinitesimal process. Both (1) and (3) are true.

9. Temperature of 1 mol of a gas is increased by 1°C at constant pressure. Work done is (1) R (2) 2R (3) R/2 (4) 3R 10. Work done in reversible isothermal process by an ideal gas is given by (1) 2.303 nRT log

V2 nR (2) (T2 - T1 ) (g - 1) V1

(3) 2.303 nRT log

V1 nR (4) (T2 - T1 ) (g + 1) V2

11. What is true for the reaction? PCl 5 (g ) → PCl 3 (g ) + Cl 2 (g ) (1) ∆H = ∆U (2) ∆H > ∆U (3) ∆H < ∆U (4) none 12. If 50 cal are added to a system and system does work of 30 cal on surroundings, the change in internal energy of system is (1) 20 cal (2) 50 cal (3) 40 cal (4) 30 cal 13. Work done in reversible adiabatic process by an ideal gas is given by (1) 2.303 nRT log

V1 nR (2) (T1 - T2 ) (g - 1) V2

(3) 2.303 nRT log

V2 nR (4) (T2 - T1 ) (g - 1) V1

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OBJECTIVE CHEMISTRY FOR NEET

14. When 1 mol gas is heated at constant volume, temperature is raised from 298 to 309 K. Heat supplied to the gas is 500 J. Then which statement is correct?

Compound H2(g) N2(g) NH3(g)

(1) q = w = 500 J, ∆U = 0 (2) q = ∆U = 500 J, w = 0 (3) q = w = 500 J, ∆U = 0 (4) ∆U = 0, q = w = −500 J

15. 1 mole of an ideal gas at 25°C is subjected to expand reversibly ten times of its initial volume. The change in entropy of gas is (1) 19.15 JK−1 mol–1 (2) 16.15 kJ−1 mol–1 (3) 22.15 JK−1 mol–1 (4) None

24. The standard Gibbs free energy of formation is non zero for

(1) C(graphite) (2) O2(g) (3) O3(g) (4) S (rhombic) 25. An endothermic reaction is spontaneous only if

16. Which of the following thermodynamic quantities is an outcome of the second law of thermodynamics?

(1) (2) (3) (4)

(1) Enthalpy (2) Internal energy (3) Work (4) Entropy 17. Entropy change of vaporization at constant pressure is given by (1) ∆S( vapor ) = (3) ∆S( vapor ) =

T ∆H vap ∆T

(2) ∆S( vapor ) =

∆U vap T



(1) The reaction is spontaneous at this temperature (298 K). (2) If temperature is decreases forward reaction is favored. (3) The reaction is spontaneous in forward direction only at temperature above 1000 K. (4) The reaction is spontaneous in forward direction only at temperature above 1114.375 K.

(1) ∆H < 0, ∆S(surrounding) > 0 (2) ∆H < 0, ∆S(system) < 0

Gibbs Free Energy and Spontaneity of a Reaction

27. Which of the following is true of this reaction? 2 N 2O5(g )  4 NO2(g ) + O2(g ) ∆H o = 110 kJ



19. For spontaneous reaction the value of change of Gibbs free energy, ∆G is

(1) Both ∆Ho and ∆So favor the reaction’s spontaneity (2) Both ∆Ho and ∆So oppose the reaction’s spontaneity. (3) ∆Ho favors the reaction, but ∆So opposes it. (4) ∆Ho opposes the reaction, but ∆So favors it.

(1) negative. (2) positive. (3) greater than one. (4) one. 20. For a reaction at equilibrium (1) ∆G = ∆G o ≠ 0 (2) ∆G o = 0 (3) ∆G = ∆G o = 0 (4) ∆G = 0, ∆G o ≠ 0

Thermochemistry, Hess’ Law and Kirchoff’s Equation 28. If H2(g) → 2H(g); ∆H = 104 kcal, then heat of atomization of hydrogen is

21. In which case, a reaction is possible at any temperature? (1) ∆H < 0, ∆S > 0 (2) ∆H < 0, ∆S < 0 (3) ∆H > 0, ∆S > 0 (4) ∆H > 0, ∆S < 0

∆H° = 178.3 kJ and ∆S° = 160 JK–1

Select correct statement:

(4) None

(3) Both (1) and (2) (4) None of these

the entropy of the surrounding increases. entropy of the system increases. total entropy decreases. None of these.

26. For the reaction at 298 K, CaCO3(s) → CaO(s) + CO2(g)

18. Select the correct option for the exothermic process.

(1) 52 kcal (2) 104 kcal (3) 208 kcal (4) 84 kcal 29. For the combustion reaction at 298 K

22. Thermodynamic equilibrium involves (1) chemical equilibrium. (2) thermal equation. (3) mechanical equation. (4) all the three. 23. Use the following data to calculate ∆So for the reaction:



Chapter 4_Thermodynamics.indd 92

2NH3(g) →  N2(g) + 3H2(g)

−98.74 −114.99 −304.99

(1) −198.8 JK−1 (2) −91.26 JK−1 (3) −106.22 JK−1 (4) 198.8 JK−1

Entropy and Second Law of Thermodynamics

∆H vap

So (J mol−1 K−1)

1 H 2 (g ) + O 2 (g ) → H 2O ( l ) 2







Which of the following alternative(s) is/are correct? (1) ∆H = ∆U (2) ∆H > ∆U

1/4/2018 5:10:51 PM

Thermodynamics (3) ∆H < ∆U (4) ∆H and ∆U has no relation with each other.

(1) (2) (3) (4)

30. ∆H of sublimation of a solid is equal to (1) ∆Hfusion + ∆Hcondensation (3) ∆Hcondensation –∆Hfusion (4) ∆Hvaporization + ∆Hfusion 31. For the two reactions given below

(1) 464 kJ mol–1 (2) 488.5 kJ mol–1 (3) 232 kJ mol–1 (4) –232 kJ mol–1

1 H 2(g ) + O 2(g ) → H 2O(g ) + X 1 kJ 2 1 H 2(g ) + O2(g ) → H 2O( l ) + X 2 kJ 2

39. If

S + O 2 ® SO 2 ; DH = -298.2 kJ mol -1

1    SO2 + O2 ® SO3 ; DH = -98.7 kJ mol -1 2       SO3 + H 2O ® H 2SO4 ; DH = -130.2 kJ mol -1

Select the correct answer. (1) X1 > X2 (2) X1 < X2 (3) X1 = X2 (4) X1 + X2 = 0

32. Given N2(g) + 3H2(g) → 2NH3(g); ∆Ho = –22 kcal. The standard enthalpy of formation of NH3 gas is (1) –11 kcal mol−1 (2) 11 kcal mol−1 (3) –22 kcal mol−1 (4) 22 kcal mol−1 33. The temperature of a 5 mL of strong acid increases by 5°C when 5 mL of a strong base is added to it. If 10 mL of each are mixed, temperature should increase by (1) 5°C (2) 10°C (3) 15°C (4) cannot be known. 34. The bond enthalpies of H   H, Cl   Cl and H   Cl are 435, 243 and 431 kJ mol−1, respectively. The enthalpy of formation of HCl(g) will be (1) 92 kJ mol (2) −92 kJ mol (3) 247 kJ mol−1 (4) 770 kJ mol−1 −1

The reaction of 0.624 mol of Al. The formation of 0.624 mol of Al2O3. The reaction of 0.312 mol of Al. The formation of 0.150 mol of Al2O3.

38. The enthalpy of combustion of H2(g) at 298 K to give H2O is –298 kJ mol–1 and bond enthalpies of H   H and O   O are 433 kJ mol–1 and 492 kJ mol–1 respectively. The bond enthalpies of O   H is (Given: DH vap of water = 42 kJ mol -1)

(2) ∆Hcondensation + ∆Hsublimation



93

−1

1           H 2 + O2 ® H 2O; DH = -287.3 kJ mol -1 2

The enthalpy of formation of H2SO4 at 298 K will be (1) – 814.4 kJ mol–1 (2) +814.4 kJ mol–1 (3) –650.3 kJ mol–1 (4) –433.7 kJ mol–1

40. For which of the following equations, will ∆H be equal to ∆U? 1 (1) H 2 (g ) + O2 (g ) → H 2O( l ) 2 (2) H2(g) + I2(g) → 2HI(g) (3) 2NO2(g) → N2O4(g) (4) 4NO2(g) + O2(g) → 2N2O5(g) 41. Heat of combustion of CH4, C2H4, C2H6 are –890, –1411 and –1560 kJ mol−1 respectively. Which of the given hydrocarbons has the lowest calorific fuel value in kJ g−1?

(1) CH (2) C2H4 35. For the combustion of n-octane C18H18(g ) + O2(g ) → CO2(g ) + H 2O(l)4 (3) C2H6 (4) All same C18H18(g ) + O2(g ) → CO2(g ) + H 2O(l) at 25°C (ignoring resonance in CO2), which of the following statements is correct? 42. The enthalpy of combustion of carbon and carbon monoxide are –390 kJ and –278 kJ respectively. The enthalpy (1) ∆H = ∆U - 5.5 × 8.31 × 0.298 in kJ mol -1 of formation of CO in kJ is -1 (2) ∆H = ∆U + 4.5 × 8.31 × 0.298 in kJ mol (1) 668 (2) 112 (3) ∆H = ∆U - 4.5 × 8.31 × 0.298 in kJ mol -1 (3) –112 (4) –668 (4) ∆H = ∆U - 4.5 + 8.31 × 0.298 in kJ mol -1 43. ∆H for CaCO3(s) → CaO(s) + CO2(g) is 176 kJ mol–1 at 36. Enthalpies of formation of CO2(g), CO(g), N2O(g) and 1240 K. The ∆U for the reaction is equal to NO2(g) in kJ mol−1 are respectively −393, −110, 81 and 34 (1) 160 kJ (2) 165.6 kJ respectively. Calculate ∆rH in kJ of the following reaction. (3) 186.3 kJ (4) 180.0 kJ 2NO2(g ) + 2CO(g ) ® N 2O(g ) + 3CO2(g ) 44. In which case of mixing of a strong acid and a strong base each of 1 N concentration, temperature increase is highest? (1) 836 (2) 1460 (3) −836 (4) −1460 (1) 20 mL acid –30 mL alkali (2) 10 mL acid –40 mL alkali 37. The standard heat of combustion of Al is –837.8 kJ (3) 25 mL acid –25 mL alkali –1 mol at 25°C. Which of the following releases 250 kcal (4) 35 mL acid –15 mL alkali of heat?

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45. In the reaction CO2(g) + H2(g) → CO(g) + H2O(g); ∆H = 2.8 kJ; ∆H represents (1) heat of reaction. (2) heat of combustion. (3) heat of formation. (4) heat of solution. 46. The heat of formation H2O(l) is –68.0 kcal, the heat of formation of H2O(g) is likely to be (1) –68.0 kcal (2) –69.4 kcal (3) 80.0 kcal (4) –58.3 kcal 47. Which of the following corresponds to the definition of enthalpy of formation at 298 K? 1 (1) C (graphite) + 2H 2(g ) + O 2( l ) → CH 3OH(g ) 2 1 (2) C (diamond ) + 2H 2(g ) + O 2(g ) → CH 3OH( l ) 2 (3) 2C (graphite ) + 4H 2(g ) + O2(g ) → 2CH 3OH(g )

 ∂H  (3) For an ideal gas   =0  ∂p  T (4) All of these. 6. During the adiabatic expansion of an ideal gas against atmospheric pressure, the internal energy will (1) increase. (2) decrease. (3) stay the same. (4) impossible to say. 7. Two moles of helium gas undergo a reversible cyclic process as shown in below figure. Assuming gas to be ideal, what is the net work involved in the cyclic process? p

2 atm

1 (4) C (graphite) + 2H 2(g ) + O 2(g ) → CH 3OH( l ) 2

1 atm

1. Which of the following is true regarding reversible adiabatic expansion of an ideal gas? (1) Plot of T vs V is a straight line with slope equal to g. (2) Plot of ln T vs ln V is a straight line with slop equal to g. (3) Plot of ln T vs ln V is a straight line with slope equal to −g. (4) Plot of ln T vs ln V is a straight line with slope (1 − g ). 2. Calculate the work done when 1 mol of an ideal gas is compressed reversibly from 1 bar to 5 bar at a constant temperature of 300 K. (1) −14.01 kJ (2) 16.02 kJ (3) −4.01 kJ (4) −8.02 kJ 3. Out of boiling point (I), entropy (II), pH (III) and emf of a cell (IV) intensive properties are (1) I, II (2) I, II, III (3) I, III, IV (4) All of these 4. A gas expands against a constant external pressure of 2 atm, increasing its volume by 3.40 L; simultaneously the system absorbs 400 J of heat from its surroundings. What is DU, in joules, for this gas? (1) –689 (2) –289 (3) +400 (4) +289 5. Which of the following statements is wrong? (1) Change in internal energy of an ideal gas on isothermal expansion is zero. (2) In a cyclic process w ¹ q

Chapter 4_Thermodynamics.indd 94

A 300 K

Level II Heat, Work, Internal Energy, Enthalpy, Heat Capacity and First Law of Thermodynamics

B

C D 400 K

T

(1) –100 R ln4 (2) +100 R ln4 (3) +200 R ln4 (4) –200 R ln4 8. An ideal gas expands from volume V1 to V2. This may be achieved by any of the three processes: isobaric, isothermal and adiabatic. Let ∆U be the change in internal energy of the gas, q be the quantity of heat added to the system and w be the work done by the system on the gas. Identify which of the following statements is not true for ∆U ? (1) (2) (3) (4)

∆U is the least in the adiabatic expansion. ∆U is the greatest in the adiabatic expansion. ∆U is greatest under isobaric process. ∆U in isothermal process lies in-between the value obtained under isobaric and adiabatic process.

9. One mole of ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27°C. If the work done by the gas in the process is 3 kJ, the final temperature will be equal to (CV = 20 JK−1 mol) (1) 100 K (2) 450 K (3) 150 K (4) 400 K 10. The increase in internal energy of 1 kg of water at 100°C when it is converted into steam at the same temperature and at 1 atm (100 k Pa) will be (The density of water and steam are 100 kg m−3 and 0.6 kg m−3 respectively. The latent heat of vaporization of water is 2.25 × 106 J kg−1.) (1) 2.08 × 106 J (2) 4 × 107 J (3) 3.27 × 108 J (4) 5 × 109 J 11. The factor that does not influence the heat of reaction is (1) the physical state of reactants and products. (2) the temperature.

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Thermodynamics (3) the pressure or volume. (4) the method by which the final products are obtained. 12. One mole of ideal gas goes through the following transformations:  Step (I):  Isochoric cooling to 1/3rd of the initial temperature. Step (II):  Adiabatic compression to its initial pressure.  Step (III):  Isobaric expansion back to the initial state.

If the initial temperature and pressure are 600 K and 3.00 atm and if the ideal gas is monoatomic (CV = 3/2R), what is the work for the third step (going from state 3 to state 1)?



Given: (16.4)

5/3

=105.86 and ( 35.3)

3/5

= 8.48

(1) −23.8 L atm (2) −9.11 L atm (3) 92.7 L atm (4) 9.11 L atm 5   13. Two moles of an ideal gas  CV = R  was compressed  2  adiabatically against constant pressure of 2 atm which was initially at 350 K and 1 atm pressure. The work involved in the process is equal to (1) 250 R (2) 300 R (3) 400 R (4) 500 R 14. The internal energy of a monoatomic ideal gas is 1.5 nRT. 1 mol of helium is kept in a cylinder of cross section 8.5 cm2. The cylinder is closed by a light frictionless piston. The gas is heated slowly in a process during which a total of 42 J heat is given to the gas. If the temperature rises through 2°C, the distance moved by the piston is (Atmospheric pressure being 100 k Pa.) (1) 10 cm (2) 20 cm (3) 25 cm (4) 30 cm 15. (∆H − ∆U) for the formation of NH3 from N2 and H2 is (1) RT (2) 2RT (3) −RT (4) −2RT 16. A system undergoes a two-step process. In step 1, 15 J of work is done on the system as its internal energy increases by a total of 30 J. In step 2, at constant volume, the system decreases its internal energy by 20 J. Which statement is true about these changes? (1) The system loses 5 J as a result of the net heat flow in steps 1 and 2. (2) 20 J of work is done on the surroundings in step 2. (3) q = + 45 J in step 1 (4) q = + 20 J in step 2

Entropy and Second Law of Thermodynamics 17. The statement “If an object A is in thermal equilibrium with an object B, and B is in thermal equilibrium with an object C, then C is also in thermal equilibrium with A” is an example of which of the following laws?

Chapter 4_Thermodynamics.indd 95

(1) (2) (3) (4)

95

The zeroth law of thermodynamics The first law of thermodynamics The second law of thermodynamics None of these

18. The entropy change when two moles of ideal monoatomic gas is heated from 200°C to 300°C reversibly and isochorically is (1)

3 5  300  æ 300 ö (2) R ln ç R ln   200  è 200 ÷ø 2 2

3 æ 573 ö æ 573 ö (3) 3R ln ç (4) R ln ç è 473 ÷ø è 473 ÷ø 2 19. The area under a plot of Cp/T versus the absolute temperature (T) is directly related to (1) ∆S (2) ∆U (3) ∆H (4) q 20. The entropy change during an isothermal expansion of an ideal gas from V1 to V2 at temperature T is given by V2 V1 V2 V2 (3) ∆S = 2.303RT log (4) ∆S = 2.303R log V1 V1 (1) ∆S = 0 (2) ∆S = 2.303nR log

21. A certain process releases 64.0 kJ of heat, which is transferred to the surroundings at a constant pressure and a constant temperature of 300 K. For this process ∆Ssurr is (1) 64.0 kJ (2) −64.0 kJ (3) −213 JK−1 (4) 213 JK−1 22. Which of the following statement is true? (1) For every process, in an isolated system ∆T = 0. (2) For every process in an isolated system, ∆S = 0. (3) For every process in an isolated system that has no macroscopic kinetic or potential energy, ∆U = 0. (4) If a closed system undergoes a reversible change for which ∆V = 0, then the pV work = 0. 23. A heat engine operating between 227°C and 27°C absorbs 2 kcal of heat from the 227°C reservoir reversibly per cycle. The amount of work done in one cycle is (1) 0.4 kcal (2) 0.8 kcal (3) 4 kcal (4) 8 kcal 24. Select the correct alternate for the endothermic change. (1) ∆H > 0, ∆S (system ) < 0 (2) ∆H > 0, ∆S (system ) > 0 (3) ∆H < 0, ∆S (surrounding ) < 0 (4) ∆H > 0, ∆S (surrounding ) > 0 25. Arrange the following reactions in order of decreasing ∆So values. (I) 2NO(g) → N 2(g)+O2(g) (II) 2CO(g)+O2(g )→ 2CO2(g) (III) MgCO3 (s)→ MgO(s) + CO2(g )

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OBJECTIVE CHEMISTRY FOR NEET (1) (2) (3) (4)

∆So(I) > ∆So(II) > ∆So(III) ∆So(I) > ∆So(III) > ∆So(II) ∆So(II) > ∆So(III) > ∆So(I) ∆So(III) > ∆So(I) > ∆So(II)

26. When one mole of an ideal gas is compressed to half its initial volume and simultaneously heated to twice its initial temperature, the change in entropy (∆S) is

(1) ∆H > 0, ∆S > 0 (2) ∆H > 0, ∆S < 0 (3) ∆H < 0, ∆S > 0 (4) ∆H < 0, ∆S < 0 33. For the reaction at 300 K



∆U = –3.0 kcal and ∆S = – 10.0 cal K−1. The value of ∆G will be

(1) CV ln 2 (2) Cp ln 2 (3) R ln 2 (4) (CV – R) ln 2 27. The maximum efficiency of a heat engine operating between 100°C and 25°C is (1) 20% (2) 22.2% (3) 25% (4) 24.5%

(1) –600 cal (2) –6600 cal (3) –6000 cal (4) None 34. The entropy change for the reaction given below is



Then ∆G o , ∆H o and ∆S o for the reaction at 298 K will be (1) (2) (3) (4)

∆G° = 6.25 kJ, ∆H° = 18.29 kJ, ∆S° = 45 JK–1 mol–1 ∆G° = 7.323 kJ, ∆H°= 19.37 kJ, ∆S°= 48 JK–1 mol–1 ∆G° = 8.729 kJ, ∆H° = 20.22 kJ, ∆S° = 51.23 JK–1 mol–1 ∆G° = 5.185 kJ, ∆H° = 17.11 kJ, ∆S° = 40 JK–1 mol–1

29. If So for H2, Cl2 and HCl are 0.13, 0.22 and 0.19 kJK–1 mol–1 respectively. The total change in standard entropy for the reaction H2 + Cl2 → 2HCl is (1) 30 JK–1 mol–1 (2) 40 JK–1 mol–1 (3) 60 JK–1 mol–1 (4) 20 JK–1 mol–1 30. For this reaction, F2C = CF - CF = CF2 → Product, ∆H = –49.0 kJ mol−1 and ∆S = –40.2 JK−1. Upto what temperature is the forward reaction spontaneous?

2H2(g) + O2(g) → 2H2O(l)







is _________ at 300 K. Standard entropies of H2(g), O2(g) and H2O(l) are 126.6, 201.20 and 68.0 JK–1 mol–1 respectively.

Gibbs Free Energy and Spontaneity of a Reaction 28. The equilibrium constant for the reaction, A → B was measured over a range of temperature and can be represented by the equation, ln Kp = 4.814 – (2059/T).

A(g) + B(g) → C(g)



(1) –318.4 JK–1 mol–1 (2) 318.4 JK–1 mol–1 (3) 31.84 JK–1 mol–1 (4) None 35. ∆G° for the reaction 2NO(g) + O2(g)→N2O4(g) at 298K and 1 atm pressure, will be [Given that the enthalpy of formation of NO(g) is 90.5 kJ mol–1. The enthalpy of formation of N2O4(g) is 9.7 kJ mol–1. The standard entropy of NO (g) is 210 JK–1 mol–1, O2 (g) is 205 JK–1 mol–1 and N2O4 (g) is 304 JK–1 mol–1.] (1) – 52.92 kJ (2) – 75.94 kJ (3) – 64.24 kJ (4) – 83.27 kJ 36. Consider the reaction below at 298 K



C(graphite) + 2H2(g) → CH4(g); DH fo = -74.9 kJ mol -1

So +5.6 (JK−1 mol−1)

+130.7

+186.3

Which of the following statements is correct?

o = -50.8 kJ and (1) ∆G reaction enthalpy only. o = -50.8 kJ and (2) ∆G reaction 1 31. The temperature at which the reaction Ag 2O(s) → 2 Ag(s)+ O2(g ) entropy only. 2 o 1 = +50.8 kJ and (3) ∆G reaction Ag 2O(s) → 2 Ag(s)+ O2(g ) is at equilibrium is 2 entropy only. o = -50.8 kJ and (4) ∆G reaction (Given: ∆H = 30.5 kJ mol–1 and ∆S = 0.066 kJK–1 mol–1) enthalpy and entropy. (1) 462.12 K (2) 362.12 K

(1) 1492°C (2) 1219°C (3) 945°C (4) 1080°C

(3) 262.12 K (4) 562.12 K 32. What can be concluded about the values of ∆H and ∆S from the below graph?

DG kJ mol-1

+50 0 -50

the reaction is driven by the reaction is driven by





2H2(g) + O2(g) → 2H2O(g)

∆Ho = −ve and ∆So = −ve ∆Ho = −ve and ∆So = +ve ∆Ho = +ve and ∆So = −ve ∆Ho = −ve and ∆So = 0

38. At 25°C, which of the following substances has the highest molar entropy?

-100 100 200 300 400 500 T (K)

Chapter 4_Thermodynamics.indd 96

the reaction is driven by

37. Use your understanding of thermodynamics to predict the signs of ∆Ho and ∆So for the following reaction. (1) (2) (3) (4)

+100

the reaction is driven by

(1) Al(s) (2) C6H6(l) (3) C2H6(g) (4) CH4(g)

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97

Thermodynamics 39. At what temperature will the following process would not be spontaneous?



(1) (2) (3) (4)

A(g) → A(l); ∆H = −20 kJ and ∆S = −50 JK−1

(1) >200 K (2) >300 K (3) >350 K (4) >400 K 40. Which of the following statements is false regarding the reaction A(g)  B(g) ? (I) The reaction can achieve the lowest possible free energy by going to completion. (II) At equilibrium ∆G = 0 (III) At equilibrium, GA = GB (1) (I) and (II) only (2) (III) only (3) (I) only (4) (II) and (III) only

47. An athlete is given 100 g of glucose (C6H12O6) of energy equivalent to 1560 kJ. He utilizes 50% of this gained energy in the event. In order to avoid storage of energy in the body, the weight of water he would need to perspire is (The enthalpy of evaporation of water is 44 kJ mol−1.) (1) 319 g (2) 422 g (3) 293 g (4) 378 g 48. Which of the following is the heat of combustion? 1 (1) C (graphite) + O2(g ) → CO(g ) + x cal 2 (2) C (diamond ) + O2(g ) → CO2 (g ) + y cal

Thermochemistry, Hess’ Law and Kirchoff’s Equation

1 (3) C (graphite) + O2(g ) → CO(g ) + z cal 2 (4) None

41. The enthalpy of neutralization of the given reaction is



H2SO4 + 2NaOH → Na2SO4 + 2H2O + y kcal

(1) y kcal (2) –y kcal (3) +y/2 kcal (4) –y/2 kcal 42. When solute remains in equilibrium with given solvent then

49. The heats of neutralization of four acids a, b c and d when neutralized against a common base are 13.7, 9.4, 11.2 and 12.4 kcal respectively. The weakest among these acids is

(1) ∆Hhydration = Lattice energy

(1) c (2) b (3) a (4) d

(2) ∆Hhydration< Lattice energy (3) ∆Hhydration> Lattice energy (4) None 43. Given H2(g) + Br2(g) → 2HBr(g), ∆H 1o and standard enthalpy of condensation of bromine is ∆H 2o , standard enthalpy of formation of HBr at 25°C is (1)

DH 1o DH 1o (2) + DH 2o 2 2

(3)

DH 1o DH 1o - DH 2o - DH 2o (4) 2 2

44. The heats of combustion of yellow phosphorus and red phosphorous are –9.19 kJ and –8.78 kJ respectively, then heat of transition of yellow phosphorus to red phosphorous is (1) –18.69 kJ (2) +1.13 kJ (3) +18.69 kJ (4) –0.41 kJ 45. Heat of reaction at constant p or constant V varies with temperature as given by Kirchoff’s equation is/are (1) (2) (3) (4)

∆H2 = ∆H1 + ∆Cp (T2 – T1) ∆U2 = ∆U1 + ∆Cp (T2 – T1) H2 = H1 + ∆Cp (T2 – T1) ∆U2 = ∆U1 + Cp (T2 – T1)

boiling points. melting points. enthalpies of neutralization. enthalpies of combustion.

50. Trans-2-butene is more stable as compared to cisbutene. When a given sample of 2-butene is heated at constant volume, the concentration of trans-2-butene (1) increases. (2) decreases. (3) remains same. (4) changes unpredictably. 1 51. If H 2 + O2 ® H 2O ; DH = -68.39 kcal 2 1   K(s) + H 2O(l)® KOH(aq) + H 2 ; DH = -48 kcal 2   KOH + water → KOH(aq); ∆H = - 14 kcal

The heat of formation of KOH is (1) –68.39 + 48 – 14 (2) –68.39 – 48 + 14 (3) 68.39 – 48 + 14 (4) +68.39 + 48 – 14

52. Which of the following plot represents an exothermic reaction? (1)

(2)

R

H P

46. The enthalpies of formation of organic compounds are conveniently determined from their

R

t

t

(3)

H

Chapter 4_Thermodynamics.indd 97

P

H

(4)

R

P

H R

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(1)

(2)

R

H

P

H P R OBJECTIVE CHEMISTRY FOR NEET

98

t

t

(3)

59. The bond energies of C   C, C   H, H   H and C   C are 198, 98, 103, 145 kcal respectively. The enthalpy change of the reaction CH   CH + H2 → CH2   CH2 is

(4)

H

P

R

H R

t

t

53. In the reaction CS2(l) + 3O2(g) → CO2(g) + 2SO2(g); ∆H = –265 kcal.

(1) –152 kcal (2) 96 kcal (3) 48 kcal (4) –40 kcal

P

The enthalpies of formation of CO2 and SO2 are both negative and are in the ratio 4:3. The enthalpy of formation of CS2 is +26 kcal mol−1. Calculate the enthalpy of formation of SO2. (1) –90 kcal mol−1 (2) –52 kcal mol−1 (3) –78 kcal mol−1 (4) –71.7 kcal mol−1

60. Under the same conditions how many mL of 1 M KOH and 0.5 M H2SO4 solutions, respectively when mixed for a total volume of 100 mL produce the highest rise in temperature. (1) 67:33 (2) 33:67 (3) 40:60 (4) 50:50 61. X is a metal that forms an oxide X2O



When a sample of metal X reacts with one mole of oxygen, what will be the DH in that case? (1) 480 kcal (2) –240 kcal (3) –480 kcal (4) 240 kcal

54. Heat of hydrogenation of ethene is x1 and that of benzene is x2. Hence resonance energy is (1) x1 – x2 (2) x1 + x2 (3) 3x1 − x2 (4) x1 − 3x2 55. AB, A2 and B2 are diatomic molecules. If the bond enthalpies of A2, AB and B2 are in the ratio 1:1:0.5 and enthalpy of formation of AB from A2 and B2 –100 kJ mol–1. What is the bond enthalpy of A2?

Previous Years’ NEET Questions 1. Consider the following reactions: (I) H+(aq ) + OH - (aq ) = H 2O( l ), DH = - X 1kJ mol -1 (II) H 2 (g ) +

(1) 13.7 kcal (2) 27.4 kcal (3) 6.85 kcal (4) 3.425 kcal 57. Given the following reactions: N2(g) + 2O2(g) → 2NO2(g), ∆H1 = 16.18 kcal N2(g) + 2O2(g) → N2O4(g), ∆H2 = 2.31 kcal

Based on the above facts (1) NO2 is more stable than N2O4 at low temperature. (2) N2O4 is more stable than NO2 at low temperature. (3) Both are equally stable at low temperature. (4) None of the above.

58. The enthalpy of neutralization of a strong acid is –13,700 cal. A certain monobasic weak acid is 14% ionized in a molar solution. If the enthalpy of ionization of the weak acid is +366 cal mol−1, what is the enthalpy of neutralization of one molar solution of the weak acid? (1) –13,649 cal (2) –13,385 cal (3) –14,066 cal (4) –13,334 cal

Chapter 4_Thermodynamics.indd 98

5 (IV) C 2H 2(g )+ O2(g ) = 2CO2(g ) + H 2O(l), DH = + X 4 kJ mol -1 2

-

56. If H + OH → H 2O + 13.7 kcal then the enthalpy change for complete neutralization of 1 mol of H2SO4 by base will be

1 O2(g ) = H 2O(l), DH = - X 2 kJ mol -1 2

(III) CO2(g ) + H 2 (g ) = CO(g ) + H 2O( l ), DH = - X 3 kJ mol -1

(1) 400 kJ mol–1 (2) 200 kJ mol–1 –1 (3) 100 kJ mol (4) 300 kJ mol–1 +

1 1 X 2O ® X + O2 ; DH = 120 kcal 2 4





The enthalpy of the formation of H2O(l) is (1) +X1 kJ mol−1 (3) +X3 kJ mol−1

(2) −X2 kJ mol−1 (4) −X4 kJ mol−1 (AIPMT 2007)

2. Given that bond energies of H   H and Cl   Cl are 430 kJ mol−1 and 240 kJ mol−1, respectively, and D r H o for HCl is −90 kJ mol−1. The bond enthalpy of HCl is (1) 245 kJ mol−1 (3) 380 kJ mol−1

(2) 290 kJ mol−1 (4) 425 kJ mol−1 (AIPMT 2007)

3. Bond dissociation enthalpies of H2, Cl2 and HCl are 434, 242 and 431 kJ mol−1, respectively. Enthalpy of formation of HCl is (1) 245 kJ mol−1 (2) 93 kJ mol−1 (3) −245 kJ mol−1 (4) –93 kJ mol−1 (AIPMT 2008) 4. For the gas-phase reaction, PCl 5(g )  PCl3 (g ) + Cl 2(g )

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Thermodynamics

Which of the following conditions is correct? (1) DH > 0 and DS < 0 (2) DH = 0 and DS < 0 (3) DH > 0 and DS > 0 (4) DH < 0 and DS < 0

10. For vaporization of water at 1 atm pressure, the values of ∆H and ∆S are 40.63 kJ mol−1 and 108.8 JK−1 mol−1, respectively. The temperature when Gibbs energy change (∆G) for this transformation will be zero, is

(AIPMT 2008)

(1) 393.4 K (2) 373.4 K (3) 293.4 K (4) 273.4 K

5. Which of the following are not state functions?  (I) q + w  (II)  q (III) w (IV)  H–TS (1) (II) and (III) (2) (I) and (IV) (3) (II), (III) and (IV) (4) (I), (II) and (III)

(AIPMT MAINS 2010) 11. Three moles of an ideal gas expanded spontaneously into vacuum. The work done will be (1) 3 J (2) 9 J (3) zero. (4) infinite.

(AIPMT 2008) 6. Standard free energies of formation (in kJ mol−1) at 298 K are −237.2, −394.4 and −8.2 for H2O(l), CO2(g) and peno tane (g), respectively. The value of E cell for the pentane oxygen fuel cell is (1) 0.0968 V (2) 1.968 V (3) 2.0968 V (4) 1.0968 V (AIPMT 2008) 7. From the following bond energies

H   H bond energy: 431.37 kJ mol−1



C   C bond energy: 606.10 kJ mol−1



C   C bond energy: 336.49 kJ mol−1



C   H bond energy: 410.50 kJ mol−1



Enthalpy for the reaction



H

H

C

C+H

H

H

H

H

(1) 553.0 kJ mol (3) −243.6 kJ mol−1

12. The following two reactions are known

Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g); ∆H = −26.8 kJ



FeO(s) + CO(g) → Fe(s) + CO2(g); ∆H = −16.5 kJ



Correct target equation is



Fe2O3(s) + CO(g) → 2FeO(s) + CO2(g), ∆H = ?

(AIPMT MAINS 2010) 13. Match List-I (Equations) with List-II (Type of processes) and select the correct option.

H

H

C

C

H

H

List-I Equations H

(2) 1523.6 kJ mol (4) −120.0 kJ mol−1

8. The values of DH and DS for the reaction, C(Graphite) + CO2(g ) → 2CO(g ) are 170 kJ and 170 JK−1, respectively. This reaction will be spontaneous at (1) 510 K (2) 710 K (3) 910 K (4) 1110 K (AIPMT 2009) 9. Standard entropies of X2, Y2, and XY3 are 60, 40 and 50 JK–1 mol–1 respectively. For the reaction 1 3 X 2 + Y2  XY3 , ∆H = - 30 kJ to be at equilibrium, the 2 2 temperature should be (1) 1000 K (2) 1250 K (3) 500 K (4) 750 K (AIPMT PRE 2010)

List-II Type of processes

(a)  Kp > Q

 (i)  Non spontaneous

(b)  ∆Go < RT lnQ

(ii) Equilibrium

(c)  Kp = Q

(iii)  Spontaneous and endothermic

−1

(AIPMT 2009)

Chapter 4_Thermodynamics.indd 99

(AIPMT MAINS 2010)

(1) −43.3 kJ (2) −10.3 kJ (3) +6.2 kJ (4) +10.3 kJ

will be −1

99

(d)  T >

(iv)  Spontaneous

∆H ∆S

Options: (a)

(b)

(c)

(d)

(1)

(iii)

(iv)

(ii)

(i)

(2)

(iv)

(i)

(ii)

(iii)

(3)

(ii)

(i)

(iv)

(iii)

(4)

(i)

(ii)

(iii)

(iv) (AIPMT MAINS 2010)

14. Which of the following is correct option for free expansion of an ideal gas under adiabatic condition? (1) q = 0, DT < 0, w ≠ 0 (2) q = 0, DT ≠ 0, w = 0 (3) q ≠ 0, DT = 0, w = 0 (4) q = 0, DT = 0, w = 0 (AIPMT PRE 2011)

1/4/2018 5:11:03 PM

100

OBJECTIVE CHEMISTRY FOR NEET

15. If the enthalpy change for the transition of liquid water to steam is 30 kJ mol -1 at 27°C, the entropy change for the process would be (1) 100 J mol -1K -1 (2) 10 J mol -1K -1 (3) 1.0 J mol -1K -1 (4) 0.1 J mol -1K -1

21. The Gibb’s energy for the decomposition of Al2O3 at 500°C is as follows:



The potential difference needed for the electrolytic reduction of aluminium oxide (Al2O3) at 500°C is at least

(AIPMT PRE 2011)

(1) 4.5 V (2) 3.0 V (3) 2.5 V (4) 5.0 V

16. Enthalpy change for the reaction, 4H(g) → 2H2(g) is –869.6 kJ. The dissociation energy of H   H bond is (1) +217.4 kJ (2) –434.8 kJ (3) –869.6 kJ (4) +434.8 kJ (AIPMT PRE 2011) 17. Consider the following processes:



∆H(kJ mol−1) 1 A → B   +150 2 3B → 2C + D   −125







For B + D → E + 2C, ∆H will be



2 4 Al 2O3 → Al + O2 ; ∆ r G = 960 kJ mol -1 3 3



(AIPMT MAINS 2012) 22. Equal volumes of two monoatomic gases, A and B, at same temperature and pressure are mixed. The ratio of specific heats Cp/CV of the mixture will be (1) 0.83 (2) 1.50 (3) 3.3 (4) 1.67 (AIPMT MAINS 2012) 23. A reaction having equal energies of activation for forward and reverse reactions has

E + A → 2D   +350

(1) ∆S = 0 (2) ∆G = 0 (3) ∆H = 0 (4) ∆H = ∆G = ∆S = 0

(1) 325 kJ mol−1 (2) 525 kJ mol−1 (3) −175 kJ mol−1 (4) –325 kJ mol−1

(NEET 2013)

(AIPMT MAINS 2011) 18. In which of the following reactions, standard reaction entropy change ( ∆S o ) is positive and standard Gibb’s energy change ( ∆G o ) decrease sharply with increasing temperature? 1 (1) C (graphite) + O2(g ) → CO(g ) 2 1 (2) CO(g ) + O2(g ) → CO2(g ) 2 1 (3) Mg(s) + O2(g ) → MgO(s) 2 1 1 1 (4) C (graphite) + O2(g ) → CO2(g ) 2 2 2 (AIPMT PRE 2012) 19. Standard enthalpy of vaporization ∆vapH° for water at 100°C is 40.66 kJ mol−1. The internal energy of vaporization of water at 100°C (in kJ mol−1) is (1) +37.56 (2) −43.76 (3) +43.76 (4) +40.66

24. For the reaction



X 2O4(l) → 2XO 2(g)

∆U = 2.1 k cal, ∆S = 20 cal K−1 at 300 K. Hence, ∆G is (1) 2.7 kcal (2) −2.7 kcal (3) 9.3 kcal (4) −9.3 kcal (AIPMT 2014) 25. The heat of combustion of carbon to CO2 is –393.5 kJ mol−1. The heat released upon formation of 35.2 g of CO2 from carbon and oxygen gas is (1) –630 kJ (2) –3.15 kJ (3) –315 kJ (4) +315 kJ (RE-AIPMT 2015) 26. The correct thermodynamic conditions for the spontaneous reaction at all temperatures is (1) ∆H < 0 and ∆S > 0. (2) ∆H < 0 and ∆S < 0. (3) ∆H < 0 and ∆S = 0. (4) ∆H > 0 and ∆S < 0. (NEET-I 2016)

(AIPMT PRE 2012) 20. The enthalpy of fusion of water is 1.435 kcal mol . The molar entropy change for the melting of ice at 0°C is −1

(1) 10.52 cal mol−1 K−1 (2) 21.04 cal mol−1 K−1 (3) 5.260 cal mol−1 K−1 (4) 0.526 cal mol−1 K−1

27. For a sample of perfect gas when its pressure is changed isothermally from pi to pf, the entropy change is given by p  p  (1) ∆S = nR ln  i  (2) ∆S = nRT ln  f   pf   pi 

(AIPMT PRE 2012)

Chapter 4_Thermodynamics.indd 100

1/4/2018 5:11:04 PM

Thermodynamics p  p  (3) ∆S = RT ln  i  (4) ∆S = nR ln  f   pf   pi  (NEET-II 2016)

101

29. A gas is allowed to expand in a well-insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy ∆U of the gas in joules will be (1) −500 J (2) −505 J (3) +505 J (4) 1136.25 J

28. For a given reaction ∆H = 35.5 kJ mol−1 and ∆S = 83.6 JK−1 mol−1. The reaction is spontaneous at (Assume that ∆H and ∆S do not vary with temperature.)

(NEET 2017)

(1) T > 425 K (2) all temperatures. (3) T > 298 K (4) T < 425 K (NEET 2017)

Answer Key Level I  1. (1)

 2. (3)

 3. (1)

 4. (4)

 5. (4)

 6. (3)

 7. (2)

 8. (4)

 9. (1)

10. (3)

11. (2)

12. (1)

13. (4)

14. (2)

15. (1)

16. (4)

17. (1)

18. (1)

19. (1)

20. (4)

21. (1)

22. (4)

23. (4)

24. (3)

25. (2)

26. (4)

27. (4)

28. (1)

29. (3)

30. (4)

31. (1)

32. (1)

33. (1)

34. (2)

35. (1)

36. (3)

37. (2)

38. (2)

39. (1)

40. (2)

41. (2)

42. (3)

43. (2)

44. (3)

45. (1)

46. (1)

47. (4)

 1. (4)

 2. (3)

 3. (3)

 4. (2)

 5. (2)

 6. (2)

 7. (1)

 8. (2)

 9. (3)

10. (1)

11. (4)

12. (1)

13. (4)

14. (2)

15. (4)

16. (1)

17. (1)

18. (3)

19. (1)

20. (2)

Level II

21. (4)

22. (4)

23. (2)

24. (2)

25. (4)

26. (4)

27. (1)

28. (4)

29. (1)

30. (3)

31. (1)

32. (1)

33. (1)

34. (1)

35. (2)

36. (1)

37. (1)

38. (3)

39. (4)

40. (3)

41. (4)

42. (1)

43. (4)

44. (4)

45. (1)

46. (4)

47. (1)

48. (4)

49. (2)

50. (2)

51. (2)

52. (1)

53. (4)

54. (3)

55. (1)

56. (2)

57. (2)

58. (2)

59. (4)

60. (4)

61. (3)

Previous Years’ NEET Questions  1. (2)

 2.(4)

 3. (4)

 4. (3)

 5. (1)

 6. (4)

 7. (4)

 8. (4)

 9. (4)

10. (2)

11. (3)

12. (3)

13. (2)

14. (4)

15. (1)

16. (4)

17. (3)

18. (1)

19. (1)

20. (3)

21. (3)

22. (4)

23. (3)

24. (2)

25. (3)

26. (1)

27. (1)

28. (1)

29. (2)

Hints and Explanations Level I

T1V1g -1 = T2V2g -1 æV ö T (V )g -1 = ( 4T )ç ÷ è 32 ø

3. (1) As gas is expanding into vacuum, therefore, pext = 0

1 æ 1ö =ç ÷ 4 è 32 ø

Work done is given by w = - pext ´ DV = 0

6. (3) For any adiabatic process, TV g -1 = constant

g -1

T (V )

æV ö = ( 4T )ç ÷ è 32 ø

1 æ 1ö =ç ÷ 4 è 32 ø Chapter 4_Thermodynamics.indd 101

-2

g -1

2 -2 = 2 -5(g -1) Þ

T1V1g -1 = T2V2g -1



g -1

g -1

2 7 = g -1Þg = 5 5

So it is a diatomic gas.

g -1

-5(g -1)

2

7

1/4/2018 5:11:05 PM

102

OBJECTIVE CHEMISTRY FOR NEET w = - pext dV = -nRT

9. (1) We know   

w =R





We know

DH = DU + Dng RT



Therefore,

DH > DU

26. (4) For all spontaneous reaction ∆G < 0

(1)

Substituting q = +50 cal and w = −30 cal in Eq. (1), we get ∆U = w + q = 50 − 30 = 20 cal

 R  nR 3. (4)   w = nCV ∆T = n  1 ∆T = (T2 - T1 )  g -1  g - 1 w = - pext dV



For isochoric process, ∆V = 0, therefore, w = 0.



From the first law of thermodynamics ∆U = q + w. As w = 0, therefore, ∆U = q = 500 J

T  V  15. (1) Entropy is given by ∆S = nCV ln  2  + nR ln  2   T1   V1 

       = 0 + 1(R) ln(10)

    = 19.15 J K−1 mol−1 18. (1) For exothermic process, ∆H < 0, therefore, q < 0. DSsystem

DH sys T

Therefore, DH < 0, DSsystem < 0, DSsurrounding > 0





Given that ∆H > 0 [Positive sign indicates that reaction oppose the spontaneity]



Also, from the reaction, ∆ng > 0



Therefore, ∆S = +ve [Favors spontaneity of reaction].

104 = 52 kcal 2 1 9. (3) For the reaction H 2(g ) + O2(g ) ® H 2O( l ) 2 2 1ö 3 æ Dng = nP - nR = ç 0 - 1 - ÷ = è 2ø 2 28. (1) Heat of atomization =



We know DH = DU + Dng RT 3 Therefore, DH = DU - RT 2 Hence, ∆H < ∆U

30. (4) The sublimation process is

    A(s) ® A(g); DH sublimation (1)

= + ve

DH 178.3 ´ 103 or T > or T > 1114.375 K DS 160

27. (4) For the spontaneous reaction, ∆G < 0

q = = - ve T

   DSsurrounding = -

¯

+ ve

Only for high value of ∆S (+ve), ∆G < 0

DH - T DS < 0 Þ T >

12. (1) Internal energy can be expressed as ∆U = w + q

14. (2) We know

DH - T DS < 0   ¯

Dng = np - nR = 2 - 1 = 1



or

+ ve

11. (2) For the reaction





w = -1 ´ R(1)

    A(s) ® A( l); DH fusion 

    A(l) ® A(g); DH vaporization (3)

From Hess’ law, (1) = (2) + (3)



Therefore, DH sublimation = DH fusion + DH vaporization

21. (1) Gibbs free energy is given by ∆G = ∆H - T∆S.

31. (1) H 2O( l )® H 2O(g)





Therefore, X1 = X2 + ∆H vaporization



Hence, X1 > X2

For a reaction to be spontaneous at all temperatures, the condition should be ∆H < 0 and ∆S > 0. When ∆H > 0, ∆S < 0, then ∆G > 0 always.

23. (4) Entropy can be calculated as D S = SN 2 + 3SH2 - 2SNH3 o

     = −114.99 + 3 × (−98.74) −2 × (−304.99)   

25. (2) For an endothermic reaction, ∆H > 0

For spontaneous reaction, ∆G < 0

Chapter 4_Thermodynamics.indd 102

DH vaporization = + ve

32. (1) From the reaction, N2(g) + 3H2(g) → 2NH3(g) ∆H o = 2 ∆H f, NH3 - ∆H f, N 2 - 3∆H f, H2 (1)



We have



Substituting ∆H f, N 2 = ∆H f, H2 = 0 and ∆H f, NH3 = -22 kcal in Eq. (1), we get





  = 198.8 J K-1

24. (3) The standard free energy of formation of any element in its stable form is zero. Ozone is not the stable form of oxygen.

(2)

DH f, NH3 = -11 kcal mol -1

34. (2) The reaction involved is 1 1 H 2(g ) + Cl 2(g ) ® HCl(g ) 2 2

1/4/2018 5:11:10 PM

Thermodynamics 40. (2) We know DH = DU + Dng RT .

1 1 DH = ( 435) + ( 243) - 431 2 2 DH = -92 kJ mol -1



35. (1) The combustion reaction is 25 C8H18(g ) + O2(g ) ® 8CO2(g ) + 9H 2O( l ) 2 25   ∆ng = (nP - nR ) =  8 - 1 = -5.5   2

Only for option (2), Dng = 0, therefore, DH = DU .

41. (2) Calorific value =

   DH = DU -

890 kJg -1 = 55.625 kJg -1 16 1411 C 2H 4 = kJg -1 = 50.39 kJg -1 28 1560 C 2H6 = kJg -1 = 52 kJg -1 30

5.5 ´ 8.314 ´ 298 kJ mol -1 1000

42. (3) We have C(s) + O2(g ) ® CO2(g ); DH 1 (1)

36. (3) For the reaction 2NO2(g ) + 2CO(g ) ® N 2O(g ) + 3CO2(g )

    

= 81 + 3 ´ ( -393) - 2 ´ ( 34) - 2 ´ ( -110)

  



1 ∆H for the reaction C(s) + O2(g ) ® CO(g) can be obtained by Eq. (1) - Eq. (2) 2

= -836 kJ

37. (2) Given DH combustion = -837.8 kJ mol -1 or 837.8 kJ energy is released x × 837.8 = 250 × 4.18 Þ x = 1.248



DH = DH1 - DH 2 = -390 - ( -278) = -112 kJ 43. (2) We know  DH = DU + Dng RT DU = DH - 1 ´ R ´ T

3 2 Al(s) + O2(g) ® Al 2O3(s) 2

= 176 ´ 103 - 8.314 ´ 1240 = 165.6 kJ

    

Therefore, no. of moles of Al2O3      =

1 CO(g) + O2(g ) ® CO2(g ); DH 2 (2) 2



We have DH = DH f, N 2 O + 3DH f, CO2 - 2 DH f, NO2 - 2 DH f, CO



DcH Molecular wt.

CH 4 = -

We know DH = DU + Dng RT



46. (1) The reactions involved are as follows;

No. of moles of Al = 0.624 mol Al 2O3 2

1 38. (2) The reaction is H 2(g ) + O2(g ) ® H 2O( l ) 2 1    DH reaction = +(H - H ) + (O = O) - 2(O - H ) - 42 2 1 - 298 = 433 + ( 492) - 2 × B.E(O - H ) - 42   2 B.E(O - H ) = 488.5 kJ mol -1





1 H 2(g ) + O2(g ) ® H 2O(g ) 2

x1 (1)





1 H 2(g ) + O2(g ) ® H 2O( l ) 2

x 2 (2)



From Eq. (1) and Eq. (2), we get x1 = x 2 + DH vap  + ve



39. (1) The reaction involving the formation of H2SO4 is

Therefore, x1 = -68 + ( + ve quantity )





H 2 + S + 2O2 ® H 2SO4 ; DH 5 (1)

Level II





S + O2 ®SSO + O2     ;  ;2 ® DH SO 1 2 ; DH 1 

1. (4) We know









(2)

TVg −1 = k

1 ;   ;  DH 2 (3) SO2 + O2 ® SO3    2 SO3 + H 2O ® H 2SO4  ;  ;  DH 3 (4)



Taking natural logarithm on both sides, we get





logT + (g – 1) lnV = constant

1 ; DH 4 (5) H 2 + O2 ® H 2O      ; 2





logT = (1 – g ) lnV + constant







Eq. (1) = Eq. (2) + Eq. (3) + Eq. (4) + Eq. (5)

2. (3) For a reversible process,



Therefore, DH 5 = DH 1 + DH 2 + DH 3 + DH 4







Also, for an isothermal process,

    

Chapter 4_Thermodynamics.indd 103

work done= - 2.303 nRT log

= -298.2 - 98.7 - 130.2 - 287.3 = -814.4 kJ mol -1

103

V2 (1) V1

V2 p1 = (2) V1 p2

1/4/2018 5:11:22 PM

104

OBJECTIVE CHEMISTRY FOR NEET

Therefore, from Eq. (1) and Eq. (2), we get



æ 1ö w = -2.303 ´ 1 ´ 8.314 ´ 300 log ç ÷ è 5ø

10. (1) The reaction is H 2O(water) ® H 2O(steam)

= -4.01 kJ 4. (2) We know w = - pext ´ DV



We know w = - pDV = - p(Vsteam - VH2 O )

= ( -2 atm) ´ ( 3.40 L) ´ 101.3 J        = - 688.4 J

Therefore, 1 × 20(T2 - 300) = -3 × 103 ⇒ T2 = 150 K

      

From first law of thermodynamics, we have



1   1 = -100 × 103  J = - 1.66 × 105 J  0.6 100 

From first law of thermodynamics, we have DU = q + w

  ∆U = q + w

DU = 2.25 ´ 106 - 1.66 ´ 105

     = −688.84 + 400 = −288.84 J or −289 J

= 2.08 ´ 106 J 12. (1) Step (I):

5. (2) Option (1): DU = nCV DT = 0 [ As DT = 0]

T1 = 600 K , p1 = 3, V1 =

RT 1 ´ R(600) Þ V1 = = 200R = 16.4L p1 3



Option (2): In cyclic process, DU = 0



From first law, we know DU = w + q



Step (II): T2 = 200 K , p2 = 1, V1 = 200R = 16.4 L



Therefore, w = q



Step (III): T3 = 3(1-4 ) × 200 K , p2 = 3, V2 = 8.48 L



 ∂H  Option (3): For ideal gas  = 0 { As H = f (T ) only }  ∂p  T

6. (2) For an adiabatic process, q = 0 

(1)

(1)

Step 1



(2)

Step 2

Step 3



Also, from first law of thermodynamics ∆U = q + w(2)



Work done in step 3 = −p2 (V1 − V2)



From Eq. (1) and Eq. (2), we get



   

= −3(16.4 – 8.48)



   

= −23.8 L atm



∆U = w



Since, for an adiabatic expansion ∆w = −ve or ∆U = −ve



Therefore, internal energy will decrease.

13. (4) From first law of thermodynamics, DU = q + w 

As the process is adiabatic, therefore, q = 0



From Eq. (1), we get DU = w

7. (1)

 1 = - 600 R ln    2



(1)



From Eq. (2) and Eq. (3), we get  nRT2 nR × 350  5  -2   = n  2 R  (T2 - 350) p1  p2

w BC = Isobaric expansion = - pDV

 = -nR DT = -2 R ´ 100 = -200 R (2)

 T 350  5 -2  2  = (T2 - 350) 2 1  2

 2 w CD = Isothermal expansion = - 2R( 400)ln    1 = -800 R ln 2





Net work = (1) + (2) + (3) + (4)



    = 600 R ln2 − 200 R − 800 R ln2 + 200 R

    -T2 + 700 = 2.5T2 - 875 ⇒ T2 = 450 K

(3)

w DA = Isobaric compression = - nR DT = - 2R( 300 - 400) = +200 R  (4)



Therefore, w = 2 ´

14. (2) We know 

5 R( 450 - 350) = 500 R 2

DU = q + w DU = q + ( - pext DV )

    = − 200 R ln2

1.5nR( DT ) = 42 - 100(8.5 ´ x ) ´ 10 -6 ´ 103

    = − 100 R ln4

1.5 ´ 8.314 ´ 2 = 42 - 100(8.5 ´ x ) ´ 10 -6 ´ 103 x = 20 cm

9. (3) Work done in reversible adiabatic process = nCV DT

As expansion process indicates negative work done, therefore, w = −3 × 103 J

Chapter 4_Thermodynamics.indd 104

(1)

5      w = DU = nCV DT = n R(T2 - 350) (2) 2 w = - pext (V2 - V1 ) (3) Also,



 1 w AB = Isothermal compression = - 2R( 300)ln    2

(3)

15. (4) The reaction is N 2(g ) + 3H 2(g ) ® 2NH 3(g ) ∆ng = nP - nR = 2 - 4 = -2

1/4/2018 5:11:27 PM

Thermodynamics DS ° = 2SHCl - SH2 - SCl 2

DH = DU + Dng RT DH - DU = -2RT

= ( 2 ´ 0.19 - 0.13 - 0.22) ´ 103 JK -1mol -1 = 40 JK -1mol -1

16. (1) Step 1: Given that, w = 15 J, DU = 30 J DU = q + w 30 = q + 15 Þ q = 15 J

Step 2: Given that, w = 0, DU = -20 J DU = q + w -20 = q + 0 Þ q = -20 J



30. (3) For reaction to be spontaneous, DG < 0

or

DH - T DS < 0



or

T>

T>

Therefore, ( DU )net = 30 - 20 = 10 J and qnet = 15 - 20 = -5 J

T  V  18. (3) We know ∆S = nCV ln  2  + nR ln  2   T1   V1 

21. (4) DSsurr =

DH surroundings T

=

64 ´ 103 = 213 J K -1 300

300 w 4 T = Þ w = = 0.8 kcal 23. (2) h = 1 - 2 = 1 500 2 5 T1 25. (4) DS ° µ Dng T  V  26. (4) We know ∆S = nCV ln  2  + nR ln  2   T1   V1   1 = CV ln 2 + R ln    2

-49 ´ 103 Þ T < 1218.90 K or T < 945.90 °C 40.2



Therefore, DH - T DS = 0 30.5 DH = T ÞT = = 462.12 K DS 0.066

32. (1) ∆G is positive at low temperature and reaction is feasible at high temperature only. Therefore DH > 0 and DS > 0 33. (1) For the reaction A(g) + B(g) → C (g) Dng = np - nR = [1 - (1 + 1)] = -1

We know DH = DU + Dng RT       = ( -3 ´ 103 ) + ( -1)´ 2 ´ 300 = -3600 cal



Also,     DG = DH - T DS = -3600 - 300( -10) = -600 cal

= (CV - R )ln 2 27. (1) Efficiency of engine is given by h = 1-

298 T2 = 1= 0.2 or 20% 373 T1

34. (1) For the reaction 2H2(g) + O2(g) → 2H2O(l) ∆S = 2 SH2 O, l - 2SH2, g - 2SO2, g

28. (4) Gibbs free energy is given by DG o = - RT ln K p



2059   = - RT  4.814   T 





= 8.314 ´ ( -4.814 ´ 298) + 2059 ´ 8.314 = 5.185 kJ



æ DH ° ö æ 1 ö æ DS ° ö We know ln K = ç ÷ ÷ç ÷ +ç è R ø èT ø è R ø



On comparing the above equation with lnKp = 4.814 – (2059/T), we get

   

Also, -

DH ° = -2059 Þ DH ° = 2059 ´ 8.314 = 17.11 kJ R

29. (2) The reaction is H 2(g ) + Cl 2(g ) ® 2HCl(g )

Chapter 4_Thermodynamics.indd 105

= 2(68) - 2(126.6 ) - 201.20 = -318.4 JK -1 mol -1

35. (2) For the reaction 2NO(g) + O2(g)→N2O4(g) o ∆H ° = ∆H No 2 O4 - 2 ∆H NO [ As, ∆H f,o O2 = 0]

= (9.7 - 2 ´ 90.5) ´ 103 J = -171.3 ´ 103 J

DS ° = 4.814 R DS ° = 4.814 ´ 8.314 = 40 JK -1 mol -1

DH DS

31. (1) At equilibrium, DG = 0

3  573   573  ∆S = 2 × R ln  ⇒ ∆S = 3R ln   473   473  2 V  V  20. (2) We know ∆S = nR ln  2  = 2.303nR log  2   V1   V1 

105

Entropy is given by DS ° = SN 2 O4 - 2SNO - SO2        



= 304 - 2 ´ 210 - 205 = -320 J mol -1K -1

Gibbs free energy can be calculated as DG ° = DH ° - T DS °        

= -171.3 ´ 103 - 298( -320) = -75940 J or - 75.94 kJ

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OBJECTIVE CHEMISTRY FOR NEET

36. (1) Entropy is given as

DH f = ( 2 ´ DH SO2 , f + DH CO2, f ) - DH CS2, l

o DSreaction = SCH4 - SC - 2SH2

  -265 = 4 x + 2( 3x ) - DH CS2, l

= +186.3 - 5.6 - 2(130.7 ) = -80.7 J

    -265 = 10x - 26

Gibbs free energy can be calculated as



o o o DG reaction = DH reaction - T DSreaction

   DH f, so2 = 3x = 3( -23.9) = -71.7 k cal mol -1

55. (1) The reaction is

= -74.9 ´ 103 - 298 ´ ( -80.7 ) = -50.8 kJ

  

     x = -23.9 kcal mol-1

1 1 -100 = ( x ) + (0.5x ) - x 2 2 x = 400 kJ mol -1

37. (1) The given reaction is a combustion reaction, therefore, ∆H < 0. Also, from the reaction, we know

∆ng < 0, hence, ∆S < 0.

56. (2) Two moles of H+ and OH− requires 13.7 × 2 = 27.4 kcal

38. (3) As we know entropy of gas > liquid > solid.

58. (2) We know DH neutralization = -13.7 kcal mol -1 + q

Also, as atomicity or number of atoms increase, entropy increases. Therefore, order of molar entropy C 2H6 (g) > CH 4 (g) > C6 H6 (l) > Al(s) .

20 ´ 103 Þ T < 400 K 50

41. (4) As energy is released and 1 mol of H+ and OH− is to be -y taken into account. Therefore, DH neutralization = 2 Br2(g) ® Br2(l) DH 2o = 0 - DH f,o Br2 (g) (1)



H 2(g) + Br2(g) ® 2HBr(g) DH = 2 DH o 1

o f, HBr(g)

- DH

o f, Br2 (g)

(2)

o o o    DH 1 = 2 DH f, HBr + DH 2

∆H 1o - ∆H 2o 2 4. (4) The reactions involved are as follows: 4

∆H fo, HBr =

Therefore,

P4

+ 5O2 → P4O10 ; ∆H = -9.19 kJ (1)







 



The enthalpy for the P4(yellow) ® P4(red) can be obtained from Eq. (1) - Eq. (2)



Therefore, ∆H = −9.19 + 8.78 = −0.41 kJ

(yellow)

P4 + 5O2 → P4O10 ; ∆H = -8.78 kJ (2)

(red)

49. (2) We have DH neutralization = -13.7 + q , where q = heat of dissociation

Lowest value of DH neutralization implies highest value of heat of dissociation or weakest acid. Therefore, b is the weakest acid.

53. (4) From the reaction, we have

Chapter 4_Thermodynamics.indd 106

DH neutralization = -13700 + 366 ´ 0.86 (As 86% is to be neutralized)



Therefore, reaction is not spontaneous when T > 400 K.

43. (4)

where q is the heat of dissociation of weak acid.

59. (4) Enthalpy of reaction = Σ Bond energies of reactants – Σ Bond energies of products

DH - T DS < 0





= -13, 385 cal

39. (4) For spontaneous process, ∆G < 0

-20 ´ 103 + T (50) < 0 Þ T
DH , T >



1 1 H 2(g) + Cl 2(g) ® HCl(g) 2 2

is, T >

We know,

å Bond enthalpies

Dr H ° =

reactant

- å Bond enthalpies product

= - 93 kJ mol -1



Entropy for the reaction is ∆S = Sproducts - Sreactants





4. (3) Since the gaseous product is forming, hence ∆Stotal will be positive. As the given reaction represents the decomposition, hence ∆H is positive. 5. (1) The change in internal energy is given by ∆U = q + w

170 ´ 103 J > 1000 K 170 JK -1

9. (4) At equilibrium, ∆G = 0 and ∆H = -30 kJ

1 ( ∆H H-H + ∆H Cl -Cl ) - ( ∆H H-Cl ) 2 1 1  =  × 434 + × 242 - 431 2  2

∆r H ° =

1 3   =  50 -  × 60 + × 40   = 40 JK -1 mol -1   2 2  

Therefore, T =

DH -30 ´ 103 J = = 750 K DS 40 JK -1 mol -1

10. (2) According to Gibb’s equation, we have ∆G = ∆H - T∆S

q and w will depend on how the change takes place. So, these are not state functions. However, the change in internal energy ∆U does not depend on how a change take place and will depend only on the initial and final states of the system, therefore, is a state function. Enthalpy (H), temperature (T), entropy (S) and pressure (p) are also state functions.

When ∆G = 0, it becomes ∆H = T∆S



Substituting the values, we get

6. (4) The reaction taking place in the fuel cell is

12. (3) We have

C5H12(g) + 8 O2(g) ® 5 CO2(g) + 6 H 2 O (g)

We know, DG ° =

å D G °(products) - å D G °(reactants) f

f

= { D f G °[5CO2(g)] + D f G ° [6H 2O(g)]} - { D f G °[C5H12(g)] + D f G ° [8O2(g)]}

T =



Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g); ∆H1 = -26.8 kJ (1)



FeO(s) + CO(g) → Fe(s) + CO2(g); ∆H2 = -16.5 kJ(2)



On multiplying Eq. (2) by 2 and subtracting it from Eq. (1), we get the required equation



Fe2O3(s) + CO(g) → 2FeO(s) + CO2(g) DH = DH 1 - 2 ´ DH 2 = ( -26.8) - 2 ´ ( -16.5) = + 6.2 kJ

= - 3387 kJ mol -1 or - 3387 ´ 103 J mol -1

- 3387 ´ 10 J mol 3

-1

13. (2) (a)→(iv) : When Kp > Q

= - 32 ´ 96500 E °

E° =

-3387 ´ 10 J mol -32 ´ 96500 3

-1

å Bond enthalpies

reactant

-



This implies rate of forward reaction > rate of backward reaction. Therefore, reaction is spontaneous.



(b)→(i) : When ∆G° < RT lnQ



This implies, ∆G° is positive, reverse reaction is feasible, thus reaction is non-spontaneous.

= 1.09 V

7. (4) For any reaction Dr H ° =

Work done is given by w = -pext ∆V = 0



Substituting the given values of heat of formation of compounds, we have

We know, ∆G° = -nFE°

DH 40.63 ´ 103 J mol -1 = = 373.43 K DS 108.8 JK -1 mol -1

11. (3) As ideal gas is expanding into vacuum, therefore, pext = 0

DG ° = {5 ´ ( -394.4) + 6 ´ ( -237.2)} - {1 ´ ( -8.2) + 8 ´ 0 }

DH , that DS

å Bond enthalpies

product

= (1 ´ DH C ¾ C + 4 ´ DH C ¾ H + DH H ¾ H ) - ( DH C ¾ C + 6 ´ DH C ¾ H )

(c)→(ii) : When Kp = Q

= (606.10 + 4 ´ 410.5 + 431.37 ) - ( 336.49 + 6 ´ 410.5)

This implies, rate of forward reaction = rate of backward reaction. Therefore, reaction is in equilibrium.

= 2679.37 - 2799 = 119.6 kJ mol -1

Chapter 4_Thermodynamics.indd 107



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(d)→(iii) : When T∆S > ∆H



Given that  ∆H = 40.66 kJ mol-1 = 40660 J mol-1, so



This implies ∆G will be negative only when ∆H = +ve, that is, endothermic.



   40660 = ∆U + 1 × 8.314 × 373

14. (4) For an adiabatic process, q = 0

    ⇒ ∆U = 37560 J mol-1 = 37.56 kJ mol-1 20. (3) According to the definition of entropy, we have

For free expansion of an ideal gas, pext = 0. Therefore, w = - å pext ´ DV = 0



From first law of thermodynamics, ∆U = q + w = 0



We have,

∆U = nCV ∆T



Therefore,

∆T = 0

15. (1) For the reaction

DS =



DH 1.435 ´ 103 = = 5.260 cal K -1 mol -1 T 273

21. (3) We know

∆G° = -nFE°cell(1)



where

n=



On substituting the values in Eq. (1), we get 960 × 1000 = -4 × 96500 × E°cell ⇒ E°cell = -2.5 V

2 ´ 2 ´ ( +3) = 4  3



H2O(l) → H2O(g) ; ∆HTransition = 30 kJ mol-1 and T = 300 K





Therefore,

22. (4) Using the thermodynamic relation ∆H Trans T 30 × 103 = = 100 J mol -1 K -1 300

∆S =

Cp CV

Given Ea(forward) = Ea(backward). Therefore, ∆H = 0



2H2 → 4H ; ∆H = +869.6 kJ



Therefore, for the reaction

24. (2) We know,    ∆H = ∆U + ∆ngRT = 2.1 + 2 ´

+869.6 H 2 ® 2H ; DH = = + 434.8 kJ 2



DG = DH - T DS ∆H (kJ mol-1) + 150





3B → 2C + D

− 125

(2)





E + A → 2D

+ 350

(3)



From Eq. (1) × 2 + Eq. (2) − Eq. (3), we get the required equation



        

(1)





 = −175 kJ mol-1

1 8. (1) For the reaction C(s) + O2 (g) → CO(g), the degree 1 2 of randomness increases as the phase is changed from solid to gas. According to the relation ∆Go = ∆H - T∆So, as S increases and T also increases; thus, Gibbs energy decreases.

∆H = ∆U + ∆ngRT, where ∆ng = 1 as



H2O(1)  H2O(g)

Chapter 4_Thermodynamics.indd 108

26 = - 2.7 kcal 1000



Heat released due to formation of 44 g of CO2 = −393.5 kJ mol−1



Therefore, heat released due to formation of 35.2 g of -393.5 CO2 = ´ 35.2 = - 315 kJ 44

26. (1) For a spontaneous reaction, ∆G is negative from the following equation ∆G = ∆H − T∆S







Using this, we can say that ∆G will always be negative, when ∆H < 0 and ∆S is > 0. 27. (1) The entropy change is given by

19. (1) Using the thermodynamic relationship

= 3.3 - 300 ´

25. (3) We have C(s) + O2 (g) → CO2 (g)

∆H = [(150) × 2 + (-125)] – 350

 

2 ´ 300 1000

= 3.3 kcal



17. (3) Given that 1 A ® B 2

(5 / 2)R 5 = = 1.67 ( 3 / 2)R 3

23. (3) We know ∆H = Ea(forward) = Ea(backward)

16. (4) For the reaction

=

DS = nC p ln

T2 p + nR ln 1 T1 p2

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Thermodynamics

where Cp is the specific heat at constant pressure, p1 and p2 are initial and final pressures, respectively and T1 and T2 are initial and final temperatures, respectively. For an isothermal process, T1 = T2, so the equation reduces to T DS = nR ln



w = - pDV = - 2.5( 4.5 - 2.5) = - 5 L atm = - 5 ´ 101.3 J = - 506.5 J

pi pf

Substituting ∆H = 35500 J mol-1, ∆S = 83.6 JK-1 mol-1 and ∆G = 0 (At equilibrium) in Eq. (1), we get DG = DH - T DS 0 = 35500 - T ´ 83.6 35500 T = = 424.64 K » 425 K 83.6

Chapter 4_Thermodynamics.indd 109

Since, both ∆H and ∆S are positive, thus, reaction will be spontaneous at high temperature, that is, when T > 425 K.

29. (2) Work done by a gas in an irreversible expansion is given by

28. (1) Gibbs free energy is given by ∆G = ∆H − T∆S (1)

109



Since the system is well insulated, q = 0. From first law of thermodynamics, we have



∆E = q + w



∆E = 0 - w = -506.5 J

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5

Chemical Equilibrium

Chapter at a Glance 1. Equilibrium An equilibrium system is the system in which the macroscopic properties of the system, such as temperature, pressure, concentration, etc., do not change. Any increase in these properties is balanced by a decrease in that property within the system. There are two kinds of equilibrium (a) Physical equilibrium and (b) Chemical equilibrium. 2. Physical Equilibrium It is an equilibrium which exists between different physical states or when phases undergo transition. The physical equilibrium can be attained in closed system. The rate of change between the phases is the same at equilibrium. Thus, the equilibrium is stable yet dynamic because two opposing processes take place simultaneously.



(a) Types of physical equilibrium (i) Solid–liquid equilibrium: The state when solid and liquid phases of a substance coexist is called solidliquid equilibrium. Solid  Liquid (ii) Liquid–vapor equilibrium: When a liquid is transformed into a gas is known as evaporation or vaporization. Water(liquid)  Water(vapor) (iii) Solid–vapor equilibrium: The process by which a solid is directly converted into vapor state is known as sublimation. Iodine(solid)  Iodine(vapor) (b) Equilibrium in dissolution of solids and gases in liquids and Henry’s law (i) Solids in liquids: When the maximum amount of solid is dissolved in a given amount of liquid at a specific temperature, the solution is said to be a saturated solution. Sugar(solution)  Sugar(solid) (ii) Gases in liquids: At a given pressure, equilibrium exists between the gas (e.g., CO2) molecules dissolved in the liquid and those in gaseous state, is represented as CO2 (gas)  CO2 (in solution)

(iii) Henry’s law    It states that the concentration of a gas dissolved by a given volume of a liquid at constant temperature is directly proportional to the pressure of the gas over the solution. • Henry’s law quantitatively correlates pressure and solubility C gas ∝ pgas C gas = K H pgas    where Cgas is the concentration of the gas in the solution, pgas is the partial pressure of the gas above the solution and the proportionality constant KH is called the Henry’s constant. The unit of KH depends on the units used for concentration and pressure.

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OBJECTIVE CHEMISTRY FOR NEET

• An alternate and commonly used form of Henry’s law is C1 C 2 = p1 p2

where the subscripts 1 and 2 refer to initial and final conditions. Henry’s law can also be expressed in terms of mole fraction as p = K Hx where p is the partial pressure, x is the mole fraction and KH is the Henry’s constant.

3. Chemical Equilibrium The state at which the concentration of reactants and products do not change with time is called a state of chemical equilibrium. It is a dynamic system in which two or more opposing chemical reactions are going on at the same time and at the same rate. Chemical equilibrium can be attained only in a closed system. (a) Law of mass action and equilibrium constant (i) The law of mass action states that, at a constant temperature and pressure, the rate of a chemical reaction is directly proportional to the product of the molar concentration of the reactants each raised to a power equal to the corresponding stoichiometric coefficient, which appears in the balanced chemical equation. (ii) Consider a reversible reaction, aA + bB  cC + dD



At equilibrium, the two rates become equal, that is, Rate of forward reaction (Rf ) = Rate of reverse reaction (Rb). Rf Equilibrium Rf = Rb

Rate Rb

Reaction

(iii) At equilibrium, kf [ A]a [B]b = kb [C]c [D]d kf [C]c [D]d = = KC kb [ A]a [B]b The ratio of kf /kb is the equilibrium constant KC . The above equation is called law of chemical equilibrium or the equilibrium law. The concentration of the product C (or D) at equilibrium is called its active mass. (b) Characteristics of equilibrium constant (i) For any reaction, KC is constant for a specific temperature. (ii) It is independent of the initial concentrations of the reactants. (iii) It is independent of presence of catalyst and inert materials in the reaction. (iv) It is independent of the direction from which the equilibrium is attained. (v)  It is dependent on the representation of the reaction. The equilibrium of the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction.

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Chemical Equilibrium

A +B  C + D

KC =

C + D  A +B

K C′ =

Therefore, K C′ =

113

[C][D] [ A][B] [ A][B] [C][D] 1 KC

(f ) I f a chemical equation is multiplied by a common factor (n), then the new equilibrium constant should also reflect the change. For example, A +B  C + D

KC =

nA + nB  nC + nD

K C′ =

[C][D] [ A][B] [C] n[D]n n

[ A] [B]n

= ( K C )n

(c) Equilibrium constant for gaseous reactions (Kp) For gaseous reactions, since gas pressures are more conveniently measures, the equilibrium law expressions for gaseous reactions are written using partial pressures instead of molar concentrations. Therefore, [ p ]c ´ [ pD ]d Kp = C a [ pA ] ´ [ pB ]b where, pA, pB, pC and pD the partial pressures of A, B, C and D, respectively and Kp is called pressure equilibrium constant. (d) Relationship between Kp and KC

Dng

K p = K C ´ ( RT )

Dng = total number of moles of gaseous products – total number of moles of gaseous reactants. (i) For Dng= 0; Kp = KC (ii) For Dng> 0; Kp > KC (iii) For Dng< 0; Kp < KC (e) Units of KC and Kp (i) When the total number of moles of reactants and products are equal, for such reactions, KC and Kp have no units. (ii) When the total number of moles of reactants and products are unequal, KC will have units (mol L–1)Δn and Kp will have the units (atm)Δn. where Δng = total number of moles of products – total number of moles of reactants. (f ) Homogeneous and heterogeneous equilibria (i) A homogeneous equilibrium is established in a system where all reactants and products are in the same phase. SO2 (g ) + O2 (g )  2SO3 (g ) (ii)  A heterogeneous equilibria is established in a system where the reactants and products are in more than one phase, such as between solid and liquid, liquid and gas, solid and gas, etc. CaCO3 (s )  CaO(s ) + CO2 (g ) (iii)  Simultaneous equilibrium occurs when more than one equilibrium exists in the same container s­ imultaneously. CaCO3 (s )  CaO(s ) + CO2 (g ) CO2 (g ) + C(s)  2CO(g )

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OBJECTIVE CHEMISTRY FOR NEET

(g) Reaction quotient    In a reversible reaction at any stage other than equilibrium, the ratio of molar concentration of product to the molar concentration of reactant is called reaction quotient. It is represented by QC. (i) For the reaction aA + bB  cC + dD QC = or in terms of pressure, we have Q=

[C] c[D]d [ A]a [B]b

( pC )c ( pD )d ( pA )a ( pB )b

(ii) Relation between QC and KC values and the direction of reaction are as follows: QC and KC values

Relative concentration

Direction of net direction

QC > KC

Products more than reactants

Reverse

QC < KC

Reactants more than products

Forward

QC = KC

Equilibrium concentrations

No reaction

(h) Relation between Gibbs energy change and chemical equilibrium (i) The relation between ΔG o and the position of equilibrium is ΔG = ΔG o + RT ln Q

(ii) For a reaction in equilibrium, ΔG = 0 and Q = KC , so ΔG = ΔG o + RT ln K = 0 ΔG o = - RT ln K ⇒ ln K = -

Therefore,     K = e - ΔG

o

ΔG o RT

/ RT

• If ΔG o < 0, then, -ΔG o/RT is positive, and therefore, > 1 or K > 1. Reaction is spontaneous and it proceeds in the forward direction to an extent that products are present predominantly. • If ΔG o > 0, then, -ΔG o/RT is negative, and therefore, < 1 or K < 1. Reaction is non-spontaneous and it proceeds in the forward direction to a small extent, so small amount of products are present. 4. Le Chatelier’s Principle It states that if a system at equilibrium is subjected to a change in one or more variables such as, pressure, temperature, and/or concentration, then the equilibrium shifts in such a way so as to undo the effect of the change. The factors affecting equilibria are as follows: (a) Temperature: If temperature is increased, additional amount of heat is given to the system, then an endothermic reaction will proceed faster in the forward direction. aA + bB  cC + dD + Heat However, if reaction is exothermic reaction, it will be slow. For example, 2NO2 (g )  N 2 O4 (g ), DH = -5.72 kJ mol -1 is an exothermic reaction. On increasing the temperature, the reaction will proceed in the reverse direction. (b) Concentration: If at the point of equilibrium, the concentration of product is increased, reaction will proceed in reverse direction. (i) If product is removed, reaction will proceed in the forward direction. (ii) Likewise is the case if the reactant is added.

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Chemical Equilibrium

115

(c) Pressure: There is no effect of pressure, if the reaction has the same number of moles in reactants and products. For example: N 2 (g ) + O2 (g )  2NO(g ) (i) Pressure effect is seen, if the number of moles of reactants and products are different. (ii) The reaction will proceed in the direction where number of moles is decreased. For example, N 2 (g ) + 3H2 (g )  2NH3 (g ) If we increase pressure, reaction will proceed in forward direction. (d) Addition of catalyst: A catalyst can increase the rate of reaction (it reduces the time taken to reach at equilibrium), however the point of equilibrium will remain the same and the value of equilibrium constant will not be affected. (e) Addition of inert gas (i) At constant volume: If the number of moles of reactant and product are same and by adding inert gas there is no change in volume, there is no effect of adding inert gas. (ii) At constant pressure: If volume changes, we follow pressure rule, that is, the reaction will proceed in a direction where the number of moles is less.

Solved Examples 1. Which of the following reactions represent a heterogene­ ous equilibrium? (1) H 2(g ) + I 2(g )  2HI(g) (2) CaCO3(s)  CaO(s) + CO2(g ) (3) N 2O4(g )  2NO2(g ) (4) 2O3(g )  3O2(g )

Solution (4) KC or Kp do not depend on concentration, but only on temperature, therefore, equilibrium constant will remain constant. 4. Kp for the reaction N 2(g ) + 3H 2(g )  2NH 3(g ) at 400°C is 1.64 × 10–4. Calculate KC. (1) 0.3 mol2 L–2 (2) 0.4 mol2 L–2 (3) 0.5 mol2 L–2 (4) 0.6 mol2 L–2

Solution (2) In a heterogeneous equilibrium, physical state of all the reactants and products are not the same. 2. Reaction between iron and steam is reversible if it is carried out 3Fe + 4H 2O ¾® ¾ Fe3O4 + 4H 2 (1) at constant T. (2) at constant p. (3) in an open vessel. (4) in a closed vessel.

Solution (3) For reaction N 2(g ) + 3H 2(g )  2NH 3(g );

Dn ng = nP - nR = 2 - 4 = -2



We know



Substituting the values in Eq. (1), we get

3. In a reversible reaction, two substances are in equilib­ rium. If the concentration of each one is doubled, the equilibrium constant will be (1) reduced to half, its original value. (2) becomes (original)/4. (3) doubled. (4) constant.

Chapter 5_Chemical Equilibrium.indd 115



(1)

1.64 ´ 10 -4 = K C ´ (0.0821 ´ 673)-2 K C = 1.64 ´ 10 -4 ´ (0.0821 ´ 673)2

Solution (4) For a reaction to be reversible, it should be carried out in a closed vessel.

Δng

K p = K C ×( RT )

= 0.5 mo ol 2 L-2 5. For the reaction PCl 3(g ) + Cl 2(g )  PCl 5(g )

The value of KC at 250°C is 26. The value of Kp at this temperature will be (1) 0.61 (2) 0.57 (3) 0.83 (4) 0.46

Solution Dng

(1) We know K p = K C ´ ( RT )

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OBJECTIVE CHEMISTRY FOR NEET

Substituting K C = 26, R = 0.0821, T = 523 K and ∆ng = (1 - 2) = -1 in above equation, we get K p = 26 ´ (0.0821 ´ 523)-1 = 0.61



6. The equilibrium constant for the reaction, N 2(g ) + O2(g )  2NO is 4 × 10−4 at 2000 K. In presence of a catalyst, equilibrium is attained ten times faster. Therefore, the equilibrium constant, in presence of the catalyst at 2000 K is

Solution (2) Equilibrium constant is constant at constant tempera­ ture for a reaction. 7. For the reaction A(g)+B(g)  3 C(g) at 250°C, a 3 L ves­ sel contains 1, 2, 4 mol of A, B and C respectively. If KC for the reaction is 10, the reaction will proceed in





   = 5.1 × 10−2 atm−2 9. The vapor density of undecomposed N2O4 is 46. When heated, vapor density decreases to 24.5 due to its dis­ sociation to NO2. The percentage dissociation of N2O4 at the final temperature is (1) 80 (2) 60 (3) 40 (4) 70 Solution (1) The reaction involved is



8. When 1.0 mol of N2 and 3.0 mol of H2 were heated in a vessel at 773 K and a pressure of 3.55 atm, 30% of N2 is converted into NH3 at equilibrium. Calculate Kp for the reaction.

46 1 + a ⇒ 1 + a = 1.8 ⇒ a = 0.8 or 80% = 1 25.4 10. 2SO2(g) + O2(g)  2SO3(g)

(2) 4.1 × 10–2 atm–2 (4) 6.1 × 10–2 atm–2

+

N2



1

2NH 3

3

1 –x

0 2x

3 – 3x

Given: Degree of dissociation = 0.3 = Kp =





3H 2

( pNH3 )2 ( pN 2 ) ( pH2 )3

Partial pressures can be calculated as pNH3 = xNH3 ´ pTotal

Chapter 5_Chemical Equilibrium.indd 116

Solution (2) For reaction 2SO2(g) + O2(g)  2SO3(g)

x Þ x = 0.3 1 (1)

(1 - x ) ´ 3.55 0.7 = ´ 3.55 = ( 4 - 2x ) 3.4

If the partial pressure of SO2, O2 and SO3 are 0.559, 0.101 and 0.331 atm respectively. What would be the partial pressure of O2 gas, to get equal number of moles for SO2 and SO3? (1) 0.188 atm (2) 0.288 atm (3) 0.388 atm (4) 0.488 atm

Solution (3) At t = 0 At t = t eq

(Vapor density)initial ninitial = nfinal (Vapor density)final

We know

Since, QC > KC, therefore, reaction will proceed in back­ ward direction.



N 2O4  2NO 2 1 0 1 –a 2a

At t = 0 At t = t eq

Final moles = 1 − a + 2a = 1 + a

[C]3 (4)3 ´ 3 ´ 3 = = 10.67 [A][B] (3)3 ´ 1 ´ 2

(1) 3.1 × 10–2 atm–2 (3) 5.1 × 10–2 atm–2

3 - 3x ´ 3.55 2.1 = ´ 3.55 4 - 2x 3.11

æ 0.6 ö ´ 3.55 ÷ ç (0 0.6 )2( 3.4)2 3.4 è ø Kp = = 3 2 2 æ 0.7 ö æ 2.1 ö (0.7 )( 2.1) ( 3.55) ´ 3.55 ÷ ç ´ 3.55 ÷ ç è 3.4 ø è 3.4 ø

Solution

QC =

pH2 = xH2 ´ pTotal =

Substituting the values of partial pressure in Eq. (1), we get

(1) forward direction. (2) backward direction. (3) either direction. (4) equilibrium.

(2) Reaction quotient can be calculated as

2x 0.6 ´ 3.55 = ´ 3.55 4 - 2x 3.4

2

(1) 40 × 10−4 (2) 4 × 10−4 (3) 4 × 10−5 (4) difficult to compute

pN 2 = xN 2 ´ pTotal =

Kp =

( pSO3 )2 2

( pSO2 ) ( pO2 )

=

(0.331 atm )2 = 3.47 atm -1 (0.559 atm )2(0.101 atm )



If number of moles of SO2 = SO3, then, partial pressure of SO2 = partial pressure of SO3



Also, Kp would remain same as temperature is constant. 3.47 =

2 pSO 2 3

2 pSO pO2 2

= 1/pO2 or pO2 = 0.288 atm

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117

Chemical Equilibrium 2 AB(g) 11. The equilibrium constant of the reaction A 2(g) + B2(g)  Solution A 2(g) + B2(g)  2 AB(g) at 100°C is 50. If 1 L flask containing one (3) mole of A2 is connected to a 3 L flask containing two moles of B2 the number of moles of AB formed at 373 K will be

Initial moles Conc.at equil.

(1) 1.886 (2) 2.317 (3) 0.943 (4) 18.86

Solution

We know

(1) A 2 (g) At t = 0 At t = t eq

+

1 1+ 3 1- x 4

( 2x / 4)

B2(g)



[(1 - x )/ 4][( 2 - x )/ 4]



0 2x 4

2

= 50 Þ

=

2AB(g); K C = 50

2 1+ 3 2-x 4

4x 2 = 50 (1 - x )( 2 - x )

199x2 + 5x – 6 = 0 On solving, we get x = 0.1615 Out of 3 mol, 0.1615 mol of I2 is converted into HI. Therefore, percentage of I2 converted into HI is 0.1615 ´ 100 = = 5.38 % 3



Therefore, number of = 2 x = 2 ´ 0.943 = 1.886

moles

of

AB

Solution (2) The reaction involved is NH 4HS(s)  NH 3(g) + H 2S(g) Moles at equilibrium

Solution (2)          CO +   Cl2    COCl2



Moles at t = 0



Moles at equilibrium

2

(2 – 1) 1 Conc. at equilibrium (V = 5 L) 5 Moles at K C =

x

x

Since moles of NH3 and H2S are equal, their partial pres­ sure would also be equal. Let the partial pressure of NH3 = H2S = p at equilibrium Given: 2p = 1.12 or p = 0.56. The equilibrium constant is given by K p = p 2 = (0.56 )2 = 0.3136 atm 2

3

0

15. If pressure is applied to the following equilibrium, liquid  vapors the boiling point of liquid

(3 – 1) 2 5

1 1 5

(1) will increase. (2) will decrease. (3) may increase or decrease. (4) will not change.

[COCl 2 ] (1/ 5) 5 = = or 2.5 [CO][Cl 2 ] (1/ 5)´ ( 2 / 5) 2

13. A mixture of H2 and I2 in molecular proportion of 2 : 3 was heated at 440°C till the reaction H2 + I2  2HI reached equilibrium state. Calculate the percentage of iodine converted into HI. (KC at 440°C is 0.02) (1) 3.38% (2) 4.38% (3) 5.38% (4) 6.38%

Chapter 5_Chemical Equilibrium.indd 117

NH 4HS(s)  NH 3(g) + H 2S(g). If the observed pressure of the mixture is 1.12 atm at 106°C. What is the Kp of the reaction? (1) 0.2136 (2) 0.3136 (3) 0.4136 (4) 0.5126

formed

(1) 2 (2) 2.5 (3) 3.0 (4) 4



( 2 x /V )2 4x = = 0.02 ( 2 - x /V )( 3 - x /V ) ( 2 - x )( 3 - x )

2 x 2 = 50 + 25x 2 – 75x

12. At a certain temperature 2 mol of carbon monoxide and 3 mol of chlorine were allowed to reach equilibrium according to the reaction CO + Cl 2  COCl 2 in a 5 L ves­ sel. At equilibrium, if 1 mol of CO is present then equi­ librium constant for the reaction is



[HI]2 [H 2 ][I 2 ]

14. Ammonium hydrogen sulphide dissociates according to the equation,

x = 2.32 or x = 0.943 ( rejected as x £ 1)



KC =

I 2  2HI 3 0 3- x 2x V V

2 x 2 = 25 ( 2 + x 2 – 3x ) 23x 2 – 75x + 50 = 0

H2 + 2 2-x V

Solution (1) Boiling point of a liquid is the temperature at which vapor pressure became equal to atmospheric pressure. If the pressure is applied to the above equilibrium the reaction will go to the backward direction, that is, vapor pressure decreases hence the boiling point increase. 16. The volume of a closed reaction vessel in which the equilibrium 2SO2(g) + O2(g)  2SO3(g) sets is halved. Now

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118

OBJECTIVE CHEMISTRY FOR NEET (1) The rates of forward and backward reactions will remain the same. (2) The equilibrium will not shift. (3) The equilibrium will shift to the right. (4) The rate of forward reaction will become more than that of reverse reaction and the equilibrium will shift to the right.

Solution (4) In this reaction, 2SO2(g ) + O 2(g )  2SO3(g ) three moles (or volumes) of reactants are converted into two moles (or volumes) of products, that is, there is a decrease in volume and so if volume of the reaction vessel is halved, the equilibrium will be shifted to the right, that is, more product will be formed and the rate of for­ ward reaction will increase, that is, double than that of reverse reaction. 17. The enthalpies of two reaction are DH1 and DH2 (both positive) with DH2 > DH1. If the temperature of reacting system is increased from T1 to T2, predict which of the following alternatives is correct? (1)

K 1¢ K 2¢ K 1¢ K 2¢ = (2) > K1 K 2 K1 K 2

K¢ K¢ (3) 1 < 2 (4) None of these K1 K 2

Solution (2) The equilibrium constant varies only with temperature. A catalyst can increase the rate of reaction (it reduces the time taken to reach at equilibrium), however the point of equilibrium will remain the same and the value of equilibrium constant will not be affected. 19. The reaction 3O2  2O3 DH = +69,000 calories is favored by (1) (2) (3) (4)

Solution (2) According to Le Chatelier’s principle, formation of ozone is favored by high temperature (endothermic reaction) and high pressure. 20. The equilibrium composition for the reaction is PCl 3(g ) + Cl 2(g ) → PCl 5(g ) 0.20 0.05 0.40 mol L-1

K 1¢ K 2¢ < K1 K 2

(2) 0.48 mol L−1 (4) 1.20 mol L−1

Solution (1) From the given data, equilibrium constant can be calcu­ lated as 0.4 KC = = 40 0.2 × 0.05

18. The equilibrium constant at 323°C is 1000. What would be its value in the presence of a catalyst in the forward reaction?

If 0.25 mol of Cl2 is added, the equilibrium will go to the forward direction as per Le Chatelier’s principle to con­ sume Cl2. PCl 3 At t = t At t = t eq

A + B  C + D + 38 kcal (1) 1000 × concentration of catalyst (2) 1000 1000 (3) concentration of catalyst (4) Cannot be predicted

If 0.25 mol L−1 of Cl2 is added at same temperature. Find equilibrium concentration of PCl5 (1) 0.525 mol L−1 (3) 0.56 mol L−1

Solution (3) As the temperature of reacting system is increased, the ­equilibrium constant of reaction will also increase for endothermic reactions. So, for the given two reactions on increasing the temperature by equal amounts, the correct expression is

high temperature and low pressure. high temperature and high pressure. low temperature and high pressure. low temperature and low pressure.



+

Cl 2

0.2 0.05 + 0.25 0.2 – x 0.3 – x



PCl 5 0.4 0.4 + x

KC would remain same as temperature is constant (0.4 + x ) = 40 Þ x = 0.125 (0.2 - x )(0.3 - x )



Therefore, concentration of PCl5 is = 0.4 + 0.125 = 0.525 mol L-1

Practice Exercises Level I Law of Mass Action 1. The rate of the elementary reaction H 2(g ) + I 2(g )  2HI(g) at 25°C is given by 1.7 × 10−18 [H2] [I2]. The rate of

Chapter 5_Chemical Equilibrium.indd 118

decomposition of gaseous HI to H2(g) and I2(g) at 25°C is 2.4 × 10−21 [HI]2. Equilibrium constant for 1 1 H 2(g ) + I 2(g )  HI(g) is 2 2 (1) 708 (2) 354 (3) 0.0014 (4) 26.6

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Chemical Equilibrium 2. Which one of the following statements is correct? (1) (2) (3) (4)

Reversible reactions do not proceed to completion. Irreversible reactions do not proceed to completion. Both proceed to completion. Neither of the two proceeds to completion.

3. According to law of mass action, the rate of a reaction is directly proportional to (1) (2) (3) (4)

molarities of the reactants. normalities of the reactants. molalities of the reactants. mole fractions of the reactants.

4. In the chemical reaction, N 2(g ) + 3H 2(g )  2NH 3(g) at equilibrium point, state whether (1) equal volumes of N2 and H2 are reacting. (2) equal masses of N2 and H2 are reacting. (3) the reaction has stopped. (4) the same amount of ammonia is formed as is decomposed into N2 and H2.

Relation between Kp and KC, Characteristics of Equilibrium Constant K 5. For the reaction: NH +4 + 2H 2O  NH 4OH + H 3O+ ; K 1 = 5.56 × 10 -10 H 2O + H 2O  H 3O+ + OH - ; K 2 = 1 × 10 -14

Hence K3 of the reaction NH 4OH  NH +4 + OH - is (1) 1.8 × 105 (2) 1.8 × 10−5 (3) 5.56 × 104 (4) 5.56 × 10−4

6. What will be the equilibrium constant at 717 K for the 1 1 reaction HI(g)  H 2(g) + I 2(g) if its value for the 2 2 reaction H 2(g)+I 2(g)  HI(g) at 717 K is 64?

119

9. If K1 and K2 are the equilibrium constants of the equi­ libria (I) and (II) respectively, what is the relationship between the two constants? 1 (I) SO2(g ) + O2(g )  SO3(g ); K 1 2 (II) 2SO 3(g )  2SO 2(g ) + O 2(g ); K 2 (1) ( K 1 )2 = (3) K 1 =

1 (2) K 2 = ( K 1 )2 K2

1 (4) K1 = K2 K2

10. In which one of the following reactants, Kp is less than KC? (1) 2SO3(g)  2SO 2(g)+O 2(g ) (2) N 2(g)+3H 2(g)  2NH 3(g) (3) PCl 5(g)  PCl 3(g)+Cl 2(g ) (4) H 2(g) + I 2(g)  2 HI(g ) 11. For which one of the following reactions, value of equi­ librium constant depends upon the units of concentra­ tion? (1) COCl 2(g)  CO(g)+Cl 2(g ) 1 1 (2) NO(g)  N 2(g)+ O2(g ) 2 2 (3) Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) (4) CH3COOH(l) + C2H5OH(l)  CH3COOC2H5(l) + H2O(l) 12. For a homogeneous chemical reaction, Kp = KC when (1) ∆n = 0 (2) ∆n = 1 (3) ∆n = 2 (4) ∆n = ∞ 13. For the following equilibrium N2O4 (g)  2NO2(g), Kp is found to be equal to KC. This is attained when (1) T = 1 K (2) T = 12.18 K (3) T = 27.3 K (4) T = 273 K

14. The equilibrium constant for the reaction, Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s) and Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag(s) are K1 and K2 respectively. Then the 1 7. The ratio of Kp/KC for the reaction, CO(g) + O2(g)  CO2(g) equilibrium constant for the reaction Zn(s) + 2Ag+(aq) 2 1  Zn2+(aq) + 2Ag(s) will be CO(g) + O2(g)  CO2(g) is 2 (1) K1 + K2 (2) K1K2 (1) 1 (2) RT (3) K1 / K2 (4) K1 − K2 (3) (RT)1/2 (4) (RT)−1/2 (1) 64 (2) 8 (3) 1/64 (4) 1/8

8. For the equilibrium: (I) C6H5COOH + H2O  C6H5COO– + H3O+ ; K1 = 6.30 × 10-5 (II) C6H5COOH + OH−  C6H5COO– + H2O ; K2 = 6.30 × 109 (III)  H2O + H2O  H3O+ + OH− ; K3 = ?

K3 using above equilibria is (1) (6.30)2 × 104 (2) 1.0 × 10−14 (3) 1 × 1014 (4) (6.30)–2 × 10–4

Chapter 5_Chemical Equilibrium.indd 119

15. In which of the following reactions, the equilibrium constant will have no units? 1 1 (1) NO(g)  N 2(g ) + O2(g ) 2 2 (2) H 2(g) + I 2(g)  2 HI(g ) (3) CO(g) + H 2O(g)  CO2(g ) + H 2(g) (4) All of the above

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120

OBJECTIVE CHEMISTRY FOR NEET

Reaction Quotient (Q), Extent of Reaction 16. A small value of equilibrium constant shows that (1) the reaction is more in the forward direction and less in the backward direction. (2) the reaction is less in the forward direction and more in the backward direction. (3) the reaction is very little in the forward as well as backward direction. (4) the reaction is taking place at high temperature. 17. In which of the following cases, does the reaction go fur­ thest to completion? (1) K = 103 (2) K = 10-2 (3) K = 105 (4) K = 1

18. At 448°C, the equilibrium constant (KC) for the reaction

H 2(g ) + I 2(g )  2HI(g) is 50.5. If we start with 2 × 10−2 mol of HI, 1 × 10−2 mol of H2 and 3 × 10−2 mol I2 in a 2 L vessel, then at 448°C (1) (2) (3) (4)

reaction is in equilibrium. reaction will proceed in the forward direction. reaction will proceed in the backward direction. no reaction will take place.

Calculation of K in Different Reactions and Degree of Dissociation 19. For the following equilibrium 2Fe(s) + 3H2O(g)  Fe2O3(s) + 3H2(g), KC = 27. Hence ratio of molar concen­ tration of H2(g) and H2O(g) is (1) 3 (2) 1 (3) 9 (4) 27 20. The following reaction was allowed to come to equili­ brium: A(g) + 2B(g)  C(g). The initial amounts of reac­ tants placed into a 5 L in both vessels were 1 mol of A and 1.8 mol of B. After the reaction reached equilibrium, 1 mol of B was found. Calculate the value of KC for this reaction. (1) 0.06 (2) 5.1 (3) 17 (4) 19 21. The decomposition of N2O4 into NO2 is carried out at 280 K in chloroform. When equilibrium has been estab­ lished, 0.2 mol of N2O4 and 2 × 10−3 mol of NO2 are pre­ sent in 2 L of the solution. The equilibrium constant for the reaction, N 2O4(g)  2NO2(g), is (1) 1 × 10−2 (2) 2 × 10−3 (3) 1 × 10−5 (4) 2 × 10−5 22. An equilibrium mixture for the reaction 2H2S(g)  2H2(g) + S2(g) had one mole of hydrogen sulphide 0.2 mol of H2 and 0.8 mol of S2 in a 2 L vessel. The value of KC in mol L−1 is (1) 0.004 (2) 0.016 (3) 0.080 (4) 0.160

Chapter 5_Chemical Equilibrium.indd 120

23. Two samples of HI each 5 g were taken separately in two vessels of volume 5 L and 10 L respectively at 27°C. The extent of dissociation of HI will be (1) more in 5 L vessel. (2) more in 10 L vessel. (3) equal in both vessel. (4) nill. 24. At a certain temperature, the degree of dissociation of PCl5 is a. Starting with 1 mol of PCl5, the total number of moles present at equilibrium will be (1) 1 − a (2) 1 + a (3) 2a (4) 1 + 2a 25. 2.2 mol of phosphorous pentachloride were taken in a closed vessel and dissociated into phosphorous trichlo­ ride and chlorine. At equilibrium, the total number of moles of the reactant and the products was 2.53. The degree of dissociation is (1) 0.33 (2) 0.165 (3) 0.15 (4) 0.30 26. The equilibrium constant for the reaction, H2(g) + I2(g)  2HI(g) is 64. If the volume of the container is reduced to half of the original volume, the value of the equilib­ rium constant will be (1) 16 (2) 32 (3) 64 (4) 128 27. 2 mol of N2 and 2 mol of H2 are taken in a closed ves­ sel of 5 L capacity and suitable conditions are provided. When the equilibrium is reached, it is found that 0.6 mol of H2 is used up. The equilibrium concentration of NH3 is (1) 0.4 (2) 0.08 (3) 0.8 (4) 0.24 28. The normal vapor density of PCl5 is 104.25. At 250°C, its vapor density is 57.9. The degree of dissociation of PCl5 at this temperature will be (1) 40% (2) 60% (3) 80% (4) 90% 29. 10 mol of hydrogen and 3 mol of nitrogen are sealed in a 4 L vessel. The vessel is heated at constant temperature. At equilibrium the pressure has fallen to 2/3rd of its origi­ nal value. The numerical value of KC for N2 + 3H2  2NH3 is (1) 1/2 (2) 1 (3) Cannot be calculated. (4) None of these. 30. A reaction takes place in two steps with equilibrium con­ stants 10−2 for slow step and 102 for fast step. The equilib­ rium constant of the overall reaction will be (1) 104 (2) 10−4 (3) 1 (4) 10−2

1/4/2018 5:11:32 PM

Chemical Equilibrium 31. 1 mol of ethanol is treated with 1 mol of ethanoic acid at 25°C. One-fourth of the acid changes into ester at equi­ librium. The equilibrium constant for the reaction will be (1) 1/9 (2) 4/9 (3) 9 (4) 9/4 32. In the reaction, A(s) + B(g) + Heat  2C(s) + 2D(g), equilibrium is established. The pressure of B is doubled to re-establish the equilibrium. The factor by which D is changed is

(3) no effect. (4) reaction stops. 40. According to Le Chatelier’s principle, if heat is given to solid–liquid system, then (1) (2) (3) (4)

(1) a catalyst is used. (2) an adsorbent is used to remove SO3 as soon as it is formed. (3) a low pressure is used. (4) small amounts of reactants are used.

33. For the reaction, 2NH3(g)  N2(g) + 3H2(g), the units of Kp will be

Thermodynamics of Equilibrium 1 3 N (g) + H2 (g)  NH3 (g) 2 2 2 is –16.5 kJ mol–1. Hence, thermodynamic equilibrium constant is

42. Consider the reaction, PCl5(g)  PCl3(g) + Cl2(g) in a closed container at equilibrium. At a fixed temperature what will be the effect of adding more PCl5 on the equi­ librium concentration of Cl2(g)? (1) (2) (3) (4)

34. ∆Go at 25°C for the reaction

(1) 1.08 (2) 7.80 × 102 (3) 4.57 × 106 (4) 7.98 × 1034 35. When equilibrium is attained,

(1) –115 kJ (2) +115 kJ (3) 166 kJ (4) –166 kJ

Le Chatelier’s Principle

(1) increases. (2) decreases. (3) remains unaffected. (4) first increases and the decreases. 44. Consider the following equilibrium in a closed container N2O4(g)  2NO2(g). At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statements holds true regarding the equilib­ rium constant (Kp) and degree of dissociation (a)? (1) Neither Kp nor a changes. (2) Both Kp and a change. (3) Kp changes but a does not. (4) Kp does not change but a changes.

37. In which of the following reaction, pressure has no effect on equilibrium? (1) N2O4(g)  2NO2(g) (2) 2SO2(g) + O2(g)  2SO3(g) (3) CO2(g) + H2(g)  CO(g) + H2O(g) (4) N2(g) + 3H2(g)  2NH3(g) 38. For the reaction H2(g) + I2(g)  2HI(g), the equilibrium constant Kp changes with (1) total pressure. (2) catalyst. (3) the amounts of H2 and I2 present. (4) temperature. 39. The effect of increasing the pressure on the following gase­ ous equilibrium reaction 2A(g) + 3B(g)  3A(g) + 2B(g) is (1) forward reaction will be favored. (2) backward reaction will be favored.

Chapter 5_Chemical Equilibrium.indd 121

It decreases. It increases. It remains unaffected. It cannot be predicted without the value of Kp.

43. At a constant temperature, if a small amount of one of the products is removed, the equilibrium constant

(1) ∆Go = 0 (2) E o = 0 (3) K = 0 (4) None of these 36. The equilibrium constant for a reaction is 1022 at 300 K, the standard free energy change for this reaction is

quantity of solid will reduce. quantity of liquid will reduce. temperature will increase. temperature will decrease.

41. The equilibrium 2SO2(g) + O2(g)  2SO3(g) shifts forward if

(1) 2 (2) 3 (3) 2 (4) 3

(1) atm (2) (atm)3 (3) (atm)−2 (4) (atm)2

121

45. In which manner will increase of pressure affect the fol­ lowing equation?

(1) (2) (3) (4)

C(s) + H2O(g)  CO(g) + H2(g)

Shift in the forward direction. Shift in the reverse direction. Increase in the yield of hydrogen. No effect.

46. The equilibrium constant K for the reaction 2HI(g)  H2(g) + I2(g) at room temperature is 2.85 and that at 698 K is 1.4 × 10−2. This implies that the forward reaction is (1) exothermic. (2) endothermic. (3) exergonic. (4) unpredictable.

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OBJECTIVE CHEMISTRY FOR NEET

47. In which of the following reactions, the equilibrium is shifted to the right by increasing the temperature? (1) N2(g) + 3H2(g)  2NH3(g); ∆H = −ve (2) CaCl2(g) + aq  CaCl2(aq) + Heat (3) 2NO(g) − Heat  N2(g) + O2(g) (4) NH4Cl(s) + H2O  NH4+(aq) + Cl−(aq); ∆H =+ve

(1) [Cl2] > [PCl3] (2) [Cl2] > [P4] (3) [P4] > [Cl2] (4) [PCl3] > [P4] 55. The vessel containing the following equilibrium N2O4  2NO2 is compressed. This will result in (1) increase in the concentration of N2O4. (2) increasing in the formation of NO2. (3) no change in the concentration of NO2 or N2O4. (4) change in the equilibrium constant.

48. An exothermic reaction which proceeds with decrease in volume will give maximum yield at (1) (2) (3) (4)

low pressure, low temperature. high pressure, high temperature. high pressure, low temperature. low pressure, high temperature.

56. For equilibrium, 2NO2(g)  N2O4(g) + 14.6 kcal, increase of temperature would (1) (2) (3) (4)

49. In the reaction CO(g) + H2O(g)  CO2(g) + H2(g) + x kJ, the equilibrium will shift in the forward direction if (1) (2) (3) (4)

temperature is increased. pressure is increased. temperature is decreased. pressure is decreased.

50. Which of the following factors will change the value of equilibrium constant for the reaction between N2 and O2? (1) (2) (3) (4)

increasing concentration of N2 or O2. increasing the pressure. increasing the temperature. adding a catalyst.

51. For the chemical reaction 3X(g) + Y(g)  X3Y(g) the amount of X3Y at equilibrium is affected by (1) (2) (3) (4)

(1) (2) (3) (4)

57. For the reaction I2(g)  2I(g), KC = 37.6 × 10−6 at 1000 K. If 1 mol of I2 is introduced into a 1 L flask at 1000 K, at equilibrium (1) [I2] > [I] (2) [I2] = [I] (3) [I2] < [I] (4) unpredictable.

Level II Law of Mass Action 1. A reversible reaction is said to have attained equilibrium when (1) the backward reaction stops. (2) both backward and forward reactions stop. (3) both backward and forward reactions starts taking place at equal speed. (4) the concentrations of each of the reactants and products becomes equal.

temperature and pressure. pressure only. temperature only. temperature, pressure and catalyst.

52. Consider the equilibrium reaction: N2(g) + 3H2(g)  2NH3(g): ∆H = −98 kJ. Apply Le Chatelier’s principle and check which of the following is the correct effect of the increase in temperature? Concentration of N2 decreases. Concentration of H2 decreases. Concentration of NH3 increases. Concentration of NH3 decreases.

53. With increase in temperature, equilibrium constant of a reaction (1) always increases. (2) always decrease. (3) may increase or decrease depending upon whether np > nr or np < nr. (4) may increase or decrease depending upon whether reaction is exothermic or endothermic. 54. The equilibrium P4(g) + 6Cl2(g)  4PCl3(g) is attained by mixing equal moles of P4 and Cl2 in an evacuated vessel. Then at equilibrium

Chapter 5_Chemical Equilibrium.indd 122

favor the formation of N2O4. favor the decomposition of N2O4. not after the equilibrium. stop the reaction.

2. For the reaction A + 2B → C + D if active mass of A is kept constant and active mass of B is tripled, the rate of reaction will become (1) three times. (2) six times. (3) eight times. (4) nine times. 3. A liquid is in equilibrium with its vapor at its boiling point. On the average, the molecules in the two phases have equal (1) (2) (3) (4)

intermolecular forces. potential energy. kinetic energy. both kinetic energy and potential energy.

Relation Between Kp and KC and Characteristics of Equilibrium Constant K 4. For the following gaseous equilibria X, Y and Z at 300 K

X: 2SO2 + O2  SO3

1/4/2018 5:11:35 PM

Chemical Equilibrium 10. Which oxide of nitrogen is most stable at 2739 K?



Y: PCl 5  PCl 3 + Cl 2



Z: 2HI  H 2 + I 2



ratio of Kp and KC in the increasing order is (1) X = Y = Z (2) X < Y < Z (3) X < Z < Y (4) Z < Y < X

5. Select the reaction for which the equilibrium constant is written as [MX3]2 = K[MX2]2 [X2] 1 (1) MX 3  MX 2 + X 2 (2) 2MX 3  2MX 2 + X 2 2 1 (3) 2MX 2 + X 2  2MX 3 (4) MX 2 + X 2  MX 3 2 6. Consider the equilibrium reactions, K1 +   H 3PO 4    H + H 2PO 4 K2 2+   H 2PO 4-    H + HPO4 K

3+ 3   HPO24-    H + PO4

The equilibrium constant, K for the following dissocia­ K 3+  tion H 3PO 4    3H + PO 4 K1 (2) K 1K 2 K 3 K 2K 3

(1)

K2 (3) K 1K 3



(4) K1 + K2 + K3

7. Hydrogen is produced industrially by the steam hydro­ carbon reforming process as given by H2O(g) + CH4(g)  CO(g) + 3H2(g). If KC = 3.8 × 10-3 at 1000 K, what is the value of Kp at the same temperature? (1) 22.8 (2) 25.61 (3) 9.8 (4) 37.8 8. For 2A + B2  2AB; KC = K1 and 2AB + B2  A2B4; KC = K2. 1 The value of equilibrium constant for A + B2  A 2B4 is 2 (1) K1K2 (2 K1/K2 (3)

K 1K 2 (4) None of these

Reaction Quotient (Q) and Extent of Reaction 9. For the reversible reaction 2NO(g) + O2(g)  2NO2(g), net rate is æ dx ö 3 2 2 ç ÷ = 2.6 ´ 10 [NO] [O2] – 4.1 [NO2] è dt ønet



 



If a reaction mixture contain 0.01 mol each of NO and O2 and 0.1 mol of NO2 in 1 L closed flask, then above reac­ tion is (1) (2) (3) (4)

shifted in forward direction. shifted in backward direction. In equilibrium. given values are incomplete.

Chapter 5_Chemical Equilibrium.indd 123

123

(1) N2O5  2N2 + 5O2; KC = 1 × 1034 (2) 2N2O  2N2 + O2; KC = 1 × 1032 (3) 2NO  N2 + O2; KC = 1 × 1035 (4) NO2  N2 + 2O2; KC = 1 × 1015 11. For the following reactions, occurring at 500°K, arrange them in order of increasing tendency to proceed to com­ pletion (least → greatest tendency) (I) 2NOCl  2NO + Cl2; Kp = 1.7 × 10−2, (II) 2SO3  2SO2 + O2; Kp = 1.3 × 10−5, (III) 2NO2  2NO + O2; Kp = 5.9 × 10−5 (1) II < I < III (2) I < II < III (3) II < III < I (4) III < II < I 12. When two reactants A and B are mixed to give products C and D, the reaction quotient Q at the initial stages of the reaction (1) is zero. (2) decrease with time. (3) is independent of time. (4) increases with time.

Calculation of K in Different Reactions and Degree of Dissociation 13. 1 mol of N2 and 3 mol of H2 are mixed in 1 L flask. If 50% N2 is converted into ammonia by the reaction, N2(g) + 3H2(g)  2NH3(g) then the total number of moles of gas at equilibrium is (1) 1.5 (2) 3.0 (3) 4.5 (4) 6.0 14. The equilibrium constant (Kp) for the reaction 2SO2 + O2  2SO3 at 1000°K is 3.5. What would be the partial pres­ sure of O2 gas with equal moles of SO2 and SO3? (1) 0.29 atm (2) 3.5 atm (3) 0.53 atm (4) 1.87 atm 15. The value of K for the reaction 2A(g)  B(g) + C(g) at 750 K and 10 atm, pressure is 2.86. The value of K at 750 K and 20 atm is (1) 28.6 (2) 5.72 (3) 2.86 (4) 11.44 16. XY2 dissociates as XY2(g)  XY(g) + Y(g) when the initial pressure of XY2 is 600 mm of Hg, the total equilibrium pressure is 800 mm of Hg. K for the reaction assuming that the volume of the system remains unchanged is (1) 50 (2) 100 (3) 166.6 (4) 400 17. For the reaction N2O4(g)  2NO2(g) the reaction con­ necting the degree of dissociation (1) of N2O4(g) with its equilibrium constant Kp is Kp /p Kp (1) a = (2) a = 4 + Kp /p 4+ Kp

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124

OBJECTIVE CHEMISTRY FOR NEET 1/2

é Kp /p ù (3) a = ê ú êë 4 + K p / p úû

1/2

é Kp ù (4) a = ê ú êë 4 + K p úû

18. There is 50% dimer formation of benzoic acid (C6H5COOH) in benzene solution 2C6H5COOH  (C6H5COOH)2,







hence abnormal molecular weight of benzoic acid ( the­ oretical value = 122 g mol−1) is (1) 61 g mol−1 (2) 244 g mol−1 (3) 163 g mol−1 (4) 81 g mol−1

(1) 87% (2) 13% (3) 17% (4) 29% 20. Which one of the following statements is incorrect about chemical equilibrium? (1) Chemical equilibrium is attained whether we start with reactants or products. (2) Chemical equilibrium is dynamic in nature. (3) Chemical equilibrium CaCO3(s)  CaO(s) + CO2 is attained when CaCO3(s) is heated in an open vessel. (4) At equilibrium, the concentration of each of the reactants and products becomes constant. 21. Pure ammonia is placed in a vessel at a temperature where its dissociation is appreciable. At equilibrium, Kp does not change significantly with pressure. a does not change with pressure. concentration of NH3 does not change with pressure. concentration of H2 is less than that of N2.

22. In a closed container at 1 atm pressure, 2 mol of SO2(g) and 1 mol of O2(g) were allowed to react to form SO3(g) under the influence of a catalyst. Reaction 2SO2(g) + O2(g)  2SO3(g) occurred. At equilibrium it was found that 50% of SO2(g) was converted to SO3(g). The partial pressure of O2 (g) at equilibrium will be (1) 0.66 atm (2) 0.493 atm (3) 0.33 atm (4) 0.20 atm

27. The equilibrium constant, Kp for the reaction A  2B, is related to degree of dissociation of a and total pressure p as (1)

4a 2 p 4a 2 p 2 (2) 2 1-a 1-a 2

(3)

4a 2 p 2 4a 2 p (4) 1-a 1-a 2

28. A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. The value of K if the total pressure at equilibrium is 0.8 atm is (1) 1.8 atm (2) 3 atm (3) 0.3 atm (4) 0.18 atm 29. In the dissociation of N2O4 into NO2, (1 + x) varies with the vapor densities ratio as given by ( D / d ) , where, x is degree of dissociation, D and d are initial and final vapor densities respectively (1)

(1 + x)

(D/d)

(3)

(D/d)

(4) (1 + x)

(1 + x) (D/d)

30. In the above question, x varies with ( D / d ) according to (1)

(2) x

(2) p² atm (4) 2p atm

(1) 53.3% (2) 106.6% (3) 26.7% (4) None

(2)

(1 + x)

x

2

24. The vapor density of N2O4 at a certain temperature is 30. What is the percentage dissociation of N2O4 at this tem­ perature?

Chapter 5_Chemical Equilibrium.indd 124

26. Vapor density of PCl5 is 104.16 but when heated at 230°C its vapor density is reduced to 62. The degree of dissocia­ tion of PCl5 at this temperature will be

(D/d)

23. On decomposition of NH4HS, the following equilibrium is established: NH4HS(s)  NH3(g) + H2S(g). If the total pressure is p atm, then the equilibrium constant Kp is equal to (1) p atm (3) p²/4 atm2

(1) 1 (2) 1 + a (3) 2 (4) 2 + a

(1) 6.8% (2) 68% (3) 46% (4) 64%

19. At 30°C, Kp for the dissociation reaction, SO2Cl2(g)  SO2(g) + Cl2(g) is 2.9 × 10–2 atm. If the total pressure is 1 atm, the degree of dissociation of SO2Cl2 is

(1) (2) (3) (4)

25. If fraction a of HI dissociates at equilibrium in the reac­ tion 2HI  H2 + I2 then, starting with 2 mol of HI, the total number of moles of the reactants and products at equilibrium is equal to

(D/d)

(D/d)

(4)

(3) x

x (D/d)

(D/d)

1/4/2018 5:11:40 PM

Chemical Equilibrium 31. If 0.2 mol of H2(g) and 2 mol of S(s) are mixed in a 1 dm3 vessel at 90°C, the partial pressure of H2S(g) formed according to the reaction, H2(g) + S(s)  H2S, Kp = 6.8 × 10−2 would be (1) 0.19 atm (3) 0.6 atm



(2) 0.36 atm (4) 0.072 atm

32. 1 mol of N2O4(g) at 300 K is kept in a closed container under 1 atm. It is heated to 600 K when 20% by mass of N2O4(g) decomposes to NO2(g). The resultant pressure is (1) 1.2 atm (2) 2.4 atm (3) 2.0 atm (4) 1.0 atm 33. 3 mol ammonia is heated in presence of 1 mol of neon. At equilibrium, the mole fraction of hydrogen is found to be 1 3 0.5. The value of (KC) × (V ) for NH 3(g )  N 2(g ) + H 2(g ) 2 2 3 (1) 27 (2) 8 (3) 3 3 (4)

Le Chatelier’s Principle 38. Consider the reaction, CaCO3(s)  CaO(s) + CO2(g) in a closed container at equilibrium. What would be the effect of addition of CaCO3 on the equilibrium concentration of CO2? (1) Increases (2) unpredictable (3) Decreases (4) remains unaffected 39. Consider the reaction, PCl5(g)  PCl3(g) + Cl2(g) in a closed container at equilibrium at a fixed temperature. What will be the effect of adding more PCl5 on the equi­ librium concentration of Cl2(g)? (1) (2) (3) (4)

34. For an isomerization reaction A  B, the tempera­ ture dependence of equilibrium constant is given by 2000 log K = 4 . The value of ∆So at 400 K is, therefore T (1) 4R (2) 5R (3) 400 R (4) 2000 R 35. Ice and water are in equilibrium at 273 K. Which of the following statements is correct? (1) G(ice) > G(H2O) (2) G(ice) < G(H2O) (3) G(ice) = G(H2O) = 0 (4) G(ice) = G(H2O) ≠ 0

(1) (2) (3) (4)

(1) (2) (3) (4)

the equilibrium will shift in the forward direction. the equilibrium will shift in the backward direction. the equilibrium will remain undisturbed. the equilibrium constant will change.

42. For which one of the following reactions is a combina­ tion of high pressure and high temperature helpful in obtaining a high equilibrium yield? (1) 2NF3(g)  N2(g) + 3F2(g) −54.40 kcal (2) N2(g) + 3H2(g)  2NH3(g)+ 22.08 kcal (3) Cl2(g) + 2O2(g)  2ClO2(g) −49.4 kcal (4) 2Cl2O7(g)  2Cl2(g) + 7O2(g) + 126.8 kcal

log10 K

43. Given the reaction 2X(g) + Y(g)  Z(g) + 80 kcal. Which combination of pressure and temperature gives the highest yield of Z at equilibrium? (1) 1000 atm and 500°C (2) 500 atm and 500°C (3) 1000 atm and 100°C (4) 500 atm and 100°C

1/T

(1) +4.606 cal (2) –4.606 cal (3) 2 cal (4) –2 cal 37. What are the signs for ∆H, ∆S and ∆G for the freezing of liquid water at –10°C?   ∆H  ∆S ∆G     ∆H  ∆S  ∆G (1) +  −  + (2) −  −  0 (3) −  +  − (4) −  −  −

Chapter 5_Chemical Equilibrium.indd 125

system in equilibrium. system not in equilibrium. homogeneous reaction. heterogeneous reaction.

41. 1 mol of helium is added to the dissociation equilibrium N2O4(g)  2NO2(g) − x kJ set up in a cylinder fitted with a piston in such a way that the piston is moved outwards to keep the total equilibrium pressure constant

36. Variation of log10 K with 1/T is shown by the following graph in which straight line is at 45°, hence ∆Ho is



It decreases. It increases. It remains unaffected. It cannot be predicted without the value of Kp.

40. Le Chatelier’s principle is applicable only to a

3 5

Thermodynamics of Equilibrium

125

44. Consider the reversible reaction,

HCN  H ++(aq ) + CN --(aq )



At equilibrium, the addition of CN−(aq) would (1) (2) (3) (4)

reduce HCN(aq) concentration. decrease H+ ion concentration. increase equilibrium constant. decrease equilibrium constant.

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126

OBJECTIVE CHEMISTRY FOR NEET

45. Calcium carbonate, CaCO3, dissociates in the following manner CaCO3(s)  CaO(s) + CO2(g ); ∆H = 110 kJ in a closed vessel. The pressure of carbon dioxide (1) (2) (3) (4)

increases if temperature is raised. increases on adding a catalyst. decreases if temperature is raised. increases if an inert gas is pumped keeping the tem­ perature constant.

46. Which one of the following will favor the reverse reaction in a chemical equilibrium? (1) (2) (3) (4)

Removal of one of the products regularly. Increasing the concentration of one of the reactants. Increasing the concentration of one of the products. Adding negative catalyst.

47. Dilute HCl is added to the following equilibrium at constant temperature

(1) (2) (3) (4)

CH3COOH  CH3COO− + H+

Concentration of CH3COO− will increase. Concentration of CH3COO− will decrease. Equilibrium constant will increase. Equilibrium constant will decrease. -

48. The equilibrium constant Br2  2 Br at 500 K and 700 K are 10–10 and 10–5 respectively. Hence the reaction is

(1) (2) (3) (4)

53. Consider following equilibria at 300 K and 400 K with their equilibrium constants: 300 K 400 K (I) A(g)  2B(g) Keq = 10 Keq = 5    (II) C(g)  D(g)

(1) increase. (2) decrease. (3) remains unchanged. (4) depend upon the amount of helium added. 50. The dissociation of PCl 5(g )  PCl 3(g ) + Cl 2(g ) is an endothermic reaction. Which of the following will favor the forward reaction? (1) Adding Cl2(g) to the equilibrium mixture at a con­ stant volume. (2) Compressing the gaseous mixture. (3) Increasing the volume of the gaseous mixture. (4) Decreasing the temperature. 51. If pressure is applied to the liquid vapor equilibria liquid  vapor , the boiling point of the liquid (1) will decrease. (2) will increase. (3) may increase or decrease. (4) will not change. 52. For the manufacture of ammonia by Haber’s process, that is, N 2(g ) + 3H 2(g )  2NH 3(g ); ΔH = -22 kcal the favorable conditions are

Chapter 5_Chemical Equilibrium.indd 126

Keq = 2 Keq = 5

Thus, (1) (2) (3) (4)

I is endothermic, II is exothermic. I is exothermic, II is endothermic. I and II both are endothermic. I and II both are exothermic.

54. In the lime kiln, the reversible reaction,

CaCO3(s)  CaO(s) + CO2(g) proceeds to completion because of (1) high temperature. (2) CO2 escapes. (3) CaO is removed. (4) low temperature.

55. If pressure is increased on the equilibrium Ice  Water which of the following occurs? (1) (2) (3) (4)

(1) endothermic. (2) exothermic. (3) fast. (4) slow. 49. The total pressure at equilibrium for the dissociation PCl 5(g )  PCl 3(g ) + Cl 2(g ) is 2 atm. 1 mol of helium is now introduced into the same vessel containing the reaction mixture. The equilibrium constant will

high temperature, high pressure. low temperature, low pressure. high temperature, low pressure. low temperature, high pressure.

More of the ice melts. More of the ice is formed. There is no change in the amounts of ice and water. Some water gets vaporized.

56. The value of KC for the reaction 2SO3(g)  2SO2(g)+O2(g) is 10 mol L−1 at a certain temperature T°K. The number of molecules of O2 present at equilibrium in a 10 L ves­ sel such that the active masses of SO3 and SO2 is same at equilibrium (1) 10 (2) 10 NA (3) 100 (4) 100 NA

(where NA = Avagadro Number)

57. Consider the water gas equilibrium reaction

C(s) + H2O(g)  CO(g) + H2(g). Which of the following statement is true at equilibrium? (1) If the amount of C(s) is increased, less water would be formed. (2) If the amount of C(s) is increased, more CO and H2 would be formed. (3) If the pressure on the system is increased by halving the volume, more water would be formed. (4) If the pressure on the system is increased by halving the volume, more CO and H2 would be formed.

58. Densities of diamond and graphite are 3.5 g cm−3 and 2.3 g cm−3 respectively. Increase of pressure on the equilibrium C(Diamond)  C(Graphite)

1/4/2018 5:11:43 PM

Chemical Equilibrium (1) (2) (3) (4)



favors backward reaction. favors forward reaction. have no effect. increase the reaction rate.

(1) 25 (2) 100 (3) 56 (4) 0.25

Previous Years’ NEET Questions 1. The following equilibrium constants are given:

1 1 (2) 8 16 1 (3) (4) 16 64 (AIPMT 2008) 5. The values of Kp1 and Kp2 for the reactions



X  Y + Z 

(1)





A  2B 

(2)



are in the ratio of 9:1. If degree of dissociation of X and A be equal, then total pressure at equilibrium (1) and (2) are in the ratio

N 2 + 3H 2  2NH 3 ; K 1 N 2 + O2  2NO ; K 2

(1) 1:1 (2) 3:1 (3) 1:9 (4) 36:1

1 H 2 + O 2  H 2O ; K 3 2

(AIPMT 2008)

The equilibrium constant for the oxidation of the NH3 by oxygen to give NO is (1)

K 1K 2 K 2 K 33 (2) K3 K1

K K2 (3) 2 3 K1



K 23 K 3 (4) K1

6. The dissociation constants for acetic acid and HCN at 25°C are 1.5 ´ 10 -5 and 4.5 ´ 10 -10 , respectively. The equilibrium constant for the equilibrium

2. The dissociation equilibrium of a gas AB2 can be repre­ sented as 2 AB2(g )  2 AB(g ) + B2(g )

(AIPMT 2009) 7. In which of the following reaction equilibrium K C and K p are not equal? (1) SO2(g ) + NO2(g )  SO3(g ) + NO (g )

The degree of dissociation is ‘x’ and is small compared to 1. The expression relating the degree of dissociation (x) with equilibrium constant Kp and total pressure p is

(2) H 2(g ) + I 2(g )  2HI (g) (3) 2C(s) + O2(g )  2 CO2(g ) (4) 2 NO2  N 2(g)+O2(g )

æ Kp ö æ 2K p ö (1) ç ÷ (2) ç ÷ è p ø è p ø 1/3

(AIPMT PRE 2010)

1/2

æ 2K p ö æ 2K p ö (3) ç ÷ (4) ç ÷ p è è p ø ø

(AIPMT 2008) −

3. If the concentration of OH ions in the reaction Fe(OH )3(s)  Fe3+ (aq ) + 3OH - (aq ) is decreased by 1/4 times, then equilibrium concentration of Fe3+ will be increased by (1) 4 times. (2) 8 times. (3) 16 times. (4) 64 times. (AIPMT 2008) 4. The value of equilibrium constant of the reaction 1 1 HI(g )  H 2(g ) + I 2(g ) 2 2

Chapter 5_Chemical Equilibrium.indd 127

CN - + CH 3COOH  HCN + CH 3COO - would be (1) 3.0 ´ 104 (2) 3.0 ´ 105 (3) 3.0 ´ 10 -5 (4) 3.0 ´ 10 -4

(AIPMT 2007, NEET 2017)



is 8.0. The equilibrium constant of the reaction H 2(g ) + I 2(g )  2HI(g ) will be (1)

59. The equilibrium constant for the reaction, H2 + I2  2HI at 700 K is 56. If 0.5 mol of hydrogen and 1 mol of iodine are added to the system at equilibrium, the value of equi­ librium constant will be



127

8. The reaction 2A(g)+ B(g)  3C(g) + D(g)







is begun with the concentrations of A and B both at an initial value of 1.00 M. When equilibrium is reached, the concentration of D is measured and found to be 0.25 M. The value for the equilibrium constant for this reaction is given by the expression. (1) [(0.75)3 (0.25)] ÷ [(0.50)2 (0.75)] (2) [(0.75)3 (0.25)] ÷ [(0.50)2 (0.25)] (3) [(0.75)3 (0.25)] ÷ [(0.75)2 (0.25)] (4) [(0.75)3 (0.25)] ÷ [(1.00)2 (1.00)] (AIPMT MAINS 2010)

9. For the reaction N2(g) +O2(g)  2NO(g), the equilibrium constant is K1. The equilibrium constant

1/4/2018 5:11:49 PM

128

OBJECTIVE CHEMISTRY FOR NEET

is K2 for the reaction 2NO(g) + O2(g)  2NO2(g). What is K for the reaction?



(1) 2.0 (2) 1.9 (3) 0.62 (4) 4.5

1 NO2(g )  N 2(g ) + O2(g ) 2

(AIPMT MAINS 2012)

1 1 (2) ( 2 K 1K 2 ) ( K 1K 2 )

(1)

 1  1 (4)   ( 4 K 1K 2 )  K 1K 2 

(3)

13. For the reversible reaction

1/2

N 2(g) + 3H 2(g)  2NH 3(g) + Heat

(AIPMT PRE 2011) 10. The value of ∆H for the reaction X2(g) + 4Y2(g)  2XY4(g)







is less than zero. Formation of XY4(g) will be favored at (1) (2) (3) (4)

high pressure and low temperature. high temperature and high pressure. low pressure and low temperature. high temperature and low pressure. (AIPMT PRE 2011)

11. Given that the equilibrium constant for the reaction 2SO2(g) + O2(g)  2SO3(g) has a value of 278 at a par­ ticular temperature. What is the value of the equilib­ rium constant for the following reaction at the same temperature? (1) 1.8 × 10−3 (2) 3.6 × 10−3 −2 (3) 6.0 × 10 (4) 1.3 × 10−5 (AIPMT MAINS 2012) 12. Given the reaction between two gases represented by A2 and B2 to give the compound AB(g). A 2(g ) + B2(g )  2 AB(g)

At equilibrium, the concentration



of A2 = 3.0 × 10−3 M



of B2 = 4.2 × 10−3 M



of AB = 2.8 × 10−3 M

If the reaction takes place in a sealed vessel at 527°C, then the value of KC will be

The equilibrium shifts in forward direction (1) (2) (3) (4)

by increasing the concentration of NH3(g). by decreasing the pressure. by decreasing the concentrations of N2(g) and H2(g). by increasing pressure and decreasing temperature. (AIPMT 2014)

14. For a given exothermic reaction, Kp and Kp′ are the equi­ librium constants at temperatures T1 and T2 respectively. Assuming that heat of reaction is constant in tempera­ tures range between T1 and T2, it is readily observed that (1) Kp > Kp′ (2) Kp < Kp′ (3) Kp = Kp′ (4) Kp = 1/Kp′ (AIPMT 2014) 15. If the value of an equilibrium constant for a particular reac­ tion is 1.6 × 1012, then at equilibrium the system will contain (1) (2) (3) (4)

mostly reactants. mostly products. similar amounts of reactants and products. all reactants.

16. If the equilibrium constant for N 2(g) + O2(g)  2NO(g) is K, then, the equilibrium constant for 1 1 N 2(g) + O2(g)  NO(g) will be 2 2 (1) K (2) K2 1 (3) K1/2 (4) K 2 (RE-AIPMT 2015)

Answer Key Level I  1. (4)

 2. (1)

 3. (1)

 4. (4)

 5. (2)

 6. (4)

 7. (4)

 8. (2)

 9. (1)

10.  (2)

11.  (1)

12.  (1)

13.  (2)

14.  (2)

15.  (4)

16.  (2)

17.  (3)

18.  (2)

19.  (1)

20.  (3)

21.  (3)

22.  (2)

23.  (3)

24.  (2)

25.  (3)

26.  (3)

27.  (2)

28.  (3)

29.  (4)

30.  (3)

31.  (1)

32.  (3)

33.  (4)

34.  (2)

35.  (4)

36.  (1)

37.  (3)

38.  (4)

39.  (3)

40.  (1)

41.  (2)

42.  (2)

43.  (3)

44.  (4)

45.  (2)

46.  (1)

47.  (4)

48.  (3)

49.  (3)

50.  (3)

51.  (1)

52.  (4)

53.  (4)

54.  (3)

55.  (1)

56.  (2)

57.  (1)

Chapter 5_Chemical Equilibrium.indd 128

1/4/2018 5:11:50 PM

Chemical Equilibrium

129

Level II  1. (3)

 2. (4)

 3. (3)

 4. (3)

 5. (3)

 6. (2)

 7. (2)

 8. (3)

 9. (2)

10.  (4)

11.  (3)

12.  (4)

13.  (2)

14.  (1)

15.  (3)

16.  (2)

17.  (3)

18.  (3)

19.  (3)

20.  (3)

21.  (1)

22.  (4)

23.  (3)

24.  (1)

25.  (3)

26.  (2)

27.  (1)

28.  (1)

29.  (1)

30.  (2)

31.  (2)

32.  (2)

33.  (3)

34.  (1)

35.  (4)

36.  (2)

37.  (4)

38.  (4)

39.  (2)

40.  (1)

41.  (1)

42.  (3)

43.  (3)

44.  (2)

45.  (1)

46.  (3)

47.  (2)

48.  (1)

49.  (3)

50.  (3)

51.  (2)

52.  (4)

53.  (2)

54.  (2)

55.  (1)

56.  (4)

57.  (3)

58.  (1)

59.  (3)

 7. (3)

 8. (1)

 9. (4)

Previous Years’ NEET Questions  1. (2)

 2. (3)

 3. (4)

 4. (3)

 5. (4)

 6. (1)

11.  (3)

12.  (3)

13.  (4)

14.  (1)

15.  (2)

16.  (3)

10.  (1)

Hints and Explanations Kp

Level I



1. (4) For the reaction H 2(g ) + I 2(g )  2HI(g); K = K f Kb

8. (2) K3 for the equilibria H2O + H2O  H3O+ + OH− can be obtained from Eq. (I) − Eq. (II). Therefore,



Therefore, for the reaction 1 H 2(g ) + 1 I 2(g )  HI(g) 2 2  1.7 × 10 -18  K′= K =   2.4 × 10 -21 

Therefore,

KC

K3 =

1/2



NH +4 + 2H 2O  NH 4OH + H 3O+ ; K 1 = 5.56 × 10 -10 (1)



 



K3 of the reaction NH 4OH  NH +4 + OH obtained from Eq. (2) − Eq. (1). Therefore,

H 2O + H 2O  H 3O+ + OH - ; K 2 = 1 × 10 -14 (2)

K2 for the reaction HI(g)  obtained from −Eq. (1)/2  1 K2 =    K1 

1/2

1 1 = K 2 or K 12 = 2 (K 1 ) K2

Chapter 5_Chemical Equilibrium.indd 129

13. (2) We know

Δng

K p = K C ( RT )

(1)



For the reaction N2O4(g)  2NO2(g); ∆ng = np - nR = 2 - 1 = 1



From Eq. (1), we get   Kp = KC(RT )1

1 1 = = 12.18 K R 0.0821 4. (2) Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s); K1(1) 1

Therefore, RT = 1 ⇒ T =

1 1 H 2(g) + I 2(g) can be 2 2



   Cu(s) + 2Ag+(aq)  Cu2+(aq) + 2Ag(s); K2(2)



K3 for the reaction Zn(s) + 2Ag+(aq)  Zn2+(aq) + 2Ag(s) can be obtained from Eq. (1) − Eq. (2). Therefore, K 3 = K 1 × K 2

1/2

=

1 8

1 O2(g)  CO2(g) 2

1  1 Δng = np - nR = 1 -  1 +  =  2 2 We know   K p = K C ( RT )Δng

10. (2) For reaction, N2(g) + 3H2(g)  2NH3(g) ∆n < 0, there­ fore, Kp < KC

(1)

 1 =   64 

7. (4) For the reaction CO(g) +



can be

K2 10 -14 = = 1.8 × 10 -5 K 1 5.56 × 10 -10

6. (4) We know H 2(g) + I 2(g)  HI(g); K1 = 64

K 1 6.30 × 10 -5 = = 10 -14 K 2 6.30 × 109

9. (1) The relationship between K1 and K2 can be obtained from −2 Eq. (I) = Eq. (II)

= 26.6

5. (2) We have

K3 =

= ( RT )-1/2

18. (2) For the reaction H 2(g ) + I 2(g )  2HI(g); K = 50.5 2

 2 × 10 -2   2  4 [HI]2 = = 1.33 Q= = -2 [H 2 ][I 2 ]  1 × 10   3 × 10 -2  3  2   2 

1/4/2018 5:11:53 PM

130

OBJECTIVE CHEMISTRY FOR NEET

As Q < K, therefore, reaction will proceed in the forward direction.



Given that: 3x = 0.6 ⇒ x = 0.2



Therefore, [NH 3 ] =

19. (1) For the reaction 2Fe(s) + 3H2O(g)  Fe2O3(s) + 3H2(g)  

Therefore,  

KC =

28. (3) The reaction involved is

[H 2 ]3 = 27 [H 2O]3

PCl 5  PCl 3 + Cl 2 At t = 0 At t = t eq

[H 2 ] = ( 27 )1/3 = 3 [H 2O]

20. (3) The reaction involved is A(g) + 2B (g)  C(g)         At t = 0 1 1.8 0 At t = t eq 1 - x 1.8 - 2 x x

Given that 1.8 – 2x = 1 ⇒ x = 0.4 KC =

[C] (0.4 / 5) = = 16.67 or 17(approx.) 2 2 [ A ][B]  0.6   1     5   5

21. (3) For the reaction N 2O4(g )  2NO 2(g ), equilibrium constant can be calculated as



  



pinitial pequilibrium

=

ninitial nequilibrium   

KC =

 0.2   0.8  [H 2 ]2[S 2 ]  2   2  22. (2)     KC = = = 0.016 2 [H 2S]2  1   2



0 a

0 a

104.25 = 1 + a ⇒ a = 0.80 or 80% 57.9

N 2 + 3H 2  2NH 3 At t = 0 3 10 0 At t = t eq 3 - x 10 - x 2-x

2

1 1-a

0 a

29. (4) The reaction involved is

2

PCl 5  PCl 3 + Cl 2

0 a

( V .D)normal nobserved 1 + a = = 1 ( V .D)observed nnormal

 2 × 10 -3   2     K = [NO ] =  = 10 -5 0.2 [ N 2O 4 ] 2

24. (2) The dissociation reaction involved is

1 1-a

Given that nnormal = 1 and nobserved = 1 + a

2 2 eq

At t = 0 At t = t eq

2 x 2 × 0.2 = 0.08 = 5 5

=

=

(10 + 3) 13 = ( 3 - x ) + (10 - 3x )+ 2 x 13 - 2 x

3 13 13 ⇒x= = 2 13 - 2 x 6

[NH 3 ]2 [N 2 ][H 2 ]3 ( 2 x/4)2  3 - x   10 - 3x     4  4 

3

=

4 x 2 × 16 = 8.4 ( 3 - x )(10 - 3x )3

30. (3) K = Kfast × Kslow = 102 × 10−2 = 1 31. (1) The reaction involved is C 2H5OH + CH 3COOH  CH 3COOC 2H5 + H 2O 1 At t = 0 At t = t eq 1 - x

Therefore, the total number at equilibrium = (1 – a) + a + a=1+a

1 1- x

0 x

25. (3) Let the degree of dissociation be a.



Given that x = 0.25





The equilibrium constant can be calculated as

The reaction involved is PCl 5  PCl 3 + Cl 2 At t = 0 At t = t eq

2.2 2.2 - x

0 x

0 x



At equilibrium, (2.2 – x) + x + x = 2.53 ⇒ x = 0.33



Therefore, degree of dissociation a =

Moles dissociated Initial moles

                 =

x 0.33 = = 0.15 2.2 2.2

27. (2) The reaction involved is

Chapter 5_Chemical Equilibrium.indd 130

(0.25)2 1 x2 = = (1 - x )2 (0.75)2 9

32. (3) We have K = constant =

( pD )2 (As temperature is (pB ) constant)

Given that: (pB)new = 2pB, then ( pD )new = 2 pD so that K remains constant.

34. (2) We know ΔG° = −RT lnK = −2.303 RT log K   −16.5 × 103 = −2.303 × 8.314 × 298 log K

N 2 + 3H 2  2NH 3 2 At t = 0 2 At t = teq 2 - x 2 - 3x

K=

0 x

0 2x



 

 K = 7.80 × 102

35. (4) At equilibrium ΔG = 0, E = 0

1/4/2018 5:11:55 PM

Chemical Equilibrium 36. (1) ΔG° = −2.303 RT log K

131



Equilibrium constant, K for the reaction K 3+  H 3PO 4    3H + PO4 can be obtained from Eq. (1) + Eq. (2) + Eq. (3)

39. (3) If pressure increases reaction goes in the direction where pressure or number of moles decreases. But in this case Δng = 0, therefore, number of moles are same in both the directions, hence, there will be no effect of pressure on equilibrium.



Therefore, K = K1 × K2 × K3

40. (1) When the solid  liquid system is heated, the quan­ tity of the solid will reduce because conversion of solid to liquid is an endothermic reaction. When heated, the solid melts and equilibrium shifts in the forward reaction.



   = −2.303 × 8.314 × 300 log (10 ) × 10 kJ    = −115 kJ 22

−3

43. (3) Equilibrium constant depends only on temperature and thus will remain unaffected. P4





t = 0

a   a     0





t = teq a − x  a − 6x   4x





8. (3) We have 2A + B2  2AB; K1(1)







57. (1) The reaction involved is I 2  2I 1 0 1 - x 2x

K C = 37.6 × 10 -6 =

1 Equilibrium constant, K for the reaction A + B2  A 2B4 2 1 1 can be obtained from Eq.(1) + Eq.( 2) 2 2 Therefore, K = ( K 1 × K 2 )1/2 = K 1K 2





2

[2x ] ⇒ x ≈ 0.03 (1 - x )

    [I] = 0.06 or [I2] > [I]

Level II

3

At t = teq   1 − x

3 − 3x 2x



Given: x = 0.5



Total moles at equilibrium = (1 – x) + (3 – 3x) + 2x



    = 4 – 2x = 4 – 2 × (0.5) = 3

Kp =

Rate ∝ [Active mass of A] [Active mass of B]2





Therefore,



(Rate)2 [ Active mass of A ] [ 3 Active mass of B]2 9 = = [ Active mass of A ][ Active mass of B]2 (Rate)1 1

For Z:



Kp KC Kp KC

= ( RT )-1; For Y:

Kp KC

= ( RT )1;

Therefore, X < Z < Y

K1 +   6. (2) We have H 3PO 4    H + H 2PO 4 (1)





  H 2PO    H + HPO (2)





K3 3+   HPO24-    H + PO4 (3)

Chapter 5_Chemical Equilibrium.indd 131

K2

( pSO2 )2 ( pO2 )

As pSO3 = pSO2 (as no. of moles are equal), therefore, pO2 =

1 1 = = 0.2857 or 0.29 atm K p 3.5

16. (2) We have              XY2  XY + Y

= ( RT )0 = 1

4

( pSO3 )2

15. (3) K is independent of pressure and depends only on tem­ perature. Thus, if temperature is constant, K is constant.

4. (3) We know K p = K C ( RT )Δng



0

14. (1) For the reaction 2SO2 + O2  2SO3



For X:

N 2 + 3H 2  2NH 3

At t = 0    1

2. (4) According to law of mass action,



[NO2 ]2 (0.1)2 = 2 [NO] [O 2 ] (0.01)2(0.01)

3 Also,    K = K f = 2.6 × 10 Kb 4.1 since Q > K, therefore, reaction will go in backward direction.

13. (2) We have

[I 2 ] @ 1



2AB + B2  A2B4; K2(2)



Q=

Therefore, [P4] > [Cl2]

At t = 0 At t = t eq

   = (3.8 × 10−3) (0.0821 × 1000)2 = 25.61

9. (2) For the reaction 2NO(g) + O2(g)  2NO2(g)

+ 6Cl2  4PCl3

54. (3)

7. (2) We know K p = K C ( RT )Dng

+

24





At t = 0     600    0   0





At t = teq    600 − x 



Given: (600 – x) + x + x = 800 or x = 200



Therefore, K =

x   x

( 200)2 x2 = = 100 (600 - x ) 400

1/4/2018 5:11:57 PM

132

OBJECTIVE CHEMISTRY FOR NEET 25. (3) The reaction is    2HI    H2 + I2

17. (3) We have N 2O4(g )  2NO2(g ) At t = 0 At t = t eq

Therefore, K p =

1 1 -a

( pNO2 )2 ( pN 2 O4 )

=

  

0 2a



( xNO2 × p )2

p

=

Total moles at equilibrium = (2 – 2x) + x + x = 2

26. (2) We have PCl 5  PCl 3 + Cl 2

( xN2 O4 × p )

2

Kp



At t = 0

1



At t = teq

1−a a   a



 K p /p  Kp Kp 2 4a 2 + ⇒ a = 4a 2 ⇒ a =   2 p p (1 - a )  4 + K p /p 

1/2

18. (3) The reaction involved is At t = 0

1

At t = t eq

1-a

A  2B

27. (1)

At t = 0

1    0



At t = teq

1 − a  2a 2

 2a  × p  ( pB )2  1 + a Kp = = (pA )  1 - a    ×p 1+ a 

0 a 2  a  1 -  2 = 1

( V .D.)normal nobserved = ( V .D.)observed nnormal

2     K p = 4a × p (1 - a 2 )

28. (1) The reaction involved is CO2(g ) + C(s)  2CO(g ) At t = 0 0.5 0 At t = t eq 0.5 - x 2x

122/2 (1 - 0.5/2) ⇒ M. wt. = 162.66 or 163 g mol -1 = M. wt./2 1 19. (3)     SO2Cl 2  SO2 + Cl 2



At t = 0

1

0   0





At t = teq

1 − a

a  a



Therefore, degree of dissociation can be calculated as



Given: (0.5 – x) + 2x = 0.8 ⇒ x = 0.3       K = ( 2 ´ 0.3) = 1.8 atm (0.2) 2

29. (1) The dissociation reaction is N 2O4  2NO2 At t = 0 1 0 At t = t eq 1 - x 2x

a2 = 2.9 × 10 -2 ⇒ a = 17% 1-a

         D = nfinal = 1 + x ⇒  D  = 1 + x d d ninitial 1

23. (3) The reaction is NH 4HS(s)  NH 3(g ) + H 2S(g ) p1

p1

We know that: 2p1 = p or p1 = p/2



Therefore, K p = ( pNH ) ( pH S ) = p = p 3 2 4 2 1

2

24. (1) The reaction is  N 2O4  2NO 2



At t = 0 1

0





At t = teq 1 − a

2a

30. (2) x = 31. (2)



n 1+a ( V .D.)normal (92/2) = observed = ⇒ = 1 + a ⇒ a = 53.3% nnormal 1 30 ( V .D.)observed

Chapter 5_Chemical Equilibrium.indd 132

0   0

104.16 ( V .D.)normal nobserved 1 + a = = ⇒ - 1 = a ⇒ a = 68% 1 62 ( V .D.)observed nnormal

2C6 H5COOH  (C6 H5COOH )2



0   0

      At t = teq 2 − 2x x   x

 2a   (1 + a ) × p 4a 2 = p = (1 - a 2 )  1-a  × p  1 + a 



At t = 0 2

D -1 d H 2(g ) + S(s)  H 2S(g ) At t = 0 0.2 At t = t eq 0.2 - x

0 x

Δng



We know  K p = K C ( RT )



For the reaction Δng = np - nR = 1 - 1 = 0



Therefore, K P = K C = 6.8 × 10 -2 =



Total number of moles at equilibrium = 0.2 − x + x = 0.2

x ⇒ x = 0.012 (0.2 - x )

1/4/2018 5:12:00 PM

Chemical Equilibrium

From ideal gas equation pV = nRT



     ptotal × 1 = (0.2) × (0.0821) × (363)



ptotal = 6 atm pH 2 S

38. (4) CaCO3 is solid, therefore, its activity is 1. Hence, there would be no effect of adding CaCO3 in the equilibrium. 41. (1) In the reaction N 2O4(g )  2NO2(g ) - heat , if volume increases, reaction will go in a direction where the num­ ber of moles increases or in forward direction.

0.012 = × 6 = 0.36 atm 0.2

32. (2) The decomposition reaction is

48. (1)    log  K 2  = ΔH  1 - 1   K  2.303 R  T T  1 1 2

N 2O4(g )  2NO2(g ) At t = 0 1 0 At t = t eq 1 -a 2a

Total number of moles at equilibrium = 1 + a = 1.2 mol

 10 -5  ΔH  1 1  log  -10  =   700  10  2.303 R 500      

pf Vf nf R(600) × = × ⇒ pf = 1 × 1.2 × 2 or pf = 2.4 atm pi Vi ni R( 300)

55. (1) Increase in pressure will favor the more dense state, therefore, the ice will melt to become a liquid.



33. (3) The reaction involved is

xH 2 =

nH2 nNH3 + nN 2 + nH2 + nNe

=

KC =

3/2 x 3/2 x = x 3 (3 - x) + + x + 1 4 + x 2 2

3/2





1/2

3/2

 3   x  x    2 2 = V ×1

1/2

Hence,   K C × V = 3 3



ΔH ° ΔS ° (1) + RT R 2000 On comparing the Eq. (1) with log K = 4 , we get T ΔS ° = 4 ⇒ ΔS ° = 4R R



Therefore, number of moles of O2 = 10 × 10 = 100 mol



Hence, number of molecules of O2 = 100 NA C(Diamond)  C(Graphite)



34. (1) We know    ΔG° = ΔH° − TΔS° = −RT loge K

[SO2 ]2[O2 ] = [O2 ] = 10 [SO 3 ]2

58. (1)

1.5x 0.5 = ⇒ 4 + x = 3x ⇒ x = 2 4+ x  3/ 2   x / 2 x ×    V  V  Therefore, K C =  3- x  V 

+ ve

+ ve

+ ve

56. (4) For the reaction 2SO3(g)  2SO2(g) + O2(g)

1 3 NH 3(g )  N 2(g ) + H 2(g ) 2 2 0 0 At t = 0 3 x 3x At t = t eq 3 - x 2 2

133

log e K = -

  35. (4) Ice    Water



3.5 g cm -3

r

2.3 g cm -3

If pressure increases reaction goes to a direction where volume decreases or in backward direction.

59. (3) Equilibrium constant = 56 = constant

(As temperature is the same.)

Previous Years’ NEET Questions 1. (2) We have

  

N 2 + 3H 2  2NH 3 ; K 1 =



  

N 2 + O 2  2NO; K 2 =



  

H2 +



The required equation

[NH 3 ]2 (1) [N 2 ][H 2 ]3

[NO]2 (2) [N 2 ][O2 ]2

273 K



At equilibrium ΔG = 0 or GH2 O - Gice = 0



or GH2 O = Gice ≠ 0

36. (2) We have log K =

 -ΔH °  1 ΔS ° + 2.303 R  2.303 R  T

-ΔH ° =1 2.303 R Therefore,    ΔH° = −2.303 × 2 cal = −4.606 cal m=



-10°C   37. (4) H 2O( l )    H 2O(Ice)



Since, it is spontaneous reaction, therefore, ΔG < 0, ΔH < 0 and ΔS < 0.

Chapter 5_Chemical Equilibrium.indd 133

2NH 3 +

1 [H 2O] (3) O 2  H 2O ; K 3 = [H 2 ][O2 ]1/2 2

5 [NO]2[H 2O]3 O2 → 2NO + 3H 2O; K = 2 [NH 3 ]2[O2 ]5/2

can be obtained from Eq. (2) + Eq. (3) × 3 − Eq. (1) [NO]2 [H 2O]3 [N 2 ][H 2 ]3 K 2 K 33 = × × [N 2 ][O2 ]2 [H 2 ]3[O 2 ]3/2 K1 [NH 3 ]2 =

[NO]2[H 2O]3 [NH 3 ]2[O 2 ]5/2

=K

1/4/2018 5:12:02 PM

134

OBJECTIVE CHEMISTRY FOR NEET

2. (3) At initial stage

2 AB2(g )  2 AB(g ) + B2(g ) 2 mol 0 0

At equilibrium ( 2 - 2 x )mol 2 x mol x mol Mole fraction 2x x 2 - 2x 2+ x 2+ x 2+ x 2 p AB ´ p B2 Kp = [ partial pressure = mole fraction ´ pTotal ] 2 p AB 2 2

Kp =

 2x   x   2 + x p  2 + x p  2 - 2x   2 + x 

K p2 =

On solving, we get K p1 K p2

[H + ][CN - ] = 4.5 ´ 10 -10 [HCN] CH 3COOH  CH 3COO - + H + K1 =

K2 =

4x ×p 4x 3 × p 1 2+x     K p = = × 2 4(1 - x ) 2 4



2K p  On solving, we get x =   p 



[H + ][CH 3COO - ] = 1.5 ´ 10 -5 [CH 3COOH]

Therefore, the equilibrium constant K eq =

1/3

K 2 1.5 ´ 10 -5 = = 3.33 ´ 104 K 1 4.5 ´ 10 -10

7. (3) The relation between K C and K p is

3. (4) We have

K p = K C ( RT )Δn Fe(OH)3(s)  Fe3+(aq ) + 3 OH - (aq )

At initial stage At equilibrium

0 x

0 x

0 3x

- 3

K C = [Fe ][OH ] 3+

3

3  = x × ( 3x )3 = x ′  x  ⇒ x ′ = 64 x 4 

Equilibrium constant K C will be the same, so, Fe3+ con­ centration is increased by 64 times. 1 1 1 H 2(g ) + I 2(g )  HI(g ); K = 2 2 8 On multiplying above reaction with 2, it becomes

For reactions (1), (2) and (4) Δn = 0. Therefore, K p = K C . For reaction (3), Dn = 1, hence, K p ≠ K C . 8. (1) We have   2 A(g ) + B(g )  3C(g) + D(g)   Initial moles 1 1 0 0 Moles at eq. 1 - ( 2 ´ 0.25) 1 - 0.25 3 ´ 0.25 0.25 = 0.5 = 0.75 = 0.75

Equilibrium constant is given by K=

4. (3) We have

H 2(g ) + I 2(g )  2 HI(g )

p1 9 = or p1 : p2 = 36 : 1 4 p2 1

HCN  H + + CN -



3

[As (1 − x) ≈1 and similarly 2 + x = 2]

=

6. (1) The dissociation constants for acetic acid and HCN are given by

2



[( 2 x /1 + x ) p2 ]2 4x 2 = p (1 - x /1 + x ) p2 (1 + x )(1 - x ) 2

The new equilibrium constant would be

9. (4) We have









  





2

1 æ 1ö K¢= K2 =ç ÷ = è 8ø 64

5. (4) We have

Initial conc. At equilibrium K p1 =

[( x /1 + x )p1 ]2 x2 = p1 (1 - x /1 + x )p1 (1 + x )(1 - x )

Initial conc. At equilibrium

Chapter 5_Chemical Equilibrium.indd 134

X  Y + Z; 1 0 0 1- x x x

A  2B ; 1 0 1 - x 2x

[C]3[D] (0.75)3(0.25) = [ A ]2[B] (0.5)2(0.75)

1 1 2NO2(g2)NO  2NO (g )2NO + O2((gg))     +KO′ 2=(g ) K ′ = 2 (g )  K2 K2 1 1 2NO(g )  2NO N 2(g )  + ON g      )K+′′O =2(g ) K ′′ = 2 (2g() K1 K1 1 . 1 1 . 1 2NO2(g2)NO  2N (g2)(g ) +N22O (g2()g+) 2KO¢¢¢2(=g ) K ¢¢¢ = K1 K 2 K1 K 2

Therefore, for the reaction NO2(g )  K=

1 N 2 (g ) + O 2 (g ) 2

1 K1 × K 2

10. (1) A decrease in temperature will favor the reaction in the forward direction as the reaction is exothermic. For the given reaction, ∆ng > 0, therefore, on increasing the pressure, the reaction will move in a direction where n increases, that is forward direction. Therefore, formation of XY4 will be favored at high pressure and low temperature.

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Chemical Equilibrium 11. (3) The chemical reaction is

14. (1) Using Arrhenius equation

2SO2(g ) + O 2(g )  2SO3(g )

Given that KC = 278. On reversing the reaction, the new equilibrium constant is K C′ =

1  1  =  K C  278 

1/2

= 6 × 10 -2

12. (3) The equilibrium constant for the reaction is [ AB]2 KC = [A 2 ][B2 ]



Given that [A2] = 3.0 × 10−3 M, [B2] = 4.2 × 10−3 M and [AB] = 2.8 × 10−3 M. Substituting all the values, we get KC =

( 2.8 × 10 -3 )2 = 0.62 ( 3.0 × 10 -3 )( 4.2 × 10 -3 )

13. (4) According to Le Chatlier’s principle, on increasing the pressure, the equilibrium shifts in the direction in which there is decrease in the number of moles, that is, in for­ ward direction.

For an exothermic reaction, on increasing the tempera­ ture the equilibrium will shift in the backward direction, so temperature should be kept low for facilitating for­ ward direction.

Chapter 5_Chemical Equilibrium.indd 135

135

log

K p′ Kp

=

1 1 Ea  -  2.303R T1 T2 

If T1 > T2, then Kp >> Kp′ 

15. (2) For a general reversible reaction aA + bB  cC + dD



The equilibrium constant K is given by K=

[C]c[D]d = 1.6 × 1012 [ A ]a[B]b

The value of K is very high which implies that at equilib­ rium the system contains mostly products.

16. (3) For the reaction N 2(g ) + O2(g )  2NO(g ); K

On multiplying both the sides of reaction with (1/2), we get 1 1 N 2(g ) + O2(g )  NO(g ) 2 2



Let the equilibrium constant for the above reaction be K ′.



Then, K′= (K)1/2

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Chapter 5_Chemical Equilibrium.indd 136

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6

Ionic Equilibrium

Chapter at a Glance 1. Ionic Equilibrium in Solutions (a) Electrolytes: Any chemical substance which can conduct electricity in its aqueous state or in its molten state is called electrolyte. Depending upon the degree of ionization we can classify the electrolytes as: (i) Strong electrolytes: These are those electrolytes which get 100% ionized (i.e. a = 1). For example, HNO3, HCl, H2SO4, NaOH, etc. (ii) Weak electrolytes: These are those electrolytes which get ionized upto 5% (i.e. a = 0.05). For example, CH3COOH, HNO2, NH4OH all organic acids and bases etc. (b) Ionic equilibrium: It is the point/state/stage at which the concentration of dissociated ions and undissociated electrolyte is in a state of equilibrium. It is shown by weak electrolytes. For example, CH 3COOH + H 2O  CH 3COO − + H 3O+

2. Ostwald’s Dilution Law The expression that correlates the variation of the degree of dissociation of an electrolyte with dilution is known as Ostwald’s dilution law.     KC =

a2 (1 − a )V

where C is the initial concentration of the electrolyte in mol L−1 and a is the degree of dissociation. (a) For weak electrolytes: The degree of dissociation is small, that is, a 7 NH4OH + NH4Cl

(a) Buffer: It is a weak acid + salt of that weak acid with strong base or weak base + salt of that weak base with strong acid. (b) Buffer capacity (f ): Every buffer has the capacity to resist the change in concentration of solution by addition of number of moles of acids or base in one liter solutions so to change the pH by one unit.

Buffer capacity(f )=

Number of moles of acid or base added to one liter buffer solution Change of pH

(c) pH of Buffer solutions: Henderson – Hasselblach equation (i) pH of an acidic buffer pH = pK a + log

[Salt] [ Acid]

(ii) pH of a basic buffer pOH = pK b + log

[Salt] [Base]

(iii) Salt buffer: When a solution of a salt itself acts as a buffer, it is known as a salt buffer. These are salts of a weak acid and weak base. (iv) Buffer index: The buffer index is given by B.I. = 2.303 ×

y(x − y ) x

              where x is number of moles of acid and y is number of moles of salt. (v) Buffer range: It is generally accepted that a solution has useful buffer index provided that the value of [Salt]/[Acid] lies within the range 10–0.1. Hence, from Henderson equation we have pKa + log(0.1) < pH < pKa + log10

Chapter 6_Ionic Equilibrium.indd 139

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140

OBJECTIVE CHEMISTRY FOR NEET

or (pKa− 1) < pH < (pKa + 1) Beyond this range, the buffer index is very small for any practical application. 7. Hydrolysis of Salts and the pH of their Solutions Salt

Degree of hydrolysis

Hydrolysis Constant

(i) NaCl (Strong acid + Strong base)

No hydrolysis

(ii) CH3COONa (Weak acid + Strong base)

h=

Kw K aC

Kh=

(iii) NH4Cl (Strong acid + Weak base)

h=

Kw K bC

Kh =

(iv) CH3COONH4 (Weak acid +Weak base)

h=

Kw KaKb

Kh =

pH

pH =

1 [pKw + pKa + logC ] 2

Kw Kb

pH =

1 [pKw – pKb – logC ] 2

Kw KaKb

pH =

1 [pKw + pKa – pKb] 2

Kw Ka

8. Common Ion Effect: It is defined as suppression of the dissociation of a weak electrolyte by the solution of strong electrolyte having a common ion. 9. Solubility Equilibria of Sparingly Soluble Salts (a) Consider the dissociation of any sparingly soluble electrolyte, AxBy A x B y  xA y + + yBx −

 

 Applying the law of mass action, we get K=

 

In saturated solution, [A x B y ] = constant = K ′

 

Therefore, K sp = [ A y + ]x[Bx − ]y

[ A y + ]x [Bx − ]y [A x B y ]

(b) E  ffect of common ion on solubility: The presence of a common ion in the solution decreases the solubility of a given substance. (c) Effect of pH on solubility: When pH is lowered, the concentration of anion is lowered due to its protonation which in turn increases the solubility of the salt. For a given electrolyte, solubility product is always constant at a particular temperature. (d) Ionic product (IP) and relation with solubility product (i) If ionic product = solubility product, the solution is in equilibrium. (ii) If ionic product < solubility product, the solution is unsaturated and more salt can be dissolved in it. (iii) If ionic product > solubility product, then the solution is containing more salt than it can dissolve, therefore, precipitation starts and continues till ionic product becomes equal to solubility product.

Solved Examples 1. H3BO3 is_________acid. (1) monobasic (2) dibasic (3) tribasic (4) none

H 3BO3 + H 2O → [B(OH)4 ]− + H + 2. Which of the following is the strongest acid?

Solution

(1) HClO4 (2) HBrO4

(1) H3BO3 is monobasic acid.

(3) HIO4 (4) HNO3

Chapter 6_Ionic Equilibrium.indd 140

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Ionic Equilibrium Solution



(1) The acidic character of oxyacids decreases down the group and increases along the period. Also, acidity increases with increase in oxidation number of central atom. 3. The degree of dissociation of 0.05 M NH3 at 25°C in a solution of pH = 11 will be

7. The pH of an aqueous solution of 0.1 M solution of a weak monoprotic acid which is 1% ionized is (1) 1 (2) 2 (3) 3 (4) 11

Solution (3) Initial conc.

(1) 0.02 (2) 0.04 (3) 0.01 (4) 0.8

NH 4OH  NH +4 + OH Initial conc. C 0 0 At eq. C(1 - a ) Ca Ca

Therefore,

[NH +4 ][OH − ] [NH 4OH]

Therefore,    K b =



Given that pH = 11 and using pH + pOH = 14, we get pOH = 3.



[OH−] = 10−3 = Ca



Substituting C = 0.05 M, we get a = 0.02.

2− (3) HPO2− 4 (4) HPO 3

(1) (2) (3) (4)

(4) Conjugate acid and base differs by one proton hence 2− the conjugate acid of PO3− 4 is HPO 3 . 5. The conjugate base of hydrazoic acid is

+

H → N 3− (3) N 3H −

6. At −50°C, the self-ionization constant (ion product) of NH3 is K NH3 = [NH +4 ][NH 2− ] = 10 −30 M 2. How many amide ions are present per mm3 of pure liquid ammonia? (1) 600 ions mm−3 (2) 6 × 106 ions mm−3 (3) 6 × 104 ions mm−3 (4) 60 ions mm−3 Solution



In option (2), 10 mL of M/10 HCl will be left out after neutralization.







10. Calculate the pH of a solution prepared by mixing 2.0 mL of a strong acid (HCl) solution of pH 3.0 and 3.0 mL of a strong base (NaOH) of pH 10.0.

Solution

K NH3 = [NH +4 ][NH 2− ] = x 2 = 10 −30

(2) Number of milliequivalents of HCl = 2 × 10−3

− 2

M = [NH ]

10 [NH 2− ] = 10−15 mol L−1 = mol mm − 3 = 10 −21 mol mm − 3 106 − 2

[NH ] = 10

Chapter 6_Ionic Equilibrium.indd 141

−21

× 6 × 10 ions mm = 600 ions mm 23

−3

10 × ( M /10) = 0.01 M 100 Therefore,    pH = 2 [H+] =

(1) 2.5    (2) 3.5    (3) 4.5    (4) 6.5

x

−15





2NH 3( l ) NH 4+ + NH 2−

Thus,      x = 10

100 mL of M/10 HCl + 100 mL of M/10 NaOH 55 mL of M/10 HCl + 45 mL of M/10 NaOH 10 mL of M/10 HCl + 90 mL of M/10 NaOH 75 mL of M/5 HCl + 25 mL of M/5 NaOH

(4) The answer has to be either (2) or (4) since only in these two choices, HCl is greater than NaOH (only when HCl is in greater amount, pH can be acidic).

Solution



1 1 [HA] = = = 1.0 × 106 [ A − ][H + ] K a 1.0 × 10 −6

Solution

(1) NH3 (2) N3H (3) N3– (4) N2–

−15

[H+] = 10–3 ⇒ pH = 3

9. Which of the following solution will have pH value close to 1.0?

Solution



1 × 0.1 100

(1) The reaction is the reverse of the ionization reaction of HA, hence the equilibrium constant is the reciprocal of Ka.

(1) H3PO4 (2) H 2PO 4−



1 × 0.1 100

A− 0

Solution

K=

x

+

(1) 1.0 × 106 (2) 1.0 × 10−8 8 (3) 1.0 × 10 (4) 1.0 × 10−6

4. The conjugate acid of PO3− 4 is

At eq. 1 − x

H+ 0



8. Consider the reaction, A− + H3O+  HA + H2O. The Ka value for acid HA is 1.0 × 10−6. What is the value of K for this reaction?



(1)

HA 0.1

1   0.1 1 −  100 

At eq.

Solution (1)

141

−3



We know that



   10 + pOH = 14 ⇒ pOH = 4



Number of milliequivalents of NaOH = 3 × 10−4



The resulting solution is acidic.

pH + pOH = 14

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142

OBJECTIVE CHEMISTRY FOR NEET ( 2 × 10 −3 − 3 × 10 −4 )meq. = 3.4 × 10−4 M ( 2 + 3)mL







or   pH = −log[H ] = −log(3.4 × 10 ) = 3.5

[H+] =

+

−4

11. Which salt among the following undergoes hydrolysis? (1) CH3COONa (2) KNO3 (3) NaCl (4) K2SO4

Solution (2) Precipitation occurs when the ionic product (IP) exceeds Ksp value. When equal volumes of two solutions are mixed, the concentration of each is reduced to half. Therefore,

In first case,







As, IP < Ksp, therefore, no precipitation occurs.



In second case,







As, IP > Ksp, therefore, precipitation occurs.



In third case,







As, IP < Ksp, therefore, no precipitation occurs.



In fourth case,







As, IP < Ksp, therefore, no precipitation occurs.

2

Solution (1) Salts of strong acid and strong base does not undergo hydrolysis. 12. Calculate the pH at the equivalence point in the titration of 25 mL of 0.1 M formic acid with a 0.1 M NaOH solution (given that pKa of formic acid = 3.74). (1) 4.74 (2) 8.22 (3) 8.37 (4) 6.06 Solution

1  1  1 IP =  × 10 −4   × 10 −4  = × 10 −12 = 1.25 × 10 −13  8 2  2

2

1  1  1 IP =  × 10 −2   × 10 −3  = × 10 −7 = 1.25 × 10 −8  8 2  2

2

(2) At the equivalence point, 0.05 M of HCOONa (sodium formate) is formed. This problem is solved by the considering the concept of hydrolysis equilibria. 1 1 1 pK w + pK a + log C 2 2 2 1  1  1 =  × 14 +  × 3.74 + log(0.05) 2  2  2 = 8.22

pH =

13. A solution of benzoic acid is titrated with NaOH. The pH of the solution is 4.2, when half of the acid is neutralized. Dissociation constant of the acid is

1  1  1 IP =  × 10 −5   × 10 −3  = × 10 −13 = 1.25 × 10 −14 2  2  8

2

1  1  1 IP =  × 10 −4   × 10 −5  = × 10 −13 = 1.25 × 10 −14 2  2  8

15. A weak acid HX has Ka = 1.0 × 10−5 and it forms a salt NaX when treated with caustic soda. The percentage of hydrolysis of a 0.1 M NaX solution is (1) 0.1% (2) 0.01% (3) 0.001% (4) 1%

1) 6.31 × 10–5 (2) 3.2 × 10–5 (3) 8.7 × 10–8 (4) 6.42 × 10–4

Solution

Solution

(2) We have      h =

(1) The reaction involved is C6 H5COOH + NaOH  C6 H5COONa + H 2O At half neutralization

0.5

0.5

It is a buffer solution of weak acid and its salt. Therefore, [salt ] pH = pK a + log [acid] pH = pK a 4.2 = pK a ⇒ K a = 6.31 × 10 −5

14. The precipitate of Ag2CrO4 (Ksp = 1.9 × 10−12) is obtained when equal volumes of which of the following are mixed? (1) 10−4 M Ag+ + 10−4 M CrO 2− 4 (2) 10−2 M Ag+ + 10−3 M CrO 2− 4 (3) 10−5 M Ag+ + 10−3 M CrO 2− 4 (4) 10−4 M Ag+ + 10−5 M CrO 2− 4

Chapter 6_Ionic Equilibrium.indd 142

      

    =

Kw Ka ×C 10 −14 = 10 −4 10 −5 × 0.1

Percentage of hydrolysis is = 10−4 × 100 = 0.01%

16. In which of the following solvents, will AgBr have highest solubility? (1) 10−3 M NaBr (2) 10−3 M NH4OH (3) Pure water (4) 10−3 M HBr Solution (2) With ammonia, Ag+ forms a complex of the type, [Ag(NH3)2]+. This increases its solubility compared to pure water. In NaBr and HBr, AgBr will encounter a common ion effect, which would reduce its solubility compared to pure water.

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Ionic Equilibrium 17. Solubility of AgCl in water, 0.01 M CaCl2, 0.01 M NaCl and 0.05 M AgNO3 are S1, S2, S3 and S4 respectively then (1) S1 < S2 < S3 < S4 (3) S1 > S2 = S3 > S4

(2) S1 > S3 > S2 > S4 (4) S1 > S3 > S4 < S2

Solution (3) Number of moles of CaCO3 in the residue is 7 × 10 −3 = = 7 × 10 −5 100

Moles of CaCO3 in 1 liter solution = 7 × 10−5

Solution

CaCO3(s)  Ca 2+

(2) The reactions involved are as follows:

7 × 10 −5

K sp = [Ca ] × [CO23− ]

NaCl  Na + + Cl − 0.01 0.01 0.01

  

19. [M(OH)x] has Ksp = 4 × 10−12 and solubility 10−4 M. Find the value of x. (1) 1 (2) 2 (3) 3 (4) 4

0.05

Common ion effect is maximum in AgNO3. Therefore, S1 > S 3 > S2 > S4.

18. One liter of a saturated solution of CaCO3 is evaporated to dryness due to which 7.0 mg of residue is left. The solubility product for CaCO3 is (1) 4.9 × 10−8 (2) 4.9 × 10−5 (3) 4.9 × 10−9 (4) 4.9 × 10−7

= 7 × 10 −5 × 7 × 10 −5 = 4.9 × 10 −9

AgNO3  Ag + + NO3−

CO23−

2+

CaCl 2  Ca 2+ + 2 Cl − 0.01 0.01 2 × 0.01

0.05

+

7 × 10 −5

AgCl  Ag + + Cl −

0.05

143

Solution (2) [M(OH)x] will ionize as [M(OH)x ]  M x + + x OH −

10 −4 x 10 −4 x+ Therefore,      Ksp = [M ] [OH−] x

4 × 10−12 = (10−4) (x × 10−4)x By inspection, we can find that the relation will hold good when x = 2.

Practice Exercises Level I Strength of Acids/Bases and Conjugate Acid-Base Pairs 1. Which among the following is the strongest base? (1) Be(OH)2 (2) Mg(OH)2 (3) Ba(OH)2 (4) Sr(OH)2 2. The conjugate acid of NH 2− is (1) NH4 + (2) NH3 (3) NH2OH (4) N2H4

3. The conjugate base of H3O+ is (1) H2O (2) OH– (3) H+ (4) None of these

4. Which of the following species is an acid and also a conjugate base of another acid? (1) HSO4− (2) H2SO4 (3) OH– (4) H3O+ 5. In the reaction I 2 + I − → I 3− . The Lewis base is (1) I2 (2) I– (3) I3– (4) None of these

Chapter 6_Ionic Equilibrium.indd 143

6. The compound that is not a Lewis acid is (1) BF3 (3) NF3

(2) AlCl3 (4) SnCl4

7. Which of the following is a Lewis base? (1) BF3 (2) Na+ (3) NH4+ (4) CH3OH 8. The dissociation constants of acids HA, HB, HC and HD are 2.6 × 10−3, 5.3 × 10−9, 1.1 × 10−2, 7.5 × 10−5 respectively. The weakest acid among these acids is (1) HA (2) HB (3) HC (4) HD

pH for Weak/Strong (Monobasic/Monoacidic) Acids/Bases, Kw , Common Ion Effect and Mixtures of Acids and Bases 9. Which of the following substances when added into a solution of acetic acid will not suppress its degree of dissociation? (1) dil. HCl (2) CH3OONa (3) water (4) dil. H2SO4

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OBJECTIVE CHEMISTRY FOR NEET

10. If an aqueous solution at 25°C has twice as many OH- as pure water, its pOH will be (1) 6.699 (2) 7.307 (3) 7 (4) 6.98 11. The pH of a soft drink is 3.82. Its hydrogen ion concentration in mol L–1 will be (1) 1.96 × 10–2 (2) 1.96 × 10–3 (3) 1.5 × 10–4 (4) 1.96 × 10–1 12. pH of a 10–10 M NaOH is nearest to

21. 100 mL of 0.2 M HCl are mixed with 100 mL of 0.2 M CH3COOH, the pH of the resulting solution would be nearly (1) 1 (2) 0.7 (3) 1.3 (4) 1.6

Salt Hydrolysis 22. Which of the following, when mixed, will give a solution with pH greater than 7? (1) (2) (3) (4)

(1) 10 (2) 7 3) 4 (4) 10 13. Let Kw at 100°C be 5.5 × 10−13 M2. If an aqueous solution at this temperature has pH 6, its nature will be (1) acidic. (2) alkaline. (3) neutral. (4) can’t say.

23. If a salt of strong acid and weak base is hydrolyzed, which of the following formulae is to be used to calculate degree of hydrolysis X ?

14. The pH of a solution is increased from 3 to 6; its H+ ion concentration will be (1) reduced to half. (2) doubled. (3) reduced by 1000 times. (4) increased by 1000 times. 15. One liter of water contains 10–7 mol of H+ ions. Degree of ionization of water is (1) 1.8 × 10–7 % (2) 0.8 × 10–9 % (3) 3.6 × 10–9 % (4) 3.6 × 10–7 %. 16. Which of the following will occur if a 0.1 M solution of a weak acid is diluted to 0.01 M at constant temperature? (1) [H+] will decrease to 0.01 M. (2) pH will decrease. (3) Percentage ionization will increase. (4) Ka will increase.

(1) 5 times. (2) 1000 times. (3) 105 times. (4) 4 times. 18. If the degree of ionization of water is 1.8×10−9 at given temperature, its ionic product will be (1) 1.8 × 10−16 (2) 1 ×10−14 (3) 1 × 10−16 (4) 1.67 × 10−14 19. The pH of an aqueous solution of 0.1 M solution of a weak monoprotic acid which is 1% ionized is

(1) 1 (2) 2 (3) 3 (4) 11 20. A monoprotic acid in a 0.1 M solution ionizes to 0.001%. Its ionization constant is

Chapter 6_Ionic Equilibrium.indd 144

(1) X =

Kw (2) X = Ka × C

Kw Kb × C

(3) X =

Kw (4) None of these Ka × Kb

24. Ionization constants of weak acid HA and weak base BOH are 3 × 10–7 each at 298 K. Percentage hydrolysis of their salt at 0.1 M concentration is (1) 40% (2) 50% (3) 75% (4) 33.33% 25. Equilibrium constant for the following reaction is 1.8 × 109.

− − CH CH33COOH COOH++OH OH −  CH CH33COO COO − ++ H H22O O

Hence the equilibrium constant for CH 3COOH + H 2O  CH 3COO − + H 3O+ CH 3COOH + H 2O  CH 3COO − + H 3O+ is

17. The pH of a solution is 7.00. To this solution sufficient base is added to increase the pH to 12.0. The increase OH– ion concentration is

(1) 1.0 × 10–3 (2) 1.0 × 10–6 (3) 1.0 × 10–8 (4) 1.0 × 10–11

0.1 M HCl + 0.2 M NaCl 100 mL 0.2 M H2SO4 + 100 mL 0.3 M NaOH 25 mL of 0.1 M HNO3 + 25 mL 0.1 M NH3 100 mL 0.1 M CH3COOH + 100 mL 0.1 M NaOH

(1) 1.8 × 1023 (2) 1.8 × 105 (3) 1.8 × 10–5 (4) 5.55 × 10–10 26. Which of the following salts undergoes anionic hydrolysis? (1) Cu(NO3)2 (2) NH4Cl (3) AlCl3 (4) K2CO3 27. Assuming 100% ionization, which will have maximum pH? (1) 0.01 M NH4Cl (2) 0.01 M (NH4)2SO4 (3) 0.01 M (NH4)3PO4 (4) equal 28. The pH of a solution obtained by mixing 100 mL of 0.2 M CH3COOH is with 100 mL of 0.2 M NaOH would be (pKa for CH3COOH = 4.74 and log 2 = 0.301) (1) 4.74 (2) 8.87 (3) 9.10 (4) 8.57 29. Which of the following salts when dissolved in water will hydrolyze? (1) NaCl (2) KCl (3) NH4Cl (4) Na2SO4

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Ionic Equilibrium 30. Which one of the following anions does not hydrolyze? (1) H (2) CN (3) NO 3− (4) S2– –

(1) 5 (2) 9 (3) 4.74 (4) 8.26

–

40. The pH of a buffer containing equal molar concentrations of a weak base and its chloride (Kb for weak base = 2 × 10–5, log 2 = 0.3) is

31. The following reaction occurs in the body CO2 + H 2O  H 2CO 3  H + + HCO3−



(1) 5 (2) 9 (3) 7 (4) 9.3

If CO2 escapes from the system (1) pH will decrease. (2) hydrogen ion concentration will decrease. (3) H2CO3 concentration remain unaltered. (4) forward reaction will be promoted.

41. pH of a mixture containing 0.10 M X– and 0.20 M HX is [pKb (X–) = 4] (1) 4 + log 2 (2) 4 –log 2 (3) 10 + log 2 (4) 10 –log 2

32. The compound whose 0.1 M solution is basic is (1) ammonium acetate. (2) ammonium chloride. (3) ammonium sulphate. (4) sodium acetate.

42. When sodium acetate is added to an aqueous solution of acetic acid its pH (1) increases. (2) decreases. (3) remains constant. (4) none of these.

33. An aqueous solution of CuSO4 is

43. Which of the following will not function as buffer solution?

(1) acidic. (2) basic. (3) neutral. (4) amphoteric.

(1) NaOH + NH4OH (2) NaH2PO4 + Na2HPO4

34. The dissociation constant of an acid HA is 1 × 10 , the pH of 0.1 M solution of acid will be approximately –5

(1) 3 (2) 5 (3) 1 (4) 6 35. 100 mL of 0.2 M HCl is mixed with 100 mL of 0.2 M CH3COONa, the pH of the resulting solution would be nearly (pKa for CH3COOH = 4.74) (1) 1 (2) 0.7 (3) 2.875 (4) 1.6 36. A 0.1 M solution of HCN is 0.01% ionized, the ionization constant for HCN is (1) 10–9 (2) 10–7 (3) 10–5 (4) 10–3 37. The aqueous solution of potash alum is acidic due to hydrolysis of (1) K+ (2) Al3+ (3) SO2− 4 (4) presence of acid in its crystal as impurity.

Buffer Solutions 38. To prepare a buffer of pH = 8.26, amount of (NH4)2SO4 to be added into 500 mL of 0.01 M NH4OH solution [pKa (NH +4 ) = 9.26] is (1) 0.05 mol (2) 0.025 mol (3) 0.10 mol (4) 0.005 mol 39. What is the pH of solution which has 1 mL NH4OH (conc. 0.1 M) 1 mL and (NH4)2SO4 concentration 0.05 M. Given that Kb(NH4OH) = 10–5.

Chapter 6_Ionic Equilibrium.indd 145

145

(3) Borax + boric acid (4) CH3COOH + CH3COONa 44. pH of following buffer solution is:

2 × 10−2 M CH3COOH + 3 × 10−2 M (CH3COO)2Ca. (pKa of CH3COOH = 4.75) (1) 4.27 (2) 8.77 (3) 5.23 (4) 7.0

45. There is no effect of dilution on pH of which of the following? (1) 0.01 M CH3COOH + 0.01 M CH3COONa (2) 0.01 M H3COONH4 (3) 0.01 M NH4OH + 0.01 M NH4Cl (4) In all the cases 46. The pH of a buffer is 6.745. When 0.01 mol of NaOH is added to 1 L of it, the pH changes to 6.832. Its buffer capacity is (1) 0.187 (2) 0.115 (3) 0.076 (4) 0.896 47. pH of 0.01 (NH4)2SO4 and 0.02 M NH4OH buffer (pKa[NH4+] = 9.26) is (1) 9.26 (2) 9.26 + log2 (3) 4.74 (4) 4.74 + log2 48. Which of the following mixture will be a buffer solution when dissolved in 500 mL of water? (1) (2) (3) (4)

0.20 mol of aniline and 0.20 mol of HCl. 0.20 mol of aniline and 0.20 mol of NaOH 0.20 mol of NaCl and 0.20 mol of HCl 0.20 mol of aniline and 0.1 mol of HCl

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OBJECTIVE CHEMISTRY FOR NEET

Solubility and Solubility Product 49. The precipitate of CaF2 (Ksp = 1.7 × 10–10) is obtained when equal volume of following are mixed

4. A compound having the formula NH2CH2COOH may behave (1) only as an acid. (2) only as a base. (3) both as an acid and base. (4) neither acid nor base.

(1) 10–4 M Ca2+ + 10–4 M F– (2) 10–2 M Ca2+ + 10–3 M F– (3) 10–5 M Ca2+ + 10–3 M F–

(4) 10–3 M Ca2+ + 10–5 M F–

50. When HCl gas is passed through a saturated solution of common salt, pure NaCl is precipitated because

(1) HCl is highly ionized in solution. (2) HCl is highly soluble in water. (3) the solubility product of NaCl is lowered by HCl. (4) the ionic product of [Na+] exceeds the value of solubility product of NaCl. 51. What is correct representation for the solubility product of SnS2?

pH for Weak/Strong (Monobasic/Monoacidic) Acids/ Bases, Kw , Common Ion Effect and Mixtures of Acids and Bases 5. Equal volumes of two solutions of a strong acid having pH 3 and pH 4 are mixed together. The pH of the resulting solution will then be equal to (1) 3.5 (2) 3.26 (3) 7 (4) 1.0 6. The pH of 10–8 M HCl is (1) 7 (2) 8 (3) 7.02 (4) 6.98

2+ − 2 (1) [Sn2+] [S 2− 2 ] (2) [Sn ] [S ]

(3) [Sn2+] [S−] (4) [Sn4+] [S−]2

7. An acid solution of pH 6 is diluted thousand times. The pH of solution becomes approx.

52. Which of the following is most soluble? (1) Bi2S3 (Ksp = 1 × 10–70) (2) MnS (Ksp = 7 × 10–16)

(1) 7 (2) 6 (3) 4 (4) 9

(3) CuS (Ksp = 8 × 10–37) (4) Ag2S (Ksp = 6 × 10–51) 53. The solubility product of BaSO4 is 1.5 × 10−9. The precipitation in a 0.01 M Ba2+ ions solution will start on adding H2SO4 of concentration

8. Which of the following will occur if a 0.1 M solution of a weak acid is diluted to 0.01 M at constant temperature? (1) [H+] will decrease to 0.01 M. (2) pH will decrease. (3) Ka will decrease. (4) Percentage ionization will increase.

(1) 10 M (2) 10 M (3) 10−7 M (4) 10−6 M −9

−8

54. The molar solubility of calcium phosphate is S. What is its solubility product? (1) S5 (2) 108 S 5 (3)

S5 108 (4) 5 108 S

H+ + OH −  H 2O

55. The solubility of A2X5 is x mol dm−3. Its solubility product is (1) 36 x6 (2) 64 × 104 x7 (3) 126 x 7 (4) 1.25 × 104 x7

Strength of Acids/Bases and Conjugate Acid-Base Pairs 1. Which of the following are amphiprotic in nature? (1) OH– (2) H 2PO2− (3) HSO3− (4) HF 2. The conjugate base of [Cr(H 2O)5Cl ]

(1) [Cr(H 2O)5 ] (2) [Cr(H 2O)4Cl ]

2+

(3) [Cr(H 2O)4(OH)Cl ] (4) [Cr(H 2O)4(OH)Cl ] 3+

3. Which one of following is the strongest acid? (1) H3PO4 (2) H3PO2 (3) H3PO3 (4) Equally acidic

Chapter 6_Ionic Equilibrium.indd 146

0

10

25

40

50

0.114 × 10–14

0.292 × 10–14

1.008 × 10−14

2.91 × 10–14

5.474 × 10–14

(1) exothermic. (2) endothermic. (3) can’t say. (4) ionization. 10. The dissociation constant of acetic acid at a given temperature is 1.69 × 10–5. The degree of dissociation of 0.01 M acetic acid in the presence of 0.01 M HCl is equal to (1) 0.41 (2) 0.13 (3) 1.69 × 10–3 (4) 0.013.

is

3+

Temp.°C Kw

Level II

2+

9. If the ionic product of water varies with temperature as follows and the density of water is nearly constant for this range of temperature, then the process is

+

11. If pKb of F– ion at 25°C is 10.83, the ionization constant of HF in water at this temperature is (1) 1.75 × 10–5 (2) 3.52 × 10–3 (3) 6.75 × 10–4 (4) 5.38 × 10–2 12. If Ka = 10–5 for a weak acid, then pKb for its conjugate base would be

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Ionic Equilibrium (1) 10–10 (2) 9 (3) 10–9 (4) 5

(3) pH =

pK1 − pK 2 2



(4) pH = 7 −

147

pK1 + pK 2 2

13. At 25°C, the pH of pure water is 7. It dissociates H 2O(l) + H 2O(l)  H 3O+ + OH −

21. The aqueous solution of NH4CN is slightly alkaline because

H3O+(aq) + OH–(aq)  2H2O(l) at 25°C; ∆H° = –13.7 k cal mol–1. pH of water of at 37°C is expected to be

(1) CN– ion hydrolyzes to a greater extent than NH4 + ion. (2) NH4 + ion hydrolyzes to a greater extent than CN– ion (3) both hydrolyze to an equal extent. (4) it is a salt.



(1) greater than 7. (2) less than 7. (3) equal to 7. (4) none of these is true. 14. A dilute HCl solution saturated with H2S (0.1 M) has pH value 3. The [S2–] is (Given the dissociation constants of H2S are K1 = 1 × 10–7, K2 = 1.3 × 10–13.) (1) 2 × 10–13 (2) 2.4 × 10–13 (3) 3 × 10–15 (4) 1.3 × 10–15 M 15. Dissociation constant of two acids HA and HB are respectively 4 × 10−10 and 1.8 × 10−5. The pH value of which acid will be higher for a given molarity of their aqueous solution? (1) HA (2) HB (3) Both are same (4) Cannot be determined 16. The following equilibrium exists in aqueous solution CH3COOH  CH3COO- + H+. If dil. HCl is added without change in temperature, the (1) (2) (3) (4)

concentration of CH3COO- will increase. concentration of CH3COO- will decrease. the equilibrium constant will increase. the equilibrium constant will decrease.

17. In an aqueous solution of triprotic acid H3A, which of the following is true? (1) [H+] = 3[A3–] (2) [H+] > 3[A3–] (3) [H+] < 3[A3–] (4) [H+] = [OH–] 18. The dissociation constants of HCOOH and CH3COOH are 1.8 × 10–4 and 1.8 × 10–5 respectively. [H3O+] in a solution which is 0.1 M CH3COOH and 0.01 M HCOOH is approximately (1) 1.9 × 10–3 (2) 3.6 × 10–3 (3) 1.9 × 10–6 (4) 3.6 × 10–6

19. The dissociation constant of a weak acid is 1.0 × 10–5, the equilibrium constant for its reaction with strong base is (1) 1.0 × 10–5 (2) 1 × 109 (3) 1.0 × 107 (4) 1.0 × 1014 20. Calculate the pH of 0.1 M NaHCO3 solution. (Dissociation constants of H2CO3 are K1 and K2.)

Chapter 6_Ionic Equilibrium.indd 147

pK1 + pK 2 2

(2) pH = 7 +

(1) 2 (2) 1 (3) 4 (4) none of these 23. H 2O + H 3PO4  H 3O+ + H 2PO4− ; pK 1 = 2.15

H 2O + H 2PO4−  H 3O+ + HPO4− ; pK 2 = 7.20



pH of 0.01 M NaH2PO4 is (1) 9.350 (2) 4.675 (3) 2.675 (4) 7.350

24. Calculate the pH of ‘C ’ M Na2S solution. (K1, K2 of H2S are 5.0 × 10–5, 2.5 × 10–11.) pK1pK 2 1 + logC 2 2 pK 1 1 (2) pH = 7 + + logC 2 2 pK 2 1 (3) pH = 7 + + logC 2 2 pK 1 1 + logC (4) pH = 7 + 2pK 2 2 (1) pH = 7 +

25. pH of 0.5 M Ba(CN)2 solution ( pKb of CN− is 9.3) is (1) 8.35 (2) 3.35 (3) 9.35 (4) 9.50

Buffer Solutions 26. When 100 mL of 0.4 M CH3COOH is mixed with 100 mL of 0.2 M NaOH, the [H3O+] in the solution is approximately [Ka(CH3COOH) = 1.8 × 10–5]

Salt Hydrolysis

(1) pH =

22. The pH of 10−2 M aqueous solution of barium acetate (CH3COO)nBa is 8.52. Calculate the value of n. (pKa of acetic acid = 4.75.)

pK1 − pK 2 2

(1) 1.8 × 10–6 (2) 1.8 × 10–5 (3) 9 × 10–6 (4) 9 × 10–5 27. pKa for acetic acid is 4.75. What should be the ratio of concentrations of acetic acid and acetate ions to have a solution with pH 5.05? (1) 1 : 10 (2) 10 : 1 (3) 1 : 2 (4) 2 : 1 28. What will be the final pH if 0.2 mol of NaOH is added to 1 L of buffer solution consisting of 0.4 M CH3COOH and 0.3 M CH3COONa? (pKa of CH3COOH = 4.75)?

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148

OBJECTIVE CHEMISTRY FOR NEET (1) 4.75 (2) 5.15 (3) 10.53 (4) 13.4

(1) AgCl (2) AgBr (3) AgI (4) All will precipitate simultaneously

Solubility and Solubility Product 29. During salt analysis, the medium is made acidic in the precipitation of sulphides of Group II to (1) increase the S2– ion concentration. (2) decrease the S2– ion concentration. (3) dilute the solution. (4) make the solution viscous. 30. The solubility of CH3CO2Ag would be the least amongst the following solvents in (1) (2) (3) (4)

acidic solution of pH = 3. basic solution of pH = 8. neutral solution of pH = 7. pure water.

31. Why only As gets precipitated as As2S3 and not Zn as ZnS when H2S is passed through an acidic solution containing As3+ and Zn2+? 2+

Solubility product of As2S3 is less than that of ZnS. Enough As3+ is present in acidic medium. Zinc salt does not ionize in acidic medium. Solubility product changes in presence of an acid.

32. The molar solubility of AgCl in 1.8 M AgNO3 solution is (Ksp(AgCl) = 1.8 × 10–10) (1) 10–10 (2) 10–5 (3) 1.82 × 10–10 (4) None of these 33. What is the solubility of Al(OH)3, Ksp = 1 × 10–33, in a solution having pH 4? (1) 10–3 M (2) 10–6 M (3) 10–4 M (4) 10–10 M 34. The best explanation for the solubility of MnS in dil. HCl is that (1) solubility product of MnCl2 is less than that of MnS. (2) concentration of Mn2+ is lowered by the formation of complex ions with chloride ions. (3) concentration of sulphide ions is lowered by oxidation to free sulphur. (4) concentration of sulphide ions is lowered by the formation of the weak acid H2S. 35. A solution is saturated with respect to AgSCN as well as AgBr. The concentration of Ag+ in the solution would be (Ksp for AgSCN = 1.2 × 10–12 and for AgBr = 5 × 10–13) (1) 1.3 × 10–5 (2) 1.3 × 10–12 (3) 1.3 × 10–6 (4) 1.3 × 10–8.

38. The solubility of CaF2 (Ksp, = 3.4 × 10–11) in 0.1 M solution of NaF would be

39. The solubility product of a sparingly soluble salt AB at room temperature is 1.21 × 10–6. Its molar solubility is (1) 1.21 × 10–6 (2) 1.21 × 10–3 (3) 1.1 × 10–4 (4) 1.1 × 10–3 40. The solubility product of BaCrO4 is 2.4 × 10–10 M2. The maximum concentration of Ba(NO3)2 possible without precipitation in a 6 × 10–4 M K2CrO4 solution is (1) 4 × 10–7 M (2) 1.2 × 10–10 M (3) 6 × 10–4 M (4) 3 × 10–4 M. 41. The solubility of sparingly soluble substance AgCl can be increased by the addition of (1) aq. NH3 (2) aq. NaCN (3) Both (4) None of these 42. The solubility product of Hg2I2 can be best represented as (1) [Hg + ]2[I − ]2 (2) [Hg +2 ]2[I − ]2 (3) [Hg 22+ ][I − ]2 (4) [Hg 22+ ][I 22− ]

Previous Years’ NEET Questions 1. Calculate the pOH of a solution at 25°C that contains 1 × 10−10 M of hydronium ions, that is, H3O+. (1) 1 (2) 7 (3) 4 (4) 9 (AIPMT 2007) 2. A weak acid HA has a Ka of 1.00 × 10−5. If 0.100 mol of this acid is dissolved in 1 L of water, the percentage of acid dissociated at equilibrium is closest to (1) 0.100% (2) 99.0% (3) 1.00% (4) 99.9% (AIPMT 2007)

36. To a solution having equal concentration of Cl , Br and I–, solid AgNO3 is slowly added. Which one will precipitate out first? (Given Ksp (AgCl) > Ksp (AgBr) > Ksp (AgI)) –

Chapter 6_Ionic Equilibrium.indd 148

(1) Both M(OH)2 and X(OH)3 will precipitate together. (2) M(OH)2 will precipitate first. (3) X(OH)3 will precipitate first (4) None of these will precipitate with NH4OH solution.

(1) 3.4 × 10–12 M (2) 3.4 × 10–10 M (3) 3.4 × 10–9 M (4) 3.4 × 10–13 M

3+

(1) (2) (3) (4)

37. There is a solution which is one molar w.r.t. each M2+ and X3+ ions present in it. The Ksp of M(OH)2 and X(OH)3 are 4.0 × 10–10 and 2.7 × 10–14 respectively . If NH4OH solution is added gradually to the above solution which of the following will happen?

–

3. Equal volumes of three acid solutions of pH 3, 4 and 5 are mixed in a vessel. What will be the H+ ion concentration in the mixture?

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Ionic Equilibrium (1) 1.11 × 10−3 M (2) 1.11 × 10−4 M (3) 3.7 × 10−4 M (4) 3.7 × 10−3 M

149

(1) 1.1 × 10-5 (2) 1.8 × 10-5 (3) 9.0 × 10-6 (4) 3.5 × 10-4

(AIPMT 2008)

(AIPMT PRE 2010)

4. Equimolar solutions of the following were prepared in water separately. Which one of the solutions will record the highest pH?

11. An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to

(1) CaCl2 (3) BaCl2

(2) SrCl2 (4) MgCl2 (AIPMT 2008)

5. The ionization constant of ammonium hydroxide is 1.77 × 10−5 at 298 K. Hydrolysis constant of ammonium chloride is (1) 5.65 × 10−12 (3) 6.50 × 10−12

(2) 5.65 × 10−10 (4) 5.65 × 10−13 (AIPMT 2009) −

6. What is the [OH ] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of 0.10 M Ba(OH)2? (1) 0.12 M (2) 0.10 M (3) 0.40 M (4) 0.0050 M (AIPMT 2009) 7. The dissociation constants for acetic acid and HCN at 25°C are 1.5 × 10 −5 and 4.5 × 10 −10, respectively. The equilibrium constant for the equilibrium would be CN − + CH 3COOH  HCN + CH 3COO − (1) 3.0 × 10 (2) 3.0 × 10 (3) 3.0 × 10-5 (4) 3.0 × 10-4 4

5

(AIPMT 2009) 8. If pH of a saturated solution of Ba(OH)2 is 12, the value of its K sp is (1) 4.00 × 10-7 M3 (2) 5.00 × 10-6 M3 (3) 5.00 × 10-7 M3 (4) 4.00 × 10-6 M3 (AIPMT PRE 2010, 2012) 9. In a buffer solution containing equal concentration of B– and HB, the Kb for B– is 10–10. The pH of buffer solution is (1) 7 (2) 6 (3) 4 (4) 10 (AIPMT PRE 2010, 2012) 10. What is[H + ] in mol L−1 of a solution, that is, 0.20 M in CH 3COONa and 0.10 M in CH 3COOH ? Ka for CH 3COOH = 1.8 × 10 −5

Chapter 6_Ionic Equilibrium.indd 149

(1) 100% ionization of electrolyte at normal dilution. (2) increase in both, that is, number of ions and ionic mobility of ions. (3) increase in number of ions. (4) increase in ionic mobility of ions. (AIPMT PRE 2010) 12. A buffer solution is prepared in which the concentration of NH3 is 0.30 M and the concentration of NH4Cl is 0.20 M. If the equilibrium constant, Kb for NH3 equals 1.8 × 10 −5, what is the pH of this solution? (1) 8.73 (2) 9.08 (3) 9.43 (4) 11.72 (AIPMT PRE 2011) 13. In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag+ and Pb2+ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl− concentration is 0.10 M. What will the concentrations of Ag+ and Pb2+ be at equilibrium? (Ksp for AgCl = 1.8 × 10−10, Ksp for PbCl2 = 1.7 × 10−5) (1) [Ag+] = 1.8 × 10−11 M; [Pb2+] = 1.7 × 10−4 M. (2) [Ag+] = 1.8 × 10−7 M; [Pb2+] = 1.7 × 10−6 M. (3) [Ag+] = 1.8 × 10−11 M; [Pb2+] = 8.5 × 10−5 M. (4) [Ag+] = 1.8 × 10−9 M; [Pb2+] = 1.7 × 10−3 M. (AIPMT MAINS 2011) 14. Buffer solutions have constant acidity and alkalinity because (1) these give unionized acid or base on reaction with added acid or alkali. (2) acids and alkalies in these solutions are shielded from attack by other ions. (3) they have large excess of H+ or OH− ions. (4) they have fixed value of pH. (AIPMT PRE 2012) 15. Equimolar solutions of the following substances were prepared separately. Which one of these will record the highest pH value? (1) BaCl2 (2) AlCl3 (3) LiCl (4) BeCl2 (AIPMT PRE 2012)

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OBJECTIVE CHEMISTRY FOR NEET

16. Which of the following salts will give highest pH in water? (1) KCl (2) NaCl (3) Na 2CO3 (4) CuSO4 (AIPMT 2014) 17. Using the Gibbs change, ∆G° = +63.3 kJ, for the following reaction,

Ag 2CO3 (g )  2 Ag + (aq ) + CO23− ( aq )

Ksp of Ag2CO3(s) in water at 25°C is (R = 8.314 JK−1 mol−1) (1) 3.2 × 10 (2) 8.0 × 10 (3) 2.9 × 10−3 (4) 7.9 × 10−2 (AIPMT 2014) −26

−12

18. The Ksp of Ag2CrO4, AgCl, AgBr and AgI are respectively, 1.1 × 10−12, 1.8 × 10−10, 5.0 × 10−13 and 8.3 × 10−17. Which one of the following salts will precipitate last if AgNO3 solution is added to the solution containing equal moles of NaCl, NaBr, NaI and Na2CrO4? (1) AgCl (2) AgBr (3) Ag2CrO4 (4) AgI

(1) 7.0 (2) 1.04 (3) 12.65 (4) 2.0 (RE AIPMT 2015) 22. MY and NY3, two nearly insoluble salts, have the same Ksp values of 6.2 × 10−13 at room temperature. Which statement would be true in regard to MY and NY3? (1) The molar solubilities of MY and NY3 in water are identical. (2) The molar solubility of MY in water is less than that of NY3. (3) The salts MY and NY3 are more soluble in 0.5 M KY than in pure water. (4) The addition of the salt of KY to solution of MY and NY3 will have no effect on their solubilities. (NEET I 2016) 23. The solubility of AgCl(s) with solubility product l.6 × 10−10 in 0.1 M NaCl solution would be

(AIPMT 2015)

19. Which one of the following pairs of solution is not an acidic buffer? (1) H2CO3 and Na2CO3 (2) H3PO4 and Na3PO4 (3) HClO4 and NaClO4 (4) CH3COOH and CH3COONa

21. What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed?

(1) 1.6 × 10−9 M (2) 1.6 × 10−11 M (3) zero (4) 1.26 × 10−5 M (NEET II 2016) 24. The percentage of pyridine (C5H5N) that forms pyridinium ion (C5H5N+H) in a 0.10 M aqueous pyridine solution (Kb for C5H5N = 1.7 × 10−9) is

(RE AIPMT 2015)

(1) 0.013% (2) 0.77% (3) 1.6% (4) 0.0060%

20. Aqueous solution of which of the following compounds is the best conductor of electric current?

(NEET II 2016) 25. Concentration of the Ag+ ions in a saturated solution of Ag2C2O4 is 2.2 × 10−4 mol L−1. Solubility product of Ag2C2O4 is

(1) Ammonia, NH3 (2) Fructose, C6H12O6

(1) 2.66 × 10−12 (2) 4.5 × 10−11 (3) 5.3 × 10−12 (4) 2.42 × 10−8

(3) Acetic acid, C2H4O2 (4) Hydrochloric acid, HCl

(NEET 2017)

(RE AIPMT 2015)

Answer Key Level I  1. (3)

 2. (2)

 3. (1)

 4. (1)

 5. (2)

 6. (3)

 7. (4)

 8. (2)

 9. (3)

10.  (1)

11.  (3)

12.  (2)

13.  (1)

14.  (3)

15.  (1)

16.  (3)

17.  (3)

18.  (2)

19.  (3)

20.  (4)

21.  (1)

22.  (4)

23.  (2)

24.  (4)

25.  (3)

26.  (4)

27.  (1)

28.  (2)

29.  (3)

30.  (3)

31.  (2)

32.  (4)

33.  (1)

34.  (1)

35.  (3)

36.  (1)

37.  (2)

38.  (2)

39.  (2)

40.  (4)

41.  (4)

42.  (1)

43.  (1)

44.  (3)

45.  (4)

46.  (2)

47.  (1)

48.  (4)

49.  (2)

50.  (4)

51.  (1)

52.  (2)

53.  (4)

54.  (2)

55.  (4)

Chapter 6_Ionic Equilibrium.indd 150

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Ionic Equilibrium

151

Level II  1. (3)

 2. (4)

 3. (2)

 4. (3)

 5. (2)

 6. (4)

 7. (1)

 8. (4)

 9. (1)

10.  (3)

11.  (3)

12.  (2)

13.  (2)

14.  (4)

15.  (1)

16.  (2)

17.  (2)

18.  (1)

19.  (2)

20.  (1)

21.  (1)

22.  (1)

23.  (2)

24.  (3)

25.  (3)

26.  (2)

27.  (3)

28.  (2)

29.  (2)

30.  (2)

31.  (1)

32.  (1)

33.  (1)

34.  (4)

35.  (3)

36.  (3)

37.  (2)

38.  (3)

39.  (4)

40.  (1)

41.  (3)

42.  (3)

Previous Years’ NEET Questions  1. (3)

 2. (3)

 3. (3)

 4. (3)

 5. (2)

 6. (2)

 7. (1)

 8. (3)

 9. (3)

10. (3)

11. (4)

12. (3)

13. (4)

14. (1)

15. (1)

16. (3)

17. (2)

18. (3)

19. (3)

20. (4)

21. (3)

22. (2)

23. (1)

24. (1)

25. (3)

Hints and Explanations Level I

 10 −10  pH = − log x = − log  10 −7 − = 7.002 2  

8. (2) Lower the value of Ka, weaker is the acid. 9. (3) Other than H2O, all are strong electrolytes. 10. (1) [OH ] in pure H2O = 10 −





At neutral point, [H + ] = [OH − ]



Therefore, pH at neutral point =



pH at neutral point =



Hence, at pH = 6, the solution is acidic.

−7

Therefore, [OH−] in aqueous solution = 10−7 × 2



We know pOH= - log[OH - ]





= - log(10 -7 ´ 2) = 7 - log 2 = 6.699

Therefore,   3.82 = − log[H + ] ⇒ [H + ] = 1.5 × 10 −4

12. (2) Since concentration Ksp or [Ca 2+ ][F - ]2 > K sp, that is, 2

 10 −2   10 −3  −10  2   2  > 1.7 × 10

    K sp = [Sn 2+ ][S 22- ]

53. (4) If ionic product > solubility product, then the solution is containing more salt than it can dissolve, therefore, precipitation starts and continues till ionic product becomes equal to solubility product.

41. (4) [X - ] [H + ] é 0.1 ù = (14 - 4) + log ê ë 0.2 úû = 10 - log 2

pH = pK a + log

[Ba 2+ ][SO 24- ] > 1.5 ´ 10 -9 [0.01][SO 24- ] > 1.5 ´ 10 -9 [SO 24- ] > 1.5 ´ 10 -7 54. (2) The reaction is -

[CH 3COO ] [CH 3COOH]

( 3 ´ 102 ´ 2) ( 2 ´ 102 )        = 4.75 + log 3 = 5.23

Ca 3(PO 4 )2

+ 2PO 342S

55. (4) The reaction is A 2 X 5  2 A 5+ + 5 X − 2x

Moles of NaOH added 6. (2) Buffer capacity = 4 Change in pH

Chapter 6_Ionic Equilibrium.indd 153

 3Ca 2+ 3S

K sp = [Ca 2+ ]3[PO34- ]2 = ( 3S )3( 2S )2 = 108S 5

= 4.75 + log

[NH +4 ] [NH 4OH]

0.2 mol

Option (4):



pH = 14 - pOH = 9 + log 2 = 9.3

47. (1) pOH = pK b + log

0.2 mol

51. (1) The reaction is SnS 2  Sn 2+ + S 22-

pOH = pK b = 5 - log 2

0.01   = (6.832 - 6.745) = 0.115

0.2 mol

Basic buffer

[NH +4 ] [NH 4OH]

44. (3)     pH = pK a + log

0.2 mol

0.2 mol

Option (3):

2a (14 − 8.26 ) = (14 − 9.26 ) + log (500 × 0.01 × 10 −3 ) 2a = 10 ⇒ a = 2.5 × 10 −2 or 0.025 mol (500 × 10 −3 )



æ 0.01 ´ 2 ö = pK b pOH = pK b + log ç è 0.02 ÷ø       pH = 14 − pOH = 14 − pKb = pKa = 9.26

0.2 mol

+ 4

153

5x

K sp = ( 2 x ) (5x ) = 1.25 ´ 104 x 7 5

5

Level II 2. (4) Conjugate base of [Cr(H 2O)5Cl]2+ = [Cr(H 2O)5Cl]2+ - H + or [Cr(H 2O)4(OH)Cl]+. 5. (2) We know pH = - log[H + ]

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154

OBJECTIVE CHEMISTRY FOR NEET -3 -4      = - log éê10 V + 10 V ùú = 3.26 2V ë û

6. (4) Since, the concentration of HCl £ 10 -6 , contribution of H2O is to be taken into account. HCl 10 −8 0

At t = 0 At t = t eq H 2O

+

H+ 0 10 −8 + x  H+ 10 −8 + x 

Cl − 0 10 −8 + OH − x +

5 + pK b = 14 Þ pK b = 9 13. (2) It is given that DH is negative, therefore as the temperature increases, Kw increases or pKw decreases.

Hence, new neutral point or

H 2S  2H + + S 2- ; K a = K a1 ´ K a 2 = 1.3 ´ 10 -20 [H + ]2[S 2- ] Ka = [H 2S]

(10 -8 + x )( x ) = 10 -14 x 2 + 10 -8 x - 10 -14 = 0 -10 -8 + 4 ´ 10 -14 10 -8 x= = 10 -7 2 2 -8 æ ö 10 [H + ] = 10 -8 + x = 10 -7 + ç è 2 ÷ø æ 10 -8 ö pH = - log[H + ] = - log ç 10 -7 + @ 6.98 2 ÷ø è 7. (1) The solution remains acidic but is near pH = 7 as new concentration is 10−9. x + 10 ( x + 10 )( x ) = 10 x=

+ OH -9

1.3 ´ 10 -20 =

[10 -3 ]2[S 2- ] Þ [S 2- ] = 1.3 ´ 10 -15 M (0.1)

15. (1) We know    pH = - log[H + ] = - log( K a ´ C ) ( pH )HA = - log( 4 ´ 10 -10 ´ C ) ( pH )HB = - log( 1.8 ´ 10 -5 ´ C )

Therefore, ( pH )HA > ( pH )HB

16. (2) CH3COOH  CH3COO− + H+

x

-14

-10 -9 + 4 ´ 10 -14 10 -9 = 10 -7 2 2

[H + ] = 10 −7 +     

pK w > 3[ A 3- ] = 3z

CH 3COOH C C(1 − a ) ≈ C

 CH 3COO − 0 Ca

-

+

H+ 0 Ca + 0.01 ≈ 0.01 (Due to common ion effect)

+

[CH 3COO ][H ] (Ca ) ´ (0.01) = C [CH 3COOH]



Therefore, K a =



-5 -3    1.69 ´ 10 = a ´ 0.01 Þ a = 1.69 ´ 10

11. (3) We have ( pK b )F- + ( pK a )HF = pK w

Chapter 6_Ionic Equilibrium.indd 154

c-x

x-y

y -z

x+ y+z

H+

x+ y+z x+ y +z

x-y

+ HA 2-; Ka2 y -z

z

x  y  z [ As K a1  K a 2  K a 3 ]

18. (1) For the mixture of two weak acids (HCOOH + CH3COOH), we have [H + ] = K a1C1 + K a 2C 2 = 1.8 ´ 10 -4 ´ 0.01 + 1.8 ´ 10 -5 ´ 0.1 0 -3 = 3.6 ´ 10 -3 = 1.9 ´ 10 19. (2) HA + OH −  A − + H 2O; K =

1 K = b Ka Kw

1/4/2018 5:12:53 PM

Ionic Equilibrium



Therefore, K =

28. (2) The reaction is

10 -5 = 109 10 -14

CH 3COOH + NaOH ® CH 3COONa + H 2O 0.4 0.2 0.3 0.2 0 0.5

20. (1) NaHCO3 is an amphiprotic salt. +H+

H2CO3(K1)

pH = pK a + log

3 

HCO

-H+



CO3 2-(K2)

= 4.75 + log( 2.5) = 5.15

1 Therefore, pH = ( pK 1 + pK 2 ) 2

32. (1) We have

     AgCl 

1 pH = [ pK w + pK a + log C ] 2                   8.52 = 1 [14 + 4.75 + log(10 -2 ´ n )] 2 n=2



1 pH = ( pK 1 + pK 2 ) 2 1 = ( 2.15 + 7.20) 2 = 4.675



S=

   10 -33 = [ Al 3+ ](10 -10 )3 [ Al 3+ ] = 10 -3 M 35. (3)

25. (3) [CN ] = 0.5 ´ 2 = 1 M 1 pH = ( pK w + pK a + log C ) 2 1    = (14 + (14 - 9.3) + log 1) 2 = 9.35 6. (2) The reaction is 2 CH 3COOH + NaOH  CH 3COONa + H 2O 100 ´ 0.4 100 ´ 0.2 100 ´ 0.2 0 100 ´ 0.2

AgSCN 

Ag + S1 + S2

+ SCN - ; K sp1 S1

AgBr 

Ag + S1 + S2

+ Br − ; K sp 2 S2

  

K sp1 = [ Ag + ][SCN - ] = (S1 + S2 )(S1 ) (1)

  

K sp 2 = [ Ag + ][Br - ] = (S1 + S2 )(S2 ) (2)



On adding Eq. (1) and Eq. (2), we get K sp1 + K sp 2 = (S1 + S2 ) = [ Ag + ] 1.2 ´ 10 -12 + 5 ´ 10 -13 = [ Ag + ]

The pH value for acidic buffer can be calculated as

[ Ag + ] = 1.3 ´ 10 -6

[Salt] [Acid]

36. (3) The one with lower Ksp will require lesser [Ag+] to precipitate and thus will precipitate first.

(100 ´ 0.2) (100 ´ 0.2)

+

+

- log[H ] = - log[1.8 ´ 10 ] Þ [H ] or [H 3O ] = 1.8 ´ 10

38. (3) The reactions are as follows: -5

CaF2

27. (3) pH can be calculated as

[CH 3COOH] 1 = [CH 3COO - ] 2

Chapter 6_Ionic Equilibrium.indd 155

[CH 3COO - ] [CH 3COOH]

 Ca 2+ S

NaF  Na + 0.1 0 0 0.1

[CH 3COO - ] pH = pK a + log [CH 3COOH] 5.05 = 4.75 + log

1.8 ´ 10 -10 = 10 -10 1.8

K sp = [ Al 3+ ][OH - ]3

-

-5

S  1.8 1.8 + S » 1.8

33. (1) The reaction is Al(OH)3  Al 3+ + 3OH -

     pH = 1 ( pK w + pK 2 + log C ) 2

+

(1)

Therefore, from Eq. (1), we get

24. (3) In Na2S, S2− gets hydrolyzed.

= - log(1.8 ´ 10 -5 ) + log

Ag + + Cl 1.8 + S S

Ksp = (1.8 + S)(S)   



23. (2) NaH2PO4 is an amphiprotic salt.

pH = pK a + log

[CH 3COO - ] [CH 3COOH]

æ 0.5 ö = pK a + log ç è 0.2 ÷ø

22. (1) We have [CH 3COO - ] = 10 -2 ´ n



155

+ +

2F 2S + 0.1 F0 0.1

K sp = [Ca 2+ ][F - ]2 = S(0.1)2 (As 2S Z  > X  (2) Z > X  > Y  − − − − − − (3) Y  > X  > Z  (4) X > Z  > Y  14. A gas X at 1 atm is bubbled through a solution containing a mixture of 1 MY- and 1 MZ- ions at 25°C. If the reduction potentials of Z > Y > X, then (1) (2) (3) (4)

Y will oxidize X but not Z. Y will oxidize both X and Z. Y will oxidize Z but not X. Y will reduce both X and Z.

15. In a series of chemical test reactions, the following observations were made when several metallic element samples were treated with 3.00 M hydrochloric acid solution. Zn: metal dissolves, hydrogen gas is emitted. Cu: no reaction. Ag: no reaction. Mg: metal dissolves, hydrogen gas is emitted. Mn: metal dissolves, hydrogen gas is emitted. Which of the following statements is true? (1) Mg, Mn, Zn are more reactive than Ag, Cu; but less reactive than H2. (2) Mg, Mn, Zn are less reactive than Ag, Cu; but more reactive than H2. (3) Mg, Mn, Zn are more reactive than Ag, Cu; and more reactive than H2. (4) Mg, Mn, Zn are less reactive than Ag, Cu; and less reactive than H2.

(1) 17.0 (2) 35.0 (3) 51.0 (4) 68.0 21. Number of moles of electrons change per mole of Pb(N3)2 in Pb(N3)2 + Co (MnO4)3 ® CoO + MnO2 + Pb3O4 + NO is 42 44 (2) 3 3 (3) 10 (4) 44 (1)

22. In an experiment 50 mL of 0.1 M solution of a salt reacted with 25 mL of 0.1 M solution of sodium sulphite. The half equation for the oxidation of sulphite ion is SO23- (aq ) + H 2O( l ) ® SO24- (aq ) + 2H + + 2e

23. 2H3PO4 + 3Ca(OH)2 ® Ca3(PO4)2 + 6H2O

The values of x, y and z in the reaction are, respectively (1) 2, 5 and 8 (2) 2, 5 and 16 (3) 5, 2 and 8 (4) 5, 2 and 16

17. In the reaction,

Zn + NaNO3 + NaOH ® Na2ZnO2 + NH3 + H2O

The molar coefficients of Zn and NaNO3 are

Chapter 7_Redox Reactions.indd 173

Equivalent weight of H3PO4 in this reaction is (1) 98 (2) 49 (3) 32.66 (4) 24.5

24. The oxide of an element possesses the formula M2O3. If the equivalent weight of the metal is 9, then the atomic weight of the metal will be (1) 9 (2) 18 (3) 27 (4) 35

16. Consider the following reaction z xMnO 4 - + yC 2O 24- + zH + → xMn 2+ + 2 yCO2 + H 2O 2

If the oxidation number of metal in the salt was 3, what would be the new oxidation number of metal? (1) zero (2) 1 (3) 2 (4) 4

Balancing of Redox Reactions



173

25. The equivalent weight of Fe3O4 in the reaction,



Fe3O4 + KMnO4 ® Fe2O3 + MnO2 would be

(1) M/6 (2) M (3) 2M (4) M/3 26. 0.534 g Mg displaces 1.415 g Cu from the salt solution of Cu. Equivalent weight of Mg is 12. The equivalent weight of Cu would be (1) 15.9 (2) 47.7 (3) 31.8 (4) 8.0

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174

OBJECTIVE CHEMISTRY FOR NEET

27. Approximate atomic weight of an element is 29.89. If its equivalent weight is 8.9, the exact atomic weight is (1) 26.89 (2) 8.9 (3) 17.8 (4 26.7 28. Equivalent weight of KMnO4when it is converted into MnSO4 is (1) M/5 (2) M/6 (3) M/3 (4) M/2 29. A bivalent metal has the equivalent weight of 12. The molecular weight of its oxide will be (1) 24 (2) 34 (3) 36 (4) 40 30. What is the equivalent mass of HCl in the given reaction? 2KMnO4 + 16HCl ® 2KCl + 2MnCl2 + 5Cl2 + 8H2O M M (2) 1 10 8M M (3) (4) 5 5 31. Equivalent weight of H3PO2 when it disproportionate into PH3 and H3PO3 is (Mol. wt. of H3PO2 = M) (1)

(1) M (2) M/2 (3) M/4 (4) 3M/4 32. In the following reaction, MnO2 + 4HCl ® MnCl2 + Cl2 + 2H2O

n-factor of HCl is (1) 1/2 (2) 1 (3) 2 (4) 4

Redox Titrations 33. In the titration of K2Cr2O7 and ferrous sulphate, following data is obtained: V1 mL of M1 K2Cr2O7 requires V2 mL of M2 FeSO4. Which of the following relations are true? (1) 6M1V1 = M2V2 (2) M1V1 = 6 M2V2 (3) N1V1 = 2N2V2 (4) M1V1 = M2V2 34. An element A in a compound ABD has oxidation number A n-. It is oxidized by Cr2O72- in acidic medium. In the experiment, 1.68 × 10–3 mol of K2Cr2O7 were used for 3.26 × 10-3 mol of ABD. The new oxidation number of A after oxidation is (1) 3 (2) 3 – n (3) n – 3 (4) +n 35. A solution is containing 2.52 g L–1 of a reductant. 25 mL of this solution required 20 mL of 0.01M KMnO4 in acid medium for oxidation. Given that each of the two atoms which undergo oxidation per molecule of reductant, suffer an increase in oxidation state by one unit. The molecular weight of reductant is

Chapter 7_Redox Reactions.indd 174

(1) M = 126 (2) M = 130 (3) M = 128 (4) M = 127 36. 10 mL of H2O2 solution (volume strength = x) required 10 mL of 0.1 N MnO -4 solution in acidic medium. Hence, x is (1) 0.56 (2) 5.6 (3) 0.1 (4) 10.0 37. In the reaction, P + Cr2O7 2- + H+ ® PO4 3- + Cr+3 + H2O if 0.2 moles of Cr2O72- are taken then moles of ‘P’ reacted is (1) 0.34 (2) 0.14 (3) 0.24 (4) 0.04 38. 3 mol of a mixture of FeSO4 and Fe2(SO4)3 required 100 mL of 2 M KMnO4 solution in acidic medium. Hence, mol fraction of FeSO4 in the mixture is (1) 1/3 (2) 2/3 (3) 2/5 (4) 3/5

Previous Years’ NEET Questions 1. The number of moles of KMnO4 that will be needed to react with one mole of sulphite ion in acidic solution is 3 (1) 1 (2) 5 4 2 (3) (4) 5 5 (AIPMT 2007) 2. Number of moles of MnO -4 required to oxidize one mole of ferrous oxalate completely in acidic medium will be (1) 0.2 mol (2) 0.6 mol (3) 0.4 mol (4) 7.5 mol (AIPMT 2008) 2– 3. Oxidation numbers of P in PO3– 4 , of S in SO 4 and that of 2– Cr in Cr2O7 are respectively

(1) -3, +6 and +6. (2) +5, +6 and +6. (3) +3, +6 and +5. (4) -5, +3 and +6. (AIPMT 2009) 4. Oxidation states of P in H 4P2O5, H 4P2O6 , H 4P2O7 , are respectively (1) +5, +3, +4 (2) +5, +4, +3 (3) +3, +4, +5 (4) +3, +5, +4 (AIPMT PRE 2010) 5. When Cl2 gas reacts with hot and concentrated sodium hydroxide solution, the oxidation number of chlorine changes from (1) zero to +1 and zero to –5. (2) zero to –1 and zero to +5.

1/4/2018 5:13:37 PM

Redox Reactions (3) zero to –1 and zero to +3. (4) zero to +1 and zero to –3. (AIPMT PRE 2012) 6. In which of the following compounds, nitrogen exhibits highest oxidation state?

175

8. Assuming complete ionization, same moles of which of the following compounds will require the least amount of acidified KMnO4 for complete oxidation? (1) FeC2O4 (2) Fe(NO2)2 (3) FeSO4 (4) FeSO3 (RE-AIPMT 2015)

(1) N2H4 (2) NH3 (3) N3H (4) NH2OH (AIPMT PRE 2012) 7. A mixture of potassium chlorate, oxalic acid and sulphuric acid is heated. During the reaction which element undergoes maximum change in the oxidation number? (1) S (2) H (3) Cl (4) C

9. Which one of the following statements is correct when SO2 is passed through acidified K2Cr2O7 solution? (1) The solution turns blue. (2) The solution is decolorized. (3) SO2 is reduced. (4) Green Cr2(SO4 )3 is formed. (NEET I 2016)

(AIPMT PRE 2012)

Answer Key Level I  1. (2)

 2. (2)

3. (2)

 4. (2)

 5. (4)

 6. (1)

 7. (4)

 8. (1)

 9. (1)

10.  (2)

11. (3)

12. (4)

13. (3)

14. (3)

15. (4)

16. (2)

17. (4)

18. (4)

19. (1)

20. (3)

21. (3)

22. (2)

23. (3)

24. (1)

25. (3)

26. (3)

27. (3)

28. (1)

29. (2)

30. (3)

31. (4)

32. (2)

33. (1)

34. (1)

35. (2)

36. (4)

37. (4)

38. (1)

39. (3)

40. (2)

41. (2)

42. (3)

43. (4)

44. (2)

45. (3)

46. (2)

47. (4)

48.   (2)

49. (2)

50. (2)

51. (4)

52. (1)

53. (3)

54. (2)

55. (3)

56. (4)

Level II  1. (3)

 2. (2)

 3. (2)

 4. (2)

 5. (4)

 6. (2)

 7. (2)

 8. (3)

 9. (1)

10. (3)

11. (3)

12. (2)

13. (1)

14. (1)

15. (3)

16. (2)

17. (2)

18. (2)

19. (3)

20. (2)

21. (2)

22. (3)

23. (3)

24. (3)

25. (2)

26. (3)

27. (4)

28. (1)

29. (1)

30. (3)

31. (4)

32. (1)

33. (1)

34. (2)

35. (1)

6. (3)

7. (3)

8. (3)

9. (4)

Previous Years’ NEET Questions 1. (4)

2. (2)

3. (2)

4. (3)

5. (2)

Hints and Explanations Level I 1. (2) Oxidation state of carbon can be calculated as:

CH3Cl : x + 3(1) + 1(−1) = 0 Þ x = −2



CCl4 : x + 4(−1) = 0 Þ x = 4



CHCl3 : x + 1(1)+3(−1) = 0 Þ x = 2



CH2Cl2 : x + 2(1) + 2(−1) = 0 Þ x = 0

Chapter 7_Redox Reactions.indd 175

2. (2) The reaction can be represented as 0 +1 –1 Cl2 + H2O2(aq) Oxidizing Reducing agent agent

+1 –1 0 2HCl + O2

Oxidation by 1 unit

1/4/2018 5:13:37 PM

176

OBJECTIVE CHEMISTRY FOR NEET

4. (2) A species can act as a reducing agent if the central atom is not present in its highest oxidation state. 5. (4) The oxidation state of nitrogen in N2O5 and HNO3 is

N2O5: 2x + 5(−2) = 0 Þ x = 5



HNO3: 1 + x + 3(−2) = 0 Þ x = 5

6. (1) 2HI + H 2SO 4 ® I 2 + SO2 + 2H 2O

On left hand side oxidation state of sulphur = 6



On right hand side oxidation state of sulphur = 4



There is a decrease of oxidation state of sulphur from +6 in sulphuric acid to +4 in sulphur dioxide. It means that H2SO4 is reduced and iodide ions are oxidized to iodine by conc. H2SO4.

7. (4) MnO 4- : x + 4(−2) = −1 Þ x = 7



H2SO3 itself gets oxidized hence acts as a reducing agent while Sn4+ getting reduced to +2 oxidation state acts as an oxidizing agent.

21. (3) K is higher up in activity series in comparison to other elements (Mg, Na, Br2). Therefore, it will be the most strongly reducing substance. 23. (3) The order of reactivity of these metals is Mg > Mn > Zn > Cu > Ag. Out of these metals, Cu and Ag are least reactive and their reaction with acid will not produce hydrogen gas.

More reactive metal can substitute the less reactive metal from their aqueous solution, so, Mn2+ will substitute Zn2+ from aqueous solution.

24. (1) The reaction involved is

Manganese has highest oxidation state of 7. Therefore, it can only reduce itself and oxidize others.

9. (1) Statement (1) is not correct. For example, during neutralization and precipitation reactions oxidation number does not change. Hence, these reactions are not redox reactions. +3 -2 2

+1

+5 -2 3

-2

+

10. (2) N O + H 2 O → N O + 2H + 2e -3

-2

-

-1

2IO-3 + 10e - + 12H + ® I 2 + 6H 2O [ I - ® I - + 2e - ] ´ 5 2IO3- + 10I - + 12H + ® 6I 2 + 6H 2O

26. (3) The balanced reaction is

+1

16. (2) Suppose oxidation state of I is x, then 3x = -1 Þ x = -1/3 17. (4) No change in the oxidation state of chromium takes place. K 2 Cr2 O7

Potassium dichromate

Alkaline medium →

IO-3 + 5I - + 6H + ® 3I 2 + 3H 2O

25. (3) The balanced reaction is Zn + 2OH - ® ZnO22- + H 2

13. (3) The order is N H 4+ < N 2 H 4 < N H 2OH < N 2 O.

+6

or

60H - + ClO4- + 3MnO 2 ® ClO - + 3MnO24- + 3H 2O

Therefore, x = 1, y = 3 and z = 6

27. (3) Reduction 3e-

+6

K 2 Cr O 4

Cr2O72- + Fe2+ + C2O42+6 +2 +3

Potassium chromate

19. (1) Sum of oxidation numbers of all the elements in the molecular formula of any compound is zero. Putting the oxidation number of all the elements in the molecular formula of all the compounds, we find that in molecular A3(BC4)2 the sum of oxidation number of all the elements comes out to zero.

Cr3+ + Fe3+ + CO2 +3 +3 +4 1e1e-

2 × 3e- = 6eCr2O72- + Fe2+ + C2O42-

2Cr3+ + Fe3+ + 2CO2

3 ´ ( +2) + 2 ´ ( +5 + 4 ´- 2) = 0 2 × 1e- = 2e-

20. (3)

2 × 1e- × 2 Oxidation (2e–) H2SO3 + Sn + H2O 4+

+4

Sn + HSO4 + 3H

+

The balanced reaction is Cr2O72- + 2Fe2+ + 2C 2O 24- ® 2Cr 3+ + 2Fe3+ + 4CO2

+6 Reduction (2e–)

Chapter 7_Redox Reactions.indd 176

–

2+



Total 6e- are involved in given redox reaction.

1/4/2018 5:13:39 PM

Redox Reactions 28. (1) Eq. wt. of metal = Eq. wt. of H2

39. (3) When the metal is burnt, its oxide is formed.

0.1 34.2 = ´ 2 Þ Eq. wt. = 32.7 Eq. wt. 22400



Let the weight of metal be x.



Eq. wt. of metal = Eq. wt. of metal oxide x 1.24 x = Þ (Eq. wt.)Metal = 33.3 (Eq. wt.)Metal (Eq. wt.)Metal + 8

29. (2) Eq. wt. of metal = Eq. wt. of metal oxide

Therefore,

+3

n-factor of As2S3 = 2(5 – 3) = 4

30. (3) 35.5 g of Cl implies 1 eq. wt. of Cl

M 4 1 5 1. (2) The reaction is 4e - + 4H + + HNO3 ® N 2O + H 2O 4 2 2

(Eq. wt.)Metal + (Eq. wt.)Cl = 74.5 Þ (Eq. wt.)Metal = 39 +5

+5

40. (2) As2 S 3 ® AsO 34-

0.5 0.79 = Þ (Eq. wt.)Metal = 14 (Eq. wt.)Metal (Eq. wt.)Metal + 8



+5

31. (4) In the reaction NaOH + H 3 P O4 ® NaH 2 P O4 + H 2O, no change in oxidation state of phosphorus occurs. So, equivalent weight = molecular mass of H3PO4 = 3 × 1 + 31 × 1 + 16 × 4 = 98 g.

Therefore, Eq. wt. of As2S3 =

42. (3) The reactions involved are +7

Acidic medium : MnO4- + 8H + + 5e - → Mn+2 + 4H 2O +7



Therefore, M + x(35.5) = 80

(1)



M Also,        = 4.5 x

(2)



From Eq. (1) and Eq. (2), we get



4.5 + 35.5 =



(Eq. wt.)Metal = (Eq. wt.)Chloride



Therefore,

56 = 18.6 3

35. (2) Eq. wt. of M = Eq. wt. of Cu

Therefore,

1.5 4 = ÞM = 24 ( M/2) (64/2)

37. (4) Eq. wt. of metal hydride = Eq. wt. of H2 0.84 0.042   (Eq. wt.)M  19 (Eq. wt.)M  1 2/2

Therefore, (Eq. wt.)Metal hydride = 19 + 1 = 20

38. (1) Let the metal oxide be M2Ox. Therefore, 60 =



2M ´ 100 2 M + 16 x

60 M /x M = Þ = 12 = (Eq. wt.)Metal x 100 M /x + 8

Chapter 7_Redox Reactions.indd 177

-2

+5

+5

+6

+2



n-factor of As2S3 = 2(5 – 3) + 3(6 – (−2)) = 28



n-factor of HNO3 = 1 × (5 – 2) = 3



Eq. wt. of As2S3 = Eq. wt. of HNO3

Therefore, number of moles of As2S3 × (n factor)As2S3 = moles of As2S 3 ´ ( n factor )AsNumber = No. of moles of HNO 3 ´ ( n factor )HNO3 2 S3

x 0.475 Þ x = 0.12 g = 12 12 + 35.5

Therefore, Eq. wt. =

+4

43. (4) The reaction is +3

34. (1) Oxidation number of iron in Fe2O3 is +3.

+7

Neutral medium : MnO4- + 2H 2O + 3e - → MnO2 + OH -

As 2 S3 + H N O3 ® H 3 As O 4 + H 2 S O 4 + N O

80 Þx=2 x Also, from Eq. (2), M = 9

33. (1) Let the weight of metal be x

+6

Basic medium : MnO4- + e - → MnO24-

32. (2) Let the formula be MClx



177



No. of moles of As2S 3 =

1´ 3 3 = 28 28 +2

+7

+3

+2

44. (2) The reaction is Mn O4- + Fe2+ ® Fe3+ + Mn 2+ n-factor of MnO -4 = 1 ´ (7 - 2) = 5 n-factor of Fe2+ = 1 × 1 = 1 Eq. wt. of MnO -4 = Eq. wt. of Fe2+



Therefore, No. of moles of MnO -4 × 5 = No. of moles of Fe2+ × 1 1 × V × 5 = nFe2+



The reaction is Cr2 O72- + Fe2+ ® Fe3+ + Cr 3+



+6

+2

+3

+3

n-factor of Cr2O72- = 2(6 - 3) = 6 n-factor of Fe2+ = 1 × 1 = 1

Eq. wt. of Cr2O72- = Eq. wt. of Fe2+



Therefore, no. of moles of     Cr2O72- × 6 = no. of moles of Fe2+ × 1







Hence, Fe(II) will be oxidized more by K2Cr2O7.

1 × V × 6 = nFe2+

1/4/2018 5:13:42 PM

178

OBJECTIVE CHEMISTRY FOR NEET

45. (3) Let the weight of FeSO4·7H2O be x in 1 L.

No. of moles of FeSO 4 × 7H 2O =



x in 1 L 278

+3

Therefore, no. of moles of FeSO4·7H2O in 25 mL 25 x = ´ 278 1000 Eq. wt. of KMnO4 = Eq. wt. of FeSO4·7H2O 1 20 ´ ´ 10-3 = No. of moles of FeSO4·7H2O × 1 10 (As n-factor = 1 of FeSO4·7H2O) 1 x 1 ´ 10-3 = ´ Þ x = 278 × 40 × 2 × 10−3 10 278 40

20 ´



% of FeSO 4 × 7H 2O =

278 ´ 40 ´ 2 ´ 10-3 ´ 100 = 89 % 25

Therefore, n-factor of H3PO3 = 2



Normality = Molarity × 2



or Normality = 0.1 × 2 = 0.2 N

+4

+2

n-factor of NaHC2O4 = 2(4 – 3) = 2

Eq. wt. of NaHC2O4 = Eq. wt. of KMnO4



  



Dividing Eq. (1) by Eq. (2), we get





b ´ 2 = (100 ´ 10-3 ´ 0.2)´ 5  (Mol. wt.)NaHC2 O4

a ´1 1 a 2 = or = b´2 5 b 5





3 5 = Þ (Eq. wt.)Metal = 33 (Eq. wt.)Metal + 8 (Eq. wt.)Metal + 35.5

53. (3) n-factor of NaHC2O4 with NaOH = 1

  Eq. wt. of NaHC2O4 = Eq. wt. of NaOH





47. (4) The reaction is Mn O2 + HCl ® Mn + Cl 2



   V = 0.1



n-factor of NaHC2O4 with KMnO4 = 2(4 – 3) = 2

+4

+2

-1

2+

0

n-factor of MnO2 = 1 × (4 – 2) = 2

(10 × V × 10−3) × 1 = (10 × 0.1 × 10−3) × 1

n-factor of Cl2 = 2 × (1 – 0) = 2





   Eq. wt. of NaHC2O4 = Eq. wt. of KMnO4

Eq. wt. of MnO2 = Eq. wt. of Cl2







No. of moles of MnO2 × 2 = 0.1



   M = 0.04



x 8.7 ´ 2 = 0.1 or x = 87 2 8.7/2 Therefore, % purity = ´ 100 = 43.5% 10

(10 × V × 10−3) × 2 = (10 × 10−3 × M) × 5

54. (2) Eq. wt. of FeCl3 = Eq. wt. of Fe(OH)3 100 × 10 -3 × N =

 

48. (2)

1.425 Þ N = 0.48 (89/3)

55. (3) MnO -4 + C 2O 24- + H + ® Mn 2+ + CO 2 + H 2O

Eq. wt. of Acid = Eq. wt. of NaOH

       

0.52 = 100 ´ 0.1 ´ 10-3 Þ (Eq. wt.)Acid = 52 (Eq. wt.)Acid

  Eq. wt. of Na2SO3 = Eq. wt. of K2Cr2O7



 (M × 20 × 10 ) × 2 = (30 × 0.01 × 10 ) × 6 −3

MnO -4 + Cl - ® Cl 2 + Mn 2+

MnO -4 gets easily reduced to Mn2+ in the presence of HCl, hence not used.



49. (2) The reaction is SO23- + Cr2O72- ® Cr 3+ + SO24-

56. (4) n-factor of H3PO2 = 1 and n-factor of NaOH = 1

−3







          1×1=x+1

50. (2) Applying M1V1 = M2V2, we get



   n-factor of H3PO3 = 2 and n-factor Ca(OH)2 = 2

0.4 × V = 50 × 0.2 Þ V = 25 mL





       1 × 2 = y × 2



     n-factor of H3PO4 = 3 and n-factor of Al(OH)3 = 3





  M = 0.045 M



(2)

52. (1) Eq. wt. of metal oxide = Eq. wt. of metal chloride

46. (2) Number of H+ ion replaced in H3PO3 = 2

+7

Na HC 2O4 + K Mn O4 ® CO2 + Mn 2+





When NaHC2O4 reacts with KMnO4, it is a redox reaction

51. (4) NaHC2O4 when reacts with NaOH has n-factor = 1

Therefore, no. of replaceable H+ = 1



Eq. wt. of NaHC2O4 = Eq. wt. of NaOH



a     ´ 1 = (100 ´ 10-3 ´ 0.2)´ 1  (Mol. wt.)NaHC2 O4

Chapter 7_Redox Reactions.indd 178

Eq. wt. of H3PO2 = Eq. wt. of NaOH

Eq. wt. of H3PO3 = Eq. wt. of Ca(OH)2



(2)

Eq. wt. of H3PO4 = Eq. wt. of Al(OH)3

         1×3=z×3 (1)

(1)

(3)

Therefore, from Eq. (1), (2) and (3), we get



x:y:z=1:1:1

1/4/2018 5:13:44 PM

Redox Reactions

Level II



179

Net electrons transferred/SCN− = 8e−s lost – 2e− gained

(S2- to S+6) (C+4 to C+2) 1. (3) Marshall’s acid is represented as O HO

S



S

O

O

OH

Hence, oxidation state of sulphur = +6

2. (2) The molecular formula of carbon suboxide is C3O2. Let the oxidation number of carbon be x

Therefore, no. of electrons transferred in Al(SCN)3 = 6 × 3 = 18

10. (3) C changes from CN− to CO

O

O



    = 6e−s/SCN−

O



Oxidation state of C in CN− = +1



Oxidation state of C in CO = +2

11. (3) Cl2 disproportionates as follows: Cl2

Oxidation

n factor = n

3(x) + 2(−2) = 0 Þ x = 4/3

-2

O−

S

Na+

+6

5. (4) Suppose oxidation state of oxygen = x, then we have B4O10

CsO2

4 × 1 + 10(x) = 0 ⇒ 4 + 10x = 0 ⇒ 4 = –10x 4 ⇒ x=10

KO3

+1 × 2(x) = 0 ⇒ 1 × 2x = 0 ⇒ 1 = -2x ⇒ x = -

+1 × 3(x) = 0 ⇒ 1 = –3x 1 ⇒ x=3

1 2

6. (2) Let the oxidation state of oxygen be x

2 × x = +1 Þ x = 1/2



7. (2) The structure of CrO5 has two peroxylinkages. O

O

Cr O

Therefore, n =



Hence,



O

O−





Na+

S



Cl 2 ® ClO-3 + Cl -

If

Difference in maximum and minimum oxidation state = 5 – (−1) = 6

14. (1) Higher the reduction potential, stronger the oxidizing agent. Since reduction potentials decrease in the order Z > Y > X, their oxidizing powers also decrease in the ­similar order. Hence, Y is a stronger oxidizing agent than X but weaker than Z. Therefore, Y can oxidize X but not Z. 15. (3) The increasing reactivity order of given metals is Ag < Cu < Zn < Mn < Mg.

2MnO -4 + 5C 2O 24 - + 16H + ® 2Mn 2+ + 10CO2 + 8H 2O

x + 5(−2) = 0 Þ x = 10

Hence, x = 2, y = 5, z = 16

17. (2) The reaction is

But this is greater than maximum oxidation state of Cr. Hence Cr must be present in its maximum oxidation state 6.

+5

0

+2

-3

5H 2O + 4 Zn + N O-3 ® 4 Zn O22- + N H 3 + 7H + 18. (2) The balanced reaction is

8. (3) SO42-

SCN(-1) S



n1 = 10, n2 = 2

16. (2) The balanced equation is

O

Let the oxidation state of Cr be x



n1 ´ n2 10 = (Given) (n1 + n2 ) 6

13. (1) As TX does not react with Y or Z solutions, X is least oxidizing, and since TZ reacts with X, Z is more oxidizing than X. And Y is most oxidizing as it reacts with both X and Z solutions.

O



n factor = n2



-1 l- + KSO4

+5 KlO3 + SO2

4. (2)

Redu

ction

3. (2) The reaction involved is

n factor = n1

-2

Chapter 7_Redox Reactions.indd 179

C

N

+4

-3

+6

4Mg + 10HNO3 ® 4Mg(NO3 )2 + N 2O + 5H 2O

+ CN-

19. (3) The balanced reaction is (-1) C +2

N -3

+5

-2

+6

+5

0

+3

22H + + 3 As2 S 5 + 5 Cr2 O72- ® 6 AsO 34- + 15 S+ 10 Cr 3+ + 11 H 2O

Therefore, coefficient of S = 15

1/4/2018 5:13:46 PM

180

OBJECTIVE CHEMISTRY FOR NEET

20. (2) Let the formula be M2(SO4)x

29. (4)

2 Mol. wt. × 100 42.2 = 2 Mol. wt. × 96 x

30. (3) Only 10 mol of electrons are transferred by 16 mol of HCl. 10 5 n-factor = = 16 8

42.2 2 Mol. wt./x = 100 2 Mol. wt./x + 48 Eq. wt. = +2

-1/3

+3

2 Mol. wt. = 35 x

+7

+2

+4

8/3

+2

21. (2) Pb(N 3 )2 + Co(Mn O4 )3 → CoO + Mn O2 + Pb3 O4 + N O

Mol. wt. = 12 Þ Mol. wt. = 24 2

8M 5

31. (4) The disproportionation reaction is n1 =

 8   1  n-factor of Pb(N 3 )2 = 1 ×  - 2 + 6  2 -  -   3   3  

        =

Eq. wt. =

Therefore,

PH3

4

-3

H3 PO2

2 42 44 + = 3 3 3

n2 =

+1

22. (3)    Eq. wt. of salt = Eq. wt. of sulphite

2

H3 PO3

(50 × 0.1) × n-factor = (25 × 0.1) × 2 +3



   (n-factor)salt = 1



Since, SO23 gets oxidized, thus metal in the salt solution must be reduced.



Therefore, 3 – n = 1 Þ n = 2

1 1 1 = + n n1 n2 n=

23. (3) n-factor of H3PO4= 3 (no. of replaceable H+)

Eq. wt. =



98 = 32.66 3

24. (3) Eq. wt. of metal (Molecular wt.)metal = Þ Mol. wt. = 3 ´ 9 = 27 3 25. (2) Fe3 O4 ® Fe2 O3



8ö æ n-factor of Fe3O4 = 3 ç 3 - ÷ = 1 3ø è M Therefore, Eq. wt. of Fe3O4 = 1

26. (3) We have Eq. wt. of Mg = Eq. wt. of Cu 0.534 1.415 = Þ (Eq. wt.)Cu = 31.8 12 (Eq. wt.)Cu Atomic weight 7. (4) We have Eq. wt.= 2 n factor n factor » 3 =

29.89 8.9



Given



Therefore, Exact atomic weight = 8.9 × 3 = 26.7 +7

Therefore, n- factor =



2 1 = 4 2

33. (1) Eq. wt. of K2Cr2O7 = Eq. wt. of FeSO4

M1V1 × 6 = M2V2 × 1



or         6M1V1 = M2V2

34. (2) Eq. wt. of ABD = Eq. wt. of K2Cr2O7





(3.26 × 10−3) [x – (−n)] = 1.68 × 10−3 × 6

    x + n = 3   x = 3 – n 35. (1) Let the molecular weight of reductant be x Molarity =

2.52 x



n-factor of reductant = 2 × (1) = 2









Eq. wt. of reductant = Eq. wt. of KMnO4 25 ´

2.52 ´ 2 ´ 10-3 = ( 20 ´ 0.01 ´ 10-3 )´ 5 Þ x = 126 x

+2

28. (1) The reaction is Mn O4- ® Mn 2+

n-factor of KMnO4 = 1 × (7 – 2) = 5



Therefore, Eq. wt. = M/5

Chapter 7_Redox Reactions.indd 180

3M 4

32. (1) 2 mol of electrons are transferred by 4 mol of HCl.

+3

8/3



Therefore, Eq. wt. =



4´ 2 8 4 = = ( 4 + 2) 6 3

36. (1) Normality of x volume of H2O2 =



x 5.6

Eq. wt. of H2O2 = Eq. wt. of MnO -4

1/4/2018 5:13:50 PM

Redox Reactions



10 ´



5. (2) The chemical reaction is Cl 2 + NaOH ® NaCl + NaClO3

x = 10 ´ 0.1 Þ x = 0.56 5.6



37. (3) Eq. wt. of P = Eq. wt. of Cr2O72

No. of moles of P × 5 = 0.2 × 6



No. of moles of P = 0.24

38. (1) Let the no. of moles of FeSO4 be a and Fe2(SO4)3 be 3 – a

Eq. wt. of FeSO4 = Eq. wt. of KMnO4 (As Fe2(SO4)3 will not oxidize)



6. (3) In N2H4, the oxidation state of nitrogen is -2; in NH3, the oxidation state of nitrogen is -3; in N3H, the oxidation state of nitrogen is -1/3 and in NH2OH, the oxidation state of nitrogen is -1. 7. (3) The reaction is

KClO 3 + H 2C 2O4 + H 2SO4 ® K 2SO4 + KCl + CO2 + H 2O

b=2



The oxidation state of Cl in KClO3 is +5 while in KCl it is -1.

Therefore, mole fraction of FeSO4 = 1/3

8. (3) KMnO4 + FeC 2O 4 ® Mn 2+ + Fe3+ + 2CO2 n - factor for KMnO 4 =[( +7 )´ 1 - ( +2)´ 1] = 5 n - factor for FeC 2O4 =[( +2)´ 1 - ( +3)´ 1]+ [( +3)´ 2 - ( +4)´ 2]= 3

1. (4) The balanced chemical equation is 2MnO -4 + 6H + + 5SO 23- ® 2Mn 2+ + 5SO24- + 3H 2O



According to the reaction, 5 mol of SO

2. (2) The reaction involved is +

3+

5 mol of FeC2O4 requires 3 mol of MnO -4 . Therefore, 1 mol of FeC2O4 requires 3/5 = 0.6 mol of MnO -4 .

3. (2) In PO3– 4 , the oxidation state of P atom is    x - 8 = -3 ⇒ x = +5 In SO2– 4 , the oxidation state of S atom is

 

x - 8 = -2 ⇒ x = +6

2– 7



In Cr2O , the oxidation state of Cr atom is





2 x - 14 = -2 ⇒ x = +6

4. (3) The oxidation state of P in the given compounds are as follows: In H 4P2O5 : 0 = 4 ´ ( +1) + 2 ´ ( x ) + 5 ´ ( -2) Þ x = +3

In H 4P2O6 : 0 = 4 ´ ( +1) + 2 ´ ( x ) + 6 ´ ( -2) Þ x = +4

In H 4P2O7: 0 = 4 ´ ( +1) + 2 ´ ( x ) + 7 ´ ( -2) Þ x = +5

Chapter 7_Redox Reactions.indd 181



Let’s the number of reacting moles of FeC2O4 be x. There3x fore, number moles of KMnO4 = 5 Similarly, for reaction











Number moles of KMnO4 =



For reaction KMnO4 + FeSO 4 → Mn 2+ + Fe3+



n - factor for KMnO 4 = 5; n - factor for FeSO4 = 1



Number moles of KMnO4 =



For reaction KMnO4 + FeSO 3 ® Mn 2+ + SO 24- +Fe3+



n - factor for KMnO 4 = 5; n - factor for FeSO3 = 3



Number moles of KMnO4 =



Hence, FeSO4 required least amount of acidified KMnO4 for complete oxidation.

KMnO4 + Fe(NO 2 )2 ® Mn 2+ + Fe3+ + NO -3 n - factor for KMnO 4 = 5 ; n - factor for Fe(NO2 )2 = 5

2+

5FeC 2O4 +3MnO + 24 H ® 5 Fe + 10 CO 2 + 3 Mn + 12 H 2O





23

requires 2 mol 2 of MnO4 −. Therefore, 1 mol of SO3 2− will require moles 5 of MnO4 −

4



In Cl2, the oxidation state of Cl is 0; in NaCl, the oxidation state of Cl is -1 and in NaClO3, the oxidation state of Cl is +5.

a × 1 = 100 × 2 × 10 × 5 Þ a = 1 −3

Previous Years’ NEET Questions



181

5x =x 5

x 5

3x 5

9. (4) When SO2 is passed through acidified K2Cr2O7, the following reaction takes place K 2Cr2O7 + 3SO2 + H 2SO4 → K 2SO4 + Cr2(SO4 )3 + 3H 2O (green)

1/4/2018 5:13:54 PM

Chapter 7_Redox Reactions.indd 182

1/4/2018 5:13:54 PM

8

Solid State

Chapter at a Glance 1. Characteristics of a Solid (a) (b) (c) (d)

Particles are closely packed and are held together by strong intermolecular inter-ionic force of attraction. Have only vibrational motion. Have definite shape and volume. Have high density and low compressibility.

2. Types of Solid (a) Depending upon the arrangement of particles, the solids are classified as: (i) Amorphous solids (e.g., glass, plastic, rubber, etc.) (ii) Crystalline solids (e.g., metals, graphite, diamond, NaCl, etc.,). (b) Differences between amorphous and crystalline solids: Property Particles arrangement

Amorphous solid Random

Crystalline solid Proper

Melting point

Not sharp

High and sharp

Nature

Isotropic

Anisotropic

Planes

Cut along random direction and have irregular surfaces.

Cut along specific crystal planes.

Heat of fusion

Do not have definite heat of fusion.

Characteristics heat of fusion.

(c)  Anisotropy is the property by which the magnitude of physical properties such as refractive index, thermal conductivity, etc. show a variation with the direction in which it is measured. Isotropy is the property by which the substances exhibit the same value of any physical property in all directions. (d) Classification of crystalline solid Ionic solids Have ions as constituent particles. Held together by electrostatic forces. Hard and highly brittle, have low volatility, high melting and boiling points.

Covalent solids Have same element atoms or different element atoms as constituent particles. Held together by strong covalent bonds. Hard and have high melting point.

In pure state, they are insulators, but good conductors in molten state.

Non-conductors of ­electricity.

For example, NaCl, KCl, CsCl, BaCl2

For example, Diamond, SiC

Chapter 8_Solid State.indd 183

Metallic solids Have positively charged metal ions as constituent in a sea of electrons. Held together by metallic bonds. Hard, malleable and ductile and possess luster. Good conductors in both solid and liquid states. For example, Au, Ag

Molecular solids Have molecules as constituent particles. Held together by van der Waals forces. Soft, have low density and vaporize easily. Bad conductor of ­electricity. For example, HCl, solid CO2

1/4/2018 5:14:01 PM

184

OBJECTIVE CHEMISTRY FOR NEET

3. Crystal Lattice and Unit Cells (a) Crystal lattice: The characteristic features of a crystal are:  (i) Faces (ii) Edges (iii) Interfacial angle (iv) Zone and zone axis (b) Unit cell: The arrangement of particles in three dimensional shapes is known as space lattice and the smallest representation of the space lattice which repeats itself and forms the complete lattice is known as unit cell. The relation between number of faces (F ), edges (E ) and interfacial angle (C ) for a crystal is given by: F+C=E+2 4. Classification of Unit Cell: Bravais classified the unit cell in two types: (a) Primitive unit cells: They are simple cubic (sc) in structure and atoms (lattice points) are present only at the corners of the unit cells. There are seven unique and basic unit-cell shapes (primeline unit cells) with varying elements of symmetry in a three-dimensional space as shown below: Crystal system

Edge lengths

Angles

Cubic Tetragonal Orthorhombic Monoclinic Triclinic Rhombohedral Hexagonal

a=b=c a=b≠c a≠b≠c a≠b≠c a≠b≠c a=b=c a=b≠c

a = b = g = 90° a = b = g = 90° a = b = g = 90° a = b = 90°; g ≠ 90° a ≠ b ≠ g ≠ 90° a = b = g ≠ 90° a = b = 90°; g = 120°

Examples NaCl, ZnS, Cu White Sn, SnO2,TiO2, CaSO4 Rhombic sulphur, BaSO4. Monoclinic sulphur K2Cr2O7,CuSO4×5H2O CaCO3, HgS Graphite, ZnO, AgI

(b) Non-primitive or centered: In these unit cells, atoms are present not only at the corners of the unit cell but also at some other places such as center of face, edge, etc. They are further classified as follows: Non-primitive or centered

Body centered (bcc)

Face centered cubic (fcc)

End-centered

5. (a)  Number of atoms   (i) An atom at the corner of a unit cell is shared by eight unit cells. Hence each atom contributes 1/8 to the unit cell. (ii) An atom at the face is a shared by two unit cells, contributing 1/2 to the unit cell. (iii) An atom within the body of the unit cell is shared by no other unit cell. Hence, each atom contributes 1 to the unit cell. (iv) An atom present on the edge is shared by four unit cells. Hence, each atom on the edge contributes 1/4 to the unit cell. (b) Density calculations

Chapter 8_Solid State.indd 184

Density (r ) =

(number of atoms/cell ) × (atomic mass) zM atomic = (Volume of unit cell) × (Avogadro’s constant) Vc × N A

1/4/2018 5:14:02 PM

Solid State

185

6. Characteristics of Cubic System Simple cubic

Body centered (bcc)

Face centered (fcc)

Edge length: a = 2r, where r is the atomic radius.

Edge length: a = 4/ 3 (since diagonal of a cube = 3 times the length of one of its sides). The diagonal of the cube covers R + 2R + R = 4R (as the middle sphere actually touches the two corner spheres). Eight atoms at the corners = 8 × 1/8 = 1 atom Atoms at the body center = 1 atom Number of atoms in a body-centered cubic unit cell = 1 + 1 = 2 atoms or rank (z) = 2 Number of nearest neighbors = 8

Edge length: a = 2r 2 [since (4r)2 = a2 + a2 ⇒ (a2 = 8r2)]

Number of atoms per unit cell = (8 × 1/8) = 1 atom or rank (z) = 1

Number of nearest neighbors = 6

One atom at each corner = (8 × 1/8 each) = 1 atom Atoms at each of the six face centers = (6 × 1/2 each) = 3 atoms Number of atoms in a face-centered cubic unit cell = 4 atoms or rank (z) = 4 Number of nearest neighbors = 12

7. Close-Packed Structures (a) One dimension: This has only one way of arranging spheres and the coordination number is 2. (b) Two dimensions: This has two ways of arranging spheres and the coordination number is 4 (square close ­packing) or 6 (hexagonal close packing). (c) Three dimensions: The following are the ways of arranging the spheres:   (i) The close packing structure obtained from square close packing in two dimensions is a lattice with     AAA… type of pattern. The lattice generated is a simple cubic lattice. (ii) There are two types of close packing structures obtained from the two-dimensional arrangement of hexagonal close packing. • Hexagonal close packing or hcp: The pattern in this type of packing is ABABAB… . Examples of hcp are Mg, Zn, Cd. • Cubic close packing or ccp: The pattern in this type of packing is ABCABCA… . Examples of ccp are Cu and Ag. 8. Packing Efficiency: It is the fraction of the total volume of the unit cell actually occupied by atoms. Packing efficiency =

Packing Simple cubic (sc) structure Body-centered cubic (bcc) structure Cubic close packing or face-centered cubic (fcc) arrangement Hexagonal close packing (hcp)

Chapter 8_Solid State.indd 185

Volume occupied by atoms/spheres × 100 Total volume of the unit cell

Packing efficiency Packing efficiency =

1 × ( 4/3)p R 3 p × 100 = × 100 = 52.35% 6 ( 2R )3

Packing efficiency =

2 × ( 4 / 3)p R 3 (8/3)p R 3 × 100 × 100 = = 68% 64/3 3R 3 ( 4/ 3R )3

Volume of four spheres in the unit celll ×100 Total volume of the unit cell 16 / 3p r 3 p = × 100 = × 100 = 74% 3 16 2r 3× 2

Packing efficiency =

Volume of six spheres in the unit celll ×100 Total volume of the unit cell 8p r 3 p = × 100 = × 100 = 74% 24 2r 3 3× 2

Packing efficiency =

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9. Void: The interstitial space between the particles is known as void.

Tetrahedral void

Voids

Features

Radius

• The four spheres touch each other in a manner that a tetrahedron is formed when their centers are joined and the space in the center is called a tetrahedral site/hole or void. • The number of tetrahedral voids per unit cell is eight.

Tetrahedral void

The radius of the tetrahedral void is r = 0.225 R

Octahedral void

A

• An octahedral site/hole or void is formed by six spheres touching each other. • The total number of octahedral void per unit cell is 4.

B

B

A

A B

The radius of the octahedral void is r = 0.414 R 10. Radius-Ratio Rule The ratio of radii of cation (r+) to radii of anion (r  ) is known as radius ratio-rule, it determines the shape of void, coordination number, etc. −

r+ r−

Coordination number

Type of void

Example

0 - 0.155

2

Linear

Not known

0.155 - 0.225

3

Triangular

B2O3

0.225 - 0.414

4

Tetrahedral

ZnS

0.414 - 0.732

6

Octahedral

NaCl

0.732 - 1.000

8

Cubic

CsCl



11. Bragg’s Law According to this law nl = 2d sinq where n is an integer (1, 2, 3, 4, …) representing the serial order of diffracted beams, d is the distance between the successive atomic planes and l is the wavelength.

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187

12. Simple Ionic Compounds Simple ionic compounds

Rock salt (or NaCl) • fcc packing • (6, 6)-coordination • 8 Cl− at the 8 corners and 6 Cl− at 6 faces So. total Cl−ions per unit cell are 4. • 12Na+ at the edge centers and 1 Na+ at body center. So, total Na+ ions per unit cell are 4. • The total NaCl units in one unit cell are 4.

Sphalerite or zinc blende (ZnS) • fcc packing • (4, 4)-coordination • 8 S2− at the 8 corners and 6 S2− at the faces. So, total S2− ions per unit cell are 4. • Since Zn2+ ions occupy only half of the available tetrahedral sites, the number of Zn2+ ions per unit cell will be 4. • Total number of ZnS per unit cell is 4.

Caesium chloride (CsCl) • Simple cubic array of cations with the anions at the center of the cube. • (8, 8)-coordination • With 8 Cl- ions at corners contributing one Cl- and one Cs+ ion in the body center contributing one Cs+. it has one CsCl per unit cell.

Fluorite structure (CaF2) • fcc lattice • Ca2+ ions at the corners and the face centres, and the fluoride F− anions occupy all of the tetrahedral voids. • (8, 4)-coordination.

Perovskite structure • General formula ABX2. • Simple cubic arrangement of B atoms with the A atoms located at the body center of the cube and the X atoms the center of the 12 edges of the simple cube.

Antifluorite structure (Na2O) • (4, 8)coordination. • Number of Na2O per unit cell is 4.

Normal spinel • General formula AB2O4. • O2− ions are cubic-closed packed. whereas 1/8th of the tetrahedral holes are occupied by A2+ cations and 1/2 of the octahedral holes are occupied by cations B2+.

Spinel structure

Inverse-spinel • Formula Fe3O4 • Ratio of Fe2+ to Fe2− = 2:1. • Fe2+ ions and half of the Fe2+ ions occupy octahedral sites. The other half of Fe2+ ions are occupied by tetrahedral sites.

13. Imperfections in Crystalline Solids (a) D  efect refers to a disruption in the periodic order of a crystalline material. These occur in crystals because they lower the energy of a crystal to make it more thermodynamically stable. (b) Defects in solids are further classified as follows: (i) Point defects: Point defects are imperfect point-like regions in the crystal. Based on their nature, point defects are classified into three types: stoichiometric defects, impurity defects and non-stoichiometric defects.

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Ionic compounds

Vacancy defects

Interstitial defects

Frenkel defect

Schottky defect

• These result when some of the lattice sites in the crystal are vacant. • These develop during formation of crystals or when substance is heated. • Density of the solid decreases as a result of vacancy defect. • These arise when some constituent particles (atoms or molecules) occupy interstitial sites in the crystal. • The density of the solid increases as a result of interstitial defect. • When an ion is displaced from its regular position to an interstitial position, it creates a vacancy; this pair of vacancy–interstitial is called Frenkel defect. • This defect does not result in any change in density of the substance. • Examples are ZnS, AgCl, AgBr and AgI • When a pair of one cation and one anion of equal valence are missing from an ionic crystal, the condition of charge neutrality is still maintained. The pair of vacant sites, thus formed, is called Schottky defect. • It decreases density of the solid. • Examples are NaCl, KCl, KBr, AgBr and CsCl.

Defects in ionic crystals introduced by adding impurities in which the ions are in different valence state than the constituent ions of the crystal.

Metal excess defect

• In these defects, a negative ion may be absent from its lattice site, leaving a “hole” which is occupied by an electron, thereby maintaining the electrical balance. • Anion sites occupied by electrons in this way are called F-centers.

Due to anionic vacancies

Due to the presence of extra cations in the interstitial sites

Metal deficiency

Impurity defects Non-stoichiometric defects (when the ratio of cation and anion is not unity) Chapter 8_Solid State.indd 188

Non-ionic Solids

OBJECTIVE CHEMISTRY FOR NEET

Stoichiometric defects (when the ratio of cation and anion is unity)

188

• These defects occur when an extra positive ion occupies an interstitial position in the lattice, and electrical neutrality is maintained by the inclusion of an interstitial electron. • Example, ZnO, CdO, Fe2O3 and Cr2O3

Metal-deficient compounds may be represented by the general formula A1–d Χ. Crystals with metal deficiency defects are semiconductors.

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(c) Line defects: These defects, also known as single dimensional or linear defects, are groups of atoms in irregular positions. Linear defects are commonly called dislocations. 14. Electrical Properties of Solids Solids may be classified in terms of their conductivity as conductors, insulators and semiconductors. Conduction band (CB)

CB

CB Lesser gap

Valence band (VB) VB Conductor

Semiconductor

More gap VB Insulator

15. Magnetic Properties of Solids Unpaired electrons

Type

Magnetization

Diamagnetic substance

Weakly magnetized in a direction opposite to that None of an applied magnetic field.

1 or more

Paramagnetic substances

Magnetized along the direction of the applied magnetic field.

Net magnetic moment

Examples

Remarks

0

NaCl, KCl, TiO2

Weakly repelled when placed near a magnet. Diamagnetism is independent of temperature.

Present (arises from the electron spin angular momentum)

Organic free radicals and gaseous nitric oxide Paramagnetism is shown by these KMnO4, Na, K, substances only in the Cu2+ and Fe3+ ions presence of a magnetic field.

Atoms and ions with unfilled inner electron shells

Strongly attracted by Ferromagnetic applied magnetic field and can be permanently substances magnetized; spontaneous magnetization.

Present (due to their angular momentum)

Iron, nickel, cobalt CrO2

AntiferromagZero net magnetization netic substance

No net total moment in zero applied magnetic field

Metals, alloys and salts of transition elements such as MnO, MnSe, etc.

Ferrimagnetic substances

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The atomic dipoles are arranged antiparallel to one another and the number of magnetic moments aligned in one direction is more than that aligned in the other direction; net magnetization is observed

At sufficiently high temperatures, all these substances become paramagnetic.

Fe3O4, and ferrites • Such materials are like MgFe2O4 and weakly attracted by ZnFe2O4. magnetic fields; • These lose ferrimagnetism on heating and become paramagnetic.

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Solved Examples 1. A compound XY crystallizes in BCC lattice with unit cell edge length of 480 pm. If the radius of  Y− is 225 pm, then the radius of X+ is (1) 190.68 pm (2) 225 pm (3) 127.5 pm (4) None of these Solution 2(rc + ra ) = 3a

(1) For bcc structure,

where rc and ra represents radius of cation and anion respectively.



So, 2 (rc + 225) = 3 × 480



Therefore,   rc = 190.68 pm

2. A solid crystal is composed of  X, Y and Z atoms. Y atoms are occupying 50% of octahedral voids, whereas X atoms are occupying 100% tetrahedral voids with Z atoms in ccp array arrangement. Then the rational formula of the compound in the given crystal is (1) X8Y2Z4 (2) X5Y10Z8 (3) X4YZ2 (4) X16Y4Z8 Solution

rA+ rB



Number of tetrahedral voids = 8 Number of octahedral voids = 4 Thus, number of atoms of X = 8 Also, number of atoms of Y = 0.5 × 4 = 2

Hence, the molecular formula of the compound is X8Y2Z4 .

3. If the edge length of the unit cell of sodium chloride is 600 pm, and the ionic radius of Cl− ion is 190 pm, then the ionic radius of Na+ ion is (1) 310 pm (2) 110 pm (3) 220 pm (4) None of these Solution

a 2 600 + 190 = = 300 2

(2) As rNa + + rCl − =

Therefore,

rNa +





rNa + = 300 − 190 = 110 pm

4. An ionic compound AB has ZnS type structure. If the radius of A+ is 22.5 pm, then the ideal radius of B− would be (1) 54.35 pm (2) 100 pm (3) 145.16 pm (4) None of these Solution (2) Since compound is ionic and AB has ZnS type structure, therefore it has tetrahedral voids, for which ideal radius ratio is 0.225.

Chapter 8_Solid State.indd 190

= 0.225 ⇒ rB− =

22.5 = 100 pm 0.225

5. Select the incorrect statement among the following. 3a (1) For CsCl unit cell (edge length = a), rc + ra = 2 a (2) For NaCl unit cell (edge length = a), rc + ra = 2 (3) The void fraction in a BCC unit cell is 0.68. (4) The percent void space in face−centered unit cell is 26%. Solution (3) In CsCl structure,

rc + ra =



rc + ra =

In NaCl structure,

3a 2 a



2 Packing fraction of a bcc unit cell is 0.68, so the void fraction would be 1− 0.68 = 0.32 In fcc unit cell, packing percentage = 74%



Therefore, void percentage = 100 − 74 = 26%



6. In the diamond type cubic unit cell of silicon (a = 5.43 Å), the number of atoms per meter along the body diagonal are

(1) Number of atoms of z = 4

−

(1) 8.5 × 109 m–1 (2) 2.1 × 109 m–1 (3) 2.6 × 109 m–1 (4) 1.84 × 109 m–1 Solution (2) In diamond type cubic unit cell, carbon atoms are present as fcc and in alternate tetrahedral voids. On body diagonal ( 3a ) , there will be two atoms.

Therefore, number of atoms per meter along the body diagonal



=

2 = 2.1 × 109 m–1 3 × 5.43 × 10 −10

7. When heated above 916°C, iron changes its crystal structure from bcc to ccp structure without any change in the radius of atom. The ratio of density of the crystal before heating and after heating is (1) 1.089 (2) 0.918 (3) 0.725 (4) 1.231 Solution (2)

r bcc packing efficiency of bcc 67.92 = = 0.918 = r ccp packing efficiency of ccpp 74.02

8. The packing efficiency of a simple cubic crystal with an interstitial atom exactly fitting at the body center is (1) 0.48 (2) 0.52 (3) 0.73 (4) 0.91

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Solid State Solution

Solution

(3) In a simple cubic crystal, a = 2r.

(2) Edge length = 2(r + + r - )



Let the radius of the interstitial atom in simple cubic structure be r0, then



2 (r + r0) =



Therefore, r0 =

3a =

  400 = 2(75+ r − ) ⇒ r − = 125 pm 12. The stacking, which has one stacking fault in the fcc lattice is

3 × 2r 3r − r = 0.732 r

4 3 4 4 3 p r + p (0.732r )3 p r [1 + (0.732)3 ] 3 3 3 Packing fraction = = ( 2r )3 8r 3 p 3 = [1 + (0.732) ] = 0.73 6 −

9. A binary solid has zinc blend structure with B ions constituting the lattice and A + ions occupying 25% ­tetrahedral voids. The formula of solid is (1) AB (2) A2B (3) AB2 (4) AB4 Solution 1 1 (3) Number of B− ions in unit cell = 8× + ×6 = 4 8 2 Now A + ion occupies 25% of tetrahedral voids. 8 ´ 25 =2 Therefore, number of A + = 100 + − Thus ratio of A to B is 1:2. Hence formula is AB2. 10. The diameter of the largest sphere that fits the void at the center of a cube edge of a bcc crystal of lattice parameter, a is (1) 0.293 a (2) 0.414 a (3) 0.134 a (4) 0.336 a Solution

(1) … ABCABABCABABCAB … (2) … ABCABABCABCABC … (3) … ABABABCABABABCABABABC … (4) … ABCABCABABCABCABABCABCAB… Solution (2) Stacking faults are planar surface imperfections created by a fault in the stacking sequence of atomic planes in crystals. Consider the stacking arrangement in an fcc crystal. ↓ ….ABCABCABCABC…..

If an A plane indicated by an arrow above is missing, the stacking sequence becomes



.…ABCABCBCABC…..



The stacking in the missing region is …BCBC… which is hcp stacking.



In option (1), there are 2 stacking faults. In option (2), there is only one, option (3) has 6 and option (4) has 2 faults.



The stacking, which has only one stacking fault is …ABCABABCABCABC…

13. A substance has density of 2 kg dm−3 and it crystallizes to fcc lattice with edge-length equal to 700 pm, then the molar mass of the substance is

(3) Let the radius of sphere that fits the void at the center of edge of bcc lattice be r0, then a = 2r + 2r0



Also,

3a = 4r

3a a= + 2r0 2  3a 3 Therefore, 2r0 = a − = a 1 − = 0.134 a 2  2 

11. A binary solid (A+ B− ) has a rock salt structure. If the edge length is 400 pm, and radius of cation is 75 pm the radius of anion is (1) 100 pm (2) 125 pm (3) 250 pm (4) 325 pm

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(1) 74.50 g mol−1 (2) 103.30 g mol−1 (3) 56.02 g mol−1 (4) 65.36 g mol−1 Solution (2) We know    r =

z×M  N A × a3



For fcc, number of atoms = 4



From Eq. (1), we get

(1)

4´M 6.023 ´ 1023 mol -1 ´ 700 pm 4´M 2 g cm -3 = 23 6.023 ´ 10 mol -1 ´ (7 ´ 10 -8 cm)3

2 kg dm -3 =

Þ M = 103.03 g mol -1

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14. Which of the following statement for crystals having Frenkel defect is not correct? (1) Frenkel defects are observed where the difference in sizes of cations and anions is large. (2) The density of crystals having Frenkel defect is lesser than that of a pure perfect crystal. (3) An ionic crystal may have Frenkel defect along with the Schottky defect.

(4) Pure alkali halides normally do not have Frenkel defect. Solution (2) The density of crystals having Frenkel defect is unaltered as the defect involves displacement of an atom or ion from its lattice position to an interstitial position, so the mass and volume remains unaffected.

Practice Exercises Level I Types of Unit Cell, Simplest Formula, Coordination Number, Packing Fraction 1. The unit cell with crystallographic dimensions a = b ≠ c, a = b = d = 90° is (1) cubic. (2) tetragonal. (3) monoclinic. (4) hexagonal. 2. An alloy of copper, silver and gold is found to have copper constituting the ccp lattice. If silver atoms occupy the edge centers and gold is present at body center, the alloy has a formula (1) Cu4Ag2Au (2) Cu4Ag4Au (3) Cu4Ag3Au (4) CuAgAu 3. If the coordination number of an element in its crystal lattice is 8, then packing is (1) fcc. (2) hcp. (3) bcc. (4) None of the above. 4. If a be the edge length of the unit cell, r be the radius of an atom, then for face-centered cubic lattice, the correct relation is (1) 4r = 3a (2) 4r = 2a (3) 4a = 3r (4) None of these. 5. In a face centered cubic arrangement of A and B atoms in which A atoms are at the corners of the unit cell and B atoms at the face centers, one of the A atoms is missing from one corner in unit cell. The simplest formula of compound is (1) A7B3 (2) AB3 (3) A7B24 (4) A7/8B3 6. Metallic gold crystallizes in fcc lattice with edge length 4.070 Å. Closest distance between gold atoms is (1) 2.035 Å (2) 8.140 Å (3) 2.878 Å (4) 1.357 Å 7. Which of the following crystalizes in bcc structure? (1) ZnS (2) Na2S (2) NaCl (4) CsCl

Chapter 8_Solid State.indd 192

8. A certain metal fluoride crystalizes in such a way that F atoms occupy simple cubic lattice sites, while metal atoms occupy the body center of the cubes. The formula of metal fluoride is (1) M2F (2) MF (3) MF2 (4) MF8 9. A compound formed by elements A and B crystalizes in the cubic structure, when atoms of A are at the corners of the cube and B atoms are at the face center. The formula of the compound is (1) AB3 (2) AB (3) A3B (4) None of these 10. Body-centered cubic and face-centered cubic unit cells have n1 and n2 effective number of atoms. Which one of the following (n1, n2) combinations is correct? (1) 1, 4 (2) 2, 4 (3) 4, 2 (4) 4, 1 11. The radius of the atom is 100 pm. If it crystallizes with fcc lattice, the length of side of unit cell is (1) 1.414 pm (2) 141.4 pm (3) 1,414 pm (4) None of these 12. Platinum has fcc crystal structure with a unit cell length of 392 pm. What is the radius of the Pt atom? (1) 98 pm (2) 139 pm (3) 69 pm (4) 196 pm 13. The coordination number of the body-centered cubic lattice is (1) 10 (2) 12 (3) 16 (4) 8

Density of Unit Cell 14. Silver iodide has the same structure as zinc sulphide, and its density is 5.67 g cm–3. The edge length of the unit cell is (1) 4.50 Å (2) 5.50 Å (3) 6.50 Å (4) 7.50 Å

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Solid State 15. If the unit cell length of sodium chloride crystal is 600 pm, then its density will be (1) 2.165 g cm−3 (2) 3.247 g cm−3 (3) 1.79 g cm−3 (4) 1.082 g cm−3 16. The edge length of the unit cell of PbS is 500 pm. The density is (At. wt. Pb = 207 and S = 32, z = 4) (1) 13.8 g cm−3 (2) 6.35 g cm−3 (3) 13.0 g cm−3 (4) 12.7 g cm−3 17. The density of a face centered cubic element (atomic mass 60.2) is 6.25 g cm−3. Calculate the edge length of the unit cell (z for fcc is 4). (1) 700 pm (2) 200 pm (3) 400 pm (4) 500 pm 18. NaBr has same crystal structure as NaCl. If density is 3.203 g cm−3, calculate the unit cell side length (At. wt. Na = 23, Br = 79.9). (1) 5.976 Å (2) 6.976 Å (3) 7.976 Å (4) 4.976 Å

Close-Packed Structures 19. The coordination number of metal crystallizing in a hexagonal close packed structure is (1) 12 (2) 4 (3) 8 (4) 6 20. The rank of atoms in the hexagonal until cell is (1) 4 (2) 3 (3) 5 (4) 6 21. In a hexagonal closest packing in two layers one above the other, the coordination number of each sphere will be (1) 4 (2) 6 (3) 8 (4) 12 22. A metallic element crystallizes into a lattice containing a sequence of layers of ABABAB … Any packing of sphere leaves out voids in the lattice. What percentage by ­volume of this lattice is empty space? (1) 74% (2) 26% (3) 50% (4) None of these 23. If the coordination number of an element in its crystal lattice is 8, then the packing is (1) fcc (2) simple cubic. (3) bcc (4) none of these. 24. CsI lattice has structure of two interpenetrating simple cubic structures, the I− ions being at the center of lattice of Cs+ ions. The unit cell side length is 4.562 Å. Calculate density (Cs = 132.9, I = 126.9). (1) 4.002 g cm−3 (2) 4.996 g cm−3 (3) 4.546 g cm−3 (4) 4.732 g cm−3

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Tetrahedral and Octahedral Voids 25. The anions (A) form hexagonal close packing and cations (C) occupy only 2/3 of octahedral voids in it. Then the general formula of the compound is (1) CA (2) C2A2 (3) C2A3 (4) C3A2 26. The number of tetrahedral and octahedral voids in hexagonal primitive unit cell are (1) 8, 4 (2) 2, 1 (3) 12, 6 (4) 6, 12 27. Lithium selenide can be described as a closest−packed array of selenide ions with lithium ions in all of the tetrahedral holes. Formula of lithium selenide is (1) Li2Se (2) LiSe (3) LiSe2 (4) Li3Se 28. In fcc lattice, A, B, C, D atoms are arranged at corner, face center, octahedral void and tetrahedral void respectively. Then the body diagonal contains (1) 2A, C, 2D (2) 2A, 2B, 2C (3) 2A, 2B, D (4) 2A, 2D

Types of Ionic Crystals, Radius Ratio 29. A compound XY crystallizes in bcc lattice with unit cell edge-length of 480 pm. If the radius of Y− is 225 pm, then the radius of X+ is (1) 190.70 pm (2) 225 pm (3) 127.5 pm (4) None 30. The number of molecules in a unit cell of fluorite is (1) 2 (2) 4 (3) 6 (4) 8 31. If the edge length of the unit cell of sodium chloride is 600 pm, and the ionic radius of Cl− ion is 190 pm, then the ionic radius of Na+ ion is (1) 310 pm (2) 110 pm (3) 220 pm (4) 350 pm 32. The structure of CsCl crystal is (1) bcc lattice. (2) fcc lattice. (3) octahedral. (4) none of these. 33. A binary solid (A+B−) has a rock salt structure. If the edge length is 400 pm and radius of cation is 75 pm, then the ­radius of anion is (1) 100 pm (2) 125 pm (3) 250 pm (4) 325 pm 34. The 8:8 coordination number type of packing is present in (1) CsCl (2) KCl (3) NaCl (4) MgF2

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35. A solid AB has NaCl type structure. If the radius of the cation A is 100 pm, then the radius of the anion B will be (1) 241 pm (2) 414 pm (3) 225 pm (4) 44.4 pm 36. Which of the following does not crystallize in the rock-salt structure? (1) NaCl (2) KCl (3) NaBr (4) CsCl 37. In the zinc blende structure (ZnS), S2− ions adopt ccp arrangement and Zn2+ occupy (1) (2) (3) (4)

octahedral sites. hexagonal sites. tetrahedral sites. both octahedral and tetrahedral sites.

38. The intermetallic compound LiAg crystallizes in cubic lattice in which both Li and Ag have coordination number of eight. The class of crystals is (1) simple cubic. (2) body centered cubic. (3) face centered cubic. (4) none of these. 39. The structure of Na2O crystal is (1) CsCl type. (2) NaCl type. (3) ZnS type. (4) anti-fluorite. 40. CuCl has zinc blend structure. Its density is 3.4 g cm−3. ­Calculate the length of unit cell edge. (1) 578 pm (2) 662 pm (3) 732 pm (4) 802 pm 41. Which of the following expressions is correct for NaCl unit cell with lattice parameter a? (1) rNa + + rCl − =

a (2) rNa + + rCl − = 4a 2

(3) rNa + + rCl − =

a 3 (4) rNa + + rCl − = a 4 4

42. 4:4 coordination is observed in

(1) NaCl. (2) CsCl. (3) ZnS. (4) All of these. 43. For an ionic crystal of general formula AX and coordination number 6, the value of the radius ratio will be (1) (2) (3) (4)

greater than 0.73. in between 0.73 and 0.41 in between 0.41 and 0.22 less than 0.22

44. In calcium fluoride structure, the coordination numbers of calcium and fluoride ions are respectively (1) 8 and 4. (2) 6 and 8. (3) 4 and 4. (4) 4 and 8.

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45. The radius of Ag+ ion is 126 pm while that of I− ion is 216 pm. The coordination number of Ag in AgI is (1) 2 (2) 4 (3) 6 (4) 8 46. In antiflouride structure, the negative ions (1) (2) (3) (4)

occupy tetrahedral voids. occupy octahedral voids. are arranged in ccp. are arrange in hcp.

47. Each Cs+ is CsCl lattice is surrounded by (1) 8 Cl− ions. (2) 6 Cl− ions. (3) 4 Cl− ions. (4) 1 Cl− ion. 48. The radius of A+ is 95 pm and that of B− ion is 181 pm. Hence the coordination number of A+ will be (1) 4 (2) 6 (3) 8 (4) unpredictable. 49. For tetrahedral coordination, the radius ratio (r+/r-) should be (1) 0.155 to 0.225 (2) 0.225 to 0.414 (3) 0.414 to 0.732 (4) 0.732 to 1.000 50. Sodium metal crystallizes in bcc lattice with the cell edge (a) = 4.99 Å. The radius of Na atom is (1) 186 pm (2) 6.07 Å (3) 214.5 pm (4) None of these.

Imperfections in Solids 51. In a solid lattice, the cation and anion both have left a lattice site. The lattice defect is known as (1) interstitial defect. (2) valency defect. (3) frenkel defect. (4) schottky defect. 52. When an ion leaves its correct lattice site and occupies interstitial sites in its crystal lattice, it is called (1) crystal defect. (2) Frenkel defect. (3) Schottky defect. (4) none of these. 53. KBr shows which of the following types of defects? (1) Frenkel defect (2) Schottky defect (3) Metal excess defect (3) Metal deficiency 54. Frenkel defect appears in (1) AgI (2) ZnS (3) AgBr (4) All 55. Non-stoichiometric metal deficiency is shown in the salts of (1) all metals. (2) alkali metals. (3) alkaline earth metals. (4) transition metals.

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Solid State 56. Schottky defect is observed in (1) NaCl. (2) CsCl. (3) KCl. (4) All of these.

Level II Types of Unit Cell, Simplest Formula, Coordination Number, Packing Fraction 1. If three elements X, Y and Z crystallized in cubic solid lattice with X atoms at corners, Y atoms at cube center and Z atoms at the edges, then the formula of the compound is (1) XYZ (2) XY3Z (3) XYZ3 (4) X3YZ 2. A solid has a structure in which W atoms are located at the corners of cubic lattice, O atoms at the center of edges and Na atom at the center of the cube. The formula of the compound is (1) NaWO2 (2) NaWO3 (3) Na2WO3 (4) NaWO4 3. In hexagonal primitive unit cell, the corner is shared by (1) 4 unit cells. (2) 6 unit cells. (3) 3 unit cells. (4) 5 unit cells.

Density of Unit Cell 4. CsBr has CsCl structure (body-centered cubic type of ­lattice). Its density is 4.49 g cc−1. Calculate the side of unit cell. (1) 5.225 Å (2) 4.285 Å (3) 3.225 Å (4) 3.285 Å 5. Vanadium crystallizes with a body-centered cubic structure with a unit cell of side length 3.011 Å. Calculate the atomic radius and density (At.wt. V = 50.94). (1) 2.609 Å, 16.199 g cm−3 (2) 1.303 Å, 16.99 g cm−3 (3) 2.609 Å, 7.200 g cm−3 (4) 1.303 Å, 6.199 g cm−3 6. Silver (At.wt. 107.88) crystallizes with fcc lattice for which the side length of the unit cell is 4.0774 Å. Density of Ag is 10.53 g cm−3. Calculate the Avogadro’s number. (1) 6.023 × 1023 (2) 6.045 × 1023 (3) 6.045 × 1022 (4) 6.023 × 1022

Close-Packed Structures 7. Which one of the following schemes of ordering close packed sheets of equal sized sphere do not generate close packed lattice? (1) ABACABAC…. (2) ABCBCABCBC….. (3) ABCABC…. (4) ABBAABBA…..

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195

8. The number of octahedral voids per atom in a crystal ­lattice of close packing is (1) 4 (2) 2 (3) 1 (4) 8

Tetrahedral and Octahedral Voids 9. In the closest packing of atom A of radius ra, the maximum radius of atom B that can be fitted in tetrahedral void is (1) 0.225 ra (2) 0.155  ra (3) 0.414 ra (4) 0.732  ra 10. A binary solid (A+B−) has zinc blende structure with B− ions constituting the lattice and A+ ions occupying 25% tetrahedral holes. The formula of solid is (1) AB (2) A2B (3) AB2 (4) AB4 11. In closest packing of atoms (1) the size of tetrahedral void is greater than that of the octahedral void. (2) the size of the tetrahedral void is smaller than that of the octahedral void. (3) the size of tetrahedral void is equal to that of the octahedral void. (4) the size of tetrahedral void may be larger or smaller or equal to that of the octahedral void depending upon the size of atoms. 12. A solid is formed and it has three types of atoms X, Y and Z, X forms a fcc lattice with Y atoms occupying all the tetrahedral voids and Z atoms occupying half the octahedral voids. The formula of the solid is (1) X2Y4Z (2) XY2Z4 (3) X4Y2Z (4) X4YZ2

Types of Ionic Crystals, Radius Ratio 13. Select the correct statement: 3 a 2 l (2) For NaCl unit cell (edge-length = l ), rc + ra = 4 (3) The void space in a bcc unit cell is 0.68. (1) For CsCl unit cell (edge-length = a), rc + ra =

(4) The void space % in a face-centered unit cell is 46%. 14. The C–C and Si–C interatomic distances are 154 pm and 188 pm. The atomic radius of Si is (1) 77 pm (2) 94 pm (3) 114 pm (4) 111 pm 15. A common example of spinel structure is (1) Fe2O3 (2) FeO (3) Fe3O4 (4) FeCl3

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196

OBJECTIVE CHEMISTRY FOR NEET

16. A crystal of lead (II) sulphide has NaCl structure. In this crystal the shortest distance between a Pb2+ ion and S2− ion is 297 pm. What is the volume of unit cell in lead sulphide? (1) 209.6 × 10−24 cm3 (2) 207.8 × 10-23 cm3 (3) 22.3 × 10−23 cm3 (4) 209.8 × 10-23 cm3 17. If the positions of Na+ and Cl− are interchanged in NaCl, having fcc arrangement of Cl− ions then in the unit cell of NaCl, (1) Na ions will decrease by 1 while Cl ions will increase by 1. (2) Na+ ions will increase by 1 while Cl− ions will decrease by 1. (3) number of Na+ and Cl− ions will remain the same. (4) the crystal structure of NaCl will change. +

−

18. 6:6 of NaCl coordinate changes to 8: 8 coordination on (1) (2) (3) (4)

applying high pressure. decrease in temperature. both (1) and (2). no effect on coordination.

(1) 4 (2) 8 (3) 6 (4) 12 20. Which of the following statements is incorrect? (1) The coordination number of each type of ion in CsCl crystal is 8. (2) A metal that crystallizes in bcc structure has a coordination number of 12. (3) A unit cell of an ionic crystal shares some of its ions with other unit cells. (4) The length of the unit cell in NaCl is 552 pm rNa + = 95 pm ; rCl − = 181 pm .

)

21. CaO and NaCl have the same crystal structure and nearly the same ionic radii. If u is the lattice energy of NaCl, the lattice energy of CaO is very nearly (1) 2u. (2) u. (3) 4u. (4) u/2. 22. Pick out the statement which is not true. (1) NaCl structure on heating transforms to CsCl structure. (2) In CaF2 structure, each F− ion in coordinated by four Ca2+ ions. (3) NaCl has 6:6 coordination; while CsCl is with 8:8 coordination. (4) In Na2O, each oxide ion is coordinated by eight Na+ ions and each Na+ ion by four oxide ions.

Chapter 8_Solid State.indd 196

23. The second order Bragg diffraction of X-rays with l = 1.0 Å from a set of parallel planes in a metal occurs at an angel of 60°. The distance between the scattering planes in the crystal is (1) 0.575 Å (2) 1.00 Å (3) 2.00 Å (4) 1.15 Å

Imperfections in Solids 24. Select the correct statements, if CaCl2 doped with NaCl solid solution results in (1) (2) (3) (4)

substitutional cation vacancy. Frenkel’s defect. Schottky defect. no change in density.

25. When NaCl is doped with MgCl2, the nature of defect produced is (1) interstitial defect. (2) Schottky defect. (3) Frenkel defect. (4) none of these.

19. In the sodium chloride structure, each Na+ ion is surrounded by six Cl− ions nearest neighbors and ___ Na+ ion next nearest neighbors.

(

Bragg’s Law

26. Select the correct statement(s): (1) Schottky defect is shown by CsCl. (2) Frenkel defect is shown by ZnS. (3) Hexagonal close packing (hcp) and cubic close packing (ccp) structures have the same coordination number 12. (4) All of these. 27. To get n type doped semiconductor, impurity to be added to silicon should have the following number of valence electrons. (1) 2 (2) 5 (3) 3 (4) 1 28. The yellow color of ZnO and conducting nature produced on heating is due to (1) (2) (3) (4)

metal excess defects due to interstitial cation. extra positive ions present in an interstitial site. trapped electrons. all of these.

29. Which of the following defects in the crystal does not lowers its density? (1) (2) (3) (4)

Metal deficiency defects F centers Schottky defect Interstitial defect

30. The NaCl is doped with 2 × 10−3 mol % SrCl2, the concentration of cation vacancies is (1) 6.02 × 1018 mol−1 (2) 12.04 × 1018 mol−1 (3) 3.01 × 1018 mol−1 (4) 12.04 × 1020 mol−1

1/4/2018 5:14:16 PM

Solid State 31. At zero Kelvin, most of the ionic crystals possess (1) Frenkel defect. (2) Schottky defect. (3) Metal excess defect. (4) no defect. 32. As a result of Schottky defect (1) (2) (3) (4)

5. If ‘a’ stands for the edge length of the cubic system: ­simple cubic (sc), body centered cubic (bcc) and face centered cubic (fcc), then the ratio of radii of the spheres in these systems will be respectively: (1) 1a : 3a : 2a (2)

density of the crystal decreases. density of the crystal increases. there is no effect on the density. any of the three can happen.

(3)

1 3 1 a: a: a 2 4 2 2

1 1 1 3 2 a : 3a : a (4) a : a: a 2 2 2 2 2 (AIPMT 2008)

33. Superconductors are derived from compounds of (1) p-block elements. (2) lanthanoides. (3) actinoides. (4) transition elements.

Electrical Properties 34. A hole in a semiconductor differs from an electron in that (1) (2) (3) (4)

it has mass equal to that of a position. it has zero mass. it is a positively charged vacancy. it has a mass equal to that of proton.

6. Lithium metal crystallizes in a body-centered cubic crystal. If the length of the unit cell of lithium is 351 pm, the atomic radius of the lithium will be (1) 300.5 pm (3) 151.8 pm

1. The fraction of the total volume occupied by the atoms present in a simple cube is p p (1) (2) 6 4 p p (3) (4) 4 2 3 2 (AIPMT 2007) 2. If NaCl is doped with 10−4 mol% of SrCl2, the concentration of cation vacancies will be (NA = 6.02 × 1023 mol−1)

(AIPMT 2009) 7. Copper crystallizes in a face-centered cubic lattice with a unit cell length of 361 pm. What is the radius of copper atom in pm?

(2) 6.02 × 1015 mol−1 (4) 6.02 × 1017 mol−1

(AIPMT 2009) 8. A solid compound XY has NaCl structure. If the radius of the cation is 100 pm the radius of the anion (Y−) will be (1) 165.7 pm (2) 275.1 pm (3) 322.5 pm (4) 241.5 pm (AIPMT PRE 2010) 9. The number of octahedral void(s) per atom present in a cubic close-packed structure is (1) 1 (2) 3 (3) 2 (4) 4 (AIPMT MAINS 2011)

(AIPMT 2007) 3. Percentage of free space in a body-centered cubic unit cell is (1) 28% (2) 30% (3) 32% (4) 34%

10. A metal crystallizes with a face-centered cubic lattice. The edge of the unit cell is 408 pm. The diameter of the metal atom is (1) 288 pm (2) 408 pm (3) 144 pm (4) 204 pm

(AIPMT 2008) 4. Which of the following statements is not correct? (1) The number of Bravais lattices in which a crystal can be categorized is 14. (2) The fraction of the total volume occupied by the atoms in a primitive cell is 0.48. (3) Molecular solids are generally volatile. (4) The number of carbon atoms in a unit cell of diamond is 8. (AIPMT 2008)

Chapter 8_Solid State.indd 197

(2) 240.8 pm (4) 75.5 pm

(1) 108 (2) 128 (3) 157 (4) 181

Previous Years’ NEET Questions

(1) 6.02 × 1014 mol−1 (3) 6.02 × 1016 mol−1

197

(AIPMT PRE 2012) 11. Structure of a mixed oxide is cubic close-packed (ccp). The cubic unit cell is mixed oxide is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal B. The formula of the oxide is (1) ABO2 (3) A2B3O4

(2) A2BO2 (4) AB2O2 (AIPMT PRE 2012)

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OBJECTIVE CHEMISTRY FOR NEET

12. The number of carbon atoms per unit cell of diamond unit cell is (1) 4 (2) 8 (3) 6 (4) 1

(3) Schottky defects have no effect on the density of crystalline solids. (4) Frenkel defects decrease the density of crystalline solids.

(AIPMT MAINS 2012)

(RE-AIPMT 2015)

13. A metal has fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm−3. The molar mass of the metal is

18. Lithium has a bcc structure. Its density is 520 kg m−3 and its atomic mass is 6.94 g mol−1. Calculate the edge length of a unit cell of lithium metal. (NA = 6.02 × 1023 mol−1)



(1) 154 pm (3) 527 pm

(Avogadro’s constant = 6.02 × 1023 mol−1) (1) 40 g mol (2) 30 g mol (3) 27 g mol−1 (4) 20 g mol−1 −1

−1

(NEET I 2016)

(NEET 2013) 14. If a is the length of the side of a cube, the distance between the body centered atom and one corner atom in the cube will be 2 4 (1) a (2) a 3 3 (3)

3 a (4) 4

(2) 352 pm (4) 264 pm

3 a 2 (AIPMT 2014)

15. A given metal crystallizes out with a cubic structure having an edge length of 361 pm. If there are four metal atoms in one unit cell, what is the radius of one atom? (1) 127 pm (2) 80 pm (3) 108 pm (4) 40 pm

19. The ionic radii of A+ and B− ions are 0.98 × 10−10 m and 1.81 × 10−10 m respectively. The coordination number of each ion in AB is (1) 6 (2) 4 (3) 8 (4) 2 (NEET I 2016) 20. In calcium fluoride, having the fluorite structure, the coordination numbers for calcium ion (Ca2+) and fluoride ion (F−) are (1) 6 and 6 (3) 4 and 8

(2) 8 and 4 (4) 4 and 2 (NEET II 2016)

21. Which is the incorrect statement?

16. The vacant space in bcc lattice unit cell is (1) 23% (2) 32% (3) 26% (4) 48% (RE-AIPMT 2015) 17. The correct statement regarding defects in crystalline solids is

(1) Density decreases in case of crystals with Schottky’s defect. (2) NaCl(s) is insulator, silicon is semiconductor, silver is conductor, quartz is piezo electric crystal. (3) Frenkel defect is favored in those ionic compounds in which sizes of cation and anions are almost equal. (4) FeO0.98 has non-stoichiometric metal deficiency defect.

(1) Frenkel defect is a dislocation defect. (2) Frenkel defect is found in halides of alkaline metals.

(NEET 2017)

Answer Key Level I  1. (2)

 2. (3)

 3. (3)

 4. (2)

 5. (3)

 6. (3)

 7. (4)

 8. (2)

 9. (1)

10.  (2)

11.  (4)

12.  (2)

13.  (4)

14.  (3)

15.  (3)

16.  (2)

17.  (3)

18.  (1)

19.  (1)

20.  (4)

21.  (4)

22.  (2)

23.  (3)

24.  (3)

25.  (3)

26.  (3)

27.  (1)

28.  (1)

Chapter 8_Solid State.indd 198

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Solid State 29.  (1)

30.  (2)

31.  (2)

32.  (1)

33.  (2)

34.  (1)

35.  (1)

36.  (4)

37.  (3)

38.  (2)

39.  (4)

40.  (1)

41.  (1)

42.  (3)

43.  (2)

44.  (1)

45.  (3)

46.  (3)

47.  (1)

48.  (2)

49.  (2)

50.  (4)

51.  (4)

52.  (2)

53.  (3)

54.  (4)

55.  (4)

56.  (4)

 1. (3)

 2. (2)

 3. (2)

 4. (2)

 5. (4)

 6. (2)

 7. (4)

 8. (3)

 9. (3)

10.  (3)

11.  (2)

12.  (1)

13.  (1)

14.  (4)

15.  (3)

16.  (1)

17.  (3)

18.  (3)

19.  (4)

20.  (2)

21.  (1)

22.  (1)

23.   (4)

24.  (1)

25.  (4)

26.  (4)

27.  (2)

28.  (4)

29.  (4)

30.  (2)

31.  (4)

32.  (1)

33.  (4)

34.  (2)

199

Level II

Previous Years’ NEET Questions  1. (2)

 2. (4)

 3. (3)

 4. (2)

 5. (2)

 6. (3)

 7. (2)

 8. (4)

 9. (1)

10.  (1)

11. (4)

12.  (2)

13.  (3)

14.  (4)

15.  (1)

16.  (2)

17.  (1)

18.  (2)

19.  (1)

20.  (2)

21. (3)

Hints and Explanations Level I 2. (3) Number of Ag atoms =

1 × 12 = 3 4



Number of Au atoms = 1 × 1 = 1



Number of Cu atoms = 4



Hence, the formula is Cu4Ag3Au. 1 7 ×7 = 8 8 1    Number of B atoms = × 6 = 3 2

5. (3) Number of A atoms =

Therefore, the formula is A 7/8B3 = A 7 B24

6. (3) For an fcc structure 4r = 2a o 2 2 a= × 4.07 ⇒ 2r = 2.878 A 2 2 1 8. (2) Number of atoms of F = × 8 = 1 8



Therefore, 2r =



Number of atoms of M = 1 × 1 = 1



Therefore, the formula is MF

1 9. (1) Number of atoms of A = × 8 = 1 8

Chapter 8_Solid State.indd 199

1 ×6 = 3 2



Number of atoms of B =



Therefore, the formula is AB3.

10. (2) n1 (bcc) = 2 and n2 (fcc) = 4 11. (4) For an fcc structure 4r = 2a

Therefore, a =

4 4 ×r = × 100 = 282.84 pm 2 2

2 × 392 = 139 pm 4 z ×m 4. (3) We have r = 3 1 a × NA 12. (2) 4r = 2a ⇒ r =

o 4 × 97.3 ⇒ a = 6.50 A 23 a × (6.023 × 10 ) z ×m 5. (3) We have r = 3 1 a × NA

  5.67 =

r=

3

4 ´ 58.5 = 1.79 g cm -3 (600 ´ 10 ) ´ (6.023 ´ 1023 ) -10 3

16. (2) Density can be calculated as

   r =

z ×m a3 × N A

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200

OBJECTIVE CHEMISTRY FOR NEET

=

4 × 239 (500 × 10 −10 )3 × (6.023 × 1023 )

40. (1) Density can be calculated as r =

= 6.35 g cm − 3 17. (3)       r =

6.25 =

3.4 =

z´M a3 ´ N A

4 × 99 ⇒ a = 578 pm a 3 × (6.023 × 1023 )

43. (2) The ratio of cationic to anionic radii is rAg +

4 × 60.2 ⇒ a = 400 pm a 3 × 6.023 × 1023

1 1 20. (4) z = 12 ´ + 3 ´ 1 + 2 ´ = 6 6 2

rI−

22. (2) ABAB is hcp type packing

Therefore, packing fraction = 0.74 or 74%



Hence, empty space = 0.26 or 26%

25. (3) Number of atoms of A = 6 2 × No. of octahedral voids 3

=

126 = 0.5833 216

The radius ratio is in between 0.414 and 0.732, thus ionic solid has octahedral geometry and thus coordination number of each ion in AX is 6.

49. (2) For tetrahedral, the radius ratio should be 0.225 to 0.414 and the coordination number is 4, for example, in CuCl, CuBr, etc 50. (4) For body-centered cubic cell 4r = 3a

Therefore, r =

o 3 × 4.99 = 2.16 A 4



Number of atoms of C =



Number of octahedral voids = z = 6

Level II



2 Thus, number of atoms of C = × 6 = 4 3

1. (3) Number of X atoms =



Hence, the formula is C4A6 or C2A3.

26. (3) For hcp, we know z = 6

Therefore, number of octahedral voids = z = 6



Number of tetrahedral voids = 2z = 12

27. (1) Number of selenide ions = 4

Number of lithium ions = 4 × 2 = 8



Therefore, the formula of the compound is Li8Se4 or Li2Se.

28. (1) Body diagonal covers 2 corners, 0 face center, 1 octahedral void (body center in fcc) and 2 tetrahedral voids. Therefore, body diagonal contains 2A, C, 2D.

2(rX + + rY − ) = 3a ⇒ rX + = 190.70 pm

31. (2) Atoms touch along the edge. Therefore,

2(rNa + + rCl − ) = a ⇒ rNa + = 110 pm

33. (2) Atoms touch along the edge. Therefore,

2(rA+ + rB− ) = a ⇒ rB− = 125 pm

35. (1) Coordination number is 6 in NaCl type structure. rA+ rB−

Chapter 8_Solid State.indd 200

= 0.414 ⇒ rB− =

100 = 241 pm 0.414

1 ×8 =1 8



Number of Y atoms = 1 × 1 = 1



Number of Z atoms =



Therefore, the formula is XYZ3.

1 × 12 = 3 4

2. (2) Number of W atoms =

1 ×8 =1 8

1 × 12 = 3 4



Number of O atoms =



Number of Na atoms = 1 × 1 = 1



Therefore, the formula is NaWO3.

4. (2) We know r = 4.49 =

29. (1) Atoms touch along the body diagonal. Therefore,

z´M a3 ´ N A

z×M a3 × N A

o 1 × 213 ⇒ a = 4.285 A 23 a × (6.023 × 10 ) 3

5. (4) For body-centered cubic cell 4r = 3a

o 3 3 ×a = × 3.011 = 1.303 A 4 4 z´M Density can be calculated as r = 3 a ´ NA

Therefore, r =



=

2 × 50.94 = 6.199 g cm − 3 (1.303 × 10 −8 )3 × (6.023 × 1023 )

6. (2) We have r = 10.53 =

z×M a3 × N A

4 × 107.88 ⇒ N A = 6.045 × 1023 ( 4.0774 × 10 −8 )3 × N A

1/4/2018 5:14:20 PM

Solid State 8. (3) Number of atoms = z

Therefore, number of octahedral voids = z



Hence, number of octahedral voids/atom =

z =1 z

9. (1) Coordination number = 4, Radius ratio = 0.225 – 0.414

27. (2) Impurity should have excess electrons than Si (4), therefore, to get n type doped semiconductor valence electrons should be 5. 30. (2) 1 Sr2+ replaces 1Na+ and creates 1 cationic vacancy.

Therefore, 2 × 10−3 mol% SrCl2 will create 2 × 10 −3 × 6.02 × 1023 × 10 −2 = 12.04 Maximum that can be fit will be such that radius ratio is−3 2 × 10 × 6.02 × 1023 × 10 −2 = 12.04 × 1018 cationic vacancies. just less than 0.414. rB < 0.414 ⇒ rB < 0.414 rA Previous Years’ NEET Questions rA

10. (3) Number of B− ions = 4

Number of tetrahedral holes = 2 × 4 = 8



Number of A+ ions = 0.25 × 8 = 2



Therefore, the formula is A2B4 or AB2.

1. (2) For simple cubic structure arrangement, the unit cell edge is a = 2R, since two spheres just touch each other along the edge. Here R is the radius of the sphere. There is one atom per unit cell of a simple cubic structure. Therefore, Packing fraction =

12. (1) Number of X atoms = z = 4

Number of Y atoms = 2z = 8



Number of Z atoms =



Therefore, the formula is X4Y8Z2 or X2Y4Z.

14. (4) We know 

1 × ( 4/3) p R 3 ( 2R 3) p = 6 2. (4) As each Sr2+ ion provides one cation vacancy, hence the concentration of cation vacancies created is

rC + rSi = 188 (2)

10 −4 × NA 100 −4 10 = × 6.023 × 10 −23 100 = 6.023 × 1017 mol −1 =

From Eq. (1) and Eq. (2), we get rSi = 188 − 77 = 111 pm

15. (3) Spinel formula = M 2+ (M 3+ )2 O4

Therefore, Fe2+ (Fe3+ )2 O 4 or Fe3O4

16. (1) We have 2(rPb2+ + rS2− ) = a a = 2 × 297 = 594 pm

3. (3) We know, Packing efficiency =

Volume occupied atoms/sphere × 100 Total volume of unit cell



For bcc structure, the spheres touch each other along the body diagonal. Hence unit cell edge a is 4R 3a a= or R = 4 3



In bcc structure, the total number of atoms per unit cell is 2. 4 Volume of two spheres is 2 × p R 3 3

Volume = a3 = (594 × 10 −10 )3 cm 3 = 209.6 × 10 −24 cm 3

17. (3) For NaCl crystal, the ratio is 6:6 or 1:1.

Volume occupied by atoms/spheres Total volume of the unit cell

=

1 ×z=2 2

2rC = 154 ⇒ rC = 77 pm (1)



201

qq 21. (3) Lattice energy ∝ 1 2 or L.E. ∝ q1q2 (As radius are (r1 + r2 ) same)



u ∝ 1 × 1 (1)







u ′ ∝ 2 × 2 (2)

4  Volume of the cube is a 3 =  p R 3 



Therefore,



From Eq. (1) and Eq. (2) u ′ ∝ 4u



Packing efficiency=

23. (4) We know         

Chapter 8_Solid State.indd 201

nl = 2d sin q 2 × 1 = 2 × d × sin 60° = 2d × 0.8666 o

d = 1.15 A

3

2 × ( 4/3 )p R 3 × 100 = 68% and % of ( 4/ 3 R )3 free space = 100 - 68 = 32%

4. (2) For primitive cell, the packing efficiency is 52.35%.

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202

OBJECTIVE CHEMISTRY FOR NEET

5. (2) The radii of spheres for the given systems are

Tetrahedral voids

a 2



For simple cubic, r =



For body centered cubic, r =

3 a 4 a For face centered cubic, r = 2 2



1 3 1 a. a: a: 2 4 2 2 6. (3) In case of bcc, the body diagonal = 4r. Therefore, 3a = 4r

Hence, the ratio is



or

r=

3a = 4

fcc sites

13. (3) We know r = 2.72 =

3 × 351 pm = 151.98 pm 4



or

2 × 361 pm = 127.632 pm  128 pm 4 r+ = 0.414 r− 100 r− = = 241.5 pm 0.414

  

9. (1) In case of ccp arrangement, there are 4 atoms present per unit cell (same as fcc) and as the number of octahedral voids are equal to the number of atoms present, so the number of octahedral voids are 4. Hence, the number of octahedral voids present per unit atom = 4/4 = 1. 10. (1) For fcc arrangement, the relationship between r (radius) and a (edge length) is: 2a = 4r . Given that a = 408 pm. So, r=



3 a and d = 2r where r is 4 the radius of the sphere and d is the distance between 3 the body-centered atom and corner atom. So, d = a. 2

11. (4) In case of ccp arrangement, there are four atoms present per unit cell

Octahedral voids = number of atoms = 4



Tetrahedral voids = 2 × number of atoms = 8



So, A = (1/4) × 8 = 2 and B = 4 × 1 = 4



Number of O22-  ions = 4 (8 × 1/8 (corners) + 6 × 1/2 (faces) = 4



So, the formula is A2B4O4 or AB2O2

12. (2) In diamond, carbon atoms occupy half of tetrahedral voids and occupy fcc sites.

Number of atoms/unit cell = 8

Chapter 8_Solid State.indd 202

15. (1) There are four atoms in one unit cell, hence, it is an fcc structure. 2a 4r = 2a ⇒ r = 4 2 × 361 r= = 127 pm 4      16. (2) As we know, packing efficiency in bcc lattice is 68%.

Therefore, vacant space in bcc lattice = 100 – 68 = 32%

17. (1) The correct statements are as follows

Option (1): When an ion is displaced from its regular position to an interstitial position, it creates a vacancy; this pair of vacancy-interstitial is called Frenkel defect. This defect is also known as dislocation effect.



Option (2): This type of defect is favored by a large difference in size between the positive and negative ions. For example, ZnS, AgCl, AgBr and AgI.



Option (3): When a pair of one cation and one anion of equal valence is missing from an ionic crystal, the condition of charge neutrality is still maintained. The pair of vacant sites, thus formed, is called schottky defect. This decreases the density of the solid.



Option (4): Frenkel defect does not decrease the density of crystals.

2 × 408 = 144.23 pm 4

Therefore, diameter = 2r = 2 × 144.23 = 288.4 pm.

2.72 × ( 4.04)3 × 6.02 × 10 −1 4 = 27 g mol − 1

14. (4) For bcc arrangement, r =

8. (4) For NaCl structure



4×M ( 4.04 × 10 −8 ) 3 × 6.02 × 1023

M=

7. (2) In an fcc structure, face diagonal = 4r 2a = 4r

z´M V ´ NA

18. (2) The density of bcc unit cell is given by r=

z´M a3 ´ N A

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Solid State

For bcc structure, z = 2 and given, r = 520 kg m−3. Therefore, edge length can be determined as a3 =

2 × 6.94 = 43.5 × 10 −24 cm3 0.53 × 6.02 × 1023

a =

3

43.5 × 10 −24 = 3.52 × 10 −8 cm = 352 pm

19. (1) The ratio of cationic to anionic radii is r + 0.98 × 10 −10 = = 0.54 r − 1.81 × 10 −10

203

20. (2) For CaF2, in fluorite structure, each calcium atom is surrounded by 8 F− ions and each fluoride ion is surrounded by 4 Ca2+ ions. So, the coordination number for calcium ions is 8 and for fluoride ions are 4. 21. (3) When an ion is displaced from its regular position to an interstitial position, it creates a vacancy at the original site and a defect at the interstitial site; this pair of vacancy-interstitial is called Frenkel defect. This defect is exhibited by the compounds in which metal ions are generally smaller than the anions. Examples are ZnS, AgCl, AgBr and AgI.

The radius ratio is in between 0.414 and 0.732, thus ionic solid has octahedral geometry and thus coordination number of each ion in AB is 6.

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9

Solutions

Chapter at a Glance 1. Solution A solution is a system in which one or more substances are homogeneously mixed or dissolved in another substance. A simple solution has two components: a solute and a solvent. The solute is the component that is dissolved or is the less abundant component in the solution. The solvent is the dissolving agent or the more abundant component in the solution. 2. Expressing Concentration of Solutions The concentration of a solution can be expressed in a number of ways. (a) Mass percentage or percentage by mass: Mass fraction of solute = =

Mass of solute(g) Mass of solute(g) + Mass of solvent(g) Mass of solute(g) Mass of solution(g)

Mass percent of solute = Mass fraction of solute × 100 g solute × 100 = g solution (b) Volume percentage (V/V ): In case of solutions formulated by two liquids, volume percent with respect to solute is expressed as Volume of liquid present as solute     × 100 Volume percent = Total volume of solution (c) Mass by volume percentage (m/V ): It expresses concentration as grams of solute per 100 mL of solution.

Mass by volume percent 1 r M B g solute = = = m M 1000 g solute



(d) Strength of solution (m/V ): Strength of solution =

Mass of solute in grams Volume of solution in liter

(e) Mole fraction of solute (xA):

The mole fraction of any substance A is given by xA =

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nA nA + nB + nC + ⋅⋅⋅⋅

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where nA, nB, nC, are the number of moles of component A, B, C ..., respectively. For a solution of substance A (solute) and substance B (solvent), mole fraction of solute is given by xA =

  

nA (number of moles of solute ) = nB (number of moles of solvent ) =

nA (unit-less) nA + nB w A (Given mass of solute ) M A ( Molar mass of solute) wB (Given mass of solute ) M B ( Molar mass of solute)

If xB is the mole fraction of the solvent, then x A + xB = 1

(f ) Molarity (M): It is the number of moles of solute present in one liter of solution. M=

Number of moles of solute L of solution

=

w A ´ 1000 M A ´ V (mL )



where wA = given mass of solute (g), MA = Molar mass of solute (g mol−1) and V = volume of solution (mL)



Units: mol L−1 or M or Molar

(g) Molality (m): It is the number of moles of solute present in one kilogram of solvent. For solute A, m=

mol of solute n w × 1000 = A = A kg of solvent w( kg ) M A × W ( g )

where wA = given mass of solute (g), MA = molar mass of solute (g mol−1) and w = mass of solvent (g)

Units: Molal; mol kg–1 (h) Relation between mole fraction, molarity and molality For a solution of solute A in solvent B, mole fraction of solute is given by xA =

nA nA + nB

  (i) Relation between mole fraction and molality Mole of solute w A × M B xA m = = = x B N Mole of solvent wB × M A

 

x A × 1000 w A × 1000 x × 1000 = =m⇒ A =m xB × M B wB × M A (1 − x A )M B

(ii) Relation between mole fraction and molarity M=

where r is the density of the solution.

x B × 1000 × r x A M A + xB M B

1 r MB = = m M 1000 (iii) Normality (N ): It is number of gram equivalents of solute present per liter of solution. N=

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Number of gram equivalent of solute w A × 10000 × n = Volume of the solution M A × V (mL )

where n = number of electrons participating (acidity or basicity)

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wA = weight of solute (g),

207

MA = equivalent weight of solute, V = Volume of solution (liter) n N = M ×n





−1

Unit: g equiv L or Normal (iv) Parts per million (ppm): It is applied whenever concentration of solution is very low. ppm =



massof solute ×106 mass of solution

Similarly concentration can also be expressed in terms of parts per billion (ppb).

3. Solubility (a) T  he term solubility is the amount of one substance (solute) that will dissolve in a specified amount of another substance (solvent) under stated conditions. (b) Factors affecting solubility (i) Particle size: Since surface-to-volume ratio increases as size decreases, smaller crystals dissolve faster than the larger ones. (ii) Nature of solute and solvent. (iii) Concentration of the solution: As the concentration of the solution increases and the solution becomes more nearly saturated with the solute, the rate of dissolution decreases. (iv) Agitation or stirring. (v) Temperature: In most cases, the rate of dissolution of solid increases with temperature. For gases, the solubility decreases with increase in temperature. (vi) Pressure: This is generally applicable to solubility of gases and is governed by Henry’s law 4. Henry’s Law

I t states that the concentration of a gas dissolved by a given volume of a liquid at constant temperature is directly proportional to the pressure of the gas over the solution. (i) Henry’s law quantitatively correlates pressure and solubility C gas ∝ pgas C gas = K H pgas



where Cgas is the concentration of the gas in the solution, pgas is the partial pressure of the gas above the solution and the proportionality constant KH is called the Henry’s constant. The unit of KH depends on the units used for concentration and pressure. (ii) An alternate and commonly used form of Henry’s law is C1 C 2 = p1 p2





where the subscripts 1 and 2 refer to initial and final conditions. Henry’s law can also be expressed in terms of mole fraction as p = K Hx



where p is the partial pressure, x is the mole fraction and KH is the Henry’s constant.

5. Vapor Pressure of Liquid Solutions



(a) R  aoult’s law: The vapor pressure for liquid solutions is expressed in terms of Raoult’s law, according to which, the partial pressure of a particular component is directly proportional to the component’s mole fraction in the solution. pA = x A pAo

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where xA is the mole fraction of A, pA is the partial pressure of this fraction and pAo is the partial pressure of liquid A when pure. And, by the same argument, pB, the partial pressure of component B, is pB = x B pBo

(b) Vapor pressure of liquid in liquid solution: For a binary solution of two components A and B, to calculate partial pressures, Raoult’s law equation is used for each component in solution because according to Dalton’s law of partial pressures, the total vapor pressure will be the sum of the partial pressures. Therefore, ptotal = pA + pB = x A pAo + x B pBo

Using the relation x A + x B = 1 or x A = 1 − x B in the above equation, we have ptotal = (1 − x B ) pAo + x B pBo = pAo + ( pBo − pAo )x B

(c) Vapor pressure of solid in liquid o pSolution = x solvent psolvent



 he effect of the solute’s mole fraction concentration, x solute , on the vapor pressure can be expressed in terms of T change in vapor pressure as o Dp = x solute psolvent o Dp = psolvent − psolution



Here,



So,      x solute =

o − psolution ) ( psolvent o psolvent

6. Ideal and Non-Ideal Solutions (a) Comparison between ideal and non-ideal solutions is given in the table below. Ideal solution

Non-ideal solution

DVmix = 0

DVmix ≠ 0

DH mix = 0

DH mix ≠ 0

It obeys Raoult’s law.

It does not obey Raoult’s law

Example: Benzene and toluene, n-hexane and n-heptane Example: NaCl in water and sugar in water (b) Positive and negative deviation from Raoult’s law in non-ideal solutions is given in the table below.

Obeys Raoult’s law

Non-ideal solution Positive deviation from Raoult’s law Negative deviation from Raoult’s law Does not obey Raoult’s law Does not obey Raoult’s law

DHmix = 0

DHmix > 0

DHmix < 0

DVmix = 0

DVmix > 0

DVmix < 0

p = pA + pB

pA > pA oxA and pB > pB o xB

pA < pA oxA and pB < pB o xB

p = pA ox + pB o xB

pA + pB > pA oxA + pB o xB

pA + pB < pA oxA + pB o xB

A – A, A – B or B – B intertactions should be the same For example, Benzene + Toluene and n-hexane + n-heptane

A – B attractive forces are weaker than A – A or B – B interactions For example, Acetone + Ethanol Acetone + CS2 Water + CH3OH CCl4 + CH3OH

A – B attractive forces are stronger than A – A and B – B interactions For example, Acetone + Aniline, Acetone + Chloroform CH3OH + CH3COOH

Ideal solution

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209

7. Graphical Representation pAo

p

B

Vapor pressure

= pA

p = pA + pB

pBo

=

po Bx B o xA A p

Vapor pressure xA = 0 xB = 1

xA = 1 xB = 0

pBo

pAo

Vapor pressure

xA = 1 xB = 0

xA = 0 xB = 1

xA = 1 xB = 0

Positive deviation

Ideal

pBo

pAo

xA = 0 xB = 1

Negative deviation Non-ideal

8. Azeotropic (Constant Boiling) Mixtures (a) S ome liquid solutions of two (or more) components have the same composition in solution as well as in vapor phase. Such mixtures are azeotropic or constant boiling mixtures and are known as azeotropes. (b) The azeotropes have a characteristic boiling point which is either lower (positive deviation) or higher (negative deviation) than any of its constituents. For example, ethanol, obtained on distillation [an azeotrope of ethanol (95%) + water (5%)] is a minimum boiling (positive deviation) azeotrope. Similarly, mixture of nitric acid (68%) + water (32%) by mass is a maximum boiling (negative deviation) azeotrope. 9. Colligative Properties and Determination of Molar Mass The colligative properties are properties that depend only on the number of solute particles in a solution, but are independent of the nature of those particles. The four colligative properties of solutions are vapor pressure lowering, boiling point elevation, freezing point depression and osmotic pressure.

Elevation of Boiling Point

Relative Lowering of Vapor Pressure

Colligative property

Definition

Mathematical relation

The presence of a o DpA p A − pA non-volatile solute in = = xB o o pA pA a solvent lowers the equilibrium vapor For dilute solutions nA + nB is almost equal to nA pressure from that of o p A − pA nB wB × M A the pure solvent. = = o nA M B × w A pA where wA and wB are the masses of solvent and solute taken and MA and MB are the molar masses of the solvent and solute. Addition of a solute raises the boiling point of a solution. Since boiling point elevation is a ­colligative property, its magnitude is proportional to the relative amount of solvent and solute.

The difference in the boiling point of a solution containing a nonvolatile solute and the boiling point of the pure solvent is known as boiling-point elevation (∆Tb). DTb = Tb − Tbo where ΔTb is the number of celsius (or kelvin) degrees that the boiling point is raised, Tb is the boiling point of the solution and Tbo is the boiling point of pure solvent. Relation with molality: DTb = K bm where, Kb is a constant characteristic of the solvent (for water Kb = 0.512 molal is molality) The constant Kb is called boiling point elevation constant or Ebullioscopic constant. (Continued)

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OBJECTIVE CHEMISTRY FOR NEET

Colligative property Definition

Mathematical relation Determination of molar mass MB =

1000 × wB × K b DTb × w A

Relation between Kb and enthalpy of vaporization

Depression in Freezing Point

RTb*2 M A  ∂T  = Kb =  b  ∂m  m→0 1000 × DH vap Addition of a solute lowers the freezing point of a solution. Since freezing point depression is a colligative property, its magnitude is proportional to the relative amount of solvent and solute.

The difference in the freezing point of a solution ­containing a non-volatile solute and the freezing point of the pure solvent is known as freezing point depression (∆Tf ) is given by the equation DTf = Tf o − Tf where ΔTf the number of Celsius (or Kelvin) degrees that the freezing point is depressed, Tf is the freezing point of the solution and o T f is the freezing point of pure solvent. Relation with molality DTf = K f m where Kf is constant characteristic of the solvent and m is molality. The constant Kf is called freezing point depression constant or cryoscopic constant. Determination of molar mass from the definition of molality MB =

1000 × wB × K f DTf × w A

Relation between Kf and enthalpy of fusion

Osmotic Pressure

RTf *2 M A  ∂T  = Kf =  f   ∂m  m→0 1000 × DH fusion The extra pressure required to establish an osmotic equilibrium is known as the osmotic pressure. Osmotic pressure is a colligative property and is dependent only on the concentration of the solute particles and is independent of their nature.

The osmotic pressure P is given by P = CRT where C is the concentration in mol−1. In terms of mole per liter, osmotic pressure is given as: n P = B RT V where nB is the number of moles of solute and V is the volume of the solution. w PV = B RT MB Mass of an unknown solute in the solution is MB =

Chapter 9_Solutions.indd 210

wB RT P V

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Solutions

211

10. Abnormal Colligative Properties (a)  When a solute undergoes association on dissolving in a solvent, the number of particles of solute is reduced. As a result, the experimentally determined colligative property is lower than that expected from Raoult’s law and ideal behavior. In case of association of solute particles, the molar mass determined experimentally is always higher than the actual value. Similarly, when a solute dissociates, the number of particles increases and the experimentally determined colligative property is higher. The molar mass determined experimentally is lower than the actual value. These values of molar mass that are higher or lower than the true value are known as abnormal molar mass. (b)  Abnormal colligative properties can be corrected by van’t Hoff factor. It is defined as the ratio of experimentally determined value of colligative property to the calculated value of colligative property        i =



It can also be expressed in terms of i=



Observed value of colligative property Calculated value of colliggative property

i=

or

Number of moles of solute after association/dissociation Number of moles of solute before association/dissociation Normal molar mass Abnormal molar mass

(c)  For solutes that associate or dissociate in solution, the equation of colligative properties is modified to include the van’t Hoff factor. Relative lowering of vapor pressure: Dp =

pAo − pA n =i B o pA nA

Elevation of boiling point: DTb = iK bm Depression of freezing point: DTf = iK f m Osmotic pressure: P = i

nB RT V

(d) Association and dissociation of solute (i) Dissociation of solute a=





where a = degree of dissociation, i = van’t Hoff factor, n = number of ions present (ii) Association of solute nA  An (1-a )





Chapter 9_Solutions.indd 211

i −1 n −1

a n

Total number of particles in the solution is 1 − a +

van’t Hoff factor,

a 1  = 1 +  − 1 a n  n

1  1 +  − 1 a n  i= 1

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OBJECTIVE CHEMISTRY FOR NEET

Solved Examples 1. The ratio of the value of any colligative ­property for KCl solution to that for sugar is nearly _____ times. (1) 1 (2) 0.5 (3) 2 (4) 2.5

For B; 2p = 0.1 × R × T p1 = 3p

(3) KCl dissociates into K+ and Cl− ions whereas sugar does not dissociate. Therefore, the ratio of van’t Hoff factor of KCl solution to that of sugar is 2. 2. The van’t Hoff factor for 0.1 M La(NO3)3 solution is found to be 2.74. The percentage dissociation of the salt is (2) 58% (3)  65.8%

(4) For ternary electrolyte; p1 = CRT = 0.05 × 3 × S × T

Solution

(1)  85%

Solution

(4) 56.8%

Solution

5. Osmotic pressure of blood is 7.65 atm at 310 K. An aqueous solution of glucose that will be isotonic with blood is__________(w/V). (1) 5.41% (2) 3.54% (3) 4.53% (4) 53.4% Solution

La(NO 3 )3  La 3+ + 3NO 3Initial moles 1 0 0 Molesat equilibrium 1 -a a 3a

Total number of moles at equilibrium = 1 − a + a + 3a = i



1 + 3a = 2.74 ⇒ a = 0.58



Therefore, % dissociation of salt is 58%.



Substituting P = 7.65 atm, T = 310 K, R = 0.0821 L atm mol−1K−1 and M = 180 g mol−1 in Eq. (1), we get w ´ 0.0821 ´ 310 180 w 7.65 ´ 180 = = 54.1 g L-1 V 0.0821 ´ 310

7.65 ´ V =



The expression for mass by volume percentage of solute is

3. The values of observed and calculated formula mass of silver nitrate are 92.64 and 170 respectively. The degree of dissociation of silver nitrate is (1)  60% (3)  46.7%

(2) The reaction involved is AgNO3  Ag + + NO 3Initial moles 1 Moles at equilibrium 1 -a

0 a

0 a

Total number of moles at equilibrium        1 − a + a + a = 1+ a = i    (1) We know

Calculated formula mass i= Observed formula mass =

170 = 1.835 92.64

From Eq. (1), we get

1 + a = 1.835 ⇒ a = 0.835

Therefore, degree dissociation of salt is 83.5%.

4. Solute A is a ternary electrolyte and solute B is non-electrolyte. If 0.1 M solution of solute B produces an osmotic pressure of 2p, then 0.05 M solution of A at the same temperature will produce an osmotic pressure equal to (1) p (2) 1.5p (3) 2p (4) 3p

Chapter 9_Solutions.indd 212

Mass of solute × 100 Volume of solution(mL) 54.1 = × 100 = 5.41% 1000

=

(2) 83.5% (4)  60.23%

Solution



w RT (1) M

For isotonic solution, PGlucose = PBlood

(2) The reaction involved is



PV =

(1) We know

6. A solution containing 8.6 g urea in one liter was found to be isotonic with a 5% (w/V) solution of an organic non-volatile solute at same temperature. The molecular weight of latter is (1) 348.9 g mol−1 (2) 34.89 g mol−1 (3) 362.5 g mol−1 (4) 861.2 g mol−1 Solution (1) Isotonic solutions have the same osmotic pressure and hence the same concentration of the solute in the solution. Therefore, (8.6 g/60 g mol −1 ) (5 g/ M ) = 1L 0.1 L 5 × 60 M= = 348.8 g mol −1 8.6 × 0.1 7. The relationship between the values of osmotic pressures of 0.1 M solutions of KNO3 (P1) and CH3COOH (P2) is (1) P1 > P2 (2) P2> P1 (3) P1 = P2 (4) P1/(P1 + P2) = P2/(P1 + P2) Solution (1) For equimolar solution, osmotic pressure depends on the value of van’t Hoff factor. Therefore, osmotic pressure ∝ i

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Solutions

For KNO3 (being strong electrolyte), i1 = 2



But for CH3COOH, i2 = 1 + a, since a < 1, therefore, i2 < 2



Thus, i1 > i2, hence, P1 > P2

8. 20 g of a binary electrolyte (Mol. wt. = 100) is dissolved in 500 g of water. The freezing point of the solution is –0.74°C, Kf = 1.86 K m–1. The degree of ionization of the electrolyte is

D Tf = K f mGlucose + K f mKCl + K f mUrea é 1000 ´ mGlucose 1000 ´ mKCl 1000 ´ mUrea ù = Kf ê + ´i + ú M ´ M ´ m M m water KCl water Urea ´ mwater û ë Glucose é1000 ´ 10 1000 ´ 10 ´ 2 1000 ´ 5 ù = Kf ê + + 74.5 ´ 90 60 ´ 95 úû ë 180 ´ 90 = 1.86 ´ 4.476 = 8.325 K

(1) 50% (2) 75% (3) 100% (4) 0 Solution (4) The freezing point depression is given by DTf = iK f m We have



van’t Hoff factor is calculated as i=



(1) –1.86°C (2) 1.86°C (3) –3.92°C (4) 2.42°C Solution (1) Depression in freezing is given by D Tf =

0.74 × 500 = 0.99 ≈ 1 1.86 × (20 /100) × 1000

       

For binary electrolyte, 1 + a = i ⇒ 1 + a = 1 ⇒ a = 0

9. Addition of 0.643 g of a compound to 50 mL of benzne (density = 0.879 g mL–1) lowers the freezing point from 50.51°C to 50.03°C. If Kf for benzene is 5.12, the molecular weight of the compound is (1) 156 g (2) 312 g (3) 78 g (4) 468 g Solution (1) Weight of benzene = V × r = 50 × 0.879 = 43.95 g

11. The molal freezing point constant for water is 1.86°C m–1. If 34.2 g of cane sugar (C12H22O11) is dissolved in 1000 g of water, the solution will freeze at

Substituting the values, we get i=



D Tf × w H2O DTf = K f × m K f × nelectrolyte × 1000

Freezing point depression is given by D Tf = Tfo − Tf 8.325 = 273 - Tf Þ Tf = 273 - 8.325 = 264.67 K

DTf = 0 − ( −0.74) = 0.74



Freezing point depression is





We have DTf =

M unknown compound ´ w Benzene 0.48 =

M unknown compound

5.12 ´ 1000 ´ 0.643 M unknown compound ´ 43.95

5.12 ´ 1000 ´ 0.643 = 0.48 ´ 43.95 = 156 g

10. The freezing point of an aqueous solution containing 5% by mass urea, 10% by mass KCl and 10% by mass of glucose, is (Kf for water 1.86 K m–1) (1) 290.2 K (2) 285.5 K (3) 269.93 K (4) 264.67 K Solution (4) We have

Chapter 9_Solutions.indd 213

DTf = DTf(Glucose) + DTf(KCl) + DTf(Urea)

K f × 1000 × wcanesugar M canesugar × w water

1.86 × 1000 × 34.2 = 342 × 100 = 1.86°C

1.86 = 0 − Tf ⇒ Tf = −1.86°C

We have

12. 1.0 molal aqueous solution of an electrolyte X3Y2 is 25% ionized. The boiling point of the solution is (Kb for H2O = 0.52 K kg mol−1) (1) 375.5 K (2) 374.04 K (3) 377.12 K (4) 373.25 K Solution (2) The reaction involved is X 3 Y2  3 X 2+ + 2 Y 3−

DTf = (50.51 - 50.03)°C = 0.48°C K f ´ 1000 ´ w unknown compound

213

(1 − a )



3a

2a

Total number of moles at equilibrium = 1 − a + 3a + 2a = 1 + 4a  = i

   i = 1 + 4 × 0.25 = 2

We know

DTb = iK bm

        DTb = 2 ´ 0.52 ´ 1 = 1.04



Boiling point elevation is given by D Tb = Tb - Tbo 1.04 = Tb - 373 Þ Tb = 373 + 1.04 = 374.04 K

13. A 0.2 molal aqueous solution of a weak acid is 20% ionized. The freezing point of this solution is (Kf = 1.86 K kg mol−1 for water) (1) −0.45°C (2) −0.9°C (3) −0.31°C (4) −0.53°C

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OBJECTIVE CHEMISTRY FOR NEET (1) −0.544°C (2) −0.512°C (3) −0.272°C (4) −1.86°C

Solution (1) The reaction involved is HX  H+ + X − Initial moles 1 0 0 Moles at equilibrium 1 −a a a

Therefore, i = 1 − a + a + a = 1 + a = 1 + 0.2 = 1.2

We know

Solution (3) For same solution, we have K DTf K f = or DTf = f ´ DTb DTb K b Kb DTf =

DTf = iK f m

        = 1.2 ´ 1.86 ´ 0.2 = 0.45 = 1 . 2 ´ 1 . 86 ´ 0 . 2       = 0.45

Freezing point depression is given by



     DTf = Tfo − Tf = 0 − 0.45 = − 0.45°C

14. NaCl is added to 1 L water to such an extent that ∆Tf/Kf becomes equal to 1/500. Calculate the weight of NaCl added.



Diluting the solution with equal amount of water reduces the freezing point of solution by half, as 1 DTf ∝ weightof solvent



Therefore,



Freezing point depression is given by

(3) We know, for NaCl, i = 2.

Depression of freezing point can be expressed as



   DTf = iK f m

(1)

4K bY M (2) K bY M

(3)

K bY K bY (4) 4M M

Solution (2) Boiling point elevation is given by K × 1000 × wsolute DTb = b M solute × w Benzene

15. The amount of ice that will separate on cooling a solution containing 50 g of ethylene glycol in 200 g water to –9.3°C is (Kf = 1.86 K m–1) (1) 38.71 g (2) 38.71 mg (3) 42 g (4) 42 mg Solution (1) Depression in freezing is given by DTf =

1.86 × 1000 × 50 9.3 = ⇒ w ice = 161.29 g 62 × w ice

Therefore, amount of ice separated = (200 − 161.29) g = 38.71 g

16. The boiling point of an aqueous solution of a non-­ volatile solute is 100.15°C. What is the freezing point of an aqueous obtained by diluting the above solution with an equal volume of water? The values of Kb and Kf for water are 0.512°C kg m−1 and 1.86°C kg mol−1 respectively

Chapter 9_Solutions.indd 214

DTb =

K b × 1000 × Y 4 K bY = M solute × 250 M

18. The vapor pressure of a solvent decreased by 10 mm Hg when a non-volatile solute was added to the solvent. The mole fraction of solute in solution is 0.2, what would be mole fraction of the solvent if decrease in vapor pressure is 20 mm of Hg. (1) 0.8 (2) 0.6 (3) 0.4 (4) 0.2

K f × 1000 × w ethylene glycol M ethylene glycol × w ice

0.545 = 0.2725 2

17. Y g of non-volatile organic substance of molecular mass M is dissolved in 250 g benzene. Molal elevation constant of benzene is Kb. Elevation in its boiling point is given by

DTf = 2´m Kf 1 1 1 = 2´m Þm = mol kg -1 water or mol L-1 water 500 1000 1000 1 m= ´ 58.5 g NaCl = 0.0585 g NaCl 1000

DTf =

DTf = Tfo − Tf 0.2725 = 0 − Tf ⇒ Tf = −0.2725°C

(1) 5.85 g (2) 0.585 g (3) 0.0585 g (4) 58.5 g Solution

0.15 × 1.86 = 0.545 0.512

Solution (2) From Raoult’s law, we have o o o psolvent − psolution = xsolute psolvent

o (1) 10 = 0.2 × psolvent



o 20 = x × psolvent



Similarly



From Eq. (1) and Eq. (2), we get



(2)

xsolute = 0.4

Therefore, mole fraction of solvent

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Solutions xsolvent = 1 − xsolute = 1 − 0.4 = 0.6 19. Which of the following plots represents the behavior of an ideal binary liquid solution? (1) Plot of ptotal vs yA (mole fraction of A in vapor phase) is linear. (2) Plot of ptotal vs yB is linear. (3) Plot of 1/ptotal vs yA is linear. (4) Plot of 1/ptotal vs yB is non-linear.

22. If the total vapor pressure of the liquid mixture A and B is given by the equation p = 180 x A + 90 , then the ratio of the vapor pressure of the pure liquids A and B is given by (1) 3:2 (2) 4:1 (3) 3:1 (4) 6:2 Solution (3) We know      p = x A p Ao + xB pBo     p = x A p Ao + pBo(1 - x A )

Solution (3) From Dalton’s law, we have p A = y A pTotal where yA is the mole fraction and pA is the partial pressure of A. p yA = A pTotal

Thus, graph of y A vs

1 pTotal

is linear.

20. The vapor pressure of a solvent is 20 torr, while that of its dilute solution is 17 torr. The mole fraction of the solvent is (1) 0.6 (2) 0.85 (3) 0.5 (4) 0.7 Solution (2) We know    xsolute =

o (psolvent − psolution ) (1) o psolvent

o o = 20 torr and psolution = 17 torr in  Eq. (1), Substituting psolvent we get

xsolute =

20 − 17 3 = = 0.15 20 20



p = ( p Ao - pBo )x A + pBo (1)



Given  p = 180 x A + 90 (2)



Comparing Eq. (1) and Eq. (2), we have





p Ao − pBo = 180 and pBo = 90 p Ao = 180 + 90 = 270



Therefore,

p Ao 270 3 = = = 3 :1 pBo 90 1

23. The degree of dissociation of Ca(NO3)2 in dilute aqueous solution containing 7 g of the salt per 100 g of water at 100°C is 70%. If the vapor pressure of water at 100°C is 760 mm Hg. The vapor pressure of the solution is (1) 746.3 mm Hg (2) 1492.6 mm Hg (3) 373.2 mm Hg (4) 74.63 mm Hg Solution (1) The dissociation of Ca(NO3)2 is Ca(NO3 )2 ® Ca 2+ + 2NO3-

   xsolvent = 1 − xsolute = 1 − 0.15 = 0.85 21. The vapor pressure of pure benzene and toluene are 160 and 60 torr respectively. The mole fraction of toluene in vapor phase in contact with equimolar solution of benzene and toluene is (1) 0.5 (2) 0.6 (3) 0.27 (4) 0.73







Solution (4) As the solution is equimolar, therefore, xBenzene = x Toluene = 0.5 o pBenzene = xBenzene pBenzene

= 0.5 ´ 160 = 80

o o We know    pTotal = xBenzene pBenzene + x Toluene pToluene

pTotal = 0.5 ´ 160 + 0.5 ´ 60



    



Therefore, mole fraction of toluene in vapor phase y toluene =

Chapter 9_Solutions.indd 215

= 80 + 30 = 110 torr ptoluene 80 = = 0.73 ptotal 110

215



Initial conc. Final conc.

1 1 -a

0 a

0 2a

Total number of moles (N) = 1 − a + a + 2 a 7 = 1 + 2 a = (1 + 2 × 0.7 ) × = 0.1 164 100 Number of moles of solvent (H2O) = = 5.55 18 From Raoult’s law, we have

o nH2O (psolvent − psolution ) = o N + nH2O psolvent

760 - psolution 0.1 = Þ psolution = 746.3 mm Hg 100 760 0.1 + 18 24. The vapor pressure of benzene at 80°C is lowered by 10 mm by dissolving 2 g of a non-volatile substance in 78 g of benzene. The vapor pressure of pure benzene at 80°C is 750 mm. The molecular weight of the substance will be (1) 15 (2) 150 (3) 1500 (4) 148

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OBJECTIVE CHEMISTRY FOR NEET

o pwater − pwater 1 1 = = o 1+ 2 3 pwater

Solution (4) We have

p

−p

o Benzene Benzene o Benzene

p

=

1−

w non-volatile × M Benzene M non-volatile ×w Benzene

10 2 ´ 78 = (750 - 10) M non-volatile ´ 78 2 ´ 78 ´ 740 10 2 ´ 78 = 148 = Þ M non-volatile = 10 ´ 78 740 M non-volatile ´ 78 25. A mixture of two immiscible liquids nitrobenzene and water boiling at 99°C has a partial vapor pressure of water 733 mm and that of nitrobenzene 27 mm. The ratio of the weights of nitrobenzene to the water in the distillate is (1) 2:1 (2) 4:1 (3) 3:1 (4) 1:4 Solution

pwater 1 2 =1− = o pwate 3 3 r 27. Dry air was passed successively through a solution of 5 g of a solute in 180 g of water and then through pure water. The loss in weight of solution was 2.50 g and that of pure solvent is 0.04 g. The molecular weight of the solute is (1) 31.25 (2) 3.125 (3) 312.5 (4) 302.5 Solution (1) p0 − ps ∝ loss in weight of water chamber ps ∝ loss in weight of solution chamber

o xsolvent (4) We know     psolution = psolvent



  w1 /123 For nitrobenzene 27 = 760  (1)  w 2 /18 + w1 /123 



For water



Therefore, from Eq. (1) and Eq. (2), we get

  w 2 /18 733 = 760  (2)  w 2 /18 + w1 /123  w1 1 = w2 4



26. One mole of non-volatile solute is dissolved in two moles of water. The vapor pressure of the solution relative to that of water is

p

− pwater 1 1 = = o 1+ 2 3 pwater 1−

pwater 1 = o pwater 3

p Practice Exercises

water

Level I

o pwate r

=1−

(1) 0.467 (2) 0.502 (3) 0.513 (4) 0.556 Solution (3) We have

o o pTotal = pmethanol x methanol + pethanol xethanol

16 / 32 46 / 46     + 42 ×  pTotal = 88.5 ×   16 / 32 + 46 / 46   16 / 32 + 46 / 46  = 29.5 + 28 = 57.5

o Now, pmethanol x methanol = pTotal y methanol

29.5 = 57.5 y methanol ⇒ y methanol =

29.5 = 0.513 57.5

1 2 = 3 3

Concentration Terms 1. 8 g NaOH is dissolved in one liter of solution. Its molarity is (1) 0.8 M (2) 0.4 M (3) 0.2 M (4) 0.1 M 2. Mole fraction of C3H5(OH)3 in a solution of 36 g of water and 46 g of C3H5(OH)3 is (1) 0.46 (2) 0.36 (3) 0.20 (4) 0.40

Chapter 9_Solutions.indd 216

o pwater − pwater wsolute × M water = o pwater M solute × w water

28. The vapor pressures of ethanol and methanol are 42.0 mm Hg and 88.5 mm Hg respectively. An ideal solution is formed at the same temperature by mixing 46.0 g of ethanol with 16.0 g of methanol. The mole fraction of methanol in the vapor is

Solution

o water

We have

0.04 5 × 18 5 × 18 × 2.50 ⇒ M solute = = 31.25 = 2.50 M solute × 180 0.04 × 180

(1) 2/3 (2) 1/3 (3) 1/2 (4) 3/2

p o − pwater = x non-volatile (1) We have    water o pnon-volatile

pwater 1 = o pwater 3

3. Which of the following concentration terms is independent of temperature? (1) Molarity (2) Normality (3) Molality (4) Both (1) and (2) 4. How many grams of cane sugar are present in 534.2 g of its aqueous solution having molality 0.2? (1) 3.42 (2) 34.2 (3) 18 (4) None of these

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Solutions

5. What weight of glucose dissolved in 100 g of water will produce the same lowering of vapor pressure as one gram of urea dissolved in 50 g of water, at the same temperature? (1) 3 g (2) 5 g (3) 6 g (4) 4 g 6. The mole fraction of toluene in vapor phase which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene 0.500 is (Vapor pressure of pure benzene and pure toluene are 119 torr and 37.0 torr respectively at the same temperature.) (1) 0.5 (2) 0.763 (3) 0.237 (4) 1 7. The vapor pressure of pure benzene at 50°C is 268 torr. How many moles of non-volatile solute per mol of benzene are required to prepare a solution of benzene having a vapor pressure of 167 torr at 50°C? (1) 0.377 (2) 0.605 (3) 0.623 (4) 0.395 8. At 35°C, the vapor pressure of pure chloroform is 0.359 atm and that of pure acetone is 0.453 atm. A solution containing 1 mol of chloroform and 4 mol of acetone has a vapor pressure of (in atm)

(3) ideal behavior. (4) all of the above. 14. The relative lowering of the vapor pressure of an aqueous solution containing a non-volatile solute is 0.0125. The molality of the solution is (1) 0.70 (2) 0.50 (3) 0.80 (4) 0.40 15. The vapor pressure of a dilute aqueous solution of glucose is 750 mm of mercury at 373 K. The mole fraction of solute is 1 1 (2) 10 7.6 1 1 (3) (4) 35 76 (1)

16. The vapor pressure of a pure liquid A is 70 torr at 27°C. It forms an ideal solution with another liquid B. The mole fraction of B is 0.2 and total pressure of the solution is 84 torr at 27°C. The vapor pressure of pure liquid B at 27°C is (1) 14 (2) 56 (3) 140 (4) 70 17. The diagram given below is a vapor pressure composition diagram for a binary solution of A and B. Vapor pressure

Vapor Pressure, Raoult’s Law and Relative Lowering of Vapor Pressure

(1) 0.400 (2) 0.812 (3) 0.094 (4) 0.434 9. For A and B to form an ideal solution, which of the following conditions should be satisfied? (1) ∆H (mixing) = 0 (2) ∆V (mixing) > 0 (3) ∆S (mixing) = 0 (4) All three conditions. 10. Lowering of vapor pressure due to a solute in 1 molal aqueous solution at 100°C is (1) 13.43 torr (2) 14.12 torr (3) 312 torr (4) 352 torr 11. At a given temperature, total vapor pressure in torr of a mixture of volatile components A and B is given by pTotal = 120 – 75xB. Hence, vapor pressure of pure A and B respectively (in torr) are (1) 120, 75 (2) 120, 195 (3) 120, 45 (4) 75, 45 12. When a solution of CHCl3 is mixed with a solution of O   C(CH3)2, ∆Vmix is (1) positive. (2) negative. (3) zero. (4) can’t be predicted. o 13. Mixture of 1 mol benzene ( pbenzene = 42 mm ) and 2 mol o toluene ( ptoluene = 36 mm ) will have

(1) total vapor pressure 38 mm. (2) mole fraction of vapors of benzene above liquid mixture is 7/19.

Chapter 9_Solutions.indd 217

217

C

B

A

XB

D

In the solution, A−B interactions are (1) similar to A−A and B−B interactions. (2) greater than A−A and B−B interactions. (3) smaller than A−A and B−B interactions. (4) unpredictable. 18. When the azeotropic mixture of water (b.p. = 100°C) and HCl (b.p = 85°C) is distilled, it is possible to obtain (1) (2) (3) (4)

pure HCl. pure water. pure HCl as well as water. neither HCl nor H2O in pure form.

19. A binary liquid solution of n-heptane and ethyl alcohol is prepared. Which of the following statements correctly represents the behavior of this liquid solution? (1) The solution formed is an ideal solution. (2) The solution formed is a non-ideal solution with positive deviation from Raoult’s law. (3) The solution formed is non-ideal solution with negative deviation from Raoult’s law. (4) n-Heptane exhibits positive deviation, whereas ethyl alcohol exhibits negative deviation from Raoult’s law.

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OBJECTIVE CHEMISTRY FOR NEET

2 0. Mole fraction of A vapors above the solution in mixture of A and B (xA = 0.4) will be (Given: p Ao = 100 mm Hg and pBo = 200 mm Hg) (1) 0.4 (2) 0.8 (3) 0.25 (4) None of these

Elevation in Boiling Point and Depression in Freezing Point 2 1. 0.48 g of a substance is dissolved in 10.6 g of C6H6. The freezing point of benzene is lowered by 1.8°C. What will be the molecular weight of the substance (Kf for benzene = 5)? (1) 250.2 (2) 90.8 (3) 125.79 (4) 102.5 2 2. 0.01 M solution each of urea, common salt and Na2SO4 are taken, the ratio of depression of freezing point is (1) 1:1:1 (2) 1:2:1 (3) 1:2:3 (4) 2:2:3 2 3. An aqueous solution boils at 100.50°C. The freezing point of the solution would be (For water, Kb = 0.51°C m−1 and Kf = 1.86° C m−1) [Assume no association or dissociation] (1) 0°C (2) –1.86°C (3) –1.82°C (4) +1.82°C 2 4. 1.0 molal aqueous solution of an electrolyte A2B3 is 60% ionized. The boiling point of the solution at 1 atm is ( K b (H2 O ) = 0.52 K kg mol -1 ) (1) 274.76 K (2) 377 K (3) 376.4 K (4) 374.76 K 2 5. Molal depression constant is given by the expression ∆Tf (2) ∆Tf × m m ∆Tf (3) ∆Tf × N (4) M 2 6. The freezing point of equimolal aqueous solution will be highest for (1)

(1) C6H5NH3Cl (2) Ca(NO3)2 (3) La(NO3)3 (4) C6H12O6 2 7. Glucose is added to 1 liter water to such an extent that DTf /K f becomes equal to 1/1000. The weight of glucose added is (1) 180 g (2) 18 g (3) 1.8 g (4) 0.18 g 28. A solution containing 3.3 g of a substance in 125 g of benzene (b.p = 80°C) boils at 80.66°C. If Kb for benzene is 3.28 K kg mol−1, the molecular mass of the substance will be (1 130.20 (2) 129.20 (3) 132.20 (4) 131.20 2 9. 100 mL of liquid A was mixed with 25 mL of liquid B to give a non-ideal solution of A–B mixture. The volume of this mixture would be

Chapter 9_Solutions.indd 218

(1) (2) (3) (4)

75 mL 125 mL close to 125 mL but not exactly 125 mL just more than 125 mL

3 0. The freezing point of 0.05 m solution of a non-electrolyte in water is ( K f(H2O) = 1.86°Cm −1 ) (1) –1.86°C (2) –0.93°C (3) –0.093°C (4) 0.93°C 3 1. Which solution will have the highest boiling point? (1) 1 m C6H12O6 solution (2) 1 m NaCl solution (3) 1 m BaCl2 solution (4) 1 m Co(NH2)2 solution 3 2. The molal elevation constant of water is 0.51. The boiling point of 0.1 molal aqueous NaCl solution is (1) 100.05°C (2) 100.1°C (3) 100.2°C (4) 101.0°C 3 3. Calculate the ebullioscopic constant for water if its enthalpy of vaporization is 40.685 kJ mol−1. (1) 0.512 K kg mol−1 (2) 1.86 K kg mol−1 (3) 5.12 K kg mol−1 (4) 3.56 K kg mol−1

Osmosis, Reverse Osmosis and Osmotic Pressure 3 4. The process of getting fresh water from sea water is known as (1) osmosis. (2) filtration. (3) desaltation. (4) reverse osmosis. 3 5. Equal volumes of two 0.1 M glucose solutions are mixed. The resultant solution will have (1) (2) (3) (4)

lower osmotic pressure. same osmotic pressure. higher osmotic pressure. none of these.

3 6. In which of the following cases osmosis takes place, if the solutions are separated by a semi permeable membrane (1) 0.1 M NaCl and 0.2 M glucose. (2) 0.1 M sucrose and 0.1 M fructose. (3) 0.05 M K4[Fe(CN)6] and 0.1 M CaCl2. (4) 10−3 M CaCl2 and 1.5 × 10−3 M NaCl. 3 7. If the observed and normal osmotic pressures of 1% NaCl solution are 5.7 and 3.0 atm, the degree of dissociation of NaCl is (1) 0.9 (2) 1.0 (3) 0.57 (4) 0.3

van’t Hoff Factor and Abnormal Colligative Properties 3 8. The degree of dissociation (a ) of a weak electrolyte AxBy is related to van’t Hoff factor (i) by the expression. (1) a =

i +1 i −1 (2) a = x + y −1 x + y −1

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Solutions

(3) a =

x + y −1 x + y +1 (4) a = i −1 i −1

3 9. Which salt may show the same value of van’t Hoff factor as that of K4[Fe(CN)6] in very dilute solution state? (1) Al2(SO4)3 (2) NaCl (3) Al(NO3)3 (4) Na2SO4 4 0. Benzoic acid dissolved in benzene will show a molecular mass of (assume complete dimerization) (1) 122 (2) 61 (3) 244 (4) 366 4 1. If in a solvent, n simple molecules of solute combine to form an associated molecule, x is the degree of association, the van’t Hoff factor i is equal to (1)

1 1 − x + nx (2) 1 1 − nx

1 − x + x /n x /n − 1 + x (3) (4) 1 1 4 2. The ratio of the value of any colligative property for K4 [Fe(CN)6] solution to that of Fe4[Fe(CN)6]3 solution is nearly (1) 1 (2) 0.71 (3) 1.4 (4) Less than 1

Level II Concentration Terms 1. 1000 g aqueous solution of CaCO3 contains 10 g of calcium carbonate. Concentration of the solution is (1) 10 ppm (2) 100 ppm (3) 1000 ppm (4) 10000 ppm 2. The molarity of a 0.2 N Na2CO3 solution will be (1) 0.05 M (2) 0.2 M (3) 0.1 M (4) 0.4 M 3. What will be the molarity of a 200 mL solution containing 1.2 g of urea? (1) 0.05 M (2) 0.2 M (3) 0.1 M (4) 0.4 M

Vapor Pressure, Raoult’s Law and Relative Lowering of Vapor Pressure 4. A sample of 20 g of a compound (Mol.wt. = 120) which is a non-electrolyte is dissolved in 10.0 g of ethanol (C2H5OH). If the vapor pressure of pure ethanol at the temperature is 0.250 atm. What is the vapor pressure of the solution? (1) 0.250 atm (2) 0.83 atm (3) 0.125 atm (4) 0.141 atm

Chapter 9_Solutions.indd 219

219

5. Which of the following azeotropic solutions has the boiling point less than boiling point of the constituents A and B? (1) CHCl3 and CH3COCH3 (2) CS2 and CH3COCH3 (3) CH3CH2OH and CHCl3 (4) CH3CHO and CS2 6. Vapor pressure of methyl alcohol and ethyl alcohol solution is represented by p = 115xA + 140 where xA is the mole fraction of methyl po alcohol. The value of lim B is x A →0 x B (1) 255 (2) 115 (3) 140 (4) 135 7. An ideal solution has equal mole fractions of two volatile components A and B. In the vapor above the solution, the mole fractions of A and B (1) (2) (3) (4)

are both 0.50. are equal but necessarily 0.50. are not very likely to be equal. are 1.00 and 0.00 respectively.

8. Which of the following statements is correct, if the intermolecular forces in liquids A, B and C are in the order A < B < C? (1) (2) (3) (4)

B evaporates more readily than A. B evaporates less readily than C. A and B evaporate at the same rate. A evaporates more readily than C.

9. Benzene (Mol. wt. = 78 g mol−1) and toluene (Mol.wt. = 92 g mol−1) form an ideal solution. At 60°C the vapor pressure of pure benzene and pure toluene are 0.507 atm and 0.184 atm, respectively. The mole fraction of benzene in a solution of these two chemicals that has a vapor pressure of 0.350 atm at 60°C (1) 0.514 (2) 0.690 (3) 0.486 (4) 0.190 10. Dry air was successively passed through a solution of 5 g of a solute in 180 g of water and then through pure water. The loss in weight of solution was 2.50 g and that of pure, solvent 0.04 g. The molecular weight of the solute is (1) 31.25 g (2) 3.125 g (3) 312.5 g (4) 3125 g 1 1. The vapor pressure of ethyl alcohol and methyl alcohol are 45 mm and 90 mm. An ideal solution is formed at the same temperature by mixing 60 g of C2H5OH with 40 g of CH3OH. Total vapor pressure of the solution is approximately (1) 70 mm (2) 35 mm (3) 105 mm (4) 140 mm

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OBJECTIVE CHEMISTRY FOR NEET

1 2. The vapor pressure in mm of Hg of an ideal solution of A and B at 25°C is given by pAB = 33x + 94, whereas that of an ideal solution of A and C at 25°C is given by pAC = 81x + 46; x being mole fraction of A in the solution. The vapor pressure of solution containing 2 mol A, 3 mol B and 4 mol of C will be (1) 80 (2) 90 (3) 100 (4) 70 1 3. At 323 K, the vapor pressure in millimeters of mercury of a methanol-ethanol solution is represented by the equation p = 120xA + 140, where xA is the mole fraction of p methanol. Then the value of lim A is x A →1 x A (1) 120 mm (2) 140 mm (3) 260 mm (4) 20 mm 1 4. For an ideal binary liquid solution with p Ao > pBo for which relation between xA (mole fraction of A in liquid phase) and yA (mole fraction of A in vapor phase) is correct? (1) yA < yB (2) xA < xB (3)

yA xA y x > (4) A < A y B xB y B xB

1 5. The vapor pressure of a solution of a non-volatile electrolyte B in a solvent A is 95% of the vapor pressure of the solvent at the same temperature. If the molecular weight of the solvent is 0.3 times the molecular weight of solute, the weight ratio of the solvent and solute are (1) 0.15 (2) 5.7 (3) 0.2 (4) 4.0 1 6. The vapor pressure of a solvent decreased by 10 mm of Hg when a non-volatile solute was added to the solvent. The mole fraction of solute is 0.2. What would be mole fraction of the solvent if decrease in the vapor pressure is 20 mm of Hg? (1) 0.2 (2) 0.4 (3) 0.6 (4) 0.8

Elevation in Boiling Point and Depression in Freezing Point 1 7. Equimolal solutions of A and B show depression in freezing point in the ratio 2:1. If A remains in its normal state in solution, then B will be in which state (1) normal. (2) dissociated. (3) associated. (4) hydrolysis. 1 8. 20 g of a binary electrolyte AB (Mol.wt. = 100) are dissolved in 500 g of water (Kf = 1.86). The freezing point of the solution is –0.74°C. The degree of ionization of the electrolyte is (1) 50% (2) 75% (3) 10% (4) 0%

Chapter 9_Solutions.indd 220

1 9. A solution containing 28 g phosphorous in 315 g CS2 (b.p. = 46.3°C) boils at 47.9°C ( K b( CS2 ) = 2.34). What will be molecular formula of phosphorus? (Assuming complete association.) (1) P4 (2) P8 (3) P2 (4) None 2 0. An aqueous solution of a solute which neither associates nor dissociates has a freezing point depression of X°C. An equimolal solution of a second salt has a freezing point depression of 3X°C. Assuming 100% dissociation of salt, the second solution could be a salt of formula (1) AB3 (2) AB2 (3) A3B (4) A2B3 2 1. When mercuric iodide is added to the aqueous solution of potassium iodide, the (1) (2) (3) (4)

freezing point is raised. freezing point is lowered. freezing point does not change. cannot predict.

2 2. The normal boiling point of toluene is 110.7°C, and its Kb is 3.32 K kg mol−1. The enthalpy of vaporization of toluene is nearly (1) 17.0 kJ mol−1 (2) 34.0 kJ mol−1 (3) 51.0 kJ mol−1 (4) 68.0 kJ mol−1

Osmosis, Reverse Osmosis and Osmotic Pressure 23. The weight of urea dissolved in 100 mL solution which produce an osmotic pressure of 20.4 atm at 25°C, will be (1) 5 g (2) 4 g (3) 3 g (4) 6 g 2 4. The relationship between osmotic pressure at 273 K when 10 g of glucose (p1), 10 g urea (p2) and 10 g sucrose (p3) are dissolved in 250 mL of water is (1) p1 > p2 > p3 (2) p3 > p1 > p2 (3) p2 > p1 > p3 (4) p2 > p3 > p1 2 5. A 5.85% (w/V) NaCl solution will exert an osmotic pressure closest to which one of the following? (1) (2) (3) (4)

5.85% (w/V) sucrose solution. 5.85% (w/V) glucose solution. 2 m sucrose solution. 1 m glucose solution.

2 6. Which one of the following pairs of solution can we expect to be isotonic at the same temperature? (1) (2) (3) (4)

0.1 M urea and 0.1 M NaCl 0.1 M urea and 0.2 M MgCl2 0.1 M NaCl and 0.1 M Na2SO4 0.1 M Ca(NO3)2 and 0.1 M Na2SO4

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Solutions 27. The osmotic pressure of a 5% (w/V) solution of cane sugar (Mol.wt. = 342 g) at 15°C is (1) 3.46 atm (2) 3.64 atm (3) 4.0 atm (4) 2.45 atm

4. Mole fraction of the solute in a 1.00 molal aqueous ­solution is (1) 1.7700 (2) 0.1770 (3) 0.0177 (4) 0.0344 (AIPMT PRE 2011, RE-AIPMT 2015)

van’t Hoff Factor and Abnormal Colligative Properties 28. If the van’t Hoff factor of a 0.005 M aqueous solution of KCl is 1.95, then the degree of dissociation of KCl is (1) 0.95 (2) 0.97 (3) 0.94 (4) 0.96 29. If a solute undergoes dimerization and trimerization, the minimum values of the van’t Hoff factors are (1) 0.50 and 1.50 (2) 1.50 and 1.33 (3) 0.50 and 0.33 (4) 0.25 and 0.67 30. The values of observed and calculated molecular mass of Ca(NO3)2 are 65.4 and 164 respectively. What is the degree of ionization of the salt?

5. The van’t Hoff factor i for a compound which undergoes dissociation in one solvent and association in other ­solvent is respectively (1) (2) (3) (4)

greater than one and greater than one. less than one and greater than one. less than one and less than one. greater than one and less than one. (AIPMT PRE 2011)

6. The freezing point depression constant for water is 1.86°C m−1. If 5.00 g Na2SO4 is dissolved in 45.0 g H2O, the freezing point is changed by –3.82°C. Calculate the van’t Hoff factor for Na2SO4. (1) 0.381 (2) 3.11 (3) 2.63 (4) 3.11

(1) 0.25 (2) 0.50 (3) 0.60 (4) 0.75 31. A compound X undergoes tetramerization in a given organic solvent. The van’t Hoff factor is (1) 4.0 (2) 0.25 (3) 0.125 (4) 2.0

(AIPMT PRE 2011) 7. A 0.1 molal aqueous solution of a weak acid is 30% ionized. If Kf for water is 1.86°C m−1, the freezing point of the solution will be (1) −0.24°C (2) −0.18°C (3) −0.54°C (4) −0.36°C

Previous Years’ NEET Questions 1. 0.5 molal aqueous solution of a weak acid (HX) is 20% ionized. If Kf for water is 1.86 K kg mol−1, the lowering in the freezing point of the solution is (1) −0.56 K (2) −1.12 K (3) 0.56 K (4) 1.12 K (AIPMT 2007) 2. A 0.0020 m aqueous solution of an ionic compound [Co(NH3 )5 (NO2 )]Cl freezes at –0.00732°C. Number of moles of ions per mole of ionic compound in water will be

(AIPMT MAINS 2011) 8. 200 mL of an aqueous solution of a protein contain its 1.26 g. The osmotic pressure of this solution at 300 K is found to be 2.57 × 10−3 bar. The molar mass of protein will be (R = 0.083 L bar mol−1K−1) (1) 61038 g mol−1 (3) 122044 g mol−1

(2) 51022 g mol−1 (4) 31011 g mol−1 (AIPMT MAINS 2011)

9. pA and pB are the vapor pressure of pure liquid components, A and B, respectively of an ideal binary solution. If xA represents the mole fraction of component A, the total pressure of the solution will be

(Given: K f = 1.86°C m −1 ) (1) 1 (2) 2 (3) 3 (4) 4 (AIPMT 2009)

(1) pA + xA(pB−pA) (2) pA + xA(pA−pB)

3. A solution of sucrose (molar mass = 342 g mol ) has been prepared by dissolving 68.5 g of sucrose in 1000 g of water. The freezing point of the solution obtained will be (Kf for water = 1.86 K kg mol–1)

(3) pB + xA(pB−pA) (4) pB + xA(pA−pB)

–1

(1) −0.520°C (2) +0.372°C (3) −0.570°C (4) −0.372°C (AIPMT PRE 2010)

Chapter 9_Solutions.indd 221

221

(AIPMT PRE 2012) 10. Which of the following compounds can be used as antifreeze in automobile radiators? (1) Methyl alcohol (2) Glycol (3) Nitrophenol (4) Ethyl alcohol (AIPMT MAINS 2012)

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222

OBJECTIVE CHEMISTRY FOR NEET

11. Vapor pressure of chloroform (CHCl3) and dichloromethane (CH2Cl2) at 25°C are 200 mm Hg and 41.5 mm Hg respectively. Vapor pressure of the solution obtained by mixing 25.5 g of CHCl3 and 40 g of CH2Cl2 at the same temperature will be: (Molecular mass of CHCl3 = 119.5 u and molecular mass of CH2Cl2 = 85 u.) (1) 173.9 mm Hg (3) 347.9 mm Hg

(2) 615.0 mm Hg (4) 285.5 mm Hg (AIPMT MAINS 2012)

12. How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0 M HNO3? The concentrated acid is 70% HNO3. (1) 45.0 g conc. HNO3 (2) 90.0 g conc. HNO3 (3) 70.0 g conc. HNO3 (4) 54.0 g conc. HNO3 (NEET 2013) 13. 6.02 × 1020 molecules of urea are present in 100 mL of its solution. The concentration of solution is (1) 0.02 M (2) 0.01 M (3) 0.001 M (4) 0.1 M (NEET 2013) 14. Of the following 0.10 m aqueous solutions, which one will exhibit the largest freezing point depression? (1) KCl (2) C6 H12O6

17. Which one is not equal to zero for an ideal solution? (1) ΔSmix (3) Δp = pobserved − pRaoult

(2) ΔVmix (4) ΔHmix (AIPMT 2015)

18. Which of the following statements about the composition of the vapor over an ideal 1:1 molar mixture of benzene and toluene is correct? Assume that the temperature is constant at 25°C. (Given, vapor pressure data at 25°C, benzene = 12.8 kPa, toluene = 3.85 kPa.) (1) The vapor will contain a higher percentage of ­benzene. (2) The vapor will contain a higher percentage of ­toluene. (3) The vapors will contain equal amounts of benzene and toluene. (4) Not enough information is given to make a prediction. (NEET I 2016) 19. At 100°C the vapor pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm. If Kb = 0.52, the boiling point of this solution will be (1) 102°C (2) 103°C (3) 101°C (4) 100°C (NEET I 2016)

(3) Al 2 (SO4 )3 (4) K 2SO4 (AIPMT 2014) 15. Which one of the following electrolytes has the same ­value of van’t Hoff factor (i) as that of the Al2(SO4)3 (if all are 100% ionized)? (1) K3[Fe(CN)6] (2) Al(NO3)3 (3) K4[Fe(CN)6] (4) K2SO4

20. Which one of the following is incorrect for an ideal solution? (1) (2) (3) (4)

ΔUmix = 0 Δp = pobs − pcalculated by Raoult’s law = 0 ΔGmix = 0 ΔHmix = 0 (NEET II 2016)

(AIPMT 2015) 16. The boiling point of 0.2 mol kg−1 solution of X in water is greater than the equimolal solution of Y in water. Which one of the following statements is true in this case? (1) Molecular mass of X is greater than the molecular mass of Y. (2) Molecular mass of X is less than the molecular mass of Y. (3) Y is undergoing dissociation in water while X undergoes no change. (4) X is undergoing dissociation in water.

21. The van’t Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is (1)  1

(2)  2 (3)  3 (4) 0 (NEET II 2016)

22. If molality of the dilute solution is doubled, the value of molal depression constant (Kf ) will be (1) halved. (2) tripled. (3) unchanged. (4) doubled. (NEET 2017)

(AIPMT 2015)

Chapter 9_Solutions.indd 222

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Solutions

223

Answer Key Level I  1. (3)

 2. (3)

 3. (3)

 4. (2)

 5. (3)

 6. (3)

 7. (2)

 8. (4)

 9. (1)

10.  (1)

11.  (3)

12.  (2)

13.  (4)

14.  (1)

15.  (4)

16.  (3)

17.  (3)

18.  (4)

19.  (2)

20.  (3)

21.  (3)

22.  (3)

23.  (3)

24.  (4)

25.  (1)

26.  (4)

27.  (4)

28.  (4)

29.  (3)

30.  (3)

31.  (3)

32.  (2)

33.  (1)

34.  (4)

35.  (2)

36.  (3)

37.  (1)

38.  (2)

39.  (1)

40.  (3)

41.  (3)

42.  (2)

 1. (4)

 2. (3)

 3. (3)

 4. (4)

 5. (1)

 6. (3)

 7. (3)

 8. (4)

 9. (1)

10.  (1)

11.  (1)

12.  (1)

13.  (3)

14.  (3)

15.  (2)

16.  (3)

17.  (3)

18.  (4)

19.  (1)

20.  (2)

21.  (1)

22.  (2)

23.  (1)

24.  (3)

25.  (3)

26.  (4)

27.  (1)

28.  (1)

29.  (3)

30.  (4)

31.  (2)

Level II

Previous Years’ NEET Questions  1. (4)

 2. (2)

 3. (4)

 4. (3)

 5. (4)

 6. (3)

 7. (1)

 8. (1)

 9. (4)

10.  (2)

11.  (None)

12.  (1)

13.  (2)

14.  (3)

15.  (3)

16.  (4)

17.  (1)

18.  (1)

19.  (3)

20.  (3)

21.  (3)

22.  (3)

Hints and Explanations Level I

6. (3) The total vapor pressure of the solution of liquids benzene and toluene is given by

1. (3) Molarity =

Moles of solute Volume of solution in liter

(8/40 ) = 0.2 M 1 nC3 H5 (OH )2 2. (3) We know  xC3 H5 (OH )2 = nH2 O + nC3 H5 (OH )2



       =



xC3 H5 (OH)2 =

( 46/92 ) = 0.2 ( 36/18 ) + ( 46/92 )

4. (2) Molality(m ) =

Moles of solute Mass of solvent in kg





Moles of cane sugar Wt. of solvent (kg)



Let the weight of cane sugar be x





0.2 =

0.2 =

( x /342 ) ⇒ x = 34.2 g (534.2 − x)/1000

5. (3) Relative lowing of vapor pressure = xsolute × i xGlucose ´ i1 = xUrea ´ i2

Chapter 9_Solutions.indd 223

( x/180 ) ´ 1 (1/60) ´ 1 = ( x/180 ) + (100 /18 ) (1/60 + 50/18) x=6g

ptotal = xB pBo + x T pTo







Therefore, mole fraction of toluene in vapor phase

= 0.5 ´ 119 + 0.5 ´ 37 = 78 mm Hg

yT =





7. (2) We know  

pTo x T 37 ´ 0.5 = = 0.237 pTotal 78

p Ao - p A = i ´ xsolute p Ao

 268 − 167   1 × nsolute  = Therefore,  × 1 ⇒ nsolute = 0.605  268   1 + nsolute 

o o 8. (4)    pTotal = ( pChloroform ´ xChloroform ) + ( pacetone ´ xacetone )

1 4 + 0.453 ´ = 0.434 atm 5 5 9. (1) For ideal solution: DH mix = 0 and DVmix = 0

Therefore, pTotal = 0.359 ´

Moles of solute × 1000 Wt. of solvent Moles of solute × 1000 1= Moles of solvent × 18       18 8 nsolute = nsolvent 1000 10. (1) We know Molality =

1/4/2018 5:15:09 PM

224



OBJECTIVE CHEMISTRY FOR NEET

Therefore,   xsolute =

18 1018

po − p 18 Also,   o = p 1018 18 o     p - p = ´ 760 = 13.43 torr 1018 xB = 0, pTotal = p A = 120 torr 11. (3) When

When

xB = 1, pTotal = pB



Therefore, 

pB = 120 - 75 = 45 torr

12. (2) Hydrogen bonding (intermolecular) occurs, thus negative deviation. 13. (4) We know ptotal = xB pBo + x T pTo pTotal =



1 2 ´ 42 + ´ 36 = 38 mm 3 3



Elevation in boiling point is given by DTb = iK bm







Depression in freezing point is given by DTf = iK f m







Dividing Eq. (2) by Eq. (1), we get

14. (1) We have nsolute 0.0125 = xsolute = nsolute + nsolvent ( nsolute /nsolvent ) 0.0125 = ( nsolute /nsolvent ) + 1

(n × 18 × 1000)/(nsolvent × 18 × 1000) 0.0125 = solutee  nsolute × 18 1000  +1 ×  n 1000  solvent × 18

0.0125 =

(m × 18/1000) ⇒ m = 0.7 (m × 18/1000) + 1

p Ao - p A = xsolute ´ i p oA 760 - 750 1 = xsolute ´ 1 Þ xsolute = 760 76

DTf = i ´ 1.86 ´ m (2)

DTf 1.86 = Þ DTf = 1.82 0.5 0.51



Also,   DTf = Tfo - Tf

  1.82 = 0 - Tf Þ Tf = -1.82°C 24. (4) The ionization reaction is



 2 A 3+ 0 2a

A 2B3 At t = 0 1 At t = t 1 − a



o Also, y Benzene pTotal = xBenzene ´ pBenzene

1 ´ 42 7 y Benzene = 3 = 38 19

0.5 = i ´ 0.51 ´ m (1)

van’t Hoff factor can be calculated as i = 1 + 4a

i = 1 + 4 ´ 0.6 = 3.4 Elevation in boiling point is given by DTb = iK bm DTb = 3.4 ´ K b ´ m

Tb − 373 = 3.4 × 0.52 × 1

[Since, at 1 atm Tbo = 373 K ]

Tb = 374.76 K  374.8 K

26. (4) Depression in freezing point is given by DTf = iK f m

or   Tfo - Tf = i ´ K f ´ m



Thus, lower the value of i, higher is Tf , hence C6H12O6 will have highest freezing point.

15. (4) We know

27. (4) We know DTf = iK f m

 

    

16. (3)  

pTotal = p Ao x A(l) + pBo xB(l)

84 = 70 ´ 0.8 + pBo ´ 0.2 Þ pBo = 140 torr        

17. (3) The graph represents the positive deviation. Therefore, A—B interaction is smaller than A—A and B—B interaction. 20. (3) We have y A ´ pTotal = x A(l) ´ p Ao

Therefore, y A =

(0.4 ´ 100) = 0.25 (0.4 ´ 100 + 0.6 ´ 200)

21. (3) Depression in freezing point is given by DTf = iK f m 1.8 = 1 ´ 5 ´ ( 0.48/M ) Þ M = 125.79 (10.6 /1000 ) 23. (3) We know DTb = Tb - T

o b



   = 100.50 - 100 = 0.50°C

Chapter 9_Solutions.indd 224

+ 3B2 − 0 3a

( x /180) DTf = 1´ Þ x = 0.18 g 1 Kf

28. (4)  

DTb = iK bm



0.66 = 1 ´ 3.28 ´

( 3.3 /M ) Þ M = 131.20 (125/78 )

29. (3) As the solution is non-ideal solution, therefore, DVmix ¹ 0. DTf = iK f m

30. (3) We know

    Tf - Tf = i ´ K f ´ m o

    

(0 - Tf ) = 1 ´ 1.86 ´ 0.05 Tf = -0.093°C

31. (3) We know that

    DTb = iK bm



Since m is the same, i in case of BaCl2 will be the highest. Therefore, ∆Tb will be the highest in case of BaCl2.

1/4/2018 5:15:15 PM

Solutions 32. (2) Elevation in boiling point is given by DTb = iK bm



In the given options, i = 5 for Al2(SO4)3.

40. (3) The dimerization reaction is

Tb - Tbo = i ´ K b ´ m Tb - 100 = 2 ´ 0.51 ´ 0.1 = 0.102

2C6 H5COOH  (C6 H5COOH )2

Tb = 100.1°C 33. (1) Kb =

RTb2 M 1000 ´ DH

0.00831 kJ K -1 mol -1 ´ ( 373 K )2 ´ 18 kg mol -1 = 1000 ´ 40.685 kJ mol -1



Complete dimerization involves (C 6 H5COOH)2 only.



Therefore, molecular mass = 244.

41. (3) The reaction is

= 0.512 K kg mol -1

At t = 0 At t = t

36. (3) For solution to be isotonic P1 = P2. Osmotic pressure is given by P = iCRT.

Option (1): i1C1 = i2C 2 (0.1 ´ 2 = 0.2 ´ 1). Therefore, solution is isotonic.



Option (2): i1C1 = i2C 2 (0.1 ´ 1 = 0.1 ´ 1). Therefore, solution is isotonic.



Option (3):  P1 = iC1RT = 5 × 0.05 RT = 0.25 RT



P2 = iC2RT = 3 × 0.1 RT = 0.3 RT



P1 ¹ P2. Therefore, in this case osmosis will take place.



Option (4): i1C1 = i2C 2 ( 3 ´ 10-3 = 2 ´ 1.5 ´ 10-3 ). Therefore, solution is isotonic.

Observed colligative property 7. (1) We know i = 3 Normal colligative property

5.7 = 1.9 3







The dissociation reaction is

38. (2) The dissociation reaction is



1 1−a

0 xa

0 ya

Therefore, i = (1 - a ) + xa + ya

i = 1 + ( x + y - 1)a Þ a =      



K 4[Fe(CN )6 ]  4K + At t = 0 1 0 At t = t 0 4 Therefore, i for K 4[Fe(CN)6 ] = 5

Chapter 9_Solutions.indd 225

0 x n

Therefore,    i = 1 - x +

K 4[Fe(CN)6 ]  4K + At t = 0 1 0 At t = t 0 4





+ [Fe(CN)6 ]4 0 1

Therefore, i for K 4[Fe(CN)6 ] = 5 Fe4[Fe(CN)6 ]3 At t = 0 1 At t = t 0

 4Fe3+ 0

+ 3[Fe(CN)6 ]4 − 0

4



Therefore, i for Fe4[Fe(CN)6 ] = 7



Ratio can be calculated as

iK 4 [Fe(CN)6 ]3 iFe4 [Fe(CN)6 ]3

3

=

5 = 0.71 7

i -1 x + y -1

+ [Fe(CN)6 ]0 1

Wt. of CaCO3 ´ 106 Wt. of H 2O

10 ´ 106 = 104 = 10000 ppm 1000 2. (3) n factor for Na2CO3 is 2. =







We know

N = M × n factor

0.2 = 0.1 M 2 Moles of solute 3. (3) Molarity = Volume of solution in liter (1.2/60 ) = = 0.1 M ( 200/1000) Therefore,  M =

4. (4) We know

39. (1) The reaction is

1− x

A

x n 2. (2) The reaction involved is 4



A x B y  xA + + yB− At t = 0 At t = t



1. (4) Concentration in ppm =

+ Cl 0 a

van’t Hoff factor can be calculated as i = 1 + a Þ a = 0.9



nA 1

Level II

NaCl  Na + 0 At t = 0 1 a At t = t 1 - a



i=

225



p Ao - p A = i ´ xsolute p Ao

æ 20 ö çè ÷ 0.25 - p A 120 ø =1 ´ Þ p A = 0.141 attm 0.25 æ 20 ö æ 10 ö + çè ÷ ç ÷ 120 ø è 46 ø

6. (3) Given that    p = 115x A + 140

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226

OBJECTIVE CHEMISTRY FOR NEET 17. (3) Depression in freezing point is given by

pBo 140 = = 140 xA ® 0 x 1 B



lim

9. (1) We have pTotal = p     

o Benzene

xBenzene + p

DTf = iK f m o Toluene

0.350 = 0.507 xBenzene + 0.184(1 - xBenzene )

Loss in weight of solvent µ p Ao - p A 2.5 µ p A





Therefore, B is in associated state.

18. (4) The ionization of electrolyte is AB  A+ 20 At t = 0 0 100 0.5a At t = t 0.5(1 - a )

0.04 µ p Ao - p A p Ao - p A 0.04 5/ M = = Þ M = 31.25 g p Ao 2.5 + 0.04 æ 5 180 ö + çè ÷ M 18 ø



11. (1) We know o o pTotal = pethylalcohol xethylalcohol + pmethyl alcohol x methyl alcohol

pTotal

  

(60/46 ) ( 40/32) = 45 ´ + 90 ´ (60/46 ) + ( 40/32) (60/46 ) + ( 40/32) = 70 mm

12. (1) p Ao = 33 + 94 = 127 mm of Hg



xBenzene = 0.514

10. (1) Loss in weight of solution µ p A

( DTf )B iB = ( DTf )A i A 1 iB = Þ iB = 0.5 or iB < 1 2 1

(1 - xBenzene )



pBo = 94 mm of Hg and pCo = 46 mm of Hg pTotal = p Ao x A + pBo xB + pCo xC 2 3 4 + 94 ´ + 46 ´ 9 9 9 = 80 mm of Hg = 127 ´





    p Ao =







     pBo =



Since, p Ao > pBo



Therefore,

y A ´ pTotal (1) xA

y B ´ pTotal = pBo ´ xB y B ´ pTotal (2) xB

y A yB y x > or A > A x A xB y B xB

15. (2) We know

p - pA = xSolute p Ao o A

p Ao - 0.95p Ao xB/M B = o x x pA B + A MB M A     xB/M B x 0.05 = Þ A = 5.7 xB xA xB + M B 0.3M B po - p 16. (3) We have A o A = xSolute pA 10 10    o = 0.2 Þ p Ao = (1) pA 0.2 20 20 = xsolute Þ ´ 0.2 = xsolute Þ xsolute = 0.4 p Ao 10

Therefore, xsolvent = 0.6

Chapter 9_Solutions.indd 226

0 0.5a

0.5(1 + a ) i= = 1+a 0.5



  



Depression in freezing point is given by DTf = iK f m 0.74 = (1 + a ) × 1.86 ×



( 20/100 ) (500/1000 )

a = 0.0053 ≈ 0 19. (1) Elevation in boiling point is given by

DTb = iK bm or Tb - Tbo = iK bm (1)



xP  Px At t = 0 1 0 1 At t = t 0 x



14. (3) We know y A ´ pTotal = p Ao ´ x A

B-

+

1 x



van’t Hoff factor i =



Substituting the values in Eq. (1), we get



( 47.9 - 46.3) =

1 ( 28/31x ) ´ 2.34 ´ Þx=4 x ( 315/1000 )

20. (2) We have    

i1 ( DTf )1 = i2 ( DTf )2



1 x = Þ i2 = 3 i2 3 x AB2 At t = 0 1 At t = t 0







A 2+ 0 1

+ 2B − 0 2

For AB2 salt, van’t Hoff factor is 3.

21. (1) The reaction is HgI 2 ¯ + 2I - ® [HgI 4 ]2( ppt. )



Soluble

2-

[HgI 4 ] is consumed, therefore molality decreases and DTf decreases or Tf increases.

23. (1) Osmotic pressure can be calculated as P = iCRT 20.4 = i ´ C ´ R ´ T ( x / 60) 20.4 = 1 ´ ´ 0.0821 ´ 298 Þ x = 5 g 100 ( /1000)

1/4/2018 5:15:25 PM

Solutions 24. (3) Osmotic pressure can be calculated as P = iCRT .

i = 1 for all cases, therefore, P µ C .



C1 =



Therefore, C 2 > C1 > C 3 or p2 > p1 > p3

10 10 10 ; C2 = ; C3 = 180 60 342



Total number of moles at equilibrium = 0.8 + 0.2 + 0.2 = 1.2 = i



Therefore, depression in freezing point is



DTf = iK f m = 1.2 ´ 1.86 ´ 0.5 = 1.116 K » 1.12 K

2. (2) We know that DTb = iK f ´ m

27. (1) We know P = iCRT (5 /342)      P = 1 ´ (100/1000) ´ 0.0821 ´ 288 = 3.46 atm 28. (1) The dissociation reaction is +

+



KCl  K At t = 0 0.005 0 At t = t 0.005(1 - a ) 0.005a



van’t Hoff factor can be calculated as

-

Cl 0 0.005a

0.005(1 + a ) = 1+a 0.005 1.95 = 1 + a Þ a = 0.95 i=



29. (3) The reactions are 2A  At t = 0 1



At t = t

 



0

1 or 0.5 2 3A  At t = 0 1

A2 0 1 2

 

0

At t = 0 At t = t

Therefore, number of ions per mole of ionic components = 2.

3. (4) The depression in freezing point can be calculated as DTf = K f ´ m



A3 0 1 3

 Ca 2+ 0 a

4A  1 0

Therefore,      DTf =



The freezing point of the solution is Tf = 0 - 0.372 = - 0.372°C

Molality ´ Mol. wt. of H 2O w H2 O

1 ´ 18 1000 = 0.0177 =



5. (4) Let the dissociation reaction be

+ 2NO30 2a

A4 0 1 4

van’t Hoff factor i = 0.25

1. (4) The reaction involved is



Chapter 9_Solutions.indd 227

1.86 ´ 68.5 = 0.372°C 342 ´ 1



xsolute =

Previous Years’ NEET Questions





+   [Co(NH 3 )5 NO2 ]Cl    [Co(NH 3 )5 NO 2 ] + Cl



+   HA    H +A Initial conc. 1 0 0 Final conc. 1 -a a a



van’t Hoff factor i = (1 - a + a + a ) = (1 + a ) Þ i > 1



Let the association reaction be

    i =







4. (3) Mole fraction can be calculated as

(1 + 2a ) 164 = Þ a = 0.75 1 65.4 1. (2) The reaction is 3

In the aqueous solution, the compound dissociates as

i=

Ca(NO3 )2 At t = 0 1 At t = t 1-a







1 or 0.33 3 0. (4) The ionization reaction is 3



0.00732 = i ´ 1.86 ´ 0.002 Þ i = 1.967 » 2

i=

At t = t

227

+   HX    H +X Initial moles 1 0 0 Moles at equilibrium 1 - 0.2 0.2 0.2

Initial conc.

Final conc.

  2HA    (HA)2 1 0 a 1- a 2

van’t Hoff factor i = 1 - a +

a a =1+ Þi ( DTb )Y , thus, iX > iY, indicates X undergoes dissociation in water. 17. (1) For an ideal solutions, ∆Smix ≠ 0. On mixing of solute and solvent to form an ideal solution, ∆S remains non-zero. 18. (1) By Raoult’s Law o pT = pCo6 H6 xC6 H6 + pToluene x Toluene 







o = 3.85 kPa Given, pCo6 H6 = 12.8 kPa; pToluene

(1)

Since, it is a 1:1 molar mixture of benzene and toluene, thus xC6 H6 = 0.5; x Toluene = 0.5

nMoles of HNO3 = M ´ V 250 ´ 2 = 0.5 mol 1000

=

Molar concentration ( M ) =

12. (1) We know

=

number of molecules NA



o o 9. (4) According to Raoult’s law, p A = p A x A and pB = pB xB .



Number of moles =



Substituting these values in Eq. (1), we get pT = (12.8 ´ 0.5) + ( 3.85 ´ 0.5) = 6.2 + 1.925 = 8.125 kPa

1/4/2018 5:15:31 PM

760 - 732 w B ´ MA = 760 M B ´ wA 28 6.5 ´ 18 = Solutions 760 M B ´ 100 M B = 31.75 g mol DTb = K b ´ m

Mole fraction of benzene in vapor form y C 6 H6 =



pCo6 H6 xC6 H6 pT

=

12.8 ´ 0.5 = 0.75 8.125

Mole fraction of toluene in vapor form y toluene = 1 - y C6 H6 = 1 - 0.75 = 0.25



The component having higher vapor pressure will have higher percentage in vapor phase.

19. (3) Let A be solvent and B be solute.

p o - ps nB = po nA

At 100°C, p° = 760 mm 760 - 732 w B ´ MA = 760 M B ´ wA 28 6.5 ´ 18 = 760 M B ´ 100

M B = 31.75 g mol -1 DTb = K b ´ m Molality, m =

Chapter 9_Solutions.indd 229

Molality, m = Therefore, DTb = 0.52 ´

229

-1

w B ´ 1000 MB ´ wA

6.5 ´ 1000 = 1.06 31.75 ´ 100

Boiling point = 100 + 1.06 = 101.06°C. 20. (3) For an ideal solution; ∆Hmix = 0; ∆Umix = 0 and ∆Smix ¹ 0

We know that, ∆G = ∆Hmix − T∆Smix ⇒ ∆G ¹ 0

21. (3) Ba(OH)2 is strong electrolyte, so 100% dissociation occurs in solution. Ba(OH)2 → Ba2+ + 2OH−(aq)







The number of ions formed is 3, therefore, van’t Hoff factor, i = 3.

22. (3) Freezing point depression is given by ∆Tf = Kfm, where Kf is constant characteristic of the solvent. The value of Kf for a given solvent corresponds to the number of degrees of freezing point lowering for each molal unit of concentration. Hence, it is independent of molality.

w B ´ 1000 MB ´ wA

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10

Electrochemistry

Chapter at a Glance 1. Electrochemistry is the branch of chemistry dealing with production of electricity from energy released from spontaneous chemical reactions which in turn can be used to bring about non-spontaneous chemical changes. 2. Electrochemical Cells The devices which convert electrical energy into chemical energy or vice versa are called electrochemical cells. These devices are classified into two main categories: (a) Electrolytic cells: The devices in which chemical changes occur in the presence of applied electrical energy. (b) Galvanic (or voltaic) cells: The devices in which electrical energy is generated on account of the chemical reactions occurring in them. These are further divided into chemical cells and concentration cells. 3. Galvanic Cells (a)  A galvanic cell consists of two electrodes immersed in solutions of their respective salts. (b)  The electrode at which reduction (or electron gain) takes place is known as the cathode and the electrode at which oxidation (loss of electrons) takes place is known as the anode. (c)  A commonly used galvanic cell is Daniell cell. It is built by two electrodes, Zn rod dipped in ZnSO4 solution and Cu rod dipped in CuSO4 solution, connected by a salt bridge.   The half-cell reactions are: Zn(s)  Zn 2+ (aq) + 2e Cu 2+ (aq) + 2e -  Cu(s)   The overall cell reaction is

Zn(s) + Cu 2+ (aq) ® Zn 2+ (aq ) + Cu(s )

(d) Salt bridge: A salt bridge is usually used to connect the electrodes of a galvanic cell and consists of a gel and a suitable electrolyte drawn into a clean U-tube. It is useful in the following ways: (i) It brings about internal contact between the electrodes. (ii) It minimizes the liquid junction potential. (iii) It minimizes polarization. 4. Electrode Potential It is the potential difference developing between the electrode and electrolyte and is measured by voltmeter. The electrode potential which is determined when the concentration of all solutes is unity (1 M) at standard conditions of temperature (298 K) and pressure (1.00 bar) is known as standard electrode potential. 5. Potential of a Cell or EMF (a) Cell potential: It is the maximum potential that a given cell can generate and is represented by Ecell. (b) Standard cell potential: When the system is at standard state, the potential of a galvanic cell is the standard cell o potential, symbolized by E cell . (c) Electromotive force (emf or EMF): It is the potential difference between the two terminals of the cell when no current is drawn from it. It is measured by using potentiometer. 6. IUPAC Cell Notation

(a) The interface across which a potential develops is denoted by either a single vertical line (|) or a semicolon (;). Cu CuSO4 (1 M) or Cu ;CuSO4 (1 M)

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(b) The salt bridge used to connect the electrodes is represented by double vertical lines. Anode; Anode electrolyte (conc.) || Cathode; Cathode electrolyte (conc.); Cu CuSO4 (1 M)

AgNO3 (1 M) Ag

(c)  The emf of a cell is represented as E cell = E right - E left , where E right and E left are the standard reduction potentials of RHS and LHS electrodes, respectively. (d)  The electrode which has higher reduction potential is cathode and one which has lower electrode potential is anode. Therefore, for the cell, Salt bridge Anode Cu(s) Cu2 + (aq) Anode electrode

Anode electrolyte

Anode half-cell

Cathode Ag+ (aq) Ag(s) Cathode electrolyte

Cathode electrode

Cathode half-cell

o o o o o E cell = E cathode - E anode = E Ag - ECu + 2+ /Ag / Cu

7. Measurement of Reduction Potential Reference electrodes: The potential of an electrode can be measured if the electrode is connected to another electrode of known potential. Such an electrode is termed as reference electrode. (a) Standard hydrogen electrode (SHE/NHE) is one in which pressure of hydrogen gas is maintained at 1 atm and the concentration of hydrogen ions in solutions is 1 M. Pt, H2 (1 atm ) (1.0 M) HCl If reduction occurs at the electrode, the reaction taking place will be as follows: 2H+ (aq ) + 2e -  H2 (g ) o The emf of the cell (E cell ) is determined at a standard condition and since E o of SHE is assumed to be zero, the potential of the second electrode can be obtained. (b)  Calomel electrode and silver–silver chloride electrodes are commonly used secondary reference electrodes. They are standardized using standard hydrogen electrode (SHE) and then used as reference electrodes for measuring potential of other electrodes.

8. Electrochemical Series and their Applications (a)  Electrochemical series is made up of the standard reduction potentials of half-reactions arranged in decreasing order – the half-reactions at the top have the greatest tendency to occur as reduction, while those at the bottom have the least tendency to occur as reduction. (b) Applications: The series are used to (i) Identify the oxidizing and reducing agents. (ii) Calculate the emf of the cell using the reaction. (iii) Compare the reactivities of metals. Metal with lower reduction potential can displace the metal with higher reduction potential. (iv) Predict if a metal can displace hydrogen to produce H2 gas. Metals with lower reduction potential than hydrogen can displace H2 gas from its acid. (v)  Predict spontaneity of a reaction. Reactions for which E cell is negative or zero are not feasible. For a spontaneous reaction, the cell potential is always positive.

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Electrochemistry

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9. Nernst Equation It defines the relationship of cell potential with ion concentration. For the reaction Mn + + ne - ® M , the Nernst equation is RT [ M] o log      E = E - 2.303 nF [ M]n+ 10. Applications of Nernst Equation

(a) Equilibrium constant of electrochemical cell The emf of a galvanic cell is given by a general expression: o E cell = E cell -

where Q is the reaction quotient. For a reaction:

2.303 RT nF

log Q

M(s) + N n + (aq )  Mn + (aq ) + N(s )

[ Mn+ ][ N] [ Mn+ ] = [ M][ N n+ ] [ N n+ ] 2.303 RT [ Mn + ] o log n + The Nernst equation is    E cell = E cell nF [N ] At equilibrium, the reaction quotient Q becomes equal to the equilibrium constant Keq. Q =

2.303RT log K eq nF (b) Gibb’s energy of the reactions: Relation between electrochemical cell and Gibbs energy of reaction: o E cell =

 DG o = -nFE o and DG o = - RT ln K 11. Concentration Cells A concentration cell is a galvanic cell which generates electrical energy at the expense of chemical energy. These cells are classified into two types: (a) Electrode concentration cells: In these cells, the concentration of the electrolyte is the same. The two electrodes contain the same substance but with different concentration. For example, two hydrogen electrodes having hydrogen gas at different pressures, dipped in the same electrolytic solution. The cell can be represented as Pt, H2 ( p1 ) HCl(x M) H2 ( p2 ), Pt where p1 and p2 are the pressures at which H2 gas is maintained in the cell. (b) Electrolyte concentration cells: In these cells, the electrodes contain different concentration of the same electrolyte and are connected directly through a diffusion membrane (no salt bridge is required). Pt, H2 (1 atm ); HCl(a1 ) HCl(a2 ); H2 (1 atm ), Pt 12. Conductance of Electrolytic Solutions (a) Electrical resistance: Electrolytic conductors also offer some resistance to the flow of current. The resistance R (unit is ohm) offered by is given as l R = r. A where l is length, A is area of cross-section of the conductor, and r is the proportionality constant called resistivity or specific resistivity. It is the resistance offered by a conductor of 1 m length and 1 m2 area of cross-section. In SI system, it has unit of Ω m (ohm-meter). l A (b) Electrical conductance: The reciprocal of resistance is known as conductance and is represented by G. R = r.

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G =

1 A = R lr

The SI unit of conductance is Siemens, represented by the symbol S and is equal to ohm−1(Ω−1), also called mho. (i) The reciprocal of specific resistance is called the specific conductance (k ) or conductivity.    Therefore, 1 1 l k = = ´ A r R The ratio of length (l ) to the cross-sectional area (A) is called the cell constant (G*) (c) Molar conductance: It is the conductance of an electrolytic solution due to all the ions obtained from 1 mol of an electrolyte at a given concentration. It is denoted by the symbol Λ m . k Λm = C where k is the conductivity of the solution and C is the molar concentration of the solution. (i) If k is expressed in Sm-1 or Scm-1, the unit of molar conductance will be: Λ m (Sm 2 mol -1 ) = Λ m (Scm 2 mol -1 ) =

k (Sm -1 ) 1000 (L m -3 ) × C (mol L-1 ) k (Scm -1 ) × 1000 (cm3 L-1 ) Molarity (mol L-1 )

(ii) As the concentration approaches zero, the molar conductance of the solution reaches the maximum value. This value is called limiting molar conductivity and is denoted by L0m. (d) Equivalent conductance: It is the conductance of an electrolytic solution due to all the ions obtained from 1 gram-equivalent (g-equiv.) mass of the electrolyte at a given concentration. Therefore, L = kV where k is the conductivity of the solution and V is the volume of the solution containing 1 g-equiv. of the substance. In terms of concentration, k L= 1000 ´ C Equivalent conductance has unit of S m2g-equiv−1. 13. Variation in Molar Conductivity with Concentration (a) Strong electrolytes (i)  The variation in molar conductivity with concentration for strong electrolytes is given by an empirical relation proposed by Kohlrausch, L m = L0m - k C where k is a constant. (ii) Kohlrausch’s law of independent migration of ions: It states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte. Mathematically, it can be stated as L0m = v+ l 0+ + v- l 0where L0m, l+0 and l-0 are limiting molar conductivities of the electrolyte, cations and anions respectively and v+ and v - are the number of cations and anions. (b) Weak electrolytes The degree of dissociation is given by Number of moles of electrolyte dissociated a = Initial number of moless of electrolyte Λ The degree of dissociation in terms of molar conductivity is a = 0m Λm

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Electrochemistry

235

(i) For a weak electrolyte, the dissociation constant at concentration C is given by K = 14. Laws of Electrolysis

(C a )(C a ) C a 2 = 1-a C (1 - a )

(a) Faraday first law: The amount of substance deposited/dissolved liberated is directly proportional to the amount of charge (coulomb) allowed to flow through the electrolytic conductor, that is, W µQ Here W = Amount of substance deposited/dissolved/liberated (gases); Q = amount of charge flow through conductor (coulomb) W = ZQ Equivalent weight Here Z = Electrochemical equivalent = 96500 Also Q = It Here I = Current in ampere, t = time in seconds

MA ´ I ´ t Therefore, W = n ´ 96500 MA ´ Q     W = n ´ 96500 Here, MA = molar mass of substance, n = number of electrons participating in the reaction. (b) Faraday second law: When the same amount of charge is allowed to flow through two electrolytic conductors, the amount of substances deposited/dissolved/liberated are in the ratio of their equivalent weight. Mathematically, second law can be written as W1 Z E = 1 = 1 (Q is the same) W2 Z 2 E 2 Here, E1 and E2 are equivalent weights of substances 1 and 2. 15. Batteries It is a combination of cells, either in series or parallel, or both, in order to generate required amount of electrical energy. Its components are anode, cathode, electrolyte and separator. (a) Primary cells: These undergo redox reaction only once and cannot be reversed. (i) Dry cell (Leclanche cell) • Anode: Zinc rod • Anode electrolyte: 1M ZnCl2 • Cathode: Graphite • Cathode electrolyte: Paste of MnO2 and NH4Cl 2MnO2 + 2NH+4 + 2e - ® Mn 2 O3 + 2 NH3 + H2 O



At the cathode:



2+ At the anode:        Zn → Zn + 2e 2MnO2 + 2 NH4 Cl + Zn → Mn 2 O3 + Zn(NH3 )2 Cl 2 + H2 O Overall reaction:  

E cell = 1.5 V (maximum)



(ii) Mercury cell (Ruben–Mallory cell) • Anode: Zn • Anode electrolyte: KOH • Cathode: Graphite • Cathode electrolyte: HgO At the cathode:   HgO + H2 O + 2e ® Hg + 2OH At the anode:    Zn + 2OH- ® ZnO + H2O + 2e-

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Overall reaction:

HgO + Zn ® Hg + ZnO E cell = 1.35 V (maximum)

(b) Secondary cell: These can be recharged by passing current through them in the opposite direction. (i) Lead storage battery • Anode: Pb • Anode and cathode electrolyte: 38% H2SO4 (density 1.38 g/cc) • Cathode: PbO2 + 2At the cathode:   PbO2 + 4H + SO4 + 2e ® PbSO 4 + 2H2 O

At the anode:

2     Pb + SO4 ® PbSO 4 + 2e

+ 2Overall cell reaction: Pb + PbO2 + 4H + 2SO4 ® 2PbSO4 + 2H2 O (ii) Nickel-Cd battery • Anode: Cadmium • Anode electrolyte: Cd(OH)2 • Cathode: Nickel (NiO) • Cathode electrolyte: Ni(OH)2

At the cathode:   2NiO(OH) + 2H2O + 2e - ® 2Ni(OH)2 + 2OHAt the anode:     Cd + 2OH ® Cd(OH)2 + 2e Overall cell reaction: Cd + 2NiO(OH) + 2OH ® Cd(OH)2 + 2Ni(OH)2 E cell = 1.4 V (maximum)

16. Fuel Cells (a)  These are devices used for converting the chemical energy of fuel directly into electrical energy through catalytically activated redox reactions. A fuel cell can be represented as Fuel Electrode Electrolyte Electrode Oxidant   The chemical reactions involved at the electrodes are: Fuel ® Oxidation product + ne At the anode: At the cathode:     Oxidant + ne ® Reduction product Overall cell reaction: Fuel + Oxidant ® Oxidation product + Reduction product (b) Hydrogen-oxygen fuel cell: This fuel cell combines hydrogen and oxygen to produce electricity, heat and water. As long as hydrogen is supplied, it continues to produce electricity without being discharged. The chemical reactions involved at the electrodes are: + At the anode:   2H2 ® 4H + 4e + At the cathode: O2 + 4H + 4e ® 2H2 O Overall reaction:   2H2 + O2 ® 2H2 O

17. The thermodynamic efficiency (h) of a fuel cell is given by the ratio of electrical work done to the enthalpy change associated with the reaction. Mathematically, nFE DG Efficiency of cell = = DH DH

Solved Examples 1. The standard reduction potentials for Fe2+ Fe and Sn 2+ Sn are –0.44 V and −0.14 V respectively. The standard emf for cell Fe2+ + Sn ® Sn 2+ + Fe is

Chapter 10_Electrochemistry.indd 236

(1) +0.30 V (2) –0.58 V (3) +0.58 V (4) –0.30 V

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Electrochemistry Solution

(1)

(4) We have Sn → Sn 2+ + 2e - ; E o = + 0.14 V Fe

2+

-

+ 2e → Fe; E = - 0.44 V o

Sn + Fe2+ → Sn 2+ + Fe ; E o = - 0.3 V 2. The logarithm of the equilibrium constant, log K, of the cell reaction of the cell Zn(s) Zn 2+ Cu 2+ Cu(s) DE o = 1.1 V is 1.1 1.1 (1) (2) 0.0291 0.0291 0.0291 1.1 (3) (4) 1.1 0.0592 Solution (2) The reaction involved is

Zn(s) + Cu2+ (aq) → Cu(s) + Zn2+ (aq)



o From Nernst equation,   E = E cell



At equilibrium, E = 0, so the equation becomes

2.303 RT log K nF

2.303 RT log K nF 2.303 × 8.314 × 298 1.1 = log K 2 × 96500

o E cell =-

1.1 = - 0.02956 log K ⇒ log K = -

Given

CuI + e - → Cu + I - ; E o = - 0.17 V





Cu + + e - → Cu; E o = -0.53 V (1) −67.55 (2) −35.0 (3) −52.63 (4) −17.5

Solution (2) For reaction  CuI + e - ® Cu + I DG1o = - 1 × F × ( -0.17 V) = 0.17 V







For reaction  Cu + + e - ® Cu







Standard free energy change for reaction Cu + + I - ® CuI can be calculated as

 

 

We know 1F = 96500 J DG o = -0.36 ´ 96500 J = - 34740 J or - 34.74 kJ  - 35.0 kJ

o E Cu 2+ /Cu +

o o o E cell = E Cu + E Cu + + /Cu 2 + /Cu

= - 0.153 + 0.53 = 0.377 V 5.  In electrolysis of NaCl when Pt electrode is taken then H2 is liberated at cathode while with Hg cathode it forms sodium o = - 2.71 V, E Ho + /H = 0 V ) amalgam because ( E Na + /Na 2 (1) Hg is more inert than Pt. (2) more voltage is required to reduce H+ at Hg than at Pt. (3) Na is dissolved in Hg while it does not dissolve in Pt. (4) concentration of H+ ions is larger when Pt electrode is taken. Solution (2) When electrolysis of aqueous solution of NaCl is carried out, the products obtained are NaOH, Cl2 and H2. This is because H+ and OH− ions obtained from electrolysis of water (solvent) molecules are also present along with Na+ and Cl− ions. The two competing reactions at the cathode are: o Na + (aq ) + e - → Na(s); E cell = -2.71 V 1 + o H (aq ) + e → H 2(g ); E cell = 0.0 V 2



The reaction with higher value of E o is preferred, and therefore, the reaction at the cathode during electrolysis is H + (aq ) + e - ®





1 H 2 (g ) 2

At the anode, the following reaction takes place 1 Cl - (aq ) ® Cl 2(g ) + e 2 If Hg electrodes are used during the electrolysis of aq. NaCl, the products we get at anode and cathode are different.



At anode: Cl− oxidizes to give Cl2.



At cathode: Hg selectively increases the overvoltage of H2O giving H2. Thus we get the reduction of Na+ to give Na at cathode.



This Na combines with Hg to form Na(Hg) amalgam which can be easily separated.

DG 2o = -1 ´ F ´ (0.53 V)= - 0.53 V

DG o = DG 2o - DG1o = -0.53 F - ( -0.17 F ) = -0.36 F

(4) The reaction involved is 2Cu+ → Cu2+ + Cu

1.1 0.02956



+ 0.683 V (2) – 0.367 V (3) + 0.3415 V (4) + 0.367 V

Solution



3. Calculate the standard free energy change in kJ for the reaction: Cu + + I - ® CuI

237

6. A solution containing one mole per liter of each Cu(NO3)2; AgNO3; Hg2(NO3)2; Mg(NO3)2 is being electrolyzed using inert electrodes. The values of standard electrode potentials (reduction potentials) in volts are



Ag/Ag+ = 0.80 V, 2Hg/Hg 22+ = 0.79 V

Cu/Cu 2+ = + 0.24 V , Mg/Mg 2+ = - 2.37 V 4. Cu+ is not stable and undergoes disproportionation. o o E° for Cu+ disproportionation is (E Cu = + 0.153 V , E Cu = + 0.53 V 2+ + /Cu + /Cu With increasing voltage, the sequence of deposition of o = + 0.153 V , E Cu +/Cu = + 0.53 V ) metals on the cathode will be

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238

OBJECTIVE CHEMISTRY FOR NEET (1) Ag, Hg, Cu. (2) Cu, Hg, Ag. (3) Ag, Hg, Cu, Mg. (4) Mg, Cu, Hg, Ag.

Solution (1) Greater the value of standard reduction potential, greater will be its tendency to undergo reduction. So the sequence of deposition of metals on cathode will be Ag, Hg, Cu. Here, magnesium will not be deposited because its standard reduction potential is negative. So it has strong tendency to undergo oxidation. Therefore, on electrolysis of Mg(NO3)2 solution, H2 gas will be evolved at cathode. 7. Calculate the emf of the electrode concentration cell





at 25°C, if the concentration of the zinc amalgam are 2 g per 100 g of mercury and 1 g per 100 g of mercury in anode and cathode half-cell respectively.

Hg-Zn (C1 M) | Zn (C M) | Hg-Zn (C2 M) 2+

(1) 6.8 × 10−2 V (2) 8.8 × 10−3 V (3) 5.7 × 10−2 V (4) 7.8 × 10−3 V Solution (2) The net cell reaction is

W =

9. The emf of a Daniell cell at 298 K is E1.



Since, it is a concentration cell, therefore, E



From Nernst equation, we have

Zn ZnSO 4(0.01 M) CuSO4 (1.0 M) Cu When the concentration of ZnSO4 is 1.0 M and that of CuSO4 is 0.01 M, the emf is changed to E2. What is the relationship between E1 and E2? (1) E1 > E 2 ( 3 ) E1 = E2

Zn(s) + Cu 2+(aq ) ® Cu(s) + Zn 2+ (aq ) 0.059 [ Zn 2+ ] log n [Cu 2+ ] 0.059 0.01 o E1 = E cell log 2 1



o From Nernst equation  E cell = E cell -







When the concentration of Zn2+ is 1.0 M and ­con­cen­­t­ration of Cu2+ is 0.01 M

o E1 = E cell + 0.059

0.059 C log 2 2 C1 0.059 C1 0.059 ( 2 / 65.4) log log = = (1/ 65.4) 2 C2 2 -3

= 8.8 × 10 V 8. Copper sulphate solution is electrolyzed between two platinum electrodes. A current is passed until 1.6 g of oxygen is liberated at anode. The amount of copper deposited at the cathode during the same period is

0.059 [ Zn 2+ ] log n [Cu 2+ ] 0.059 1 o E 2 = E cell log 2 0.01

o - 0.059 (2)     E 2 = E cell



From Eq. (1) and Eq. (2), E1 > E2

10. What is the volume of gas, measured at STP, liberated at anode from the electrolysis of Na2SO4 solution by passage a current of 5.0 A passed for 3 min 13 s? (1) 56 mL (2) 112 mL (3) 224 mL (4) None of these Solution (1) Reaction that takes place at anode is 4O2- ® 2O2 + 4e -

(1) 6.36 g (2) 63.6 g (3) 12.7 g (4) 3.2 g



Q = It

We know

= (5 A) ´ (193 s) = 965 C

Solution (1) From the Faraday’s first law, we have Q =



W ×n×F M

where Q is the amount of charge flowing through conductor, W is the amount of substance deposited/liberated, n is number of electrons participating in the reaction and M is the molar mass. Anode reaction: 4O2- ® 2O2 + 4e Q=

1.6 ´ 4 ´ F = 0.2 F 32 2+



Cathode reaction: Cu + 2e ® Cu



Amount of copper deposited can be calculated as

Since, 96500 C = 1F, therefore, Q=





32 ´ 965 M ´Q = = 0.08 g 4 ´ 96500 n´F

Amount of O2 liberated in moles n=



965 C ´ 1 F = 0.01 F 96500 C

Amount of O2 liberated in grams W =

-



Chapter 10_Electrochemistry.indd 238

(1)

o E 2 = E cell -

= 0.

E = Eo -

(2) E1 < E 2 ( 4 ) E2 = 0 ¹ E1

Solution (1) The cell reaction involved is

Zn(C1 ) + Zn 2+ (C )  Zn 2+ (C ) + Zn(C 2 ) o cell

M × Q 63.5 × 0.2 F = = 6.35 g 2 ×F n×F

0.08 g = 0.0025 mol 32 g mol -1

Volume of O2 at STP = 22400 ´ 0.0025 = 56 mL

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Electrochemistry 11. A vanadium electrode is oxidized electrically. If the mass of the electrode decreases by 114 mg during the passage of 650 C, what is the oxidation state of the vanadium product? (1) +1 (2) +2 (3) +3 (4) +4 Solution Q´M (3) We have n = W ´F (650 C) × (51 g mol -1 ) = =3 (114 × 10 -3 g ) × (96500 C mol -1 )



Since, 96500 C = 1F, therefore,



Q=

The reaction at cathode is 2H 2O + 2e - ® H 2 + 2OH -



From the reaction, we know 1F = 1 mol of OH -



Thus, 9.99 × 10 -4 F = 9.99 × 10 -4 mol of OH -



Therefore, concentration of OH− in solution after electrolysis is 9.99 ´ 10 -4 mol = 0.0199 M æ 50 ö çè ÷ø L 1000

15. One coulomb of charge passes through solution of AgNO3 and CuSO4 connected in series and the concentration of two solutions being in the ratio 1:2. The ratio of weight of Ag and Cu deposited on Pt electrode is

(1) 12.7 g (2) 15.9 g (3) 31.8 g (4) 63.5 g Solution (2) From Faraday’s second law, we have Eq. wt.of H 2 Wt.of H 2 = Eq. wt.of Cu Wt.of Cu (1/1) 0.504 = (63.5 / 2) WCu WCu = 16 g 13. The same amount of electricity was passed through molten cryolite containing Al2O3 and NaCl. If 1.8 g of Al was liberated in one cell, the amount of Na liberated in the other cell is (1) 4.6 g (2) 2.3 g (3) 6.4 g (4) 3.2 g Solution (1) From Faraday’s second law, we have Eq. wt.of Al wt.of Al = Eq. wt.of Na wt.of Na ( 27 / 3) 1.8 = ( 23 /1) WNa WNa =

1.8 × 3 × 23 = 4.6 g 27

14. A current of 0.20 A is passed for 482 s through 50.0 mL of 0.100 M NaCl. What will be the hydroxide ion concentration in the solution after the electrolysis? (1) 0.0159 M (2) 0.0199 M (3) 0.10 M (4) 0.030 M Solution (2) We know

(1) 107.9 : 63.54 (2) 54 : 31.77 (3) 107.9 : 31.77 (4) 54 : 63.54 Solution (3) From Faraday’s second law, we have

Eq. wt. of Ag wt. of Ag = Eq. wt. of Cu wt. of Cu





For

Ag + + e - → Ag ; E Ag =





For

Cu 2+ + 2e - ® Cu ; E Cu



Therefore, from Eq. (1), we get

(1)

107.9 1 63.54 = 2

wt. of Ag (107.9/1) 107.9 = = wt. of Cu (63.54/2) 31.77



16. When electricity is passed through a solution of AlCl3, 13.5 g of Al is discharged. The amount of charge passed is (1) 1.5 F (2) 0.5 F (3) 1.0 F (4) 2.0 F Solution (1) We have

Q=

W × n × F 13.5 × 3 × F = = 1.5 F M 27

17. If the specific resistance of a solution of concentration C g equiv L−1 is R, then its equivalent conductance is (1)

100 R RC (2) C 1000

(3)

1000 C (4) RC 1000 R

Solution Q = It          = (0.20 A) × (482 s) = 96.4 C

Chapter 10_Electrochemistry.indd 239

96.4 C = 9.99 × 10 -4 F 96500 C



As three electrons are involved in the oxidation process, therefore, oxidation state of vanadium is +3.

12. A certain current liberated 0.504 g of hydrogen in 2 h. How many grams of copper can be liberated by the same current flowing for the same time in CuSO4 solution?

239

(3) The reciprocal of specific resistance is the specific con1 ductance, that is, k = r

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OBJECTIVE CHEMISTRY FOR NEET



Equivalent conductance can be expressed as k ´ 1000 1000 L= = C RC 18. The equivalent conductivity of 0.1 N CH3COOH at 25°C is 80 and at infinite dilution 400 Ω−1. The degree of dissociation of CH3COOH is (1) 1 (2) 0.2 (3) 0.1 (4) 0.5



a=

a=

        =

L L¥ 80 = 0.2 400

19. The equivalent conductance of monobasic acid at infinite dilution is 438 W -1 cm 2 eq -1. If the resistivity of the solution containing 15 g of acid (Molecular weight = 49) in 1 L is 18.5 Ω cm. What is the degree of dissociation of acid? (1) 45.9% (2) 40.2% (3) 60.4% (4) 50.7% Solution

1 Specific conductance (k ) = (4) We know r k × 1000 Equivalent conductance is given by Λ Ε = molality L=

1 1000 ´ = 176.57 18.5 15 / 49

176.57 L = = 0.507 or 50.7% L¥ 348

20. The electrical resistance of a column of 0.05 mol L−1 NaOH solution of diameter 1 cm and length 50 cm is 5.55 × 103 Ω. Its resistivity, conductivity and molar conductivity is (1) 87.135 Ω, 0.01148 S cm−1, 229.6 S cm2 mol−1 (2) 77.123 Ω, 0.0241 S cm−1, 119.5 S cm2 mol−1 (3) 92.29 Ω, 0.0124 S cm−1, 272.3 S cm2 mol−1 (4) 83.92 Ω, 0.01192 S cm−1, 220.3 S cm2 mol−1

Solution (2) We know

The degree of dissociation can be expressed as

Solution (1) We know

ælö R = rç ÷ è Aø



3 2 æ A ö 5.55 ´ 10 ´ p (0.5) Therefore, r = R ç ÷ = èlø 50



Resistivity,     r = 87.135 Ω cm



Conductivity, k =



Molar conductivity,

1 1 = = 0.01148 Scm -1 r 87.135

Λ m (S cm 2mol -1 ) =

     

k (Scm -1 ) × 1000(cm 3L-1 ) Molarity(mol L-1 )

1000 0.05 = 229.6 S cm 2mol -1 = 0.01148 ´

Practice Exercises Level I Construction of Cells and Reference Electrodes 1. All galvanic cells do not contain (1) a cathode. (2) an anode. (3) ions. (4) a porous plate. 2. In galvanic cells (1) (2) (3) (4)

electrical energy is converted into chemical energy. chemical energy is converted into electrical energy. electrical energy is converted into heat. chemical energy is converted into heat.

3. If a salt bridge is removed between the two half cells, the voltage (1) drops to zero. (2) does not change. (3) increases gradually. (4) increases rapidly. 4. In the galvanic cell Cu Cu 2+ (1 M) Ag + (1 M) Ag , the electrons will travel in the external circuit (1) from Ag to Cu. (2) from Cu to Ag.

Chapter 10_Electrochemistry.indd 240

(3) electrons do not travel in the external circuit. (4) cannot be predicted. 5. The equation representing the process by which standard reduction potential of zinc can be defined as 2+ (1) Zn (s) + 2e ® Zn (2) Zn(g ) ® Zn 2+(g ) + 2e -

(3) Zn 2+(g ) + 2e - ® Zn (4) Zn 2+(aq ) + 2e - ® Zn(s) 6. Normal hydrogen electrode has been assigned a potential of (1) one volt. (3) hundred volts.

(2) zero volt. (4) none of the above.

7. Which of the following half-reaction is involved in the standard hydrogen electrode? (1) Pt 2+(aq ) + 2e - ® Pt(s) (2) 2H 3O+ (aq ) + 2e - ® 2H 2O( l ) + H 2(g ) (3) H 2O2(aq ) + 2H 3O+ (aq ) + 2e - ® 4H 2O( l ) (4) O2(g ) + 2H 2O( l ) + 4e - ® 4OH - (aq )

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Electrochemistry 8. The reference electrode is made from which of the following?

o (1) E cell = reduction potential of cathode + oxidation potential of anode o (2) E cell = reduction potential of cathode − oxidation potential of anode o (3) E cell = reduction potential of cathode − reduction potential of anode o (4) E cell = −oxidation potential of cathode + oxidation potential of anode

(1) ZnCl2 (2) CuSO4 (3) Hg2Cl2 (4) HgCl2

Electrochemical Series, Calculation of Eo for the Cell and Feasibility of Cell 9. Which is the best reducing agent? (1) F− (2) Cl− (3) Br − (4) I− 10. The reaction Zn2+ + 2e – → Zn has a standard potential of –0.76 V. This means

17. Two half-cells have standard electrode potentials −0.44 V and 0.799 V respectively. These two are coupled to make a galvanic cell. Which of the following will be true? (1) Electrode of half-cell potential −0.44 V will act as anode. (2) Electrode of half-cell potential −0.44 V will act as cathode. (3) Electrode of half-cell potential 0.799 V will act as anode. (4) Electrode of half-cell potential −0.44 V will act as a positive terminal.

(1) Zn can’t replace hydrogen from acids. (2) Zn is a reducing agent. (3) Zn is an oxidizing agent. (4) Zn2+ is a reducing agent. 11. A redox reaction is spontaneous in a given direction, if (1) emf is zero. (2) emf is negative. (3) emf is positive. (4) None of these. 12. Four colorless salt solutions are placed in separate test tubes and a strip of copper is dipped in each. Which solution finally turns blue? (1) Pb (NO3)2 (2) AgNO3 (3) Zn(NO3)2 (4) Cd(NO3)2

18. The standard electrode potentials for the reactions, Ag +(aq ) + e - ® Ag +(s) Sn 2+(aq ) + 2e - ® Sn(s)

13. If the half-cell reaction A + e – ® A– has a large negative reduction potential, it follows that (1) A is readily reduced. (2) A is readily oxidized. (3) A– is readily reduced. (4) A– is readily oxidized. 14. The standard electrode potentials of four elements A, B, C and D are −3.05, 1.66, −0.40 and 0.80 V respectively. The highest chemical activity will be shown by (1) A (2) B (3) C (4) D

Pb2+ + 2e - ® Pb 2+



-

19. The equilibrium constant for the reaction, Sn(s) + 2H+® Sn2+ + H2(g) at 25°C is [E ° (Sn2+ | Sn) = −0.14] (1) 5.44 ´ 104

(3) 1.1 ´ 10 (4) 1.00 20. The value of equilibrium constant for a feasible cell reaction is (1) 1

E o = - 0.13 V

Nernst Equation and Concentration Cells 21. For the half-cell reaction: Au 3+ + 3e - ® Au

Cu + 2e ® Cu

E = 0.34 V E o = 0.80 V

Pt 2+ + 2e - ® Pt

E o = 1.20 V

o

Which of the following metals is the strongest reducing agent? (2) Pb (4) Hg

16. Which one of the following is not the correct repre­ sentation?

Chapter 10_Electrochemistry.indd 241

(2) 2.3 ´ 102

2

E o = -2.76 V

Hg 22+ + 2e - ® Hg 2

(1) Ca (3) Cu

at 25°C are 0.80 V and −0.14 V respectively. The emf of cell Sn | Sn2+ (1 M) || Ag+ (1 M) | Ag is (1) 0.66 V (2) 0.80 V (3) 1.08 V (4) 0.94 V

15. Consider the following standard reduction potentials: Ca 2+ + 2e - ® Ca

241



The value of n used in Nernst equation is (1) 3 (2) 2 (3) 1 (4) 3 ´ 96500

22. The standard emf for the cell reaction Zn + Cu2+ ® Zn2+ + Cu is 1.10 V at 25°C. The emf for the cell reaction when 0.1 M Cu2+ and 0.1 M Zn2+ solutions are used at 25°C is (1) 1.10 V (3) −1.10 V

(2) 0.110 V (4) 0.110 V

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OBJECTIVE CHEMISTRY FOR NEET

23. Doubling all the coefficients in the equation for the cell reaction (1) doubles both E° and DG°. (2) doubles E°, but does not change DG°. (3) doubles DG°, but does not change E°. (4) does not change E° or DG°. 24. Cu+ (aq) is unstable in solution and undergoes simultaneous oxidation and reduction according to the reaction

Electrolysis and Faraday’s Laws of Electrolysis 29. Electrolytic cell is used to convert (1) (2) (3) (4)

30. NaOH

is manufactured by the electrolysis of brine ­solution. The products of the reaction are

2Cu + (aq ) ® Cu 2+ (aq ) + Cu (s)

(1) Cl2 and H2 (2) Cl2 and Na-Hg (3) Cl2 and Na (4) Cl2 and O2

Choose correct E° for the above reaction if o o E Cu = 0.53 V and E Cu = 0.15 V + 2+ /Cu /Cu +



31. Which of the following reactions occurs at the cathode?

(1) +0.49 V (2) +0.38 V (3) -0.19 V (4) -0.38 V 25. Calculate the cell potential E at 25°C for the reaction 2 Al(s) + 3Fe2+(aq ) ® 2 Al 3+ (aq ) + 3Fe(s) given that [Fe2+] = 0.020 M, [Al3+] = 0.10 M, and the standard reduction potential is −1.66 V for Al3+/Al and E° for Fe2+/Fe is -0.45 V. (1) +1.03 V (2) 1.45 V (3) +1.18 V (4) +1.20 V 26. How much will the potential of Zn/Zn2+ change if the solution of Zn2+ is diluted 10 times? (1) Increase by 0.03 V (3) Increase by 0.059

(2) Decrease by 0.03 V (4) Decrease by 0.059 V

27. What would you observe if you set up the following electrochemical cell?



Ag | AgNO3 (0.001 M) || AgNO3 (1 M) | Ag (1) Electrons will flow from left to right, causing a decrease in the [Ag+] concentration in the right cell. (2) Electrons will flow from right to left, causing an increase in the [Ag+] concentration in the left cell, and a decrease in the [Ag+] concentration in the right cell. (3) Electrons will flow from left to right, causing an increase in the[Ag+] concentration in the left cell, and a decrease in the [Ag+] concentration in the right cell. (4) Electrons will flow from right to left, causing a decrease in the [Ag+] concentration in the right cell.

28. Which of the following facts about the galvanic cell and concentration cell is correct? (1) Galvanic cell is non-spontaneous whereas concentration cell is spontaneous. (2) Galvanic cell has an overall cell reaction whereas concentration cell has no overall reaction. (3) Two half-cells of both the galvanic and concentration cells are chemically different. (4) Ecell equations (Nernst equation) of both the cells o have the term E cell .

Chapter 10_Electrochemistry.indd 242

chemical energy to electrical energy. electrical energy to chemical energy. chemical energy to mechanical energy. electrical energy to mechanical energy.

(1) 2OH– ® H2O + O + 2e – (2) Ag ® Ag+ + e – (3) Fe2+ ® Fe3+ + e– (4) Cu2+ + 2e– ® Cu

32.

In the electrolysis of molten NaCl (1) Cl– ion is oxidized at anode. (2) Cl– ion is reduced at anode. (3) Cl– ion is oxidized at cathode. (4) Cl– ion neither reduced nor oxidized.

33. A solution of sodium sulphate in water is electrolyzed using inert electrodes. The products at the cathode and anode are respectively (1) H2, O2 (2) O2, H2 (3) O2, Na (4) O2, SO2 34. Which of the following moves towards the anode during electrolysis of fused NaOH? (1) Na+ (2) H+ (3) OH– (4) O2– 35. The amount of ion discharged during electrolysis is not directly proportional to (1) resistance. (3) current.

(2) time. (4) chemical equivalent of the ion.

36. Charge required to liberate 11.5 g sodium is (1) 0.5 F (2) 0.1 F (3) 1.5 F (4) 96500 C 37. The unit of electrochemical equivalent is (1) g (3) g C−1

(2) g A−1 (4) C g −1

38. When the same quantity of electricity is passed through the solution of different electrolytes in series, the amounts of product obtained are proportional to their (1) atomic weights. (2) chemical equivalents. (3) gram molecular volume. (4) gram atomic ions.

39. The quantity of electricity needed to liberate one gram equivalent of an element is

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Electrochemistry (1) 1 A (2) 96500 A (3) 96500 C (4) 96500 F 40. A certain current liberated 0.504 g of hydrogen in 2 h. How many grams of copper can be liberated by the same current flowing for the same time in a copper sulphate solution? (1) 12.7 g (2) 15.9 g (3) 31.8 g (4) 63.5 g 41. The number of electrons required to deposit 1 g atom of aluminium (At.wt. = 27) from a solution of aluminium chloride will be (1) 1 NA (2) 2 NA (3) 3 NA (4) 4 NA 42. How many coulombs of electricity are consumed when 100 mA current is passed through a solution of AgNO3 for half an hour during an electrolysis experiment? (1) 108 C (2) 180 C (3) 1800 C (4) 18000 C 43. An electric current is passed through an aqueous solution of the following, which one will decompose? (1) Urea (2) Glucose (3) Silver nitrate (4) Ethyl alcohol 44. The amount of electricity that can deposit 108 g of silver from AgNO3 solution is (1) 1 A (2) 1 C (3) 1 F (4) None of the above 45. One faraday of electricity will liberate one gram moles of the metal from the solution of (1) BaCl2 (2) CuSO4 (3) AlCl3 (4) NaCl 46. On passing 0.1 F of electricity through aluminium chloride, the amount of aluminium metal deposited on cathode is (Al = 27) (1) 0.27 g (2) 0.3 g (3) 0.9 g (4) 2.7 g

Conductance, Kohlrausch’s Law, Batteries and Corrosion 47. The best conductor of electricity is a 1 M solution of (1) boric acid. (2) acetic acid. (3) sulphuric acid. (4) phosphoric acid. 48. Which of the following aqueous solutions will conduct an electric current quite well? (1) Glycerol (2) HCl (3) Sugar (4) Pure water 49. Which of the following is a secondary cell? (1) Dry cell (2) Mercury cell (3) Ni-Cd cell (4) H2-O2 cell

Chapter 10_Electrochemistry.indd 243

243

50. Conductivity of a solution is directly proportional to (1) dilution. (2) number of ions. (3) current density. (4) volume of the solution. 51. The one that is a good conductor of electricity in the following list of solids is (1) sodium chloride. (2) graphite. (3) diamond. (4) sodium bromide. 52. Electrolytes when dissolved in water dissociate into ions because (1) they are unstable. (2) the water dissolves it. (3) the forces of repulsion increase. (4) the forces of electrostatic attraction are broken down by water. 53. The distance between two electrodes of a cell is 2.5 cm and area of each electrode is 5 cm2. The cell constant is (1) 2 (2) 12.5 (3) 7.5 (4) 0.5 54. The unit of molar conductivity is (1) Ω–1 cm–2 mol–1 (2) Ω cm2 mol–1 (3) Ω–1 cm2 mol–1 (4) Ω cm2 mol 55. The molar conductance of NaCl, HCl and CH3COONa at infinite dilution are 126.45, 426.16 and 91 Ω−1cm2 mol−1 respectively. The molar conductance of CH3COOH at infinite dilution is (1) 201.28 Ω−1 cm2 mol−1 (2) 390.71 Ω−1 cm2 mol−1 (3) 698.28 Ω−1 cm2 mol−1 (4) 540.48 Ω−1 cm2 mol−1 56. Which of the following is not an essential requirement for Kohlrausch’s law of independent migration of ions to hold true? (1) (2) (3) (4)

The solution must be infinitely dilute. Electrolyte must be strong. Electrolyte must be completely ionized. There should be no inter-ionic interaction.

57. The specific conductance of a N/50 solution of KCl at 25°C is 0.002765 mho. If the resistance of a cell containing this solution is 400 Ω, what is the cell constant? (1) 2 (2) 1.106 (3) 3 (4) 3.2 58. The specific conductance of a 0.1 N KCl solution at 23°C is 0.0112 Ω−1 cm−1. The resistance of the cell containing the solution at the same temperature was found to be 55 Ω. The cell constant will be (1) 0.142 cm−1 (2) 0.918 cm−1 (3) 1.12 cm−1 (4) 0.616 cm−1

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OBJECTIVE CHEMISTRY FOR NEET

59. As a lead storage battery is charged (1) (2) (3) (4)

lead dioxide dissolves. sulphuric acid is regenerated. lead electrode becomes coated with lead sulphate. the concentration of sulphuric acid decreases.

6. When a rod of metal A is dipped in an aqueous solution of metal B (concentration of B2+ ion being 1 M) at 25°C, the standard electrode potentials are A2+/A = – 0.76 V, B2+/B = + 0.34 V. Then (1) (2) (3) (4)

Level II Construction of Cells and Reference Electrodes 1. Which of the following statements about porous disks in voltaic cells is true? (1) Free electrons flow through the porous disk to maintain electrical neutrality in the two half-cells. (2) Ions present in the two half-cells flow through the porous disk to maintain electrical neutrality in both half-cells. (3) A porous disk contains a strong electrolyte like potassium chloride (KCl). (4) A wire must be connected directly to the porous disk in order for the porous disk to be able to maintain electrical neutrality in the two half-cells. 2. Which of the following statements about a salt bridge in a voltaic cell is true? (1) Free electrons flow through the salt bridge to maintain electrical neutrality in the two half-cells. (2) The salt bridge allows the ions present in the two half-cells to mix extensively. (3) The wire must be connected directly to the salt bridge,in order to maintain electrical neutrality in the two half-cells. (4) Ions from the electrolyte in the salt bridge flow into each half-cell to maintain electrical neutrality.

o = 0.79 V and 7. For reaction Cr2O72- + I - ® I 2 + Cr 3+, E cell o o E Cr O2- = 1.33 V . Find E I2 . 2

8. The position of some metals in the electrochemical series in decreasing electropositive character is given Mg > Al > Zn > Cu > Ag. What will happen if a copper spoon is used to stir a solution of aluminium nitrate? (1) (2) (3) (4)

4. On the basis of position in the electrochemical series, the metal which does not displace hydrogen from water and acids is (1) Hg (2) Al (3) Pb (4) Ba 5. Which one is the wrong statement about electrochemical series? (1) (2) (3) (4)

Active metals have negative reduction potentials. Active non-metals have positive reduction potentials. Metals above hydrogen liberate hydrogen from acids. Metals below hydrogen are strong reducing agents.

Chapter 10_Electrochemistry.indd 244

The spoon will get coated with aluminium. An alloy of copper and aluminium is formed. The solution becomes blue. There is no reaction.

9. If a strip of copper metal is placed in a solution of ferrous sulphate (1) (2) (3) (4)

copper will precipitate out. iron will precipitate out. copper and iron both will be dissolved. no reaction will take place.

10. When a piece of copper wire is immersed in a solution of silver nitrate, the color of the solution becomes blue. This is due to the (1) (2) (3) (4)

Electrochemical Series, Calculation of E for the Cell and Feasibility of Cell (1) Cu gets oxidized. (2) Cu gets reduced. (3) Al gets oxidized. (4) CuSO4 gets decomposed.

7

(1) 0.54 V (2) −0.054 V (3) +0.18 V (4) −0.18 V

o

3. CuSO4 is not stored in aluminium bottles because

A will not dissolve. B will deposit on A. No reaction will occur. Water will decompose into H2 and O2.

oxidation of Cu. reduction of Cu. formation of a soluble complex. oxidation of Ag.

11. From the following E o values of half cells,     (I)  A + e - → A - ; E o = -0.24 V   (II)  B- + e - → B2- ; E o = +1.25 V (III) C - + 2e - → C 3- ; E o = -1.25 V (IV)  D + 2e - → D2- ; E o = + 0.68 V

Which combination of two half cells would result a cell with the largest potential? (1) II and III (2) II and IV (3) I and III (4) I and IV

12. The standard reduction potentials, E o, for the half- reactions are as Zn → Zn 2+ + 2e - ; E o = + 0.76 V Fe → Fe2+ + 2e - ; E o = + 0.41 V The emf for the cell reaction, Fe2+ + Zn  Zn 2+ + Fe is

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Electrochemistry (1) -0.35 V (2) +0.35 V (3) +1.17 V (4) -1.17 V

18. Ag + (aq ) + e - → Ag(s); E o = +0.800 V

13. What is the cell potential (standard emf E ) for the reaction below? o



AgBr(s) + e - → Ag(s) + Br - (aq ); E o = +0.071 V



Br2( l ) + 2e - → 2Br - ; E o = +1.066 V



Use some of the data above to calculate Ksp at 25°C for AgBr.

2Fe(s) + O2(g ) + 2H 2O(l) → 2Fe2+(aq) + 4OH - (aq )

(1) 6.3 ´ 10-2 (2) 4.4 × 10-13 (3) 1.9 ´ 10-15 (4) 2.4 ´ 10-34

[ E o(Fe2+ (aq )/ Fe) = - 0.44 V and E o(O2(g )/ H 2O/OH - ) = + 0.4 V ]   o o = - 0.48 V (2) E cell = -0.04 V (1) E cell

(3) E

o cell

= + 0.84 V (4) E

o cell

19. The emf of the following three galvanic cells:

= +1.28 V

Zn Zn 2+ (1M) Cu 2+ (1M) Cu

14. If the following half cells have the E  values as o



3+

-

2+

Zn Zn 2+ (0.1M) Cu 2+ (1M) Cu

Fe + e → Fe ; E = +0.77 V and 2+

-

o

Zn Zn 2+ (1 M) Cu 2+ (0.1 M) Cu

Fe + 2e → Fe ; E = -0.44 V ; the E of the half-cell o

o

Fe3+ + 3e - → Fe will be



are represented by E1, E2, E3. Which of the following statement is true? (1) E1 > E2 > E3 (2) E3 > E2 > E1 (3) E3 > E1 > E2 (4) E2 > E1 > E3

(1) 0.33 V (2) 1.21 V (3) −0.036 V (4) 0.605 V

20. The standard reduction potentials for two reactions are given below:

Nernst Equation and Concentration Cells

AgCl(s) + e - → Ag(s) + Cl - (aq ); E o = 0.22 V

15. What is Eo for the following half reaction? -

-

MX 2(s) + 2e → M(s) + 2 X ; E = ? -

o

Ag +(aq) + e - → Ag(s); E o = 0.80 V

M (s) + 2e → M(s); E = 0.100 V 2+

o



MX 2(s)→ M 2+ + 2 X - ; K sp = e -10 (1) E o = 0.100 -

(1) 1.6 ´ 10-5 (2) 1.5 ´ 10-10 (3) 3.2 ´ 10-10 (4) 3.2 ´ 10-8

0.0592 1 log (2) E = 0.100 2 K sp 0.0592 log K sp 2 0.0592 1 log (4) E o = 0.100 + 2 K sp (3) E o = 0.100 -

21. What is the cell potential at 25°C for the reaction



3Ag(s) + NO-3 + 4H+ → 3Ag+ + NO(g) + 2H2O



when [NO-3] = 2.0 M, [Ag+] = 0.010 M, the pH is 1.00, and the pressure of NO is 0.20 atm? [E o( Ag + / Ag) = 0.799 V, E o(NO3- / NO) = 0.9 o + E ( Ag / Ag) = 0.799 V, E o(NO3- / NO) = 0.96 V ]

16. The cell potential for the electrochemical reaction shown below depends upon the Cl− and Cu2+ concentrations. Calculate the cell potential (in V ) at 25oC if [Cu2+] = 3.5 M and [Cl−] = 1.7 M. -

Cu (aq) + 2Cl (aq ) + 2 Ag(s) → Cu(s) + 2AgCl(s); E = 0.12 V o

(1) 0.15 V (2) −0.15 (3) 0.30 V (4) −0.30 V 17. The correct statement among the following is (1) The emf of a cell (E) is an extensive property and so is the DG of the cell reaction. (2) The emf of a cell (E) is an intensive property and so is the DG of the cell reaction. (3) The emf of a cell (E) is an intensive property whereas DG of the cell reaction is an extensive property. (4) The emf of a cell (E) is an extensive property but DG of the cell reaction is intensive.

Chapter 10_Electrochemistry.indd 245

The solubility product of AgCl under standard conditions of temperature (298 K) is given by

0.0592 1 ln 2 K sp

o

2+

245

(1) 0.10 V (2) 0.16 V (3) 0.22 V (4) 0.30 V 22. The potential of the cell below was 1.05 V. What is the o pH? E Ag = 0.799 V + /Ag



Pt | H2 (1 atm) | H+ (aq) || Ag+ (0.1 M) | Ag (1) 10.50 (2) 5.25 (3) 1.00 (4) 7.00

23. Calculate the cell potential (in V) at 25°C for a copper concentration cell in which the Cu2+ concentration in one half-cell is equal to 0.050 M and the Cu2+ concentration in the other half-cell is equal to 1.5 M. Note: The number of moles of electrons transferred is equal to 2. (1) 0.40 V (2) 0.80 V (3) 0.044 V (4) 0.34 V

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OBJECTIVE CHEMISTRY FOR NEET

24. The cell shown below generates a potential of 0.643 V at 300 K. Find out the Ksp for AgBr.



Ag | AgBr (s) | NaBr (0.1 M) || AgNO3 (0.1 M) | Ag (1) 1.15 × 10–9 (2) 2.00 × 10–30 (3) 1.26 × 10–13 (4) 3.5 × 10–11

25.

On the basis of information available from the reaction 4 2 Al + O2 → Al 2O3 , DG = -827 kJ mol -1 3 3



The minimum emf required to carry out electrolysis of Al2O3 is (1) 2.14 V (2) 4.28 V (3) 6.42 V (4) 8.56 V

26. Calculate Ksp for PbI2(s) at 25°C.

Pb2+ (aq ) + 2e - → Pb(s); E o = -0.126 V (1) 4.5 × 10–13 (2) 9.1 × 10–5 (3) 2.5 × 10–17 (4) 8 × 10−9 27. The cell Pt | H2 (1 atm) |H+, pH = X || Normal calomel electrode has emf of 0.4747 V at 25°C. The standard oxidation potential of calomel electrode is −0.28 V, then pH of solution will be (1) 6.6 (2) 3.3 (3) 13.2 (4) 1.1

2 Ag +(aq ) + H 2(g ) → 2 Ag(s) + 2H+(aq ) Assume that the silver ion remains constant at [Ag+] = 1.0 M and the H2 pressure remains constant at 1 atm.

30. In electrolysis of NaCl, when Pt electrode is taken, then H2 is liberated at cathode while with Hg cathode it forms sodium amalgam. This is because (1) (2) (3) (4)

Hg is more reactive than Pt. more voltage is required to reduce H+ at Hg than Pt. Na is dissolved in Hg while it does not dissolve in Pt. concentration of H+ ions is larger when Pt electrode is taken.

31. The passage of current liberates H2 at cathode and Cl2 at anode. The solution is (3) H2SO4 (4) water. 32. Zn gives H2 gas with H2SO4 and HCl but not with HNO3 because (1) Zn acts as oxidizing agent when reacts with HNO3. (2) HNO3 is a weaker acid than H2SO4 and HCl. (3) in electrochemical series, Zn is above hydrogen. (4) NO3- is reduced in preference to hydronium ion. 33. On electrolyzing a solution of dilute H2SO4 between platinum electrodes, the gas evolved at the anode is (3) O2 (4) H2 34. Copper sulphate solution is electrolyzed using copper electrodes, the reaction taking place at anode is + (1) H + e → H

(2) SO24- (aq ) → SO 4 + 2e 2+ (3) Cu + 2e → Cu

(4) Cu(s) → Cu 2+(aq ) + 2e -

(2)

35. The number of faradays needed to reduce 4 g equivalents of Cu2+ to Cu metal will be

Ecell

Ecell

Electrolysis and Faraday’s Laws of Electrolysis

(1) SO2 (2) SO3

28. Which of the following graph correctly illustrates the dependence of the cell voltage (Ecell) on the pH for the reaction?

(1)

(4) Zn | Zn+2 (10 M) || H+ (10M) | H2 (1 atm), Pt

(1) copper chloride in water. (2) NaCl in water.

PbI 2(s) + 2e - → Pb(s) + 2I - (aq ); E o = -0.365 V



(3) Zn | Zn+2 (1 M) || H+ (1M) | H2 (1 atm), Pt

(1) 1 (2) 2 pH

pH

(4)

36. The quantity of electricity required to reduce 12.3 g of nitrobenzene to aniline assuming 50% current efficiency is

Ecell

Ecell

(3)

pH

(1) 115800 C (2) 57900 C pH

29. A chemist found that the standard electrode potential o E Zn , of the zinc electrode is −0.763 V. Which cell will 2+ /Zn give this value of emf? (1) Pt, H2 (1 atm) | H+ (10M) || Zn+2 (10 M) | Zn (2) Pt, H2 (1 atm) | H (1M) || Zn (1 M) | Zn +

Chapter 10_Electrochemistry.indd 246

+2

(3) 1/2 (4) 4

(3) 231600 C (4) 28950 C 37. The amount of charge that must be passed through a solution containing Cu2+ in order to deposit 1 g atom of copper (63.55) is (1) 1520 C (2) 3040 C (3) 96500 C (4) 193000 C

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Electrochemistry 38. An electrolytic cell contains a solution of Ag2SO4 and has platinum electrodes. A current is passed until 1.6 g of O2 has been liberated at anode. The amount of silver deposited at cathode would be (1) 107.88 g (2) 1.6 g (3) 0.8 g (4) 21.60 g 39. If a direct current deposits 19.5 g of potassium (At. wt. 39) in one minute, the number of grams of aluminium (At. wt. 27) deposited by the same current during the same time interval would be (1) 27.0 (2) 13.5 (3) 9.0 (4) 4.5 40. Chromium plating is applied by electrolysis to objects suspended in a dichromate solution, according to the following (unbalanced) half reaction:

Cr2O72- (aq ) + e - + H + (aq ) → Cr(s) + H 2O(l)

How many hours would it take to apply a chromium plating of thickness 2.0 × 10−2 mm to a car bumper of surface area 0.25 m2 in an electrolysis cell carrying a current of 75 A? (The density of chromium is 7.19 g cm−3.) (1) 2.2 h (2) 1.5 h (3) 3.0 h (4) 0.25 h 41. An electric current is passed through silver voltameter connected to a water voltameter. The cathode of the silver voltameter weighed 0.108 g more at the end of the electrolysis. The volume of oxygen evolved at STP is (1) 56 mL (2) 550 mL (3) 5.6 mL (4) 11.2 mL 42. When an electric current is passed through acidulated water, 112 mL of hydrogen gas at STP is collected at the cathode in 965 s. The current passed in amperes, is (1) 1.0 (2) 0.5 (3) 0.1 (4) 2.0 43. On passing I ampere of electricity through an electrolyte solution for t second, m g metal deposits on cathode. The equivalent weight E of the metal is

247

45. 3 F electricity was passed through an aqueous solution of iron (II) bromide. The weight of iron metal (At. wt. = 56) deposited at the cathode (in gram) is (1) 56 (2) 84 (3) 112 (4) 168 46. A current of strength 2.5 A was passed through CuSO4 solution for 6 min 26 s. The amount of copper deposited is (At. wt. of Cu = 63.5) (1) 0.3175 g (2) 3.175 g (3) 0.635 g (4) 6.35 g 47. A certain quantity of electricity is passed through an aqueous solution of AgNO3 and cupric salt solution connected in series. The amount of Ag deposited is 1.08 g, the amount of copper deposited is (At. wt. Cu = 63.5; Ag = 108). (1) 0.6454 g (2) 6.354 g (3) 0.3177 g (4) 3.177 g 48. A current of 2.0 A passed for 5 h through a molten metal salt deposits 22.2 g of metal (At.wt. = 177). The oxidation state of the metal in the metal salt is (1) +1 (2) +2 (3) +3 (4) +4 49. The cost of electricity required to deposit 1 g of Mg is Rs 5. How much would it cost to deposit 10 g of Al ? (At. wt. Al = 27, Mg = 24) (1) Rs 10.00

(2) Rs 27.00

(3) Rs 44.44

(4) Rs 66.67

50. In electrolysis of alkaline water, a total of 1 mol of gases is evolved. The amount of water decomposed would be (1) 1 mol (2) 2 mol (3) 1/3 mol (4) 2/3 mol 51. Three faraday of electricity are passed through molten Al2O3, aqueous solution of CuSO4 and molten NaCl taken in different electrolytic cells. The amount of Al, Cu and Na deposited at the cathodes will be in ratio of (1) 1 mol : 2 mol : 3 mol (2) 3 mol : 2 mol : 1 mol (3) 1 mol : 1.5 mol : 3 mol (4) 1.5 mol : 2 mol : 3 mol

(1) E =

I ×t I ×m (2) E = t × 96500 m × 96500

Conductance, Kohlrausch’s Law, Batteries and Corrosion

(3) E =

96500 × m I × t × 96500 (4) E = t×I m

52. Electrolytic conduction differs from metallic conduction. In the case of electrolytic conduction

44. When during electrolysis of a solution of AgNO3, 9650 C of charges pass through the electroplating bath, the mass of silver deposited on the cathode will be (1) 1.08 g (2) 10.8 g (3) 21.6 g (4) 108 g

Chapter 10_Electrochemistry.indd 247

(1) the resistance increases with increasing temperature. (2) the resistance decreases with increasing temperature. (3) the flow of current does not generate heat. (4) the resistance is independent of the length of the conductor.

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248

OBJECTIVE CHEMISTRY FOR NEET

53. The highest electrical conductivity of the following aqueous solution is of (1) (2) (3) (4)

0.1 M acetic acid. 0.1 M chloroacetic acid. 0.1 M fluoroacetic acid. 0.1 M difluoroacetic acid.

(1) Molar conductivity increases linearly on increasing concentration for both types of electrolytes. (2) Molar conductivity decreases linearly on increasing concentration for both types of electrolytes. (3) Molar conductivity increases linearly on dilution for strong electrolyte but it is constant for weak ­electrolyte. (4) On dilution, molar conductivity increases linearly for strong electrolyte but for weak electrolyte, the increase is gradual in higher concentration range but very rapid in lower concentration range. 55. When a solution of an electrolyte is heated the conductance of the solution (1) increases because electrolyte conducts better. (2) decreases because of the increased heat. (3) decreases because of the dissociation of the electrolyte is suppressed. (4) increases because the electrolyte is dissociated more. 56. Which one of the following statements is not applicable to electrolytic conductors? (1) New products show up at the electrodes. (2) Ions are responsible for carrying the current. (3) Show a positive temperature coefficient for conductance. (4) A single stream of electrons flows from cathode to anode. 57. Which of the following plots represents correctly variation of equivalent conductance with dilution for strong electrolytes? (2) Λ

(1) increases on dilution. (2) remains constant. (3) decreases on dilution. (4) depends on density. 59. The conductivity of a saturated solution of BaSO4 is 3.06 × 10 -6 Ω -1 cm -1 and its equivalent conductance is 1.53 Ω -1 cm 2 equiv -1. The Ksp for BaSO4 will be

54. Which of the following statements concerning concentration dependence of molar conductivity for a strong and a weak electrolyte is true?

(1)

58. The molar conductivity of a strong electrolyte

Λ

(1) 4 ´ 10-12 (2) 2.5 ´ 10-9 (3) 2.5 ´ 10-13 (4) 4 ´ 10-6 60. On increasing the dilution, the specific conductance (1) increases. (3) remains constant.

61. Consider the following standard reduction potentials: Half reaction

(3)

Dilution

(4) Λ

Λ

Dilution

Chapter 10_Electrochemistry.indd 248

Dilution

(V) o

Ni (aq) + 2e  → Ni(s)

E  = -0.23 V

Fe (aq) + 2e  → Fe(s)

E  = -0.41 V

Mn2+(aq) + 2e - → Mn(s)

E  = -1.03 V

Co2+(aq) + 2e - → Co(s)

E  = -0.28 V

Cr3+(aq) + 3e - → Cr(s)

E  = -0.74 V

-

2+

o

-

2+

o o o

Which of the following metals could be used successfully to galvanize steel? (1) Ni only

(2) Ni and Co

(3) Fe only

(4) Mn and Cr

62. Which of the following statements concerning conductance and molar conductance is true? (1) Both conductance and molar conductance increase on increasing concentration. (2) Both conductance and molar conductance decrease on increasing concentration. (3) Conductance increases but molar conductance decrease on increasing concentration. (4) Conductance decreases but molar conductance increases on increasing concentration. 63. Equivalent conductance of BaCl2, H2SO4 and HCl are x1, x2 and x3 S cm2 equiv-1 at infinite dilution. If specific conductance of saturated BaSO4 solution is y W -1cm -1, then Ksp of BaSO4 will be (1)

Dilution

(2) decreases. (4) none of these.

(3)

103 y 106 y (2) 2 2 ( x1 + x 2 - 2 x 3 ) 4 ( x1 + x 2 - x 3 ) 106 y 2

4 ( x1 + x 2 - 2 x 3 )

2

(4)

x1 + x 2 - 2 x 3 106 y 2

64. An aqueous solution of a weak acid (HA) has a molar conductance of 19 S cm2 mol−1 and it is only 1% ionized in the given condition. The limiting value of molar conductance of the above acid under similar experimental condition is

1/4/2018 5:15:38 PM

Electrochemistry (1) 1900 S cm2 mol−1

(2) 950 S cm2 mol−1

(3) 475 S cm2 mol−1

(4) Infinite

(2) finite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever be the nature of the other ion of the electrolyte. (3) infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte depending on the nature of the other ion of the electrolyte. (4) infinite dilution, each ion makes definite contribution to conductance of an electrolyte whatever be the nature of the other ion of the electrolyte.

65. Ionization constant of a weak acid (HA) in terms of Λ m and Λ∞m is (1) K a =

C Λ∞m C Λ2 (2) K a = ∞ ∞ m ∞ (Λ m - Λ m ) Λ m(Λ m - Λ m )

(3) K a =

C( Λ∞m )2 (4) none of these Λ ( Λ∞m - Λ m ) ∞ m

66. Given the following molar conductivity at infinite dilution and 25°C,

HCl: Λ∞m = 426.2 S cm2 mol−1



KCl: Λ∞m = 271.5 S cm2 mol−1



CH3COOK: Λ∞m = 114.42 S cm2 mol−1



The molar conductance at infinite dilution and at 25°C, for acetic acid solution is (1) 583.28 S cm2 mol−1

(2) 269.12 S cm2 mol−1

(3) 289.63 S cm2 mol−1

(4) 172.94 S cm2 mol−1

1. The equilibrium constant of the reaction

Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s); E° = 0.46 V at 298 K is (1) 3.63 × 1015 (3) 2.0 × 1010

(2) 2.4 × 1010 (4) 4.0 × 1010 (AIPMT 2007)

2. The efficiency of a fuel cell is given by

3. On the basis of the following Eo values, the strongest oxidizing agent is:

Fe2+ → Fe3+ + e ;

E o = -0.35 V E o = -0.77 V

(1) [Fe(CN )6 ]3- (2) [Fe(CN )6 ]4(3) Fe2+ (4) Fe3+ (AIPMT 2008) 4. Kohlrausch law states that at (1) infinite dilution, each ion makes definite contribution to equivalent conductance of an electrolyte, whatever is the nature of the other ion of the electrolyte.

Chapter 10_Electrochemistry.indd 249

(1) 0.0968 V (2) 1.968 V (3) 2.0968 V (4) 1.0968 V (AIPMT 2008) 6. Given (I)  Cu 2+ + 2e – → Cu, E o = 0.337 V (II)    Cu 2+ + e – → Cu + , E o = 1.153 V Electrode potential, E º for the reaction, Cu+ + e - → Cu, will be (1) 0.38 V (2) 0.479 V (3) 0.90 V (4) 0.30 V (AIPMT 2009) 7. The equivalent conductance of M / 32 solution of a weak monobasic acid is 8.0 mhos cm2 and at infinite dilution, it is 400 mhos cm2. The dissociation constant of this acid is (1) 1.25 × 10−4

(2) 1.25 × 10−5

−6

(4) 6.25 × 10−4 (AIPMT 2009)

(AIPMT 2007)



5. Standard free energies of formation (in kJ mol−1) at 298 K are −237.2, −394.4 and −8.2 for H2O(l), CO2(g) and peno tane (g), respectively. The value of E cell for the pentane oxygen fuel cell is

(3) 1.25 × 10

DS DH (1) (2) DG DG DG DG (3) (4) DS DH

[Fe(CN )6 ]4- → [Fe(CN )6 ]3- + e ;

(AIPMT 2008)



Previous Years’ NEET Questions

249

8. Al2O3 is reduced by electrolysis at low potentials and high currents. If 4.0 × 104 A of current is passed through molten Al2O3 for 6 h, what mass of aluminium is produced? (Assume 100% current efficiency, atomic mass of Al = 27 g mol−1.) (1) 4.0 × 104 g (2) 9.0 × 103 g (3) 8.1 × 104 g (4) 2.4 × 105 g (AIPMT 2009) 9. For the reduction of silver ions with copper metal the standard cell potential was found to be +0.46 V at 25°C . The value of standard Gibbs energy, DG° will be (F = 96500 C mol–1) (1) –89.0 J (2) –44.5 kJ (3) –98.0 kJ (4) –89.0 kJ (AIPMT PRE 2010)

1/4/2018 5:15:40 PM

250

OBJECTIVE CHEMISTRY FOR NEET

10. An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to (1) 100% ionization of electrolyte at normal dilution. (2) increase in both, that is, number of ions and ionic mobility of ions. (3) increase in number of ions. (4) increase in ionic mobility of ions. (AIPMT PRE 2010)

11. Which of the following expressions correctly represents the equivalent conductance at infinite dilution of Al2(SO4)3? Given that Λ0Al 3+ and Λ0SO2- are the equiva4 lent conductances at infinite dilution of the respective ions (1) Λ

0 Al 3+

(3)

+Λ

0 SO24 -

(

(2) Λ

0 Al 3+

+Λ

0 SO24 -

)×6

1 0 1 Λ 3+ + Λ0SO2- (4) 2Λ0Al 3+ + 3Λ0SO24 4 3 Al 2 (AIPMT MAINS 2010)

12. Consider the following relations for emf of an electrochemical cell: I Emf of cell = (Oxidation potential of anode) − (Reduction potential of cathode) II Emf of cell = (Oxidation potential of anode) + (Reduction potential of cathode) III Emf of cell = (Reduction potential of anode) + (Reduction potential of cathode) IV Emf of cell = (Oxidation potential of anode) − (Oxidation potential of cathode) Which of the above relations are correct?

o 15. If the E cell for a given reaction has a negative value, then which of the following gives the correct relationships for the values of ∆Go and Keq?

(1) ∆Go > 0; Keq < 1

(2) ∆Go > 0; Keq > 1

(3) ∆Go = 0; Keq < 1

(4) ∆Go < 0; Keq < 1 (AIPMT PRE 2011)

16. Standard electrode potential of three metals X, Y and Z are –1.2 V, +0.5 V and –3.0 V respectively. The reducing power of these metals will be (1) X > Y > Z (2) Y > Z > X (3) Y > X > Z (4) Z > X > Y (AIPMT PRE 2011) 1 7. A solution contains Fe2+, Fe3+ and I− ions. This solution was treated with iodine at 35°C. E° for Fe3+ / Fe2+ is +0.77 V and E° for I2/2I− is 0.536 V. The favorable redox reaction is (1) Fe2+ will be oxidized to Fe3+. (2) I2 will be reduced to I−. (3) there will be no redox reaction. (4) I− will be oxidized to I2. (AIPMT MAINS 2011) 18. Limiting molar conductivity of NH4OH (i.e. Λ0m (NH 4OH ) is equal to (1) Λ0m (NH 4Cl ) + Λ0m (NaCl ) - Λ0m (NaOH) (2) Λ0m (NaOH) + Λ0m (NaCl ) - Λ0m (NH 4Cl ) (3) Λ0m (NH 4OH ) + Λ0m (NH 4Cl ) - Λ0m (HCl )

(1) I and II (2) III and IV

(4) Λ0m (NH 4Cl ) + Λ0m (NaOH) - Λ0m (NaCl )

(3) II and IV (4) III and I (AIPMT MAINS 2010) -

13. The electrode potentials for Cu (aq) + e → Cu (aq) and Cu +(aq) + e - → Cu(s) are +0.15 V and +0.50 V respec­ o tively. The value of E Cu will be 2+ /Cu 2+

+

(1) 0.150 V (2) 0.500 V (3) 0.325 V (4) 0.650 V (AIPMT PRE 2011) 14. Standard electrode potential for Sn 4+/Sn 2+ couple is +0.15 V and that for the Cr 3+/Cr couple is –0.74 V. These two couples in their standard state are connected to make a cell. The cell potential will be (1) +1.83 V (2) +1.19 V (3) +0.89 V (4) +0.18 V (AIPMT PRE 2011)

Chapter 10_Electrochemistry.indd 250

(AIPMT PRE 2012) 19. Standard reduction potentials of the half reactions are given below:

F2(g ) + 2e - ® 2F - (aq);



Cl 2(g ) + 2e - → 2Cl - (aq); E o = +1.36 V



Br2( l ) + 2e - → 2Br - (aq); E o = +1.06 V



I 2( l ) + 2e - → 2I - (aq); E o = +0.53 V



The strongest oxidizing and reducing agents respectively are:

E o = +2.85 V

(1) F2 and I− (2) Br2 and Cl− (3) Cl2 and Br− (4) Cl2 and I2 (AIPMT MAINS 2012)

1/4/2018 5:15:42 PM

Electrochemistry 20. Molar conductivities (Λ0m) at infinite dilution of NaCl, HCl and CH3COONa are 126.4, 425.9 and 91.0 S cm2 mol−1 respectively. Λ0m for CH3COOH will be

26. A device that converts energy of combustion of fuels like hydrogen and methane, directly into electrical energy is known as (1) electrolytic cell. (2) dynamo. (3) Ni-Cd cell. (4) fuel cell.

(1) 425.5 S cm2 mol−1 (2) 180.5 S cm2 mol−1 (3) 290.8 S cm2 mol−1

(4) 390.5 S cm2 mol−1

(AIPMT 2015)

(AIPMT MAINS 2012, NEET II 2016) 21. A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passing hydrogen gas around the platinum wire at 1 atm pressure. The oxidation potential of electrode would be

27. The pressure of H2 required to make the potential of H2-electrode zero in pure water at 298 K is (1) 10−14 atm (2) 10−12 atm (3) 10−10 atm (4) 10−4 atm

(1) 0.059 V (2) 0.59 V

(NEET I 2016)

(3) 0.118 V (4) 1.18 V (NEET 2013) 22. At 25°C molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is 9.54 ohm–1 cm2 mol–1 and at infinite dilution its molar conductance is 238 ohm–1 cm2 mol–1. The degree of ionization of ammonium hydroxide at the same concentration and temperature is

28. The number of electrons delivered at the cathode during electrolysis by a current of 1 A in 60 s is (charge on electron = 1.60 × 10−19 C) (1) 6 × 1020 (2) 3.75 × 1020 (3) 7.48 × 1023 (4) 6 × 1023 (NEET II 2016) 29. The molar conductivity of a 0.5 mol dm−3 solution of AgNO3 with electrolytic conductivity of 5.76 × 10−3 S cm−1 at 298 K is

(1) 2.080% (2) 20.800% (3) 4.008% (4) 40.800% (NEET 2013)

(1) 11.52 S cm2mol−1 (2) 0.086 S cm2mol−1 (3) 28.8 S cm2mol−1 (4) 2.88 S cm2mol−1

23. A button cell used in watches functions as following

Zn(s) + Ag 2O(s) + H 2O( l ) → 2 Ag(s) + Zn 2+ (aq ) + 2OH - (aq )



If half-cell potentials are





251

Zn 2+ (aq ) + 2e - → Zn(s); E ° = -0.76 V

Ag 2O(s) + H 2O( l ) + 2e - → 2 Ag(s) + 2OH - (aq ); E° = 0.34 V The cell potential will be

(NEET II 2016) 30. During the electrolysis of molten sodium chloride, the time required to produce 0.10 mol of chlorine gas using a current of 3 A is (1) 108 min. (2) 220 min. (3) 330 min. (4) 55 min.

(1) 1.10 V (2) 0.42 V (3) 0.84 V (4) 1.34 V

(NEET II 2016) (NEET 2013)

24. When 0.1 mol MnO24 is oxidized the quantity of electricity required to completely covert MnO24 to MnO 4 is (1) 96500 C (2) 2 × 96500 C (3) 9650 C (4) 96.50 C (AIPMT 2014) 25. The weight of silver (At. wt. = 108) displaced by a quantity of electricity which displaces 5600 mL of O2 at STP will be (1) 5.4 g (2) 10.8 g (3) 54.0 g (4) 108.0 g

31. In the electrochemical cell

Zn | ZnSO4(0.01 M) || CuSO4 (1.0 M) | Cu, the emf of this Daniel cell is E1. When the concentration of ZnSO4 is changed to 1.0 M and that of CuSO4 changed to 0.01 M, the emf changed to E2. From the followings, which one is RT the relationship between E1 and E2 (Given: = 0.059)? F (NEET 2017) (1) E1 < E2 (3) E2 = 0 ≠ E1

(2) E1 > E2 (4) E1 = E2

(AIPMT 2014)

Chapter 10_Electrochemistry.indd 251

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252

OBJECTIVE CHEMISTRY FOR NEET

Answer Key Level I  1. (4)

 2. (2)

 3. (1)

 4. (2)

 5. (4)

 6. (2)

 7. (2)

 8. (3)

 9. (4)

10.  (2)

11.  (3)

12.  (2)

13.  (4)

14.  (1)

15.  (1)

16.  (2)

17.  (1)

18.  (4)

19. (1)

20.  (4)

21.  (1)

22.  (1)

23.  (3)

24.  (2)

25.  (3)

26.  (1)

27.  (3)

28. (2)

29.  (2)

30.  (1)

31.  (4)

32.  (1)

33.  (1)

34.  (3)

35.  (1)

36.  (1)

37.  (3)

38.  (2)

39.  (3)

40.  (2)

41.  (3)

42.  (2)

43.  (3)

44.  (3)

45.  (4)

46.  (3)

47.  (3)

48.  (2)

49.  (3)

50.  (2)

51.  (2)

52.  (4)

53.  (4)

54.  (3)

55.  (2)

56.  (2)

57.  (2)

58.  (4)

59.  (2)

Level II  1. (2)

 2. (4)

 3. (3)

 4. (1)

 5. (4)

 6. (2)

 7. (1)

 8. (4)

 9. (4)

10.  (1)

11.  (1)

12.  (2)

13.  (3)

14.  (3)

15.  (2)

16.  (1)

17.  (3)

18.  (2)

19.  (4)

20.  (2)

21.  (3)

22.  (2)

23.  (3)

24.  (3)

25.  (1)

26.  (4)

27.  (2)

28.  (2)

29.  (3)

30.  (2)

31.  (2)

32.  (4)

33.  (3)

34.  (4)

35.  (4)

36.  (1)

37.  (4)

38.  (4)

39.  (4)

40.  (2)

41.  (3)

42.  (1)

43.  (3)

44.  (2)

45.  (2)

46.  (1)

47.  (3)

48.  (3)

49.  (4)

50.  (4)

51.  (3)

52.  (2)

53.  (4)

54.  (4)

55.  (4)

56.  (4)

57.  (2)

58.  (1)

59.  (4)

60.  (2)

61.  (4)

62.  (3)

63.  (3)

64.  (1)

65.  (2)

66.  (2)

Previous Years’ NEET Questions  1. (1)

 2. (4)

 3. (4)

 4. (1)

 5. (4)

 6. (2)

 7. (2)

 8. (3)

 9. (4)

10.  (4)

11.  (1)

12.  (3)

13.  (3)

14.  (3)

15.  (1)

16.  (4)

17.  (4)

18.  (4)

19.  (1)

20.  (4)

21.  (2)

25.  (4)

26.  (4)

27.  (1)

28.  (2)

22.  (3)

23.  (1)

24.  (3)

29.  (1)

30.  (1)

31.  (2)

Hints and Explanations Level I 17. (1) We know E

o Cell

= SRPCathode - SRPAnode

(1)

= 0.80 - ( -0.14) = 0.94 V



   



At the cathode: ( Ag + + 1e - ® Ag ) ´ 2



2+ At the anode: Sn ® Sn + 2e



o For the cell to be spontaneous, E cell > 0.



Cell reaction: 2 Ag + (1M) + Sn ® Sn 2+ (1M) + 2 Ag



We have SRPCathode = 0.799 V and SRPAnode = - 0.44 V .



o Nernst equation is given by E Cell = E Cell -



o Therefore, from Eq. (1), we get E Cell = 0.799 - ( -0.44) > 0

o 18. (4) E Cell = SRPCathode - SRPAnode

Chapter 10_Electrochemistry.indd 252

o E Cell = E Cell -

0.059 [Sn 2+ ] log n [ Ag + ]2

0.059 (1) log 2 (1) 2

1/4/2018 5:15:44 PM

Electrochemistry

30. (1) The reactions involved are as follows:

o = 0.94 V Therefore, E Cell = E Cell

o 19. (1) We know DG ° = - RT ln K = -nFE cell

  -2.303 ´ 8.314 ´ 298 log K = -2 ´ 96500 [0 - ( -0.14)]  

-2.303 × 8.314 × 298 log K = -2 × 96500 × 0.14 K = 5.44 × 104

20. (4) Gibbs energy can be expressed as DG ° = - RT ln K (1)







For a feasible reaction, DG° < 0



Therefore, from Eq. (1), we get K > 1.



At the cathode: 2e - + 2H 2O ® H 2 + 2OH -



At the anode: Cl - ®



At the cathode: 2H 2O + 2e - ® H 2 + 2OH -



At the anode: 2H 2O ® O2 + 4e - + 4H +

36. (1) According to Faraday’s law of electrolysis, we have M ×Q n×F 23 11.5 = ×Q 1× F F = Q ⇒ Q = 0.5 F 2 W=

0.059 [ Zn 2+ ] log 2 [Cu 2+ ] 0.059 æ 0.1ö = 1.1 log ç è 0.1÷ø 2 = 1.1 V

o E cell = E cell -

23. (3) Gibbs energy is given by DG ° = -nFE

40. (2) The reaction is 2e- + 2H+ → H2

According to Faraday’s first law

o

Doubling the coefficient, doubles DGo value.



Also, E o = SRPCathode - SRPAnode



Since, E o value is independent of reaction coefficient; therefore, it would not change.

WH2



At the anode: Cu + ® Cu 2+ + e -



+

E

=E

-E

  WCu =



On dividing Eq. (2) by Eq. (1), we get WCu 63.5 /( 2 × 96500) 63.5 = = 2 /( 2 × 96500) 2 WH2

o Cu 2+ /Cu +

= 0.53 - (0.15) = 0.38 V



25. (3) According to Nernst equation, we have o E cell = E cell -

42. (2) We know Q = It

0.059 [ Al ] log 6 [Fe2+ ]3



(0.1) 0.059 g log (0.02)3 6

æ1 ö = 100 ´ 10 -3 ´ ç ´ 3600÷ = 180 C è2 ø

(

)

44. (3) The reaction is Ag + + e - ® Ag W Ag = Z Ag × Q

2+

26. (1) The reaction involved is Zn ® Zn + 2e

-

0.059 log[Zn 2+ ] (1) 2 0.059 [ Zn 2+ ] (2) log 2 10





o ( E cell )1 = E cell -





o ( E cell )2 = E cell



Subtracting Eq. (2) from Eq. (1), we get

Chapter 10_Electrochemistry.indd 253



2

= 1.18 V

( E cell )2 - ( E cell )1 =

63.5 × 0.504 = 15.9 g 2

Therefore, WCu =

3+ 2

= ( -0.45 + 1.66 ) -

63.5 × I × ( 2 × 3600) (2) 2 × 96500



2+

o Cu + /Cu

2 × I × ( 2 × 3600) (1) 2 × 96500

  WCu = Z Cu × I × ( 2 × 3600)

Cell reaction: 2 Cu ® Cu + Cu o cell

0.504 =

Cu 2+ + 2e - ® Cu

24. (2) The reactions taking place are as follows:



M × I ×t n×F = Z H2 × I × ( 2 × 3600)

W=



At the cathode: Cu + + e - ® Cu

1 Cl 2 + e 2

33. (1) The reactions take place is as follows:

22. (1) For the given cell reaction, Nernst equation can be expressed as



253

0.059 = 0.03 V 2

108 =

108 × Q ⇒ Q = 96500 C or 1 F 1 × 96500

46. (3) The reaction is Al3+ + 3e- → Al W Al = Z Al × Q 27 × 0.1 F = = 0.9 g 3×F 53. (4) Cell constant =

L 2.5 = = 0.5 A 5

1/4/2018 5:15:48 PM

254

OBJECTIVE CHEMISTRY FOR NEET

55. (2) Using Kohlrausch’s law

16. (1) From the Nernst equation, we have

Λ∞CH3 COOH = Λ∞CH3 COONa - Λ∞NaCl + Λ∞HCl

o E cell = E cell

= 91 - 126.45 + 426.16 = 390.71 Ω -1 cm 2 mol -1 58. (4) Electrical resistance is given by R = r ´

l A

Level II o o 7. (1) We have      E cell = E Cr - E Io /IO2 - /Cr 3+ 7

= 0.15 V 18. (2) We have Ag + + e - → Ag; E1o = 0.8

E1o =







On subtracting Eq. (1) from Eq. (2), we get

E 2o = +

æ (0.8 - 0.071)ö -ç ÷ = log K sp è 0.059 ø

( )

(K )

12. (2) From the Nernst equation, we have 0.059 [ Zn 2+ ] log 2 [Fe2+ ] 0.059 æ 1ö = ( -0.41 + 0.76 ) log ç ÷ è 1ø 2 = 0.35 V

o E cell = E cell -

sp



 

Fe + e ® Fe , E = 0.77 V , n1 = 1 (1)



 

Fe2+ + 2e - ® Fe, E 2o = -0.44 V , n2 = 2 (2)



 

Fe3+ + 3e - ® Fe, E 3o = ?, n3 = 3 (3)



From Eq. (3) = Eq. (1) + Eq. (2), we get DG 3o = DG1o + DG 2o n3 E 3o = n1E1o + n2 E 2o 1 ´ 0.77 + 2 ´ ( -0.44) E = = - 0.036 V 3 o 3

15. (2) We have E cell = ( E o - 0.1) - 0.059 log[M 2+ ][ X - ]2 2

At equilibrium, E cell = 0 and [M 2+ ][ X - ]2 = K sp 0.059 log K sp 2      0.059 1 E o = 0.1 log K sp 2 0 = E o - 0.1 -

Chapter 10_Electrochemistry.indd 254

0.22 =

(1) (2)

(0.22 - 0.80) = log[ Ag + ][Cl - ] 0.059 K sp(AgCl) = [ Ag + ][Cl - ] = 1.5 ´ 10 -10

14. (3) The reactions involved are as follows:

= 4.4 ´ 10 -13

On subtracting Eq. (1) from Eq. (2), we get

   = 0.4 - ( -0.44) = 0.84 V

o 1

AgBr

AgBr

0.059 log[Cl - ] 1 0.059 1    0.80 = log [ Ag + ] 1

20. (2) We have  

= SRPCathode - SRPAnode

2+

(2)

E1o - E 2o = -0.059 log[ Ag + ][Br - ]

2

-

(1)

2

2

3+

0.059 log[Br - ] 1



E Io /I- = 0.54 V

13. (3) We know E

0.059 1 log [ Ag + ] 1

AgBr(s) + e - ® Ag + Br - ; E 2o = 0.071 V , n2 = 1

0.79 = 1.33 - E Io /I-

o cell

0.059 1   log    (3 3.5)(1.7 )2  2

= 0.12 -

1 1 æl ö =k = ´ç ÷ r R è Aø 1 æl ö æl ö -1 0.0112 = ç ÷ Þ ç ÷ = 0.616 cm 55 è A ø è Aø

2

0.059 1   log   [Cu 2+ ][Cl - ]2  2

o 21. (3) E cell = E cell -



æ ( p )[ Ag + ]3 ö 0.059 log ç NO + 4 ÷ 3 è [NO3 ][H ] ø

= (0.96 - 0.799) -

é 0.2 ´ (0.01)3 ù 0.059 log ê -1 4 ú 3 ë ( 2) ´ (10 ) û

= 0.22 V 22. (2) The reactions are

Anode: H 2 ® 2H + + 2e -



Cathode: 2 Ag + + 2e - ® 2 Ag

Cell reaction: 2Ag + + H 2 → 2H + + 2Ag o E cell = E cell -

0.059 [H + ]2 log 2 [ Ag + ]2( pH2 )

1.05 = (0.799 - 0) - 0.059 log

[H + ] (0.1)

(1.05 - 0.799) = - log[H + ] + log(0.1) 0.059 - log[H + ] = pH = 5.25

1/4/2018 5:15:51 PM

Electrochemistry 23. (3) The reaction involved is Cu 2+ (C 2 = 1.5 ) + Cu ® Cu + Cu 2+ (C1 = 0.05 ) 0.059 C o E cell = E cell log 1 2 C2



Overall reaction: Hg 22+ + H 2 ® 2H + + Hg 0.059 log[H + ]2 2 -0.28 = 0.4747 + 0.059 pH pH = 3.3 o E cell = E cell -

0.059 (0.05) log = 0.044 V 2 (1.5)

28. (2) We have E cell = E o + - 0.059 log[H + ]2 Ag /Ag 2

24. (3) At the anode: Ag ® Ag + + e C=

o E cell = E Ag + 0.059 pH + /Ag 

K sp( Ag Br)

Constant

[Br - ]

At the cathode: Ag + (0.1M) + e - ® Ag o E cell = E cell



At the anode: H 2 ® 2H + + 2e -

o = 0. Therefore, For concentration cell, E cell

E cell = -





35. (4) We know W =

 K sp(AgBr) /[Br - ] 0.059 log   1 [ Ag + ]  

(Eq. wt.)Cu ×Q F

W × F = Q ⇒ Q = 4F (Eq. wt.)Cu 36. (1) The reaction is

For concentration cell, E cell = 0 . Therefore, o

NO2

K sp(AgBr) -0.059 0.643 = log 1 (0.1)(0.1)

NH2 6e−

+

K sp(AgBr) = 1.26 ´ 10 -13 25. (1) The reactions are as follows:

Al ® Al 3+ + 3e

We know      W = 12.3 =

37. (4) From Faraday’s Law, we have WCu =

26. (4) PbI 2 (s) + 2e - ® Pb(s) + 2I - (aq ), E1o = - 0.365 V

0.059 log[I - ]2 2 Pb2+ + 2e - ® Pb, E 2o = -0.126 V -0.365 =

-0.126 =

0.059 1 log 2+ [ Pb ] 2







On subtracting Eq. (1) from Eq. (2), we get ( -0.365 + 0.126 ) = -

0.059 log [Pb2+ ][I - ]2 2

0.239 × 2 = log K sp(PbI2) 0.059 K sp(PbI2 ) = 8 × 10-9

27. (2) The reactions are

At the cathode: Hg 22- + 2e - ® Hg

Chapter 10_Electrochemistry.indd 255

M ×Q n×F

123 ´ Q 6 ´ 96500 Q = 115800 C

4 4 4 Al ® Al 3+ + 3 ´ e 3 3 3 4 Therefore, n = 3 ´ = 4 3 We know DG = -n FE Also,

-827 ´ 103 = -4 ´ 96500 ´ E E = 2.14 V



255

nCu =

(1)

Mol. wt. ×Q 2 × 96500

1 WCu = ×Q (Mol.wt.)Cu 2 × 96500

nCu ´ 2 ´ 96500 = Q 1 ´ 2 ´ 96500 = Q Q = 193000 C

(2)

38. (4) The reaction is Ag + + e - ® Ag

W Ag =



108 × Q (1) 1× F

2H 2O ® O2 + 4H + + 4e -



WO2 =

32 × Q (2) 4×F







On dividing Eq. (1) by Eq. (2), we get W Ag 1.6

=

108 × 4 ⇒ W Ag = 21.6 g 32

1/4/2018 5:15:54 PM

256

OBJECTIVE CHEMISTRY FOR NEET 96500 × m I ×t

39. (4) We have    WK =

39 × Q (1) 1× F



W Al =

27 × Q (2) 3 ×F

44. (2) We have W =



  



On dividing Eq. (2) by Eq. (1), we get 45. (2) WFe =

40. (2) Mass of metal deposited =  Volume × density = (l × b × h) × r For the given reaction, n = 6



Also,



where E is the number of equivalents =

(l × b × h ) × r =

I ×t × E 96500 M n-factor

  



  



On dividing Eq. (1) by Eq. (2), we get W Ag

52 ´ 75 t ´ 6 ´ 96500 3600 t = 1.5 h

WCu

−

108 × Q (1) 1 × 96500



  2H2O → O2 + 4e− + 4H+







WO2 =

(Mol. wt.)O2 ´ Q

22.2 =

0.108

=

108 ´ 4

10 -3 = 0.00025 mol Therefore, nO2 = 4

V at STP = 0.00025 ´ 22.4 ´ 10 O2

L = 5.6 mL or 5.6 cm

WH2 (Mol. wt.)H2

(Mol. wt.)H2 n×F

43. (3) We know W = ZIt =

Therefore, oxidation state of metal is +3.

(Mol. wt.) × I ×t nF



   1 =

24 ´ Q1 2 ´ 96500 Þ Q1 = 2 ´ 96500 24



which costs Rs 5.



Amount of electricity required for Al = Q2

3

10 =

× I ×t

I nH2 = 2 × 100 112 I= × 2 × 100 = 1 A 22400

where E =

Chapter 10_Electrochemistry.indd 256

m n

177 ´ 2 ´ (5 ´ 3600) Þn = 3 n ´ 96500

We know W =



I × 965 = 2 × 96500

(2)

(Mol. wt.) × I ×t nF



42. (1) From Faraday’s law, we have WH2 =

63.5 × I × t 2×F

49. (4) Amount of electricity required for Mg = Q1

(2)

(Mol. wt.)O2 ´ 1

-3

(1)

1.08 ´ 63.5 æ 108 ö æ 2 ö =ç Þ WCu = = 0.3175 g è 1 ÷ø çè 63.5 ÷ø 2 ´ 108

From Eq. (1) and Eq. (2), we get WO2



4 ´ 96500



(W )Cu =

48. (3) Using the equation W =

41. (3) The reaction involved is Ag + e → Ag 0.108 =

(Mol. wt.) × I ×t nF 108 (W )Ag = × I × t 1× F



(7.19) ´ (0.25 ´ 104 )( 2 ´ 10 -3 ) =

+

56 ´ 3F = 84 g 2F

47. (3) We have     W =

(W )Cr = 7.19 ´ (0.25 ´ 104 ) ´ ( 2 ´ 10 -2 ´ 10 -1 ) g

(Mol. wt.) × I ×t nF 108 = ´ 9650 = 10.8 g 1 ´ 96500

W Ag

27 27 W Al = ⇒ W Al = × 19.5 = 4.5 g 3 × 39 3 × 39 WK



Therefore, E =



27 ´ Q2 30 ´ 96500 Þ Q2 = 3 ´ 96500 27

5 ´ Q2 Q1 5 ´ 24 30 ´ 96500 ´ = Rs 66.67       = 2 ´ 96500 27 Therefore, cost =

50. (4) The reaction is 2H 2O → 2H 2 (g) + O2(g)

2 mol of H2O gives 3 mol of gas.



Therefore, 1 mol of gas will be obtained by

51. (3) We have nAl =

m´ I ´t E ´ I ´t = 96500 n ´ 96500

nNa =

2 mol of H2O. 3

Q Q ;n = ; 3 ´ 96500 Cu 2 ´ 96500

Q 1 ´ 96500

Therefore, nAl = nCu = nNa = 1 : 1.5 : 3

1/4/2018 5:15:57 PM

Electrochemistry

59. (4) Conductivity is given by Λ BaSO = k × 4

We know C = K sp

1000 C

257

2.303 × 8.314 × 298 log K eq 2 × 96500 0.46 × 2 × 96500 = 15.56 = 2.303 × 8.314 × 298

0.46 V = log K eq

1000 k ´ = 1.53 K sp

K eq = anti log(15.56 ) = 3.63 × 1015

2

æ 3.06 ´ 10-6 ´ 1000 ö ç ÷ = K sp 1.53 è ø K sp = 4 ´ 10-6 61. (4) In galvanize steel, SOPMetal > SOPFe. Therefore, Mn and Cr can be used. 63. (3) We have   Λ∞BaSO = Λ∞BaCl + Λ∞H SO - 2Λ∞HCl 4 2 2 4

Therefore,   Λ∞BaSO4 = ( x1 + x 2 - 2 x 3 ) (1)



y × 1000 (2)    Λ BaSO4 = 2 K spBaSO4



For sparingly soluble salts, Λ = Λ∞



Therefore, from Eq. (1) and Eq. (2), we get ( x1 + x 2 - 2 x 3 ) =

y × 1000 2 K spBaSO4

           K sp =

y 2 × 106 4( x1 + x 2 - 2 x 3 )2

Λ 4. (1) The degree of dissociation is given by a = ∞ 6 Λ 19 0.01 = ¥ Þ L¥ = 1900 Scm 2 mol -1 L

2. (4) Fuel cells have a working efficiency of about 70–75%. The thermodynamic efficiency (h) of a fuel cell is given by the ratio of electrical work done to the enthalpy change associated with the reaction. Mathematically, Efficiency of cell =



- nFE DG = DH DH

3. (4) Fe3+ shows the maximum tendency to undergo reduction and is hence the strongest oxidizing agent. 4. (1) Kohlrausch’s law states that at infinite dilution, when all the interionic effects disappear, each ion makes a definite contribution towards molar conductivity of the electrolyte irrespective of the nature of the other ion with which it is associated. Mathematically, the law can be stated as Λ0m = v + l+0 + v - l-0. 5. (4) The reaction taking place in the fuel cell is C5H12(g) + 8O 2(g) ® 5CO2(g) + 6 H 2O(g)

We know, D G ° = å D f G ° ( products) - å D f G ° (reactants)

{ {D G ° éëC H

}

= D f G ° éë5 CO2 (g )ùû + D f G ° éë6 H 2 O (g )ùû f

5

12

(g )ùû + D f G ° éë8O2 (g )ùû }

65. (2) The reaction is HA  At t = 0 C At t = t eq C(1 - a )

H+ + A 0 0 Ca Ca



D G° = {5 × ( - 394.4) + 6 × ( - 237.2) } - {1 × ( - 8.2) + 8 × 0 } = - 3387 kJ mol -1 or - 3387 × 103 J mol -1

2

Λ  C  ∞m   Λm  Ca C Λ2 Ka = = ⇒ Ka = ∞ ∞ m (1 - a )  Λ m  Λ m (Λ m - Λ m )  1 - Λ∞    m 2

66. (2) We have Λ∞CH3 COOH = Λ∞HCl - Λ∞KCl + Λ∞CH3 COOK   

L¥CH3 COOH = 426.2 - 271.5 + 114.42



Eo =



2.303 RT log K eq (1) nF

o = 0.46 V , R = 8.314 JK-1mol-1, T = 298 K, Substituting E cell n = 2 and F = 96500 C in Eq. (1), we get

Chapter 10_Electrochemistry.indd 257

-3387 ´ 103 J mol -1 = 1.09 V -32 ´ 96500

6. (2) The electrode potential and Gibbs energy for the reaction are related by

= 269.12 Scm 2mol -1

o E cell =

We know,    ∆Go = -nFE o -3387 ´ 103 J mol -1 = - 32 ´ 96500 E o

Cu 2+ + 2e - ® Cu, E1o = 0.337 V; DG1o = - 2 FE1o Cu 2+ + e - → Cu + , E 2o = 1.153 V; DG 2o = - 1FE 2o

Previous Years’ NEET Questions 1. (1) We have

Substituting the given values of Gibbs energy of formation of compounds, we have

Cu + + e - ® Cu, DG 3o = - n3 FE 3o ; DG 3o = - 1FE 3o

On solving, we get D G 3o = D G1o - D G 2o - 1 FE 3o = - 2 F ( 0.337 ) + F (1.153) E 3o = 0.479

1/4/2018 5:16:01 PM

258

OBJECTIVE CHEMISTRY FOR NEET

7. (2) The degree of dissociation is given by a =

ΛΕ 8.0 = = 2 × 10 -2 ∞ Λ Ε 400

Cu + + e - → Cu

 H+ + A 0 0

At t = t eq C (1 - a ) Ka =

a

a

2 1 Ca 2 @ Ca 2 = ´ ( 2 ´ 10-2 ) = 1.25 ´ 10-5 1-a 32

-n3 FE 3o = - n1FE o - n2 FE 2o



  



On solving, we get E 3o = 0.325 V

14. (3) We have o o o E cell = E cathode - E anode

= 0.15 - ( -0.74) = 0.89 V

8. (3) According to Faraday’s law of electrolysis

=

15. (1) DG° > 0 , K eq < 1

Elt 96, 500 27 ´ 4 ´ 104 ´ 6 ´ 60 ´ 60 3 ´ 96, 500

= 80, 580.31 g  8.1 ´ 104 g 9. (4) For the reaction 2Ag+  + Cu → Cu+ + 2Ag; Eo = +0.46 V; n = 2







We know ∆Go = -nFE o D G ° = - nFE o = - 2 × 96, 500 × 0.46 = - 88780 J or - 88.780 kJ

10. (4) Strong electrolytes have a = 1, therefore, at high dilution the ionic mobility increases, resulting in increase in equivalent conductance. 11. (1) At infinite dilution, when dissociation is complete, each ion makes a definite contribution towards the conductance of the electrolyte irrespective of the nature of the other ion with which it is associated. Hence, Λ0Al 2 (SO4)3 = Λ0Al 3+ + Λ0

2-

SO4

12. (3) The cathode is the electrode at which reduction (electron gain) occurs. The anode is the electrode at which oxidation (electron loss) occurs.

EMF of a cell = Reduction potential of cathode− Reduction potential of anode



= Reduction potential of cathode + Oxidation potential of anode

= Oxidation potential of anode − Oxidation potential of cathode

Chapter 10_Electrochemistry.indd 258

o E o = 0.50 V ; DG 2o = - n 2FE Cu + /Cu

DG 3o = DG1o + DG 2o

  (As a X > Y. 17. (4) The reactions involved are

At the cathode: 2 ´ (Fe3+ + e - ® Fe2+ ); E o = 0.77 V



At the anode: 2I - → I 2 + 2e - ; E o = 0.536 V



Overall reaction: 2Fe3+ + 21- → 2Fe2 + I 2



o o - E anode Electrode potential is given by E o = E cathode



   = 0.77 – 0.536



   = 0.234 V



Since, E o is positive, therefore, DG = - nFE o is negative, hence, the reaction is spontaneous or feasible.

18. (4) Using Kohlrausch’s law, we have L0m ( NH 4OH ) = L0m ( NH 4Cl ) + L0m ( NaOH ) - L0m ( NaCl ) 19. (1) Lower the reduction potential, greater is the reducing ability and higher is the reduction potential greater is the oxidizing property. Therefore, looking at the reactions, we see that F2 is the strongest oxidizing agent while I- is the strongest reducing agent. 20. (4) Using Kohlrausch’s law

L0m ( CH 3COOH ) = L0m ( CH 3COONa ) + L0m ( HCl ) - L0m ( NaCl ) = 91 + 425.9 - 126.4

= 390.5 Scm 2 mol -1

1/4/2018 5:16:03 PM

Electrochemistry

21. (2) The reaction involved is





-10

As, pH = 10; [H ] = 10

M



E cell = -

( ) -10

0.059 10 log 1/2 1 (1)

= 0.59 V 22. (3) Degree of ionization =

        

=

23. (1) We know o o o E cell = E cathode - E anode

Therefore,    pH2 = (10-7 ) = 10-14 atm 2

28. (2) Given that I = 1 A; t = 60 s; e = 1.6 × 10−19 C From Faraday’s first law of electrolysis,

Q = ne or It = ne

Λm × 100 Λ∞m 9.54 × 100 = 4.008% 238

[H + ]2 =1 pH 2

Thus

0.059 [H + ] log 1/2 n pH2

o E cell = E cell -

+

1 H2 (g ) ® H+ + e 2

where Q is the charge or quantity of electricity passed through the electrolyte and n is the number of electrons. Substituting the values, we get n=

I ´t 1 ´ 60 = = 3.75 ´ 1020 e 1.6 ´ 10-19

29. (1) Given that: C = 0.5 mol dm-3; k = 5.76 × 10−3 S cm−1 and T = 298 K.

= 0.34 - ( -0.76 )

We know that, molar conductivity is given by

(

)

L m Scm 2mol -1 =

= 1.1 V 24. (3) We have, MnO 24- → MnO -4 + e

1 mol of MnO4 2- requires 1F conversion to MnO4-



1 F = 96500 C



Thus, 0.1 mol requires 0.1 ´ 96500 C = 9650 C

=

WAg

1 4 ´ = WO2 ´ M O2 108

WO2

= nO2 =



Given,



Therefore,  WAg ×

)

-1

5.76 ´ 10 -3 = 11.52 S cm 2 mol -1 0.5

M O2

At the anode: 2 Cl - → Cl 2 (g ) + 2e -



At the cathode: Na + e - → Na



From first law of electrolysis, we have

5600 1 = 22400 4

1 1 = × 4 ⇒ W Ag = 108 g 108 4

27. (1) The half reaction for hydrogen electrode is 2H + ( aq ) + 2e -  H 2 ( g ) At 25°C, [H+] = 10–7 M, then using Nernst equation 0.0591 [H + ]2 Eo = log pH 2 2 o For hydrogen electrode, Pt, H2(g)/H+, E H2 /H+ = 0.0 V

Chapter 10_Electrochemistry.indd 259

) ( M (mol L )



W =

M I ×t n×F

Substituting values as F = 96500 C, n = 1, I = 3 A and W = 0.1 × 71 g, we get

26. (4) Devices that convert the chemical energy of fuel directly into electrical energy through catalytically redox reactions are known as fuel cells.



(

k Scm -1 ´ 1000 cm 3L-1

30. (1) On electrolysis of brine solution, the following reactions take place

25. (4) According to Faraday’s law of electrolysis



259

35.5 ´3´t 96500 t = 6433.38 s = 107.22 min  108 min 0.1 ´ 71 =



31. (2) The overall cell reaction is Zn(s) + Cu 2+ (aq) ® Zn 2+ (aq) + Cu(s)

According to Nernst equation o E cell = E cell -

2.303 RT log Q nF







For the cell: Zn | ZnSO4(0.01 M)|| CuSO4(1 M)|Cu

1/4/2018 5:16:07 PM

260

OBJECTIVE CHEMISTRY FOR NEET 0.059 [ Zn 2+ ] log 2 [Cu 2+ ] 0.01 0.059 o log = E cell 2 1 o = E cell + 0.059 log 2 (1)







For the cell: Zn | ZnSO4(1 M)|| CuSO4(0.01 M) | Cu

Chapter 10_Electrochemistry.indd 260

0.059 [ Zn 2+ ] log 2 [Cu 2+ ] 0.059 1 log 2 0.01

o E 2 = E cell -

o E1 = E cell -

o = E cell

o = E cell - 0.059 log 2







From Eq. (1) and Eq. (2), we get





(2)

E1 > E2

1/4/2018 5:16:07 PM

11

Chemical Kinetics

Chapter at a Glance 1. Chemical kinetics deals with the study of the rate of chemical reactions and with the elucidation of the mechanisms by which they proceed. 2. Rate of a Chemical Reaction For a given chemical change, rate of reaction is the speed with which the reactants disappear and the products form. For a general reaction A → B, the rate of reaction can be mathematically expressed as: Decrease in concentration of A

∆A Time ∆t Increase in concentration of B ∆B =+  Rate = Time ∆t Units of rate of reaction: The unit of rate of reaction is mol L−1 s−1. Rate =

=-

(a)  Average rate: The rate of reaction can be measured as rate of disappearance of reactant A or rate of formation of product B. It can be expressed mathematically as Decrease in concentration of A ∆A =     Average rate of reaction(ravg. ) = Time ∆t Increase in concentration of B ∆B Average rate of reaction(ravg. ) = =+ or Time ∆t (b) Instantaneous rate: For the reaction A → B, the instantaneous rate is - ∆A ∆B =    ravg = ∆t ∆t - d A dB = As ∆t → 0, ravg → rinst       rinst = dt dt 3. Dependence of Relative Rates of Reaction on Coefficients in the Equation C3 H8 (g ) + 5O2 (g ) → 3CO2 (g) + 4H2 O(g )

 

 

Rate of reaction = -

D[C3 H8 ] 1 D[O2 ] 1 D[CO2 ] 1 D[H2 O] == = Dt 5 Dt 3 Dt 4 Dt

4. Factors Affecting the Rate of Reactions The rate of reaction depends upon the probability of effective collision. (a) Nature of reactant: This is the primary factor that determines the rate of reaction. The nature and the strength of the bonds in the reactants significantly affect the rate of its reaction. Surface area also affects the rate of reaction. (b) Concentration of reactant: With increase in concentration of reactant, probability of effective collision increases, thus the rate of reaction also increases. (c) Pressure: With increase in pressure or (decrease in volume) the reactant particles come close to each other and they collide with a greater force. Hence the probability of effective collision increases and the rate of reaction also increases.

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(d) Temperature: Reaction rates are normally favored by increase of temperature. (e) Catalyst: A catalyst does not undergo any change in the reaction but provides a reaction path for which the energy of activation is lower. This results in more molecules in the system colliding and hence increase in the rate of reaction. 5. Rate Law The mathematical expression that relates the rate of reaction to the concentration of either reactants or products is known as the rate law. For a general reaction, aA + bB → Products



The rate of the reaction is expressed as

d[A] = k[ A ]x [B]y dt where the exponents x and y may or may not be equal to the stoichiometric coefficient (a and b of the reactants). This is known as differential rate equation and k is the constant of proportionality, known as the rate constant.

6. Elementary and Complex Reactions (a) T  he reactions that proceed in a single step are known as elementary reactions. Elementary reactions can be two types: (i) If only one type of reactant is present, that is, nA → Products



(ii) If more than one type of reactants is present, then usually coefficient of all reactants should be equal to one (unless otherwise mentioned), then such reactions are elementary. (b) Most of the reactions involve more than one step, that is more than one elementary reaction, and are known as complex reactions. The complex reactions can be of the following types. (i) Consecutive reactions: These are reactions taking place in a series of steps, represented as A1 → A 2 → A 3



(ii) Reverse reactions: These occur in forward and backward steps and are represented as

AB (iii) Parallel reactions: In these reactions, some side reactions take place leading to the formation of some by products along with the main product. A B C

7. Order of Reaction It is the sum of power raised on the concentration terms in order to write rate expression. It is an experimental concept.

aA + bB → cC + dD



Rate = k[ A ]x [B]y Order of reaction due to A is x and to B is y. The overall order is x + y. (a) Order of reaction can be zero, integer or fractional. (b) The value of order of reaction can be maximum limited to three.

8. Molecularity of Reaction It is defined as the total number of molecules of reactants taking part in an elementary reaction. (a) Molecularity can neither be zero nor fractional. (b) Reactions having molecularity more than or equal to three are very rare. (c) It is not an experimental concept. 9. Differential Rate Laws A rate law is a mathematical equation that describes the progress of reaction. Differential rate laws can take on different forms, especially for complicated chemical reactions. The form of differential rate law depends on the order of reaction.

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(a) For zero order reactions, differential rate law r = k (b) For first order reactions, differential rate law r = k[A] (c) For second order reactions, differential rate law r = k[A]2 10. Integrated Rate Laws (a) For zero-order reactions

   kt = [A]0 - [A]

where [A]0 or [A0] is the initial concentration of reactant and [A] is concentration after time t. (b) For first order reactions A → Products For general reaction of the type: [A] kt = ln 0 [A] 2.303 a k= log t a−x

where a = initial concentration of reactant; a−x = remaining concentration after time t (i) Examples of first order reactions • Inversion of sugar cane

+

C12 H22 O11 + H2 O H→ C 6 H12 O6 + C 6 H12 O6

• Decomposition of H2O2 H2 O 2 → H2 O +



• Decomposition of N2O5

1 O2 2

N 2 O5 → 4 NO2 + O2 11. Rate Constant Units The unit of rate constant depends upon the order of reaction æ mole ö Unit of rate constant= ç è liter ÷ø



1-n

´

1 time unit

where, n is order of reaction. (a) Zero order reaction: Here n = 0, 1  dx  k=  × = mol L−1 time −1  dt  [Concentration]0 (b) First order reaction: Here n = 1, mol L−1 1 1  dx  k = × = × = time         1 [Concentration] dt time mol L−1 12. Graphical Representation of Integrated Rate Laws [A]0

In [A]0

Slope = − k

[A ]

Time (t ) (a) Zero order

Slope = − k

In [A]

Time (t ) (b) First order

13. Half-Life of Reactions It is the time needed to reduce the reactant concentration to half of its initial value. The equations for half-lives depends on the order of reaction.

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(b) For first-order reaction:     Half-life of nth order reaction:

[ A ]0 2k ln 2 0.693 = = k k

t1/2 =

(a) For zero-order reaction:

t1/2 µ

 

t1/2

1 [ A ]n0 -1

where [A]0 is the initial concentration and n is the order of reaction. 14. Pseudo Unimolecular Reactions There are certain reactions whose molecularity is more than one, but follow first order kinetics due to presence of one of the reactants in excess. For example hydrolysis of ester in acidic medium. CH3 COOC 2 H5 + H2 O → CH3 COOH + C 2 H5OH



Here a large excess of water is used and the rate law can be written as

Rate = k[CH3 COOH][H2 O] = k′[CH3 COOH]

15. Collision Theory (a)  According to this theory, in a bimolecular reaction, two molecules should collide effectively and acquire sufficient energy on collision to bring about the reaction. (b)  An effective collision is one in which molecules collide with sufficient kinetic energy and proper orientation in a manner that the bonds between reacting molecules are broken and new bonds are formed to yield the products. A

A

B

B

A

A (Orientation)

B

B

(c)  The rate of a reaction is proportional to the number of effective collisions per second among the reactant molecules. (d)  The number of collisions per second per unit volume of the reaction mixture is known as collision frequency (Z). For a bimolecular reaction A + B → Products, the rate of reaction can be expressed as Rate = Z AB e − Ea / RT    where ZAB represents the collision frequency of reactant, A and B and e − Ea / RT represents the fraction of molecules with energy equal to greater than Ea (energy of activation). (e) To account for effective collisions, another factor P, called the probability or steric factor is introduced. It takes into account the fact that in a collision, molecules must be properly oriented, that is,

− E / RT    Rate = PZ AB e a

16. Arrhenius Equation Arrhenius proposed an equation that gives the relation between temperature and rate constant for a reaction. − E a / RT    k = Ae Here, k is the rate constant for reaction, R is ideal gas constant, T is temperature in Kelvin and A is known as Arrhenius factor or frequency factor. It is also called exponential factor and is specific for the reaction. E The equation can be written as     ln k = ln A − a RT If k1 and k2 are the rate constant values at two temperatures T1 and T2, respectively, then  

Ea  1 k 1 log 2 = or        −  2.303 T1 T2  k1  

Arrhenius plot of ln k vs 1/T is a straight line.

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265

Intercept = log A

log K Slope, tanq = −

Ea 2.303 R

1/T

Solved Examples 1. Consider a gaseous reaction, the rate of which is given by k[A][B], the volume of the reaction vessel containing these gases is suddenly reduced to 1/4 th of the initial volume. The rate of reaction relative to the original rate would be (1) 16/1 (2) 1/16 (3) 8/1 (4) 1/8 Solution (1) By reducing volume to 1/4 the concentration will become 4 times hence rate 16 times.

4. When concentration of reactant in reaction A → B is increased by 8 times, the rate increases only 2 times. The order of reaction would be 1 (1) 2 (2) 3 1 (3) 4 (4) 2 Solution (2) We know         Rate = k[ A ]n (1)

th

Solution



Dividing Eq. (2) by Eq. (1), we get

(2)

2 = 8n ⇒ 21 = 23n ⇒ n =

1 3

5. For the following first order reaction A → Products, which one of the following is correct plot of log[A] versus time? (1) (2) log[A]

(1) the rate of disappearance of B is one fourth of the rate of disappearance of A. (2) the rate of formation of C is one-half of the rate of consumption of A. (3) the rate of appearance of D is half of the rate of disappearance of B. (4) the rate of formation of C and D are equal.

Given that

log[A]

2. For the reaction 4A + B → 2C + 2D, which of the following statement is not correct?

2(Rate) = k[8 A ]n 



(3) The rate of appearance of D is double the rate of disappearance of B. t

Solution (1)    Rate ∝ concentration Rate = k concentration k=

Rate 4 ´ 10 -6 mol L-1s -1 = 1/2 ( 4 ´ 10 -4 )1/2 mol L-1 (Concentration)

-6          = 4 ´ 10 = 2 ´ 10 -4 mol1/2 L-1/2s -1 2 ´ 10 -2

Chapter 11_Chemical Kinetics.indd 265

t

(4) log[A]

(1) 2 × 10–4 mol1/2 L–1/2 s–1 (2) 1 × 10–2 s–1 (3) 2 × 10–4 mol–1/2L1/2 s–1 (4) 25 mol–1 L min–1

(3) log[A]

3. The rate of a reaction increases four fold when concentration of reactant is increased 16 times. If the rate of reaction is 4 × 10–6 mol L–1 s–1 when concentration of the reactant is 4 × 10–4 mol L–1, the rate constant of the reaction will be

t

t

Solution (2) Plot of log [A] vs time is linear with negative slope. 6. A first order reaction is 87.5% complete in an hour. The rate constant of the reaction is (1) 0.0346 min–1 (2) 0.0693 h–1 (3) 0.0693 min–1 (4) 0.0346 h–1

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Solution (1) 87.5% change takes place in three half lives. Hence half1 60 time of the reaction will be h= = 20 min, since 3 3 half-time of a first order reaction is a constant inde-

10. A tangent drawn on the curve obtained by plotting concentration of product (mol L–1) of a first order reaction vs time (min) at the point corresponding to time 20 min makes an angle to 30° with concentration axis. Hence the rate of formations of product after 20 min will be (1) 0.580 mol L–1 min–1 (2) 1.732 mol L–1 min–1 (3) 0.290 mol L–1 min–1 (4) 0.866 mol L–1 min–1

pendent of initial concentration. min min min 100% 20 → 50% 20 → 25% 20 → 12.5%

(unreacted)

0.693 0.693 k= = = 0.0346 min −1 t1/2 20 7. The half-life of a first order reaction is 24 h. If we start with 10 M initial concentration of the reactant then concentration after 96 h will be (1) 6.25 M (2) 1.25 M (3) 0.125 M (4) 0.625 M

Solution (2) Tangent makes an angle of 30° with concentration axis so it must make an angle of 60° with the time axis (–ve direction). The slope of the tangent will be tan 60° i.e., 1.732. 11. For a first order reaction with half-life of 150 s, the time taken for the concentration of the reactant to fall from M/10 to M/100 will be approximately

Solution

96 (4) Number of half-lives = =4 24

Concentration remaining after 96 h =

(1) 1500 s (2) 498 s (3) 900 s (4) 598 s 10 M = 0.625 M ( 2)4

Solution k=

0.693 −1 0.693 = 0.00462 s −1 s = t 150

Also,  t =

2.303 (1/10) = 498.48 s log 0.00462 (1/100)

(2) We have

8. In the formation of sulphur trioxide by contact process, 2SO2 + O2 ® 2SO3, the rate of reaction was measured

−d [O 2 ] = 2.5 × 10 −4 mol L−1 s −1. The rate of reaction dt expressed in terms of SO3 will be as

(1) -5 ´ 10 -4 mol L-1 s -1 (2) -1.25 ´ 10 -4 mol L-1 s -1



12. For reaction 3A → Products , it is found that the rate of reaction increases four fold when concentration of A is increased 16 times keeping the temperature constant. The order of reaction is (1) 2 (2) 1 (3) 1 (4) 0.5

(3) 2.5 ´ 10 -4 mol L-1 s -1 (4) 5.0 ´ 10

-4

-1

mol L s

-1

Solution

Solution (4) From the reaction, we have

−

d [O2 ] 1 d [SO 3 ] = dt 2 dt

d [SO 3 ] = 5.0 ´ 10 -4 mol L-1 s -1 dt 9. A first order reaction is half-completed in 45 min. How long does it need for 99.9% of the reaction to be completed? (1) 20 h (2) 10 h (3) 7.5 h (4) 5 h Solution

(4) The rate data suggests, the rate law as follows





Therefore, the order of the reaction is 0.5.

13. If t1/2 does not change with initial concentration, then order of the reaction is (1) zero. (2) second. (3) first. (4) third. Solution (3) We have

(3) We have 0.693 min −1 = 0.0154 min −1 45 a 2.303 k= log t a−x a 2.303 log       t = a − 0.999 a 0.0154 k=

=

Chapter 11_Chemical Kinetics.indd 266

2.303 log 103 = 448.6 min or 7.48 h 0.0154

Rate ∝ A or Rate ∝( A )1/2

t1/2 ∝

1 a n−1

Therefore, for first order reaction t1/2 is independent of initial concentration.

14. If initial concentration is reduced to 1/4th in a zero order reaction, the time taken for half the reaction to complete (1) remains same. (2) becomes 4 times. (3) becomes one-fourth. (4) doubles.

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Chemical Kinetics Solution

(2) increases the activation energy. (3) alters the reaction mechanism. (4) increases the frequency of collisions of reacting species.

[A] (3) For zero order reactions, t1/2 = 0 2k

When [A]0 is reduced to 1/4th, t1/2 will become 1/4th.

15. The half- life of a reaction is halved as the initial concentration of the reactant is doubled. The order of reaction is (1) 0.5 (2) 1 (3) 2 (4) 0 Solution (3) We know

t1/2 µ

Solution

n −1 n −1

⇒2=2

⇒n = 2

(2) We have

16. Two substance A and B are present [A] = 4[B]. The halflife of A is 5 min and that of B is 15 min. If they start decaying at the same time following first order kinetics how much time later the concentration of both of them would be same?

Solution  1 (1) Amount of A in n1 halves  =    2

n1

æ 1ö Amount of B in n2 halves  = ç ÷ è 2ø

n2



Since,    [ A ] = 4 [B]



2n1 Therefore,   n2 = 2n1 −n2 ⇒ (n1 − n2 ) = 2 2



Now

Substituting Ea ′ = 10 kJ mol-1, T1 = 300 K, Ea = 20 kJ mol-1 in Eq. (1), we get





[B ]

Solution

0

(1)

(2) We know

k = Ae -Ea /RT



T → ∞, k = A = 6 × 1014

n1 = 3 (2) n2



From Eq. (1) and Eq. (2), we get







Therefore,

 

 

n1 = 3, n2 = 1 t = n1 × t1/2(1) = 3 × 5 = 15 min

(1) (2) (3) (4)

Chapter 11_Chemical Kinetics.indd 267

is negative of Ea. is always less than Ea. can be less than or more than Ea. is always double of Ea.

Solution (3) We know

∆H = E a (forward ) − E a (reverse)



For an exothermic reaction, DH is negative, therefore, Ea(forward) < Ea(reverse)



For an endothermic reaction, DH is positive, therefore, Ea(forward) > Ea(reverse)



Hence, the activation energy for reverse reaction can be less than or more than Ea(forward).

17. A catalyst (1) increases the average kinetic energy of reacting molecules.

when

20. The activation energy for a simple chemical reaction A → B is Ea in forward direction. The activation energy for reverse reaction

t = n1 × t1/2(1) = n2 × t1/2( 2 )



10 20 = Þ T2 = 600 K or 327°C 300 T2

(1) 2 × 1018 s–1 (2) 6 × 1014 (3) a (4) 3.6 × 1030 s–1

0

     n1 × 5 =1 n2 × 15

E a¢ E a = (1) T1 T2



[A ]

             n2 = (n1 − 2)

 

19. The rate constant, the activation energy and Arrhenius parameter of a chemical reaction at 25°C are 3 × 10–4 s–1, 104.4 kJ mol–1 and 6 × 1014 s–1 respectively. The value of the rate constant at T → ∞ is

(1) 15 min (2) 10 min (3) 5 min (4) 12 min



(3) A catalyst alters the reaction mechanism.

(1) –123°C (2) 327°C (3) –23°C (4) +23°C

(t ) [ A ]n−1 Therefore, 1/2 1 = 0 n2 −1 (t1/2 )2 [ A 0 ]1

Solution

18. A catalyst lowers the activation energy of a reaction from 20 kJ mol–1 to 10 kJ mol–1. The temperature at which the unanalyzed reaction will have the same rate as that of the catalyzed at 27°C is

1 [ A 0 ]n-1

t  2a  =  t /2  a 

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21. The temperature dependence of rate constant (k) of a chemical reaction is written in terms of Arrhenius equation, k = Ae−Ea/RT. Activation energy (Ea) of the reaction can be calculated by plotting 1 (1) k vs T (2) k vs logT 1 1 (3) logk vs (4) logk vs T logT Solution



(3)

EP

E

∆H ER

k = Ae -Ea /RT

(3) We have

Solution

If we take the natural logarithm of both sides, we obtain E log k = log A - a RT



Let us rewrite the equation as







To determine the activation energy, we can make a æ1ö graph of log k vs ç ÷. The graph is straight line whose èT ø slope = −Ea/R and intercept = ln A.

 E   1 log k = log A −  a  ×    R  T 

22. The rate of chemical reaction doubles for every 10°C rise in temperature. If the temperature is increased by 60°C, the rate of reaction increased by about (1) 20 times. (2) 32 times. (3) 64 times. (4) 128 times.

Reaction coordinate

24. The rate of reaction is doubled for every 10°C rise in temperature. The increase in rate as result of increase in temperature from 10°C to 100°C is (1) 112 (2) 512 (3) 400 (4) 256 Solution (2) We have

kT +10 rT +10 = =2 kT rT

For increases of temperature from 10°C to 90°C rates will increase by (2)9 times that is, 512 times.

25. The chemical reaction 2O3 → 3O2 proceeds as following: O3  O2 + O(fast) O+O3 ® 2O2 (slow)

The rate expression should be (1) r = k1 [O3]2 (2) r = k1 [O3]2 [O2]−1 (3) r = k1 [O3] [O2] (4) unpredictable

Solution (3) 10°C increases has been made 6 times, therefore, the rate will increase 26 = 64 times.

Solution

23. For an endothermic reaction where DH represents the enthalpy of the reaction, the minimum value for the energy of activation will be



Rate = k [O][O3]



Therefore,





(1) less than DH. (2) zero. (3) more than DH. (4) equal to DH.

kC =

(2)

[O2 ][O2 ] [O 3 ]

Rate =

Þ [O] =

kC [O3 ]

[O 2 ]

k kC [O3 ][O3 ] 2 -1 = k1 [O 3 ] [O 2 ] [O 2 ]

Practice Exercises Level I Rate of the Reaction 1. For a hypothetical reaction A + B → C + D, the rate = k[A]1/2[B]3/2. On doubling the concentration of A and B the rate will be (1) 4 times. (2) 2 times. (3) 3 times. (4) none of these. 2. Rate of formation of SO3 in the following reaction 2SO 2 + O2 → 2SO 3 is 100 kg min–1. Hence rate of disappearance of SO2 will be

Chapter 11_Chemical Kinetics.indd 268

(1) 100 kg min–1 (2) 80 kg min–1 (3) 64 kg min–1 (4) 32 kg min–1 3. For a reaction A + 2B → C + D, the following data were obtained Initial concentration (mol L-1) [A] [B] 0.1 0.1 0.3 0.2

0.3 0.4

0.4 0.1

Initial rate of ­formation of D (mol L-1) 6.0 × 10-3 7.2 × 10-2 2.88 × 10-1 2.4 × 10-2

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Chemical Kinetics

12. Which of the following is true for order of reaction?

The correct rate law expression will be (1) rate = k[A][B] (2) rate = k[A][B] (3) rate = k[A]2[B]2 (4) rate = k[A]2[B]

(1) It is equal to the sum of exponents of the molar concentrations of the reactants in the rate equation. (2) It is always a whole number. (3) It is never zero. (4) It can be determined theoretically.

2

4. The specific rate constant for a first order reaction is 60 × 10–4 s–1. If the initial concentration of the reactant is 0.01 mol L—1, the initial rate is (1) 60 × 10–6 mol s–1 (2) 36 × 10–4 mol s–1 (3) 60 × 10–2 mol s–1 (4) 36 × 10–1 mol s–1 5. In the reaction 2A + B → Products, the order w.r.t. A is found to be one and w.r.t. B is equal to 2. Concentration of A is doubled and that of B is halved, the rate of reaction will be (1) doubled. (2) halved. (3) remains unaffected. (4) four times.

Order and Molecularity 6. What is the order of a reaction which has a rate expression rate = k[A]3/2 [B]−1? (1) 3/2 (2) 1/2 (3) zero (4) None of these 7. In a particular reaction the time required to complete half of the reaction was found to increase 9 times when the initial concentration of the reactants was reduced to one third. What is order of the reaction? (1) 0 (2) 1 (3) 2 (4) 3 8. For a reaction pA + qB → Products, the rate law exp­ ression is r = k[A]m[B]n then (1) (p + q) ≠ (m + n) (2) (p + q) = (m + n) (3) (p + q) may or not be equal to (m + n) (4) (p + q) > (m + n) dx 9. For reaction A + B → Products, = k[ A ]a[B]b dt dx = k, the order is if dt (1) 4 (2) 2 (3) 1 (4) 0

13. For the reaction A + B → Products, it is found that the order of A is 2 and of B is 3 in the rate expression. When concentration of both is doubled the rate will increases by (1) 10 (2) 6 (3) 32 (4) 16

Integrated Rate Law and Half-Life (t1/2) 14. The half-life a first order reaction is 24 h. If we start with 10 M initial concentration of the reactant then concentration after 96 h will be (1) 6.25 M (2) 1.25 M (3) 0.125 M (4) 0.625 M 15. During a particular reaction 20% of the reactant decomposes in 1 h, 40% in 2 h, 60% in 3 h and so on. The unit of rate constant is (1) h–1 (2) L mol–1 h–1 (3) mol L–1 h–1 (4) mol h–1 16. Rate of the chemical reaction nA → Products, is doubled when the concentration of A increased four times. If the half-time of the reaction at any temperature is 16 min then time required for 75% of the reaction to complete is (1) 24 min (2) 27.3 min (3) 48 min (4) 49.4 min 17. If a is the initial concentration and k is the rate constant of a zero order reaction, the time for the reaction to go to completion will be (1) k/a (2) a/k (3) a/2k (4) k/2a d[A] = k and at differ 18. For the reaction A → Products, − dt ent time interval, [A] values are

10. Which of the following rate law has an overall order of 0.5 for reaction involving substances X, Y and Z? (1) (2) (3) (4)

rate = k(CX) (CY ) (CZ) rate = k (CX)0.5 (CY )0.5 (CZ)0.5 rate = k (CX)1.5 (CY )−1 (CZ)0 rate = k (CX) (CZ)0.5/(CY )2 d[ A ] d[B] =2 1. If for the reaction nA → B, − 1 then rate law dt dt is (if the reaction is elementary) d[A] d[B] = k[ A ]2 (2) = k[ A ] dt dt d[B] d[A] = k[B]2 (3) = k[ A ]2 (4) − dt dt (1) −

Chapter 11_Chemical Kinetics.indd 269

269



Time

5 min

10 min

15 min

20 min

[A]

20 mol

18 mol

16 mol

14 mol

At 20 min, rate will be (1) 12 mol min−1 (2) 10 mol min−1 (3) 8 mol min−1 (4) 0.4 mol min−1

19. The rate constant of a zero order reaction is 0.2 mol dm−3 h−1. If the concentration of the reactant after 30 min is 0.05 mol dm−3, then its initial concentration would be (1) 0.01 mol dm −3 (2) 0.15 mol dm −3 (3) 0.25 mol dm −3 (4) 4.00 mol dm −3

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20. A reaction of first-order completed 90% in 90 min, hence, it is completed 50% in approximately (1) 50 min. (2) 54 min. (3) 27 min. (4) 623 min.

(1) 3.75 × 10−4 s−1 (2) 2.5 × 10−4 s−1 (3) 1.5 × 10−3 s−1 (4) 3 × 104 s−1 29. For a first order reaction t3/4 is 1200 s. The specific rate constant is (in s−1)

21. If half-life period is 100 years for a first order reaction, average life is nearly (1) 100 years. (2) 70 years. (3) 144 years. (4) 90 years. d[A] = k[ A ]. dt 1 At a time when t = concentration of the reactant is (C0 k is initial concentration)

22. For a reaction 2A + B → Products, rate law is −

(1) (3)

(1) 10−3 (2) 10−2 (3) 10−9 (4) 10−5

Methods to Find Order of the Reaction 30. For the reaction, A + B → C

C0 (2) C 0e e C0 1 (4) C0 e2

23. A first order reaction is half-completed in 45 min. How long does it need for 99.9% of the reaction to be completed? (1) 20 h (2) 10 h (3) 7.5 h (4) 5 h 24. Concentration of the reactant in first-order is reduced to 1 1 of its original value after (Natural life = ) e2 k (1) one natural life-time. (2) two natural life-time. (3) three natural life-time. (4) four natural life-time. 25. For the first order reaction, half-life is 14 s. The time required for the initial concentration to reduce to 1/8th of its value is (1) 28 s (2) 42 s (3) (14)2 s (4) 14 s 26. For a reaction, graph between logt50 (y-axis) and log a (x-axis) is a straight line parallel to x-axis. Hence, order is (1) 0 (2) 1 (3) 2 (4) 3



d[ A ] dt

[A]

[B]

1.

1.0 M

1.0 M

0.25 M min−1

2.

2.0 M

1.0 M

0.50 M min−1

3.

1.0 M

2.0 M

0.25 M min−1

-

Hence, rate law is (1) k[A][B] (2) k[A][B]1/2 (3) k[A] (4) k[B]

31. For a first order reaction the plot of log [A]t vs t is linear with a (1) (2) (3) (4)

positive slope and zero intercept. positive slope and non-zero intercept. negative slope and zero intercept. negative slope and non-zero intercept.

32. For a zero order reaction, the plot of concentration vs time is linear with (1) (2) (3) (4)

positive slope and zero intercept. negative slope and zero intercept. positive slope and non-zero intercept. negative slope and non-zero intercept.

33. For the reaction, of the first order, variation of log t50 with log a (where t50 is half-life period and a is the initial concentration) is given by (1) log t50

(2) 45°

log t50

27. The rate constant for the reaction

2N 2O5 ® 4NO2 + O2 is 2 ´ 10 -5 s -1. If the rate is 1.2 × 10-5 mol L-1 s-1, then concentration of N2O5 in mol L−1 is

(3)

(1) 1.4 (2) 1.2 (3) 0.04 (4) 0.6

log t50

28. The rate of the first order reaction X → Products is 7.5 × 10−4 mol L−1 s−1 when the concentration of X is 0.5 mol L−1. The rate constant is

Chapter 11_Chemical Kinetics.indd 270

log a

(4) log t50

log a

45° log a 45°

log a

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Chemical Kinetics

Replacement of Concentration Terms by Other Variables 34. The reaction A(g) + 2B(g) → C(g) + D(g) is an elementary process. In an experiment, the initial partial pressure of A and B are pA = 0.40 and pB = 0.60 atm. When pC = 0.2 atm and the rate of reaction relative to the initial rate is

(1) k = Ae − Ea /RT (2) A (3) e − Ea /RT (4) None of these. 41. In the reaction, rate at a given temperature becomes slower, that means (1) the energy of activation is higher. (2) the energy of activation is lower. (3) the entropy changes. (4) the initial concentration of the reactants remains constant.

(1) 1/48 (2) 1/24 (3) 9/16 (4) 1/18 35. At 373 K, a gaseous reaction A → 2B + C is found to be of first order. Starting with pure A, the total pressure at the end of 10 min was 176 mm and after a long time when A was completely dissociated, it was 270 mm. The pressure of A at the end of 10 min was (1) 94 mm (2) 47 mm (3) 43 mm (4) 90 mm 36. For reaction A (g) → B (g) + C (g),

42. The plot of log k versus 1/T is linear with a slope of (1) Ea/R (2) −Ea/R (3) Ea/2.303 R (4) −Ea/2.303 R 43. The chemical reactions in which reactants require high amount of activation energy are generally (1) slow. (2) fast. (3) instantaneous. (4) spontaneous.

d[ A ] = k[ A ]. dt

At the start pressure is 100 mm and after 10 min, pressure is 120 mm hence rate constant (min−1) is 1 a (1) ln t a−x a 2 2.303 100 (3) log t1/2 80

Mechanism of Complex Reactions 44. Consider the hypothetical reaction A + 2B →AB2. It follows rate law, rate = k[A][B]

Hence, rate determining step is

(2)

(4)

0.693 k

271

(1) A + B → AB (2) A + 2B → AB2 (3) AB + B → AB2 (4) AB + A → A2 +B 45. A two step mechanism has been suggested for the reaction of nitric oxide and bromine:



NO(g)+Br2(g )  NOBr2(g )

Arrhenius Equation and Collision Theory





NOBr2(g )+ NO(g) → 2NOBr(g)

37. The rate constant of a reaction at temperature 200 K is 10 times less than the rate constant at 400 K. What is the activation energy (Ea) of the reaction?



Observed rate law is, rate = k[NO]2[Br2]. Hence, rate determining step is (1) NO(g) + Br2(g) → NOBr2(g) (2) NOBr2 (g) + NO (g) → 2NOBr(g) (3) 2NO(g) + Br2(g) → 2NOBr(g) (4) None of these

(1) 1842.4 R (2) 921.2 R (3) 460.6 R (4) 230.3 R

39. Which one of the following statements is incorrect? (1) The temperature coefficient of a reaction is the ratio of the rate constant at any two temperatures. (2) The temperature coefficient of a reaction is the ratio of the rate constant at 298 K and 308 K. (3) The temperature coefficient of most of the reactions lies between 2 and 3. (4) In an endothermic reaction, activation energy of reactants is more than that of the products. 40. In Arrhenius equation, the fraction of effective collision is given by

Chapter 11_Chemical Kinetics.indd 271

46. A reaction takes place by the following mechanism A + BC → AC + B AC + D → A + CD







The potential energy profile for this is shown below Potential energy

38. For a reaction, rate constant k = 1.2 × 103 mol–1 L s–1 and Ea = 2.0 × 10² kJ mol–1. When T → ∞, the value of A is (1) 2.0 × 10² kJ mol–1 (2) 1.2 × 103 mol–1 L s–1 (3) 1.2 × 103 mol L–1 s–1 (4) 2.4 × 103 kJ mol–1 s–1

II IV III I

V Reaction progress



Transition states are shown by (1) I, V (2) II, IV (3) II, III, IV (4) III only

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272

OBJECTIVE CHEMISTRY FOR NEET

Level II



(1) zero. (2) one. (3) two. (4) three.

Rate of the Reaction 1. In the reaction of A + 2B → C + 2D, the initial rate, d[A]/dt at t = 0 was found to be 2.6 × 10−2 mol s−1. What is the value of –d[B]/dt at t = 0 in mol s−1? (1) 2.6 × 10−2 (2) 5.2 × 10−2 (3) 1.0 × 10−1 (4) 6.5 × 10−3

then, x, y and z are

(1) L2 mol−2 s−1 (2) mol L−1 s−1 −1 −1 (3) L mol s (4) s−1

d[ A ] d[B] d[C] =− = 1.5 dt dt dt

3. If concentration is measured in mol L−1 and time in minutes, the unit for the rate constant of n th order reaction is (1) mol (3) mol

n −1

L

1−n

L

−1

n- 1

min (2) L −1

mol

1- n

min

-1

min (4) min

−1

4. In a reaction 2A → Products; the concentration of A decreases from 0.5 mol L–1 to 0.4 mol L–1 in 10 min. The rate of the reaction during this interval is (1) 0.05 M min–1 (2) 0.005 M min–1 (3) 0.5 M min–1 (4) 5 M min–1 5. For the reaction 2NH 3 → N 2 + 3H 2 ,

d[NH 3 ] d[N 2 ] d[H 2 ] = k1[NH 3 ], − = k2[NH 3 ], = k3[NH 3 ] dt dt dt Then relation between k1 , k2 and k3 is −

(1) 1.5k1 = 3k2 = k3 (2) 2k1 = k2 = 3k3 (3) k1 = k2 = k3 (4) k1 = 3k2 = 2k3

Order and Molecularity 6. For the reaction A → B, when concentration of A is made 1.5 times, the rate of reaction becomes 1.837 times. The order of reaction is (1) 1 (2) 1.5 (3) 2 (4) 2.5 7. The rate of reaction 2X + Y → Products is given by d[ Y ] 2 = k [ X ] [ Y ], if X is present in large excess, the order dt of the reaction is (1) zero. (2) two. (3) one. (4) three.  dx  8. The rate of a reaction at different times  −  is found  dt  as follows Time (in min)

Rate (in mol L−1 s−1)

0

2.80 × 10-2

10

2.78 × 10-2

20

2.81 × 10-2

30

2.79 × 10-2

Chapter 11_Chemical Kinetics.indd 272

10. The rate constant of a reaction is 10.8 ´ 10 -5 mol dm−3 s−1. The order of the reactions is (1) zero. (2) 1 (3) 2 (4) 3

(1) 1, 1, 1 (2) 3, 2, 3 (3) 3, 3, 2 (4) 2, 2, 3

−(n − 2)

9. The rate of reaction Cl3CCHO + NO → CHCl3 + NO + CO is given by equation, Rate = k[Cl3CCHO][NO]. If concentration is expressed in mol L−1, the units of k is

Integrated Rate Law and Half-Life (t1/2) 11. A substance A decomposes in solution following the first order kinetics. Flask I contains 1 L of 1 M solution of A and flask II contains 100 mL of 0.6 M solution. After 8 h, the concentration of A in flask I becomes 0.25 M. What will be time for concentration of A in flask II to become 0.3 M? (1) (2) (3) (4)

0.4 h 2.4 h 4h unpredictable as rate constant is not given.

12. For a reaction, it is observed that t15/16 = 4 × t1/2 . What is the order of the reaction? (1) 0 (2) 1 (3) 2 (4) 3 13. The accompanying figure depicts the change in concentration of species A and B for the reaction A → B, as a function of time the point of intersection of the two curves represents (1) t1/2 (2) t3/4 (3) t2/3 (4) data insufficient to predict. Concentration

2. For the reaction, xA +yB → zC if −

n−2

The order of reaction is

Time

14. Select the correct statements out of I, II and III for zero order reaction. (I) Quantity of the products formed is directly proportional to time. (II) Larger the initial concentration of the reactant, greater the half-life period. (III) If 50% reaction takes place in 100 min, 75% reaction takes place in 150 min. (1) (I) only (2) (I) and (II) only (3) (II) and (III) only (4) (I), (II) and (III)

1/4/2018 5:16:22 PM

Chemical Kinetics 15. The half-life period of a first order reaction is 10 min. If initial amount is 0.08 mol L−1 and concentration at some instant is 0.01 mol L−1, then t is equal to

(II) if concentration of B is halved, concentration of A and C constant, the rate of reaction remains unaffected. (III) if concentration of C is made 1.5 times, the rate of reaction becomes 2.25 times.

(1) 10 min (2) 30 min (3) 20 min (4) 40 min 16. If a is the initial concentration of the reactant and (a − x) is the concentration at time t for the first-order reaction (rate constant k). Then which of the following statements are correct. (I) x = a(1 − e k1t ) y t æ 1ö (II) (a - x ) = a ç ÷ where y = è 2ø t 50



22. For the reaction, 2N2O5(g) → 4NO2(g) + O2(g) which of the following graph would yield a straight line (take first order) (1) log pN 2 O5 vs time with negative slope. (2) log pN 2 O5 vs time with positive slope. (3) ( pN 2 O5 )−1 vs time.

(1) I, II (2) II, III (3) I, III (4) I, II, III

(4) log pN 2 O5 vs time.

17. Which graph represents first-order reaction out of I, II and III? (I)

(II)

(III)

log (a−x)

a log a−x

t →

The order of reaction is (1) 1 (2) 2.5 (3) 3 (4) 3.5

2.303  a  (III) k1 = log   a − x  t



23. Graph between concentration x of the product and time of the reaction A → B is straight line with positive cond[A] stant slope. Hence, graph between − and time will dt be of the type (1)

t50 t →

-

a (conc.) →

d[A] dt

t

(3)

0.0693 mol L−1 0.0693 × 2.5 mol L−1 min−1 0.0693 × 5 mol L−2 min−1 0.0693 × 10 mol L−1 min−1

-

(4)

d[A] dt

-

d[A] dt

t

19. Half-lives of first order and zero order reaction are same. Ratio of rates at the start of reaction is (1) 0.693 (2) 1/0.693 (3) 2 × 0.693 (4) 2/0.693 a 0. In a first order reaction the 2 was found to be 8 after a−x 10 min. The rate constant is (2) ( 2.303 × 2 log 3) 10 (4) 10 × 2.303 × 3 log 2

Methods to Find Order of the Reaction 21. In the reaction aA + bB + cC → Products, (I) if concentration of A is doubled, keeping concentration of B and C constant, the rate of reaction becomes doubled.

Chapter 11_Chemical Kinetics.indd 273

-

t

18. Half-life period (t50) of first order reaction is 10 min. Starting with 10 mol L−1, rate after 20 min is

(1) ( 2.303 × 3 log 2) 10 (3) 10 × 2.303 × 2 log 3

(2)

d[A] dt

(1) I, II and III (2) I and II (3) II and III (4) I and III

(1) (2) (3) (4)

273

t

Replacement of Concentration Terms by Other Variables 24. Acid hydrolysis of ester is first-order reaction and rate constant is given by

k=

2.303 V − V0 log ∞ t V∞ − Vt

where V0, Vt and V∞ are the volume of standard NaOH required to neutralize acid present at a given time, if ester is 50% neutralized then (1) V∞ = Vt (2) V∞ = Vt − V0 (3) V∞ = 2Vt − V0 (4) V∞ = 2Vt + V0

Arrhenius Equation and Collision Theory 25. In a reaction, the threshold energy is equal to (1) average energy of the reactants.

1/4/2018 5:16:23 PM

274

OBJECTIVE CHEMISTRY FOR NEET (2) activation energy. (3) activation energy + average energy of the reactants. (4) activation energy − average energy of the reactants.

26. The activation energy of exothermic reaction A → B is 80 kJ. The heat of reaction is 200 kJ. The activation energy for the reaction B → A will be (1) 60 kJ (2) 120 kJ (3) 280 kJ (4) 200 kJ 27. If a reaction A + B → C is exothermic to the extent of 30 kJ mol−1 and the forward reaction has activation energy of 70 kJ mol−1, the activation energy for the reverse reaction is (1) 30 kJ mol−1 (2) 40 kJ mol−1 (3) 70 kJ mol−1 (4) 100 kJ mol−1 28. For an exothermic chemical process occurring in two step as A + B slow → X fast → AB

The progress of the reaction can be best described by (1)

(2)

X



(1) I, II, III (2) I, III (3) II, III (4) I, II

Mechanism of Complex Reactions 3 1 2 32. In the sequence reaction, A k → B k → C k → D, k3 > k2 > k1, then the rate determining step of the reaction is

(1) A → B (2) B → C (3) C → D (4) A → D 33. Slow step (rate-determining step) involves going into a transition state (in above question). It is (1) A + BC → AC + B (2) AC + D → A + CD (3) AC + B → BC + A (4) B + AD → BD + A

Previous Years’ NEET Questions 1. In a first order reaction A → B, if k is the rate constant and initial concentration of the reactant A is 0.5 M then the half-life is

X A+B

A +

AB

(3)

ln2 0.693 (2) k 0.5k

(3)

log 2 log 2 (4) k k 0.5 (AIPMT 2007)

X

2. The reaction of hydrogen and iodine monochloride is given as

AB

29. Milk turns sour at 40°C three times as faster as 0°C. Hence, Ea in cal of turning of milk sour is (1)

2.303 ´ 2 ´ 313 ´ 273 log 3 40

(2)

2.303 × 2 × 313 × 273 1 log 40 3

(3)

2.303 ´ 2 ´ 40 log 3 274 ´ 313



H 2(g ) + 2 ICl(g ) → 2 HCl(g)+I 2(g)



This reaction is of first order with respect to H2(g) and ICl(g), the following mechanisms were proposed:



Mechanism A: H 2(g ) + 2 ICl(g ) → 2 HCl(g)+I 2(g)



Mechanism B: H 2(g ) + ICl(g ) → HCl(g)+HI(g); Slow HI(g ) + ICl(g ) → HCl(g)+I 2(g); Fast

(4) None of these. 30. For a reaction A + B → C + D, ∆H = −30 kJ mol−1 the activation energy for the forward reaction is 65 kJ mol−1. The activation energy for the backward reaction is



31. Rate constant k varies with temperature as given by 2000 equation log k (min −1 ) = 5 − K. Consider following T about this equation, (I) pre-exponential factor is 10 . (II) Ea is 9.212 kcal. (III) variation of log k with 1/T is linear.

Which of the above mechanism(s) can be consistent with the given information about the reaction? (1) A only (2) B only (3) A and B both (4) Neither A nor B

(1) 35 kJ mol−1 (2) 95 kJ mol−1 (3) 85 kJ mol−1 (4) 40 kJ mol−1

(AIPMT 2007) 3. If 60% of a first order reaction was completed in 60 min, 50% of the same reaction would be completed in approximately (1) 40 min (2) 50 min (3) 45 min (4) 60 min

5

Chapter 11_Chemical Kinetics.indd 274

(1) AB

(4) All are correct

A+B

Select correct statement



(log 4 = 0.60, log 5 = 0.69) (AIPMT 2007)

1/4/2018 5:16:24 PM

Chemical Kinetics 4. The bromination of acetone that occurs in acid solution is represented by the equation



The rate of this reaction is given by (1) rate = k [A][B] (2) rate = k [A]2[B] (3) rate = k [A][B]2 (4) rate = k [A]2[B]2

CH 3COCH 3(aq ) + Br2(aq ) ® CH 3COCH 2Br(aq ) + H + (aq ) + Br - (aq )



These kinetic data were obtained for given reaction concentrations: Initial concentrations (M)

(AIPMT 2009) 8. In the reaction BrO3− (aq) + 5Br − (aq) + 6H+ → 3Br2(l) + 3H 2O (l)

+

[CH 3COCH 3 ]

[Br2 ]

[H ]

0.30

0.05

0.05

0.30

0.10

0.05

0.30

0.05

0.10

0.40

0.05

0.20

The rate of appearance of bromine (Br2) is related to rate of disappearance of bromide ions as following: d[Br2 ] 3 d[Br - ] d[Br2 ] 3 d[Br - ] = (2) =dt dt 2 dt 5 dt d[Br2 ] 5 d[Br - ] d[Br2 ] 5 d[Br - ] (3) = (4) = dt 3 dt dt 3 dt (1)

(AIPMT 2009) 9. Half-life period of a first order reaction is 1386 s. The ­specific rate constant of the reaction is

Initial rate of disappearance of Br2 (Ms −1 ) 5.7 × 10 −5

(1) 5.0 × 10−2 s−1 (3) 0.5 × 10−2 s−1

5.7 × 10 −5

(2) 5.0 × 10−3 s−1 (4) 0.5 × 10−3 s−1

1.2 × 10 −4 3.1 × 10

(AIPMT 2009)

−4

Based on these data, the rate of equation is (1) (2) (3) (4)

275

Rate = k [CH 3COCH 3 ] [Br2] [H + ] Rate = k [CH3COCH3][H+ ] Rate = k [CH3COCH3] [Br2] Rate = k [CH3COCH3] [Br2][H+ ]2

1 10. For the reaction N 2O5(g ) → 2NO2(g ) + O2(g ) the value 2 of rate disappearance of N 2O5 is given as 6.25 × 10–3 mol L–1s–1. The rate of formation of NO2 and O2 is given respectively as

(AIPMT 2008)

5. The rate constants k1 and k2 for two different reactions are 1016 ⋅ e −2000/T and 1015 ⋅ e −1000/T , respectively. The temperature at which k1 = k2 is 1000 (1) K (2) 1000 K 2.303 2000 (3) K (4) 2000 K 2.303 (AIPMT 2008)

(1) 1.25 × 10 −2 mol L−1s −1 and 3.125 × 10 −3 mol L−1s −1 (2) 6.25 × 10 −3 mol L−1s −1 and 3.125 × 10 −3 mol L−1s −1 (3) 1.25 × 10 −2 mol L−1s −1 and 6.25 × 10 −3 mol L−1s −1 (4) 6.25 × 10 −3 mol L−1s −1 and 6.25 × 10 −3 mol L−1s −1 (AIPMT PRE 2010) 11. During the kinetic study of the reaction, 2A + B → C + D ; following results were obtained; −1

Initial rate of formation of D mol L–1 min–1

Run [ A ] (mol L ) [B] (mol L ) d [NH 3 ] −4 −1 −1 = 4 × 10 mol L s 6. For the reaction, N 2 + 3H 2 → 2NH 3 , if dt I 0.1 0.1 6.0 × 10–3 −d [ H 2 ] d [NH 3 ] −4 −1 −1 would be = 4 × 10 mol L s , the value of II 0.3 0.2 7.2 × 10–2 dt dt III 0.3 0.4 2.88 × 10–1 (1) 1 × 10 −4 mol L−1 s −1 (2) 3 × 10 −4 mol L−1 s −1 IV 0.4 0.1 2.40 × 10–2 (3) 4 × 10 −4 mol L−1 s −1 (4) 6 × 10 −4 mol L−1 s −1 Based on the above data which of the following is correct? (AIPMT 2009) (1) rate = k[A][B] (2) rate = k[A]2[B]2 7. For the reaction A + B → Products, it is observed that (3) rate = k[A][B]2 (4) rate = k[A]2[B] −1

(I) on doubling the initial concentration of A only, the rate of reaction is also doubled and (II) on doubling the initial concentration of both A and B, there is a change by a factor of 8 in the rate of the reaction.

Chapter 11_Chemical Kinetics.indd 275

(AIPMT PRE 2010) 12. For an endothermic reaction, energy of activation is E a and enthalpy of reaction is ∆H (both of these in kJ mol−1). Minimum values of E a will be

1/4/2018 5:16:34 PM

276

OBJECTIVE CHEMISTRY FOR NEET (1) equal to ∆H. (2) more than ∆H. (3) equal to zero. (4) less than ∆H. (AIPMT PRE 2010)

13. The rate of the reaction 2NO + Cl2 → 2NOCl is given by the rate equation





the value of rate constant can be increased by (1) (2) (3) (4)

rate = k[NO] [Cl2] 2

increasing the concentration of NO. increasing the concentration of the Cl2. increasing the temperature. doing all of these.

18. In a zero-order reaction for every 10° rise of temperature, the rate is doubled. If the temperature is increased from 10°C to 100°C, the rate of the reaction will become (1) 256 times. (2) 512 times. (3) 64 times. (4) 128 times. (AIPMT PRE 2012) 19. In a reaction, A + B → Product, rate is doubled when the concentration of B is doubled, and rate increases by a factor of 8 when the concentrations of both the reactants (A and B) are doubled, rate law for the reaction can be written as (1) Rate = k[A][B]2 (2) Rate = k[A]2[B]2 (3) Rate = k[A][B] (4) Rate = k[A]2[B]

(AIPMT MAINS 2010) 14. Which one of the following statements for the order of a reaction is incorrect? (1) Order of reaction is always whole number. (2) Order can be determined only experimentally. (3) Order is not influenced by stoichiometric coefficient of the reactants. (4) Order of reaction is sum of power to the concentration terms of reactants to express the rate of reaction.

(AIPMT PRE 2012) 20. Activation energy (Ea) and rate constants (k1 and k2) of a chemical reaction at two different temperatures (T1 and T2) are related by (1) ln

k2 E 1 1 k E  1 1 = − a  −  (2) ln 2 = − a  −  k1 R  T1 T2  k1 R  T2 T1 

(3) ln

k2 E  1 1 k E 1 1 = − a  +  (4) ln 2 = a  −  k1 R  T2 T1  k1 R  T1 T2 

(AIPMT PRE 2011) 15. The unit of rate constant for a zero-order reaction is (1) s−1 (3) L mol−1 s−1

(2) mol L−1 s−1 (4) L2 mol−2 s−1 (AIPMT MAINS 2011)

16. The half-life of a substance in a certain enzyme-catalyzed reaction is 138 s. The time required for the concentration of the substance to fall from 1.28 mg L−1 to 0.04 mg L−1 (1) 276 s (2) 414 s (3) 552 s (4) 690 s (AIPMT MAINS 2011) 17. The rate of the reaction 2N2O5 → 4NO2 + O2 can be written in three ways d[NO2 ] d[O2 ] −d[N 2O5 ] = k[N 2O5 ]; = k ′[N 2O5 ]; = k ′′[N 2O5 ] dt dt dt

The relationships between k and k′ and between k and k′′ are (1) k ′ − k; k ′′ = k (2) k ′ = 2k; k ′′= k (3) k ′ = 2k; k ′′ = k/2 (4) k ′ = 2k; k ′′ = 2k (AIPMT MAINS 2011)

Chapter 11_Chemical Kinetics.indd 276

(AIPMT MAINS 2012) 21. What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20°C to 35°C? (R = 8.314 J mol–1 K–1) (1) 342 kJ mol–1 (2) 269 kJ mol–1 (3) 34.4 kJ mol–1 (4) 15.1 kJ mol–1 (NEET 2013) 22. The activation energy of a reaction can be determined from the slope of which of the following graphs?

1 ln k vs.T (2) ln k vs. T T 1 T vs. (4) ln k vs. T (3) ln k T (AIPMT 2015) (1)

23. When initial concentration of a reactant is doubled in a reaction, its half-life period is not affected. The order of the reaction is (1) first. (2) second. (3) more than zero but less than first. (4) zero. (AIPMT 2015)

1/4/2018 5:16:35 PM

Chemical Kinetics 24. The rate constant of the reaction A → B is 0.6 × 10−3 mol s−1. If the concentration of A is 5 M, then concentration of B after 20 min is

27. Mechanism of a hypothetical reaction X2 + Y2 → 2XY is given below: (I) X2 → X + X (fast)

(1) 0.36 M (2) 0.72 M (3) 1.08 M (4) 3.60 M

(II) X + Y2  XY + Y (slow) (III) X + Y → XY (fast)

(RE-AIPMT 2015) 25. The rate of a first order reaction is 0.04 mol L s at 10 s and 0.03 mol L−1 s−1 at 20 s after initiation of the reaction. The half-life period of the reaction is −1

−1

(1) 24.1 s (2) 34.1 s (3) 44.1 s (4) 54.1 s (NEET-I 2016) 26. The decomposition of phosphine (PH3) on tungsten at low pressure is a first order reaction. It is because the (1) (2) (3) (4)

277



The overall order of the reaction will be (1) 2 (2) 0 (3) 1.5 (4) 1 (NEET 2017)

28. A first order reaction has a specific reaction rate of 10−2 s−1. How much time will it take for 20 g of the reactant to reduce to 5 g? (1) 138.6 s (2) 346.5 s (3) 693.0 s (4) 238.6 s (NEET 2017)

rate is inversely proportional to the surface coverage. rate is independent of the surface coverage. rate of decomposition is very slow. rate is proportional to the surface coverage. (NEET-II 2016)

Answer Key Level I  1. (1)

 2. (2)

 3. (2)

 4. (1)

 5. (2)

 6. (2)

 7. (3)

 8. (3)

 9. (4)

10.  (3)

11.  (1)

12.  (1)

13.  (3)

14.  (4)

15.  (3)

16.  (2)

17.  (2)

18.  (4)

19.  (2)

20.  (3)

21.  (3)

22.  (1)

23.  (3)

24.  (2)

25.  (2)

26.  (2)

27.  (4)

28.  (3)

29.  (1)

30.  (3)

31.  (4)

32.  (4)

33.  (3)

34.  (4)

35.  (2)

36.  (3)

37.  (2)

38.  (2)

39.  (1)

40.  (3)

41.  (1)

42.  (4)

43.  (1)

44.  (1)

45.  (2)

46.  (2)

 1. (2)

 2. (3)

 3. (2)

 4. (2)

 5. (1)

 6. (2)

 7. (3)

 8. (1)

 9. (3)

10.  (1)

11.  (3)

12.  (2)

13.  (1)

14.  (4)

15.  (2)

16.  (4)

17.  (1)

18.  (2)

19.  (3)

20.  (1)

21.  (3)

22.  (1)

23.  (2)

24.  (3)

25.  (3)

26.  (3)

27.  (4)

28.  (2)

29.  (1)

30.  (2)

31.  (1)

32.  (1)

33.  (1)

Level II

Previous Years’ NEET Questions  1. (1)

 2. (2)

 3. (3)

 4. (2)

 5. (1)

 6. (4)

 7. (3)

 8. (2)

 9. (4)

10.  (1)

11.  (3)

12.  (2)

13.  (3)

14.  (1)

15.  (2)

16.  (4)

17.  (3)

18.  (2)

19.  (4)

20.  (2), (4)

21.  (3)

22.  (2)

23.  (1)

24.  (2)

25.  (1)

26.  (4)

27.  (3)

28.  (1)

Chapter 11_Chemical Kinetics.indd 277

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278

OBJECTIVE CHEMISTRY FOR NEET

Hints and Explanations Level I

6. (2) Order of the reaction =

1. (1) Given that

(Rate)i = k[ A ]1/2 [B]3/2 (1)





(Rate)f = k[ 2 A ]1/2 [ 2B]3/2 (2)



From Eq. (1) and Eq. (2), we get

2. (2) For the reaction 2SO 2 + O2 ® 2SO 3 , we have



1 d[SO2 ] d[O2 ] 1 d[SO3 ] == 2 dt dt 2 dt

dx = k. dt 0. (3) The order of reaction can be calculated as: 1 Option (1): 1 + 1 + 1 = 3 Option (2): 0.5 + 0.5 + 0.5 = 1.5 Option (3): 1.5 - 1 + 0 = 0.5 Option (4): 1 + 0.5 - 2 = -0.5

d[SO3 ] 100 ´ 103 = mol min -1 dt 80 Therefore, Given

-

1 7. (4) Time increased 9 times or rate decreases to times. 9 So order of reaction is 3. 9. (4) For zero order reaction

(Rate)f = 4(Rate)i

-

1 d[SO2 ] 1 d[SO3 ] = 2 dt 2 dt d[SO2 ] 100 = ´ 64 kg min -1 = 80 kg min -1 dt 80

3. (2) Let the order of the reaction with respect to A be m and with respect to B be n.

Thus, Rate = k[ A ]m [B]n



    6 ´ 10-3 = k[0.1]m [0.1]n (1)







      2.88 ´ 10-1 = k[0.3]m [0.4]n (3)







On dividing Eq. (3) by Eq. (2), we get



2.88 ´ 10-1 = [0.2]n Þ n = 2 7.2 On dividing Eq. (4) by Eq. (1), we get

7.2 ´ 10-2 = k[0.3]m [0.2]n (2) 2.4 ´ 10-2 = k[0.4]m [0.1]n (4)

13. (3) We know

(Rate)i = k[ A ]2 [B]2 (1)





(Rate)f = k[ 2 A ]2 [ 2B]3 (2)



From Eq. (1) and Eq. (2), we get (Rate)f = 25(Rate)i = 32 (Rate)i

10 = 0.625 M ( 2)4 5. (3) Decomposition is directly proportional to the time, 1 that is, x µ t . 14. (4) After 96 h or 4t1/2 , concentration =





2(Rate)i = k[4A]m(2)



On dividing Eq. (2) by Eq. (1), we get 2 = ( 4)m Þ m =



= 60 ´ 10 -4 ´ 0.01

= 60 ´ 10 -6 mol s -1 5. (2) Given    (Rate)i = k[ A ] [B]2 (1) 1

2

éBù (Rate)f = k[ 2 A ]1 ê ú (2) ë2û

  



From Eq. (1) and Eq. (2), we get 1 (Rate)f = (Rate)i 2

Chapter 11_Chemical Kinetics.indd 278

The reaction is nA ® Products a 0 a-x x d[A ] = k[ A ]1/2 dt dx = k(a - x )1/2 dt dx = kdt (a - x )1/2

4. (1) For first order reaction (Rate)i = k[A]i



1 2

At t = 0 At t = t

Hence, order w.r.t. A is 1 and order w.r.t. B is 2.

              

Therefore, it is zero order reaction and unit of rate constant is mol L−1 h−1.

16. (2) We have   (Rate)i = k[A]m(1)

2.4 ´ 10-2 = 4m Þ m = 1 6 ´ 10-3

3 1 -1 = 2 2

- a-x

x x =0

= k ¢t

a - a - x = k ¢t a = k ¢(16 )  2



     a -



    a - a - 0.75a = k ¢(t ) (2)

(1)

1/4/2018 5:16:38 PM

Chemical Kinetics

On dividing Eq. (2) by Eq. (1), we get



t a - a/4 = Þ t = 27.3 min 16 a - a/2 17. (2) For zero order reaction, x = kt

For completion, let time for completion be tcomplete and x = a       a = kt complete Þ t complete =





a æ ö kt = 2.303 log ç ÷ (1) è a - 0.999a ø a æ ö    k ´ 45 = 2.303 log ç ÷ (2) è a - 0.5a ø

On dividing Eq. (2) by Eq. (1), we get t 3 = Þ t = 450 min or 7.5 h 45 log 2

a k

-d[ A ] =k dt -(18 - 20) = 0.4 mol min -1 Therefore, Rate = (10 - 5) Rate =

18. (4) Given



279

24. (2) We know C = C 0 e - kt 1 C 0 = C 0 e - kt e2 2 = kt Þ t =

19. (2) For zero order reaction x = kt

2 or two natural life-time k

26. (2) As t50 is constant, therefore, the reaction is of first order.



æ1ö      = 0.2 ´ ç ÷ = 0.1 mol dm -3 è2ø Given a – x = 0.05



Therefore, a = 0.05 + 0.1 = 0.15 mol dm–3



20. (3) According to integrated equation of rate law



æ a ö   kt = 2.303 log ç ÷ èa-xø a æ ö k(90) = 2.303 log ç ÷ (1) è a - 0.9a ø



a æ ö k(t1/2 ) = 2.303 log ç ÷ (2) è a - 0.5a ø



On dividing Eq. (2) by Eq. (1), we get t1/2 log 2 = Þ t1/2 = 27 min 90 1

28. (3) We know Rate = k[X]    7.5 × 10−4 = k × 0.5

  

k = 1.5 × 10−3 s−1

2 ln 2 k 2 ´ 0.693 1200 = Þ k = 10-3 k

29. (1)  t 3/4 = 2t1/2 =

30. (3) We have Rate = k[A]m [B]n



0.25 = k[1]m [1]n

(1)





0.50 = k[2] [1] 

(2)





0.25 = k[1]m [2]n

(3)



On dividing Eq. (2) by Eq. (1), we get

m

21. (3) For first order half-life is given by ln 2 t1/2 = (1) k 1 t Avg = (2) k











On dividing Eq. (2) by Eq. (1), we get t Avg t1/2

=

1 100 Þ t Avg = = 144 years ln 2 0.693

22. (1) For the first order reaction, we have

2 = 2m Þ m = 1

On dividing Eq. (3) by Eq. (1), we get 1 = 2n Þ n = 0



Therefore, Rate = k[A]1

31. (4) We know 2.303 log

23. (3) According to integrated equation of rate law æ a ö kt = 2.303 log ç ÷ èa-xø

Chapter 11_Chemical Kinetics.indd 279

[ A ]0 = kt [ A ]t

  kt = 2.303 log[A]0 − 2.303 log[A]t k log[ A ]t = t + log[ A ]0     . 2 303     Non zero intercept

[C]   kt = ln 0 [C] C [C] æ1ö k ç ÷ = ln 0 Þ C = 0 k C] e [ è ø

n

Negative slope

32. (4) x = kt

where concentration is (a − x ) = a − kt



Therefore, (a - x ) =

k(t ) + 

Negative slope

a 

Non zero intercept

33. (3) t50 = constant for first order reaction.

1/4/2018 5:16:40 PM

280

OBJECTIVE CHEMISTRY FOR NEET

34. (4) The reaction is

A + 2B ® a b a - x b - 2x

At t =0 At t = t eq



C + D 0 0 x = 0.2 x

1

1

Ea 2.303R

45. (2) NO(g) + Br2(g)  NOBr2 (g) keq (fast )     

Initial Rate = k[ A ] [B] = k(0.4) (0.6 ) 1

Therefore,  slope = -

2

k ¾ 2NOBr(g) (slow) NOBr2 (g) + NO(g) ¾®

Final Rate = k(a - x )(b - 2 x ) = k(0.2)(0.2) 2

2

Final Rate (0.2)´ (0.2)2 1 = = Initial Rate (0.4)´ (0.6 )2 18 5. (2) The reaction involved is 3 A ® 2B + C At t =0 a 0 0 At t =10 a - x 2 x a At t = t end 0 2a a



176 µ (2a + x)





270 µ 3a



Net reaction: 2NO + Br2 ® 2NOBr



Rate = k[NOBr2] [NO] (Since, it is rate determining step)



But NOBr2 must be replaced as it is not present in the net reaction. Rate = k(keq[NO][Br2 ])[NO] = k ′[NO]2[Br2 ]



46. (2) Transition state has unstable intermediates with high potential energy.

270 - 176 µ (a - x ) 2 47 µ (a - x )

Level II 1. (2) For the reaction A + 2B → C + 2D, we have

36. (3) The reaction is

d[ A ] 1 d[B] =dt 2 dt -d[B] = 2 ´ ( 2.6 ´ 10-2 ) = 5.2 ´ 10-2 dt

A(g) ® B(g) + C(g) t =0 a t = 10 min a - x

0 x

0 x

a µ100 and a + x µ120







Therefore,

2. (3) We have

100 a = a - x 80 k=



2.303 æ 100 ö log ç ÷ 10 è 80 ø



37. (2) According to Arrhenius equation k = Ae - Ea /RT

k200 = A × e



-

Ea R( 200 )

-

Ea R( 400 )

(1)



k400 = A × e



Given

k400 = 10 (3) k200



Substituting the values of Eq. (1) and Eq. (2) in Eq. (3), we get



-

(2)

Ea é 1 1 ù ê ú

e R ë 200 400 û = 10 Ea 1 ´ = 2.303 Þ E a = 921.2R R 400

1 é d[ A ] ù 1 é d[B] ù 1 é d[C] ù = = x êë dt úû y êë dt úû z êë dt úû

x y 3 = = z z 2 Therefore, x = y = 3 and z = 2 Given that x = y and

3. (2) We know

   Rate = k[A]n

mol L-1 min -1 = k(mol )n[L- n ]

Therefore, unit of k = mol1-n Ln-1min -1

4. (2)

(0.4 - 0.5) d[ A ] =mol L-1 min -1 dt 10

1 æ d[ A] ö 1 0.1 Rate of the reaction = ç mol L-1 min ÷= ´ 2 è dt ø 2 10    = 0.005 M min -1

5. (1) We have 1 é d[NH 3 ] ù d[N 2 ] 1 d[H 2 ] = = = x (say ) dt úû dt 2 êë 3 dt

38. (2) We know k = Ae - Ea /RT



  



When T ® ¥ k ® A





Therefore, A = k = 1.2 ´ 103 mol -1 L s -1







x = k2[NH3](2)





3x = k3[NH3](3)



From Eq. (1), Eq. (2) and Eq. (3), we get

42. (4) According to Arrhenius equation k = Ae - Ea /RT

Taking log on both the sides, we get Ea log k = log A 2.303RT

Chapter 11_Chemical Kinetics.indd 280



2x = k1[NH3](1)

k1 k2 k3 = = 2 1 3

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Chemical Kinetics

or   1.5 k1 = 3k2 = k3

æ ln 2 ö æ 10 ö −1 −1 Therefore, Rate = ç ÷ ç 2 ÷ = 0.0693 × 2.5 mol L min è 10 ø è 2 ø 19. (3) Given (t1/2 )zero order = (t1/2 )1st order

6. (2) (Rate)i = k[A] 

(1)

m



   1.837 (Rate)i = k[1.5A] 



On dividing Eq. (2) by Eq. (1), we get

m

(2)

(1.5)m = 1.837 Þ m = 1.5



281

7. (3) The reaction is 2X + Y ® products

 



From Eq. (1)



k0 1 (Rate)0 k0 = = =         (Rate)1 k1a 2k0 ´ 0.693 2 ´ 0.693

( excess )



Since, [ X ] » constant, therefore, Rate=



Hence, order of the reaction is 1.

d[ Y ] = k ¢[ Y ] dt

8. (1) Rate is approximately constant with respect to time. Therefore, order of reaction is zero. 9. (3) As given reaction is of second order, therefore, unit of rate constant is mol -1L1 s-1. 11. (3) Half-life of A remains the same.

0.693 a = (1) 2k0 k1



  

= 2 ´ 0.693 20. (1) According to integrated equation of rate law æ a ö kt = 2.303 log ç ÷ èa-xø k ´ 10 = 2.303 log 8 Þ k =

2.303 ´ 3 log 2 10

21. (3) We have   (Rate)i = k[A]m [B]n [C]q

(1)

  2(Rate)i = k[2A] [B] [C]  m

n

(2)

q



In flask I concentration of A becomes 0.25 M from 1 M in 8 h or 2 t1/2 = 8 or t1/2 = 4 h

éBù   (Rate)i = k[A]m ê ú [C]q ë2û

(3)



Time taken in flask II for concentration of A to become 0.3 M from 0.6 M = 1 t1/2 or 4 h.

    2.25(Rate)i = k[A]m [B]n [1.5C]q

(4)

12. (2) t æ

1ö çè 1- 4 ÷ø 2



= 4 ´ t1/2

Therefore, reaction is of first order.

14. (4) We have      (I) x = kt



   (II) t1/2 =

a 2k

a     (III) = k ´ 100 (1) 2 3a = kt (2) 4



From Eq. (1) and Eq. (2), we get



t = 150 min

On dividing Eq. (2) by Eq. (1), we get 2 = ( 2)m Þ m = 1

(a - x ) = ae - kt Þ x = a(1 - e - kt )







(II): True. (a - x ) = ae

( - ln 2 )t t 50



On dividing Eq. (3) by Eq. (1), we get n

é1ù 1= ê ú Þn = 0 ë2û



On dividing Eq. (4) by Eq. (1), we get







Hence, the order of the reaction is 1 + 0 + 2 = 3

2.25 = [1.5] Þ q = 2 q

22. (1) pN 2 O5 µ[ A ]t

Also for first order reaction log (A)t vs t is a straight line with negative slope. Therefore, log pN 2 O5 vs t is also a straight line with negative slope. a µ V ¥ - V0

24. (3) We know

æ a ö 16. (4) (III): True. kt = 2.303 log ç ÷ èa-xø (I): True. We know that [ A ]t = [ A ]0 e - kt

y

t æ 1ö = a ç ÷ where y = è 2ø t 50

ln 2 8. (2) We know   Rate = k[A] and t1/2 = 1 k

Chapter 11_Chemical Kinetics.indd 281



a 13. (1) At intersection, a – x = x or x = 2 Therefore, it represents t1/2.



n

a - x µ V¥ - Vt

Therefore, at 50% neutralization a - x =

a 2

V ¥ - V0 2 2V¥ - 2Vt = V¥ - V0 Þ V¥ = 2Vt - V0

Thus,      V¥ - Vt =

26. (3) From the following graph, we know

E a(B® A) = E a(A®B) - DH

1/4/2018 5:16:47 PM

282

OBJECTIVE CHEMISTRY FOR NEET Activated Ea for forward complex reaction

Energy

Ea for reverse reaction DH Reactant Product

1. (1) When a reaction, overall, is first order, the rate constant is given by the equation ln

Therefore, E B® A = 80 - ( -200) = 280 kJ





The half-life of the reactant can be obtained by setting [A]t equal to one-half of [A]0 and time as t1/2. [A]t =



27. (4)

EBA = ?



Reaction coordinate



E B® A = 30 + 70 = 100 kJ mol -1



29. (1) Given that  

k40° C =3 k0° C

A × e - Ea /R( 313 ) =3 - E a /R( 273 ) A ×e e

Therefore, E a =

Ea æ 1 1 ö ç ÷ R è 273 313 ø

30. (2) We know E a(revesre) = E a(forward) - DH

Noting that the left-hand side of the equation simplifies to ln 2, and solving the equation for t1/2, we have t1/2 =



ln 2 k

2. (2) As the slowest step is the rate determining step, thus, the mechanism B will be more consistent with the given information. Also, because it involves one molecule of H2 and one molecule of ICl and it can be expressed as





this shows that the reaction is first order with respect to H2 and ICl.

=3

2.303 ´ 2 ´ 313 ´ 273 log 3 40

æ [A]0 ö ln ç ÷ = kt1/2 è [A]0 /2 ø



DH = 70 kJ mol-1

1 [A]0 2

Substituting 1/2[A]0 for [A]t and t1/2 for t in the above equation, we have

Energy Ea=30 kJ mol-1

[A]0 = kt [A]t



Reaction coordinate



Previous Years’ NEET Questions

r = k[H2][ICl]

3. (3) For the first order reaction k =

( E a )reverse = 65 - ( -30) Þ ( E a )reverse = 95 kJ mol -1

2.303 a log (1) a-x t

Given that at t = 60 min, x = (60/100)a = 0.6a,substituting in the Eq. (1), we get æ ö a 2.303 log ç ÷ t è a - 0.6 a ø 2.303 æ 1 ö log ç = ÷ 60 è 0.4 ø

k= 31. (1) We know 

log k = log A -

Ea æ 1 ö ç ÷ (1) 2.303R è T ø



2000 On comparing Eq. (1) with log k (min -1 ) = 5 K , we T get







log A = 5 ⇒ A = 105



Ea = 2000 Þ E a = 9.212 kcal 2.303R

a 2.303 log a - 0.5 a 0.01535

2.303 æ 1 ö log ç ÷ 0.01535 è 0.5 ø = 46.5 min -1

=

Therefore, k1 corresponds to the slowest step or A → B is rate determining step.

Chapter 11_Chemical Kinetics.indd 282

On substituting k = 0.01535 min-1 and x = (50/100)a = 0.5a in the Eq. (1), we can calculate t t=

32. (1) Given that k3 > k2 > k1

= 0.01535 min -1

1/4/2018 5:16:49 PM

Chemical Kinetics 4. (2) We have

9. (4) For a first order reaction

5.7 ´ 10

-5

= k[0.30] [0.05] [0.05]

5.7 ´ 10

-5

= k[0.30] [0.10] [0.05]

-4

= k[0.30]x [0.05]y [0.10]z

1.2 ´ 10



x

y

x

z

y

k=

z



On solving the above equations, we get x = 1, y = 0, z = 1. Therefore, rate = k [CH3COCH3][H+]

5. (1) We know

N 2O5(g) ® 2 NO 2(g) +



10 =

e -1000/T e -2000/T





Taking ln on both the sides of Eq. (1), we get





1 d[NO2 ] 2d[O 2 ] -d[N 2O5 ] =+ =+ 2 dt dt dt d[NO2 ] -3 = 6.25 ´ 10 ´ 2 = 1.25 ´ 10-2 mol L-1 s-1 dt d[O 2 ] 6.25 ´ 10-3 = = 3.125 ´ 10-3 mol L-1 s-1 dt 2

(1)

2.303 −1000 2000 = + T T 1 1000  T = K 2.303



6. (4) The rate of reaction can be given by -



1 d[H 2 ] 1 d[NH 3 ] =+ 3 dt 2 dt



11. (3) We have 7.2 × 10−2 = k(0.3)p (0.2)q 2.88 × 10−1 = k(0.3)p (0.1)q



On solving the above equations for q, we get q=2 6.0 ´ 10

Therefore, d[H 2 ] 3 = ´ 4 ´ 10 -4 mol L-1 s -1 dt 2 = 6 ´ 10 -4 mol L-1 s -1



7. (3) The rate of reaction in terms of [A] is given by

          rate ∝ [A]



Since, rate of reactions gets doubled by doubling the concentration of [A], therefore, x = 1

1 O2(g) 2

Rate of reaction can be expressed as

1016 e -2000/T = 1015 e -1000/T

0.693 0.693 = = 0.0005 s-1 or 0.5 ´ 10-3 s-1 t1/2 1386

10. (1) For the reaction,

3.1 ´ 10-4 = k[0.40]x [0.05]y [0.20]z



283

-3

= k(0.1)p (0.1)q

2.4 ´ 10-2 = k(0.4)p (0.2)q



On solving the above equations for p, we get

  p=1



Therefore,

rate = k[A][B]2

x

12. (2) We have

∆H = Ea(f) - Ea(b)



∆H = +ve

Given that

     Ea(f) = ∆H + Ea(b) ⇒ Ea(f) > ∆H



The rate of reaction in terms of [B] is given by

Therefore,



rate ∝ [B]y



The rate becomes 8 times by doubling the concentration of both [A] and [B]. Therefore, y = 2





13. (3) Rate constant is independent on the concentration of the reactants. It depends upon temperature. It has constant value at a particular temperature. So, the value of rate constant can be increased by increasing the temperature.

rate = [A][B]2

8. (2) The rate of appearance of bromine [Br2] is related to rate of disappearance of bromide ions by the expression -



Chapter 11_Chemical Kinetics.indd 283

14. (1) Order of a reaction can be zero or fractional but the same is not true for molecularity. 15. (2) From the rate law, we have

-

1 d[Br ] 1 d[Br2 ] = 5 dt 3 dt 3 d[Br - ] d[Br2 ] =dt 5 dt

dx = k[ c oncentration of reactant]n dt







where dx/dt = rate of reaction, k = proportionality constant, n = nth order of the reaction.

1/4/2018 5:16:50 PM

284



OBJECTIVE CHEMISTRY FOR NEET

  Rate of reaction = -



2.303 [A] log 0 k [A]

21. (3) We have é T1 - T2 ù k2 E =ê ú 2.303 R ë T1 ×T2 û k1 é 293 - 308 ù Ea log 2 = ê ú 2.303 ´ 8.314 ë 293 ´ 308 û

log

1 d[N 2O5 ] 1 d[NO2 ] d[O 2 ] =+ =+  2 dt 4 dt dt

(1)

We have,



22. (2) According to Arrhenius equation

1 1 k[N 2O5 ] = k ¢[N 2O5 ] = k ¢¢[N 2O5 ] 2 4 k k¢ On solving, we have = k ¢¢. Therefore, k′ = 2k and = 2 4 k″ = k/2





where k is the rate constant for the reaction, A is a proportionality constant that is known as Arrhenius factor and Ea is the activation energy.



Taking logarithm on both sides of Eq. (1), we get







On rewriting the Eq. (2), we have

18. (2) The ratio of the rate at 100°C to that at 10°C is



rate(100° C) = 2(100-10 )/10 = 29 = 512 times rate(10° C)



19. (4) The rate expression is rate1 = k[A]x[B]y. Given that rate2 = 2(rate1) when [B] = 2[B] and rate3 = 8(rate1) when [A] = 2[A] and B = 2[B]. Therefore, we have



(rate)2 k[A]x[ 2B]y = Þ 2 = 2y Þ y = 1 (rate )1 k[A]x[B]y





and





Chapter 11_Chemical Kinetics.indd 284

k1 = Ae -Ea/ RT1 and k2 = Ae -Ea/ RT2

E a (2) RT

æE ö æ1ö ln k = ln A - ç a ÷ ´ ç ÷  è R ø èT ø

(3)

= c +

y

m

x

To determine the activation energy, we can make a graph of ln k vs 1/T. The graph is a straight line whose slope = -Ea/R and intercept = ln A Intercept = In A Slope = -Ea/R

In k

The order with respect to [A] = 2 and with respect to [B] = 1. So, rate = k[A]2[B]1.

20. (2), (4) According to the Arrhenius equation, k = Ae -Ea/ RT . The rate constant at two different temperatures is

ln k = ln A -

æE ö æ1ö ln k = ln A - ç a ÷ ´ ç ÷ è R ø èT ø    

(rate)3 k[ 2 A]x [ 2B]y = Þ 8 = 2x ´ 2 Þ x + 1 = 3 Þ x = 2 k[A]x[B]y (rate )1

k = Ae -Ea/RT(1)



From (1) and (2), we get



0.301 ´ 2.303 ´ 8.314 ´ 293 ´ 308 15 = 34.67 kJ mol -1 » 34.7 kJ mol -1

Ea =

2N2O5 → 4NO2 + O2

-d[N 2O5 ] d[NO2 ] d[O 2 ] = k[N 2O5 ]; = k ¢[N 2O5 ]; = k ¢¢[N 2O5 ] (2) dt dt dt



Ea æ 1 1ö k1 E a æ 1 1ö = ç - ÷ Þ ln ç - ÷ R è T1 T2 ø k2 R è T1 T2 ø k 1ö 1ö E æ1 E æ1 ln k1 - ln k2 = - a ç - ÷ Þ ln 1 = - a ç - ÷ R è T2 T1 ø k2 R è T2 T1 ø

2.303 1.28 2.303 ´ 138 log = log 32 Þ 690 s k 0.04 0.693

17. (3) For the reaction



Subtracting these equations, we get

Substituting the values, we get t=

Ea E and ln k2 = ln A - a RT1 RT2

ln k1 - ln k2 =

The rate law equation for first order reaction is







0.693 0.693 = t1/2 138

t=



dx 1 ´ = mol L-1 time -1 or mol L-1 s-1 dt [concentration]0

16. (4) We have k =

Taking natural logarithm on both sides, we get ln k1 = ln A -

For zero order reaction, n = 0. Thus, k=





dx 1 ´ dt [concentration]n

Therefore, k =



0

1/T

23. (1) The t1/2 for the different order of reactions is as follows:

Zero order reaction: t1/2 =

[A]0 2k

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Chemical Kinetics





ln 2 k 1 Second order reaction: t1/2 = ka First order reaction: t1/2 =

From the above expressions, we can say that half-life period is independent on the concentration for the first order reactions.



where C is the concentration, k is rate constant and t is time.





C = (0.6 × 10−3 mol s−1) × 20 × 60 s

  = 0.72 M 25. (1) For first order reaction, rate of reaction is directly proportional to concentration of reactant.

Thus,



C 2.303 k= log 1 C2 (t 2 - t1 ) (0.04) 2.303 = log ( 20 - 10) (0.03) 2.303 ´ 0.1249 = 10   

As,   k =

0.693  t1/2



(1)



0.693 ´ 10 = 24.1 s 2.303 ´ 0.1249

26. (4) The rate of decomposition of PH3 at low pressure with tungsten as catalyst.



27. (3) The overall rate of the reaction depends on the rate of the slowest step. Since, two reagents are involved in the rate determining step; the overall rate of reaction is proportional to the concentrations of only those reagents.

  



From

X2 → X + X [X]2 [X 2 ]

[X] = keq [X 2 ]1/2 (2)





From Eq. (1) and Eq. (2), we get Rate =



keq [X 2 ]1/2 [Y2 ]

= k ¢[X 2 ]1/2 [Y2 ]



Thus, the overall order of the reaction = 1 + 0.5 = 1.5

28. (1) For first order reactions, we have



t=

2.303 [A ] log 0 (1) k [A t ]

Substituting the values in Eq. (1), we get 2.303 æ 20 ö log ç ÷ 10-2 è 5 ø 2.303 = ´ 0.60205 10-2 = 138.6 s

t=

3 ®P+ PH 3 ¾W¾ H2 2 Rate = k[PH 3 ]

Chapter 11_Chemical Kinetics.indd 285

Rate = k[X][Y2](1)



(2)

0.693 2.303 ´ 0.1249 = t1/2 10

kp 1 + kp

where q is the surface area covered by adsorption. For low pressure, the term kp can be neglected, so q = kp or rate is proportional to surface area coverage.

keq =

Therefore, from Eq. (1) and Eq. (2), we have

t1/2 =

Thus, it is a first order reaction. The rate is determined by partial pressure of PH3 which is dependent on the amount of PH3 absorbed by the surface of the catalyst. From adsorption isotherm, we have q =

24. (2) For zero-order reaction, C = kt

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12

Surface Chemistry

Chapter at a Glance 1. Terms Related to Adsorption (a) Adsorption: A surface phenomenon which involves accumulation or concentration of molecular species from gas/liquid at the surface rather than the bulk of solid or liquid. (b) Adsorbate: A substance that accumulates at the surface. (c) Adsorbent: The surface on which the substance (adsorbate) accumulates. (d) Sorption: When the absorption and adsorption both takes place together. (e) Occlusion: The adsorption of gases on the solid surface. (f ) Desorption: Removal of adsorbed gases/liquids from the adsorbent. 2. The distinguishing features of absorption and adsorption are: Absorption Bulk phenomenon Slower process

Adsorption Surface phenomenon Faster process

3. Types of Adsorption Depending upon the intermolecular force of attraction between adsorbate and adsorbent, there are two types of adsorption: (a) Physiosorption or van der Waals adsorption. (b) Chemisorption or activated adsorption.   The distinguishing features of two types of adsorption are: Physiosorption Forces of attraction involved between adsorbent and adsorbate are weak, long-range van der Waals forces. No formation of a surface complex.

Chemisorption Forces of attraction involved between adsorbent and adsorbate are strong chemical bonds. Formation of a surface complex between adsorbate and adsorbent. Reversible in nature. Irreversible in nature. Occurs at low temperatures due to low activation Occurs at high temperature due to high activation energy. energy. Magnitude of adsorption decreases with increase in Magnitude of adsorption increases with increase in temperature. temperature. Heat (enthalpy) of adsorption is low (20–40 kJ). Heat (enthalpy) of adsorption is high (80–240 kJ mol−1). Multilayered Monolayered 4. Factors Affecting Adsorption of Gases on Solids (a) Nature and surface area of adsorbent (b) Concentration (c) Heat (enthalpy) of adsorption

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OBJECTIVE CHEMISTRY FOR NEET

(d) Pressure (e) Temperature 5. Adsorption Isotherms (a) Freundlich’s adsorption isotherm: It gives the variation in the extent of adsorption with pressure at constant temperature. x = kp1/n m   w  here x is the mass of gas adsorbed on mass m of the adsorbate at pressure p; k and n are constants depending for a set of adsorbent and adsorbate, the value of n is generally greater than 1. (i) At very low pressure

x =k m

x m

Pressure (low)

(ii) At very high pressure

x = kp m

x m

Pressure (high)

(iii) At intermediate pressure log

x 1 = log k + log p n m

1 Slope = n

log x m

Intercept = log k

Pressure (intermediate)

(b) Langmuir adsorption isotherm (i)  Adsorption surface is homogeneous and has fixed number of equivalent adsorption sites, capable of adsorbing one gas molecule each.

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Surface Chemistry

289

(ii)  The rate of desorption is proportional to the concentration of adsorbed molecules, that is, to the fraction q of the surface already covered by gas molecules. At equilibrium, rate of adsorption = rate of desorption, so ka p(1 - q ) = kdq Þ q =

k1 p 1 + k1 p

x where k1 = ka / kd . Also, q is proportional to the amount of gas adsorbed per unit mass of adsorbent. So = k2q m Substituting for q, we get the expression for Langmuir’s isotherm. kk p x = 1 2 m 1 + k1 p 6. Adsorption Isobars These depict effect of temperature at constant pressure.

x m

Physcial adsorption

x m

Chemical adsorption

Temperature

Temperature

7. Catalysis (a) Catalyst: It is a substance which alters the rate of reaction without appearing in the end product of the reaction. (b)  Substances, which when present in small amounts, promote the activity of the catalyst are called promoters. On the other hand substances that deactivate the catalyst are called anticatalysts or poisons. (c) Types of catalysts: Based on activity, these may be positive catalysts (accelerate the rate) and negative catalysts (inhibit the rate). (d) Types of catalytic reactions (i) Homogeneous catalysis: The catalyst is in the same phase as reactants. SO2 (g ) + O2 (g ) ¾NO(g) ¾¾ ® SO3 (g ) (ii) Heterogeneous catalysis: The catalyst is in a different phase from the reactants. V O (s)

2 5 SO2 (g ) + O2 (g ) ¾¾¾ ® SO3 (g )

(e) Adsorption theory of heterogeneous catalysis According to this theory, the surface of the catalyst has active sites in which one or more reactants get adsorbed forming a unimolecular layer. The reaction takes place readily due to close proximity of the reactants. (f )  The ability of a catalyst to affect the art of reaction is known as its activity. For example, the metals of Group 5 to 11 show an increasing order of catalytic activity in hydrogenation reactions. The ability of the catalyst to affect the rate of certain reactions and prevent other side reactions is known as its selectivity. (g) Zeolites: These are shape selective naturally occurring catalysts, represented by general formula Na2OAl2O3·xSiO2·yH2O. They can act as catalyst for reactions in which the size of the reactants or products selectively fits into the size of the pores of the zeolite. (h) Enzyme catalysis



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Enzyme + Substrate « Enzyme - Substrate complex « Enzyme + Product

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OBJECTIVE CHEMISTRY FOR NEET

8. Colloidal State (a)  A colloid is a broad category of mixtures in which one phase (dispersed particles or dispersed phase) is suspended in the other (dispersion medium). It is a dispersion in which the dispersed particles are larger than the solute ions. (b) The characteristics of true solution, colloidal solution and suspension are tabulated below. Property Nature Particle size Filterability Visibility Charge on particle Tyndall effect Brownian movement

True solution Homogeneous

Colloidal solution Heterogeneous

Suspension Heterogeneous

Less than 10−8 cm Can pass through both animal and parchment paper Not visible even under microscope Does not carry Not exhibited Not exhibited

Between than 10−4 to 10−8 cm Can pass through only parchment paper Visible under microscope

More than 10−4 cm Neither from animal membrane nor from parchment paper Visible by naked eyes

Can carry Distinct Tyndall effect Distinct Brownian movement

May carry May show Tyndall effect May show Brownian movement

9. Classification of Colloidal Solutions (a) Physical state of dispersed phase and dispersion medium Dispersed phase Solid Solid Solid Liquid Liquid Liquid Gas Gas

Dispersion medium Solid Liquid Gas Solid Liquid Gas Solid Liquid

Type of colloid Solid sol Sol Aerosol Gel Emulsion Aerosol Solid sol Foam

Example Gem stones Paints Smoke, dust Cheese Milk Fog Foam rubber Soap lather

(b) Nature of interaction between dispersed phase and dispersion medium Depending upon the interaction between dispersed phase and dispersion medium, they are divided into two types: (i) Lyophilic: In these dispersed particles has strong affinity for dispersion medium. These can be reconstituted, so also called reversible sols. Examples are gum, gelatin, starch. (ii) Lyophobic: In these, the dispersed phase has weak interaction with dispersion medium. These cannot be reconstituted, so are irreversible. Examples are metals in water, metal sulphides in water. (c) Types of particles of the dispersed phase Multimolecular These consist of aggregate of atoms or molecules with 1 mm diameter.

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Macromolecular These are large molecular weight colloids which change to colloidal size particles in suitable solvent.

Associated Associated colloids are those substances which behave as true solution in dilute state but exhibit colloidal nature at higher concentration.

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Surface Chemistry

Multimolecular Molecules/atoms are held together by weak forces.

Macromolecular Molecules themselves acquire various shape due to flexibility.

Example: Gold sol and sulphur sol

Protein, cellulose, starch, rubber, etc.

291

Associated At high concentration ions aggregate and behave as colloids and aggregated molecules are called micelles. Sodium salts of fatty acids.

Two important characteristics of micelle formation are kraft temperature and critical micelle concentration (CMC) which is temperature and concentration, respectively above which micelle formation takes place. 10. Preparation of Sols (a) Lyophilic sols are obtained by simple mixing of dispersed phase and dispersion medium (also called intrinsic sols). For example, starch dissolved in water. (b) Lyophobic sols are obtained by dispersion method and condensation methods. Mechanical Colloidal mill (e.g., Indian ink)

Dispersion methods

Electrical disintegration Bredig’s arc method (e.g., Gold, sliver, copper sols )

Peptization A process in which freshly prepared precipitate of ionic solids is dispersed into water by addition of electrolyte containing common ion. (e.g., FeCI3 is added to freshly prepared Fe(OH)3

Oxidation Br2 + H2S → 2HBr + S ↓

Reduction 2AuCl3 + SnCl2 → 2Au ↓ + SnCl4 Chemical Hydrolysis FeCl3 + 3H2O → Fe(OH)3 + 3HCl

Condensation methods Physical

Double decomposition As2Cl3 + H2S → As2S3 + HCl Yellow sol

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OBJECTIVE CHEMISTRY FOR NEET

11. Purification of Colloidal Solutions (a) Dialysis (b) Ultrafiltration (c) Ultracentrifugation 12. Other Properties of Colloidal Solutions (a) Brownian movement: Zig-zag motion due to continuous collision of particles. (b) Tyndall effect: When a beam of light is passed through a colloidal dispersion, it becomes clearly visible and sharply outlined. (c) Electrolysis (cataphoresis): Colloidal particles are electrically charged and the particles migrate under the electrical field towards poles of opposite charge. 13. Coagulation and Flocculation (a) Coagulation: It is the property of the colloidal particles to aggregate to form a precipitate when the charge on them is removed or neutralized. Coagulation of lyophobic sols is carried out by: (i) Dialysis (ii) Mixing oppositely charged colloidal solutions. (iii) Electrophoresis (iv) Boiling (v) Addition of electrolytes (b) Coagulation or flocculation value: It is the minimum concentration of electrolyte (in milimoles) required to bring about the coagulation of one liter of colloidal solution in two hours. The reciprocal of coagulation value is coagulating power. (c) Hardy–Schulze rule: It states that greater the valence of flocculating ion added, greater is its power to cause precipitation. (d) Protection of colloids: Lyophilic colloids can be added to lyophobic colloids to protect them from coagulation in the presence of electrolytes. These form a protective layer around the lyophobic colloidal particles thus protecting them from electrolytes. (e) Gold number: It is defined as a milligram of a protective colloid added to just prevent the coagulation of 10 mL of a gold sol on the addition of 1 mL of 10% sodium chloride solution. 14. Electrical Double Layer (a)  Individual colloidal particles are charged. The charge arises due to selective adsorption of cations and anions from the solution which are common to cations and anions of the colloids. These charged colloidal particles then have tendency to acquire oppositely charged ions from the solution to form a layer around it. (b)  The first layer of charges is firmly held to the colloidal surface while the second layer is diffused. The diffuse layer in a colloidal dispersion contains a quantity of counter ions sufficient to balance the electrical charge on the particle. The colloidal dispersion thus does not have net electrical charge. The arrangement of ions for electroneutrality is called Helmholtz electrical double layer. (c)  The potential difference set up between the two layers because of fixed and diffused charges is called electrokinetic or zeta potential. − − − −

+

+

− −

+ −

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− +

+

+

+ −

−

−

+

Movable layer

−

+ +

Fixed layer

−

+

Colloidal particles with negative charge

+ +

−

−

−

Liquid medium

Surface charge

−

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Surface Chemistry

293

15. Emulsion: It is a colloidal dispersion of one liquid in another liquid that is both are dispersed phase and dispersion medium are liquids. (a) Emulsions are of two types: (i) water in oil (e.g., butter) (ii) oil in water (e.g., milk) (b) Emulsions show Brownian motion and Tyndall effect like colloidal solutions. (c) These can be precipitated by addition of electrolytes. 16. Applications of Colloids (a) (b) (c) (d) (e) (f ) (g)

Electrical precipitation of smoke. Purification of drinking water. Removal of dirt from sewage. Cleansing action of soaps and detergents. In medicines. In photographic plates and films. In rubber industry and tanning.

Solved Examples 1. Methylene blue, from its aqueous solution, is adsorbed on activated charcoal at 25°C. For this process, the correct statement is



(1) The adsorption requires activation at 25°C. (2) The adsorption is accompanied by a decrease in enthalpy. (3) The adsorption increases with increase of temperature. (4) The adsorption is irreversible.



3. A plot of log x/m versus log p for the adsorption of a gas on a solid gives a straight line with slope equal to

Solution

(1) −log k (2) n 1 (3) (4) log k n

(2) The adsorption of methylene blue on activated charcoal is physisorption, which causes decrease in enthalpy. 2. Which of the following statements is incorrect regarding physisorption? (1) It occurs because of van der Waals forces. (2) More easily liquefiable gases are adsorbed readily. (3) Under high pressure, it results in multimolecular layer on adsorbent surface. (4) Enthalpy of adsorption (ΔHadsorption) is low and positive. Solution (4)





Option (1):  When a gas is held on the surface of solid by van der Waals forces without resulting in the formation of any chemical bond between the adsorbate and adsorbant, it is called physisorption. Option (2): Easily liquefiable gases, that is, gases having high critical temperature are adsorbed more strongly because they have stronger van der Waals forces of attraction.

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Option (3): As the layers of the gas can be adsorbed one over the other by van der Waals forces, multimolecular layers are formed under high pressure. Option (4): As physisorption involves only van der Waals forces of attraction and no chemical change, the process is exothermic but the enthalpy of adsorption is quite low (20–40 kJ mol−1).

Solution (3) According to Freundlich adsorption isotherm x x µ p1/n Þ = kp1/n m m

Taking log on both sides, we get log



x 1 = log k + log p m n

On plotting the graph between log(x/m) and log p, a straight line is obtained with slope equal to 1/n and the intercept equal to log k.

4. Which one of the following forms micelles in aqueous solution above certain concentration? (1) Glucose (2) Urea (3) Dodecyl trimethyl ammonium chloride (4) Pyridinium chloride

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OBJECTIVE CHEMISTRY FOR NEET

Solution (3) Dodecyl trimethyl ammonium chloride [C12H 25N + (CH 3 )3 ]Cl - form micelles in aqueous solution above certain concentration. In this, the cation has a polar ionic head and non-polar tail, required for micelle formation. 5. According to the adsorption theory of catalysis, the speed of the reaction increases because (1) the concentration of reactant molecules at the active centers of the catalyst becomes high due to adsorption. (2) in the process of adsorption, the activation energy of the molecules becomes large. (3) adsorption produces heat which increases the speed of the reaction. (4) adsorption lowers the activation energy of the reaction. Solution (1) According to adsorption theory of catalysis, reactants are adsorbed on the surface of the catalyst and form a film. Due to high concentration of the reactants on the film, reaction proceeds at a faster rate. 6. Which of the following statements is correct for lyophilic sols? (1) (2) (3) (4)

The coagulation of the sols is irreversible in nature. They are formed by inorganic substances. They are self-stabilized. They are readily coagulated by addition of electrolytes.

Solution (3) They are fairly stable and cannot be coagulated easily. Some examples are starch, gum, gelatin, etc. dissolved in a suitable solvent. 7. Among the following, the surfactant that will form micelles in aqueous solution at the lowest molar concentration at ambient conditions is (1) C H 3(CH 2 )15N + (CH 3 )3Br (2) C H 3(CH 2 )11OSO3- Na + (3) C H 3(CH 2 )6 COO - Na + (4) C H 3(CH 2 )11 N + (CH 3 )3Br Solution (1) Critical concentration for micelle formation decreases as the molecular weight of hydrocarbon chain of surfactant grows because in this case, true solubility diminishes and the tendency of surfactant molecules to associate increases. 8. Choose the correct reason(s) for the stability of the lyophobic colloidal particles.

Chapter 12_Surface Chemistry.indd 294

(1)  Preferential adsorption of ions on their surface from the solution. (2) Potential difference between the fixed layer and the diffused layer of opposite charges around the colloidal particles. (3) Attraction between different particles having opposite charges on their surface. (4) Both (1) and (2) Solution (4) In case of lyophobic sols, the ionic colloid adsorbs ions common to its own lattice during the preparation of the sol. So they are stable and also the potential difference which arises between the fixed layer and the diffused layer makes lyophobic sols more stable. 9. The disperse phase in colloidal iron (III) hydroxide and colloidal gold is positively and negatively charged, respectively. Which of the following statements is NOT correct? (1) Magnesium chloride solution coagulates the gold sol more readily than the iron (III) hydroxide sol. (2)  Sodium sulphate solution causes coagulation in both sols. (3) Mixing the sols has no effect. (4) Coagulation in both sols can be brought about by electrophoresis. Solution (3) On mixing oppositely charged sols, their charge gets neutralized. Both the sols can be either partially or completely precipitated. 10. Which of the following is an irreversible colloid? (1) Clay (2) Platinum (3) Fe(OH)3 (4) All of these. Solution (4) All are irreversible because they are all lyophobic colloids. 11. Which of the following methods is not employed for the purification of colloids? (1) Electrodialysis (2) Dialysis (3) Ultracentrifugation (4) Peptization Solution (4) Peptization is a method of preparing colloidal sols. 12. A negatively charged suspension of gold in water will need for precipitation, minimum amount of (1) AlCl3 (2) K2SO4 (3) HCl (4) NaOH

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Surface Chemistry Solution (1) Greater the valency of effective ion (cation in this case) greater is its coagulating power as per Hardy–Schulze rule.

(1) (2) (3) (4)

13. Identify the incorrect statement among the following. (1) Brownian movement and Tyndall effect are shown by colloidal system. (2) The colloidal solution of a liquid in liquid is called a gel. (3) Hardy–Schulze rule is related with coagulation. (4) Gold number is a measure of the protective power of lyophilic colloid. Solution (2) The colloidal solution of a liquid in liquid is called emulsion. 14. Gold number represents

295

quantity of gold in alloys. percentage of gold in alloys. colloid protective power. percentage quantity of gold in colloidal solution of gold.

Solution (3) Gold number represents protective power of colloids. 15. Potential difference of the electrical double layer formed in a colloidal sol is called (1) emf. (2) zeta potential. (3) brownian potential. (4) Nernst potential. Solution (2) The potential difference set up between the two layers because of fixed and diffused charges is called electrokinetic or zeta potential.

Practice Exercises Level I Adsorption 1. Which of the following is not true for physiosorption? (1) It is reversible. (2) It occurs in the form of multimolecular layers. (3) It needs activation energy to start the process of adsorption. (4) It increases with increase in pressure. 2. Absorption is generally accompanied by (1) (2) (3) (4)

decrease in entropy of system. decrease in enthalpy. making the value of T∆S negative. all of these.

3. The order of the decomposition of ammonia gas over the surface of adsorbent is (1) 1 (2) 2 (3) 3 (4) 0 4. Which can absorbs larger volume of hydrogen gas? (1) (2) (3) (4)

Colloidal solution of palladium. Finely divided nickel. Finely divided platinum. Colloidal Fe(OH)3.

5. Which of the following is not a characteristic of chemisorption? (1) (2) (3) (4)

Adsorption is irreversible. ∆H is of the order of 900 kJ. Adsorption is specific. Adsorption increases with increase of surface area.

Chapter 12_Surface Chemistry.indd 295

6. Which forms multimolecular layer during adsorption? (1) (2) (3) (4)

Physical adsorption van der Waals adsorption Freundlich adsorption All of these

7. At extremely high temperature, the adsorption is (1) saturated. (2) undetectable. (3) surface area dependent. (4) spontaneous. 8. Adsorption is (1) an exothermic process hence increase in temperature decreases adsorption in cases where van der Waals force exists between adsorbate and adsorbent. (2) an endothermic process, hence increases in temperature increases adsorption. (3) an exothermic process, hence increase in temperature increases adsorption. (4) None of these. 9. In the adsorption of oxalic acid by activated charcoal, the activated charcoal is known as (1) adsorbent. (2) adsorbate. (3) absorber. (4) none of these. 10. Which adsorption does not take place at very low temperature? (1) Physiosorption (2) Chemisorption (3) Both (1) and (2) (4) None of these 11. Adsorption is the phenomenon in which a substance (1) accumulates on the surface of the other substance.

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OBJECTIVE CHEMISTRY FOR NEET (2) goes into the body of the other substances. (3) remains close to the other substance. (4) none of these.

12. Lake test of aluminium ion is based on absorption of blue litmus on (1) (2) (3) (4)

solid surface of Al. solid surface of Al(OH)3. solid surface of Al2O3. solid surface of AlCl3.

13. Rate of physisorption increases with (1) (2) (3) (4)



(1) C = 0 (2) C = 1 M (3) C = constant (4) C = 2 M

Catalyst 18. In which of the following reaction is a catalyst required? (1) S + O2 → SO2 (2) C + O2 → CO2 (3) 2SO2 + O2 → 2SO3 (4) All of the above 19. Following are the terms about activity and selectivity; (I) Activity is the ability of catalysts to accelerate chemical reactions and selectivity is the ability of catalysts to direct reaction to yield particular products. (II) Activity is the ability of catalysts to direct reaction to yield particular products and selectivity is the ability of catalysts to accelerate chemical reactions.

decrease in temperature. increase in temperature. decrease in pressure. decrease in surface area.

Adsorption Isotherms 14. According to Freundlich adsorption isotherm, which of the following is correct? (1)

x µ p1 m

(2)

x µ p1/n m

(3)

x µ p0 m

This is when



Select correct term: (1) I (2) II (3) I and II both (4) None of these

20. Efficiency of a catalyst depends on its (1) particle size. (2) solubility. (3) molecular weight. (4) none of these. 21. The function of zymase is to (1) (2) (3) (4)

(4) All the above are correct for different ranges of pressure. 15. A graph of extent of adsorption versus pressure at constant temperature is called (1) adsorption isobar. (2) adsorption isotherm. (3) adsorption isostere. (4) none of these.

22. The efficiency of an enzyme in catalyzing a reaction is due to its capacity (1) to form an enzyme−substrate complex. (2) to decrease the bond energies of the substrate molecule. (3) to change the shape of the substrate molecule. (4) None of the above.

16. Limitations of Freundlich’s equation is/are (1) It is valid over a certain range of pressure only. (2) The constant k and n vary with temperature. (3) Freundlich’s equation is purely empirical formula without theoretical foundation. (4) All of these. 17. For the adsorption of solution on a solid surface x æ xö = kC 1/n . Adsorption isotherm of log ç ÷ and log C è mø m was found of the type shown below.

convert starch into sugar. convert of glucose to alcohol and carbon dioxide. convert starch into malt sugar and dextrin. convert malt sugar into glucose.

23. Following reaction is catalyzed by Br−(aq) 2H2O2(aq) → 2H2O(l) + O2(g)







This is an example of (1) (2) (3) (4)



homogeneous catalysis. heterogeneous catalysis. both (1) and (2). None of these.

24. The process which is catalyzed by one of the products formed is called log

(1) acid-based catalyst. (2) auto catalyst. (3) negative catalyst. (4) positive catalyst.

x m

25. Which type of metals form effective catalysts? log C

Chapter 12_Surface Chemistry.indd 296

(1) Alkali metals (2) Transition metals (3) Alkaline earth metals (4) Radioactive metals

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Surface Chemistry 26. A catalyst is a substance which

34. The critical micellization concentration (CMC) is

(1) is always in the same phase as in the reactions. (2) alters the equilibrium in a reaction. (3) does not participate in the reaction but alters the rate of reaction. (4) participates in the reaction and provides an easier pathway for the same. 27. Which is not correct for a catalyst? (1) (2) (3) (4)

It enhances the rate of reaction in both directions. It changes enthalpy of reaction. It reduces activation energy of reaction. It is specific in nature.

Colloids, Coagulation, Flocculation and Emulsion 28. Peptization denotes (1) (2) (3) (4)

digestion of food. hydrolysis of proteins. breaking and dispersion into colloidal state. precipitation of a solid from colloidal state.

29. Gelatin protects (1) gold sol. (2) As2S3 sol. (3) Fe(OH)3 sol. (4) all of these. 30. The ability of ion to bring about coagulation of a given colloidal solution depends upon (1) (2) (3) (4)

the size of its ion. the magnitude of charge. the sign of charge. both magnitude and sign of charge.

31. Which of the following ions can cause coagulation of proteins? (1) Ag (2) Fe (3) Cu+ (4) Ni2+ +

3+

32. Lyophilic sols are more stable than lyophobic sols because (1) (2) (3) (4)

the colloidal particles have positive charge. the colloidal particles have negative charge. the colloidal particles are solvated. there are strong electrostatic repulsions between the negatively charged colloidal particles.

33. The stabilization of a dispersed phased in a lyophobic colloid is due to (1) the adsorption of charged substances on dispersed phase. (2) the large electro-kinetic potential developed in the colloid. (3) the formation of an electrical layer between two phases. (4) the viscosity of the medium.

Chapter 12_Surface Chemistry.indd 297

297

(1) the concentration at which micellization begins. (2) the concentration at which true solution is formed. (3) the concentration at which one molar electrolyte is present per 1000 g of solution. (4) the concentration at which solute and solution form equilibrium. 35. Which of the following metal sols cannot be prepared by Bredig’s arc method? (1) Copper (2) Potassium (3) Gold (4) Platinum 36. Flocculation value is expressed in terms of (1) millimole/Liter (2) mole/Liter (3) grams/Liter (4) mole/milliliter 37. In colloidal state, particle size ranges from (1) 1 to 10 Å (2) 20 to 50 Å (3) 10 to 1000 Å (4) 1 to 280 Å 38. When NaCl solution is added to Fe(OH)3 sol (1) [Fe(OH)3]Fe3+ is formed. (2) [Fe(OH)3]Cl− is formed. (3) [Fe(OH)3]Na+ is formed. (4) Fe(OH)3 is coagulated. 39. The protective power of lyophilic sol is (1) dependent on the size of colloidal particles. (2) expressed in terms of gold number. (3) expressed by x/m. (4) directly proportional to the magnitude of charge on it. 40. Tyndall effect is due to (1) reflection of light. (2) scattering of light. (3) absorption of light. (4) adsorption of light. 41. Separation of colloidal particles from those of molecular dimension with electricity is known as (1) electrolysis. (2) electrophoresis. (3) electrodialysis. (4) none of the above. 42. Colligative properties of colloidal sols of the same compound are very low in comparison to their true solutions of same concentration because (1) (2) (3) (4)

colloidal sols do not ionize. colloidal sols ionize upto lesser extent. colloidal sols do not show colligative properties. true solutions do not show colligative properties.

43. Which of the following is not a colloidal solution? (1) solid-solid (2) liquid-liquid (3) gas-gas (4) solid-liquid

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298

OBJECTIVE CHEMISTRY FOR NEET

44. Dialysis is based on the fact that (1) colloidal particles are retained by animal membrane. (2) animal membrane allows the passage of colloidal particles from low to high concentration. (3) animal membrane allows the passage of colloidal particles from high to low concentration. (4) colloidal particles have affinity for dispersion medium. 45. In which of the following respects the lyophilic sols do not differ from lyophobic sols? (1) Stability (2) Reversibility (3) Particle size (4) Behavior towards dispersion medium 46. The various factors on which the color of a sol depends is/are (1) (2) (3) (4)

size and shape of colloidal particles. method of preparation of sol. the way an observer receives the light. all of these.

47. Which one of the following is a natural colloid? (1) Sodium chloride solution (2) Cane sugar solution (3) Urea solution (4) Blood 48. When a colloidal solution is observed under an ultramicroscope, we can see (1) (2) (3) (4)

light scattered by colloidal particles. size of the particle. shape of the particle. relative size.

49. Select correct statement(s) (1) A hydrophilic colloid is a colloid in which there is strong attraction between the dispersed phase and water. (2) A hydrophobic colloid is a colloid in which there is a lack of attraction between the dispersed phase and water. (3) Hydrophobic sols are often formed when a solid crystallizes rapidly from a chemical reaction or form a supersaturated solution. (4) All of the above. 50. The stabilization of the dispersed phase in a lyophobic sol is due to (1) (2) (3) (4)

the viscosity of the medium. the surface tension of the medium. affinity for the medium. the formation of an electrical layer between the two phase.

Chapter 12_Surface Chemistry.indd 298

51. Select correct statement(s): (1) The process of dialysis is similar to osmosis, except that both solvent and small solute particles can pass through the semipermeable dialysis membrane. (2) Proteins can be purified by dialysis. (3) Hemidialysis is used to cleanse the blood of patients whose kidneys are malfunctioning. (4) All of the above. 52. Colloidon is obtained (1) when cellulose nitrate is peptized by organic solvent as ethanol. (2) when high electric current is passed between metal electrodes. (3) colloidal sol is heated. (4) aq NaOH is added to cellulose nitrate. 53. Brownian movement is _____________ property. (1) electrical (2) mechanical (3) optical (4) colligative 54. Tyndall effect is not observed in (1) suspension. (2) starch sol. (3) gold sol. (4) NaCl solution. 55. During electrophoresis of a colloidal solution, colloidal particles move towards (1) anode. (2) cathode. (3) both cathode as well as anode. (4) either cathode or anode. 56. Which of the following is kinetic phenomenon? (1) Brownian motion (2) Tyndall effect (3) Both (1) and (2) (4) None of these 57. The brownian motion is due to (1) temperature fluctuation within the liquid phase. (2) attraction and repulsion between charges on the colloidal particles. (3) impact of molecules of the dispersion medium on the colloidal particles. (4) convection currents. 58. SnO2 is taken in basic medium and current is passed. Colloidal sol migrates towards (1) anode(+plate). (2) cathode(−plate). (3) both (1) and (2). (4) none of these. 59. Colloidal solutions of gold prepared by different methods are of different colors because of (1) (2) (3) (4)

different diameters of colloidal gold particles. variable valency of gold. different concentration of gold particles. impurities produced by different methods.

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Surface Chemistry 60. Of which of the following colloidal systems, fog is an example (1) (2) (3) (4)

69. Some of the properties of emulsions are: (I) Emulsions scatter light and thus exhibit Tyndall effect. (II) Oily emulsions are more viscous than the aqueous emulsions. (III) Electrical conductance of aqueous emulsions is higher that of oily emulsion and increases on the addition of electrolyte.

liquid dispersed in gas. gas dispersed in gas. solid dispersed in gas. solid dispersed in liquid.

61. 1 mol of [AgI] Ag+ sol is coagulated by (1) 1 mol of KI. (2) 500 mL of 1 M K2SO4. (3) both (1) and (2). (4) none of these.



63. Smoke is a dispersion of (1) gas in gas. (2) gas in solid. (3) solid in gas. (4) liquid in gas.

Level II Adsorption 1. The term occlusion is restricted to

64. Which one of the following statements is correct? (1) Brownian movement is more pronounced for smaller particles than for bigger ones. (2) Sols of metal sulphides are lyophilic. (3) Schulze-Hardy law states, the bigger the size of the ion, the greater is its coagulating power. (4) One would expect charcoal to adsorb chlorine more strongly than hydrogen sulphide.

(1) sorption of gases by metals only. (2) sorption of liquid by metals only. (3) alloys. (4) all of these. 2. Surface area per gram of the adsorbent is called (1) (2) (3) (4)

65. The process of passing of a precipitate into colloidal solution, on adding an electrolyte is called (1) dialysis. (2) peptization. (3) electrophoresis. (4) electrosmosis.

the larger its value, the greater is the peptizing power. the lower its value, the greater is the peptizing power. the lower its value, the greater is the protecting power. the larger its value, the greater is the protecting power.

(1) (2) (3) (4)

68. Some of the following are suspensions: (I) Fog (III) Blood (V) Aerosol sprays

 (II)   Air (IV)   Paint (VI)   Pearl

True suspensions are (1) I, II, III (2) III, IV, V (3) I, III, V (4) I, II,V

Chapter 12_Surface Chemistry.indd 299

High pressure and high temperature. Low pressure and low temperature. Low pressure and high temperature. High pressure and low temperature.

4. Following are the events taking place to explain adsorption theory

67. Emulsions are colloids in which both the dispersed phase and dispersion medium are (1) liquids. (2) gases. (3) solids. (4) solid and liquid respectively

molar surface area. normal surface area. specific surface area. equivalent surface area.

3. Which of the following is the most favorable for van der Waals adsorption?

66. Gold number of a lyophilic sol is such property that (1) (2) (3) (4)

Select correct properties: (1) I, II (2) I, III (3) II, III (4) I, II, III

62. The diameter of colloidal particle is of the order (1) 10−3 m (2) 10−5 m (3) 10−15 m (4) 10−7 m

299

(I) Desorption (II) Diffusion of the reactants a long the surface. (III) Adsorption of the reactants. (IV) Formation of the activated surface complex.

These events are taking place in the following order: (1) I, II, III, IV (2) III, II, IV, I (3) III, IV, I, II (4) IV, III, II, I

5. Which gas is adsorbed to maximum extent on the given surface? (1) NH3 (2) H2 (3) N2 (4) O2 6. Which of the following is less than zero during adsorption? ♥

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300

OBJECTIVE CHEMISTRY FOR NEET (1) ∆G (2) ∆S (3) ∆H (4) All of the above

7. There are certain properties related to adsorption: (I) Reversible (II) Formation of unimolecular layer. (III) Low heat of adsorption. (IV) Occurs at low temperature and decreases with increasing temperature.

Which of the above properties are for physical adsorption? (1) I, II ,III (2) I, III, IV (3) II, III, IV (4) I, III

10. Which of the following is the variation of physical adsorption with temperature? (1)

(2)

Adsorption

Adsorption T

T

(3)

(4)

Adsorption

Adsorption

Adsorption Isotherms

T

8. In the following isotherm, which of the following is correct?

T

11. When saturation is attained in terms of adsorption, variæ xö ation of ç ÷ and C (concentration) is given by which è mø portion of the isotherm? C

x/m

A

B

x/m

p

O

O

B

Conc. (C)

(1) OA (2) OB (3) AB (4) BC

x µ p 0 when point B is reached. m (2) desorption may start along AB. (1)

12. 50 mL of 1 M oxalic acid (Molar mass = 126) is shaken with 0.5 g of wood charcoal. The final concentration of the solution after adsorption is 0.5 M. What is the amount of oxalic acid adsorbed per g of carbon?

x µ p1/n along OA. (3) m (4) All of these. 9. Which plot is the adsorption isobar for chemisorption? (x is the mass of gas adsorbed on the surface of m gram of adsorbent.) (1)

A

(2)

(1) 3.15 (2) 1.575 (3) 6.30 (4) 12.60

Catalyst 13. The enzyme ptyalin used for digestion of food is present in (1) saliva. (2) blood. (3) intestine. (4) adrenal glands.

x/m

x/m

14. ZMS-5 is one of the zeolite with formula

T

T

(4)

(3)

x/m

x/m

T

Chapter 12_Surface Chemistry.indd 300

(1) Hx[(AlO2)x·(SiO2)96-x]·16H2O (2) Na56[(AlO2)56·(SiO2)36]·250 H2O (3) Na2Ca(AlO2)2(SiO2)4·6H2O (4) none is correct 15. Which of the following acts as poison for Pd charcoal in Lindlar catalyst?

T

(1) BaSO4 (2) Quinoline (3) Both (1) and (2) (4) None

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Surface Chemistry 16. Catalytic poison acts by (1) coagulating the catalyst. (2) getting adsorbed on the active centers on the surface of catalyst. (3) chemical combination with any one of the reactants. (4) None of the above. 17. Zeolites are used as catalyst in (1) (2) (3) (4)

petrochemical industries during cracking. in the preparation of H2SO4. in the hydrolysis of ester. All of the above.

18. Which of the following can catalyze decomposition of ozone layer? (1) Cl from decomposition of Freon. (2) NO formed near ozone layer due to interaction of N2 and O2 or during lighting. (3) Both (1) and (2). (4) None of these. 19. In Zeigler Natta polymerization of ethylene, the active species is (1) AlCl3 (2) Et3Al (3) C2H4 (4) Ti3+ 20. The formation of diethyl ether from ethanol is catalyzed by (1) H2SO4 (2) AlCl3 (3) Cu (4) Ni

Colloids, Coagulation, Flocculation and Emulsion 21. Softening of hard water is done using sodium aluminium silicate (Zeolite). This causes (1) adsorption of Ca2+ and Mg2+ ions of hard water replacing Na+ ions. (2) adsorption of Ca2+ and Mg2+ of hard water replacing Al3+ ions. (3) both (1) and (2). (4) none of these. 22. Dirty water is passed through a tunnel fitted with metallic electrodes which are maintained at high voltage to remove dirt, rubbish etc. This is based on the principle of (1) electrolysis. (2) electrophoresis. (3) dialysis. (4) de-emulsification. 23. A sol has positively charged colloidal particles. Which of the following solutions is required in lowest concentration for coagulation? (1) NaCl (2) K4[Fe(CN)6] (3) ZnCl2 (4) Na2SO4 24. Purple of cassius is (1) colloidal solution of gold.

Chapter 12_Surface Chemistry.indd 301

301

(2) colloidal solution of silver. (3) colloidal solution of platinum. (4) oxyacids of gold. 25. On adding few drops of dil. HCl to freshly precipitated ferric hydroxide, a red colored colloidal solution is obtained. This phenomenon is known as (1) peptization. (2) dialysis. (3) protective action. (4) dissolution. 26. When a river enters the sea, a delta is formed. Formation of delta is due to (1) peptization. (2) coagulation. (3) emulsification. (4) dialysis. 27. A colloidal solution is subjected to an electrical field. The particles move towards anode. The coagulation of same sol is studied using NaCl, BaCl2 and AlCl3 solutions. Their coagulating power should be (1) NaCl > BaCl2 > AlCl3 (2) BaCl2 > AlCl3 > NaCl (3) AlCl3 > BaCl2 > NaCl (4) BaCl2 > NaCl > AlCl3 28. Cottrell precipitator acts on which of the following principle? (1) (2) (3) (4)

Hardy−Schulze rule Distribution law Le Chatelier’s principle Neutralization of charge on the colloidal particles.

29. The gold numbers of four protective colloids O, P, Q and R are 0.005, 0.01, 0.1 and 0.5 respectively. The decreasing order of their protective power is (1) R, Q, P, O (2) O, P, Q, R (3) P, Q, R, O (4) Q, R, O, P 30. Which of the following electrolytes brings about the coagulation of a gold sol quickest and in least concentration? (1) NaCl (2) AlPO4 (3) MgSO4 (4) K3[Fe(CN)6] 31. Which of the following will have the least coagulating power for arsenious sulphide sol? (1) Na+ (2) Mg2+ (3) Al3+ (4) Ca2+ 32. Colloidal sulphur particles are negatively charged with S 2O32- and other ions on the surface of sulphur. Hence, most effective electrolyte in coagulating colloidal sulphur is (1) NaCl (2) MgCl2 (3) AlCl3 (4) Na2SO4 33. A colloidal solution may be coagulated by (1) adding electrolyte. (2) heating. (3) adding oppositely charged sol. (4) all of these.

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302

OBJECTIVE CHEMISTRY FOR NEET

34. Which of the following is most effective in causing the coagulation of ferric hydroxide sol? (1) KCl (2) KNO3 (3) K2SO4 (4) K3[Fe(CN)6] 35. On adding 1 mL solution of 10% NaCl to 10 mL gold sol in the presence of 0.25 g of starch, the coagulation is just prevented. Starch has the gold number

43. An arsenious sulphide solution carries a negative charge. The maximum precipitating power of this sol is possessed by (1) K2SO4 (2) CaCl2 (3) Na3PO4 (4) AlCl3 44. A colloid of gold particles is (1) (2) (3) (4)

(1) 0.025 (2) 0.25 (3) 2.5 (4) 250 36. 1 mol of [AgI]I− requires AgNO3, Pb(NO3)2 and Fe(NO3)3 as

45. Electro-osmosis is observed when

(1) 1, 1, 1 (2) 1, 2, 3 (3) 1, 1/2, 1/3 (4) 6, 3, 2

(1) (2) (3) (4)

37. Ultrafiltration is used for (1) (2) (3) (4)

concentration of the sol. purification of the sol. both (1) and (2). none of these.

100 mL of 0.1 M AgNO3 + 100 mL of 0.1 M KI 100 mL of 0.1 M AgNO3 + 100 mL of 0.2 M KI 100 mL of 0.2 M AgNO3 + 100 mL of 0.1 M KI 100 mL of 0.15 AgNO3 + 100 mL of 0.15 M KI

39. Aluminium hydroxide forms a positively charged sol. Which of the following ionic substance should be most effective coagulating the sol? (1) NaCl (2) Fe2(SO4)3 (3) CaCl2 (4) K3PO4

(I) Casein: 0.01 (II) Hemoglobin: 0.03 (III) Gum Arabic: 0.15 (IV) Sodium oleate: 0.40 Which has maximum protective power? (1) I (2) II (3) III (4) IV 47. If a freshly prepared precipitate of SnO2 is peptized by a small amount of NaOH, these colloidal particles may be represented as 2-

(1) [SnO2] SnO3  :2 Na+ (2) [SnO2] Sn4+:O2− (3) [SnO2] Na+:OH− (4) [SnO2] Sn4+:OH− 48. The emulsifying agent in milk is (1) lactic acid. (2) fat. (3) lactose. (4) casein.

40. Colloidal particles in soap sol carry (1) (2) (3) (4)

negative charge. positive charge. no charge. either positive or negative charge.

41. To prevent possible coagulation caused by same amount of an electrolyte; quantities of protective colloids A, B, C and D are 2 g, 1.5 g, 1 g and 1.75 g; increasing order of the gold numbers of these protective colloids will be

49. Energy of activation of forward and backward reaction is equal in cases (numerical values) where (1) (2) (3) (4)

(1) 0.03 mg (2) 30 mg (3) 0.30 mg (4) 3 mg

Chapter 12_Surface Chemistry.indd 302

∆H = 0 no catalyst present. ∆S = 0 stoichiometry is the mechanism.

50. Coagulation or demulsification can be done by some of the method given below; (I) by addition of a substance which would destroy the emulsifier. (II) by addition of an electrolyte which would destroy the charge. (III) by heating, freezing and centrifuging.

(1) C < A < B < D (2) B < C < D < A (3) C < B < D < A (4) A < D < B < C 42. Gold number of hemoglobin is 0.03. Hence, 100 mL of gold sol will require how much hemoglobin so that gold is not coagulated by 1 mL of 10% NaCl solution?

dispersion medium begins to move in an electric field. dispersed phase begins to move in an electric field. in both (1) and (2). in none of these.

46. Gold number is maximum for the lyophilic sol:

38. On adding AgNO3 solution into KI solution, a negatively charged colloidal sol is obtained when they are in (1) (2) (3) (4)

hydrophobic in nature. is a sol (of solid dispersed in water). both (1) and (2). none of these.



Select correct method: (1) I, II (2) I, II, III (3) II only (4) III only

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Surface Chemistry

Previous Years’ NEET Questions 1. The Langmuir adsorption isotherm is deduced using the following assumption (1) The adsorbed molecules interact with each other. (2) The adsorption takes place in multi-layers. (3) The adsorption sites are equivalent in their ability to adsorb the particle. (4) The heat of adsorption varies with coverage. (AIPMT 2007)

5. The protecting power of lyophilic colloidal sol is expressed in terms of (1) (2) (3) (4)

coagulation value. gold number. critical micelle concentration. oxidation number. (AIPMT PRE 2012)

6. Which of the following statements is correct for the spontaneous adsorption of a gas? (1) ∆S is negative and therefore, ∆H should be highly positive. (2) ∆S is negative and therefore, ∆H should be highly negative. (3) ∆S is positive and therefore, ∆H should be negative. (4) ∆S is positive and therefore, ∆H should also be highly positive.

2. If x is amount of adsorbate and m is amount of adsorbent, which of the following relations is not related to adsorption process? x = p ´T m x (2) = f ( p ) at constant T m x (3) = f (T ) at constant p m (1)

(AIMPT 2014) 7. Which property of colloids is not dependent on the charge of colloidal particles?

æ xö (4) p = f (T ) at constant ç ÷ è mø

(1) Coagulation (2) Electrophoresis (3) Electro-osmosis (4) Tyndall effect

(AIPMT PRE 2011)

(AIPMT 2014, 2015)

3. In Freundlich adsorption isotherm, the value of 1/n is (1) (2) (3) (4)

303

8. Fog is a colloidal solution of

between 0 and 1 in all cases. between 2 and 4 in all cases. 1 in case of physical adsorption. 1 in case of chemisorption.

(1) Liquid in gas. (2) Gas in liquid. (3) Solid in gas. (4) Gas in gas. (NEET-I 2016) (AIPMT PRE 2012)

4. Which one of the following statements is incorrect about enzyme catalysis? (1) Enzymes are mostly proteinous in nature. (2) Enzyme action is specific. (3) Enzymes are denatured by ultraviolet rays at high temperature. (4) Enzymes are least reactive at optimum temperature.

9. The coagulation values in millimoles per liter of the electrolytes used for the coagulation of As2S3 are given below: (I) NaCl = 52, (III) MgSO4 = 0.22

(II) BaCl2 = 0.69,

The correct order of their coagulating power is (1) II > I > III (2) III > II > I (3) III > I > II (4) I > II > III (NEET-II 2016)

(AIPMT PRE 2012)

Answer Key Level I  1. (3)

 2. (4)

 3. (4)

 4. (1)

 5. (2)

 6. (4)

 7. (2)

 8. (1)

 9. (1)

10. (2)

11. (1)

12. (2)

13. (1)

14. (4)

15. (2)

16. (4)

17. (2)

18. (3)

19. (1)

20. (1)

21. (2)

22. (1)

23. (1)

24. (2)

25. (2)

26. (3)

27. (2)

28. (3)

29. (1)

30. (4)

31. (1)

32. (3)

33. (3)

34. (1)

35. (2)

36. (1)

37. (3)

38. (4)

39. (2)

40. (2)

41. (3)

42. (4)

43. (3)

44. (1)

45. (3)

46. (4)

47. (4)

48. (1)

49. (4)

50. (4)

51. (4)

52. (1)

53. (2)

54. (4)

55. (4)

56. (1)

57. (3)

58. (1)

59. (1)

60. (1)

61. (3)

62. (4)

63. (3)

64. (1)

65. (2)

66. (3)

67. (1)

68. (2)

69. (4)

Chapter 12_Surface Chemistry.indd 303

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OBJECTIVE CHEMISTRY FOR NEET

Level II  1. (1)

 2. (3)

 3. (4)

 4. (2)

 5. (1)

 6. (4)

 7. (2)

 8. (4)

 9. (3)

10. (2)

11. (4)

12. (3)

13. (1)

14. (1)

15. (3)

16. (2)

17. (1)

18. (3)

19. (4)

20. (1)

21. (1)

22. (2)

23. (2)

24. (1)

25. (1)

26. (2)

27. (3)

28. (3)

29. (2)

30. (2)

31. (1)

32. (3)

33. (4)

34. (4)

35. (4)

36. (3)

37. (2)

38. (2)

39. (4)

40. (1)

41. (3)

42. (3)

43. (4)

44. (3)

45. (1)

46. (1)

47. (1)

48. (4)

49. (1)

50. (2)

 5. (2)

 6. (2)

 7. (4)

 8. (1)

 9. (2)

Previous Years’ NEET Questions  1. (3)

 2. (1)

 3. (1)

 4. (4)

Hints and Explanations Level I



17. (2) According to Freundlich adsorption isotherm, we have

Number of moles of oxalic acid adsorbed per gram of 2.5 ´ 10-2 carbon = 0.5



Weight of oxalic acid adsorbed per gram of carbon = 2.5 ´ 10-2 ´ 126 = 6.30 0.5

æxö 1/n ç ÷ = k(C ) èmø

Taking log on both the sides, we get





1 æxö log ç ÷ = log k + log C (1) m n è ø



Since,

æxö log ç ÷ = constant èmø



Therefore, from Eq. (1), we get C = 1 M

27. (3) According to Hardy-Schulze rule, more is the charge of coagulating ion, higher is the coagulating power. Thus, coagulating power is Al3+ > Ba2+ > Na+ or AlCl3 > BaCl2 > NaCl 29. (2) Protective power is inversely proportional to gold number. The increasing order of gold number is O < P < Q < R, thus, increasing order of protective power is R < Q < P < O.

36. (1) Flocculation value = Number of millmoles of impurity to be added to 1 L of the colloid to cause flocculation or coagulation.

30. (2) In AlPO4 both cation and anion have the highest charge (+3 and −3 respectively) and thus causes the highest coagulation.

37. (3) Colloidal particle size is 1 to 100 nm.

31. (1) As Na+ has the least charge; therefore, its coagulating power is the least.

Level II 6. (4) The phenomenon of adsorption is spontaneous and so the free energy (DG) of the system should be negative. DG = DH - T DS

Since process of adsorption restricts the movement of gas molecules, it is always accompanied by a decrease in the entropy (DS). Since adsorption is an exothermic process, the enthalpy change (DH) of the system is also negative.

12. (3) Initial moles of oxalic acid = 50 × 1 × 10 = 5 × 10 mol −3

−2



Final moles of oxalic acid = 50 × 0.5 × 10−3 = 2.5 × 10−2 mol



Therefore, number of moles of oxalic acid adsorbed = 5 × 10−2 – 2.5 × 10−2 = 2.5 × 10−2 mol

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34. (4) [Fe(CN)6]3− has the highest charge, therefore, will cause highest coagulation. 1 1 36. (3) [AgI]I− requires 1 mol of Ag+, mol Pb2+ and mol 2 3 Fe3+. 39. (4) Positively charged sol requires negatively charged ion for coagulation.

    Charge: Cl - < SO24- < PO34-



Coagulating power: Cl - < SO24- < PO34-



Therefore, K3PO4 is the strongest coagulating agent.

41. (3) Quantity of protective colloid is C < B < D < A

Therefore, increasing order of gold number is C < B < D < A.

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Surface Chemistry 42. (3) Gold number is the minimum amount (in mg.) of protective colloid to be added in 10 mL. Gold sol to prevent coagulation by addition of 1 mL of 10% NaCl solution is 0.03 (Given).

Therefore, 100 mL gold sol will require = 0.03 × 10 = 0.3 mg

43. (4) Negatively charged sol requires positive ion for coagulation. Charge: K + = Na + < Ca 2+ < Al 3+







Therefore, precipitating power is K + = Na + < Ca 2+ < Al 3+.

46. (1) Protective power is inversely proportional to gold number.

Therefore, casein will have maximum protective power.

47. (1) The reaction is SnO2 + NaOH ® Na 2SnO3

Therefore, it is negatively charged colloid and can be represented as [SnO 2 ]SnO 32- : 2Na +

Previous Years’ NEET Questions 1. (3) Langmuir adsorption isotherm is based on the assumption that every adsorption site is equivalent and the ability of a particle to bind there is independent of whether or not nearby sites are occupied. 2. (1) According to Langmuir isotherm x = f ( p ) at constant T m = f (T ) at constant p

where x is the amount of gas absorbed and m is the mass of the adsorbent.

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305

3. (1) According to Freundlich isotherm, x/m = kp1/n where the value of n is 0 < 1/n < 1. 4. (4) Enzymes are most active at optimum temperature. At this temperature, the rate of an enzyme reaction becomes maximum. 5. (2) Gold number is the minimum weight (in milligrams) which must be added to 10 mL gold sol so that no coagulation of the gold sol takes place when 1 mL of 10% NaCl solution is rapidly added to it. 6. (2) For a spontaneous process, DG should be negative.

DG = DH - T DS

Thus, DG will be negative when DS is negative and therefore, DH should be highly negative. 7. (4) Tyndall effect is independent on the charge of the colloidal particles while depends on the size. 8. (1) Fog is colloidal solution of liquid droplets in gas (aerosol). 9. (2) The minimum concentration of the electrolyte (in millimoles) that is required to bring about coagulation of one liter of solution in two hours is called the coagulation value. The reciprocal of coagulation value is the coagulating power of the electrolyte. Coagulating power ∝

1 Coagulation value

Thus, the correct order of coagulating power is III > II > I.

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INORGANIC CHEMISTRY

Chapter 13_Classification of Elements and Periodicity in Properties.indd 307

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Chapter 13_Classification of Elements and Periodicity in Properties.indd 308

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13

Classification of Elements and Periodicity in Properties

Chapter at a Glance 1. Periodic table is the arrangement of elements in a tabular form on the basis of their properties that facilitates the systematic study of properties of elements. 2. Genesis of Periodic Classification (a)  Dobereiner’s observed that in the set of three elements having similar properties (called triads), the atomic weight of the middle element is the arithmetic mean of the atomic weights of other two elements. But, only few such triads were available at that time and day by day as many more elements were discovered, the rule could no longer be generalized (b)  Newland arranged 56 elements known then in increasing order of their atomic weight and observed that “the properties of every eighth element are similar to that of first one.” He compared this relationship to the first octave in music with eight notes and called it Newlands’ law of octaves. But, beyond Ca, this repetition was not observed. And, the inert gases were also not discovered till then. (c)  Lother Meyer plotted the atomic volume verses atomic mass curve. He proposed that the physical properties of elements are periodic functions of their atomic weights. (d)  Mendeleev’s periodic table stated that the physical and chemical properties of elements are the periodic function of their atomic weights. Study of properties of elements became more systematic and easier through this. And, its vacant positions guided discovery of new elements. But, the position of hydrogen and isotopes were uncertain. And, some of the elements were wrongly placed. (e)  Moseley’s work: He performed an experiment in which he bombarded high speed electrons on different metal surfaces and obtained X-rays. He observed that there existed a systematic mathematical relationship between the wavelengths of the X-rays produced and the atomic numbers of the elements. Mathematically, it is expressed as u ∝Z

3. Modern Periodic Law According to the modern periodic law, the physical and chemical properties of the elements are the periodic functions of their atomic number. 4. IUPAC Nomenclature of Elements with Atomic Numbers >100 The roots are put together in order of digits which make up the atomic number and “ium’ is added at the end. 0 = nil (n) 1 = un (u)

2 = bi (b) 3 = tri (t)

4 = quad (q) 5 = pent (p)

6 = hex (h) 7 = sept (s)

8 = oct (o) 9 = enn (e)

When some letters are repeated

tri + ium = trium, enn + nil = ennil Note: On 30th November 2016, the IUPAC approved the name and symbols for four elements. Nihonium (113, Nh), Moscovium (115, Mc), Tennessine (117, Ts), and Oganesson (118, Og). 5. Classification of Elements based on Electronic Configuration (a)  s-block elements: If the last electron enters into s-orbital, the elements are called as s-block elements. Their electronic configuration is ns1−2. These comprise Group 1 (ns1, alkali metals) and 2 (ns2, alkaline earth metals) elements. They are metals, good conductors or electricity, soft, highly reactive (do not occur in free-state) and generally form colorless ionic compounds (except Li and Be).

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(b)  p-block elements: If the last electron enters into the p-orbital, the elements are called as p-block elements. Their electronic configuration is ns2 np1–6. These comprise elements of Groups 13 to 18. (i)  Group 13: ns2np1; moderately reactive and compounds are borderline between ionic and covalent. Aluminium, gallium and indium are more metallic than boron.  (ii)  Group 14: ns2np2; with increasing atomic number (carbon, silicon, germanium, tin and lead) metallic character increases. (iii)  Group 15: ns2np3; also sometimes called pnictogens (nitrogen, phosphorus, arsenic, antimony and bismuth). (iv) Group 16: ns2np4; also called chalcogens (oxygen, sulphur, selenium, tellurium and polonium). (v) Group 17: ns2np5; also called halogens (fluorine, chlorine, bromine, iodine and astatine), these attain stable configuration by gaining one electron. (vi)  Group 18: ns2np6; also called noble gases (neon, argon, krypton, xenon and radon) inert gases due to stable filled outer shell configuration. (c) d-block elements: If the last electron enters into d orbital, the elements are called as d-block or transition elements (except Thorium). Their electronic configuration: ns0–2 (n − 1)d 1–10 or ns1–2 (n − 1)d 1–10 (except for palladium). These are all metals; mostly hard strong ones that well conduct heat and electricity. They form many colored and paramagnetic compounds. (d) f-block elements: If the last electron of the elements enters into f-orbital, they are considered as f-block elements. Their electronic configuration is ns2 (n − 1)d 0–1(n − 2)f 1−14. The 14 elements between lanthanum and hafnium are called the lanthanoids. Similar set of elements, between actinium and rutherfordium are called the actinoids. 6. Periodic Trends in Physical Properties (a) A  tomic radius is the distance from the nucleus to the outermost electron or up to point at which the probability of finding of electron is the maximum. There are three types of atomic radii:



van der Waals radii > Metallic radii > Covalent radii

(i) Variation in a period: There is a general decrease of atomic radius in a period for s-and p-block elements because the effective nuclear charge (Zeff ) increases across a row. For d-block elements, the atomic radius initially decreases, then remains constant and finally increases again. (ii) Variation in a group: For s- and p-block elements, the atomic radii increases down the group due to successive use of orbitals with principal quantum number (n) one higher than the last. For d-block elements, the trend in atomic radii is: r3d series < r4d series ≈ r5d series. It is because the radius increases as the shell number increases. The radius of elements of 4d series is about the same as elements of 5d series. This is due to the lanthanoid contraction. (b) Ionic radius depends upon following factors: (i) If the charge of cation increases, the ionic radius decreases. (ii) If the charge of anion increases, the ionic radius increases. (iii) For isoelectronic species, as the number of protons increases the radius of ion decreases. (c) I onization energy is the energy required to remove an electron from an isolated gaseous atom in its ground state. (i) Variation in a period: For representative (s- and p-block) elements, Zeff increases in a period, hence IE increases. For d-block elements, there is slight increase due to increase in Zeff values along the period but it is not prominent. There is a sudden jump from copper to zinc, silver to cadmium and gold to mercury because of full-filled configurations. (ii) Variation in a group: For representative (s- and p-block) elements, the IE generally decreases down a group. For d-block elements, IE1 of 3d series ≈ IE1 of 4d series while IE1 of 4d series < IE1 of 5d series due to lanthanoid contraction. (d) Electron affinity is the energy released when one mole of electron is added to one mole of an element in its isolated gaseous state (ground state). The electron affinity for the first electron is greater than that of the second electron, that is, EA1 > EA2.

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311

(i) V  ariation in a period: For representative (s- and p-block) elements, it increases from left to right as Zeff increases (except inert gases). (ii) Variation in a group: For representative (s- and p-block) elements, there is decrease of electron affinity in general but for p-block elements only the EA1 of second period element is less than EA1 of third period element. (e) Electronegativity is the tendency of the atom to attract electrons (bond pair) towards itself when combined in a compound. It is not a measurable quantity; instead the values are expressed in a number of numerical scales, such as Pauling scale, Mulliken–Jaffe scale and Allred–Rochow scale. It does not remain constant but varies depending on the element to which it is bound. It generally increases across a period with decrease in atomic radius and decreases with increase in atomic radii down the group. 7. Periodic Trends in Chemical Properties (a) P  eriodicity in oxidation states (valence): Valence is the number of electrons in the outermost shell. The periodic trends in valence can be depicted by formulas of their oxides and hydrides. Variable valence is shown to a limited extent in the p-block. In these cases, the oxidation state always changes by 2, due to a pair of electrons remaining paired and not taking part in bonding (the inert pair effect). (b) Anomalous properties of elements and diagonal relationship: The first element of each of the Groups 1, 2, 13–17 differ significantly from the rest of the elements of their respective group. This is due to their small size, large charge to radius ratio; high electronegativity and absence of d-orbitals for bonding. The chemical properties of these elements resemble the properties of second element of the following groups that is with the elements placed diagonally to them in the periodic table. 8. Periodic Trends in Chemical Reactivity (a) H  igh chemical reactivity is observed at the two extremes of the periodic table and lowest in the center. Maximum reactivity by loss of an electron to form a cation is shown by the groups at the extreme left (alkali metals) and by gain of an electron to form an anion by the groups in the extreme right (halogens). (b) The metallic character decreases and the non-metallic character increases in moving from left to right across a period. The metallic character increases down the group and the non-metallic character decreases. (c) The basic character of the oxides decreases while the acidic character increases on moving from left to right in a period. The oxides of elements in the centre are amphoteric (e.g., Al2O3, As2O3) or neutral (CO, NO, etc.) (d) The reducing nature of elements decreases from left to right in a period whereas the oxidizing nature increases. (e) Other chemical properties that show regular variations due to periodic trends are: (i) Properties of hydrides and their volatility and reactivity. (ii) Properties of halides and trends in their covalent/ionic character. (iii) Trends in electrovalence and covalence. 9. Ionic Mobility It is defined as the ability of an ion to move in an aqueous solution. Ionic mobility ∝

1 Hydrated radius of ion

Solved Examples 1. Which will have the maximum value of electron affinity OX, OY, OZ (X, Y and Z respectively are 0, −1 and −2)? (1) OX (2) OY (3) OZ (4) All have equal.

Chapter 13_Classification of Elements and Periodicity in Properties.indd 311

Solution (1) Being neutral atom oxygen will have higher electron affinity. There is electrostatic repulsion between additional electron and negative ion in case of O– and O2–.

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2. The first ionization potential of four consecutive elements, present in the second period of the periodic table is 8.3, 11.3, 14.5 and 13.6 eV respectively. Which one of the following is the first ionization potential (in eV) of nitrogen? (1) 13.6 (2) 11.3 (3) 8.3 (4) 14.5 Solution (4) The ionization energies increase regularly for the first three elements. Then there is decrease in the IE value from third to fourth element. This indicates, third element must possess stable configuration. Hence the third element is nitrogen. 3. Which one of the following order represented the correct sequence of the increasing basic nature of the given oxides? (1) MgO < K2O < Al2O3 < Na2O (2) Na2O < K2O < MgO < Al2O3 (3) K2O < Na2O < Al2O3 < MgO (4) Al2O3 < MgO < Na2O < K2O Solution (4) Because across a period, metallic strength decreases, and hence the basic character also decreases. 4. The correct sequence which shows decreasing order of the ionic radii of the elements is (1) O2− > F− > Na+ > Mg2+ > Al3+ (2) Al3+ > Mg2+ > Na+ > F− > O2− (3) Na+ > Mg2+ > Al3+ > O2− > F− (4) Na+ > F− > Mg2+ > O2− > Al3+ Solution (1) For isoelectronic species, radii would be different because of different nuclear charges. The cation with the greater positive charge will have smaller radius because of the greater attraction of electron to the nucleus. Anions with greater negative charge will have larger radius because here the net repulsion of electrons will outweigh the nuclear charge and ion will expand in size. 5. The set representing the correct order of ionic radius is (1) Li+ > Be2+ > Na+ > Mg2+ (2) Na+ > Li+ > Mg2+ > Be2+ (3) Li+ > Na+ > Mg2+ > Be2+ (4) Mg2+ > Be2+ > Li+ > Na+ Solution (1) Because in a period, the ionic radius decreases as we move from left to right because of increase in nuclear charge and also the value of principal quantum number remains the same. Whereas in a group, the size increases as we move down primarily due to an increase in the number of shells.

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6. Which one of the following constitutes a group of the isoelectronic species? (1) C 22- , O2- , CO, NO

(2) NO+ , C 22- , CN - , N 2

(3) CN - , N 2 ,O 22- ,C 22- (4) N 2 , O22- , NO + , CO Solution (2) NO+, C 22 , CN and N2. The number of electrons is

NO+ = 7 + 8 - 1 = 14 ; C 22- = 6 + 6 + 2 = 14; CN - = 6 + 7 + 1 = 14 ; N 2 = 7 × 2 = 14 7. Lithium shows diagonal relationship with (1) Be (2) B (3) Mg (4) Al Solution (3) Lithium is the smallest alkali metal, and shows diagonal relationship with magnesium (Mg) because of similar size and charge density. 8. Which of the following elements will have the lowest first ionization enthalpy? (1) Li (2) Mg (3) Ca (4) Rb Solution (4) On moving down the group, value of the principal quantum number (n) of the valence-shell electron increases, and hence, the magnitude of the ionization potential decreases. Hence, Rb has the lowest first ionization enthalpy. 9. Variable valence in general, is exhibited by (1) transition elements. (2) gaseous elements. (3) non-metals. (4) s-block elements. Solution (1) The transition elements have their valence electrons in two different sets of orbitals, that is, (n − 1)d and ns. There is very little difference in the energies of these orbitals, so both these energy levels can be used for bond formation, which is why transition elements show variable valency. 10. The basis of keeping the elements in the group of a periodic table is (1) ionization potential. (2) electronegativity. (3) electron gain enthalpy. (4) number of valence electrons. Solution (4) The basis of keeping the elements in the group of a periodic table is number of valence electrons. Moseley showed in 1911 that, since the physical and chemical

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Classification of Elements and Periodicity in Properties properties of an element depend on the number of electrons and their arrangement in different orbitals of the atom, the classification of the elements should be based on the number of these valence electrons (i.e., atomic number) and their arrangement in different orbitals.

Solution (4) We have

11. If the Aufbau principle had not been followed, Ca (Z = 20) would have been placed in the (1) s-block. (2) p-block. (3) d-block. (4) f-block.

If Aufbau principle is violated, then the electronic configuration is 1s2 2s2 2p6 3s2 3p6 3d2.

12. In Lothar Meyer’s curve, the halogens occupy positions on the descending portion of the curve. ascending portion of the curve. peak portion of the curve. no fixed position on the curve.

Solution (2) Lother Meyer plotted the atomic volumes (in cc) of different elements against their atomic numbers and obtained a graph. An inspection of this graph shows that the halogens occupy position on the ascending portion of the curve.

electron gain enthalpy increases. size increases. reactivity increases. ionization enthalpy increases

Solution (2) In going from fluorine to iodine, size increases. The size of the elements in every group of the periodic table increases as we move downwards because there is an increase in the principal quantum number and thus, increase in the number of electron shells. But at the same time, there is an increase in the atomic number or nuclear charge also. As a result, the atomic size must decrease. However, the effect of increase in the electron shells is more pronounced than the effect of increase in nuclear charge. Consequently, the atomic size or atomic radius increases down a group. 16. Which one of the following arrangements represents the correct order of electron gain enthalpy (with negative sign) of the given atomic species? (1) F < Cl < O < S (2) S < O < Cl < F (3) O < S < F < Cl (4) Cl < F < S < O

13. What is the atomic number of Ununhexium? (1) 106 (2) 96 (3) 116 (4) 118

Solution (3) The given elements are placed in the periodic table as

Solution (3) In the IUPAC nomenclature of elements which have atomic number greater than 100, the names have three roots which indicate the three digits of the atomic number of the elements. In the last root –ium is added. The roots and symbols for the digit are given below:



1 Z eff

With the increase in the effective nuclear charge, the power of nucleus to attract the valence shell electrons also increases which result in to decrease in the size of ionic radii.

(1) (2) (3) (4)

(3) Ca (Z = 20) The actual electronic configuration is 1s2 2s2 2p6 3s2 3p6 4s2.

(1) (2) (3) (4)

Ionic radii ∝

15. In going from fluorine to iodine,

Solution





Digit

0

1

2

3

4

5

6

7

8

9

Root

nil

un

Bi

tri quad pent hex sept oct enn

Then for Z = 106 (unnilhexium); Z = 96 (curium); Z = 116 (ununhexium); Z = 118 (ununoctium).

14. Ionic radii are (1) inversely proportional to square of effective nuclear charge. (2) directly proportional to effective nuclear charge. (3) directly proportional to square of effective nuclear charge. (4) inversely proportional to effective nuclear charge.

Chapter 13_Classification of Elements and Periodicity in Properties.indd 313

Group



Group 16

Group 17

Second Period

O

F

Third Period

S

Cl

The electron gain enthalpy decrease in a group but for p-block elements only the EA1 of second period element is less than EA1 of third period element. This is due to the fact that the third period elements have larger size and vacant 3d orbitals, which can accommodate incoming electrons easily. Therefore, S > O and Cl > F. On the other hand, electron gain enthalpy increases in a period, that is, from left to right as Zeff increase, thus, F > O and Cl > S. Hence, the correct order is O < S < F < Cl.

17. Which of the following has maximum ionization potential? (1) Magnesium (2) Aluminium (3) Silicon (4) Phosphorous

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Solution

Solution

(4) Ionization energy increases in the period from left to right but among Mg, Al, Si and P, phosphorous has highest ionization energy. It is because electronic configuration of phosphorous is [Ne]103s23p3.

(1) Lattice energy is the minimum amount of energy required to break the unit cell of the crystal lattice. More is the ionic character greater is the lattice energy.

18. Which of the following does not have any unit?

24. Which of the following sets has the strongest tendency to form anions?

(1) Electronegativity (2) Atomic radii (3) Ionization potential (4) None of these

(1) Ga, In and T (2) N, O and F (3) Na, Mg and Al (4) V, Cr and Mn

Solution

Solution

(1) Electronegativity does not have any unit.

(2) N, O and F have five, six and seven electrons in their valence shells. They have tendency to accept three, two and one electrons respectively in their valence shells and form anions.

19. A sudden jump between the values of second and third ionization energies of an element would be associated with which of the following electronic configuration? (1) 1s22s22p63s1 (3) 1s22s22p63s23p1

(2) 1s22s22p63s23p2 (4) 1s22s22p63s2

25. In K+F–, the ionic radius of K+ is (1) less than F–. (2) equal to F–. (3) more than F–. (4) either (1) or (2).

Solution (4) Element with 1s22s22p63s2 after losing 2 electrons will attain stable electronic configuration 1s22s22p6. Thus its third ionization energy will be too high of second ionization energy.

Solution

20. Which of the following oxide is most acidic?

26. Which of the following electronic configuration is not possible?



(1) N2O5 (3) P2O5

(2) Sb2O5 (4) As2O5

(1) In K+F–, the ionic radius of K+ is 133 pm and ionic radius of F– is 136 pm. Therefore, ionic radius of K+ is less than that of F–.

(1) 1s22s2 (3) 1s22s22p6

Solution (1) Greater the electronegativity, greater will be its acidic character. Thus, order of acidity will be



N2O5 > P2O5 > As2O5 > Sb2O5

21. Which of the following is the most basic oxide? (1) Al2O3 (3) Bi2O3

(2) Sb2O3 (4) SeO2

Solution (4) According to Aufbau principle, the 2p orbital will be filled before 3s orbital. Therefore, the electronic configuration (1s22s22p23s1) is not possible. 27. Which of the following is a correct statement about ionization potential? (1) (2) (3) (4)

Solution (3) Al2O3 and Sb2O3 are amphoteric, SeO2 is acidic and Bi2O3 is basic. 22. The element which occupy the peaks of ionization energy curve are (1) Na, K, Rb, Cs (2) Cl, Br, I, F (3) He, Ne, Ar, Kr (4) Na, Mg, Cl, I Solution

  He > Ne > Ar > Kr

23. The stability of an ionic compound is mostly due to (1) lattice energy. (2) electron affinity. (3) ionization energy. (4) electronegativity.

Chapter 13_Classification of Elements and Periodicity in Properties.indd 314

It is independent of atomic radii. It remains constant for all atomic radii. It increases with increase in atomic radii. It decreases with increase in atomic radii.

Solution (4) Ionization potential is the minimum energy required to remove electrons from valance shell in the isolated gaseous atom. It decreases with increase in the atomic radii. 28. Which of the following is the correct order of ionic size? (1) Ca2+ > K+ > Cl– > S2– (2) Ca2+ > Cl– > K+ > S2– (3) S2– > Cl– > K+ > Ca2+ (4) Cl– > K+ > Ca2+ > S2–

(3) Ionization energy of the noble gases is very high.

(2) 3d104s24p2 (4) 1s22s22p23s1

Solution (3) For isoelectronic atom, great the charge on the nucleus smaller is the radii.



S2– > Cl– > K+ > Ca2+

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Classification of Elements and Periodicity in Properties 29. Electronegativity follows the order (1) F > O > Cl > Br (2) F > Cl > Br > O (3) O > F > Cl > Br (4) Cl > F > O > Br Solution (1) Electronegativity decreases from F to I and O is next to F in order of decreasing electronegativity. 30. The tenth element in the periodic table resembles the element with atomic number

Solution (4) The d-block contains elements of Groups 3–12 of the periodic table. In case the element belongs to d block, then the group number = number of electrons in n (n − 1)d subshell and ns subshell. Therefore, the general configuration for elements of Group 9 is (n − 1)d7ns2. 32. Name an element of p-block of the periodic table in which last electron goes to the s-orbital of valence shell instead of p-orbitals.

(1) 2 as well as 30. (2) 2 as well as 54. (3) 8 as well as 18. (4) 8 only.

(1) As (3) No such element is there

Solution

Solution

(2) The tenth element in the periodic table is neon which has completely filled outer shell. It resembles the element with atomic number 2 (1s2) as well as 54 (1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p65s2 4d10 5p6).

(4) Helium: (Z = 2)1s2

31. The general configuration for elements of Group 9 is (1) (n − 1)d6 ns2 (3) (n − 1)d2 ns2 np6

315

(2)  Ga (4)  He

Helium contains only two electrons. So the electron enters into s orbital and this element is placed in the p block because it has completely filled orbital. So it is placed in noble gas group.

(2) nd7ns2 (4) (n − 1)d7 ns2

Practice Exercises Level I Development of Periodic Table and Modern Periodic Table

(3) Group 5 and 2nd period (4) Group 3 and 5th period 7. The electronic structure (n − 1)d1–10 ns0–2 is characteristic of (1) lanthanoids. (2) actinoids. (3) rare-earth. (4) transition elements.

1. Which of the following electronic configuration represent noble gas? (1) ns2np6 (2) ns2np5 (3) ns2np4 (4) ns2np3

8. Which of the following group of transition metals is called coinage metals? (1) Cu, Ag, Au (2) Ru, Rn, Pd (3) Fe, Co, Ni (4) Os, IR, Pt

2. The elements on the right side of the periodic table are (1) metals. (2) metalloids. (3) non-metals. (4) transition elements.

9. Which one of the following is a d-block element? (1) Gd (2) Hs (3) Es (4) Li

3. Which of the following element shows maximum valency of +4? (1) Carbon (2) Barium (3) Nitrogen (4) Sulphur

10. While moving down a group in the periodic table, which of the following would be true? (I) All the atoms have the same number of valence electrons. (II) Gram atomic volume increases. (III) Electronegativity decreases. (IV) Metallic character decreases and the basic nature of their oxides decrease.

4. Outermost configuration for Z = 25, is (1) 4s23d5 (2) 5s24d5 (3) 4s23d3 (4) 4s23d10 5. Atomic number of Ni and Cu are 28 and 29 respectively. Electronic configuration 1s22s22p63s23p63d10 shows (1) Ni (2) Ni2+ (3) Cu2+ (4) Cu+ 6. Element with atomic number 38 belongs to (1) Group 2 and 5th period (2) Group 2 and 2nd period

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Select the correct answer by using the following codes: (1) I, II and III (2) II, III and IV (3) II and III (4) I and III

11. Which group of the periodic table contains only metals?

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OBJECTIVE CHEMISTRY FOR NEET (1) Group 17 (2) Group 15 (3) Group 2 (4) None of these

12. Which one of the following statements relating to the modern periodic table is incorrect? (1) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbital in a p-sub shell. (2) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbital in a d-sub shell. (3) Each block contains a number of columns equal to the number of electron that can occupy that sub shell. (4) The block indicates value of Azimuthal quantum number for the last subshell that received electrons in building up the electronic configuration. 13. If the atomic number of an element is 33, it will be placed in the periodic table in (1) Group 1 (2) Group 3 (3) Group 15 (4) Group 7 14. The elements in which electrons are progressively filled in 4f-orbital are called (1) actinoids. (2) transition elements. (3) lanthanoids. (4) halogens. 15. What is the atomic number of last member of the 7th period of the extended form of periodic table? (1) 116 (2) 118 (3) 120 (4) 122 16. In modern periodic table, the element with atomic number Z = 118 will be (1) (2) (3) (4)

Uuo; Ununoctium; alkaline earth metal. Uno; Unniloctium; transition metal. Uno; Unniloctium; alkali metal. Uno; Ununoctium; noble gas.

17. Elements in the same vertical group of the periodic table have generally the same (1) (2) (3) (4)

atomic number. number of isotopes. same number of electrons. same number of electrons in the valence shell.

Shielding Effect and Effective Nuclear Charge 18. The order of screening effect of electrons of s, p, d and f orbitals of a given shell of an atom on its outer shell electrons is (1) s > p > d > f (2) f > d > p > s (3) p < d < s < f (4) f > p > s > d 19. Which of the following is/are generally true regarding effective nuclear charge (Zeff )?

Chapter 13_Classification of Elements and Periodicity in Properties.indd 316

(1) It increases on moving left to right in a period. (2) It remains almost constant on moving tip to bottom in a group. (3) For isoelectronic species, as Z increases, Zeff decrease. (4) Both (1) and (2).

Atomic and Ionic Radius 20. Identify the correct order of size of the following (1) Ca2+ < K+ < Ar < Cl− < S2− (2) Ar < Ca2+ < K+ < Cl− < S2− (3) Ca2+ < Ar < K+ < Cl− < S2− (4) Ca2+ < K+ < Ar < S2− < Cl− 21. The atom of smallest atomic radius among the following is (1) Na (2) K (3) Br (4) Li 22. On moving from left to right in a period in transition metals, their atomic size (1) decrease. (2) increase. (3) remain the same. (4) none of these. 23. The size of isoelectronic species F –, Ne and Na+ is affected by (1) (2) (3) (4)

nuclear charge (Z). valence principal quantum number (n). electron-electron interaction in the outer orbitals. none of the factors because their size is the same.

24. Increasing order of atomic radii is (1) Mg2+ < Na+ < Ne < F− < O2− (2) Na+ < Mg2+ < Ne < F < O2− (3) Na+ < Mg2+ < F− < Ne < O2− (4) Na+ < Mg2+ < F− < O2− < Ne 25. Which of the following is the correct order of size of the given species? (1) I > I– > I+ (2) I+ > I– > I (3) I > I+ > I– (4) I– > I > I+ 26. Which of the following order of radii is correct? (1) Li < Be < Mg (2) H+ < Li+ < H– (3) O < F < Ne (4) Li < Na < K < Cs < Rb 27. Which of the following ion has the highest value of ionic radius? (1) Li+ (2) F− (3) O2− (4) B3+

Ionization Enthalpy 28. Which of the following have the maximum ionization enthalpy? (1) P (2) N (3) As (4) Sb

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Classification of Elements and Periodicity in Properties 29. The set representing the correct order for first ionization potential is (1) K > Na > Li (2) Be > Mg > Ca (3) B > C > N (4) Ge > Si > C 30. The electronic configuration of the atom having maximum difference in the first and the second ionization enthalpies is (1) 1s2 2s2 2p6 3s1 (2) 1s2 2s2 2p6 3s2 (3) 1s2 2s2 2p1 (4) 1s2 2s2 2p6 3s2 3p1 31. The ionization enthalpy will be highest when the electron is to be removed from ________ if other factors are equal. (1) s-orbital (2) p-orbital (3) d-orbital (4) f-orbital 32. The atomic numbers of vanadium (V), chromium (Cr), manganese (Mn) and iron (Fe) are respectively 23, 24, 25 and 26, which one of these may be expected to have the highest second ionization enthalpy? (1) V (2) Cr (3) Mn (4) Fe 33. Which among the following elements have lowest value of IE1? (1) Pb (2) Sn (3) Si (4) C 34. The incorrect statement among the following is (1) The first ionization potential of Al is less than the first ionization potential of Mg. (2) The second ionization potential of Mg is greater than the second ionization potential Na. (3) The first ionization potential of Na is less than the first ionization potential of Mg. (4) The third ionization potential of Mg is greater than that of Al. 35. Ionization potential of Na would be numerically the same as (1) (2) (3) (4)

electron affinity of Na+. electronegativity of Na+. electron affinity of Ne. ionization potential of Mg.

36. Identify the least stable ion amongst the following (1) Li- (2) Be(3) B- (4) C 37. The first four ionization enthalpy values of an element are 191, 578, 872 and 5962 kcal. The number of valence electrons in the element is (1) 1 (2) 2 (3) 3 (4) 4

Chapter 13_Classification of Elements and Periodicity in Properties.indd 317

317

Electron Gain Enthalpy and Electron Affinity 38. Elements of which family form anions most readily? (1) Halogens (2) Alkali metals (3) Oxygen (4) Nitrogen 39. The electronic configuration of four elements are given below. Arrange these in the correct order of the magnitude (without sign) of their electron gain enthalpy.  (I) 2s2 2p5  (II)   3s2 3p5 (III) 2s2 2p4 (IV)   3s2 3p4

Select the correct answer using the codes given below. (1) (I) < (II) < (IV) < (III) (2) (II) < (I) < (IV) < (III) (3) (I) < (III) < (IV) < (II) (4) (III) < (IV) < (I) < (II)

40. Which of the following statements is incorrect? (1) The ionization potential of nitrogen is greater than that of oxygen. (2) The electron affinity of fluorine is greater than of chlorine. (3) The ionization potential of Mg is greater than aluminium. (4) The electronegativity of fluorine is greater than that of chlorine. 41. Ease of formation of the anion is favored by (1) (2) (3) (4)

lower value of ionization potential. lower value of electron gain enthalpy. higher value of electron gain enthalpy. lower value of electronegativity.

Electronegativity 42. As compared to nitrogen, oxygen is (1) (2) (3) (4)

less electronegative and less reactive. more electronegative and less reactive. more electronegative and more reactive. less electronegative and more reactive.

43. The correct order of electronegativity on Pauling scale is (1) F > Cl > O > S (2) Li > Na > K > Rb > Cs (3) Be < B < N < C (4) Both (1) and (2) 44. Which element has the highest electronegativity? (1) C (2) O (3) Mg (4) S 45. Which of the following is affected by the stable electronic configuration of an atom? (1) (2) (3) (4)

Only electronegativity. Only ionization enthalpy. Only electron gain enthalpy and ionization enthalpy. All of the above.

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OBJECTIVE CHEMISTRY FOR NEET

Level II Development of Periodic Table and Modern Periodic Table 1. The period number in the long form of the periodic table is equal to (1) magnetic quantum number of any element of the period. (2) atomic number of any element of the period. (3) maximum principal quantum number of any element of the period. (4) maximum Azimuthal quantum number of any element of the period. 2. The electronic configurations of four elements are given below. Which element does not belong to the same family? (1) [Xe]4f 145d106s2 (2) [Kr]4d105s2 2 5 (3) [Ne]3s 3p (4) [Ar]3d104s2 3. Which of the following statements is not correct regarding hydrogen? (1) It resembles halogens in some properties. (2) It resembles alkali metals in some properties. (3) It can be placed in 17th group of modern periodic table. (4) It cannot be placed in 1st group of modern periodic table. 4. Which of the following is known as the bridge element of Group 2 in Mendeleev’s table? (1) Zn (2) Sr (3) Mg (4) Hg 5. Atomic number of Ag is 47. In the same group, the atomic numbers of elements placed above and below Ag in long form of periodic table will be (1) 29, 65 (2) 39, 79 (3) 29, 79 (4) 39, 65 6. Who developed the long form of periodic table? (1) Niels Bohr (2) Moseley (3) Mendeleev (4) Lothar Meyer 7. Element with electronic configuration as [Ar] 3d 54s1 is placed in ___ in the modern periodic table. (1) (2) (3) (4)

IA (1st group), s-block IB (7th group), d-block VIB (8th group), d-block VIB (6th group), d-block

8. Of the following pairs, the one containing example of metalloid elements in the periodic table is (1) sodium and potassium. (2) fluorine and chlorine. (3) calcium and magnesium. (4) boron and silicon.

Chapter 13_Classification of Elements and Periodicity in Properties.indd 318

9. There are four elements ‘p’, ‘q’, ‘r’ and ‘s’ having atomic numbers Z – 1, Z, Z + 1 and Z + 2 respectively. If the element ‘q’ is an inert gas, select the correct answers from the following statements.

(I) p has most negative electron gain enthalpy in the respective period. (II) r is an alkali metal (III) s exists in +2 oxidation state. (1) (I) and (II) only (2) (II) and (III) only (3) (I) and (III) only (4) (I), (II) and (III)

10. Chloride of an element A gives neutral solution in water. In the periodic table, the element A belongs to (1) first group. (2) third group. (3) fifth group. (4) first transition series.

Shielding Effect and Effective Nuclear Charge 11. While dealing with multielectron atoms or ions, we must use the concept of effective nuclear charge, Zeff which is calculated by the formula (1) Zeff = Z + s

(2) Zeff = Z / s

(3) Zeff = Z - s (4) Zeff = Zs 2 12. Which of the following has a different value of effective nuclear charge than the others? (1) Mg (2) Sr (3) Ba (4) Be 13. The screening effect of d-electrons is (1) (2) (3) (4)

equal to that of p-electrons. more than that of p-electrons. same as f-electrons. less than p-electrons.

Atomic and Ionic Radius 14. Select correct statement (s) about radius of an atom: (1) Values of van der Waals radii are larger than those of covalent radii because the van der Waals forces are much weaker than the forces operating between atoms in a covalently bonded molecule. (2) The metallic radii are smaller than the van der Waals radii, since the bonding forces in the metallic crystal lattice are much stronger than the van der Waals forces. (3) Both (1) and (2) (4) None of these 15. Match the correct atomic radius with the element: S. No. (i) (ii) (iii) (iv) (v)

Element Be C O B N

Code (p) (q) (r) (s) (t)

Atomic radius (pm) 74 88 111 77 66

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Classification of Elements and Periodicity in Properties (1) (2) (3) (4)

25. Ease of formation of cation is favored by

(i) – r; (ii) – q; (iii) – t; (iv) – s; (v) – p (i) – t; (ii) – s; (iii) – r; (iv) – p; (v) – q (i) – r; (ii) – s; (iii) – t; (iv) – q; (v) – p (i) – t; (ii) – p; (iii) – r; (iv) – s; (v) – q

(1) (2) (3) (4)

16. Radius of the isoelectronic species (1) (2) (3) (4)

increases with the increase in nuclear charge. decreases with the increase in nuclear charge. first increases and then decreases. is the same for all.

(1) F < Cl < Br < I (2) Y > Sr > Rb (3) Nb > Ta (4) Li > Be > B 2+

+

18. The ionic radii of S2− and Te2− are 1.84 and 2.21 Å, respectively. The ionic radius of Se2− is (1) 1.83 Å (2) 2.22 Å (3) 4.5 Å (4) 2.02 Å

Ionization Enthalpy 21. Which one of the following statement is incorrect in relation to ionization enthalpy? (1) Ionization enthalpy increases for each successive electron. (2) The greatest increase in ionization enthalpy is experienced on removal of electron from core of noble gas configuration. (3) End of valence electron is marked by a big jump in ionization enthalpy. (4) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n values. 22. The first ionization enthalpies (in eV) of N and O are respectively given by (1) 14.6, 13.6 (2) 13.6, 14.6 (3) 13.6, 13.6 (4) 14.6, 14.6 23. The increasing order of the first ionization enthalpies of the elements B, P, S and F (lowest first) is (1) F < S < P < B (2) P < S < B < F (3) B < P < S < F (4) B < S < P < F 24. The first ionization enthalpies of Na, Mg, Al and Si are in the order (1) Na < Mg > Al < Si (2) Na > Mg > Al > Si (3) Na < Mg < Al < Si (4) Na > Mg > Al < Si

Chapter 13_Classification of Elements and Periodicity in Properties.indd 319

IE2

(1) X

500

1000

(2) Y

600

2000

(3) Z

500

7500

(4) M

700

1400

(1) IE1Ca > IE2K (2) IE1K > IE1Ca (3) IE2Ca > IE2K (4) IE2K > IE2Ca 28. Which one of the following elements has the highest ionization enthalpy? (1) [Ne]3s23p1 (2) [Ne]3s23p2 (3) [Ne]3s23p3 (4) [Ar]3d104s24p2

20. Which of the following has the largest ionic radius? (1) Li+ (2) K+ (3) Na+ (4) Cs+

IE1

27. Which of the following relation is correct with respect to first (I) and second (II) ionization enthalpies of potassium and calcium?

19. In which of the following pairs the radii of second species is greater than that of first? (1) K, Ca (2) H, He (3) Mg+, Mg2+ (4) O2–, O–

lower value of ionization potential. higher value of ionization potential. higher value of electron gain enthalpy. higher value of electronegativity.

26. Which represents alkali metals (i.e., 1st group metals) based on IE1 and IE2 values (in kJ mol−1)?

17. Which of the following order of atomic/ionic radius is correct? 3+

319

Electron Gain Enthalpy and Electron Affinity 29. The order of electron gain enthalpy (magnitude) of O, S and Se is (1) O > S > Se (2) S > Se > O (3) Se > S > O (4) S > O > Se 30. Electronic configurations of four electrons A, B , C and D are given below      (I) 1s2 2s2 2p6 (III) 1s2 2s2 2p6 3s1

 (II) 1s2 2s2 2p4 (IV) 1s2 2s2 2p5

Which of the following is correct order of increasing tendency to gain electron? (1) (I) < (III) < (II) < (IV) (2) (I) < (II) < (III) < (IV) (3) (IV) < (II) < (III) < (I) (4) (IV) < (I) < (II) < (III)

31. Which of the following statements is/ are correct? (1) Electron gain enthalpy may be positive for some elements. (2) Second electron gain enthalpy always remains positive for all the elements. (3) ∆Heg(K+) = –IE (K) (4) All of these 32. The electron gain enthalpy values for the halogens show the following trend (1) F < Cl > Br > I (2) F < Cl < Br < I (3) F > Cl > Br > I (4) F < Cl > Br < I

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OBJECTIVE CHEMISTRY FOR NEET

33. Which is chemically most active non-metal?

(4) Ar < Ca2+ < K+ < Cl− < S2−

(1) F2 (2) N2 (3) O2 (4) S 34. Which of the following will have the most negative electron gain enthalpy and which the least negative?



(3) Ca2+ < K+ < Ar < Cl− < S2−

F, P, S, Cl

(AIPMT 2007)

3. The sequence of ionic mobility in aqueous solution is (1) Rb+ > K+ > Cs+ > Na+ (3) K+ > Na+ > Rb+ > Cs+

(2) Na+ > K+ > Rb+ > Cs+ (4) Cs+ > Rb+ > K+ > Na+

(1) P, Cl (2) Cl, F (3) Cl, S (4) Cl, P

(AIPMT 2008)

Electronegativity 35. Read the following statements and answer accordingly

Statement 1: LiCl is predominanty a covalent compound.



Statement 2: Electronegativity difference between Li and Cl is too small. (1) Statement 1 is True, Statement 2 is True, and Statement 2 is correct explanation for Statement 1. (2) Statement 1 is True, Statement 2 is True, and Statement 2 is not correct explanation for Statement 1 (3) Statement 1 is True, Statement 2 is False. (4) Statement 1 is False, Statement 2 is True.

36. The electronegativity values of C, N, O and F on Pauling scale (1) decreases from carbon to fluorine. (2) increases from carbon to fluorine. (3) increases up to oxygen and then decrease up to fluorine. (4) decrease from carbon to nitrogen and then increase continuously. 37. Which of the following is not a reason for diagonal relationship? (1) (2) (3) (4)

Same size Same electronegativity Same electron affinity Same polarizability

Previous Years’ NEET Questions

(AIPMT 2009) 5. Amongst the elements with following electronic configurations, which one of them may have the highest ionization energy? (1) [Ne] 3s2 3p1 (3) [Ne] 3s2 3p2

(2) [Ne] 3s2 3p3 (4) [Ar] 3d10 4s2 4p3 (AIPMT 2009)

6. Which of the following represents the correct order of increasing electron gain enthalpy with negative sign for the elements O, S, F and Cl? (1) S < O < Cl < F (2) Cl < F < O < S (3) O < S < F < Cl (4) F < S < O < Cl (AIPMT PRE 2010) 7. The correct order of the decreasing ionic radii among the following isoelectronic species is (2) Ca2+ > K+ > S2− > Cl− (4) S2− > Cl− > K+ > Ca2+

(1) 1s2 2s2 2p6

(2) 1s2 2s2 2p5

(3) 1s2 2s2 2p3

(4) 1s2 2s2 2p5 3s1 (AIPMT 2007)

2. Identify the correct order of the size of the following. (1) Ca2+ < Ar < K+ < Cl− < S2− (2) Ca < K < Ar < S < Cl

−

Chapter 13_Classification of Elements and Periodicity in Properties.indd 320

8. Among the elements Ca, Mg, P and Cl, the order of increasing atomic radii is (1) Cl < P < Mg < Ca (2) P < Cl < Ca < Mg (3) Ca < Mg < P < Cl (4) Mg < Ca < Cl < P (AIPMT MAINS 2010)

1. With which of the following electronic configuration an atom has the lowest ionization enthalpy?

2−

(2) 3d3 4s2 (4) 3d5 4s2

(AIPMT PRE 2010)

(1) N > P > C > Si (2) C > Si > N > P (3) N < P < C < Si (4) N > C > P > Si

+

(1) 3d2 4s2 (3) 3d5 4s1

(1) K+ > Ca2+ > Cl− > S2− (3) Cl− > S2− > Ca2+ > K+

38. Correct order of electronegativity on Pauling scale is

2+

4. Which one of the elements with the following outer orbital configurations may exhibit the largest number of oxidation states?

9. Among the following which one has the highest cation to anion size ratio? (1) CsF (2) LiF (3) NaF (4) CsI (AIPMT MAINS 2010) 10. Identify the wrong statement in the following. (1) Atomic radius of the elements increases as one move down the first group of the periodic table.

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Classification of Elements and Periodicity in Properties

321

(2) Atomic radius of the elements decreases as one move across from left to right in the second period of the periodic table. (3) Amongst isoelectronic species, smaller the positive charge on the cation, smaller is the ionic radius. (4) Amongst isoelectronic species, greater the negative charge on the anion, larger is the ionic radius.

14. In which of the following options, the order of arrangements does not agree with the variation of property indicated against it?

(AIPMT PRE 2012)

(NEET-I 2016)

11. Be2+ is isoelectronic with which of the following ions? (1) H+ (3) Na+

(2) Li+ (4) Mg2+ (AIPMT 2014)

12. Which of the following orders of ionic radii is correctly represented?

(1) (2) (3) (4)

Al 3+ < Mg 2+ < Na + < F -: Increasing ionic size B < C < N < O: Increasing first ionization enthalpy I < Br < Cl < F: Increasing electron gain enthalpy Li < Na < K < Rb: Increasing metallic radius

15. The element Z = 114 has been discovered recently. It will belong to which of the following family/group and electronic configuration? (1) (2) (3) (4)

Carbon family, [Rn]5f 146d107s27p2 Oxygen family, [Rn]5f 146d107s27p4 Nitrogen family, [Rn]5f 146d107s27p6 Halogen family, [Rn]5f 146d107s27p5 (NEET 2017)

(1) H− > H+ > H (2) O2− > F− > Na+ (3) F− > O2− > Na+

(4) Al3+ > Mg2+ > N3− (AIPMT 2014)

13. The species Ar, K+ and Ca2+ contain the same number of electrons. In which order do their radii increase? (1) Ca2+ < Ar < K+ (3) K+ < Ar < Ca2+

16. Ionic mobility of which of the following alkali metal ions is lowest when aqueous solution of their salts is put under an electric field? (1) K (2) Rb (3) Li (4) Na

(2) Ca2+ < K+ < Ar (4) Ar < K+ Ca2+

(NEET 2017)

(AIPMT 2015)

Answer Key Level I  1. (1)

 2. (3)

 3. (1)

 4. (1)

 5. (4)

 6. (1)

 7. (4)

 8. (1)

 9. (2)

10.  (1)

11.  (3)

12.  (2)

13.  (3)

14.  (3)

15.  (2)

16.  (4)

17.  (4)

18.  (1)

19.  (4)

20.  (1)

21.  (4)

22.  (1)

23.  (1)

24.  (1)

25.  (4)

26.  (2)

27.  (3)

28.  (2)

29.  (2)

30.  (1)

31.  (1)

32.  (2)

33.  (2)

34.  (2)

35.  (1)

36.  (2)

37.  (3)

38.  (1)

39.  (4)

40.  (2)

41.  (3)

42.  (3)

43.  (2)

44.  (2)

45.  (3)

 1. (3)

 2. (3)

 3. (4)

 4. (3)

 5. (3)

 6. (1)

 7. (4)

 8. (4)

 9. (4)

10.  (1)

11.  (3)

12.  (4)

13.  (4)

14.  (3)

15.  (3)

16.  (2)

17.  (1)

18.  (4)

19.  (2)

20.  (4)

21.  (4)

22.  (1)

23.  (4)

24.  (1)

25.  (1)

26.  (3)

27.  (4)

28.  (3)

29.  (2)

30.  (1)

31.  (4)

32.  (1)

33.  (1)

34.  (4)

35.  (3)

36.  (2)

37.  (3)

38.  (4)

 5. (2)

 6. (3)

 7. (4)

 8. (1)

 9. (1)

10.  (3)

15.  (1)

16.  (3)

Level II

Previous Years’ NEET Questions  1. (4)

 2. (3)

 3. (4)

 4. (4)

11.  (2)

12.  (2)

13.  (2)

14.  (2) and (3)

Chapter 13_Classification of Elements and Periodicity in Properties.indd 321

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322

OBJECTIVE CHEMISTRY FOR NEET

Hints and Explanations Level I 2. (3) Metals are present on the left side as well as in the center of the periodic table. (i) Transition elements are present at the center of the periodic table. (ii) Non-metals are mostly located on the right side of the periodic table. (iii) Metalloids constitute some border line elements. 7. (4) Elements in which the last electron enters any one of the five d-orbitals of their respective penultimate shells are called d-block or transition elements. The atoms of the elements belonging to these groups usually have 1 or 2 electrons in the s-orbital of the ns orbital while the remaining electrons are progressively filled in the d-orbitals of their respective penultimate shells, that is, (n − 1) d orbitals. General outer shell electronic configuration of d-block elements is (n − 1)d1–10ns0–2, where n = 4 - 7. 10. (1) Fourth statement is not correct. The correct statement is metallic character increases and the basic nature of their oxides increases. 12. (2) The d-block has 10 columns as maximum 10 electrons can occupy all the orbitals in a d sub-shell. 13. (3) The electronic configuration of element with atomic number 33 is 1s22s22p63s23p63d104s24p3. Therefore, group number of the element = number of d electrons in the penultimate shell + valence shell = 10 + 5 = 15. 17. (4) Elements in the same vertical group of the periodic table have generally the same number of electrons in the valence shell. 22. (1) Generally, the atomic size of transition elements in a series decrease on moving from left to right in a period, but the decrease in atomic size is small after midway. 24. (1) All the given ions are isoelectronic in nature. Mg2+ having the highest nuclear charge (12 units) has the smallest size whereas O2− ion having the smallest nuclear charge (8 units) has the largest size. 27. (3) (i) The size of the anion increases with the increase in the magnitude of negative charge. (ii) The size of the cation decreases with the increase in the magnitude of positive charge. (iii) In O2−, the magnitude of negative charge is maximum, hence the ionic radius has the highest value. 29. (2) These are 2nd group elements and the order correctly reflects trend in ionization energy.

Chapter 13_Classification of Elements and Periodicity in Properties.indd 322

34. (2) Statement (2) is incorrect since Na+ has now stable octet configuration (2s22p6) and requires greater energy to remove second electron than in case of Mg+ (2p63s1). 36. (2) Be- is least stable ion because, in valance shell electronic configuration of Be, 2s orbital is completely filled and hence the addition of any extra electron from outside to this atom is not possible. 37. (3) In these values, the first three values increase slightly and 4th I.E. value increases abruptly. This means that the first three electrons present in the valence shell can be easily removed, that is, low I.E. is needed. 38. (1)

X

+ e- ® X-

( Halogens) ns 2, np 5

(Anion) ns 2, np 6

The valence-shell configuration of halogen atoms (ns2, np5) has an appetite for one electron to stabilize its configuration by attaining the stable noble gas configuration (ns2, np6). Thus, halogen atoms have a strong tendency to accept an extra electron and hence very high values of electron affinity. So, elements of halogen family form anions most readily. 41. (3) Ease of formation of the anion is favored by higher value of electron gain enthalpy. Halogens have the maximum negative electron gain enthalpies in their respective periods. So, they easily form anions. 44. (2) Electronegativity increases in moving from left to right in a period because nuclear charge increases, whereas atomic radius decreases as we move across a period. The halogens have the highest electronegativity values in their respective periods.

Level II 2. (3) [Xe]4f145d106s2: Hg belong to same group [Kr]4d105s2 : Cd and same family. 10 2 [Ar]3d 4s : Zn 2 5 [Ne]3s 3p : Cl belongs to halogen family. 4. (3) The chemistry of a first-row (second period) element often has similarities to the chemistry of the second-row (third period) element being one column to the right of it in the periodic table. For example, lithium resembles magnesium. Hence Mg acts as bridge element. 6. (1) The table which is most commonly used these days and which is based upon the electronic configuration of elements is called the long form of the periodic table. This is also called Bohr’s table since it follows Bohr’s scheme for the arrangement of various electrons around the nucleus.

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Classification of Elements and Periodicity in Properties 8. (4) Metalloids are the elements that have characteristics common to metals as well as non-metals. For example, germanium, silicon, arsenic, and tellurium. 9. (4) As ‘q’ is noble gas, p, r and s having atomic number Z – 1, Z + 1 and Z + 2 should belong to halogens, alkali metal and alkaline earth metal respectively. As halogen has one electron less than stable noble gas configuration it has greater tendency to accept an additional electron forming anion. Alkaline earth metal having valence shell configuration exists in +2 oxidation state. 10. (1) In the periodic table, the element A belongs to first group and the element A may be either sodium or potassium. 16. (2) Isoelectronic ions are those having same number of electrons but different nuclear charges. As the nuclear charge increases, the attractive forces between the nucleus and the electrons also increase. This results in decrease in ionic radius. In other words, size of the isoelectronic ions decreases with increase in the magnitude of nuclear charge. 18. (4) S → S2− Ionic radii = 1.84 Å Te → Te2− Ionic radii = 2.21 Å 1.84 + 2.21 Se → Se2− Ionic radii = = 2.02 Å 2 20. (4) The ionic radius is linked with the atomic radius and it varies accordingly. Thus, the ionic radius always increases down the group and decreases along the period provided the ions involved have the same magnitude of charge. Increase in ionic radii Li + Na + K + Rb+ Cs+

On moving down the group, two effects take place – increase in nuclear charge and addition of energy shells. Out of these, the latter is more predominant. Therefore, atomic radii as well as ionic radii increase on going down the group from top to bottom.

23. (4) Ionization energy increases as we go from left to right across a row of periodic table because the force of attraction between nucleus and an electron becomes larger as the number of protons in the nucleus of atom becomes larger. But in case of P and S, P has higher ionization enthalpy because of half-filled p-orbital. 25. (1) Ease of formation of cation is favored by lower value of ionization potential. 28. (3) Ionization enthalpy increases across the period and decreases down the group.

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32. (1) Halogens have very high values of negative electron gain enthalpies. These are the maximum in their respective periods and decrease down the group. Element

F

Cl

Br

I

At

Electron gain enthalpy −332.6 −348.5 −324.7 −295.5 −270 33. (1) F2 is chemically most active non-metal because it has high electronegativity and electron affinity value and also low bond dissociation energy since it contain single covalent bonds (F—F) in its molecule. 35. (3) Electronegativity value for Li is 0.96 whereas for Cl it is 3.15. 37. (3) Diagonal relationship are about size, electronegativity and polarizability.

Previous Years’ NEET Questions 1. (4) The atom with electronic configuration 1s22s22p63s1 would have the lowest ionization enthalpy as after losing an electron from the outermost s orbital it will acquire stable configuration 1s22s22p6. 2. (3) For isoelectronic species size of anion increases as negative charge increases whereas the size of cation decreases with increase in the positive charge, further ionic radii of atoms are more than those of cations. Hence, the correct order is Ca2+< K+ < Ar < Cl− < S2−. 3. (4) Ionic mobility is defined as the ability of an ion to move in an aqueous solution. Larger the hydrated radius (Smaller the size of the ion, more is the hydration), lesser is the ionic mobility. Therefore, the correct order is Cs+ (aq) > Rb+ (aq) > K + (aq) > Na + (aq). 4. (4) 3d 5 4s 2 has maximum number of electrons in the outermost shell; therefore, it can exhibit largest number of oxidation states. 5. (2) Half-filled electronic configuration is more stable and the ionization energy decreases down the group. Therefore, [Ne] 3s 2 3p 3 has the highest electronic configuration among the given elements. 6. (3) Electron affinity is the tendency of the atoms to gain electrons. In a group, there is decrease of electron affinity in general but for p-block elements only the EA1 of second period element is less than EA1 of third period element. The third period elements have larger size and vacant 3d orbitals, which can accommodate incoming electron easily.

Second period (O) < Third period (S)



In a period, electron affinity increases from left to right as Zeff increases. Cl has the highest electron affinity among halogens. Hence, the correct order is O < S < F < Cl.

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7. (4) Cations have smaller radii than atoms whereas anions have larger radii than the corresponding atoms. For isoelectronic species greater the anionic charge larger will be the radii. Thus, the correct order is S2− > Cl− > K+ > Ca2+. 8. (1) There is a general decrease of atomic radius in a period because the effective nuclear charge (Zeff ) increases across a row. Atomic radii increase down the group. This is due to successive use of orbitals with principal quantum number (n) one higher than the last. Therefore, the order is Cl < P < Mg < Ca.

14. (2) and (3) Given that the order for first ionization enthalpy is B < C < N < O, while the correct order is B < C < O < N. As, N has ns2 np3 configuration (which is halffilled), hence its first ionization enthalpy is greater than that of O whose electronic configuration is ns2 np4.

Given that the order for magnitude of electron gain enthalpy is I < Br < Cl < F, while the correct I < Br < F < Cl, because of the smaller size of fluorine atom, its electron gain enthalpy is less than that of Cl.



The increasing order of ionic size in option (1) is correct as on formation of positive ion, the effective nuclear charge exceeds number of electrons, so it is smaller. More is the number of electrons removed smaller is the ion.



The order of metallic radius in option (4) is correct, because on moving down the group, metallic radius increases due to successive use of orbitals with quantum number one higher than the last.

9. (1) The order of size of cations is Li+ < Na+ < Cs+

The order of size of anions is I− > F−



The cation to anion size ratio is maximum when the cation is largest and anion is smallest. So, cation to anion size ratio is maximum for CsF.

10. (3) The size of cation is always less than that of the parent atom. As the positive charge on the cation increases, its radius decreases. 11. (2) Be2+ = 1s2 = Li+ 12. (2) When a negative ion is formed, one or more electrons are added to an atom and the effective nuclear charge is reduced. The electron cloud expands and the size of negative ion is more than that of the atom and positive ion.

15. (1) The new element that has been discovered is Flerovium with atomic number Z = 114. It belongs to group 14, that is, carbon family with electronic configuration [Rn] 5f146d107s27p2 . 16. (3) Ionic mobility is defined as the ability of an ion to move in an aqueous solution. Larger the hydrated radius, lesser is the ionic mobility, that is, Ionic radius ∝

Hence, the correct order is O2- > F - > Na + .

13. (2) We know that, atomic radius or ionic radius ∝ 1/ Effective nuclear charge (Zeff ). Due to increase in the positive charge, effective nuclear charge increases resulting in the reduction of the atomic radius or ionic radius. So, for isoelectronic species the correct order is Ca2+ < K+ < Ar.

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1 Hydrated radius

Thus, the order of ionic mobility is in the reverse order of the hydrated radius order, for example Li + (aq ) < Na + (aq ) < K + (aq ) < Rb+ (aq ) < Cs+ (aq )

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14

Chemical Bonding and Molecular Structure

Chapter at a Glance 1. Whenever two same or different atoms come close to each other, there are various forces that come into existence and whenever attractive forces are more than any other kind of force, a link develops between the two which is known as bond. The redistribution of electrons to gain maximum stability and minimum energy level is known as bonding. 2. Chemical Bond Formation (a) Kössel and Lewis proposed the electronic theory of valence for explaining the formation of a chemical bond. (b) According to the theory, the formation of a chemical bond between two or more atoms can be explained as the tendency of the elements to attain stable configuration of electrons in the outermost shell. The combination of atoms leads to rearrangement of their electrons either by complete transfer of electrons from one atom to another or by sharing of electrons. 3. Lewis Representation (a) L  ewis diagrams provide a picture of bonding in molecule and ions in terms of the shared pairs of electrons and the octet rule. (b) The molecules in which if all the atoms except hydrogen have not achieved octet are called electron deficient molecule, e.g., BF3, BeCl2, etc. (c) The molecules in which the central atom is having an octet but some are bond pairs and some are lone pairs are called electron rich molecules. 4. Octet Theory (a) It states that the atoms of the molecule tend to combine in such a way that each it has eight electrons in its valence shell or attains the electronic configuration same as that of a noble gas. It is especially applicable to C, N, O, halogen, Na and Mg. (b) Drawbacks: (i) It could not explain the shape of molecules. (ii) It could not explain the relative stability of molecules, being totally silent about the energy of molecule. (iii) It could not explain the formation of electron deficient molecules. 5. Formal Charge (a) Formal charge is the apparent electronic charge of each atom in a molecule, based on Lewis diagrams. Formal charge =

Va - N a - N b 2

 where Va = Number of valence electrons when an atom is isolated.      Na = Number of antibonding electrons on the atom.      Nb = Number of bonding electrons on the atom. 6. Ionic Bonding (a) I onic solids are held together by the electrostatic attraction between the positive and negative ions. The formation of an ionic compound and its stability primarily depend on (i) the ease of formation of ions; (ii) the lattice structure of the crystal.

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(b) Formation of ionic compound will depend upon: (i) Ionization enthalpy change accompanies the removal of an electron from the outermost shell of an isolated gaseous atom to form a positive ion in gas phase. Lesser the ionization enthalpy, greater will be the ease of formation of an ionic bond. (ii) Electron gain enthalpy change accompanies the addition of an extra electron to a gaseous isolated atom to form a negative ion in gas phase. Higher the electron gain enthalpy, greater will be the energy released on formation of a negative ion and greater will be the ease of formation of an ionic bond. (iii) Lattice enthalpy is defined as the change in energy that occurs when isolated ions in the gas phase combine to form one mole of ionic compound. It is determined by the size of the ions and the charge on them. Greater the value of lattice enthalpy of the resulting ionic compound, greater will be the ease of its formation and its stability. (iv) Hydration energy is the energy released when the ions of an ionic compound are attached to water molecules. 7. Covalent Bonding (a) C  ovalent bond is formed between two atoms by mutual sharing of their outer electrons so as to complete their octets or duplets (in case of hydrogen which has only one shell). (b) Every covalent bond is characterized by two quantities: (i) The average distance between the nuclei held together by the bond. (ii) The amount of energy needed to separate the two atoms to produce neutral atoms again. (c) When two identical atoms form a covalent bond, each atom has an equal share of the bonding electron pair. The electron density at both ends of the bond is the same. The bond is known as a non-polar bond. (d) However, when different kinds of atoms combine, as in HCl, one nucleus usually attracts the electrons in the bond more strongly than the other. The result of unequal attraction for the bonding electrons causes one atom to have slightly positive charge and the other to have slightly negative charge. This is known as polar bond. 8. Fajans’ Rules (a) Fajans put forward four rules which summarize the factors favoring polarization, and hence covalency. (i) In small ions, the positive charge is concentrated over a small area. The high charge density of the small cation distorts the electron cloud of the anion to such an extent that the electron clouds of the anion and the cation overlap, making the bond resemble a covalent bond. This explains why LiCl is more covalent than KCl. (ii) Large ions are highly polarizable, that is, easily distorted by the positive ion. (iii) Large charges on either ion, or on both ions, favor covalency. This is because a high charge increases the amount of polarization. (iv) Polarization, and hence covalency, is favored if the positive ion does not have a noble gas configuration. 9. Dipole Moment (a) T  he dipole moment (m), is equal to the amount of charge on either end of the molecule q, multiplied by the distance between the charges r. m = q × r    It measures the polarity of the bond which is caused by the electronegativity difference between two atoms involved in that bond. It is measured in Debye unit (symbol D). (b) Bond moment is the dipole moment of a particular bond. (i) For diatomic heteroatomic molecules, the bond moment is equal to the dipole moment and for diatomic homoatomic molecule it is zero because there is no shifting of bond pair towards any atom. Hence q is zero in above equation. (ii) For polyatomic molecule (molecule having more than two atoms), the dipole moment is equal to the vector sum of all bond moments and lone pair moments. (c) The lone pair has no contribution to the dipole moment of the molecule, if the lone pair is present in the pure s or p orbital. However, when the lone pair is present in a hybrid orbital, the electronic density is not equal in two

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lobes of the hybrid orbital. The value of lone pair moment depends directly on the % s character of the hybrid orbital. The lone pair moment increases upto 50% and then falls down as it becomes more and more spherical. The lone pair moment order for different hybrid orbitals (consisting of s or p orbitals) is given as sp > sp2 > sp3. 10. Coordinate or Dative Bond (a) D  uring the formation of a bond by sharing of electrons between atoms, when both electrons of the shared pair are provided by one of the combining atoms, then the bond formed is called a coordinate bond or a dative bond. (b) The atom which provides the electron pair is termed as the donor atom, while the atom which accepts it is termed as the acceptor atom. (c) An arrow pointing from donor towards the acceptor atom represents a coordinate bond. An example of coordinate bond is the formation of ammonium (NH4+) ion. H ´ + ´ H N +H ´ Lone pair H of electrons Ammonia molecule

H ´ H´ N ´ H

+ H Coordinate bond

Ammonium ion

11. Resonance (a) It is the phenomenon of delocalization of p electron. It is observed if, (i) Molecule is planar. (ii) It has alternate single and double bonds. (iii) The resonating structures are unable to explain all the properties of the compound, but resonance hybrid is able to explain all the properties of the compound.

12. Bond Parameters (a) Bond order is defined as the number of pairs of electrons shared between two atoms. It is given by 1 Bond order = ( N b - N a ) 2

     where Nb and Na are the number of electrons in the bonding and antibonding molecular orbitals.

(b) B  ond length is the distance between the nucleus of atoms involved in bonding. These are also known as bond distances. The bond length decreases with increase in bond order. (c) The amount of energy required to break a bond is called bond dissociation energy or simply bond energy. B.E. = Δ = (Actual bond energy) - E A - A - EB - B (d) The relation between bond order, bond length and bond energy can be summarized as follows:     Lower bond order → Longer bond length → Lower bond energy → Weaker bond     Higher bond order → Shorter bond length → Higher bond energy → Stronger bond 13. Valence Shell Electron Pair Repulsion (VSEPR) Theory (a) The main postulates of VSPER theory are: (i) The electron pairs present on the central atom can be regarded as point charges in space. (ii) The shape of the molecule is determined by repulsions between all the electron pairs present in the valence shell. (iii) The electron pairs in the valence shell experience electrostatic repulsions due to negatively charged electron clouds. The electron pairs tend to occupy positions that minimize these repulsions.

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(iv) A  multiple bond is treated like a single bond and the electron pairs involved in multiple bonding are treated as single super pair. (v) The repulsion order in relation to the bonds is as follows: Double bond–double bond > Double bond–single bond > Single bond–single bond (vi) If a molecule has two or more resonance structures, then VSEPR model can be applied to any one structure. (vii) The magnitude of repulsions between bonding pairs of electrons depends on the electronegativity difference between the central atom and the other atoms. (viii) A lone pair (lp) of electrons takes up more space round the central atom than a bond pair (bp). The repulsion order:   Lone pair–lone pair > Lone pair–bond pair > Bond pair–bond pair  (ix) The presence of lone pairs at the central atom causes distortion in regular shape and decrease in bond angles. (b) Molecular geometry based on number of electron pairs: Number of electron pairs 2

Shape Linear

Example BeCl2

3

Planar triangular

BeCl3

4

Tetrahedral

CH4

5

Trigonal bipyramidal

PCl5

6

Octahedral

SF6

(c) Molecular geometry in presence of one or more lone pairs on central atom: Type AB2L AB3L AB2L2 AB4L AB3L2 AB2L3 AB5L AB4L2

Number of electron pairs Two bonding and one non-bonding electron pairs Three bonding and one non-bonding electron pairs Two bonding and two non-bonding electron pairs. Four bonding and one non-bonding electron pairs. Three bonding and two non-bonding electron pairs. Two bonding and three non-bonding electron pairs. Five bonding and one non-bonding electron pairs. Four bonding and two non-bonding electron pairs.

Expected geometry Trigonal planar Tetrahedral

Observed geometry Non-linear or bent or V-shaped. Trigonal pyramidal

Example SnCl2

Tetrahedral

Non-linear, bent

H2O

Trigonal bipyramidal

Trigonal pyramidal

SF4

Trigonal bipyramidal

Trigonal pyramidal

ClF3

Trigonal bipyramidal

Linear

Octahedral

Square pyramidal

BrF5

Octahedral

Square planar

XeF4

NH3

I3 –

14. Valence Bond Theory (VBT) It is one of the basic theories which was developed to use the methods of quantum mechanics in the formation of bond. According to this theory, a bond between two atoms is formed when two electrons with their spins paired are shared by two overlapping atomic orbitals, one orbital from each of the atoms joined by the bond. (a) Concept of orbital overlap: A covalent bond is formed between two atoms in a molecule, when the half-filled valence atomic orbitals (AOs) of the two atoms containing unpaired electron overlap with one another and the electrons pair up in the overlap region. When the attractive forces are stronger than the repulsive forces, energy is

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released; lowering of energy makes the molecule stable. Covalent bonds are classified into two main types– sigma (s) bonds and pi (p) bonds. (b) Types of overlapping and nature of covalent bonds: (i) s bond: It is formed by head-on overlap of orbitals. • s–s overlap, such as in H2 molecule. • s–p overlap, such as in HX molecule (X = F, Cl, Br, I). • p–p overlap, such as in X2 molecule (X = F, Cl, Br, I).



The relative strength of bond is: s–s < s–p < p–p

(ii)  p bond: It is formed by sideways overlap of orbitals and it merely shortens the bond length. The relative strength of the p bonds increases when the intermolecular distance increases and the order is:

2pp-2pp > 2pp -3dp > 2pp-3pp > 3pp-3pp (iii)  The strength of the bond depends on the extent of overlap between the orbitals participating in the bond formation. The p bond is thus weaker than the s bond. Also, p bond is always formed in addition to the s bond in molecules.

(c) H  ybridization: Concept was proposed by Pauling (VSEPR). The intermixing of pure atomic orbitals of nearly equal energies and shapes to get same number of identical new orbital is known as hybridization. The hybridized atomic orbitals can overlap more effectively than the other atomic orbitals, thus providing an overall molecular structure which has stronger bonds and lower energy. (i) All the valence orbital may or may not take part in hybridization. The number of orbitals mixed and the number of hybrid orbitals formed are the same. (ii) The hybrid orbitals have the same energy but differ in orientation. These are directed in space in directions which will lead to stable arrangement. This gives the geometry of the molecule. (iii) Hybrid orbitals must follow Hund’s rule and Pauli’s exclusion principle in filling of electrons. (iv) In the hybrid orbitals there is always a strong sigma bond; p bonds are involved in the overlap of unhybridized orbitals. Calculation of steric number For molecule/species having only one central atom, the steps involved are: 1. Calculate n, which is equal to number of total valence shell electrons of all atoms + number of negative charges (if any) – number of positive charges (if any). Note: (a) It is considered that valence shell electrons for H atom is seven (just to make calculation easier).   (b)  The value of n cannot be odd. 2. Divide n by 8, which results in Q + R (if any) where Q = number of s bond pairs, R = number of unshared electrons on the central atom. Hence, R/2 = number of lone pairs of electrons on the central atom. 3. Obtain steric number (S.N.) of the molecule, which is the number of atoms bonded to the central atom of a molecule plus the number of lone pairs on the central atom. R  Therefore, S.N. =  Q +  2

Hybridization sp sp2 sp3 sp3 sp3

Number of lone pair 0 0 0 1 2

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Geometry Linear Trigonal planar Tetrahedral Tetrahedral Tetrahedral

Shape Linear Trigonal planar Tetrahedral Pyramidal V or bent shape

Example BeF2 BF3 CCl4 NH3 H2O

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Hybridization sp3d

Number of lone pair 0

sp3d sp3d sp3d sp3d2 sp3d2 sp3d2 sp3d3 sp3d3 dsp2

1 2 3 0 1 2 0 1 0

Geometry Trigonal Pyramidal

Shape Trigonal pyramid

Trigonal Pyramidal Trigonal Pyramidal Trigonal Pyramidal Octahedral Octahedral Octahedral Pentagonal bipyramidal Pentagonal bipyramidal Octahedral

Sea saw T-shape Linear Octahedral Square pyramidal Square planar Pentagonal bipyramidal Distorted octahedral Octahedral

Example PCl5 SF4 ClF3 XeF2 SF6 XeOF4 XeF4 IF7 XeF6 [Co(NH3)6]3+

15. Molecular Orbital Theory (a) The molecular orbital theory is based on the following assumptions (i) When two atoms approach each other, their atomic orbitals lose their identity and mutually overlap to form new orbitals called molecular orbitals (MO). (ii) The molecular orbitals are polycentric and are associated with the nuclei of all the atoms constituting the molecule. (iii) Only atomic orbitals of about the same energy and same symmetry interact significantly. (iv) The total number of molecular orbitals produced is always equal to the total number of atomic orbitals. (v) When two atomic orbitals overlap, they interact to form two molecular orbitals- bonding and antibonding. (vi) Each molecular orbital can accommodate a maximum of two electrons of successively higher energy with opposite spins. (b) Formation of molecular orbitals (i) The formation of molecular orbitals can be explained by linear combination of atomic orbitals (LCAO). (ii) Consider two atoms A and B which have atomic orbitals described by the wave functions ΨA and ΨB. These combine to give rise to a pair of molecular orbitals Ψg (bonding molecular orbital [Ψg = ΨA + ΨB]) and Ψu (antibonding molecular orbital which is higher in energy [Ψu = ΨA – ΨB]). (iii) The energy of the bonding molecular orbital is lower than that of the atomic orbital by an amount Δ. This is known as the stabilization energy. Similarly the energy of the antibonding molecular orbital is increased by Δ. (c) Types of atomic and molecular orbitals Atomic orbitals Molecular orbitals

1sA + 1sB

s 1s 2pxA – 2pxB s * 2px

1sA – 1sB

s * 1s 2pyA + 2pyB p 2py

2sA + 2sB

s 2s 2pyA – 2pyB p* 2py

2sA – 2sB

2pxA + 2pxB

p 2pz

p* 2pz

s * 2s 2pzA + 2pzB

s 2px 2pzA – 2pzB

(i) Order of energy for molecular orbitals • The order of energy of molecular orbitals in simple homonuclear diatomic molecules is: p 2 px , p * 2 px ,s * 2 pz s 1s ,s * 1s ,s 2 s ,s * 2 s ,s 2 pz ,   p 2 p y , p * 2 p y Increasing energy  • F  or the lighter elements boron, carbon and nitrogen the p 2px and p 2py are probably lower than s 2pz. For these atoms the order is:

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p 2 px , p * 2 px s 2 pz ,s * 2 pz ,  s 1s ,s * 1s ,s 2 s ,s * 2 s  p 2 p y p * 2 p y In ncreasing energy  16. Back Bonding It is the bonding of p-conjugated ligands to a transition metal which involves a synergic process with donation of ­electrons from the filled p-orbital or lone electron pair orbital of the ligand into an empty orbital of the metal (donor– acceptor bond), together with the release (back donation) of electrons from a d orbital of the metal (which is of p-symmetry with respect to the metal–ligand axis) into the empty p* antibonding orbital of the ligand. 17. Metallic Bonding (a) Electron sea model (i) In the case of metals, every atom achieves a more stable configuration by sharing the outer shell electrons with the various other atoms in the metal lattice. The electrons are said to be delocalized. The metal lattice is held together by the strong forces of attraction between the positive nuclei and the delocalized electrons. This is described as “an array of positive ions in a sea of electrons”. (ii) The strength of the metallic bond depends on the number of valence electrons contributed by the atoms to the delocalized electrons and the packing arrangement of the metal atom. More number of delocalized electrons and more closely packed atoms result in a stronger bond and a higher melting point. (iii)  This explains relatively low melting points of Group I metals have compared to other metals, the properties of the metals like malleability and ductility and metallic shine and lusture. (b) Band theory: In a multi-atom system, the number of molecular orbital states will be equal to the number (n) of the atomic orbitals combining. Since the number of molecular orbitals is large, the spacing between them decreases to become almost negligible and we get a “band” of continuous energy levels. These molecular orbitals extend in all three dimensions over all the atoms in the metal piece. The molecular orbital theory can explain most of the physical properties of metals as well as the classification of materials into conductors, insulators and semiconductors. 18. Bonding in Homonuclear Diatomic Molecules Bond order 1

Configuration H2

s1s2

He2 (s 1s)2 (s *1s)2 Li2

s 1s2, s *1s2, s 2s2

C2

 p 2 px2 s 1s , s * 1s , s 2 s , s * 2 s ,  2  p 2 p y

N2 O2

2

2

2

2

 p 2 px2 s 1s , s * 1s , s 2 s , s * 2 s ,  s 2 pz2 2  p 2 p y  p 2 px2 ,  p * 2 p1x 2 2 2 2 2  s 1s , s * 1s , s 2 s , s * 2 s , s 2 pz ,  2  1  p 2 p y ,  p * 2 p y 2

2

2

2

Magnetic Bond energy Bond properties (kJ mol-1) length (pm) Diamagnetic

0

Noble gas

–

1

Diamagnetic

110

267

2

Diamagnetic

612

124

3

Diamagnetic

953

109

2

Paramagnetic

501

121

19. Bonding in Heteronuclear Diatomic Molecules (a) H  eteronuclear diatomic molecules can have non-bonding orbitals, in addition to bonding and antibonding orbitals.

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(b) I n heteronuclear diatomic molecules due to the difference in electronegativities of the two atoms, the electrons in bonding MO spend more time near the more electronegative atom, while the electrons in anti-bonding MOs are closer to the less electronegative atom. 20. Hydrogen Bonding (a) T  he electrostatic force of attraction between hydrogen atom and another electronegative atom like O, N or F is called hydrogen bonding. (b) There are two types of hydrogen bonds: (i) Intermolecular hydrogen bond: When the hydrogen bonding occurs between an H atom of one molecule and an electronegative atom of a second molecule. For example HF, alcohol, H2O, NH3 etc. (ii) Intramolecular hydrogen bond: In this type, the hydrogen bond is formed between hydrogen and an electronegative atom (F, O, N), within the same molecule. For example, intramolecular hydrogen bonds are present in molecules such as o-chlorophenol, salicylaldehyde, o-nitrobenzoic acid, etc. (c) Consequences of hydrogen bonding: (i) Physical state: Intermolecular hydrogen bonding causes two or more molecules of a compound to exist as associated or grouped molecules. This results in an increase in the size as well as the molecular mass of the compound which in turn is reflected in the physical state of the substance. (ii) Melting and boiling points: Due to intermolecular hydrogen bonding and a consequent association of molecules, larger energy is required to separate these molecules before they can melt or boil. Hence there is an elevation in the melting and boiling points of these compounds. (iii) Solubility: Covalent compounds which can form a hydrogen bond with water readily dissolve in it. For example, alcohols like ethanol, ammonia, amines, lower aldehydes and ketones are soluble in water.

Solved Examples 1. Which of the following compounds does not follow the octet rule for electron distribution? (1) H2O (2) PH3 (3) PCl3 (4) PCl5 Solution (4) P has 10 electrons in its valence shell in PCl5. Cl Cl

P Cl

Cl Cl

2. The C   C bond length is 1.54 Å, C   C bond length is 1.33 Å. What is the circumference of benzene ring? (1) 8.61 Å (2) 8.82 Å (3) 8.43 Å (4) 8.05 Å Solution (1) In benzene, due to the resonance C   C bond length is in between double bond and single bond. However, if we consider the six carbon atoms to be linked by three single and three double bonds, then the circumference of benzene ring is 3 × (C   C) + 3 × (C   C) = 3 × 1.33 + 3 × 1.54 = 8.61 Å.

Chapter 14_Chemical Bonding and Molecular structure.indd 332

3. Lattice energy of an ionic compound depends upon (1) (2) (3) (4)

charge on the ion only. size of the ion only. packing of ions only. charge and size of the ion.

Solution (4) This is because it involves both the attractive forces between ions of opposite charges and the repulsive forces between ions of like charges. 4. Amongst LiCl, RbCl, BeCl2 and MgCl2 the compounds with the greatest and the least ionic character, respectively are (1) LiCl and RbCl (2) RbCl and BeCl2 (3) MgCl2 and BeCl2 (4) RbCl and MgCl2 Solution (2) According to Fajans’ rule high positive charge, small cation and large anion favors covalency. RbCl has greatest ionic character and least covalent character among given species. As both the ions in the compound are of comparable sizes which favors the ionic character. BeCl2 has least ionic character and greatest covalent character. As Be2+ is small cation and Cl– is a large anion which favors the covalent character.

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Chemical Bonding and Molecular Structure 5. Molecule AB has a bond length of 1.617 Å and a dipole moment of 0.38 D. The fractional charge on each atom (absolute magnitude) is (e0 = 4.802 × 10−10 esu) (1) 0 (2) 0.05 (3) 0.5 (4) 1.0

333

9. The hybridization of orbital of N atom in NO3- , NO+2 and NH +4 are, respectively, (1) sp2, sp, sp3 (2) sp, sp3, sp2 (3) sp2, sp3, sp (4) sp, sp2, sp3 Solution

Solution

(1) We know n = total valence shell electrons of all atoms + number of negative charges (if any) – number of positive charges (if any). -30 -10 0.38 × 3.335 × 10 Cm = q × 1.617 × 10 m 24 For NO3-, n = 5 + 3 × 6 + 1 - 0 = 24 ⇒ = 3 ⇒ sp 2 hybridization 0.38 × 3.335 × 10 -30 Cm 8 24 -20 2 q= = 0.759 × 10 C = 3 ⇒ sp hybridization 1.617 × 10 -10 m 8 16 = 2 ⇒ sp hybridization For NO+2 , n = 5 + 2 × 6 + 0 - 1 = 16 ⇒ 0.759 × 10 -20 8 16 With respect to electron, q = = 0 . 05 = 2 ⇒ sp hybridization 1.6 × 10 -19 8 32 For NH +4 ,  n = 5 + 4 × 7 + 0 - 1 = 32 × = 4 ⇒ sp 3 hybridization 6. The bond order in NO is 2.5 whereas that in NO+ is 3. Which 8 32 3 = 4 ⇒ sp hybridization of the following statements is true for these two species? 8 10. Which one of the following has the regular tetrahedral (1) Bond length in NO+ is greater than in NO. structure? (2) Bond length is unpredictable. + (1) XeF4 (2) [Ni(CN)4]2(3) Bond length in NO in equal to that in NO. + (3) BF4- (4) SF4 (4) Bond length in NO is greater than in NO . (2)

μ=q×l

Solution

Solution (1) Bond length in NO is greater than in NO+. As we know that the bond order is inversely proportional to the bond length. So, greater is the bond order and shorter is the bond length. 7. The correct order of bond dissociation energy among N2, O2, O2- is shown in which of the following arrangements?



11. Which of the following is the hybridization state of the central atom in a molecule that has a trigonal planar shape? (1) sp3d (2) sp3 (3) sp2 (4) sp

(1) N 2 >O 2- >O2 (2) O2- >O2 >N 2 (3) N 2 >O 2 >O 2- (4) O2 >O2- > N 2

Solution

Solution (3) Bond energy µ Bond strength µ Bond order 1 Bond order = [Bonding electrons - Antibonding electrons] 2

Bond order of N2 is 3.0; bond order of O2 is 2.0; bond order of O2 – is 1.5. Thus, the decreasing order of bond energy is N 2 >O 2 >O 2-

8. In OF2, number of bond pairs and lone pairs of electrons are, respectively, (1) 2, 6 (2) 2, 10 (3) 2, 8 (4) 2, 9

(3) The shapes corresponding to the hybridization states are sp3d–Trigonal bipyramidal; sp3–Tetrahedral; sp2–Trigonal planar; sp–Planar 12. Which of the following species exhibits the diamagnetic behavior? (1) NO (2) O 22 (3) O +2 (4) O2 Solution (2) The electronic configurations are as follows: O 22  s 1s 2 s * 1s 2 s 2s 2 s * 2s 2 s 2pz2 p 2px2  p 2p y2 p * 2px2  p * 2p y2



Solution (3)

There is no unpaired electron, so it is diamagnetic. O2  s 1s 2 s * 1s 2 s 2s 2 s * 2s 2 s 2pz2 p 2px2  p 2p y2 p * 2p1x

O F



( 3 + 4 × 7 + 1 - 0) 32 = =4 8 8 Therefore, the hybridization is sp3.

(3) Hybridization of BF4 =

F



There is one unpaired electron, so it is paramagnetic.

O 2 = s 1s 2 s * 1s 2 s 2s 2 s * 2s 2 s 2pz2 p 2px2 = p 2p y2 p * 2p1x = p * 2p1y In F   O   F; Bond pairs = 2. Lone pairs around oxygen = 2, Lone pairs around 2 fluorine atoms ThereO 2 = s=13s 2×s2*=1s6.2 s 2s 2 s * 2s 2 s 2pz2 p 2px2 = p 2p y2 p * 2p1x = p * 2p1y fore, total lone pairs are 6 + 2 = 8.

Chapter 14_Chemical Bonding and Molecular structure.indd 333

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There are two unpaired electrons, so it is paramagnetic. NO is isoelectronic with O +2 , so it also has one unpaired electron, and is paramagnetic.

Solution (4)

F

13. Which one of the following properties is not shown by NO? (1) (2) (3) (4)

Cl

It is diamagnetic in gaseous state. It is a neutral oxide. It combines with oxygen to form nitrogen dioxide. Its bond order is 2.5.

F



Solution (1) The MO configuration of NO is s 1s 2 s * 1s 2 s 2s 2 s * 2s 2s 2pz2 p 2px2 p 2p y2 p * 2p1x

As can be seen it is an odd electron species and so paramagnetic in gaseous state with one unpaired electron.



Bond order = (10 - 5)/2 = 2.5.

F

The shape of the molecule is best described by T-shaped.

17. The correct order of increasing covalent character of the following is (1) KCl < CaCl2 < AlCl3 < SiCl4 (2) SiCl4 < AlCl3 < CaCl2 < KCl (3) AlCl3 < CaCl2 < KCl < SiCl4 (4) CaCl2 < SiCl4 < KCl < AlCl3 Solution

14. Which of the following is not correct regarding C2 molecule? (1) C2 molecule has been found to exist in vapor phase. (2) It has total 12 electrons, out of which 8 electrons occupy bonding orbitals, while 4 electrons occupy antibonding orbitals. (3) The molecule is paramagnetic. (4) C2 molecule contains double bond and both are p bonds.

(1) According to Fajans’ rule, as the charge on the cation increases, its effective nuclear charge as well as polarizing power increases. Hence, covalent character will also increase. The increasing order should be KCl < CaCl2 < AlCl3 < SiCl4. 18. The correct order of dipole moment is (1)  CH4 < NF3 < NH3 < H2O (3)  NH3 < NF3 < CH4 < H2O

(2) NF3 < CH4 < NH3 < H2O (4) H2O < NH3 < NF3 < CH4

Solution (1) In CH4: Net dipole moment is zero.

Solution (3) C2 = (s 1s ) (s * 1s ) (s 2s ) (s * 2s ) (p 2p = p 2p )



In NF3: Resultant dipole moment toward nucleus of nitrogen.



Four electrons are present in 2p molecular orbitals; that is why double bond contains both p bonds.



In NH3: Resultant dipole moment toward lone pair.



Molecule is diamagnetic as it does not have any unpaired electron.



In H2O: Resultant dipole moment toward oxygen.

2

2

2

2

2 x

2 y

15. Which of the following hydrogen bonds is the strongest? (1) O   H   F (2) O   H   N (3) F   H   F (4) O   H   O Solution (3) The strength of hydrogen bonding depends on the electronegativity of the other atom. The hydrogen bond in HF is the strongest because fluorine is the most electronegative element. 16. A molecule has a central atom surrounded by 2 lone pairs and 3 atoms. The best description for the shape of the molecule is (1) trigonal bipyramidal. (2) octahedral. (3) bent. (4) T-shaped.

Chapter 14_Chemical Bonding and Molecular structure.indd 334

19. Which among the following species is not perfectly planar?

.

.

(1) C  H3 (3) :CF2

(2) C  . HF2 (4) N  O2

Solution

. .

(2) C H3: sp2 hybridized C atom and the species is planar. C  HF2: The hybridization of C atom is in between sp2 and sp3. Hence, it is not perfectly planar and is pyramidal.

. 

:CF2 and N O2 must be planar because irrespective of hybridization of the central atom triatomic species must be planar. 20. Among the following, the pair in which the two species are not isostructural is (1) IO3- and XeO3 (2) BH 4- and NH +4 (3) PF6- and SF6 (4) SiF4 and SF4

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Chemical Bonding and Molecular Structure

335

Solution

22. Read the following statements and answer accordingly

(4) The structure of the given species is as follows:



Statement 1: Water is a good solvent for ionic compounds but poor one for covalent compounds.



Statement 2: Hydration energy of ions released is ­sufficient to overcome lattice energy and to break hydrogen bonds in water while covalent bonded compounds interact so weakly that even van der Waal’s force between molecules of covalent compounds cannot be broken.

S. no. Species Hybridization

1.

IO3-  

XeO3

sp3

Shape

Structure

Pyramidal O

sp3

Pyramidal

I

O-

O

Xe

O

 

O

O

  -

H

2.

BH 4-  

sp3

Tetrahedral

H

Solution

B

H

H

  +

H

 

NH +4  

sp3

Tetrahedral

H

N H

PF6-  

sp3d2

Octahedral

F

F

SF6

sp3d2

H

  -

F

F

P F

3.

(1) Statement 1 is true, statement 2 is true, and statement 2 is correct explanation for statement 1. (2) Statement 1 is true, statement 2 is true, and statement 2 is not correct explanation for statement 1. (3) Statement 1 is true, statement 2 is false. (4) Statement 1 is false, statement 2 is true.

F

F F

 

SiF4

F

sp3

F

SF4

F

sp3d

See-saw



F

F

Solution (2) Amongst hexahalides of sulphur, only SF6 is exceptionally stable for steric reason. In SCl6, smaller S cannot accommodate six larger ions.

F



PH5 does not exist as there is large difference in the energies of s, p and d orbitals of P and hence it does not undergo sp3d hybridisation.



NCl5 does not exist as there is no d orbital with N.

F

S F

F

21. In forming (I) N2 → N +2 and (II) O2 → O+2; the electrons respectively are removed from (1) (p * 2py or p * 2px) and (p * 2py or p  * 2px) (2) (p 2py or p 2px) and (p 2py or p 2px) (3) (p 2py or p 2px) and (p * 2py or p * 2px) (4) (p * 2py or p  *2px) and (p 2py or p 2px) Solution (3) In O2 molecule, the electron is removed from antibonding molecular orbital p ∗2py or p ∗2px and in N2 molecule, the electron is removed from bonding molecular orbital s  2pz. However, no such option is given. Out of the four options given, best option is (3).

Chapter 14_Chemical Bonding and Molecular structure.indd 335

SF6, BF63-, SF4, OF4, AlF63-, PH5, PCl5, NCl5, SCl6 (1) 9 (2) 5 (3) 6 (4) 8

F

Si

Tetrahedral

F

 

23. Amongst the following, the total number of species which does/do not exist is

S

Octahedral

F

4.

(1) Water is a good solvent for ionic compounds but poor one for covalent compounds because hydration energy of ions released is sufficient to overcome lattice energy and to break hydrogen bonds in water while covalent bonded compounds interact so weakly that even van der Waal’s force between molecules of covalent compounds cannot be broken.

BF63- does not exist as there is no vacant d orbital in B and hence it cannot exceed its covalency beyond four. OF4 does not exist.

24. Which of the molecules is not hypovalent but completes its octet? (1) AlCl3 (2) AlBr3 (3) AlF3 (4) All are hypovalent and completes their octet Solution (3) AlF3 is ionic and during formation of ionic bond octet of Al3+ as well as F– that of have been completed. AlCl3 and AlBr3 are covalent and hypovalent.

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25. Resonance is due to (1) (2) (3) (4)

identical arrangement of atoms. delocalization of p electrons. migration of H atoms. migration of protons.

29. What fraction of charge is present on iodine in covalently bonded Hδ+ − Iδ− if the dipole moment of HI is 0.38 D and the bond length is 1.61 Å. (1) 5% (2) 16% (3) 33% (4) 79%

Solution

Solution

(2) Resonance involves the delocalization of p electrons.

(1) Given that the dipole is 0.38 D = 0.38 × 3.34 × 10−30 Cm = 1.26 × 10−30 Cm.

26. Which of the following molecules is non-polar in nature? (1) POCl3 (2) CH2O (3) SbCl5 (4) NO2 Solution (3) POCl3 has sp3 hybridization, is tetrahedral in shape; but contains different polar bonds due to which net dipole moment is not zero and the molecule is polar.

CH2O (or HCHO) has sp2 hybridization, is triangular planar in shape; but contains one non-polar bond (i.e., C   O) due to which net dipole moment is not zero, and the molecule is polar.



SbCl5 has sp3d hybridization, is trigonal bipyramidal in shape; all Sb   Cl bonds are polar but due to regular geometry net dipole moment is zero, thus the molecule is non-polar.



NO2 has sp2 hybridization, bent shape because it contains one non-bonding electron on central atom; each N   O bond is polar, so net dipole moment is not zero and the molecule is polar.

27. The species in which central atom uses sp2 hybrid orbitals in its bonding is (1) PH3 (2) NH3 (3) CH +3 (4) SbH3 Solution (3) In CH3 +, C has sp2 hybridization. H C+ H

H

(2) ClO2 (4) SiO2

Solution (2) The number of electrons in the given species is as follows:

CO2: = 6 + 8 + 8 = 22

ClO2: = 17 + 8 + 8 = 33



SO2: = 16 + 8 + 8 = 32

SiO2: =14 + 8 + 8 = 30



Paramagnetic compounds usually contain odd number of electrons. ClO2 contains odd number of electrons. Therefore, it is paramagnetic.

Chapter 14_Chemical Bonding and Molecular structure.indd 336

If the bond was 100% ionic, the magnitude of the dipole moment will be m  qr  1.6 1019 1.611010 Cm  2.57 1029 Cm



Therefore, the fraction of charge covalently bonded is 1.26 × 10 -30 = 0.05 2.57 × 10 -29



Therefore, % of fractional charge = 5%

30. Most favorable conditions for the formation of ionic bonds are (1) (2) (3) (4)

large cation and small anion. large cation and large anion. small cation and small anion. small cation and large anion.

Solution (1) Favorable conditions for ionic bond formation are: (a) Low ionization energy (IE) values of atoms-forming cation, that is, down the group, as size of the cation increases, IE decreases. (b) High electron affinity (EA) values of atoms-forming anions, that is, down the group EA values decreases and size of the anion increases. 31. Predict the molecular geometry and polarity of the SO2 molecule by applying VSEPR theory. (1) (2) (3) (4)

Linear, nonpolar Linear, polar Bent, 109°.5′ angle, polar Bent, 120° angle, polar

Solution

28. Which one of the following oxides is expected to exhibit paramagnetic behavior? (1) CO2 (3) SO2



(4) The molecular geometry of SO2 molecule is bent. The angle between S   O   S is 120°. SO2 is a polar molecule due to asymmetrical charge distribution. 32. What is the hybridization state of C–atoms present in C2(CN)4? (1) sp and sp2 (3) sp and sp3

(2) sp2 and sp3 (4) sp and sp

Solution (1) The structure of C2(CN)4 is (C   N)2C   C(C   N)2. The C–atom forming double bond is sp2 hybridized while the C–atom forming triple bond is sp hybridized.

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Chemical Bonding and Molecular Structure 33. The correct order in which the O   O bond length increases in the following is (1) O3 < H2O < O2 (3) O2 < H2O < O3

(2) O2 < O3 < H2O2 (4) H2O2 < O2 < O3

Solution (2) Bond length is inversely proportional to the bond order, greater is the bond order lesser is the bond length and vice versa. Molecule

Structure O

O2

O

2

+

+

O

O3

O

Bond order

O -

O-

O

O 

34. Which of the following is not a correct statement? (1) The electron-deficient molecules can act as Lewis acids. (2) The canonical structures have no real existence. (3) Every AB5 molecule does in fact have square pyramid structure. (4) Multiple bonds are always shorter than corresponding single bonds. Solution (3) AB5 molecules can also have trigonal bipyramidal shape apart from square planar geometry.

Due to resonance the bond order is 1.5

B B

H

H2O2

O

O

337

1

B

B A

A

B B Trigonal bipyramidal

H

B B

B

B Square planar

Therefore, the correct order is O2 < O3 < H2O2.

Practice Exercises Level I Kossel–Lewis Approach to Chemical Bonding 1. Chemical bond implies (1) repulsion. (2) attraction. (3) attraction and repulsion balanced at a particular distance. (4) attraction and repulsion. 2. The correct sequence of increasing covalent character is represented by (1) LiCl < NaCl < BeCl2 (3) NaCl < LiCl < BeCl2

(2) BeCl2 < LiCl < NaCl (4) BeCl2 < NaCl < LiCl

3. Hydrogen chloride molecule contains (1) polar covalent bond. (2) double bond. (3) coordinate bond. (4) electrovalent bond. 4. Which one of the following is not Lewis acid? (1) AlCl3·6H2O (2) AlCl3 (3) SnCl4 (4) FeCl3 5. In PO34 , the formal charge on each oxygen atom and the P   O bond order respectively are (1) 0.75, 0.6 (2) –0.75, 1.0 (3) –0.75, 1.25 (4) –3, 1.25 6. In the N2 molecule, number of electrons pairs shared by nitrogen atom is (1) 1 (2) 2 (3) 3 (4) 5

Chapter 14_Chemical Bonding and Molecular structure.indd 337

7. Which one of the following does not follow octet rule? (1) PF3 (3) CO2

(2) BF3 (4) CCl4

8. Electron deficient molecule is (1) CCl4 (3) BF3

(2) PCl5 (4) SF6

Ionic or Electrovalent Bond 9. Elements whose electronegativities are 1.2 and 3.0. Bond formed between them would be (1) ionic. (2) covalent. (3) coordinate. (4) metallic. 10. Sodium chloride is an ionic compound whereas hydrogen chloride is mainly covalent because (1) (2) (3) (4)

sodium is less reactive. hydrogen is non-metal. hydrogen chloride is a gas. electronegativity difference in the case of hydrogen and chlorine is less than 2.1.

11. Which of the following pairs will form the most stable ionic bond? (1) Na and Cl (2) Mg and F (3) Li and F (4) Na and F 12. In which of the following the bond angle is maximum? (1) NH3 (2) SCl2 (3) NH +4 (4) PCl3

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13. As compared to covalent compounds, electrovalent compounds generally have (1) (2) (3) (4)

low melting points and low boiling points. low melting points and high boiling points. high melting points and low boiling points. high melting points and high boiling points.

14. Which of the following molecule/ion has all the three types of bonds, electrovalent, covalent and coordinate? (1) HCl (2) NH +4 (3) Cl– (4) H2O2

Bond Parameters 15. Which one of the following has the shortest C   C bond length? (1) Benzene (2) Ethene (3) Ethyne (4) Ethane 16. The relationship between the dissociation energy of N2 and N2+ is (1) dissociation energy of N2+ > dissociation energy of N2. (2) dissociation energy of N2 - dissociation energy of N +2. + (3) dissociation energy of N > dissociation energy of N 2 . 2

(4) dissociation energy of N2 can either be lower or higher than the dissociation energy of N 2- . 17. Which does not show resonance? (1) Benzene (2) Aniline (3) Ethylamine (4) Toluene 18. Which of the following bonds will be most polar? (1) N   Cl (2) O   F (3) N   F (4) N   N 19. Which one of the following molecules will have unequal M   F bond lengths? (1) NF3 (2) BF3 (3) PF5 (4) SF4 20. In X   H . . . Y, if X and Y both are electronegative elements, then, (1) electron density on X will increase and on H will decrease. (2) in both electron density will decrease. (3) in both electron density will increase. (4) electron density will decrease on X and will increase on H. 21. Mark the incorrect statement in the following. (1) The bond order in the species O2, O+2 and O2decreases as O+2 > O2 > O2-. (2) The bond energy in diatomic molecule always increases when an electron is lost. (3) Electrons in antibonding molecular orbital contribute to repulsion between two atoms.

Chapter 14_Chemical Bonding and Molecular structure.indd 338

(4) With increase in bond order, bond length decreases and bond strength increases. 22. The correct order of N   O bond lengths in NO, NO2- , NO3- , and N2O4 is (1) N2O4 > NO2- > NO3- > NO (2) NO > NO3- > N2O4 > NO2(3) NO3- > NO2- > N2O4 > NO (4) NO > N2O4 > NO2- > NO3 23. The dipole moment of diatomic molecules AB and CD are 10.41 D and 10.27 D, respectively while their bond distances are 2.82 and 2.67 Å, respectively. This indicates that (1) (2) (3) (4)

bonding is 100% ionic in both the molecules. AB has more ionic bond character than CD. AB has lesser ionic bond character than CD. bonding is nearly covalent in both the molecules.

24. N2 and O2 are converted into monocations, N2 + and O2 + respectively. Which of the following statements is wrong? (1) In N2, the N   N bond weakens. (2) In O2, the O   O bond order increases. (3) In O2, paramagnetism decreases. (4) N2+ becomes diamagnetic.

The Valence Shell Electron Pair Repulsion (VSEPR) Theory 25. Which of the following molecule is planar? (1) SF4 (3) NF3

(2) XeF4 (4) SiF4

26. Which of the following molecules does not have a linear structure? (1) CO2 (3) HgCl2

(2) SnCl2 (4) Hg2Cl2

27. In BrF3 molecule, the lone pairs occupy equatorial positions to minimize (1) lone pair – bond pair repulsion only. (2) bond pair – bond pair repulsion only. (3) lone pair – lone pair repulsion and lone pair – bond pair repulsion. (4) lone pair – lone pair repulsion only. 28. Which of the following molecules has trigonal planar geometry? (1) BF3 (2) NH3 (3) PCl3 (4) IF3 29. In a regular octahedral molecule, MX6 the number of X   M   X bonds at 180° is (1) three. (2) two. (3) six. (3) four.

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Chemical Bonding and Molecular Structure 30. The geometry of ClO3 - ion according to valence shell electron pair repulsion (VSEPR) theory will be (1) planar triangular. (2) pyramidal. (3) tetrahedral. (4) square planar. 31. The molecular shapes of SF4, CF4, and XeF4 are (1) different with 1, 0 and 2 lone pairs of electrons on the central atom, respectively. (2) different with 0, 1 and 2 lone pairs of electrons on the central atom respectively. (3) the same with 1, 1 and 1 lone pair of electrons on the central atom respectively. (4) the same with 2, 0 and 1 lone pair of electrons on the central atom, respectively. 32. The BCl3 is a planar molecule whereas NCl3 is pyramidal because (1) B   Cl bond is more polar than N   Cl bond. (2) N   Cl bond is more covalent than B   Cl bond. (3) nitrogen atom is smaller than boron atom. (4) BCl3 has no lone pair but NCl3 has a lone pair of electrons. 33. Which of the following two are isostructural? (1) NH3, BF3 (2) PCl5, ICl5 2(3) XeF2, IF2- (4) CO23 , SO 3 34. Which of the following statements is/are true? (I) PH5 and BiCl5 do not exist. (II) pp   pp bond is present in SO2. (III)   I3+ has bent geometry. (IV)   SeF4 and CH4 have same shape. (1) (I), (II), (III) (2) (I), (III) (3) (I), (III), (IV) (4) (I), (II), (IV) 35. Which of the following molecule does not have a linear arrangement of atoms? (1) H2S (2) C2H2 (3) BeH2 (4) CO2

Valence Bond Theory 36. Which of the following statements is not correct? (1) (2) (3) (4)

Double bond is shorter than a single bond. Sigma bond is weaker than a p (pi) bond. Double bond is stronger than a single bond. Covalent bond is stronger than hydrogen bond.

37. Which one of the following is the correct order of interactions? (1) Covalent < hydrogen bonding < van der Waals < dipole-dipole (2) van der Waals < hydrogen bonding < dipole-dipole < covalent (3) van der Waals < dipole-dipole < hydrogen bonding < covalent

Chapter 14_Chemical Bonding and Molecular structure.indd 339

(4) Dipole-dipole < van der Waals < hydrogen bonding < covalent 38. Which of the following statement is not correct for sigma and p bonds formed between two carbon atoms? (1) Sigma bond determines the direction between carbon atoms but a p  bond has no primary effect in this regard. (2) Sigma bond is stronger than a p  bond. (3) Bond energies of s and p  bonds are of the order of 264 kJ mol–1 and 347 kJ mol–1, respectively. (4) Free rotation of atoms about a sigma-bond is allowed but not in case of a pi-bond. 39. Which of the following overlap is correct? + (1) + + -

(2)

(3)

+

+

-

+

-

+ -

+ +

+

-

-

+

-

(4) None of the above 40. The angle between the overlapping of one s-orbital and one p-orbital is (1) 180° (2) 120° (3) 109°28′ (4) 120°, 60° 41. Linear combination of two hybridized orbitals belonging to two atoms and each having one electron leads to a (1) sigma bond. (2) double bond. (3) coordinate covalent bond. (4) p bond. 42. Among the following ions the pp-dp overlap could be present in (1) NO2- (3) PO34

(2) NO3(4) CO23

Hybridization 43. Some of the properties of the two species, NO3- and H3O+ are described below. Which one of them is correct? (1) Similar in hybridization for the central atom with different structures. (2) Dissimilar in hybridization for the central atom with different structures. (3) Isostructural with same hybridization for the central atom. (4) Isostructural with different hybridization for the central atom. 44. In which one of the following molecules the central atom said to adopt sp2 hybridization? (1) BeF2 (2) BF3 (3) C2H2 (4) NH3

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OBJECTIVE CHEMISTRY FOR NEET

45. In which of the following pairs of molecules/ions, the central atoms have sp2 hybridization?

52. Which one of the following molecules will form a linear polymeric structure due to hydrogen bonding?

(1) NO2- and NH3 (2) BF3 and NO2(3) NH 2 and H2O (4) BF3 and NH 2 46. Which of the following statements is true for an ion having sp3 hybridization? (1) All bonds are ionic. (2) H   bonds are situated at the corners of a square. (3) All bonds are coordinate covalent. (4) All atoms are situated at the corners of tetrahedron.

(1) NH3 (2) H2O (3) HCl (4) HF 53. The weakest among the following types of bonds is (1) ionic. (2) covalent. (3) metallic. (4) hydrogen bonding.

Level II Kossel–Lewis Approach to Chemical Bonding

Molecular Orbital Theory 47. The number of anti-bonding electron pairs in O22 molecular ion on the basis of molecular orbital theory is, (Atomic number of O is 8) (1) 5 (2) 2 (3) 3 (4) 4 48. Among the following which one is not paramagnetic? [Atomic numbers: Be = 4, Ne = 10, As = 33, Cl = 17] (1) Cl– (2) Be+ (3) Ne2+ (4) As+

Bonding in Some Homonuclear Diatomic Molecules 49. Match List 1 (molecules) with List 2 (bond order) and select the correct answer using the codes List 1

1. The correct order about of C   O bond length in the following is (I) CO   (II) CO2   (III) CO23 (1) II < I < III (2) III < II < I (3) I < II < III (4) I < III < II 2. Which of the following compounds contains the maximum number of lone pairs at the central atom in its best Lewis structure? (1) XeO3 (2) ClO2(3) SOCl2 (4) IO4 3. Element A has three electrons in the outermost orbit and B has six electrons in the outermost orbit. The formula of the compound will be (1) A2B3 (2) A2B6 (3) A2B (4) A3B2

List 2

(I)  Li2

(p) 3

(II)  N2

(q) 1.5

Ionic or Electrovalent Bond

(III) Be2

(r) 1.0

4. Which of the following is ionic solid?

(IV) O2

(s) 0

(1) XeF6(s) (2) PBr5(s) (3) CaC2(s) (4) All of these

 (t)  2 Codes: (1) (2) (3) (4)

(I)-(q), (II)-(r), (III)-(p),(IV)-(t) (I)-(r), (II)-(p), (III)-(s), (IV)-(t) (I)-(s), (II)-(p), (III)-(t), (IV)-(r) (I)-(r), (II)-(q), (III)-(t), (IV)-(p)

Hydrogen Bonding 50. Which one shows maximum hydrogen bonding? (1) H2O (2) H2Se (3) H2S (4) HF 51. What is the dominant intermolecular force or bond that must be overcome in converting liquid CH3OH to a gas? (1) (2) (3) (4)

Dipole-dipole interaction Covalent bonds London dispersion force Hydrogen bonding

Chapter 14_Chemical Bonding and Molecular structure.indd 340

5. The correct order of stability to form ionic compounds among Si4+, Al3+, Mg2+, and Na+ is (1) Si4+ > Al3+ > Mg2+ > Na+ (2) Al3+ > Si4+ > Mg2+ > Na+ (3) Na+ > Si4+ > Mg2+ > Al3+ (4) Na+ > Mg2+ > Al3+ > Si4+ 6. Arrange the following compounds in increasing order of their ionic character:

  SnCl2, SnCl4, SiCl4, SnF4, SnF2 (1) SnF2 < SnCl2 < SnF4 < SnCl4 < SiCl4 (2) SnF2 < SnCl2 < SnF4 < SiCl4 < SnCl4 (3) SiCl4 < SnCl4 < SnF4 < SnCl2 < SnF2 (4) SnCl4 < SnF4 < SnCl2 < SnF2 < SiCl4

7. In which of the following metal to metal bond is present? (1) Cupric chloride (2) Stannous chloride (3) Mercurous chloride (4) Mercuric chloride

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Chemical Bonding and Molecular Structure

Bond Parameters 8. H2O is dipolar, whereas BeF2 is not. It is because (1) H2O involves hydrogen bonding whereas BeF2 is a discrete molecule. (2) H2O is linear and BeF2 is angular. (3) H2O is angular and BeF2 is linear. (4) The electronegativity of F is greater than that of O. 9. The correct order of bond length is (1) SO 23- > SO 24- > SO3 (2) SO 24- > SO 23- > SO3 23

24

(3) SO3 > SO > SO (4) None of these. 10. Which is correct order for net dipole moment? (1) HF > HCl > HBr > HI (2) CH3   F > CD3   F (3) SO3 > SO2 (4) CH3   CH   CHCl (cis) > CH3   CH 

 CHCl(trans)

11. Which of the following statements is not correct? (1) Carbon-carbon bond length in CaC2 will be more than that in CH2CCH2 (2) O   O bond length in Na2O2 will be more than that in KO2. (3) O   O bond length in O2 [PtF6] will be less than that in KO2 (4) N   O bond length in NO gaseous molecule will be smaller than that bond length in NOCl gaseous molecule. 12. The correct order of C   N bond length in the given compounds is: (I) CH3CN

(II) HNCO

(III) CH3CONH2

(1) (I) > (II) > (III) (2) (I) = (II) = (III) (3) (III) > (II) > (I) (4) (III) > (I) > (II) 13. The incorrect order about bond angles is (1) H2O > H2S > H2Se > H2Te (2) C2H2 > C2H4 > CH4 > NH3 (3) SF6 < NH3 < H2O < OF2 (4) ClO2 > H2O > H2S > SF6 14. Which one of the following represents the incorrect decreasing order of bond angles? (1) CO2 > BF3 > CH4 > H2O (2) NO2+ > NO2 > NO2(3) BCl3 > PCl3 > AsCl3 > BiCl3 (4) IO3- > BrO3- > ClO3- > 15. Which of the following options with respect to increasing bond order is correct? (1) NO < C2 < O2- < B2 (2) C2 < NO < B2 < O2(3) B2 < O2- < NO < O2 (4) B2 < O2- < C2 < NO

Chapter 14_Chemical Bonding and Molecular structure.indd 341

341

The Valence Shell Electron Pair Repulsion (VSEPR) Theory 16. Which of the following is a planar molecule? (1) XeO2F2 (2) XeOF3 (3) XeF4 (4) XeF6 17. Among the following, the pair in which the two species are not isostructural is (1) IO3- and XeO3 (2) AlH 4- and PH +4 (3) AsF6- and SF6 (4) SiF4 and SeF4 18. Which of the following are isostructural?  (I)   NO3-

(II) CO23

(III)   ClO3-

(IV) SO3    (V) XeO3

(1) (I), (II) and (IV) (2) (II) and (V) (3) (III) and (IV) (4) (IV) and (V) 19. Which of the following statements is/are correct? (I) In ICl 2-, ClF3 and TeCl4, the number of lone pair(s) of electrons on central atoms are 3, 2 and 1 respectively. (II) Amongst CO, CO2, CO23 , CH3OH, the correct order from the weakest to the strongest carbon-oxygen bond is CH3OH < CO23 < CO2 < CO (III)  The hybridization of boron in BF3 is the same which nitrogen has in ClNO molecule. (1) Only (I) and (III) (2) (II) (3) Only (I) and (II) (4) (I), (II) and (III) 20. Which of the following statements are correct? (I) N2H4 is pyramidal about each N atom. (II) NH2OH is pyramidal about the N atom and bent about the O atom. (III)   CH3COCl is trigonal planar about the carbon atom (attached to O and Cl). (1) Only (I) and (III) (2) Only (II) (3) Only (I) and (II) (4) (I), (II) and (III) 21. Which of the following statements is/are true? (1) Based on VSEPR theory, the number of 90 degree F   Br   F angles in BrF5 is zero. (2) Molecular geometries of both (CH3)3N and (SiH3)3N are trigonal planar. (3) The C   C bond length in C2 is smaller than O   O bond length in O2. (4) For ozone molecule, one oxygen-oxygen bond is stronger than the other oxygen-oxygen bond.

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OBJECTIVE CHEMISTRY FOR NEET

22. The shape of SF5- can be -

F F

F

F

(II)

F

S F

(III)

-

-

+

-

(II) +

(III) +

-

-

F

-

S F

-

-

F F

(I)

-

F

+

-

F

(I)

+

+

S

S F

+

+

F

F

F

-

F

26. Which of the following atomic orbitals overlapping are not allowed?

+

+

-

F

F

(IV)

F

(IV)

(1) Only (I) (2) Only (I) and (II) (3) Only (IV) (4) (I), (II) and (III) 23. O2F2 is an unstable yellow orange solid and H2O2 is a colorless liquid, both have O   O bond and O   O bond length in H2O2 and O2F2 respectively is (1) 1.22 Å, 1.48 Å (2) 1.48 Å, 1.22 Å (3) 1.22 Å, 1.22 Å (4) 1.48 Å, 1.48 Å 24. Give the correct order of initials T or F for following statements. Use T if statement is true and F if it is false. (I) The correct order of repulsion between different pair of electrons is lp–lp > lp–bp > bp–bp. (II) In, general as the number of lone pair of electrons on central atom increases, value of bond angle from normal bond angle also increases. (III) The number of lone pair on O in H2O is 2 while on N in NH3 is 1. (IV) The structures of xenon fluorides and xenon oxyfluorides could not be explained on the basis of VSEPR theory. (1) TTTF (2) TFTF (3) TFTT (4) TFFF

(V)

(1) All (2) (I), (II), (III) (3) (I), (III),(V) (4) Only (II) 27. A sigma bond may be formed by the overlap of two atomic orbitals of atoms A and B. If the bond is formed along the x-axis, which of the following overlaps is acceptable? (1) (2) (3) (4)

s orbital of A and pz orbital of B px orbital of A and py orbital of B pz orbital of A and px orbital of B px orbital of A and s orbital of B

28. In which of the following molecule/ion all the bonds are equal? (1) SiF4 (2) IF7 (3) ClF3 (4) PCl5 29. The number and type of bonds in C 22 ion in CaC2 are (1) one s (2) one s (3) two s (4) two s

bond and one p bond. bond and two p bonds. bonds and two p bonds. bonds and one p bond.

Hybridization

Valence Bond Theory

30. The correct order of increasing s characters (in percentage) in the hybrid orbitals in below molecules/ions is (assume all hybrid orbitals are exactly equivalent):

25. Gaseous SO3 molecule

    CO23   XeF4   I 3   NCl3  BeCl2

(1) is planar triangular in shape with three s bonds from sp2-p overlap and three p-bonds formed by two pppp overlap and one pp-dp overlap. (2) is planar triangular in shape with three s bonds from sp2-p overlap and three p-bonds formed by two pppp overlap and one pp-dp overlap. (3) is a pyramidal molecule with one double bond and two single bonds. (4) planar triangular in shape with two double bonds between S and O and one single bond.

Chapter 14_Chemical Bonding and Molecular structure.indd 342

  (I)   (II)   (III)  (IV)   (V) (1) (2) (3) (4)

(II) < (III) < (IV) < (I) < (V) (II) < (IV) < (III) < (V) < (I) (III) < (II) < (I) < (V) < (IV) (II) < (IV) < (III) < (I) < (V)

31. The correct order of hybridization of the central atom in the following species NF3, BF3, PF5, and [SiF6]2- is (1) sp2, sp3, sp3d2, sp3d (2) sp3, sp2, sp3d, sp3d2 (3) sp2, sp3, sp3d, sp3d2 (4) sp3, sp2, sp3d2, sp3d3

1/4/2018 5:18:25 PM

Chemical Bonding and Molecular Structure 32. Which of the following arrangements is correct on the basis of the increasing p-character of the hybrid orbitals of the central atoms in the following? (I) ClO2-

(II) CS2

(III) SnCl2

(1) (I) > (III) > (II) (2) (II) > (I) > (III) (3) (I) > (II) > (III) (4) (III) > (I) > (II) 33. The state of hybridization of central atom in dimer of BH3 and BeH2 is (1) sp2, sp2 (2) sp3, sp2 (3) sp3, sp3 (4) sp2, sp3

Molecular Orbital Theory 34. According to molecular orbital theory which of the following is correct? (1) LUMO level for C2 molecule is a s 2p orbital. (2) In C2 molecule both the bonds are p bonds. (3) In C 22 ion there is one s  and two p bonds (4) All the above are correct. 35. In terms of the molecular orbital theory, which of the following species will most likely be the one to gain an electron to form thermodynamically more stable species? (1) CN (2) NO (3) O2 2+ (4) N2 36. After understanding the statement 1 and statement 2, choose the correct option.

Statement 1: In the bonding molecular orbital (MO) of H2, electron density is increased between the nuclei.



Statement 2: The bonding MO is yA + yB which shows destructive interference of the combining electrons waves. (1) Both statement 1 and statement 2 are correct and statement 2 is the correct explanation for the statement 1. (2) Statement 1 and statement 2 are correct, but statement 2 is not the correct explanation for the statement 1. (3) Statement 1 is correct, but statement 2 is incorrect. (4) Statement 1 is incorrect, but statement 2 is correct.

Hydrogen Bonding 37. Which of the following species shows intramolecular hydrogen bonding? (1) CH3NH2 (2) HF (4) CCl3CH(OH)2 (3) HO NO2

38. What is not true about ice? (1) It has open cage like structure. (2) It has less density than water.

Chapter 14_Chemical Bonding and Molecular structure.indd 343

343

(3) Each O atom is surrounded by 4 H atoms. (4) Each O atom has four H bonds around it. 39. The correct order of strength of H   bond in the following compound is (1) H2O > H2O2 > HF > H2S (2) HF > H2O2 > H2O > H2S (3) HF > H2O > H2S > H2O2 (4) HF > H2O > H2O2 > H2S 40. Amongst H2O, H2S, H2Se and H2Te the one with the highest boiling point is (1) H2O because of hydrogen bonding. (2) H2Te because of higher molecular weight. (3) H2S because of hydrogen bonding. (4) H2Se because of lower molecular weight.

Previous Years’ NEET Questions 1. The number of unpaired electrons in a paramagnetic diatomic molecule of an element with atomic number 16 is (1) 2 (2) 3 (3) 5 (4) 1 (AIPMT 2006) 2. In which of the following molecules all the bonds are not equal? (1) ClF3 (2) BF3 (3) AlF3 (4) NF3 (AIPMT 2006) 3. The electronegativity difference between N and F is greater than that between N and H yet the dipole moment of NH3 (1.5 D) is larger than that of NF3 (0.2 D). This is because (1) in NH3 as well as in NF3 the atomic dipole and bond dipole are in the same direction. (2) in NH3 the atomic dipole and bond dipole are in the same direction whereas, in NF3 these are in opposite directions. (3) in NH3 as well as NF3 the atomic dipole and bond dipole are in opposite directions. (4) in NH3 the atomic dipole and bond dipole are in the opposite directions whereas in NF3 these are in the same direction. (AIPMT 2006) 4. In which of the following pairs, the two species are isostructural? (1) BrO3- and XeO3 (2) SF4 and XeF4 (3) SO23- and NO3- (4) BF3 and NF3 (AIPMT 2007)

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OBJECTIVE CHEMISTRY FOR NEET

5. Which one of the following ionic species has the greatest proton affinity to form stable compound? (1) I−

(2) HS−

(3) NH 2-

(4) F−

6. Four diatomic species are listed below in different sequences. Which of these presents the correct order of their increasing bond order? (1) C 22- < He2+ < NO < O2- (2) He+2 < O2- < NO < C 22(3) O2- < NO < C 22- < He+2 (4) NO < C 22- < O2- < He+2 (AIPMT 2008) 7. The angular shape of ozone molecule (O3) consists of bond and 1p bond. bonds and 1p bond. bond and 2p bonds. bonds and 2p bonds.

(4) NH 2- and H 2O

13. In which one of the following species the central atom has the type of hybridization which is not the same as that present in the other three? (1) PCl5 (2) SF4 (3) I 3- (4) SbCl 52(AIPMT PRE 2010) 14. Which one of the following species does not exist under normal conditions? (1) Li2 (3) Be2

(2) Be+2 (4) B2 (AIPMT PRE 2010)

(AIPMT 2008)

8. According to MO theory which of the following lists ranks the nitrogen species in terms of increasing bond order? (1) N 2- < N 22- < N 2 (2) N 2- < N 2 < N 22(3) N 22- < N 2- < N 2

(3) BF3 and NO2-

(AIPMT PRE 2010) (AIPMT 2007)

(1) 1s (2) 2s (3) 1s (4) 2s

(1) BF3 and NH 2- (2) NO2- and NH 3

(4) N 2 < N 22- < N 2(AIPMT 2009)

9. What is the dominant intermolecular force or bond that must be overcome in converting liquid CH3OH to a gas? (1) London dispersion force (2) Hydrogen bonding (3) Dipole-dipole interaction (4) Covalent bonding (AIPMT 2009) 10. Which one of the following bonds produces a solid that reflects light in the visible region and whose electrical conductivity decreases with temperature and has high melting point? (1) Covalent bonding (2) Metallic bonding (3) van der Waal’s bonding (4) Ionic bonding

15. In which of the following molecules, the central atom does not have sp3 hybridization? (1) SF4 (2) BF4+ (3) NH 4 (4) CH4 (AIPMT MAINS 2010) 16. Some of the properties of the two species, NO3- and H3O+ are described below. Which one of them is correct? (1) Isostructural with same hybridization for the central atom. (2) Isostructural with different hybridization for the central atom. (3) Similar in hybridization for the central atom with different structures. (4) Dissimilar in hybridization for the central atom with different structures. (AIPMT MAINS 2010) 17. Which of the following has the minimum bond length? (1) O+2 (2) O2O2 (3) O22 (4)   (AIPMT PRE 2011)

(AIPMT PRE 2010) 11. The correct order of increasing bond angle in the following species is (1) ClO2- < Cl 2O < ClO2 (2) Cl 2O < ClO 2 < ClO22

2

(3) ClO2 < Cl 2O < ClO (4) Cl 2O < ClO < ClO2 (AIPMT PRE 2010) 12. In which of the following pairs of molecules/ions, the central atoms have sp2 hybridization?

Chapter 14_Chemical Bonding and Molecular structure.indd 344

18. The pairs of species of oxygen and their magnetic behaviours are noted below. Which of the following presents the correct description? (1) O, O22 (Both paramagnetic) (2) O2- , O22- (Both diamagnetic) (3) O+ , O22- (Both paramagnetic) (4) O+2 , O2 (Both paramagnetic) (AIPMT MAINS 2011)

1/4/2018 5:18:29 PM

Chemical Bonding and Molecular Structure 19. Which one of the following pairs is isostructural (i.e. having the same shape and hybridization)? (1) NF3 and BF3 (2) BF4- and NH 4+ (3) BCl3 and BrCl3 (4) NH3 and NO3-

28. Dipole induced dipole interactions are present in which of the following pairs? (1) SiF4 and He atoms (2) H2O and alcohol (3) Cl2 and CCl4 (4) HCl and He atoms (NEET 2013)

(AIPMT PRE 2012) 20. Which of the following species contains three bond pairs and one lone pair around the central atom? 2

(1) NH (2) PCl3 (3) H2O (4) BF3 (AIPMT PRE 2012) 21. The pair of species with the same bond order is (1) NO, CO (2) N2, O2 (AIPMT PRE 2012)

(1) XeF4 (3) SF4

(2) BF3 (4) SiF4

(NEET 2013)

30. Which of the following is paramagnetic? (1) NO+ (2) CO (3) O2- (4) CN− 31. Which of the following molecules has the maximum dipole moment? (1) CO2 (3) NH3

22. Bond order of 1.5 is shown by (1) O22 (2) O2 (3) O+2 (4) O2(AIPMT PRE 2012) 23. During change of O2 to O2- ions, the electron adds on which one of the following orbitals? (1) s* orbital (2) p orbital (3) p* orbital (4) s orbital (AIPMT MAINS 2012) 24. Four diatomic species are listed below. Identify the correct order in which the bond order is increasing in them. He+2 < O2- < NO < C 22(1) C 22- < He+2 < O 2- < NO (2) (3) NO < O2- < C 22- < He+2 (4) O2- < NO < C 22- < He+2 (AIPMT MAINS 2012) 25. Which of these is least likely to act as a Lewis base? (2) CO (4) BF3 (NEET 2013) 26. Which one of the following molecules contains no p bond? (1) NO2 (2) CO2 (3) H2O (4) SO2 (NEET 2013) 27. XeF2 is isostructural with (1) BaCl2 (2) TeF2 (3) ICl 2- (4) SbCl3 (NEET 2013)

Chapter 14_Chemical Bonding and Molecular structure.indd 345

29. Which of the following is a polar molecule?

(NEET 2013)

(3) O22- , B2 (4) O+2 , NO +

(1) PF3 (3) F−

345

(2) CH4 (4) NF3

(AIPMT 2014)

32. Which one of the following species has planar triangular shape? (1) N3 (2) NO3(3) NO2 (4) CO2

(AIPMT 2014)

33. Which of the following pairs of ions are isoelectronic and isostructural? (1) ClO3- , CO32- (2) SO23- , NO 3(3) ClO3- , SO23- (4) CO23- , SO23(AIPMT 2015) 34. Maximum bond angle at nitrogen is present in which of the following? (1) NO2- (2) NO+2 (3) NO3- (4) NO2

(AIPMT 2015)

35. Which of the following species contain equal number of s and p bonds? (1) XeO4 (3) CH2(CN)2

(2) (CN)2 (4) HCO3(AIPMT 2015)

36. In which of the following pairs, both the species are not isostructural? (1) NH3, PH3 (2) XeF4, XeO4 (3) SiCl4, PCl +4 (4) Diamond, silicon carbide (RE AIPMT 2015)

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OBJECTIVE CHEMISTRY FOR NEET

37. Which of the statements given below is incorrect? (1) ONF is isoelectronic with O2N . (2) OF2 is an oxide of fluorine. (3) Cl2O7 is an anhydride of perchloric acid. (4) O3 molecule is bent. −



(4) Lone pair-Bond pair > Bond pair-Bond pair > Lone pair-Lone pair (NEET I 2016)

41. The hybridizations of atomic orbitals of nitrogen in NO+2 , NO 3- and NH +4 respectively are

(RE AIPMT 2015) 38. Decreasing order of stability of O2, O2- , O+2 and O22 is

(1) sp2, sp3 and sp (3) sp2, sp and sp3

(2) sp, sp2 and sp3 (4) sp, sp3 and sp2 (NEET II 2016)

(1) O2 > O+2 > O22- > O2- (2) O2- > O 22- > O+2 > O2 (3) O+2 > O2 > O2- > O 22- (4) O22- > O2- > O 2 > O+2 (RE AIPMT 2015) 39. Consider the molecules CH4, NH3 and H2O. Which of the given statements is false? (1) The H   C   H bond angle in CH4, the H   N   H bond angle in NH3, and the H   O   H bond angle in H2O are all greater than 90°. (2) The H   O   H bond angle in H2O is larger than the H   C   H bond angle in CH4. (3) The H   O   H bond angle in H2O is smaller than the H   N   H bond angle in NH3. (4) The H   C   H bond angle in CH4 is larger than the H   N   H bond angle is NH3. (NEET I 2016)

42. Which of the following pairs of ions is isoelectronic and isostructural? (1) ClO3- , CO32- (3) ClO3- , SO23-

SO23- , NO3 (2) CO23- , NO 3 (4) (NEET II 2016)

43. The species having bond angles 120° is (1) ClF3 (2) NCl3 (3) BCl3 (4) PH3

(NEET 2017)

44. Which of the following pairs of species have the same bond order? (1) O2 , NO + (2) CN−, CO (3) N 2 , O2- (4) CO, NO (NEET 2017)

40. Predict the correct order among the following: (1) Lone pair-Lone pair > Lone pair-Bond pair > Bond pair-Bond pair (2) Lone pair-Lone pair > Bond pair-Bond pair > Lone pair-Bond pair (3) Bond pair-Bond pair > Lone pair-Bond pair > Lone pair-Lone pair

45. Which of the following pairs of compounds is isoelectronic and isostructural? (2) IBr2- , XeF2 (4) BeCl2, XeF2

(1) TeI2, XeF2 (3) IF3, XeF2

(NEET 2017)

Answer Key Level I  1. (3)

 2. (3)

  3. (1)

  4. (1)

  5. (3)

  6. (3)

 7. (2)

  8. (3)

  9. (1)

10. (1)

11. (2)

12. (3)

13. (4)

14. (2)

15. (3)

16. (3)

17. (3)

18. (3)

19. (3)

20. (1)

21. (2)

22. (3)

23. (3)

24. (2)

25. (2)

26. (2)

27. (3)

28. (1)

29. (1)

30. (2)

31. (1)

32. (4)

33. (3)

34. (1)

35. (1)

36. (2)

37. (2)

38. (3)

39. (1)

40. (1)

41. (1)

42. (3)

43. (2)

44. (2)

45. (2)

46. (4)

47. (4)

48. (1)

49. (2)

50. (4)

51. (4)

52. (4)

53. (4)

Level II  1. (3)

 2. (2)

 3. (1)

 4. (4)

 5. (4)

 6. (3)

 7. (3)

 8. (3)

 9. (1)

10. (1)

11. (1)

12. (3)

13. (3)

14. (4)

15. (4)

16. (3)

17. (4)

18. (1)

19. (4)

20. (4)

21. (1)

22. (4)

23. (2)

24. (2)

25. (2)

26. (2)

27. (4)

28. (1)

29. (2)

30. (1)

31. (2)

32. (1)

33. (2)

34. (4)

35. (1)

36. (3)

37. (4)

38. (4)

39. (4)

40. (1)

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347

Previous Years’ NEET Questions  1. (1)

 2. (1)

 3. (2)

 4. (1)

 5. (3)

 6. (2)

 7. (2)

 8. (3)

 9. (2)

10. (2)

11. (4)

12. (3)

13. (4)

14. (3)

15. (1)

16. (4)

17. (1)

18. (4)

19. (2)

20. (2)

21. (3)

22. (4)

23. (3)

24. (2)

25. (4)

26. (3)

27. (3)

28. (4)

29. (3)

30. (3)

31. (3)

32. (2)

33. (3)

34. (2)

35. (4)

36. (2)

37. (2)

38. (3)

39. (2)

40. (1)

41. (2)

42. (2), (3)

43. (2)

44. (2)

45. (2)

Hints and Explanations Level I

7. (2) BF3 does not follow octet rule because central atom, boron lacks an electron pair. Thus, it also acts as Lewis acid.

1. (3) At bond distance, the attractive forces overweigh the repulsive forces. 3. (1) A gaseous HCl molecule has hydrogen and chlorine linked by a covalent bond. Here electronegativity of chlorine is greater than that of hydrogen. Due to this the shared pair of electron is more attracted towards chlorine. Thus, chlorine end of molecule has higher electron density and becomes slightly negative and the hydrogen end slightly positive. Hence the covalent bond in HCl has a polar character as shown below. d

Cl

3-

P

O

P

O

O

O

O

3-

O

O

3-

O P

3-

O O

O

P

O

O

O

No. of bonds No. of resonating structures 5 = = 1.25 4

Bond order =



Three unit negative charge is being shared by four O atoms. Therefore, formal charge = –3/4 = 0.75.

6. (3) In the N2 molecule, 3 electron pairs are shared by nitrogen atoms as there are 5 electrons in its outermost shell. ´ ´

N

´ ´ F´ ´ ´ ´´

´ ´ ´ ´ F ´ ´ ´

8. (3) Structure of BF3 is F

´

´ B F ´

F

10. (2) Hydrogen is non-metal, and non-metal atoms form covalent bond.

5. (3)

O

B

d

H

O

´´ ´F ´ ´ ´´

´ ´ ´ N

Chapter 14_Chemical Bonding and Molecular structure.indd 347

N

N

11. (2) The stability of ionic bond depends upon the lattice energy which is expected to be more between Mg and F due to +2 charge on Mg atom. 12. (3) We know that bond angles of NH3 = 107o, NH4 + = 109.5o, PCl3 = 100o. Therefore bond angle of NH4+ is maximum. 13. (4) Ionic compounds generally have high melting and boiling points because of the strong electrostatic force of attraction between oppositely charged ions. Consequently, a considerable amount of energy is required to overcome strong attractive interionic forces and to break down the crystal lattice. 15. (3) The bond length decreases in the order sp3 > sp2 > sp. Because of the triple bond, the carbon-carbon bond distance in ethyne is shortest. 16. (3) Dissociation energy of any molecules depends upon bond order. Bond order in N2 molecule is 3 while bond order in N2 + is 2.5. Further we know that more the bond order, more is the stability and more is the bond dissociation energy. 17. (3) Benzene and the compounds having benzene ring show resonance while ethylamine does not show resonance.

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OBJECTIVE CHEMISTRY FOR NEET

18. (3) Polarity of the bond depends upon the electronegativity difference of the two atoms forming the bond. Greater the electronegativity difference more is the polarity of the bond.

N 





3.0 – 3.0    3.5 – 4.0    3.0 – 4.0    3.0 – 3.0



As the electronegativity difference between N and F is maximum, hence this bond is most polar.

O   F

N    F   N  N X3



19. (3) In PF5 the hybridization is sp d 3



Axial P   F bonds are longer than equatorial bonds. F F

X5

X4

P

X1

M



 Cl   

29. (1)

X6

X2

Bond angles between X4   M   X2 = 180o, X1   M   X3 = 180o and X5   M   X6 = 180o

30. (2) Hybridization is sp3 and shape pyramidal. 31. (1) SF4 – Configuration of excited S atom:

F

3s2

F

3p3

3d1

F

20. (1) In X   H-----Y, X and Y both are electronegative elements (i.e., attracts the electron pair) then electron density on X will increase and on H will decrease. 21. (2) The removal of an electron from a diatomic molecule may increase the bond order as in the conversion O2 (2) → O2+ (2.5) or decrease the bond order as in the conversion, N2 (3.0) → N2+ (2.5), As a result, the bond energy may increase or decrease. Thus option (2) is correct.

sp3d Shape - square pyramidal, one lone pair



2s1

O

Shape-tetrahedral; no lone pair



O

N

> O O-

N

> O O-

N

N

O> N

O

O

23. (3) As dipole moment = electric charge × bond length

D.M. of AB molecules = 4.8 × 10-10 × 2.82 × 10-8 = 13.53 D



D.M. of CD molecules = 4.8 × 10-10 × 2.67 × 10-10 = 12.81 D



Now,

%ionic character =

Actual dipole moment of the bond Dipole moment of pureionic compound

Then % ionic character in AB =

10.41 ´ 100 = 76.94 % 13.53

10.27 % ionic character in CD = ´ 100 = 80.23 % 12.81

24. (2) We know that in O2, the bond order is 2 and in O2bond, the order is 1.5. Therefore option (2) is incorrect.

Chapter 14_Chemical Bonding and Molecular structure.indd 348

2p3

sp3

22. (3) The N   O bond length decreases in the order O

CF4 – Configuration of excited atom:

XeF6 – have square planar geometry with 2 lone pairs.

32. (4) As there is no lone pair on boron in BCl3 therefore no repulsion takes place. But there is a lone pair on nitrogen in NCl3. Therefore repulsion takes place. Thus BCl3 is planar molecule but NCl3 is a pyramidal molecule. 33. (3) Both XeF2 and IF2- are sp3d hybridized and have planar shape. 35. (1) For linear arrangement of atoms the hybridization should be sp (linear shape, 180° angle). Only H2S has sp3 hybridization and hence has angular shape while C2H2, BeH2 and CO2 all involve sp- hybridization and hence have linear arrangement of atoms. 36. (2) Sigma bond is stronger than p bond. The electrons in the p bond are loosely held. The bond is easily broken and is more reactive than s bond. Energy released during sigma bond formation is always more than p bond because of greater extent of overlapping.

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Chemical Bonding and Molecular Structure 37. (2) The strength of the interactions follows the order van der Waals < Hydrogen bonding < Dipole-dipole < Covalent 38. (3) As sigma bond is stronger than the p (pi) bond, so it must be having higher bond energy than p (pi) bond.

349

51. (4) Due to intermolecular hydrogen bonding in methanol, it exists as associated molecule. 52. (4) HF form linear polymer structure due to hydrogen bonding. 53. (4) Hydrogen bonding is the weakest.

40. (1)

Level II s orbital



p orbital

1. (3) The order is as follows:

The overlap between s-and p-orbitals occurs along internuclear axis and hence the angle is 180o.

41. (1) Linear combination of two hybridized orbitals leads to the formation of sigma bond. 42. (3) In P—O bond, p bond is formed by the sidewise overlapping of d-orbital of P and p-orbital of oxygen. Hence, it is formed by pp- and dp- overlapping. s O−

O P

p

O−

O−

44. (2) BF3 involves sp2 hybridization. 46. (4) Ions having sp3 hybridization contain atoms at the corners of tetrahedron.

C



Antibonding electron = 8 (4 pairs)

48. (1) Paramagnetic character is based upon presence of unpaired electron Cl– = 1s2, 2s22p6, 3s23px23py23pz2



17



Be = 1s2, 2s1 4



10



33



Hence only Cl– does not have unpaired electrons.

Ne2+ = 1s2, 2s22px22py12pz1 As+ = 1s2, 2s22p6, 3s23p63d10, 4s24px14py14pz0

1 49. (2) We know, Bond order = ( N b - N a ) 2

Therefore, for Li2 = 1, N2 = 3, Be2 = 0 and O2 = 2.

50. (4) With the increase of electronegativity and decrease in size of atoms to which hydrogen is covalently linked, the strength of the hydrogen bond increases. As F is most electronegative thus HF shows maximum strength of hydrogen bond.

Chapter 14_Chemical Bonding and Molecular structure.indd 349

O HBr > HI (decreasing bond polarity)



CD3F > CH3F (D is more electropositive than hydrogen)

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OBJECTIVE CHEMISTRY FOR NEET

SO2 > SO3 (SO3 is symmetrical so dipole moment is 0) CH3

Cl C

C

H


O2 > O2- > O 22- .

O s p s Xe s O ps pp O O



39. (2) All the molecules are sp 3 hybridized. The bond angles are

It contains 4s and 4p bonds.

36. (2) The structures of the given compounds are as follows:



In CH4, H   C   H bond angle is 109°28′ (no lone pair).





In NH3, H   N   H bond angle is 107°16′ (one lone pair).



In H2O, H   O   H bond angle is 104°29′ (two lone pairs).

Option (1): Both NH3 and PH3 are pyramidal in shape.

P

N H



H

H

H

Option (2): The structure of XeF4 is square planar with hybridization sp3d2, while the structure of XeO4 is tetrahedral with sp3 hybridization. F

O

F Xe

40. (1) As per VSEPR theory, lone pair − lone pair repulsion is the highest and bond pair-bond pair repulsion is the weakest.

O

F

Molecule O

O

Option (3): Both SiCl4, PCl +4 are tetrahedral in shape. Cl

+

Cl

Si

Hybridization

Shape

+ 2 

NO

sp

Linear

NO3 −

sp2

Trigonal planar

NH4 +

sp3

Tetrahedral

42. (2), (3) The molecules that are isoelectronic and isostructural are:

P

Cl Cl

Cl

Lone pair – Lone pair > Lone pair – Bond pair > Bond pair – Bond pair

41. (2)

Xe

F



H

H

Cl Cl

Cl



Option (4): SiC has a three-dimensional structure of Si and C atoms, each atom tertrahedrally surrounded by four of the other kind.



In diamond each C atom is tetrahedrally surrounded by four other C atoms. The tetrahedra are linked together into a three-dimensional giant molecule.

Molecule

No. of electrons

Shape

3

ClO , SO  

42

Pyramidal

CO23- , NO3-  

32

Trigonal planar

23

43. (3) The shape of the BCl3 molecule is a planar triangle with bond angles of 120°. 1s

2s

2p

Electronic structure of boron atomexcited state

37. (2) Option (1): Both ONF and O2N− are isoelectronic to each other as both the species contain 24 electrons.

(sp2 hybridization)



Option (2): The electronegativity of fluorine is more than that of oxygen. Hence, OF2 is a fluoride of oxygen.





Option (3): Dichlorine heptoxide (Cl2O7) is made carefully by dehydrating perchloric acid with phosphorus pentoxide.

44. (2) The bond order of given species are as follows:

2HClO4



P4O10 H2O

Cl2O7

O

O O

Chapter 14_Chemical Bonding and Molecular structure.indd 354

O

Species

O

Bond order

+

2 and 3

CN−, CO

3 and 3

N 2 , O2-

3 and 1.5

CO, NO

3 and 2.5

O2 , NO

Option (4): The structure of O3 is angular, with an O   O   O bond angle of 116°48′. O

Three singly occupied orbitals (2s and 2p) form bonds with unpaired electrons from three chlorine atoms.

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Chemical Bonding and Molecular Structure

355

45. (2) Both XeF2 and IBr2- are linear molecules. Their structures are as follows: F

Xe

Br

I

Br

-

F

Chapter 14_Chemical Bonding and Molecular structure.indd 355

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15

Hydrogen

Chapter at a Glance 1. Hydrogen has the simplest atomic structure of all the elements, and consists of a nucleus containing one proton and one orbital electron. Hydrogen is the first element in the periodic table, the electronic structure is written as 1s1. It is the most abundant (about 70%) element in the universe. 2. Dual Nature of Diatomic Hydrogen (a) The structure of hydrogen atom is in some ways like that of the alkali metals (Group 1). (i)  Like alkali metals, it has just one electron in the outer shell. (ii)  The loss of one electron produces a cation, H+, like M+ ions formed by alkali metals, but hydrogen has a much greater tendency to pair the electron to form a covalent bond. (iii)  Like alkali metals, hydrogen forms oxides, sulphides and halides. However, it does not show metallic properties and unlike alkali metals has very high ionization enthalpy. (b) The structure of hydrogen atom is in some ways like that of the halogens (Group 17): (i) Both hydrogen and halogens is one electron short of a noble gas structure. (ii)  Halogens gain an electron to form X− ions, whereas hydrogen generally does not form H− but does form ionic hydrides M+H− for example, LiH and CaH2 with highly electropositive metals. (c) In some ways, the structure of hydrogen resembles that of the Group 14 elements: (i)  Both have a half-filled outer shell of electrons. (ii)  Hydrides and organometallic compounds show a number of similarities, the hydride is often considered as part of a series of organometallic compounds. For example, LiH, LiMe, LiEt; NH3, NMe3, NEt3; or SiH4, CH3SiH3, (CH3)2SiCl2, (CH3)3SiCl, (CH3)4Si. 3. Isotopes of Hydrogen (a)  Hydrogen has three isotopes protium ( 11 H), deuterium ( 21 Hor D) and tritium ( 31 Hor T ). (b)  These isotopes have the same electronic configuration and hence same chemical properties, which differ only in the rates of reactions and equilibrium constants. 4. Forms of Hydrogen (a)  Hydrogen exists as ortho- and para-hydrogens. If the nuclei in a molecule of hydrogen have same spin it is known as ortho-hydrogen but if they have opposite spin it is known as para-hydrogen. (b) The two forms show difference in their physical properties (boiling points, specific heats, thermal conductivity, etc.) due to differences in their internal energies. 5. Preparation of Hydrogen (a) Reaction of dilute acid with metal or an alkali with metal. For example, Zn + H2 SO4 ® ZnSO4 + H2 Zn + 2NaOH ® Na 2 ZnO2 + H2 (b)  By passing steam over red hot coke to get water gas (also known as synthesis or syn gas), which is a mixture of CO and H2. The process is also known as coal gasification. C + H2 O

1000 C ¾¾¾¾¾ ®

CO + H2  Water gas

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CO can be liquefied at a low temperature under pressure and removed. It can be mixed with steam, cooled to 400°C and passed over iron oxide in a shift converter, giving H2 and CO2 (water gas shift reaction). (c)  Steam reformer process: Light hydrocarbons such as methane are mixed with steam and passed over a nickel catalyst at 800–900°C. CH4 + H2 O ® CO + 3H2 CH4 + 2H2 O ® CO2 + 4H2      Pure hydrogen is obtained from the mixture of CO + CO2 + H2 by water gas shift reaction. (d) By the electrolysis of water using platinum electrodes in the presence of traces of acid or base. 1 O2 + 2e 2 Cathode : 2H2 O + 2e - ® 2OH- + H2

     

Anode: 2OH- ® H2 O +

   

1 Overall: H2 O ® H2 + O2 2

(e)  As a byproduct from the chlor-alkali industry, in which aqueous NaCl (brine) is electrolyzed to produce NaOH, Cl2 and H2. Anode: 2Cl - ® Cl 2 + 2e -

ì Na + + e - ® Na ï Cathode: í ï 2Na + 2H2 O ® î

2NaOH + H2

(f ) As a byproduct of cracking of natural hydrocarbon mixtures such as naptha and fuel oil. 6. Properties of Hydrogen (a)  Hydrogen gas is the lightest gas known and is colorless, odorless and almost insoluble in water. (b)  The two hydrogen atoms in dihydrogen gas are joined by very strong covalent bonds (bond energy is 435.9 kJ mol−1). This high bond energy is the cause for low reactivity of hydrogen under normal conditions. As a result many reactions are slow, or require high temperatures, or catalysts (generally transition metals). (c)  Atomic hydrogen is obtained by cleavage of H–H bond of dihydrogen at high temperature using electric arc or under ultraviolet radiations. The 1s1 electron in the orbital can then react: (i)  By forming an electron pair (covalent) bond with another atom: In reactions with non-metals, such as in formation of H2O, HCl (gas) or CH4. (ii)  By losing an electron to form H +: These always exist associated with other atoms or molecules such as H3O+, H9O4 + or H(H2O)n + ions. (iii) By gaining an electron to form H -: Formation of metal hydrides with electropositive metals. (d) Reaction with halogens (i)  With fluorine: The reaction is violent, even at low temperatures.     H2 + F2 ® 2HF (ii)  With chlorine: The reaction is slow in the dark, but is catalyzed by light. It is explosive in sunlight.    H2 + Cl 2 ® 2HCl  (iii) With bromine: The reaction proceeds at temperatures between 200°C and 400°C in the presence of platinum as catalyst.         H2 + Br2 ® 2HBr   (iv) With iodine: The reaction proceeds in the presence of catalyst or ultraviolet radiation H2 + I2 ® 2HI (e)  Reaction with dioxygen: Used in oxy-hydrogen flame. 2H2 (g ) + O2 (g ) ® 2H2 O(l); DH o = - 485 kJ mol -1

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Hydrogen

359

(f )  Reaction with dinitrogen: Used in Haber’s process for manufacture of ammonia. Fe catalyst N 2 (g ) + 3H2 (g ) ¾(673 ¾¾¾¾ ® 2NH3 (g ) DH o = - 92.22 kJ mol -1 K, 200 atm)

(g)  Reaction with metals: Leads to formation of metal hydrides. H2 (g ) + 2M(g) ® 2MH (h) Reaction with metal ions and oxides: These are reduced to corresponding metals. H2 (g) + Pd 2+ (aq) ® Pd(s) + 2H+ (aq ) MO2 + 2H2 ® M + 2H2 O M2 O3 + 3H2 ® 2M + 3H2 O MO + H2 ® M + H2 O (i) Reaction with organic compounds: These react in presence of catalysts to give hydrogenated products.

CH3 

(CH2)n 

 

CH   CH 

 

COOH + H2 → CH3 

 

(CH2)n 

 

CH2 

 

CH2 

 

COOH

 

7. Uses of Hydrogen (a)  Hydrogen is used in industries for the (i) manufacture of ammonia by Haber’s process, (ii) manufacture of hydrochloric acid, (iii) the manufacture of methyl alcohol and (iv) production of vanaspathi or margarine or dalda by the hydrogenation of vegetable oil and in the production of synthetic petrol by Fischer–Tropsch synthesis. (b)  Hydrogen is used as reducing agent in the extraction of metals like molybdenum and tungsten. (c)  Hydrogen is used as a fuel in the form of water gas, coal gas, etc., which contain hydrogen. 8. Hydrides The binary compounds of hydrogen with other elements are called hydrides and they are of three types: (a)  Saline or ionic or salt-like hydrides: These are formed with most of the s-block elements LiH, BeH2 and MgH2. They have significant covalent character. (i)  They have high melting points, insoluble in common non-aqueous solvents; react vigorously with water and acids liberating hydrogen. (ii)  Alkali metal hydrides have rock-salt structure. (b)  Metallic or interstitial hydrides: These are formed by d- and f-block elements and are formed by the presence of hydrogen atoms in the interstitial positions of the metal crystals. (i) Metallic hydrides are non-stoichiometric compounds and show electric conduction. (ii) Some metals like Pt and Pd can accommodate a very large volume of hydrogen called occluded hydrogen. (iii)  In the periodic table, the region of groups 7–9 is referred to as the hydride gap because metals in these groups do not form metallic hydrides. (c)  Molecular or covalent hydrides: These are formed by p-block elements and consist of individual, discrete covalent molecules. The molecular formula of covalent hydrides can be written as MHn or MH(8-n) where n = group number of the element in the short form periodic table. (i)  Molecular hydrides having less number of electrons for writing the conventional Lewis structure are called electron-deficient hydrides, e.g., B2H6. (ii)  Molecular hydrides formed by the elements of IV or 14 group elements in which all the valence electrons are involved for bond formation are known as electron-precise hydrides. (iii)  Molecular hydrides which contain more valence electrons, than the required for bond formation are called electron-rich hydrides. They contain lone pairs, e.g., NH3, H2O. 9. Water (a) W  ater has abnormally high freezing point, boiling point, heat of vaporization, heat of fusion, etc., due to the association of molecules through hydrogen bonds. (b) The high heat of vaporization and high heat capacity of water are responsible for moderation of the climate and body.

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(c) S olubility of ionic compounds in water is due to its polar nature while the solubility of covalent compounds like alcohols, carbohydrates, urea, etc. is due to their ability to form hydrogen bonds with water. (d) Under very high pressure and temperature water behaves like a non-polar solvent in which organic compounds are more soluble while inorganic salts are insoluble. (e) Purest water is rain water. In water, the ratio of hydrogen and oxygen is 2:1 by volume and 1:8 by weight. (f ) Dipole moment of water 1.85D while dielectric constant is 78.39. (g) Maximum density of water is at 4°C. (h) Structurally ice have nine different forms, the one which is formed at atmospheric pressure is normal hexagonal form (Ih) but at very low temperatures it condenses as cubic form. (i) Ice has lesser density than the water with which it is in equilibrium. (j) The water molecules are joined together in three dimensional network in which each oxygen atom is surrounded by four water molecules tetrahedrally and bond through four hydrogens, two by normal covalent bond and the other two by hydrogen bonds. (k) Due to the three dimensional hydrogen bonded structure ice has got open spaces due to which its density is less than water and floats over water. (l) Many salts crystallize from their aqueous solutions as hydrated salts and the water molecules associated with these salts are of five types: (i) coordinated water, (ii) hydrogen bonded water, (iii) lattice water, (iv) zeolite water and (v) clatharate water. (m) In some hydrated salts water molecules are coordinated to metal ion (complex compounds). For example, [Ni(H2O)6]Cl2; [Fe(H2O)6 ]Cl;[Cr(H2O)6]Cl3 (n) In some compounds water molecules are hydrogen bonded to oxygen containing anions. For example, in CuSO4·5H2O, four water molecules are coordinated to Cu2+ ion and one water molecule is in hydrogen bond with sulphate ions. (o) In certain compounds water is present as interstitial water, that is, it occupies the interstices in the crystal lattice, for example BaCl2·2H2O. (p) Water can act both as Brönsted acid and Brönsted base due to autoprotolysis or self -ionization and hence water is amphoteric. H2O + H2O ® H3O+ + OH(q) (r) (s) (t)

Halogen oxidizes water to oxygen. Solution of chlorine in water is called as chlorine water. Solutions of basic oxides in water are alkaline while the solutions of acidic oxides in water are acids. Salt undergoes hydrolysis in water. Water reacts with carbon dioxide in the presence of sunlight and chlorophyll to give carbohydrates (photosynthesis).

10. Hard Water The water which does not give ready and permanent lather with soap is called hard water. Hardness of water is can be of two types: (a)  Temporary hardness is due to the presence of bicarbonates of magnesium and calcium. It can be removed by boiling the water or by the addition of calculated amount of slaked lime (Clark’s method) where soluble bicarbonates converts into insoluble carbonates. (b)  Permanent hardness is due to the presence of chlorides and sulphates of magnesium and calcium. It can be removed by: (i) Addition of calculated amount of sodium carbonate (washing soda). Na 2 CO3 + MgCl2 ® MgCO3 ¯ + 2NaCl Na 2 CO3 + CaSO4 ® CaCO3 ¯ + Na 2 SO4 (ii)  Addition of calgon (sodium hexametaphosphate) which forms soluble complexes with calcium and magnesium ions at pH 10 (Calgon’s method). 2Ca 2 + + [Na 4 (PO3 )6 ]2 - → 4Na + + [Ca 2 (PO3 )6 ]2 -

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Hydrogen

361

(iii)  Percolating through the bed of zeolite (Na2Z) which are naturally occurring hydrated aluminosilicate minerals (ion exchange method or permutit process). Na 2 Z + CaSO4 (or CaCl2 ) → CaZ + Na 2 SO4 (or 2NaCl ) Na 2 Z + MgCl 2 (or MgSO4 ) → MgZ + 2NaCl (or Na 2 SO4 ) Na 2 Z + Ca(HCO3 )2 (or Mg(HCO3 )2 ) → CaZ (or MgZ ) + 2 NaHCO3 (iv)  Using synthetic ion exchange resins for removal of cations causing hardness in water (demineralization method). nRCOOH+ + Ca 2 + → n[(RCOO)2 Ca 2 + ] + 2H+ +

+

(cation exchange resin)

n [R  N(CH3 )3 ]OH- + Cl - → n [R  N(CH3 )3 ]Cl - + OH-

(anion exchange resin)

11. Heavy Water (a) It is prepared by exhaustive electrolysis of water and is obtained as a by product in some fertilizer industries. (b) The comparison of physical properties with water is given in below table. Physical constant

H2O

D2O

Freezing point (°C)

0

3.82

Boiling point (°C)

100

101.42

0.917

1.017

Density at 20°C (g cm–3)

4

Temperature of maximum density (°C)

1.0 × 10

Ionic product Kw at 25°C

11.6 –14

3.0 × 10–15

82

80.5

Solubility grams of NaCl/100 g water at 25°C

35.9

30.5

Solubility grams of BaCl2/100 g water at 25°C

35.7

28.9

Enthalpy of vaporization (at 100°C, kJ mol )

40.66

41.61

Dielectric constant at 20°C

−1

12. Various Physical Properties of Heavy Water (D2O) (a) It is colorless, odorless and tasteless. (b) All the physical constants like boiling point, density etc., are higher than that of ordinary water. (c) It is commonly used as moderator in nuclear reactors. 13. Hydrogen Peroxide (a)  Hydrogen peroxide can be considered as oxygenated water. It was first prepared by Thenard. Its commercial name is perhydrol. (b) Structure: H2O2 has a “skew” structure. The dimensions of H2O2 molecule are shown in below figure. It is the smallest molecule known to show restricted rotation, in this case about the O—O bond, and this is presumably due to repulsion between the OH groups. A similar structure is retained in the solid H2O2, but the bond lengths and angles are slightly changed because of hydrogen bonding.

95.0 pm 147.4 pm

94.8º

H

98.8 pm H 101.9º 145.8 pm H

H 111.5º (a)

90.2º (b)

Structure of H2O2 in (a) gaseous and (b) solid states.

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OBJECTIVE CHEMISTRY FOR NEET

(c)  Preparation of H2O2 (i)  H2O2 was earlier obtained by electrolysis of H2SO4 or (NH4)2SO4 at a high current density to form peroxosulphates, which were then hydrolyzed. H2 S2 O8

2SO24 - ¾Electrolysis ¾¾¾ ® S2 O82 - + 2e + H2 O ® H2 SO5 + H 2SO 4 Peroxomonosulphuric acid

Peroxodisulphuric acid

H2 SO5 + H2 O ® H2 SO4 + H2 O2 (ii)  In the laboratory, it can be prepared by adding calculated amount of sodium peroxide to ice cold dilute (20%) solution of H2SO4. Na 2 O2 + H2 SO4 ® Na 2 SO4 + H2 O2 (iii)  It can be prepared by the addition of acid to peroxide or a persulphide salt. This is known as Merck’s process BaO2 × 8H2 O(s) + H2 SO4 (aq ) ® H2 O2 (aq ) + BaSO4 (s ) + 8H2 O(l)            (iv)  It is produced on an industrial scale by a cyclic process where 2-ethyl anthroquinol is oxidized by air to the corresponding quinone and H2O2. (d)  Physical properties: Pure H2O2 is almost a colorless liquid with a tinge of pale blue and resembles water quite closely. It is more hydrogen bonded than is water and so has a higher boiling point (b.p. 152°C). It is completely miscible with water in all proportions and forms hydrate structure of the type H2O2⋅H2O. (e) Chemical properties (i)  As oxidizing agent: H2O2 acts as a strong oxidizing agent. The reaction is slow in acidic medium and fast in basic medium. 2Fe 2+ + 2H+ + H2 O2 ® 2Fe3+ + 2H2 O (Slow) 2+ 3+   2Fe + H2 O2 ® 2Fe + 2OH

(Fast)

(ii)  As reducing agent: H2O2 is reduced in presence of stronger oxidizing agents and oxygen is evolved. For example, In acidic medium: 2KMnO 4 + 5H2 O2 + 3H2 SO4 ® 2MnSO4 + K 2 SO4 + 5O2 + 8H2 O In basic medium:   2KMnO 4 + 3H2 O2 ® 2MnO2 + 3O2 + 2H2 O + 2OH

(iii) H2O2 is fairly stable and decomposes only slowly on exposure to light. 2H2 O2 ® 2H2 O + O2

(f) Uses: (i) H2O2 is a harmless strong antiseptic, so is used to clean the wounds. (ii) H2O2 is used as a mild bleaching agent to bleach silk, wool, ivory, etc. (iii) Conc. H2O2 is used as a rocket fuel. (iv) H2O2 is used in restoring the color of old and spoiled lead paintings.

Solved Examples 1. Hydrogen is not more reactive at ordinary temperature, because at ordinary temperature it is in its (1) gaseous state. (2) liquid state. (3) molecular state. (4) atomic state. Solution (3) Hydrogen is not reactive at ordinary temperature because it is in molecular state, that is, as H2 with its

Chapter 15_Hydrogen.indd 362

complete duplet which makes it stable and lesser reactive. At very high temperature, hydrogen exists in atomic form having one electron, that is, incomplete duplet which makes it less stable and highly reactive. 2. Hydrogen has high ionization energy than alkali metals because it has (1) ionic bond. (2) covalent bond.

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Hydrogen (3) large size. (4) small size. Solution (4) Hydrogen has high ionization energy in comparison with alkali metals because it has only one orbital, that is, smaller size. Due to smaller size, hold of nucleus on outermost electron is greater in case of hydrogen. 3. Which one of the following pairs of substances on reaction will not evolve H2 gas? (1) (2) (3) (4)

Iron and H2SO4 (aqueous) Iron and steam Copper and HCl (aqueous) Sodium and ethyl alcohol

(a) CaO (b) Ca(OH)2 (c) CaCO3 (d) Ca(HCO3)2 Solution (b) The chemical A is Ca(OH)2 Chemical A reacts with sodium carbonate to generate caustic soda. Ca(OH)2 + Na 2CO 3 →

Fe + dil. H2SO4 → FeSO4 + H2 ↑

(3)

6. Chemical A is used for water softening to remove temporary hardness. Chemical A reacts with sodium carbonate to generate caustic soda. When CO2 is bubbled through a solution A, it turns cloudy. What is the chemical formula of A?



Solution



2NaOH + CaCO 3

Caustic soda

When CO2 is bubbled through a solution A, it turns cloudy.

3Fe + 4H 2O → Fe3O4 + 4H 2 ↑       

Ca(OH)2 + CO2 → CaCO 3 ↓ + H 2O

(Steam)

Cloudy



       Cu + dil. HCl → No reaction



Copper does not evolve H2 from acid as it is below hydrogen in electrochemical series.





7. Which of the following statements do not define the characteristic property of water “Water is a universal solvent”? (1) (2) (3) (4)

2Na + 2C2H5OH → 2C2H5ONa + H2 ↑

4. Hydrogen will not reduce (1) (2) (3) (4)

363

heated cupric oxide. heated ferric oxide. heated stannic oxide. heated aluminium oxide.

It can dissolve maximum number of compounds. It has very low dielectric constant. It has high liquid range. None of these.

Solution

Solution (4) Hydrogen will not reduce heated Al2O3. As Al is more electropositive than hydrogen, therefore, its oxide will not be reduced by hydrogen.

(2) Water has high dielectric constant, that is, 78.39 C2 N−1m−2, high liquid range and can dissolve maximum number of compounds. That is why it is used as universal solvent. 8. Very pure hydrogen (99.9%) can be made by which of the following processes? (1) Reaction of methane with steam. (2) Cracking natural hydrocarbons of high molecular weight. (3) Electrolysis of water. (4) Reaction of salt like hydrides with water.

5. Consider the following statements: (I) Atomic hydrogen is obtained by passing hydrogen through an electric arc. (II) Hydrogen gas will not reduce heated aluminium oxide. (III) Finely divided palladium adsorbs large volume of hydrogen gas. (IV) Pure nascent hydrogen is best obtained by reacting Na with C2H5OH.

Solution -900° C  → CO + 3H 2 (3) Option (1): CH 4 + H 2O 800



In this method mixture of CO and H2 are obtained.



Option (2): High molecular weight hydrocarbon on cracking gives H2 but this mixture also contain small amount of hydrocarbon.

Solution



(3) Pure hydrogen is evolved by reacting absolute alcohol and Na.

Option (3): Electrolysis of H2O is best method formation of 99.97% pure H2. H2O is mixed with NaOH or KOH.



At the cathode



Which of the above statements is/are correct ? (1) (I) (2) (II) (3) (I), (II) and (III) (4) (II), (III) and (IV)

C 2H5OH + Na → C 2H5ONa +

Chapter 15_Hydrogen.indd 363

1 H2 2

2H 2O + 2e - → 2OH - + H 2

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OBJECTIVE CHEMISTRY FOR NEET



At the anode



1      2OH → 2H 2O + O2 + 2e 2 Overall Reaction 1 H 2O → H 2 + O 2 2



Option (4): LiH + H2O → LiOH + H2



This method is difficult to handle.

-

9. The hybrid state and oxidation state of two oxygen atoms in H2O2 are, respectively, (1) sp2, -1 (2) sp3, +1 (3) sp3, -1 (4) sp2, +2 Solution (3) The hybrid state of two oxygen atoms in H2O2 is sp3 and oxidation state is −1 due to peroxide linkage.

13. In context with the industrial preparation of hydrogen from water gas (CO + H2), which of the following is the correct statement? (1) CO and H2 are fractionally separated using differences in their densities. (2) CO is removed by absorption in aqueous Cu2Cl2 solution. (3) H2 is removed through occlusion with Pd. (4) CO is oxidized to CO2 with steam in the presence of a catalyst followed by absorption of CO2 in alkali. Solution (4) Hydrogen is made in large amounts by the steam reformer process. When water gas is passed over iron or copper catalyst at 400°C, CO gets converted into CO2. Finally CO2 is absorbed in a solution of alkali. CO + H2

+1

H

-1

O

O

+1

(1) glycerine. (2) alcohol. (3) phosphoric acid. (4) Pt powder. Solution (4) Decomposition of H2O2 can be accelerated by finely divided metals such as Ag, Au, Pt, Co, Fe etc. 11. Hydrogen peroxide acts both as an oxidizing and as a reducing agent depending upon the nature of the reacting species. In which of the following cases H2O2 acts as a reducing agent in acid medium? (1) MnO 4- (2) Cr2O72(3) SO23 (4) KI Solution (1) 2MnO 4- + 6H + + 5H 2O 2 → 2Mn 2+ + SO 2 + 8H 2O In this reaction KMnO4 is reduced from +7 to +2 oxidation state. Thus, it acts as an oxidizing agent and H2O2 act as a reducing agent in acidic medium.

12. From the following statements regarding H2O2, choose the incorrect statement. (1) It decomposes on exposure to light. (2) It has to be stored in plastic or wax lined glass bottles in dark. (3) It has to be kept away from dust. (4) It can act only as an oxidizing agent. Solution (4) The oxidation state of oxygen in H2O2 is −1 (peroxide). So, oxygen can increase and decrease its oxidation number which means it can act as a reducing as well as an oxidizing agent.

Chapter 15_Hydrogen.indd 364

CO2(Acidic oxide) + 2H2 KOH (base)

K2CO3

H

10. The decomposition of H2O2 is accelerated by



Water gas

-1

H2O

14. Permanent hardness in water cannot be cured by (1) boiling. (2) ion exchange method. (3) calgon’s method. (4) treatment with washing soda. Solution (1) Permanent hardness is due to presence of chlorides and sulphates of calcium and magnesium in water and also salts of heavier elements such as iron and aluminium. It cannot be cannot be cured by boiling. 15. At its melting point ice is lighter than water because (1) H2O molecules are more closely packed in solid state. (2) ice crystals have hollow hexagonal arrangement of H2O molecules. (3) on melting of ice the H2O molecule shrinks in size. (4) ice forms mostly heavy water on first melting. Solution (2) In the structure of ice, each molecule of H2O is surrounded by three H2O molecules in hexagonal honey comb manner. On the other hand, in water each molecule is surrounded by four neighboring molecules randomly which results an open cage like structure. As a result, there are a number of ‘hole’ or open spaces. In such a structure lesser number of molecules are packed per mL. When ice melts a large number of hydrogen bonds are broken. The molecules therefore move into the holes or open spaces and come closer to each other than they were in solid state. This result sharp increase in the density. Therefore, ice has lower density than water.

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Hydrogen 16. When a substance A reacts with water it produces a combustible gas B and a solution of substance C in water. When another substance D reacts with this solution of C, it also produces the same gas B on warming but D can produce gas B on reaction with dilute sulphuric acid at room temperature. A imparts a deep golden yellow color to a smokeless flame of Bunsen burner. A, B, C and D respectively are (1) Na, H2, NaOH, Zn (2) K, H2, KOH, Al (3) Ca, H2, Ca(OH)2, Sn (4) CaC2, C2H2, Ca(OH)2, Fe



  

(D)

(1) nuclear charge. (2) nuclear reaction. (3) electron spin. (4) proton spin.

(C)

(B)

(C)

(B)

para-hydrogen: Spin of protons or nucleus in opposite direction. 21. Which of the following could act as propellant for rockets? (1) (2) (3) (4)

Zn + dil.H 2SO4 → ZnSO 4 + H 2 ↑



20. ortho- and para-hydrogens differ in

ortho-hydrogen: Spin of protons or nucleus in same direction.

Zn + 2NaOH → Na 2 ZnO2 + H 2 ↑



(4) Heavy water declines the rate of reactions taking place in living organism. Hence, it does not support life.

(4) ortho- and para-hydrogens differ in proton spin.

2Na + 2H 2O → 2NaOH + H 2 ↑ (A)

Solution

Solution

Solution (1)

(D)

(B)

Na produces golden yellow color with smokeless flame of Bunsen burner.

17. H2 from HCl can be prepared from (1) Mg (2) Cu (3) P (4) Hg Solution (1) Hydrogen can only be displaced from dilute acids if electrons are supplied to H+ ions in the acid. Mg being an electropositive element facilitate the release of hydrogen gas. Also, Cu and Hg come after hydrogen in the reactivity series. So, both Cu and Hg cannot reduce H+ ion in HCl and phosphorus which is a non-metal does not supply electrons for the reduction of H+. Mg(s) + 2HCl(aq) → MgCl 2(aq ) + H 2 ↑ 18. Which of the following is not hard water? (1) (2) (3) (4)

Water containing CaCl2. Water containing dil. HCl Water containing MgSO4. None of these.

Solution (b) Water that does not lather with soap is known as hard water. Hardness of water is due to the presence of bicarbonates, chlorides and sulphates of calcium and magnesium in it. However, when hard water is treated with dil. HCl, the soluble bicarbonates of Ca and Mg get precipitated as CaCl2 and MgCl2, respectively, which can be removed. 19. Heavy water is injurious to (1) human beings. (2) plants. (3) animals. (4) all of these.

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365

Liquid oxygen + liquid argon Liquid nitrogen + liquid oxygen Liquid hydrogen + liquid oxygen Liquid hydrogen + liquid nitrogen

Solution (3) Liquid hydrogen has already been used as rocket fuel. The chemical reaction involved is H 2(g ) + 1 O2(g ) → H 2O(l) + 286 kJ 2

Both reactants H2 and O2 are stored as liquids in separate tanks.

22. When hydrogen gets occluded on palladium, the conductivity of the metal (1) increases. (2) becomes zero. (3) decreases. (4) first increases then decreases. Solution (3) When hydrogen gets occluded on palladium, the conductivity of the metal decreases because the adsorbed H2 forms interstitial hydrides in which H2 occupies the holes in the crystal lattice of metal. 23. Lead pipes are not used for carrying drinking water because (1) they are covered with a coating of lead carbonate. (2) they are corroded by air and moisture. (3) water containing dissolved air attacks lead, forming soluble hydroxide. (4) None of these. Solution (3) Lead pipes are not used for carrying drinking water because water containing dissolved air attacks lead, forming soluble hydroxide, which is poisonous in nature.

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366

OBJECTIVE CHEMISTRY FOR NEET 2Pb + 2H 2O + O 2 → 2Pb(OH )2



The dissolution of lead in water is known as plum­ bosolvency.

24. Limiting compositions of f-block hydrides are (1) MH2 and MH3 (3) MH2 and MH8

(2) MH3 and MH5 (4) MH2 and MH6

(3) reducing nature. (4) ability to liberate nascent hydrogen. Solution (2) The bleaching property of hydrogen peroxide is due to its ability to liberate nascent oxygen. Bleaching nature of hydrogen peroxide is due to its oxidation. H 2O 2 → H 2O +

Solution

Colored matter +

(1) General f-block elements form interstitial hydrides with the limiting composition of MH2 and MH3, because f-block elements form stable +3 oxidation state. 25. The conversion of atomic hydrogen into ordinary hydrogen is (1) (2) (3) (4)

exothermic change. endothermic change. nuclear change. photochemical change.

Solution (1) Atomic hydrogen combines to form ordinary hydrogen with the evolution of 104.2 kcal of heat (exothermic process). This recombination is catalyzed by certain metals. H + H → H 2 ; ∆H = 104.2 kcal 26. Ionization enthalpy of hydrogen is (1) (2) (3) (4)

equal to that of chlorine. lesser than that of chlorine. slightly higher than that of chlorine. much higher than that of chlorine.

Solution (3) Ionization enthalpy of hydrogen is slightly higher than that of chlorine. In hydrogen, only one electron is present but in chlorine there are 17 electrons present; hence, ionization enthalpy is slightly higher of hydrogen than that of chlorine. 27. The bleaching property of hydrogen peroxide is due to its (1) acidic nature. (2) ability to liberate nascent oxygen.

[O]

Nascent oxygen

(Oxidation)

[O]

Nascent oxygen

→ Colorless ( Bleached)

28. The maximum tolerable limit of degree of hardness in water required for our daily needs is, approximately, (1) 10–15 ppm (2) 100–150 ppm (3) 103 ppm (4) 700–800 ppm Solution (2) It may be noted that degree of hardness up to 100–150 ppm in water required for our daily needs such as cooking, bathing, washing of clothes, etc., is tolerable. However, if the degree of hardness exceeds the limit then water is not suitable for domestic use. 29. Systematic name of H2O is (1) water. (2) hydrogen oxide. (3) oxidane. (4) None of these. Solution (3) Systematic name of H2O is oxidane. 30. Nascent hydrogen consists of (1) (2) (3) (4)

hydrogen atoms with excess of energy. hydrogen molecules with excess energy. hydrogen ions in excited state. solvated protons.

Solution (2) The excess energy associated with the molecules of nascent hydrogen makes it more reactive and imparts it higher reducing strength. For example, ions such as permanganate, ferric, chromate and chlorate are not reduced by ordinary hydrogen but the reactions take place if a small piece of zinc is added to the reaction mixture.

Practice Exercises Level I Dihydrogen 1. Which of the following is correct for hydrogen? (1) It is always collected at cathode. (2) Its ionization energy is very low in comparison with alkali metals.

Chapter 15_Hydrogen.indd 366

(3) It can form bonds in +1 as well as in –1 oxidation states. (4) Its oxide is not stable. 2. Metal which does not react with cold water but evolves H2 with steam is (1) Na (2) K (3) Pt (4) Fe

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Hydrogen 3. Assertion: H2O2 is not stored in glass bottles.

Reason: Alkali metal oxides present in glass catalyze the decomposition of H2O2. (1) If both Assertion and Reason are true and Reason is a correct explanation of Assertion. (2) If both Assertion and Reason are true and Reason is not a correct explanation of Assertion (3) If Assertion is true and Reason is false. (4) If both Assertion and Reason are false.

4. The number of moles of H2 in 0.224 L of hydrogen gas at STP (273 K, 1 atm) is (1) 0.1 (2) 0.01 (3) 0.001 (4) 1 5. Which of the following statement is not correct regarding hydrogen atom? (1) (2) (3) (4)

It resembles with halogens in some properties It resembles with alkali metals in some properties. It cannot be placed in first group of periodic table. It is the lightest element.

6. Which statement is correct for hydrogen? (1) (2) (3) (4)

It has a very high ionization potential. It is always collected at cathode. It can form bonds in +1 as well as –1 oxidation state. It has same electronegativity as halogens.

7. Hydrogen can be differentiated by other alkali metals due to its (1) (2) (3) (4)

non-metallic character. affinity for non-metals. electropositive character. reducing character.

8. Hydrogen molecules differ from chlorine molecule in which of the following respect? (1) Hydrogen molecule is non-polar, but chlorine molecule is polar. (2) Hydrogen molecule is polar, while chlorine molecule is non-polar. (3) Hydrogen molecule can form intermolecular hydrogen bonds but chlorine molecule does not. (4) Hydrogen molecule cannot participate in coordination bond formation but chlorine molecule can. 9. Hydrogen is evolved by the action of cold dil. HNO3 on (1) Fe (2) Mn (3) Cu (4) Al 10. Hydrogen can behave as a metal (1) (2) (3) (4)

at very high temperature. at very low temperature. at very high pressure. at very low pressure.

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11. Hydrogen accepts an electron to form inert gas configuration. In this it resembles (1) halogen. (2) alkali metals. (3) chalcogens. (4) alkaline earth metals. 12. In Bosch’s process which gas is utilized for the production of hydrogen gas? (1) Producer gas (2) Water gas (3) Coal gas (4) None of these 13. The adsorption of hydrogen by metals is called (1) dehydrogenation. (2) hydrogenation. (3) occlusion. (4) adsorption. 14. Dihydrogen is quite stable and relatively inert at room temperature because (1) (2) (3) (4)

it takes part in hydrogen bonding. it is nonpolar. the H—H bond dissociation enthalpy is high. it has lower enthalpy of dissociation.

15. Which of the following reactions produces hydrogen? (1) Mg + H2O (2) BaO2 + HCl (3) H2S2O8 + H2O (4) Na2O2 + 2HCl 16. Which of the following gas is insoluble in water? (1) SO2 (2) NH3 (3) H2 (4) CO2 17. Hydrogen is not obtained when zinc reacts with (1) cold water. (2) dil. HCl. (3) dil. H2SO4. (4) Hot NaOH (20%).

Hydride 18. The ionization of hydrogen atom would give rise to (1) hydride ion. (2) hydronium ion. (3) proton. (4) hydroxyl ion. 19. Ionic hydrides acts as (1) strong acids. (2) weak bases. (3) weak acids. (4) strong bases. 20. Metal hydride on treatment with water gives (1) H2O2 (2) H2O (3) Acid (4) Hydrogen 21. Which hydride is an ionic hydride? (1) H2S (2) TiH1.73 (3) NH3 (4) NaH

Water 22. When two ice cubes are pressed over each other, they unite to form one cube. Which of the following forces is responsible to hold them together?

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368

OBJECTIVE CHEMISTRY FOR NEET (1) (2) (3) (4)

Hydrogen bond formation van der Waals forces Covalent attraction Ionic interaction

23. Water is (1) (2) (3) (4)

more polar than H2S. more or less identical in polarity with H2S. less polar than H2S. None of these.

24. Pure water can be obtained from sea water by (1) Centrifugation. (2) Plasmolysis. (3) Reverse osmosis. (4) Sedimentation. 25. The H–O–H angle in water molecule is about (1) 90° (2) 180° (3) 102.5° (4) 104.5° 26. The alum used for purifying water is (1) ferric alum. (2) chrome alum. (3) potash alum. (4) ammonium alum. 27. The boiling point of water is exceptionally high because (1) (2) (3) (4)

there is covalent bond between H and O. water molecule is linear. water molecules associate due to hydrogen bonding. water molecule is not linear.

33. Commercial 10 volume H2O2 is a solution with strength of approximately (1) 15% (2) 3% (3) 1% (4) 10% 34. Which of the following statements is not true for hydrogen peroxide? (1) (2) (3) (4)

Pure H2O2 is fairly stable. It sometimes acts as a reducing agent. It acts as an oxidizing agent. Aqueous solution of H2O2 is weakly basic.

35. The structure of H2O2 is (1) planar. (2) non-planar. (3) spherical. (4) linear. 36. In lab, H2O2 is prepared by (1) Cold H2SO4 + BaO2 (2) HCl + BaO2 (3) Conc. H2SO4 + Na2O2 (4) H2 + O2 37. Pure hydrogen peroxide is an unstable liquid and decomposes into water and oxygen on long standing or heating. Its decomposition is an example of (1) auto-oxidation. (2) auto-reduction. (3) disproportionation. (4) autoprotolysis. 38. Which of the following is a true structure of H2O2? 180º

(1) H

O

O

H

O

28. Water possesses a high dielectric constant, therefore, (1) (2) (3) (4)

it always contains ions. it is a universal solvent. can dissolve covalent compounds. can conduct electricity.

Hydrogen Peroxide 29. On industrial scale, H2O2 is now generally prepared by (1) (2) (3) (4)

the action of H2SO4 on barium. the action of H2SO4 on sodium peroxide. by the electrolysis of H2SO4. by burning hydrogen in an excess of O2.

30. Which of the following cannot be oxidized by H2O2? (1) Na2SO3 (2) PbS (3) KI (4) O3 31. The percentage by weight of hydrogen in H2O2 is (1) 5.88 (2) 6.25 (3) 25 (4) 50 32. The reaction H2S + H2O2 → S + 2H2O manifests (1) (2) (3) (4)

acidic nature of H2O2. alkaline nature of H2O2. oxidizing action of H2O2. reducing action of H2O2.

Chapter 15_Hydrogen.indd 368

97º

(2) H

(3)

H H

O

O

(4)

H H

O H

O

O

39. Which substance does not speed up decomposition of H2O2? (1) Glycerol (2) Pt (3) Gold (4) MnO2 40. H2O2 is a (1) weak acid. (2) weak base. (3) neutral. (4) none of these. 41. H2O2 is always stored in black bottles because (1) (2) (3) (4)

it is highly unstable. its enthalpy of decomposition is high. it undergo auto-oxidation on prolonged standing. none of these.

42. Which of the following statements is wrong about the uses of H2O2? (1) It is used as aerating agent in production of sponge rubber (2) It is used as an antichlor. (3) It is used for restoring white color of blackened lead painting. (4) None of these.

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Hydrogen 43. H2O2 → 2H+ + O2 + 2e−; E° = −0.68 V.

This equation represents which of the following behavior of H2O2? (1) Reducing (2) Oxidizing (3) Acidic (4) Catalytic

44. The O 

O 

 

H bond angle in H2O2 is

 

(1) 106° (2) 109°28¢ (3) 1200 (4) 94.8° 45. When H2O2 oxidized the product is (1) OH− (2) O2 (3) O2- (4) HO2-

Heavy Water and Hard Water 46. Which of the following terms is not correct? (1) Hydrogen molecule is diatomic (2) Hydrogen exists both as H+ and H– in different chemical compounds. (3) Hydrogen is the only species which has no neutrons in the nucleus. (4) Heavy water is unstable because hydrogen is substituted by its isotope deuterium. 47. The reagent commonly used to determine hardness of water titrimetrically is (1) (2) (3) (4)

oxalic acid. sodium thiosulphate. sodium citrate. disodium salt of EDTA.

49. Which of the following groups of ions makes the water hard? Sodium and bicarbonate Magnesium and chloride Potassium and sulphate Ammonium and chloride.

54. Which of the following is not true? (1) D2O freezes at lower temperature than H2O. (2) Reaction between H2 and Cl2 is much faster than D2 and Cl2. (3) Ordinary water gets electrolyzed more rapidly than D2O. (4) Bond dissociation energy of D2 is greater than H2. 55. Which of the following is correct about heavy water? (1) Water at 4°C having maximum density is known as heavy water. (2) It is heavier than water (H2O). (3) It is formed by the combination of heavier isotope of hydrogen with oxygen. (4) None of these. 56. When sulphur trioxide is heated with heavy water the products are (1) (2) (3) (4)

Deutero-sulphuric acid. Deuterium sulphuric acid. Deuterium and sulphuric acid. None of the above.

58. Hard water when passed through ion exchange resin containing R′COOH groups, becomes free from (1) Cl– (2) SO24 (3) H3O+ (4) Ca2+

Level II Dihydrogen 1. Which pair does not show hydrogen isotopes?

50. D2O is used in (1) motor vehicles. (2) nuclear reactor. (3) medicine. (4) insecticide. 51. The process used for the removal of hardness of water is (1) Calgon. (2) Baeyer. (3) Serpeck. (4) Hoope. 52. When zeolite (hydrated sodium aluminum silicate) is treated with hard water the sodium ions are exchanged with (1) H ions. (2) Ca ions. – (3) SO24 ions. (4) OH ions.

Chapter 15_Hydrogen.indd 369

(1) C2D2 (2) CaD2 (3) Ca2D2O (4) CD2

(1) Slaked lime (2) Plaster of Paris (3) Epsom (4) Hydrolith

(1) Na2[Na4(PO3)6] (2) Na4[Na2(PO3)6] (3) Na4[Na4(PO4)5] (4) Na4[Na2(PO4)6]

+

53. What is formed when calcium carbide reacts with heavy water?

57. Which one the following removes temporary hardness of water?

48. The structure of Calgon which is used as a water softener is

(1) (2) (3) (4)

369

2+

(1) (2) (3) (4)

Ortho and para hydrogen Protium and deuterium Deuterium and tritium Tritium and protium

2. When same amount of zinc is treated separately with excess of sulphuric acid and excess of sodium hydroxide solution the ratio of volumes of hydrogen evolved is (1) 1:1 (2) 1:2 (3) 2:1 (4) 9:4 3. Which of the following species is diamagnetic in nature? (1) H 2- (2) H +2 (3) H2 (4) He+2

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OBJECTIVE CHEMISTRY FOR NEET

4. Which of the following statements in relation to the hydrogen atom is correct? (1) 3s, 3p and 3d orbitals all have the same energy. (2) 3s and 3p orbitals are of lower energy than 3d orbital. (3) 3p orbital is lower in energy than 3d orbital. (4) 3s orbital is lower in energy than 3p orbital. 5. At room temperature, ordinary hydrogen is a mixture of (1) (2) (3) (4)

25% orthohydrogen and 75% parahydrogen 35% orthohydrogen and 65% parahydrogen 75% orthohydrogen and 25% parahydrogen 65% orthohydrogen and 35% parahydrogen

6. Which of the following statements is incorrect? (1) Dihydrogen gas is acidic to litmus. (2) Atomic hydrogen exists for less than half a second. (3) H2 is a highly combustible gas and burns in air or dioxygen with a pale blue flame. (4) It is not a supporter of combustion. 7. Which one of the following properties shows that hydrogen resembles alkali metals? (1) It shows metallic character like alkali metals. (2) It is diatomic like alkali metals. (3) Its ionization energy is of the same order as that of alkali metals. (4) When hydrogen halides and alkali metal halides are electrolyzed, hydrogen and alkali metals are liberated at the cathode. 8. A deuterium atom (1) has the same atomic mass as the hydrogen atom. (2) has the same electronic configuration as the hydrogen atom. (3) has the same composition of the nucleus as the hydrogen atom. (4) contains one proton more than a hydrogen atom. 9. Which of the following statements is most applicable to hydrogen? (1) It can act as a reducing agent only. (2) It can act as an oxidizing agent only. (3) It can act as both as oxidizing and reducing agents. (4) It can act neither as an oxidizing nor as a reducing agent. 10. Hydrogen combines with other elements by (1) (2) (3) (4)

losing an electron. gaining an electron. sharing an electron. losing, gaining and sharing of an electron.

11. Hydrogen gas will not reduce (1) (2) (3) (4)

heated cupric oxide. heated ferric oxide. heated stannic oxide. heated aluminum oxide.

Chapter 15_Hydrogen.indd 370

12. Which one of the following reactions represents watergas shift reaction? K/Iron chromate   → CO2 + H 2 (1) CO(g) + H 2O(g) 673 K  → CO(g) + H 2 (2) C(s) + H 2O(g) 1270 K/Ni ¾¾¾ ® CO(g) + 3H 2(g) (3) CH 4(g) + H 2O(g) ¾1270 (4) None of these

13. Hydrogen is evolved by the action of cold dil. HNO3 on (1) Fe (2) Mn (3) Cu (4) Al

Hydride 14. The hydride ion, H– is a stronger base than the hydroxide ion, OH–. Which one of the following reactions will occur if sodium hydride (NaH) is dissolved in water? (1) (2) (3) (4)

H - (aq ) + H 2O(l) → H 3O - (aq ) H - (aq ) + H 2O(l) → OH - (aq ) + H 2(g ) H - (aq ) + H 2O(l) → OH - (aq ) + 2H+(aq ) + 2e H– (aq) + H2O (l) → No reaction

15. The H– ion can be formed in ordinary chemical reaction under proper conditions, but the H+ ion cannot. The best explanation for this difference is due to (1) (2) (3) (4)

the radius of the H nucleus. the electronegativity of H atom. the ionization energy of the H atom. the bond dissociation energy of H–H bond.

16. Saline hydrides react explosively with water, such fires can be extinguished by (1) water. (2) carbon dioxide. (3) sand. (4) None of these. 17. Saline hydrides are formed by (1) alkali metals. (2) alkaline earth metals. (3) both (1) and (2). (4) none of these. 18. Which of the properties of interstitial hydrides is correct? (1) They generally form non-stoichiometric species. (2) The hydrogen dissolved in titanium improves its mechanical properties. (3) They give rise to metals fit for fabrication. (4) On thermal decomposition, they afford a source of pure hydrogen. 19. Which of the following statements is incorrect about metallic hydrides? (1) The hydrides are more dense than the parent metal. (2) In these hydrides, the law of constant composition does not hold good. (3) All the hydrides do not possess the lattice of the parent metal. (4) They may be regarded as solid solutions.

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Hydrogen 20. Among the hydrides given below which is reasonably electron precise hydride?

29. The atom of oxygen lost by H2O2 molecule during oxidation reaction is that which is linked through

(1) NH3 (2) SiH4 (3) NaH (4) H2S

(1) an electrovalent bond. (2) a covalent bond. (3) a coordinate bond. (4) a hydrogen bond. 30. One of the following is an incorrect statement. Point out the incorrect one.

Water 21. In the liquid state, water exists as an associated liquid compound where each O atom can form __________ H-bonds.

(1) H2O2 decomposes rapidly in presence of MnO2. (2) Ice at its melting point is lighter than water because ice crystals have hollow hexagonal arrangement of H2O molecules. (3) D2O will have maximum density at 11.5°C. (4) Water gas contains greater proportion of CO than that of H2.

(1) Four (2) Three (3) Two (4) One 22. The critical temperature of water is higher than that of O2, because H2O molecule has (1) fewer electrons than oxygen. (2) two covalent bonds. (3) V-shaped. (4) dipole moment.

31. Which of the following statements regarding hydrogen peroxide is/are incorrect? (1) The two hydroxyl groups in hydrogen peroxide lie in the different plane. (2) Aqueous solution of H2O2 turns blue litmus red. (3) When H2O2 behaves as a reducing agent, the O   O bond in its molecules is not broken down. (4) Aqueous solution of H2O2 is stored in plastic bottles and some urea, phosphoric acid or glycerol is added to that solution.

Hydrogen Peroxide 23. HCl is added to following oxides. Which one would give H2O2? (1) MnO2 (2) PbO2 (3) BaO (4) None of these 24. Which of the following is false about H2O2? (1) (2) (3) (4)

Act as both oxidizing and reducing agent. Two OH bonds lie in the same plane. Pale blue liquid. Can be oxidized by ozone.

25. In which of the following reactions, H2O2 acts as a reducing agent? (1) PbO2(s) + H2O2 (aq) → PbO (s) + H2O (l) + O2 (g) (2) Na2SO3 (aq) + H2O2 (aq) → Na2SO4 (aq) + H2O (l) (3) 2KI (aq) + H2O2 (aq) → 2KOH (aq) + I2 (s) (4) KNO2 (aq) + H2O2 (aq) → KNO3 (aq) + H2O (l) 26. Blackened oil painting can be restored into original form by the action of (1) Cl2 (2) BaO2 (3) H2O2 (4) MnO2

32. H2O2 is manufactured these days (1) (2) (3) (4)

see-saw structure. square pyramidal structure. linear structure. open book skew structure.

33. Heavy water reacts respectively with CO2, SO3, P2O5 and N2O5 to give the compounds (1) D2CO3, D2SO4, D3PO2, DNO2 (2) D2CO3, D2SO4, D3PO4, DNO2 (3) D2CO3, D2SO3, D3PO4, DNO2 (4) D2CO3, D2SO4, D3PO4, DNO3 34. D2O is preferred to H2O, as a moderator, in nuclear reactors because (1) D2O slows down fast neutrons better. (2) D2O has high specific heat. (3) D2O is cheaper. (4) None of these. 35. Which of the following chemicals is not present in clear hard water? (1) MgCO3 (2) MgSO4

28. Which of the following statements is incorrect? (1) H2O2 can act as an oxidizing agent. (2) H2O2 can act as a reducing agent. (3) H2O2 has acidic properties. (4) H2O2 has basic properties.

Chapter 15_Hydrogen.indd 371

by the action of H2O2 on BaO2. by the action of H2SO4 on Na2O2. by electrolysis of 50% H2SO4. by burning hydrogen in excess of oxygen.

Heavy Water and Hard Water

27. Hydrogen peroxide has a (1) (2) (3) (4)

371

(3) Mg (HCO3)2 (4) CaCl2 36. Choose the correct statement.

The reason for use of polyphosphates as water softening agents is, that

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372

OBJECTIVE CHEMISTRY FOR NEET (1) (2) (3) (4)

they form soluble complexes with anionic species. they precipitate out cationic species. they precipitate out anionic species. they form soluble complexes with cationic species.

37. Which one of the following processes will produce hard water? (1) (2) (3) (4)

It is less dense than common water. It is an oxide of deuterium. It has a heavy or bad taste. It has a heavier isotope of oxygen.

39. Which of the following is not correctly matched? (1) Heavy water-Used as moderator in nuclear reactors. (2) Zeolites-Used for removing permanent hardness of water. (3) Syn gas-Mixture of N2 and H2. (4) Hydrolith-On hydrolysis with water produces H2. 40. Heavy water is manufactured by (1) combination of hydrogen and heavier isotope of oxygen. (2) electrolysis of water containing heavy hydrogen dissolved in it. (3) repeated electrolysis of 3% aqueous solution of NaOH. (4) none of the above. 41. Heavy water is used in atomic reactors as (1) coolant. (2) moderator. (3) both coolant and moderator. (4) neither coolant nor moderator. 42. In the laboratory H2O2 is prepared by the action of (1) (2) (3) (4)

cold dilute. H2SO4 on hydrated BaO2. dil. HCl on MnO2. cold H2SO4 on MnO2. aqueous alkali on Na2O2.

43. Clark’s method of softening of water uses (1) (2) (3) (4)

(1) Rain water (2) Distilled water (3) Spring water (4) Demineralized water

Previous Years’ NEET Questions 1. Some statements about heavy water are given below.

Saturation of water with MgCO3. Saturation of water with CaSO4. Addition of Na2SO4 to water. Saturation of water with CaCO3.

38. Select the correct statement for heavy water. (1) (2) (3) (4)

44. Which of the following is an example of hard water?

calcium hydroxide. sodium bicarbonate. sodium chlorate. sodium hexametaphosphate.

Chapter 15_Hydrogen.indd 372

(I) Heavy water is used as a moderator in nuclear reactions (II) Heavy water is more associated than ordinary water (III) Heavy water is more effective solvent than ordinary water

Which of the above statements are correct? (1) (I), (II) and (III) (2) (II) and (III) (3) (I) and (III) (4) (I) and (II) (AIPMT MAINS 2010)

2. Role of hydrogen peroxide in the following reactions is respectively. (I) H2O2 + O3 → H2O + 2O2 (II) H2O2 + Ag2O → 2Ag + H2O + O2 (1) (2) (3) (4)

Oxidizing in (I) and reducing in (II) Reducing in (I) and oxidizing in (II) Reducing in (I) and (II) Oxidizing in (I) and (II) (AIPMT 2014)

3. The variation in the boiling points of the hydrogen halides is in the order HF > HI > HBr > HCl. What explains the highest boiling point of hydrogen fluoride? (1) The bond energy of HF molecule is greater than in other hydrogen halides. (2) The effect of nuclear shielding is much reduced is fluorine which polarizes the HF molecule. (3) The electronegativity of fluorine is much higher than for other elements in the group. (4) There is a strong hydrogen bonding between HF molecules. (RE AIPMT 2015) 4. Which of the following statements about hydrogen is incorrect? (1) Hydrogen has three isotopes of which tritium are the most common. (2) Hydrogen never acts as cation in ionic salts. (3) Hydronium ion, H3O+ exists freely in solution. (4) Dihydrogen does not act as a reducing agent. (NEET-I 2016)

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Hydrogen

373

Answer Key Level I  1. (3)

 2. (4)

 3. (1)

 4. (2)

 5. (3)

 6. (3)

 7. (1)

 8. (4)

 9. (2)

10.  (3)

11.  (1)

12.  (2)

13.  (3)

14.  (3)

15.  (1)

16.  (3)

17.  (1)

18.  (3)

19.  (4)

20.  (4)

21.  (4)

22.  (1)

23.  (1)

24.  (3)

25.  (4)

26.  (3)

27.  (3)

28.  (2)

29.  (3)

30.  (4)

31.  (1)

32.  (3)

33.  (2)

34.  (4)

35.  (2)

36.  (1)

37.  (3)

38.  (2)

39.  (1)

40.  (1)

41.  (3)

42.  (4)

43.  (1)

44.  (4)

45.  (2)

46.  (4)

47.  (4)

48.  (1)

49.  (2)

50.  (2)

51.  (1)

52.  (2)

53.  (1)

54.  (1)

55.  (3)

56.  (1)

57.  (1)

58.  (4)

 1. (1)

 2. (1)

 3. (3)

 4. (1)

 5. (3)

 6. (1)

 7. (4)

 8. (2)

 9. (3)

10.  (4)

11.  (4)

12.  (1)

13.  (2)

14.  (2)

15.  (3)

16.  (3)

17.  (3)

18.  (1)

19.  (1)

20.  (2)

21.  (3)

22.  (4)

23.  (4)

24.  (2)

25.  (1)

26.  (3)

27.  (4)

28.  (4)

29.  (3)

30.  (4)

31.  (2)

32.  (3)

33.  (4)

34.  (4)

35.  (1)

36.  (4)

37.  (2)

38.  (2)

39.  (3)

40.  (3)

41.  (3)

42.  (1)

43.  (1)

44.  (3)

Level II

Previous Years’ NEET Questions  1. (4)

 2. (3)

 3. (4)

 4.  (1), (4)

Hints and Explanations Level I

11. (1) H + e - (1s1 ) → H - (1s 2or [He])

+1 -1

1. (3) 2Na + H 2 → 2 Na H

-3 +1

   N 2 + 3H 2 → 2 N H 3

K 2. (4) 4H 2O(g) + 3Fe(s) 1000  → Fe3O4 + 4H 2

4. (2) As, 22.4 L of H2 at STP = 1 mol of H2

Therefore, 0.224 L of H2 at STP = 1/22.4 × 0.224 = 0.01 mol

5. (3) Hydrogen is in the first group of periodic table due to its much resemblance with alkali metals. 6. (3) In H2O, hydrogen has +1 oxidation state. In CaH2, hydrogen has –1 oxidation state.

F + e - ([He]2s 2 2p 5 ) → F - ([He]2s 2 2p 6or [Ne]) 12. (2)

catalyst CO+H → CO2 + 2H 2 2O   2 + H

Water gas Steam

13. (3) Occlusion is the phenomenon of adsorption of hydrogen by metal. 15. (1) Mg + 2H2O → Mg(OH)2 + H2 ↑ 16. (3) Hydrogen is the lightest gas. It is insoluble in water. 17. (1) Zinc has no action on cold water. 18. (3) H(g) → H+(g) + e –

8. (4) Chlorine has lone pair which it can donate to form coordinate bond while hydrogen cannot.

20. (4) Metal hydride + H2O → Metal hydroxide + H2

9. (2) Mn + 2HNO3 (dil.) → Mn(NO3)2 + H2

21. (4) All metal hydrides are ionic in nature.

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OBJECTIVE CHEMISTRY FOR NEET

22. (1) Two ice cubes stick to each other due to hydrogen bonding.

43. (1) As H2O2 is losing electrons so it is acting as reducing agent.

23. (1) Polarity of bond depends on difference in electronegativity of the two concerned atoms. H2O is more polar than H2S because oxygen (in O   H) is more electronegative than sulphur (in S   H).

 → H 2O + O 2 ↑ 45. (2) H 2O2 + [O] Oxidation

24. (3) Sea water is purified by reverse osmosis. 25. (4) The hybridization in water is sp3 and bond angle 104.5°. 26. (3) Potash alum is used for purifying water. 27. (3) The high boiling point of water is exceptionally high due to hydrogen bonding. 28. (2) Due to high dielectric constant, water acts as a good solvent, therefore, it is also called a universal solvent. 29. (3) By electrolysis of 50% ice cold H2SO4. H 2SO4 → 2H + + 2 SO24

At anode:

2HSO24- → H 2S 2O8 + 2e H 2S 2O8 + H 2O → H 2SO5 + H 2SO4 H 2SO5 + H 2O → H 2SO4 + H 2O2

30. (4) H2O2 cannot oxidize O3. But, O3 oxidizes H2O2 as



H2O2 + O3 → H2O + 2O2

31. (1) % hydrogen in H2O2 = 2/34 × 100 = 5.88% 32. (3) H2S is oxidized to S by H2O2. 33. (2) Strength of 10 V H 2O2 =

68 × 10 × 100 = 3.035% 22400

34. (4) It is weakly acidic in nature and pure hydrogen peroxide turns blue litmus red. (Ka = 1.57 × 10−12 at 293 K). It ionizes in two steps:

46. (4) Heavy water is stable. 47. (4) EDTA forms calcium and magnesium complex with Ca2+ and Mg2+ ions present in hard water. 48. (1) The complex salt of metaphosphoric acid sodium hexametaphosphate (NaPO3)6, is known as calgon. It is represented as Na2[Na4(PO3)6]. 49. (2) Temporary hardness is due to presence of bicarbonates of calcium and magnesium and permanent hardness is due to the sulphates and chlorides of both of calcium and magnesium. 50. (2) D2O is used in nuclear reactors as moderator. 51. (1) Calgon process is used to remove permanent hardness of water. 52. (2) Na zeolite + CaCl2 → Ca zeolite + 2NaCl 53. (1) CaC2 + 2D2O → C2D2 + Ca(OD)2 54. (1) D2O actually has higher freezing point (3.8°C) than water H2O (0°C). 55. (3) Heavy water is formed by the combination of heavier isotope (1H2 or D) with oxygen. 2D2 + O2 →



56. (1) SO3 + D2O →

HO2  H + + O22-



35. (2) Structure of H2O2 is non-planar.

D2SO4

(Deutero sulphuric acid )

57. (1) This method is known as Clark’s process. In this method temporary hardness is removed by adding lime water or milk of lime. Ca(OH)2 + Ca(HCO3 )2 → 2CaCO3 ↓ + 2H 2O

H 2O 2  H + + O 2



2D2O

Heavy water

ppt.

58. (4) An ion exchange resin containing R   COOH group exchange cations like Ca2+, Mg2+, Na+, Fe2+ with H+ when hard water is passed through it. This resin is called cation exchange resin.

 BaSO4 + H 2O2 36. (1) H 2SO4 + BaO2 → 39. (1) Glycerol, phosphoric acid or acetanilide is added to H2O2 to check its decomposition. 40. (1)

H 2O 2

Weak acid

→ H 2 O + [O ]

41. (3) H2O2 is unstable liquid and decomposes into water and oxygen either on standing or on heating. 42. (4) H2O2 show all these properties.

Chapter 15_Hydrogen.indd 374

Level II 1. (1) Ortho and para hydrogens are two forms of hydrogen (1H1 ) which differ only in the direction of spin of proton. 2. (1) Zn + H2SO4 → ZnSO4 + H2 Zn + 2NaOH → Na2ZnO2 + H2







Therefore, ratio of volumes of H2 evolved is 1:1.

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Hydrogen 8. (2) Electronic configuration of hydrogen and deuterium is the same, that is, 1s1. 9. (3) Hydrogen can act both as oxidizing and reducing agent. It acts as oxidizing agent when reacts with metal and form metal hydrides which are electrovalent compounds.

2Na + H2 → 2Na+H0





Ca + H2 → CaH2







Hydrogen also acts as reducing agent when its reacts with oxides. PbO + H2 → Pb + H2O







    Fe3O4 + H2 → 3Fe + 4H2O

CuO + H2 → Cu + H2O

25. (1) In the following reaction H2O2 acts as a reducing agent.



PbO2 (s) + H2O2 (aq) → PbO (s) + H2O (l) + O2 (g)

26. (3) The key reactions are:

PbO + H 2S → PbS + H 2O



PbS + 4H 2O 2 → PbSO4 + 4H 2O



(Black)

( White)

When blackened statues are treated with H2O2, the PbS is oxidized to PbSO4, which is colorless (white).

:

29. (3) The reaction involved is H 2O2 → H 2 O : → O: (Oxidation)



Coordinate bond

10. (4) Hydrogen can combine with other elements by losing, gaining and sharing of electrons

(i) Losing of electrons: H2 + F2 → 2H+F–





30. (4) H2 = 50 volumes, CO = 40 volumes, N2 and CO2 = 5 volumes etc.

(ii) Gaining of electrons: 2Na + H2 → 2Na+H– (iii) Sharing of electrons: N2 + 3H2 C + 2H2

       

Fe + MO 500°C High pressure 1200°C

2NH3

CH4

N H H H

14. (2) H - (aq) + H 2O(l) → OH - (aq) + H 2(g) Base 1



Acid 1

Base 2

Acid 2

In this reaction H− acts as Brönsted base as it accepts one proton (H+) from H2O and form H2.

16. (3) Fire due to action of water on saline hydrides cannot be extinguished with water or CO2. These hydrides can reduce CO2 at high temperature to produce O2. 17. (3) At high temperatures, the metals of Group 1 (alkali metals) and the heavier Group 2 metals (alkaline earth metals) Ca, Sr and Ba form ionic hydrides such as NaH and CaH2. These compounds are solids with high melting points, non-conducting in solid state and are called ionic (salt-like) hydrides. 18. (1) The interstitial hydrides are generally non-stoichiometric and their compositions vary with temperature and pressure. For example, TiH1.73, CeH2.7, LaH2.8, etc. 22. (4) Critical temperature of water is more than O2 due to its dipole moment (Dipole moment of water = 1.84 D; Dipole moment of O2 = zero). 23. (4) MnO2, PbO2 and BaO will not give H2O2 with HCl. MnO2 and PbO2 will give Cl2 and BaO will react with HCl to give BaCl2 and water. 24. (2) The value of dipole moment of H2O2 is 2.1 D, which suggest the structure of H2O2 cannot be planar. An openbook structure is suggested for H2O2 in which O   H bonds lie in different plane.

Chapter 15_Hydrogen.indd 375

: :



375

32. (3) In industrial preparation of H2O2 a 50% solution of H2SO4 is electrolyzed in a cell. As a result, peroxodisulphuric acid is formed at the anode and hydrogen is evolved at cathode. H 2SO4 → HSO4- + H +







At the anode: 2HSO4- →

+ 2e -

H 2S 2O 8 Peroxydisulphuric acid



At the cathode: 2H + 2e - → H 2



Peroxodisulphuric acid is taken out from the cell and is then hydrolyzed with water to give hydrogen peroxide as follows.



+

H 2S 2O8 + 2H 2O → 2H 2SO 4 + H 2O2

34. (4) H2O absorbs neutrons more than D2O and this decreases the number of neutrons for the fission process. 35. (1) MgCO3 is insoluble in water. 36. (4) Polyphosphates (e.g., sodium hexametaphosphate) form soluble complexes with cations such as Ca2+, Mg2+ and so they are used as water softening agents.

Na 2[Na 4(PO3 )6 ]

+ 2Ca 2+ → Na 2[Ca 2(PO3 )6 ] + 2Na +

Calgon Hardness Sodium hexa metaphosphate

Soluble

37. (2) Permanent hardness of water is due to chlorides and sulphates of calcium and magnesium,that is, CaCl2, CaSO4, MgCl2 and MgSO4. 38. (2) Heavy water contains heavy hydrogen ( 21 H), formula of heavy water is 21 H 2O. 39. (3) Syn gas is an equimolar mixture of CO and H2. 41. (3) Heavy mater is used as a coolant and moderator.

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376

OBJECTIVE CHEMISTRY FOR NEET

 BaSO4 + H 2O2 + 8H 2O 42. (1) BaO2 ⋅ 8H 2O + H 2SO4 → 43. (1) Ca(HCO3)2 + Ca(OH)2 (calculate amount) → 2CaCO3 ↓ + 2H2O Mg(HCO3)2 + 2Ca(OH)2 (calculate amount) → 2CaCO3 ↓+ Mg(OH)2 ↓+ 2H2O

Previous Years’ NEET Questions 1. (4) Heavy water is known as moderator as it is used for slowing down the speed of neutrons. Boiling point of heavy water is greater (374 K) than that of ordinary water (373 K) due to stronger bonding in case of D2O. 2. (3)

-1



There is decrease in oxidation number of oxygen from −1 to −2 in both cases, thus, hydrogen peroxide acts as reducing agent.

3. (4) Hydrogen bonding is strongest in case of HF followed by HI, HBr and HCl.

( )

4. (1), (4) Hydrogen has three isotopes protium 11 H , deuterium 21 H and tritium 31 H . The most common form is protium. Tritium, which is radioactive, is very rare to occur. Dihydrogen reduces some metal ions in aqueous solution and oxides of metals into their corresponding metals.

( )

( )

-2

H 2 O 2 + O3 → H 2 O + 2O 2 -1

-2

H 2 O 2 + Ag 2O ® 2Ag + H 2 O + O2

Chapter 15_Hydrogen.indd 376

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16

The s-Block Elements

Chapter at a Glance 1. The general electronic configuration of s-block elements is ns1-2. In this block the last electron enters the s-subshell. They include Group I (alkali metals) and Group II (alkaline earth metals) elements. 2. Diagonal Relationship: The similarity in properties between lithium (first member in Group 1) and magnesium (second element in Group 2); between beryllium (first element in Group 2) and aluminum (second element in Group 3) and between boron (first element in Group 3) and silicon (second element in Group 4) is called a diagonal relationship. It arises because of the effects of both size and charge.

Group I (Alkali Metals) 1. General Trends in Physical Properties of Elements On reaction with water, they form hydroxides which are strongly alkaline in nature. (a) Appearance: Li, Na, K and Rb are silvery, but Cs has a golden yellow appearance. (b)  Density: They have low densities due to their large sizes. The density increases as we move down the group. However, potassium has lower density than sodium. (c)  Boiling and melting points: They have low melting and boiling points because weak metallic bonding exists due to presence of single valence electron. (d)  Flame color: The elements show characteristic colors when heated in a flame. The color actually arises from electronic transition in short-lived species which are formed momentarily in the flame. Li (670.8 nm) → crimson, Na (589.2 nm) → yellow, K (766.5 nm) → Lilac, Rb (780.0 nm) → red violet, Cs (455.5 nm) → blue. When the excited electron drops back to its original energy level it gives out the extra energy it obtained corresponding to their wavelength. (e) Atomic and ionic radii (i) Have the largest size in the horizontal period they are present in. (ii)  Formation of ion by removal of electron reduces the size considerably due to removal of one orbital shell. (iii)  Ionic radii are smaller than the atomic radii but increases down the group from Li+ to Fr+ due to addition of extra shell of electrons. (f )  Ionization enthalpy: The first ionization enthalpy is low because the atoms are very large and the outer electrons are only held weakly by the nucleus. On descending the group from Li to Cs, the size of the atoms increases: the outermost electrons become less strongly held, so the ionization enthalpy decreases. Second ionization enthalpy is very high because the electron needs to be removed from stable noble gas configuration. (g)  Hydration enthalpy: The salts of alkali metals dissolve in aqueous solutions to form ions and conduct electricity. The order of ionic mobility or conductivity observed is Cs+ > Rb+ > K+ > Na+ > Li+. Li+ being the smallest ion is most heavily hydrated which increases its radius and reduces its mobility and conductivity. In general, hydration energy of alkali metals decreases with increase in ionic radii.

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2. General Trends in Chemical Properties of Elements Reaction

Observation

Comment

With water:

The hydroxides are the strongest bases known.

The reaction becomes increasingly violent on descending the group.

Monoxide is formed by Li and to a small extent by Na.

The increasing stability of the peroxides and superoxides as the size of alkali metal increases is due to stabilization of larger anions by larger cations through lattice energy effects.

M + H2O → MOH + H2 With excess dioxygen Li + O2 → Li2O

Na + O2 → Na2O2

Peroxide formed by Na and to a small extent by Li Superoxides formed by K, Rb, Cs Ionic “salt-like” hydrides

K + O2 → KO2 With hydrogen M + H2 → MH With nitrogen Li + N2 → Li3N

Nitride formed only by Li

With halogens

All the metals form halide.

M + X2 → MX where, X is F, Cl, Br, I

All the metals form amides in presence of impurities or Fe as catalyst.

With ammonia

1 M + NH3 → MNH2 + H2 2 M + (x + y )NH3 → [ M(NH3 )x ]+ + [e( NH3 ) y )]-

The ease of hydride formation decreases from lithium to caesium.

The halides are ionic in nature except lithium halides which is due to small size of lithium ion and its tendency to distort the electron cloud around the anion (polarization). The main species present in the solution are solvated metal ions and solvated electrons. The solution is paramagnetic in nature.

1 M+ (aq ) + e - (aq) + NH3 → MNH2 + H2 (g ) 2 As reducing agent M(s) → M(g)  → M (g) + e Sublimation

Ionization

M+ (g) + H2 O Hydration  → M+ (aq)

+

-

Alkali metals are strong reducing agents. The reducing character increases down the group from Li to Cs due to decreasing ionization enthalpies.

Lithium has the highest negative value for electrode potential in the group, and is the strongest reducing agent.

3. Anomalous Properties of Lithium Lithium compounds show closer similarities with Group 2 elements (particularly magnesium) than they show towards their own group. (a) Differences with alkali metals (i) Higher melting and boiling points. (ii) Forms nitride with nitrogen and ionic carbide with carbon.

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(iii) Forms only normal oxide with oxygen, does not form peroxide or superoxide. (iv)  Lithium nitrate when heated gives lithium oxide, Li2O, whereas other alkali metal nitrates decomposes to give the corresponding nitrite. 4 LiNO3 → 2 Li 2 O + 4 NO2 + O2 2 NaNO3 → 2 NaNO2 + O2



(v) Alkyls and halides of lithium show more covalent character and are hence soluble in organic solvents. (vi) LiOH is sparingly soluble and less basic than other hydroxides of the group. (b) Similarities with magnesium (i)  Both react with oxygen to form oxides but do not give any peroxides or superoxide even in presence of excess oxygen. (ii)  Both react directly with nitrogen forming nitrides (LiN3 and Mg3N2). (iii)  Carbonates decompose on heating, do not form bicarbonates. (iv)  Halides are covalent in character and soluble in organic solvents. (v) Both have tendency to form complexes with ammonia and water. 4. Compounds of Sodium Compound Sodium carbonate (Washing soda), Na2CO3·10H2O

Preparation It is prepared by the Solvay (ammonia–soda) process. Following reactions take place during the process. 2NH3 + H2 O + CO2 → ( NH4 )2 CO3 ( NH4 )2 CO3 + H2 O + CO → 2NH4 HCO3 NaCl + NH4 HCO3 → NaHCO3 + NH4 Cl

Properties It is odorless white powder that absorbs moisture from air to form colorless and transparent decahydrate salt. The anhydrous salt obtained on heating is called soda ash and forms strongly alkaline solution in water. CO32 - + H2 O → HCO3- + OH-

°C 2NaHCO3 ¾150 ¾¾ ® 2Na 2 CO3 + H2 O + CO2

Sodium hydroxide (Caustic soda), NaOH

It is produced by electrolysis of an aqueous solution of NaCl (brine) using either a diaphragm or mercury cathode (Castner–Kellner cell) cell. At the anode: 2Cl - → Cl 2 + 2e At the cathode: Na + e - ® Na + Overall cell reaction: Na + H2 O ® 2 NaOH +H2

Sodium hydrogen carbonate (Baking soda), NaHCO3

It is prepared by bubbling carbon dioxide through It causes the cakes and bread to rise because on a solution of sodium carbonate, wherein it heating it decomposes to give bubbles of carbon separates out as white crystalline solid. dioxide. Na 2 CO3 + H2 O + CO2 ® 2NaHCO3

It is white, translucent solid. It is readily soluble in water to give a strong alkaline solution. The sodium hydroxide solution at the surface reacts with the CO2 in the atmosphere to form Na2CO3.

Gentle heat

2NaHCO3  → Na 2 CO3 + H2 O + CO2

5. Biological Significance of Na and K: The different ratio of Na+ to K+ inside and outside cells produces an electrical potential across the cell membrane, which is essential for the functioning of nerve and muscle cells. K+ ions inside the cell are essential for the metabolism of glucose, the synthesis of proteins, and the activation of some enzymes.

Group II (Alkaline Earth Metals)

1. General Trends in Physical Properties of Elements They are so called because their oxides and hydroxides are alkaline in nature and these are found in earth’s crust. (a) Appearance: They are silvery white in color. (b)  Density: Group 2 metals are harder, have higher cohesive energy because they have two electrons which participate in metallic bonding.

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(c)  Melting and boiling points: They have much higher melting points and boiling points than Group 1 elements but the metals are relatively soft. (d)  Flame test: They impart characteristic color to the flame in which they are heated. Calcium-brick red, Strontium-­ crimson, Barium-apple green. (e)  Atomic and ionic radii (i)  They are smaller than the corresponding Group 1 elements due to the extra charge on the nucleus of these elements draws the orbital electrons inside atoms. (ii)  Similarly, the ions are large, but smaller than those of Group 1, especially because the removal of two orbital electrons increases the effective nuclear charge even further. (f )  Ionization enthalpies: The total energy required to produce gaseous divalent ions for Group 2 elements (first ionization energy + second ionization energy) is over four times greater than the energy required to produce M+ from Group 1 metals. (g)  Hydration enthalpy: The hydration enthalpies of the Group 2 ions are four or five times greater than for Group 1 ions. This is largely due to their smaller size and increased charge. The hydration enthalpy decreases down the group as the size of the ions increases. 2. General Trends in Chemical Properties of Elements Reactions With water: M + 2H2O → M(OH)2 + H2 With air: 2M + O2 → 2MO Ba + O2 → BaO2 3M + N2 → M3N2 With hydrogen: M + H2 → MH2

With base: Be + NaOH → Na2[Be(OH)4] + H2 With acid: M + 2HCl → MCl2 + H2 With halogens: M + X2 → MX2 where X is F, Cl, Br, I

Observations Comments Be probably reacts with steam, Mg with hot water, and Ca, Sr and Ba react rapidly with cold water. Normal oxide formed by all group members with excess dioxygen. Ba also forms the peroxide. All form nitrides at high temperatures. Ionic “salt-like” hydrides formed at All the hydrides are hydrolyzed by high temperatures by Ca, Sr and Ba. water and dilute acids with the evolution of hydrogen. CaH2, SrH2 and BaH2 are ionic, and contain the hydride ion H-. Beryllium and magnesium hydrides are covalent and polymeric. Be is amphoteric. Mg, Ca, Sr and Ba do not react with NaOH, and are purely basic. All the metals react with acids, liberating The basic properties increase on hydrogen. descending the group. All the metals form halides.

With ammonia: M + (x + y)NH3 → [M(NH3)x ]2+ +         2[e(NH3)y ]3M + 2NH3 → 2M(NH2)2

The metals all dissolve in liquid ammonia and dilute solutions are bright blue in color. All the metals form amides at high temperatures.

On evaporation of ammonia Group 2 metals give hexammoniates of the metals. These slowly decompose to give amides. M(NH3 )6 → M(NH2 )2 + 4NH3 + H2

With carbon: M + 2C→MC2

Group 2 metals typically form ionic carbides of formula MC2.

The metals Mg, Ca, Sr and Ba form carbides of formula MC2,

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THE s-BLOCK ELEMENTS

Reducing nature: These elements show a strong tendency to lose two valence shell electrons to form M2+ ions, except for beryllium. M→M2+ + 2e-

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They show strong reducing properties The reducing property increases as which are however, weaker as compared we move down the group from Be to to alkali metals. Ba because the ionization enthalpies decrease and electrode potential values become more negative with increasing atomic number.

3. Salts of Oxoacids (a)  Carbonates: These are all rather insoluble in water and the solubility products decrease with increasing size of M2+ ion. The carbonates are all ionic, but BeCO3 is unusual because it contains the hydrated ion [Be(H2O)4]2+ rather than Be2+. (b) Sulphates: The solubility of the sulphates in water decreases down the group. The order is Be > Mg >> Ca > Sr > Ba. (c)  Nitrates: These can all be prepared in solution and can be crystallized as hydrated salts by the reaction of HNO3 with carbonates, oxides or hydroxides. 4. Anomalous Properties of Beryllium (a) Differences between beryllium and other alkaline earth metals (i)  Form covalent compounds because of its small size and high electronegativity. (ii)  Can form a maximum of four conventional electron pair bonds. The other elements can have more than eight outer electrons, and may attain a coordination number of 6 using one s, three p and two d orbitals for bonding. (iii)  Beryllium salts are acidic when dissolved in pure water because the hydrated ion hydrolyzes, producing H3O+. The other Group 2 salts do not interact so strongly with water, and do not hydrolyze appreciably. (b) Similarities with aluminum (i)  Beryllium forms a protective coating of oxide on its surface like aluminum. (ii)  The chlorides of beryllium and aluminum have bridged chloride ions in their structures. (iii)  Beryllium hydroxide forms beryllate ion in excess alkali just like aluminate ion formed by aluminum hydroxide. (iv)  Both aluminum and beryllium ions have a tendency to form complexes. 5. Compounds of Calcium Compound Calcium oxide or Quick lime, CaO

Preparation It is prepared by heating CaCO3 in lime kilns at temperatures between 1070 K and 1270 K. CaCO3  CaO + CO2

Properties It is a high-melting white solid. It is a basic oxide, so it reacts with acids and acidic oxides at high temperatures to form salts. CaO + 2HCl ® CaCl2 + H2 O CaO + SO2 ® CaSO3 It absorbs water to form calcium hydroxide. The process is called slaking of lime and is important in use of lime as a binding material in construction. CaO + H2 O ® Ca(OH)2

Calcium It is prepared by adding water to quick hydroxide (Slaked lime. lime), Ca(OH)2

Reactions involved: Ca(OH)2 + CO2 → CaCO3 + H2 O CaCO3 + CO2 + H2 O → Ca(HCO3 )2 2Ca(OH)2 + 2Cl 2 → CaCl2 + Ca(OCl )2 + 2H2 O Bleaching powder

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Calcium It is prepared by passing carbon dioxide carbonate, CaCO3 through slaked lime or by the addition of sodium carbonate to calcium chloride. Ca(OH)2 + CO2 ® CaCO3 + H2 O CaCl2 + Na 2 CO3 ® CaCO3 + 2NaCl

Calcium sulphate (Plaster of Paris), CaSO4·1/2 H2O



1200 K

CaCO3 ¾¾¾ ® CaO + CO2

It reacts with dilute acid to liberate carbon dioxide. CaCO3 + 2HCl ® CaCl 2 +H2 O+CO2 CaCO3 + H2 SO4 ® CaSO4 +H2 O+CO2

It is prepared by dehydration of gypsum. 1 C CaSO4 × 2H2 O ¾150 ¾¾ ® CaSO4 × H2 O 2 Gypsum 

Plaster of Paris

1100 C

200 C

When calcium carbonate heated to 1200 K, it decomposes to evolve carbon dioxide.

® CaSO4 ¾¾¾® CaO + SO3 ¾¾¾

1 When powdered plaster of Paris, CaSO4 × H2 O , is 2 mixed with correct amount of water it sets into a solid mass of CaSO4 × 2H2O.

Anhydride

Portland cement is made from finely ground limestone and finely divided clay with composition: Lime-60–69%, Silica (SiO2) - 17–25%, Alumina (Al2O3) 3–8%, Iron oxide (Fe2O3) - 2–4%, Magnesium oxide (MgO)- 1–5%, Sulphur trioxide (SO3)- 1–3%, Alkali oxides (Na2O + K2O)- 0.3–1.5%.

Cement

When cement is mixed with water to a plastic mass, called “cement paste”, hydrating reaction begins, resulting in the formation of gel and crystalline products. The process of solidification comprises of setting, and then hardening.

6. Biological Significance of Mg and Ca (a)  Mg2+ ions form a complex with ATP, and are constituents of enzymes for reactions involving ATP and energy release. They are also essential for the transmission of impulses along nerve fibers. Mg2+ is an important constituent of chlorophyll in the green plants. (b)  Ca2+ is important in bones and teeth as apatite Ca3(PO4)2, and the enamel on teeth as fluoroapatite [3(Ca3(PO4)2) · CaF2]. Ca2+ ions are important in blood clotting, and are required to trigger the contraction of muscles and to maintain the regular beating of the heart. ∆ CaCO3 → 

Solved Examples

(X )

(2) Na2CO3 (4) Ca(HCO3)2

(1) The reactions can be represented as X

D

CO2 + Residue

H2O

D ∆ CaCO3 → 

(X )

CaO

Quick lime (Residue)

Y

(Y) Slaked lime

Ca(OH)2 + CO2 + H 2O → CaHCO3 (Y)

H2O Excess CO2

+ CO2

CaO+H 2O → Ca(OH)2

Z

(Z)

∆ CaHCO3 →  CaCO 3 + H 2O + CO2

(Z)

(X)

2. Which of the following has highest hydration energy? (1) MgCl2 (3) BaCl2

Solution

+ CO2

CaO+H 2O → Ca(OH)2

1. A solid compound ‘X’ on heating gives CO2 gas and a residue. The residue mixed with water forms ‘Y’. On passing an excess of CO2 through ‘Y’ in water, a clear solution, ‘Z’, is obtained. On boiling ‘Z’, compound ‘X’ is reformed. The compound ‘X’ is (1) CaCO3 (3) K2CO3

CaO

Quick lime (Residue)

(2) CaCl2 (4) SrCl2

Solution (1) The hydration energy of M2+ ion decreases with increase in ionic radii and in group ionic radius increases as go down the group. Mg2+ have lowest ionic radii hence higher hydration energy.

(Y) Slaked lime

Ca(OH)2 + CO2 + H 2O → CaHCO3 (Y)

(Z)

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THE s-BLOCK ELEMENTS 3. Several alkali metals and alkaline earth metals when dissolved in NH 3 produce bright blue solution. On addition of more metal to this solution which of the following characteristics get changed? (1) Electrical conductivity (2) Color (3) Magnetic behavior (4) All of these Solution (4) On addition of more metal to this solution, the solvated electrons gets paired up within the solvent cage, by which electrical conductivity decreases, color gets changed and paramagnetic behavior is changed to diamagnetic. 4. Based on lattice energy and other considerations which one of the following alkali metal chlorides is expected to have the highest melting point? (1) LiCl (2) NaCl (3) KCl (4) RbCl Solution (2) Although lattice energy of LiCl > NaCl but LiCl is covalent in nature and NaCl is ionic. Therefore, the melting point decreases as we move because the lattice energy decreases as the size of alkali metal atom increases (lattice energy is proportional to melting point of alkyl metal halide). 5. Chemically, group 1 elements are very reactive and tarnish rapidly in air due to the formation of (1) oxides. (2) hydroxides. (3) carbonates. (4) all of these. Solution (4) Alkali metals react with oxygen to form oxides which in turn react with moisture in the air to form hydroxides. Hydroxides convert into carbonates after absorbing carbon dioxide. 6. Which of the following statement is incorrect regarding the diagonal relationship between the Al and Be? (1) BeO and Al2O3 are basic in nature. (2) Carbides of both on hydrolysis produce the same gas. (3) Both can form complexes. (4) Both BeCl2 and AlCl3 can exist as dimer. Solution (1) Option (1): The hydroxides of BeO and Al2O3, that is Be(OH)2 and Al(OH)3 are amphoteric in nature.

¾ 2Be(OH )2 + CH 4 Option (2): Be2C + 2H 2O ¾®







Beryllium forms unusual carbide Be2C, which, like Al4C3 yields methane on hydrolysis.

¾ 4 Al(OH )3 + 3CH 4 Al 4C 3 + 12H 2O ¾®

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Option (3): Be forms many complexes like Al. Examples: [Be(OH)4]2-, [BeF4]2-, [Al(OH)4]-, [AlCl4]-.



Option (4): BeCl2 usually forms chains but also exists as the dimer. AlCl3 is dimeric.

7. NaOH is manufactured by electrolysis of brine solution. The products of the reaction are (1) Cl2 and H2 (2) Cl2 and Na-Hg (3) Cl2 and Na (4) Cl2 and O2 Solution (1) Electrolysis of brine solution

At the anode: 2Cl- → Cl2 + 2e−



At the cathode: 2H+ + 2e− → H2



Overall cell reaction: 2Cl− + 2H+ → Cl2 + H2

8. Which of the following can dissolve limestone? (1) CO2 + H2O (2) NaOH + H2O (3) NH3 + H2O (4) None of these Solution (1) Limestone (CaCO3) dissolves in water in the presence of CO2 due to formation of calcium bicarbonate, otherwise insoluble. CaCO3 + H2O + CO2 → Ca(HCO3)2 9. Which of the following alkali metal emits light of longest wavelength in the flame test? (1) Na (2) Li (3) K (4) Cs Solution (3) Lithium emits light of longest wavelength in the flame test because lithium gives crimson red color in the flame. 10. An amphoteric oxide dissolves HCl to form a salt. The salt does not impart any color to the flame and fumes in moist air. The oxide is (1) BaO2 (2) MgO (3) BeO (4) CaO Solution (3) Due to small size and somewhat high ionization enthalpy of Be, BeO is amphoteric and dissolves in both acids and alkalis. The flame test is not shown both by beryllium and magnesium (basic oxide) because due to smaller size of these elements, the valence shell electrons are strongly bound to the nucleus. The energy from the flame of a burner is not sufficient to raise them to the excited state. 11. Electrolysis of molten sodium chloride leads to the formation of (1) Na and H2 (2) Na and O2 (3) H2 and O2 (4) Na and Cl2

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Solution

powder, giving air-slaked like lime, which contains mostly CaCO3 (formed by action of CO2 with CaO) and very little Ca(OH)2. Hence, a mortar made from this mixture will not harden.

(4) Electrolysis of molten sodium chloride leads to the formation of Na and Cl2.

NaCl(aq)  Na + + Cl -



H 2O  H + + OH At cathode:      2H+ + 2e - → H 2



At anode:      2Cl - → Cl 2 + 2e -

12. Which out of the following will be most soluble in water? (1) Be(OH)2 (3) Ba(OH)2

(2) Ca(OH)2 (4) Mg(OH)2

Solution (3) The solubility of these hydroxides increases down the group because their lattice energies and hydration enthalpies decrease down the group but their lattice energies decrease more rapidly than their hydration enthalpies. 13. On dissolving moderate amount of sodium metal in liquid NH3 at low temperature, which one of the following does not occur? (1) Blue colored solution is obtained. (2) Na+ ions are formed in the solution. (3) Liquid NH3 becomes a good conductor of electricity. (4) Liquid ammonia remains diamagnetic. Solution (4) On dissolving moderate amount of sodium metal in liquid NH3 at low temperature, liquid ammonia does not remain diamagnetic. It changes into paramagnetic.

17. Calcium can combine both directly and indirectly with (1) oxygen. (2) nitrogen. (3) hydrogen. (4) carbon. Solution (4) Though calcium can combine directly with all the given elements; it can combine both directly and indirectly with carbon. It can form calcium carbide on heating the metal with carbon in electric furnace or when calcium oxide (quicklime) is heated with carbon. °C Ca + 2C 1100  → CaC 2 °C CaO + 3C 2000  → CaC 2 + CO

18. The composition of electrolyte for the manufacture of calcium by electrolyte method is (1) CaCO3 + CaCl2 (3) CaSO4 + CaCl2 Solution (2) Calcium is prepared by the electrolysis of fused mixture of calcium chloride (six parts) and calcium fluoride (one part). 19. Which of the following hydroxides is the strongest base? (1) Be(OH)2 (3) Ca(OH)2

14. Sodium metal cannot be stored under (1) benzene. (2) kerosene. (3) alcohol. (4) toluene. Solution (3) Sodium metal cannot be stored under alcohol because it reacts with alcohol violently and H2 gas is liberated. 15. Na2CO3·10H2O is prepared by (1) Castner process. (2) Solvay’s process. (3) Ostwald’s process. (4) Haber’s process. Solution (2) Pure Na2CO3 prepared by Solvay’s process absorbs moisture from air and normally exists as a decahydrate salt (Na2CO3·10 H2O) in form of transparent and colorless crystals. 16. Slaked lime is used in the manufacture of (1) cement. (2) fire bricks. (3) pigments. (4) medicines. Solution (1) When lime is kept exposed to air, it slowly absorbs moisture, CO2, and swells and disintegrates to fine

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(2) CaCl2 + CaF2 (4) CaCl2 + KCl

(2) Mg(OH)2 (4) Ba(OH)2

Solution (4) The basic strength of hydroxides increases from Mg to Ba, and Group 2 shows the usual trend that basic properties increase on descending the group. 20. Read the following statements and answer accordingly.

Statement 1: Thermal stability of alkaline earth metal carbonates increases from Be to Ba.



Statement 2: The basicity of metal oxides increases from Be to Ba. (1) Both statement 1 and statement 2 are true, and statement 2 is correct explanation for statement 1. (2) Both statement 1 and statement 2 are true, but statement 2 is not correct explanation for statement 1. (3) Statement 1 is true, but statement 2 is false. (4) Both statement 1 and statement 2 are false.

Solution (2) The larger compounds down the group require more heat than the lighter compounds in order to decompose, thus thermal stability increases down the group.

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THE s-BLOCK ELEMENTS 2 1. Which of the following carbonates will not decompose on heating? (1) CaCO3 (3) BaCO3

(2) Na2CO3 (4) Li2CO3

Solution (2) It is present in form of decahydrate (Na2CO3·10H2O) and will lose water molecules on heating. It first loses nine water molecules on heating to form monohydrate (Na2CO3·1H2O) and then on further heating changes to anhydrous Na2CO3 that does not further decompose. 2 2. Setting of cement is (1) (2) (3) (4)

an exothermic reaction. an endothermic reaction. neither exothermic nor endothermic. none of these.

Solution (1) When water is added to cement the setting and hardening process takes place. Chemical reaction between cement and water takes place with release of heat. Hence, it is an exothermic reaction. 2 3. Which of the following does not react with water?

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(1) BeO (2) CaO (3) MgO (4) SrO Solution (1) BeO is insoluble in water but dissolves in acids to give salts, and in alkalis to give beryllates, which on standing precipitate as the hydroxide. 2 4. The density of Ca is less than that of Mg because (1) (2) (3) (4)

nuclear charge of Ca is more than Mg. vacant 3d orbital is present in Ca. size of Ca is less than Mg. none of these.

Solution (2) Ca has less density than that of Mg because vacant 3d orbital is present in Ca. 2 5. Ground water is hard in the region of (1) coal deposits. (2) petroleum deposits. (3) limestone deposits. (4) none of these. Solution (3) Hard water contains dissolved salts of Ca, Mg, etc. In limestone deposits CaCO3 is present; hence, water becomes hard water in the ground level.

Practice Exercises Level I General Trends in Physical and Chemical Properties of Elements 1. Which one of the following metal is the most reactive? (1) Mg (2) K (3) Na (4) Pb 2. Which of the alkaline earth metal halides given below is essentially covalent in nature? (1) MgCl2 (2) BeCl2 (3) SrCl2 (4) CaCl2 3. Which of the following has the highest first ionization enthalpy? (1) Ba (2) Mg (3) Ca (4) Be 4. Magnesium burns in air to give (1) MgO (2) Mg3N2 (3) MgCO3 (4) both (1) and (2) 5. The solubility of carbonates decrease down the magnesium group due to a decrease in (1) (2) (3) (4)

lattice energies of solids. hydration energies of cations. inter-ionic attraction. entropy of solution formation.

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6. One mole of magnesium nitride on the reaction with an excess of water gives (1) (2) (3) (4)

one mole of ammonia. two moles of nitric acid. two moles of ammonia. one mole of nitric acid.

7. KOH is preferably used to absorb CO2 because (1) KOH is more soluble than NaOH in water. (2) KOH is a stronger base than NaOH. (3) KHCO3 is soluble in water and NaHCO3 is insoluble in water. (4) KOH is cheaper than NaOH. 8. Which of the following when introduced into the Bunsen’s flame gives pink violet color? (1) NaCl (2) BaCl2 (3) CsCl (4) KCl 9. Mark the false statement. (1) The electropositive character of alkali metal decreases with increase in the atomic number. (2) Lithium is a hard metal and cannot be cut with a knife. (3) Alkali metals are strong reducing agents. (4) Electronegativities of all alkali metals lie between 1.0 and 0.7.

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10. Which one of the following will react most vigorously with water? (1) V (2) Ti (3) Ca (4) Rb 11. The metal that can be extracted from seawater is (1) Cl (2) Br (3) Mg (4) All of these 12. Which of the following alkali metal is expected to melt if the room temperature rises to 30°C? (1) Na (2) K (3) Rb (4) Cs 13. Alkali metals react with water vigorously to form hydroxides and dihydrogen. Which of the following alkali metals reacts with water least vigorously? (1) Li (2) Na (3) K (4) Cs 14. Which has minimum hydration enthalpy? (1) K+ (3) Ca2+

(2) Li+ (4) Al3+

15. The wire of flash bulb is made of (1) Mg (2) Cu (3) Ba (4) Ag 16. Which of the following compounds is almost insoluble in water? (1) LiF (2) LiCl (3) LiBr (3) LiI 17. The nature of oxide of radium is (1) acidic. (2) basic. (3) neutral. (4) amphoteric. 18. Which out of the following does not exist in the solid state? (1) NaHCO3 (3) Ca(HCO3)2

(2) CaCO3 (4) BaCO3

19. Peroxide and superoxides of alkali metals are colored because of (1) defects in crystals. (2) presence of unpaired electrons. (3) partially due to the defects and partially due to presence of unpaired electrons. (4) none of these.

Anomalous Properties of the First Element of Each Group and Diagonal Relationships 20. Which of the following pairs form a diagonal relationship? (1) Be and Mg (2) Li and Al (3) Be and Al (4) Be and Na

Chapter 16_The s-Block Elements.indd 386

21. Which of the following elements does not form hydride by direct heating with dihydrogen? (1) Be (2) Mg (3) Sr (4) Ba 22. Beryllium shows diagonal relationship with aluminum. Which of the following similarities is incorrect? (1) Be forms beryllates and Al form aluminates. (2) Be(OH)2 alike Al(OH)2 is basic. (3) Be like Al is rendered passive by HNO3. (4) Be2C like Al4C3 yields methane on hydrolysis.

Preparation and Properties of Some Important Compounds of Sodium 23. Na2CO3 can be manufactured by Solvay’s process but K2CO3 cannot be prepared because (1) K2CO3 is highly soluble in water. (2) K2CO3 is less soluble in water. (3) KHCO3 is highly soluble in water as compared to NaHCO3. (4) KHCO3 is less soluble as compared to NaHCO3. 24. The principal products obtained on heating iodine with concentrated caustic soda solution are (1) NaOI + NaI (2) NaIO3 + NaI (3) NaOI + NaIO3 + NaI (4) NaIO4 + NaI 25. When NaCl is dissolved in water, the sodium ion is (1) oxidized. (2) reduced. (3) hydrolyzed. (4) hydrated. 26. Which of the following statements about Na2O2 is not correct? (1) It is diamagnetic in nature. (2) It is a derivative of H2O2. (3) Na2O2 oxidizes Cr3+ to CrO 2in acid medium. 4 (4) It is the super oxide of sodium. 27. Washing soda is (1) Na2CO3⋅7H2O (2) Na2CO3⋅10H2O (3) Na2CO3⋅H2O (4) Na2CO3 28. Baking soda is (1) Na2CO3 (2) NaHCO3 (3) Na2SO4 (4) K2CO3 29. The useful byproduct obtained in Solvay’s process of manufacture of sodium carbonate is (1) quicklime. (2) NH4Cl (3) CaCl2 (4) Ca(OH)2 30. Which one of the following is ionic? (1) Sodium chloride (2) Beryllium chloride (3) Lithium chloride (4) Carbon tetrachloride

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THE s-BLOCK ELEMENTS

Compounds of Calcium

2. The correct statement for the molecule, CsI3, is

31. Dolomite has the composition (1) KCl⋅MgCl2⋅6H2O (2) Na3AlF6 (3) CaCO3⋅MgCO3 (4) CaCl2⋅MgCl2 ×6H2O 32. Gypsum CaSO4⋅2H2O on heating to about 150°C forms plaster of Paris which has chemical composition represented by (1) 2CaSO4⋅3H2O (2) CaSO4⋅H2O 1 (3) 2CaSO4⋅ H2O (4) CaSO4 2 3. The substance not likely to contain CaCO3 is 3 (1) a marble statue. (2) calcined gypsum. (3) sea shells. (4) dolomite. 34. Commonly used laboratory desiccant is (1) calcium chloride. (2) sodium carbonate. (3) sodium chloride. (4) potassium nitrate. 35. What is X in the equation? K CaO + 3C 3273  → X + CO

(1) CaCO3 (3) CO2

(2) CaC2 (4) none of these

36. Mortar is a mixture of slaked lime and sand in the ratio (1) 3:1 (2) 1:3 (3) 2:3 (4) 1:4 37. Calcium carbide and nitrogen react to produce (1) calcium acetate. (2) calcium carbonate. (3) calcium cyanate. (4) calcium cyanamide. 38. Which is not an ore of calcium? (1) Limestone (2) Flurospar (3) Dolomite (4) Epsom salt

Biological Significance of Na, K, Mg and Ca 39. Which of the following metals is present in chlorophyll? (1) Mg (2) Be (3) Ca (4) None of these 40. Which pump is important in biological reaction in human body? (1) Ca-Mg Pump (2) K-Fe Pump (3) Na-K Pump (4) Fe-Ca Pump

Level II General Trends in Physical and Chemical Properties of Elements 1. Which of the carbonates given below is unstable in air and is kept in CO2 atmosphere to avoid decomposition? (1) BeCO3 (2) MgCO3 (3) CaCO3 (4) BaCO3

Chapter 16_The s-Block Elements.indd 387

387

(1) (2) (3) (4)

It is a covalent molecule It contains Cs+ and I 3- ions It contains Cs3+ and I- ions It contains Cs+, I- and lattice I2 molecule.

3. Of the metals Be, Mg, Ca, and Sr of Group 2 in the periodic table, the least ionic chloride would be formed by (1) Ca (2) Mg (3) Be (4) Sr 4. The reducing power of a metal depends on various factors. Suggest the factor which makes Li, the strongest reducing agent in aqueous solution. (1) (2) (3) (4)

Sublimation enthalpy Ionization enthalpy Hydration enthalpy Electron-gain enthalpy

5. Electrolysis of KH produces H2 (1) (2) (3) (4)

at the cathode. at the anode. either at the cathode or at the anode. cannot be predicted.

6. On heating sodium metal in a current of dry ammonia gas, the compound formed is (1) sodium nitrate. (2) sodium hydride. (3) sodium amide. (4) sodium azide. 7. When a substance (A) reacts with water, it produces a combustible gas (B) and a solution of substance (C) in water. When another substance (D) reacts with the solution of (C), it produces the same gas (B) on warming, but (D) can produce (B) on reaction with dilute H2SO4 at room temperature, (A) imparts a golden yellow color to flame. The compounds (A), (B), (C), and (D) are (1) K, H2, KOH, Al (2) Na, H2, NaOH, Zn (3) CaC2, C2H2, Ca(OH)2, Fe (4) Ca, H2, Ca(OH)2, Sn 8. When sodium metal is dissolved in liquid ammonia, a blue solution is formed. The blue color is due to (1) solvated Na+ ions. (2) solvated electrons. (3) solvated NH 2- ions. (4) solvated protons. 9. Sodium bicarbonate (NaHCO3) on gentle heating produces (1) Na2CO3 + CO2 (2) Na2CO3 + CO (3) Na2CO3 + CO2 + H2O (4) Na2CO3 + CO + H2O 10. Alkaline earth metals are denser than alkali metals, because metallic bonding in alkaline earth metals is (1) weaker. (2) not present. (3) stronger. (4) volatile. 11. Amphoteric hydroxides react with both alkalis and acids. Which of the following Group 2 metal hydroxides is soluble in sodium hydroxide?

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388

OBJECTIVE CHEMISTRY FOR NEET (1) Be(OH)2 (2) Mg(OH)2 (3) Ca(OH)2 (4) Ba(OH)2

12. Which of the following statements is false regarding saline hydrides? (1) (2) (3) (4)

In the molten state they conduct electricity. They dissolve in water giving off hydrogen. They are used as reducing agents. They are covalent in nature.

13. Alkali metals give color in Bunsen flame due to (1) low ionization potential. (2) low melting point. (3) softness. (4) one electron in outermost orbit. 14. Carnallite is a mineral of (1) Ca (2) Ne (3) Mg (4) Zn 15. Pure anhydrous magnesium chloride can be prepared from hydrated salt by (1) heating the hydrate to red heat in the atmosphere of HCl gas. (2) melting the hydrate. (3) heating the hydrate with coke. (4) heating the hydrate with Mg ribbon. 16. The hydride from amongst the following that cannot be obtained directly by reaction with hydrogen is (1) CaH2 (3) BeH2

(2) MgH2 (4) NaH

17. Thermal stability of alkaline earth metal carbonates decreases in order (1) BaCO3 > SrCO3 > CaCO3 > MgCO3 (2) BaCO3 > SrCO3 > MgCO3 > CaCO3 (3) CaCO3 > SrCO3 > MgCO3 > BaCO3 (4) MgCO3 > CaCO3 > SrCO3 > BaCO3

Anomalous Properties of the First Element of Each Group and Diagonal Relationships 18. Superoxide of type MO2 are formed by all except (1) potassium. (2) beryllium. (3) strontium. (4) barium. 19. Which among the following does not at all show the tendency to form peroxide? (1) Lithium (2) Magnesium (3) Beryllium (4) Barium

Preparation and Properties of Some Important Compounds of Sodium 20. Some large white transparent crystals are left out in a bowl for several days. They are then observed to have changed their form into white powder. The crystals may have been of

Chapter 16_The s-Block Elements.indd 388

(1) ammonium chloride. (2) sodium chloride. (3) sodium carbonate. (4) calcium oxide. 21. In the following reactions A + H 2O → NaOH H2 O °C A 400  → B at → NaOH + O2 25° C



B is used for oxygenating in submarines. Then A and B are respectively (1) Na 2O2 and Na 2O (2) Na 2O and Na 2O2 (3) Na 2O2 and O2 (4) Na 2O and O2

22. Sodium carbonate is manufactured by Solvay’s process. The products that are recycled are (1) CO2 and NH3 (2) CO2 and NH4Cl (3) NaCl and CaO (4) CaCl2 and CaO 23. Molten sodium is used in nuclear reactors to (1) (2) (3) (4)

absorb neutrons in order to control the chain reaction. slow down the fast moving neutrons. absorb the heat generated by nuclear fission. extract radioisotopes produced in the reactor.

Compounds of Calcium 24. In curing cement plasters water is sprinkled from time to time. This helps in (1) keeping it cool. (2) developing interlocking needle-like crystals of hydrated silicates. (3) hydrating sand and gravel mixed with cement. (4) converting sand into silicic acid 25. Dehydration of hydrates of halides of calcium, barium and strontium, i.e., CaCl2⋅6H2O, BaCl2⋅2H2O, SrCl2⋅2H2O, can be achieved by heating. These become wet on keeping in air. Which of the following statements is correct about these halides? (1) Act as dehydrating agent. (2) Can absorb moisture from air. (3) Tendency to form hydrate decreases from calcium to barium. (4) All of the above. 26. Calcium is obtained by (1) (2) (3) (4)

electrolysis of molten CaCl2. electrolysis of a solution of CaCl2 in water. reduction of CaCl2 with carbon. roasting of limestone.

27. The function of sand in mortar is (1) to decrease the hardness. (2) to prevent excessive shrinkage because of which cracks may result. (3) to increase hardness. (4) to make the mass compact.

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THE s-BLOCK ELEMENTS

Biological Significance of Na, K, Mg and Ca 28. The metal used for blood coagulation is

5. Which of the following alkaline earth metal sulphates has hydration enthalpy higher than the lattice enthalpy? (1) SrSO4 (3) BeSO4

(1) Ca (2) K (3) Na (4) Mg 29. A certain metal M is used to prepare an antacid, a medicine used in acidity. This metal accidently catches fire which cannot be put out by using CO2 based extinguisher. The metal is (1) Ca (2) Mg (3) Na (4) K

(1) Mg (2) Al (3) Ca (4) Na

(AIPMT PRE 2010) 6. Which one of the following compounds is a peroxide? (1) NO2 (3) BaO2

(AIPMT MAINS 2010)

(2) BaSO4 (4) RaSO4

8. Which of the following compounds has the lowest melting point? (1) CaCl2 (2) CaBr2 (3) CaI2 (4) CaF2

(AIPMT 2007) 2. The alkali metals form salt-like hydrides by the direct synthesis at elevated temperature. The thermal stability of these hydrides decreases in which of the following orders? (1) LiH > NaH > KH > RbH > CsH (2) CsH > RbH > KH > NaH > LiH (3) KH > NaH > LiH > CsH > RbH (4) NaH > LiH > KH > RbH > CsH (AIPMT 2008) 3. Which of the following oxides is not expected to react with sodium hydroxide? (1) BeO (2) B2O3 (3) CaO (4) SiO2



(AIPMT PRE 2011)

9. Match List-I with List-II for the composition of substances and select the correct answer using the code given below the lists: List-I

List-II

Substances

Composition

(1) Plaster of Paris

(1) CaSO4·2H2O

(2) Epsomite

(2) CaSO4·

(3) Kieserite

(3) MgSO4·7H2O

(4) Gypsum

(4) MgSO4·H2O

1 HO 2 2

(5) CaSO4 (AIPMT 2009)

4. Property of the alkaline earth metals that increases with their atomic number is (1) electronegativity. (2) solubility of their hydroxides in water. (3) solubility of their sulphates in water. (4) ionization energy.

Chapter 16_The s-Block Elements.indd 389

(AIPMT PRE 2010)

(1) Na2CO3 (2) K2CO3 (3) CaSO4 ⋅2H2O (4) CaCO3

1. In which of the following the hydration energy is higher than the lattice energy?



(2) KO2 (4) MnO2

7. The compound A on heating gives a colorless gas and a residue that is dissolved in water to obtain B. When excess of CO2 is bubbled through aqueous solution of B, C is formed which is recovered in the solid form. Solid C on gentle heating gives back A. The compound A is

Previous Years’ NEET Questions

(1) SrSO4 (3) MgSO4

(2) CaSO4 (4) BaSO4



30. A certain metal is present in the soil, plants, bones, egg shells, sea shells, and coral. It is also used to remove oxygen from molten steel and its hydroxide is used to detect CO2. The metal is

389

(AIPMT PRE 2010)

Code:

(1)

(2)

(3)

(4)

(1)

(4)

(3)

(2)

(1)

(2)

(3)

(4)

(1)

(2)

(3)

(2)

(3)

(4)

(1)

(4)

(1)

(2)

(3)

(5) (AIPMT MAINS 2011)

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390

OBJECTIVE CHEMISTRY FOR NEET

10. Which one of the alkali metals on heating in air forms only the normal oxide M2O? (1) Li (2) Na (3) Rb (4) K



(AIPMT PRE 2012)

(1) On passing H2S through acidified K2Cr2O7 solution, a milky color is observed. (2) Na2Cr2O7 is preferred over K2Cr2O7 in volumetric analysis. (3) K2Cr2O7 solution in acidic medium is orange. (4) K2Cr2O7 solution becomes yellow on increasing the pH beyond 7.

(1) MgCO3 (3) K2CO3

(1) Li+ < K+ < Na+ < Rb+ (2) Rb+ < K+ < Na+ < Li+ + + + + (3) K < Na < Rb < Li (4) Na+ < Li+ < K+ < Rb+ (AIPMT PRE 2012) 13. In the replacement reaction CF + MI

I + MF

(2) CaCO3 (4) Na2CO3 (RE AIPMT 2015)

17. Which of the following statements is false? (1) Mg2+ ions form a complex with ATP. (2) Ca2+ ions are important in blood clotting. (3) Ca2+ ions are not important in maintaining the regular beating of the heart. (4) Mg2+ ions are important in the green parts of plants.

(AIPMT PRE 2012) 12. The ease of adsorption of the hydrated alkali metal ions on an ion-exchange resin follows the order

(AIPMT 2015)

16. On heating, which of the following releases CO2 most easily?

11. Which of the statements is not true?

C

(2) K+ (4) Ca2+

(1) Mg2+ (3) Fe2+



(NEET-I 2016)

18. The product obtained as a result of a reaction of nitrogen with CaC2 is (1) Ca(CN)2 (3) CaCN3

(2) CaNCN (4) Ca2CN (NEET-I 2016)

  19. The suspension of slaked lime in water is

The reaction will be most favorable if M happens to be (1) Na (2) K (3) Rb (4) Li

(AIPMT MAINS 2012)

14. Solubility of the alkaline earth’s metal sulphates in water decreases in the sequence (1) Ca > Sr > Ba > Mg (2) Sr > Ca > Mg > Ba (3) Ba > Mg > Sr > Ca (4) Mg > Ca > Sr > Ba (AIPMT 2015) 15. The function of sodium pump is a biological process operating in each and every cell of all animals. Which of the following biologically important ions is also a constituent of this pump?

(1) (2) (3) (4)

Quick lime Milk of lime Aqueous solution of slaked lime Lime water (NEET-II 2016)

20. In context with beryllium, which one of the following statements is incorrect? (1) (2) (3) (4)

It forms Be2C. Its salts rarely hydrolyze. Its hydride is electron-deficient and polymeric. It is rendered passive by nitric acid. (NEET-II 2016)

Answer Key Level I  1. (2)

 2. (2)

 3. (4)

 4. (4)

 5. (2)

 6. (3)

 7. (3)

 8. (4)

 9. (1)

10. (4)

11. (3)

12. (4)

13. (1)

14. (1)

15. (1)

16. (1)

17. (2)

18. (3)

19. (3)

20. (3)

21. (1)

22. (2)

23. (3)

24. (2)

25. (4)

26. (4)

27. (2)

28. (2)

29. (3)

30. (1)

31. (3)

32. (3)

33. (2)

34. (1)

35. (2)

36. (2)

37. (4)

38. (4)

39. (1)

40. (3)

Chapter 16_The s-Block Elements.indd 390

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391

THE s-BLOCK ELEMENTS

Level II  1. (1)

 2. (2)

 3. (3)

 4. (3)

 5. (2)

 6. (3)

 7. (2)

 8. (2)

 9. (3)

10. (3)

11. (1)

12. (4)

13. (1)

14. (3)

15. (1)

16. (3)

17. (2)

18. (1)

19. (1)

20. (3)

21. (2)

22. (1)

23. (3)

24. (2)

25. (4)

26. (1)

27. (2)

28. (1)

29. (2)

30. (3)

Previous Years’ NEET Questions  1. (3)

 2. (1)

 3. (3)

 4. (2)

 5. (3)

 6. (3)

 7. (4)

 8. (3)

 9. (3)

10. (1)

11. (2)

12. (2)

13. (3)

14. (4)

15. (2)

16. (1)

17. (3)

18. (2)

19. (2)

20. (2)

Hints and Explanations Level I

17. (2) The basicity of oxides increases down the group.

1. (2) Alkali metals are highly reactive and reactivity increases down the group. Thus, potassium has maximum reactivity. 2. (2) BeCl2 is covalent in nature because of small size and high electronegativity of Be. 3. (4) Be has the highest first ionization enthalpy because it is the first element in Group 2 and ionization enthalpy decreases down the group. 4. (4) The reactions are 2Mg + O2 Air → 2MgO 6Mg + N 2 Air → 2Mg 3N 2 5. (2) Due to decrease in hydration energy of cation, the solubility decreases down the group. 6. (3) The reaction is Mg 3N 2 + 3H 2O → 3Mg(OH)2 + 2NH 3 7. (3) The produced insoluble NaHCO3 chokes the reactor. 8. (4) NaCl gives golden yellow, BaCl2 gives apple green, CsCl gives violet, and KCl gives pink violet color. 11. (3) It is present in seawater in amounts of about 1300 ppm and is the most commonly found cation in oceans after sodium.

18. (3) Bicarbonates of alkali metals are found in the solid state whereas bicarbonates of alkaline earth metals exist only in the solution state. 20. (3) Be and Al and Li and Mg form diagonal relationship due to similar charge by radius ratio. 22. (2) Be(OH)2 alike Al(OH)2 are amphoteric in nature. 23. (3) This is because KHCO3 formed in the process is fairly soluble in water. 24. (2) 3I 2 + 6NaOH → 5NaI + NaIO3 + 3H 2O 25. (4)

NaCl

+ H 2O ® Na+ OH - + H + Cl ( Hydrated)

26. (4) Na2O2 has peroxy linkage hence peroxide. 29. (3) 2NH4Cl + Ca(OH)2 → 2NH3 + 2H2O + CaCl2 32. (3) Calcium sulphate hemihydrate forms by partially dehydrating gypsum at 150°C. 1 °C 2CaSO4 × 2H 2O ¾150 ¾¾ ® 2CaSO4 × H 2O+ 3H 2O 2 Gypsum Plaster of paris

1 33. (2) Formula for calcinated gypsum is CaSO 4 × H 2O. 2 34. (1) Calcium chloride is used as laboratory desiccant because it is easily available, less toxic, and costs less. K 35. (2) CaO + 3C 3273  → CaC 2 + CO (X)

14. (1) K has minimum hydration enthalpy. Smaller the cation, greater is the degree of hydration.

39. (1) Mg2+ is an important constituent of chlorophyll in the green plants.

15. (1) Magnesium is used in flash bulbs for photography. In flash bulbs magnesium powder is mixed with potassium chlorate.

Level II

+

16. (1) Be is the most electronegative alkaline earth metal. Electronegativity decreases from Be to Ba.

Chapter 16_The s-Block Elements.indd 391

2. (2) CsI3 contains Cs+ and I 3- ions because Cs cannot show +3 oxidation states and it is difficult to accommodate I2 molecules into the lattice because of their large size.

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392

OBJECTIVE CHEMISTRY FOR NEET

3. (3) Be is the metal that formed least ionic chloride compound because of its more polarizing power towards anion and thus developing covalent character in molecule.

28. (1) Ca plays important role in neuromuscular function, inter-neuronal transmission, cell membrane integrity and blood coagulation.

6. (3) 2Na + NH 3 → 2NaNH2 + H 2

29. (2) A suspension of Mg(OH)2 in water is used in medicine as an antacid under the name of milk of magnesia. If magnesium catches fire accidently it cannot be put out by CO2 based extinguishers.

Sodium amide

7. (2) The reactions are

2Na + 2H 2O → 2NaOH+ H 2 (A)



(B)

2NaOH+ Zn → Na 2 ZnO2 + H 2

30. (3) The reaction is Ca(OH )2 + CO2 → CaCO 3 + H 2O

Zn + H 2SO4 → ZnSO4 + H 2

Previous Years’ NEET Questions

(C )



(C )

(D)

(B)

(D)

(B)

Sodium imparts golden color to the flame.

8. (2) Ammoniated electron (solvated electrons) is responsible for blue color solution. ∆  Na 2CO3 + CO2 + H 2O 9. (3) 2NaHCO3 →

10. (3) The metallic bonding in alkaline earth metal is stronger than alkali metal, because in alkaline earth metals there are two valence electrons per atom. 12. (4) Saline hydrides are not covalent in nature, they are ionic in nature. 13. (1) Alkali metals give color in Bunsen flame due to low ionization potential. 14. (3) Carnallite is a mineral of Mg and its formula is (MgCl2 · KCl·6H2O). Heated in  → MgCl 2 + 6H 2O 15. (1) MgCl ⋅ 6H 2O presence of HCl gas Anhydrous

16. (3) Except Be, all alkaline earth metals form hydrides (MH2) on heating directly with H2. 17. (1) The thermal stability of alkaline earth metal carbonates increases down group. 18. (2) Beryllium cannot form superoxide of type MO2. 19. (1) Due to its diagonal relationship with Mg. 20. (3) Sodium carbonate normally occurs as a crystalline heptahydrate, which readily undergo efflorescence to form a white powder.

1. (3) For a substance to dissolve, the hydration energy must exceed the lattice energy. On descending the group, the metal ions become larger and so both the lattice energy and the hydration energy decrease. A decrease in lattice energy favors increased solubility. Therefore, the solubility or hydration order is Mg >> Sr > Ba > Ra. Thus, MgSO4 is soluble and will have higher hydration energy but the sulphates of Sr, Ba and Ra are virtually insoluble which indicates their lattice energy would be higher than the hydration energy. 2. (1) Smaller cation and smaller anion form thermally stable hydrides. Therefore, correct order is LiH > NaH > KH > RbH > CsH. 3. (3) CaO is a basic oxide and hence it would not react with another base. 4. (2) The solubility of most salts decreases with increased atomic weight, though the usual trend is reversed with hydroxides. With hydroxides the lattice energy decreases more rapidly than the hydration energy, so their solubility increases on descending the group. 5. (3) For a substance to dissolve, the hydration energy must exceed the lattice energy. On descending the group the metal ions become larger and so both lattice energy and the hydration energy decrease. In case of Be, hydration enthalpy is higher than the lattice enthalpy due to the formation of strong complex [Be(H2O)4]2+.

1 → Na 2O 2 → 2NaOH + O 2 ↑ 1. (2) Na 2O  2 2 (A) (B)

6. (3) In BaO2, the oxygen exists as O22 , therefore, it is peroxide. NO2 and MnO2 are oxides and KO2 is a super­ oxide.

22. (1) CO2 and NH3 can be recycled during the Solvay process for manufacturing of sodium carbonate.

7. (4) The reactions involved are:

400° C

H 2 O at 25° C

24. (2) As setting of cement is an exothermic reaction, interlocking crystals of hydrated silicates are developed. 2+

→ Ca + 2Cl 26. (1) The reaction is CaCl 2  Fusion

-

-



At anode:

2Cl → Cl 2 + 2e



At cathode:

Ca 2+ + 2e - → Ca

Chapter 16_The s-Block Elements.indd 392

-

∆ CaCO3 →  CaO +

(A)

residue

CO2

 

(colorless gas)

CO2 D CaO + H 2O ® Ca(OH)2 ¾( excess ¾¾ ® Ca(HCO 3 )2 ¾® ¾ CaCO3   )

(B)

(C)

(A)

8. (3) According to Fajan’s rule a small positive ion and large negative ion favors covalency. Among the given compounds, CaI2 has maximum covalent character due to which its melting point is lowest.

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THE s-BLOCK ELEMENTS 9. (3) The composition of the given substances is as follows: Substance

Composition

Plaster of Paris CaSO 4 ⋅

1 H 2O 2

Epsomite

MgSO4·7H2O

Kieserite

MgSO4·H2O

Gypsum

CaSO4·2H2O

10. (1) The reaction is 1 2Li + O2 ® Li 2O 2 11. (2) Na2Cr2O7 is not used in volumetric analysis because it is deliquescent. 12. (2) Smaller is the ion, more is the ease of hydration so the order is Rb+ < K+ < Na+ < Li+. 13. (3) In case of alkali metals, as we move down the group the size of cation increases and so the ionic character also increases. Rb−F has less lattice energy and is more reactive. So, Rb will give the favorable reaction. 14. (4) The solubility of sulphates in water decreases down the group. Mg >> Ca > Sr > Ba, thus, MgSO4 is soluble but CaSO4 is sparingly soluble, and the sulphates of Sr, Ba and Ra are virtually insoluble. The significantly higher solubility of MgSO4 is due to high enthalpy of solvation of the smaller Mg2+ ion. 15. (2) K+ ions are the most abundant cations within the cell fluids they activate many enzymes. There is a very large variation in the concentration of Na+ and K+ ions found on the opposite sides of the cell membrane. These ionic gradients are called the sodium­potassium pump.

Chapter 16_The s-Block Elements.indd 393

393

16. (1) Among the given compounds, magnesium carbonate is least stable hence it dissociates readily and releases CO2.

∆ MgCO3 →  MgO + CO 2 ↑

17. (3) Magnesium ions are concentrated in animals and calcium ions are concentrated in body fluids outside the cell. Mg2+ ions form a complex with ATP and are an important component in chlorophyll, the green parts of the plants. Ca2+ ions are important in blood clotting and are required to trigger contraction of muscle and maintain regular beating of the heart. 18. (2) When CaC2 is heated in an electric furnace with atmospheric nitrogen at 1373 K, calcium cyanamide (CaNCN) is formed. It is an important method of atmospheric fixation of nitrogen. D CaC 2 + N 2 ¾® ¾ CaNCN + C

Calcium cyanamide



 

BaC2 also reacts with N2, but it forms cyanide, Ba(CN)2 not a cyanamide (NCN)2−.

19. (2) Slaked lime is calcium hydroxide. When slaked lime is dissolved in water and dissolved Ca(OH)2 filtered off, the clear solution obtained is called lime water. When excess calcium hydroxide is added to this solution, a milky suspension of calcium hydroxide particles is observed and is called milk of lime. 20. (2) Beryllium forms covalent compounds; hence its salts hydrolyzed easily. It forms unusual carbide, Be2C, brickred in color. Beryllium hydride is electron deficient, so it generally exists as clusters through the formation of three-center two electron (banana) bonds. Be is rendered passive by concentrated HNO3. This is because, HNO3 is a strong oxidizing agent and forms a thin layer of oxide on the surface of the metal, making it unreactive to further attack of acid.

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17

The p-Block Elements

Chapter at a Glance 1. p-Block Elements: The elements in which the highest energy electron enters the outermost p orbital are known as p-block elements. There are six groups (ns2 np1–6) of p-block elements in the periodic table, named as Group 13–18 as shown below. Group First element Electronic configuration Group oxidation state Other oxidation states

13 Boron (B) ns2 np1

14 Carbon (C) ns2 np2

15 Nitrogen (N) ns2 np3

16 Oxygen (O) ns2 np4

17 Fluorine (F) ns2 np5

+3

+4

+5

+6

+7

18 Helium (He) ns2 np6 (He: 1s2) +8

+1

+2, −4

+3, −3

+4, +2, −2

+5, +3, +1, −1

+6, +4, +2

2. Inert Pair Effect: This occurrence of an oxidation state two units below the Group oxidation state is sometimes attributed to the inert pair effect. The term refers to the resistance of a pair of s electrons to be lost or to participate in covalent bond formation.

Group 13 Elements

1. General Properties Elements

Occurrence

Boron (B)

Concentrated deposits of borax Na2[B4O5(OH)4]⋅8H2O and kernite Na2[B4O5(OH)4]⋅2H2O Aluminosilicate rocks, such as feldspars and micas

Aluminum (Al) Gallium (Ga) Indium (In) Thallium (Tl)

Chapter 17_p-Block Elements.indd 395

Occur as sulphides, in trace amounts

Atomic Electronic Oxidation number configuration states

Ionization enthalpy m.p. b.p. (kJ mol−1) (°C) (°C) 1st 2nd 3rd 801 2427 3659 2180 3650

7

[He] 2s22pl

3

15

[Ne] 3s23p1

(1) 3

577

1816

2744

660

2467

33

[Ar] 3d104s2 4p1 [Kr] 4d105s2 5p1 [Xe] 4f 145d10 6s26p1

1  3

579

1979

2962

30

2403

1  3

558

1820

2704

157

2080

1  3

589

1971

2877

303

1457

51 83

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OBJECTIVE CHEMISTRY FOR NEET

(a)  The metallic radii of the atoms do not increase regularly on descending the group. The ionic radii for M3+ increase down the group. (b)  The electronegativity decreases from B to Al due to difference in atomic sizes of the elements. (c)  The ionization enthalpies increase as expected (first ionization enthalpy < second ionization enthalpy < third ionization enthalpy). The ionization enthalpy values do not decrease smoothly down the group. The decrease from B to Al is the usual trend on descending a Group associated with increased size. The poor shielding by d electrons and the resulting d-block contraction affect the values for the later elements. (d)  The melting points of the Group 13 elements do not show a regular trend. (e)  Boron has an unusual crystal structure while Al, In and Tl all have close-packed metal structures. Gallium has an unusual structure with an incredibly low melting point of gallium of 30°C and also because the liquid expands when it forms the solid, that is, the solid is less dense than the liquid. (f )  The boiling point of B is unusually high, but the values for Ga, In and Tl decrease on descending the Group as expected. (g)  The densities of elements increase down the group. (h)  The electropositive or metallic nature of the elements increases from B to Al, but then decreases from Al to Tl. 2. Chemical Properties (a) Oxidation state (i)  The elements all have three outer electrons. Apart from Tl, they normally use these to form three bonds, giving an oxidation state of (+3). (ii)  The atoms in this Group have an outer electronic configuration of s2p1. Monovalency is explained by the inert pair effect. (iii)  The trivalent compounds of Group 13 elements act as electron acceptors or Lewis acids and the donor of electrons such as ammonia and water act as Lewis bases and form compounds like BF3⋅NH3. The tendency of Group 13 elements to behave as Lewis acids increases as we move down the group. (b) Reaction with water: Since the compounds in trivalent state are covalent, they are hydrolyzed in water. The trihalides on hydrolysis in water form tetrahedral species [M(OH)4]-. BCl3 + 3H2 O ® H3 BO3 + 3HCl  [B(OH)4 ]+Cl (c) Some important reactions (i) Reactions of boron Reaction 4B + 3O2 → 2B2O3

Comment At high temperature.

2B + 3S → B2S3

At 1200°C.

2B + N2 → 2BN 2B + 3X2 → 2BX3 (X = F, Cl, Br, I) 2B + 6NaOH → 2Na3BO3 + 3H2 2B + 2NH3 → 2BN + 3H2 B + M → MxBy

At very high temperature. At high temperature. Boron halides thus act as Lewis acids. The Lewis acid character, however, decreases in the order: BI3 > BBr3 > BCl3 > BF3. On fusion with alkali. At very high temperature. Many metals form borides (not Group 1) often non-stoichiometric.

(ii) Reactions of other Group 13 metals Reaction 4M + 3O2 → M2O3 2Al + N2 → 2AlN

Chapter 17_p-Block Elements.indd 396

Comment All react at high temperature. Al is very strongly exothermic. Ga only superficially oxidized. Tl forms some Tl2O as well. Only Al reacts at high temperature.

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THE p-BLOCK ELEMENTS

2M + 3X2 → 2MX3 where X = F, Cl, Br, I

2M + 6HCl → 2MCl3 + 3H2

397

All the metals  + form  Tl also formed trihalides  Trivalent iodides formed by Al, Ga, In only. Thallium (I) triiodide is formed. All react with dilute mineral acids, Al passivated by HNO3 particularly when concentrated.

Al + NaOH + H2 O ® NaAlO2 + H2 Na 3 AlO3 + H2

Al and Ga only.

M + NH3 → MNH2

All the metals form amides.

3. Boron (a) Important trends and anomalous properties of boron Boron Forms no simple B3+ cation. Exclusively forms covalent compounds.

Aluminum Electropositive and readily forms Al3+. Shows greater metallic character than boron and forms great number of ionic compounds. Forms compounds with coordination number six or more.

Gallium, indium and thallium Readily forms M3+ ions in solution. Show increasing metallic character and form coordination compounds characteristic of metals. Obeys octet rule and shows The lower oxidation state (+1) maximum covalence of four. becomes more stable especially in Tl+ and lower valent compounds are formed. Forms trivalent compounds Forms trivalent halides that dimerize Monovalent compounds are that act as Lewis acids and form to complete their octet by bridging formed and nature of covalent adducts with Lewis bases to through halogen atom. bond becomes weaker. complete their octets. Forms acidic oxide (B2O3) and Forms oxide Al2O3 and hydroxide Oxide and hydroxide of gallium hydroxide B(OH)3. Al(OH)3 that are amphoteric in are amphoteric, while that of In nature. and Tl are basic. Forms covalent halides that are Forms solid halides that are only Form trivalent halides (except easily hydrolyzed. partially hydrolyzed. from TlI3) which are aggregated in the solid state through halide ion bridge and have coordination numbers 4, 6 or higher. Forms stable hydride ion BH −4 .

Forms anionic hydride AlH −4 which is Only Ga forms hydride ion GaH −4. more reactive than BH −4 . – –

Frequently forms polyhedral compounds like boranes and borates.

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Chapter 17_p-Block Elements.indd 398

Diborane (B2H6)

Boric Acid (Orthoboric acid, B(OH)3 or H3BO3)

Borax (Na2B4O7⋅10H2O)

H

B 133 H pm

H

O

H

B

O

H

H

O

O

H

H

H

B 119 pm

H 1

H

O

O

O

O

H

H

H

O

O B

O

OH

H

H B

H

O

B

H

OH

B-

O

B

O

O

O

O

H

B

O

H

H

O

B

B

O

H

H

O

H

O

H

H

Overlap of approximately sp2 hybrid orbitals from B with an s orbital from H to give a “bananashaped” three-center two-electron bond.

H

H

H

H

HO

O

B-

OH

(b) Some important compounds of boron Orthoboric acid

2NaBO2

B 2 O3

Metaboric acid

Boron sesquioxide

(Mainly B4 H10 )

Diborane

3NaBH4 + 4BF3 ® 2B2 H6 + 3NaBF4

Boron fluoride

°C ¾¾ ® B2 H6 (g ) + 6NaF 2BF3 (g) + 6NaH ¾180

Boron trioxide

B2 O3 + 3H2 + 2Al ¾¾¾¾¾® B2 H6 + Al2 O3

750 atm, 150 ° C

Diglyme Et 2 O ⋅ BF3  + 3Na [BF4 ] → 2B2 H6 + 3NaBF4 + 4Et 2 O

4[Et 2 O × BF3 ] + 3Li[AlH4 ] ¾Ether ¾¾ ® 2B2 H6 + 3Li[AlF4 ] + 4Et 2 O

solution 2Na[BH4 ] + I2 ¾Diglyme ¾¾¾¾ ® B2 H6 + H2 + 2NaI

Mg 3 B2 + H3 PO4 → Mixture of boranes Heat  → B 2 H6

Magnesium borate

Preparation

Boric acid

B(OH) B(OH) (or3 BO H33BO ) + 32H ) +22H O 3 OH+ 3+ O+[B(OH) + [B(OH) K9a .= 259.25 3 (or3H 2 OH 4 ]    4p]K ap= + + - B(OH) HH HH B[3BO3 O )4 ])4 ]+ H pKpK= = 6.84 B(OH) ) +H 6.84 3 (or 3 BO 3 ) 3 3O 3 (OH 2O 3 (or 3 BO 3 O+ [+ 3 (OH 2O   a a °C H3 BO3 100  → HBO2 Red heat → B2 O3

Properties

Na 2 [B4 O5 (OH)4 ]× 8H2 O + 2HCl ® 2NaCl + 4H3 BO3 + 5H2 O

Preparation

Na 2 [B4 O5 (OH)4 ]× 8H2 O + 2HCl ® 2NaCl + 4H3 BO3 + 5H2 O

+

Sodium metaborate Boric anhydride Glassy borax bead

Borax is a useful primary standard for titrating against acids.

Na 2 B4 O7 ⋅10H2 O  → Na 2 B4 O7 Heat  → Heat

Na 2 B4 O7 + 7H2 O → 2NaOH + 4H3 BO3

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Chapter 17_p-Block Elements.indd 399

Tetrahydridoborates

Borazine (B3N3H6)

N

B

N

H Borazine

B

N

B

H

H B

B

H

H

H

H

H

H

H

H

B

B

B-

N

+

H

H

H

N+

B-

H

H

H

B-

N+ H

H

H H

B-

H

H

Borazine (B3N3H6) is sometimes called “inorganic benzene” because its structure shows some formal similarity with benzene, with delocalized electrons and aromatic character. The tetrahydridoborate ion [BH4]− is tetrahedral.

H

H

H

H

H

ΔH = − 2165 kJ mol −1

THF solvent

250° C , high pressure

¾¾¾¾¾¾¾¾¾ ®

Na[BH4 ] + 3Na[B(OMe)4 ]

AlCl3 + 3NaBH4 ¾Heat ¾¾ ® Al(BH4 )3 + 3NaCl

2LiH + B2 H6 ¾Ether ¾¾ ® 2LiBH4

Trimethylborate

Preparation 4B(OMe)3 + 4 NaH

Borazine

B 3 N 3 H6

3 2 6  → Higher temperature

Ratio 2NH :1B H

(BN)x

Boron nitride

3  → Higher temperature

Excess NH

3 B2 H6 + NH3  → B2 H6 ⋅ 2NH3 Low temperature

Excess NH

B2 H6 + 2CO ® 2BH3 × CO

B2 H6 + 2 Me3 N → 2[ Me3 N ⋅ BH3 ]

B2 H6 + 6ROH ® 2B(OR)3 + 6H2

B2 H6 + 6H2 O ® 2H3 BO3 + 3H2

B2 H6 + 3O2 → B2 O3 + 3H2 O

Properties

THE p-BLOCK ELEMENTS 399

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400

OBJECTIVE CHEMISTRY FOR NEET

Group 14 Elements 1. General Properties Ionization energies m.p. Atomic Electronic Oxidation (kJ mol−1) (°C) number configuration states 1st 2nd 3rd 4th Carbon (C) Carbon occurs 6 [He] 4 1086 2354 4622 6223 4100 mainly as coal, crude 2s2 2p2 oil and carbonates in rocks such as calcite CaCO3, magnesite MgCO3 and dolomite [MgCO3∙CaCO3]. Elements

Occurrence

Silicon (Si) As silica SiO2 (sand and quartz), and in a wide variety of silicate minerals and clays. Germanium As traces in some (Ge) silver and zinc ores, and in some types of coal. Tin (Sn) Tin is mined as cassiterite SnO2 Lead (Pb) Found as the ore galena PbS

b.p. (°C)

14

[Ne] 3s23p2

(2) 4

786

1573 3232 4351 1420

3280

32

[Ar] 3d104s24p2

2  4

760

1534 3300 4409 945

2850

50

[Kr] 4d105s25p2 [Xe] 4f  145d 106s26p2

2  4

707

1409 2943 3821 232

2623

2 4

715

1447 3087 4081 327

1751

82

(a)  The covalent radii increase down the group. (b)  The difference in size between Si and Ge is less than might be otherwise expected because Ge has a full 3d shell which shields the nuclear charge rather ineffectively. In a similar way, the small difference in size between Sn and Pb is because of the filling of the 4f shell. (c)  The ionization enthalpies decrease from C to Si, but then change in an irregular way because of the effects of filling the d and f shells. (d)  The increase in ionization enthalpy from tin to lead is because of their large size as well as poor shielding effect of intervening d and f orbitals. (e)  The first ionization enthalpies of Group 13 are lower than that of Group 14. The amount of energy required to form M4+ ions is extremely large, and hence simple ionic compounds are rare. (f )  The electronegativity decreases from carbon to silicon but remain the same till lead. In general, since the sizes of these elements are small, these elements are more electronegative than Group 13 elements. (g)  The melting points decrease on descending the Group because the M—M bonds become weaker as atoms increase in size. (h)  Metallic character increases down the group. 2. Chemical Properties (a) Oxidation states and trends in chemical reactivity. (i) The elements in this group are relatively unreactive but reactivity increases down the group. (ii) The most common oxidation states are +2 and +4. (iii)  The +2 oxidation state becomes increasingly stable on descending the group. The inert pair effect shows itself increasingly in the heavier members of the group. There is a decrease in the stability of the (+4) oxidation state and an increase in the stability of the (+2) state on descending the group.

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401

(b)  Reactivity towards oxygen: Carbon reacts with oxygen to form two oxides of C, that is, CO and CO2 while Si also forms two oxides of silicon, SiO and SiO2. The lower oxides GeO, SnO and PbO have layer lattices rather than the typical ionic structures. The basicity of the oxides usually increases down the group. Thus, CO2 and SiO2 are purely acidic. GeO2 is not as strongly acidic as SiO2, and SnO2 and PbO2 are amphoteric. (c)  Reactivity towards water: C, Si Ge, Pb are unaffected by water. Sn reacts with steam to give SnO2 and H2. Sn + 2H2 O ¾Heat ¾¾ ® SnO2 + 2H2



(d) Reactivity towards halogen (i)  Two series of halides, MX4 and MX2, are formed by the elements. All the tetrahalides are known except PbI4. All halides are typically covalent, tetrahedral (central atom is sp3 hydridized), and very volatile. The exceptions are SnF4 and PbF4, which have three-dimensional structures and are high melting solids. (ii)  The carbon halides are not hydrolyzed under normal conditions because they have no d orbitals, and cannot form a five-coordinate hydrolysis intermediate. (iii)  The silicon halides are rapidly hydrolyzed by water to give silicic acid. SiCl 4 + 4H2 O ® Si(OH)4 + 4HCl Silicic acid

(e) Reactivity towards acids and alkalis (i)  Si is oxidized and fluorinated by concentrated HF/HNO3. Ge dissolves slowly in hot concentrated H2SO4 and in HNO3. Sn dissolves in several concentrated acids. Pb does not dissolve in concentrated HCl because a surface coating of PbCl2 is formed. (ii)  Carbon is unaffected by alkalis. Si reacts slowly with cold aqueous solutions of NaOH and readily with hot solutions, giving solutions of silicates [SiO4]4–. Sn and Pb are slowly attacked by cold alkali, and rapidly by hot alkali, giving stannates Na2[Sn(OH)6] and plumbates Na2[Pb(OH)6]. Thus, Sn and Pb are amphoteric. 3. Important Trends and Anomalous Behavior of Carbon (a) (b) (c) (d)

Higher ionization enthalpy, being more covalent and being less metallic. Unique ability to form pp   pp multiple bonds, such as C   C, C   C, C   O, C   S and C   N. Forms many chains of great length. Allotropes of carbon

Allotrope Diamond

Graphite

Structure Tetrahedral structure; sp3 hybridization

Gemstones Abrasive powder for cutting and polishing

Two-dimensional sheets of carbon atoms.

Lubricant, electrodes, fibers, pencils, brake linings and brushes for electric motors.

(a)

Fullerenes

Chapter 17_p-Block Elements.indd 401

Uses

It consists of a fused system of five- and sixmembered rings.

(b)

Superconductor

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Si

Si

O

O

O

Si

O

Si

Si

O

O

O

O

C

O

Si

C

O

O

O

O

Si

O

Si

O

Si

and DS o = +134 kJ mol -1

Water gas

o

CaCO3 ¾¾¾¾® CaO + CO2

Strong heat

C 6 H12 O6 ¾anaerobic ¾¾¾¾¾ ® 2C 2 H5OH + 2CO2 conditions

Yeast under

CO + H2 O  CO2 + H2 CH4 + 2H2 O ® CO2 + 4H2

CaCO3 (s ) + 2HCl(aq ) ® CaCl 2 (aq ) + CO2 (g ) + H2 O( l )

Glucose

Ammonium carbamate

Urea

SiO2 + 2F2 ® SiF4 + O2

2 H2 SO4 +CaF 2 → HF +SiO  → SiF 4

+H2 O  →

 Si(OH)4 or HF +   SiO2 ⋅ 2H2 O

Silica in any form is unreactive due to high bond enthalpy of Si–O bond. SiO2 + NaOH ® ( Na 2 SiO3 )n and Na 4 SiO4

CaCO3 + CO2 + H2 O ® Ca(HCO3 )2 Soluble

Ca(OH)2 + CO2 ® CaCO3 + H2 O White precipitate

CO2 + 2NH3 ¾¾¾® NH4 CO2 NH2 ® CO(NH2 )2 + H2 O

180 ° C Pressure

6CO2 + 6H2 O ¾Sunlight ¾¾® C 6 H12 O6 + 6O2

C(s) + O2 (g ) → CO2 (g )

Carbonyl chloride (Phosphene)

COCl2

Carbonyl sulphide

CO + Cl2 ® CO2 + H2 O  H2 CO3

Producer gas

CO +

1 O2 → CO2 2 CO + S ® COS

2CO + O2 → 2CO2 ; ΔH o = −565 kJ mol −1

CH4 (g ) + 2O2 (g ) ® CO2 (g ) + 2H2 O

Air

C C + O2 + 4N 2 → CO2 + 4N 2 + → 2CO + 4 N 2

Red heat

C + H2 O ¾¾¾® CO + H2 DH = + 131 kJ mol

→ Cu(s) + CO2 (g) CuO(s) + CO(g)  -1

H ¾ COOH + H2 SO4 ® CO + H2 O

Reactions Fe 2 O3 (s) + 3CO(g) Heat  → 2Fe(s) + 3CO2 (g)

Heat

2C(s) + O2 (g) ¾¾¾ ® 2CO(g)

Preparation

(e)  Oxides of carbon and silicon

Carbon monoxide

Carbon dioxide

Chapter 17_p-Block Elements.indd 402

Silicon dioxide

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Chapter 17_p-Block Elements.indd 403

Silicones

O

Silicates

R

Si

HO

HO

R

Si

R

O

Si

O

R

Si

R

O Si

SiO4ion

Orthosilicates (neso-silicates)

R

OH

R

Si O

R

Si

R

-H2O

R

Si

R

O

OH

R

Si

OH, etc.

R

Si

R

HO

R O

Si2O67 ion

OH

O

Si

-

Si3O96- ion

-

-

-

and

-

-

12Si6O18 ion

-

Cyclic silicates

Chain silicates (SiO3 )n2 n -

(Si2O5)n2n-sheet

Sheet silicates (Si 2 O5 )n2 n −

Hydrolysis under carefully controlled conditions can produce cyclic structures, with rings containing three, four, five or six Si atoms.

°C 2 Na 2 CO3 ¾1400 ¾¾ ® CO2 + Na 2 O ¾+SiO ¾¾ ® Na 4 SiO4 , (Na 2 SiO3 )n and others

R

Si

R

Pyrosilicates (soro-silicates, disilicates)

O

R

OH + HO

HO

Si

OH +

Si

(f ) Types of silicates

HO

R

R

+2H2 O -2HCl

Si + 2CH3 Cl ¾¾¾¾¾¾¾ ® (CH3 )2 SiCl2 ¾¾¾® (CH3 )2 Si(OH)2

Cu catalyst 280 - 300 ° C

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OBJECTIVE CHEMISTRY FOR NEET

(g) Zeolites (i)  Sharing all four corners of a SiO4 tetrahedron results in a three-dimensional lattice of formula SiO2 (quartz, tridymite, cristobalite, etc.). These contain no metal ions, but three-dimensional structures can form the basis of silicate structures if there is a replacement of some of the Si4+ by Al3+ plus an additional metal ion. This gives an infinite three-dimensional lattice, and the additional cations occupy holes in the lattice. (ii) Replacing one quarter of the Si4+ in SiO2 with Al3+ gives a framework ion AlSi3O8 -. These are known as aluminosilicates. Cations are usually the larger metal ions such as K+, Na+, Ca2+ or Ba2+. Replacements of one-fourth or one-half of the Si atoms are quite common, giving structures MI [ AlSi3 O8 ] and MII [ Al 2 Si 2 O8 ]



Such replacements result in three groups of minerals: Feldspars, zeolites and ultramarines. Group 15 Elements

1. General Properties

Element

Occurrence

Nitrogen (N)

78% of the earth’s atmosphere; Nitrate deposits in desert regions; Essential constituent of proteins/ amino acids Phosphorus (P) Nucleic acids such as DNA and RNA; Fluoroapatite, hydroxyapatite and chloroapatite Arsenic (As) As sulphides Antimony (Sb) occurring as traces in other ores; Bismuth (Bi) Metallurgical byproducts from roasting sulphide ores in a smelter

Ionization energies (kJ mol−1) 1st 2nd 3rd −3, −2, –1, 1403 2857 4578 0, 1, 2, 3, 4, 5

Atomic number

Electronic configuration

Oxidation states

m.p. (°C)

b.p. (°C)

7

[He] 2s22p3

15

[Ne] 3s23p3

3, 5

1012

1897

2910

44

281

33 51 83

[Ar]3d104s24p3 [Kr]4d105s25p2 [Xe]4f  145d106s2 6p3

3, 5 3, 5 3, 5

947 834 703

1950 1590 1610

2732 2440 2467

816 631 271

615 1587 1564

–210 –195.8

(a) The atomic and ionic radii increase as one descends down the group. (b)  The ionization enthalpy of Group 15 elements is much higher than that expected because of extra stability associated with electronic configuration of half-filled p orbitals and the smaller atomic size of the elements as compared to Group 14 elements. (c)  The ionization enthalpy decreases down the Group due to increase in atomic size. The successive ionization energies increase due to greater energy required in removal of successive electrons (d) The electronegativity value decreases with increase in atomic size down the group. 2. Chemical Properties (a) Oxidation states (i)  They exhibit a maximum oxidation state of 5 towards oxygen by using all five outer electrons in forming bonds.

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THE p-BLOCK ELEMENTS

405

(ii)  The tendency for the pair of s electrons to remain inert (the inert pair effect) increases with increasing atomic weight. Thus, only the p electrons are used in bonding and trivalency results. (iii)  Valencies of 3 and 5 are shown with the halogens and with sulphur. The hydrides are trivalent. (b) Anomalous properties of Nitrogen (i)  Ability to form strong pp   pp multiple bonds itself and other elements having small size and high electronegativity. This ability to form these bonds diminishes down the group. Nitrogen forms a number of compounds like nitrates, nitrites, azides; oxides of nitrogen; cyanides; and azo and diazo compounds of which there are no P, As, Sb or Bi analogues. (ii)  The absence of lower lying 3d orbitals in nitrogen. (c) Important reactions (i)  Reactivity towards hydrogen: The elements form volatile hydrides of formula EH3. which are strong reducing agents. On descending the Group from NH3 to BiH3, we observe that • Their stability decreases. • Their reducing power increases. • The ease of replacing the hydrogen atoms by other groups such as Cl or Me decreases. • Their ability to act as lone pair donors for coordinate bond formation decreases. • The bond energy and bond angle decreases, bond length increases. • The boiling and melting points show no specific trend. (ii)  Reactivity towards oxygen: All elements of the Group form oxides of the type E2O3 and E2O5. In case of nitrogen, oxides such as N2O, NO, NO2, N2O4 are also known. (iii)  Reactivity towards halogens: All the possible trihalides of N, P, As, Sb and Bi are known. An excess of halogen gives pentahalide (EX5). •  These trihalides are predominantly covalent and, like NH3, have a tetrahedral structure with one position occupied by a lone pair. The exception includes BiF3 which is ionic and the other halides of Bi and SbF3 which are intermediate in character. •  Nitrogen is unable to form pentahalides because the second shell contains a maximum of eight electrons, that is, four bonds. The subsequent elements have suitable d orbitals, and form pentahalides. (iv) Reactivity towards metals: They react with metals to form binary compounds in which their oxidation state is −3. 3. Nitrogen (a) Dinitrogen (i)  Nitrogen is a colorless, odorless, tasteless gas which is diamagnetic and exists as diatomic molecule. (ii)  It has two stable isotopes 14N and 15N. N2 molecule contains a triple bond N   N (bond length = 1.09 pm), which is very stable with high dissociation energy. (iii)  N2 is inert at room temperature, but at elevated temperatures N2 becomes increasingly reactive. (b) Ammonia (i) Preparation • Decay of nitrogenous organic matter. NH2 CONH2 + 2H2 O ® (NH4 )2 CO3 + NH3 + H2 O + CO2 • Heating an ammonium salt with Ca(OH)2.

2NH4 Cl + Ca(OH)2 ® 2NH3 + 2H2 O + CaCl2

• Heating an ammonium salt with NaOH. This is a standard test in the laboratory for NH4 + compounds.

NH4 Cl + NaOH ® NaCl + NH3 + H2 O

• Hydrolysis of calcium cyanamide. CaNCN + 3H2 O ® 2NH3 + CaCO3 • The commercial production of ammonia is done synthetically from H2 and N2 by the Haber–Bosch process,



Chapter 17_p-Block Elements.indd 405

 N 2 + 3H2  + heat  2NH 3  4 volumes

2 volumes

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OBJECTIVE CHEMISTRY FOR NEET

The reaction is reversible, and Le Chatelier’s principle suggests that high pressure and low temperature are required to drive the reaction to the right, and thus form NH3. (ii) Properties •  The boiling and melting points of ammonia are –34.5°C and –77.8°C, respectively. NH3 has a higher boiling point and is much less volatile because it is hydrogen bonded in the liquid state. •  It dissolves very readily in water with the evolution of heat. In solution, it forms NH4OH, a weak base. •  NH3 acts as a good Lewis base and donates its lone pair of electrons to form complexes. For example, the [Co(NH3)6]3+. (iii)  Uses: As a fertilizer; making HNO3; making caprolactam; making hexamethylenediamine; transporting H2 (liquid ammonia); refrigerator coolant. (c) Oxides of nitrogen Oxides of Oxidation Shape nitrogen number N2O +1 Linear, asymmetrical molecule N

N

Bond lengths N   N = 112.6 pm N   O = 118.6 pm

O

Dinitrogen monoxide (nitrous oxide) NO

+2

N

+3

N   N = 186.4 pm in the asymmetrical form, weak

N

O N

O

O

NO2

+4

O

Nitrogen dioxide N2O4

+4

O O

N2O5

+5

Chapter 17_p-Block Elements.indd 406

N O

O

O N

O

O   N of 120 pm N   N = 164 pm

O N

O

(i)  NO + NO2 ® N 2 O3 (ii)  4NO + O2 ® 2N 2 O3 (iii)  2NO + N 2 O4 ® 2N 2 O3 N2O3 can only be obtained at low temperatures. It can be made by condensing equimolar of NO and NO2 together. It gives a blue liquid or solid which is unstable and dissociates into NO and NO2 at −30°C. Red–brown (i) Produced in large scale by oxidizing poisonous NO in the Ostwald process gas; para(ii)  2Pb(NO3 )2 ® 2PbO + 4NO2 + O2 magnetic

N O

Physical Prepartion property Stable, (i) Thermal decomposition of molten relatively ammonium nitrate at about 280°C. unreactive, NH4 NO3 ® N 2 O + 2H2 O neutral (ii) Heating a solution of NH4NO3 acidicolorless gas fied with HCl. NH4 NO3 ® N 2 O + 2H2 O Colorless (i) 3Cu + 8HNO3 → 2NO + 3Cu(NO3)2 gas, para+ 4H2O magnetic; + diamagnetic (ii) 2HNO2 + 2H → 2NO + I2 + 2H2 O in the liquid (iii) 2NaNO2 + 2FeSO4 + 3H2 SO4 ® and solid 2NO + Fe 2 (SO4 )3 + 2NaHSO 4 + 2H2 O states

O

Nitrogen monoxide (nitric oxide)

N2O3

Bond orders N   N = 2.73 and N   O = 1.61

N O

Angular Diamagnetic; with an mixed O   N   O anhydride angle of 132°. Colorless deliquescent solid, which is highly reactive; light sensitive

NO2 dimerizes into colorless N2O4

Dehydration of HNO3 with P2O5 at low temperatures treating N2O5 with O3 4HNO3 + P4 O10 → 4HPO3 + 2N 2 O5

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THE p-BLOCK ELEMENTS

407

(i)  NO readily forms coordination complexes with transition metal ions. These complexes are called nitrosyls. (ii)  NO often acts as a three-electron donor. Thus, three CO groups may be replaced by two NO groups: [Fe(CO)5 ] + 2NO ® [Fe(CO)2 ( NO)2 ] + 3CO (d) Nitric acid (i) Preparation: • HNO3 was originally made from NaNO3 or KNO3 and concentrated H2SO4. NaNO3 + H2 SO4 → NaHSO4 + HNO3

• Birkeland–Eyde process.

+O

H O

2 2 N 2 + O2 ¾spark ¾¾ ® NO ¾¾¾ ® NO2 ¾¾¾ ® 4HNO3



(Process is now obsolete, because of the high cost of electricity)

• The Ostwald process NH3 + 2O2 → HNO3 + H2O



(ii) Structure: The structure of the nitrate ion is a planar triangle 12

96

102°

pm

O

O

1p

m

H

140.6 pm

N

130°

O

(iii) Properties •  On exposure to light it turns slightly brown because of slight decomposition into NO2 and O2. 4HNO3 ® 4NO2 + O2 + 2H2 O



•  In aqueous solutions, nitric acid behaves as a strong acid and gives rise to hydronium and nitrate ions. It forms a large number of salts called nitrates, which are typically very soluble in water. HNO3 (aq) + H2 O(l) ® H3 O+ (aq) + NO3- (aq)

•  Oxidation of metals

Cold dilute < 1 M



3Cu + 8HNO3 ¾¾¾¾¾¾ ® 2NO + Cu(NO3 )2 + 4H2 O



acid Cu + 3HNO3 ¾Strong ¾¾¾ ® NO2 + Cu(NO3 )2 + H2 O

• Oxidation of non-metals

P4 + 20HNO3 ® 4H3 PO4 + 20NO2 + 4H2 O



C + 4HNO3 ® CO2 + 4NO2 + 2H2 O

(iv)  Brown ring test: The test depends on the ability of Fe+2 to reduce nitrates to nitric oxide which reacts with Fe2+ to form a brown colored complex. 2+ + ® NO + 3Fe3 + + 2H2 O   NO3 + 3Fe + 4H ¾¾ 2+ (H2 O)5 ( NO)]2 +   Fe + NO + 5H2 O ® [Fe Brown ring

4. Phosphorus (a) It is a solid at room temperature. (b) The three allotropic forms of P are white, red, and black.

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OBJECTIVE CHEMISTRY FOR NEET

Allotropic forms White phosphorus

Structure

Physical property Soft, waxy and reactive; highly toxic.

Chemiluminescence Reacts with moist air and gives out light (chemiluminescence) It ignites spontaneously in air at about 35°C forming P4O10.

Polymeric solid, which is much less reactive.

Does not ignite unless it is heated to 400°C and is not chemiluminescent.

60°

Red phosphorus (heating white P to about 250°C, or a lower temperature in sunlight)

P

P P

P P

P

P P

P

P

P P

Tetrahedral P4 molecules linked together

Black phosphorus (Heating white phosphorus under high pressure)

Thermodynamically the most stable allotrope; inert; Black, flaky, has sheet-like structure and is a good conductor of electricity; Exists in two allotropic forms, a-black P and b-black P.

Layer structure

(c) Phosphine (i) Preparation • Hydrolyzing metal phosphides such as Na3P or Ca3P2 with water or dilute acids. Ca 3 P2 + 6H2 O ® 2PH3 + 3Ca(OH)2 • Hydrolyzing white phosphorus with NaOH solution in an inert atmosphere. P4 + 3NaOH + 3H2 O ® PH3 + 3NaH2 PO2 • Reaction of phosphonium iodide with KOH. PH4 I + KOH ® PH3 + KI + H2 O (ii) Properties • Phosphine is a highly reactive and causes explosion when treated with traces of HNO3, Cl2 and Br2 vapors. • It is a very weak base, but however reacts with acids to give phosphonium compounds. PH3 + HBr → PH4 Br • Phosphine reacts with copper sulphate or mercuric chloride to form their respective phosphides. 3CuSO4 + 2PH3 → Cu 3 P2 + 3H2 SO4 • Pure PH3 is stable in air, but it catches fire when heated to about 150°C. PH3 + 2O2 ® H3 PO4 (iii) Uses: Fumigant, smoke screens, synthesis of organophosphorus compounds and some alloys. (d) Phosphorus halides: Two types of halides (X = F, Cl, Br, I), trihalides (PX3) and penathalides (PX5) formed.

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THE p-BLOCK ELEMENTS

Structure

PCl3

Cl

Reactions

P4 + 6Cl2 → 4PCl3

PCl 3 + 3H2 O → H3 PO3 + 3HCl

P4 + 8SOCl2 → 4PCl3 + 4SO2 PCl 3 + 3RCOOH ® 3RCOCl + H3 PO3 + 2S2Cl2 PCl 3 + 3ROH ® 3RCl + H3 PO3

P Cl

Preparation

409

Cl

PCl3 + Cl 2 (or SO2 Cl 2 ) ® PCl5 2PCl3 + O2 ® 2POCl3 6PCl3 + P4 O10 + 6Cl2 ® 10POCl3

PCl5

Cl 240 pm

P Cl

202

PCl5 + 4H2 O ®

4 PCl3 + Cl2 ¾CCl ¾¾ ® PCl5

Cl

H3 PO4

Phosphoric acid

+ 5HCl

PCl5 + H2 O ® POCl3 + 2HCl

pm

PCl5 ¾heat ¾® PCl3 + Cl 2

Cl

2Ag + PCl5 → 2AgCl + PCl3

Cl

PCl5 + 4RCOOH ® 4RCOCl + H3 PO4 + HCl PCl5 + 4ROH ® 4RCl + H3 PO4 + HCl 6PCl5 + P4 O10 ® 10POCl3 PCl5 + SO2 ® POCl3 + SOCl 2

5. Oxoacids of Phosphorus Orthophosphoric acids H3PO4

The acid contains three replaceable H atoms in form of three P   OH bonds, and is tribasic.

O

P OH

Oxidation state of P is (+5) compounds have oxidizing properties.

Phosphoric Acid Series

HO

Chapter 17_p-Block Elements.indd 409

OH

Pyrophosphoric acid H4P2O7 Polyphosphoric acids Tripolyphosphoric acid H5P3O10

O

HO

P

P

O

O P

O

O OH

OH OH

OH

HO

The acid contains four replaceable H atoms in form of four P   OH bonds, and is tetrabasic.

O

P

O

O OH

P

OH OH

The acid contains five replaceable H atoms in form of five P  OH bonds, and is pentabasic.

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OBJECTIVE CHEMISTRY FOR NEET

Metaphosphoric acids

O-

O O

O P

-

O

P

O O

P

O

-

O

O

P

P

O-

O

O

Metaphosphates form a family of ring compounds;

O

O-

Di-metaphosphate ion Tri-metaphosphate ion (cyclo-diphosphate) (cyclo-triphosphate) O P

O

-

O O

P

O O

P

-

O-

O O

P

O-

O O Tetra-metaphosphate ion (cyclo-tetraphosphate)

Hypophosphoric acid, H4P2O6

Tetrabasic

OH OH P

O

P

O

OH OH

These are very strong reducing agents in basic solutions. The salts are known as phosphites. The acid is monobasic and a very strong reducing agent.

H P

HO

Oxidation state (+3); Compounds are reducing agents.

Phosphorous Acid Series

Orthophosphorous acid H3PO3

OH

O

Hypophosphorous acid H3PO2

O P

H

H

OH

Pyrophosphorous acid H4P2O5

H HO

P O

It is dibasic in nature.

H O

P

OH

O

Group 16 Elements – Chalcogens 1. General Properties Atomic number

Elements

Occurrence

Oxygen (O)

It exists in the free form as 8 dioxygen molecules; Constituent of silicate minerals on the earth’s crust; Also occurs as many metal oxide ores, and as deposits of oxosalts.

Chapter 17_p-Block Elements.indd 410

Electronic configuration

Oxidation states

[He] 2s2 2p4

–2, (–1)

1st Ionization m.p. energies (°C) (kJ mol−1) 1314 –229

b.p. (°C) –183

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THE p-BLOCK ELEMENTS

Sulphur (S)

Selenium (Se) Tellurium (Te) Polonium (Po)

Sulphide ores, and as sulphates (particularly gypsum CaSO4∙2H2O Occurring as traces in ores(MgSO4⋅7H2O), galena (PbS), zinc blende (ZnS), copper pyrite (CuFeS2) Very scarce. They occur in small amounts as selenides and tellurides in sulphide ores, particularly those of Ag and Au; Pollonium occurs as the decay product of thorium and uranium minerals.

411

16

[Ne] 3s23p4

–2, (2) 4, 6

999

114

445

34

[Ar] 3d104s24p4

941

221

685

52

[Kr] 4d105s25p4 [Xe] 4f  145d106s26p4

(-2), 2 4, 6 2, 4, 6

869

452

1087

2, 4

813

254

962

84

(a)  Oxygen, sulphur, selenium and tellurium are non-metals. Collectively, they are called “the chalcogens”. (b)  The atomic and ionic radii increase down the Group. (c)  The ionization enthalpy decreases down the Group because of increase in size. The values of ionization enthalpies are however much lower than the corresponding members of Group 15 because of the stability associated with half-filled orbitals in Group 15, which make the removal of electrons difficult. (d)  The electron gain enthalpy value becomes less negative down the Group from sulphur onwards. (e)  From top to bottom in the group, the electronegativity decreases with increasing atomic size. 2. Chemical Properties (a) Oxidation states (i) The oxidation state of O is (–2) in most of its compounds. (ii) The elements S, Se and Te show oxidation states of 4 and 6, and these are more stable than the +2 state. (b) Anomalous behavior of oxygen (i)  Oxygen can use pp orbitals to form strong double bonds. The other elements can also form double bonds, but these become weaker as the atomic number increases. (ii)  Due to absence of d orbitals, oxygen cannot expand its covalency beyond four, though two is most common. In case of the other elements (S, Se, Te, Po), the valence shell can be expanded beyond four by participation of d orbitals. (c) Reactivity with hydrogen (i)  All the elements of Group 16 form covalent hydrides of the type H2E. (ii)  The hydrides decrease in stability from H2O to H2S to H2Se to H2Te because the bonding orbitals become larger and more diffuse. (iii)  The hydrides are all very weak acids and there is an increase in acidic strength from H2O to H2Te. (d)  Reactivity with oxygen: All elements form dioxides of the type EO2 and trioxides of the type EO3, where E = S, Se, Te, Po. (e) Reactivity towards halogens (i) A large number of halides of the type EX6, EX4 and EX2 are formed. (ii)  Since F is more electronegative than O, binary compounds are oxygen fluorides, whereas similar chlorine compounds are chlorine oxides. (iii)  The stability of hexahalides decreases in the order F > Cl > Br > I. (iv)  In contrast to the relatively stable hexafluorides, the tetrahalides are very sensitive to water. 3. Oxygen (a) Dioxygen (i) Preparation •  Prepared by the electrolysis of water with a trace of H2SO4 or barium hydroxide solution.

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• Thermal decomposition of KClO3: 150° C 2KClO3 ¾¾¾¾¾ ® 2KCl + 3O2 MnO2 catalyst

• Catalytic decomposition of hypochlorites: 2+

2HOCl Co  → 2HCl + O2 • Thermal decomposition of metal oxides: 2BaO2 ® 2BaO + O2 3MnO2 ® Mn3 O4 + O2 •  The decomposition of hydrogen peroxide in the presence of MnO2 and finely divided metals as catalysts 2H2 O2 ® 2H2 O + O2 (ii) Properties • Dioxygen is paramagnetic in all phases and has rather high dissociation energy of 496 kJ mol−1. • Oxygen has three isotopes: 16O (99.75%), 17O (0.0374%) and 18O (0.204%). (iii) Uses •  For respiration by both animals and plants, preparation of TiO2 from TiCl4, manufacture of HNO3 and oxirane, oxidant in rockets. (b) Simple oxides Non-metallic Oxides Metallic Oxides

These are usually covalent. They are all acidic in nature and form acids when dissolved in water. Most metal oxides are ionic and contain the O2– ion; ionic oxides typically have high melting points and are generally basic in nature.

SO3 + H2 O ® H2 SO4 B2 O3 + 3H2 O ® 2H3 BO3 Na 2 O + H2 O ® 2NaOH

(i)  Many metal oxides with formulae M2O3 and MO2, though ionic, do not react with water. Examples include Tl2O3, Bi2O3 and ThO2. These react with acids to form salts, and so are basic. (ii)  In some cases where a metal can exist in more than one oxidation state, and thus form more than one oxide, for example, CrO, Cr2O3, CrO3, PbO, PbO2, and Sb4O6, Sb4O10. (iii)  Many metals yield oxides which are amphoteric, and react with both strong acids and strong bases. Examples include BeO, Al2O3, Ga2O3, SnO and ZnO. (iv)  A few covalent oxides have no acidic or basic properties (N2O, NO, CO) and are called neutral oxides. (c) Ozone (i) Ozone O3 is the triatomic allotrope of oxygen. O O

(ii) Preparation

+

O O

-

O

+

O O

O

O

discharge 3O2 ¾Electric ¾¾¾¾ ® 2O3 DH (25°C) = +142 kJ

(iii) Properties • O3 has a characteristic sharp smell, often associated with sparking in electrical equipment. • O3 is an extremely powerful oxidizing agent, which can be attributed to the formation of nascent oxygen. • Some reactions depicting oxidizing nature of ozone are given as follows: 3PbS(s) + 4O3 (g ) → 3PbSO4 (s ) 2NO2 (g) + O3 (g) ® N 2 O5 (g) + O2 (g) • O3 adds to unsaturated organic compounds at room temperature, forming ozonides.

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THE p-BLOCK ELEMENTS

• Potassium ozonide KO3 is an orange-colored solid and contains the paramagnetic O3 − ion. 2KOH + 5O3 → 2KO3 + 5O2 + H2 O (iv) Uses: Disinfectant, mild bleaching agent and an oxidizing agent. 4. Sulphur (a) Allotropic forms Allotrope Rhombic sulphur or a-sulphur Monoclinic or b-sulphur g-Monoclinic sulphur

Melting Specific gravity Preparation point (K) (g cm–3) 385.8 2.069 Occurs naturally as large yellow crystals in volcanic areas. 393 1.94–2.01 Made by melting the rhombic sulphur in a dish and then allowing it to cool. 2.19 Made by chilling hot concentrated solutions of S in solvents such as CS2, toluene or EtOH.

Engel’s sulphur (esulphur)

Arrangement of S atoms All three forms contain puckered S8 rings with a crown conformation, and differ only in the overall packing of the rings in the crystal.

Contains S6 rings arranged in chair conformation.

(a)

S S S

105.7pm

S

102.2* S

(b)

(b)  Sulphur dioxide: SO2 gas forms discrete V-shaped molecules with bond angle is 119°30′. The molecule can be considered as resonance hybrid of the following two structures: + S -O

(i) Preparation • By burning S in air. • By burning H2S in air.

+ S O

O

O

-

S(s) + O2 (g ) → SO2 (g ) H2 S(g) + O2 (g) → SO2 (g) + H2 (g)

• By roasting various metal sulphide ores with air in smelters. 4FeS2 (s ) + 11O2 (g ) → 2Fe 2 O3 (s ) + 8SO2 (g ) • By treating a sulphite with dilute sulphuric acid. SO32− (aq ) + 2H+ (aq ) → H2 O + SO2 (g )

(ii) Properties • Sulphur dioxide is a colorless gas (b.p. −10°C, m.p. 470.5°C) and is very soluble in water. • It reacts with aqueous NaOH to form sodium sulphite and later bisulphate.

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2NaOH + SO2 ® Na 2 SO3 + H2 O Na 2 SO3 + H2 O + SO2 ® 2NaHSO3 • It reacts with chlorine in the presence of charcoal as catalyst to form sulphuryl chloride. SO2 (g) + Cl2 (g) ® SO2 Cl2 (l) • In the presence of V2O5, it reacts with oxygen to give sulphur trioxide. VO

2 5 2SO2 (g) + O2 (g) ¾¾¾ ® 2SO3 (g)

• L  aboratory test for SO2: It turns a moist acidified potassium dichromate solution green, due to the formation of Cr3+ and acts as a reducing agent. K 2 Cr2 O7 + 3SO2 + H2 SO4 ® Cr2 (SO4 )3 + K 2 SO4 + H2 O It also turns starch iodate paper blue (due to starch and I2). 2KIO3 + 5SO2 + 4H2 O ® I2 + 2KHSO4 + 3H2 SO4 (iii) U  ses: Manufacture of H2SO4, making sulphites SO32- (for bleaching, and for preserving food and wine), as a non-aqueous solvent. (c) Oxoacids of Sulphur (i) Sulphurous acid series Name Sulphurous acid Thiosulphurous acid

Formula H2SO3

Structure S HO

Di or pyrosulphurous acid

H2S2O5

+IV, −II

S HO

H2S2O4

O

HO

H2S2O2

Dithionous acid or Disulphurous acid

Oxidation number +IV

HO

S

OH

O

O

S

S

+III OH

O HO

+V, +III

O

S

S OH

O

(ii) Sulphuric acid series Name Sulphuric acid

Formula H2SO4

Structure O HO

Thiosulphuric acid

Oxidation number +VI

H2S2O3

S HO

O

+VI, −II

S HO

S

OH

O

Di or pyrosulphuric

H2S2O7

S O

Chapter 17_p-Block Elements.indd 414

OH

+VI

O

O O

S OH

O

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THE p-BLOCK ELEMENTS

415

(iii) Thionic acid series Name Dithionic acid

Formula H2S2O6

Structure S

Polythionic acid

O

O O

H2(S)n + 2O6

S O OH OH

O

+5, 0

O (S)n

S

HO

Oxidation number +5

S

OH

O

O

(iv) Peroxoacid series Name Peroxomono sulphuric acid or Caro’s acid

Formula H2SO5

Peroxo disulphuric acid or Marshall’s acid

H2S2O8

Structure

Oxidation number +6

O S

HO

O

O

O HO

OH

+6

O O

S

O

S

OH

O

O

(v) Sulphuric acid • Preparation: Contact process Air

Air

Sulphur

Sulphur Sulphur dioxide trioxide V2O5catalyst 450°C Heat is recovered

Combustion furnace Heat is recovered VO

2 5 SO2 (g) + O2 (g)  → 2SO3 (g)



Concentrated sulphuric acid Oleum Water Sulphuric acid

ΔH = − 196.6 kJ mol −1

The SO3 is passed into 98% H2SO4, forming pyrosulphuric acid H2S2O7, sometimes called oleum. The oleum is diluted with water to give concentrated sulphuric acid which is a 98% mixture with water (an 18 M solution). H2 S2 O7 + H2 O ® 2H2 SO4 • Properties  It is strongly hydrogen bonded, and in the absence of water it reacts with metals to produce H2.  Strong acidic character: K

> 10

K

= 1.2 × 10−2

a1 H2 SO4 (aq) + H2 O(l)   → H3 O+ (aq) + HSO 4− (aq) a2 HSO4− (aq) + H2 O(l)  → H3 O+ (aq) + SO24 − (aq )



Strong affinity for water: HNO3 + 2H2 SO4 → NO2+ + H3 O+ + 2HSO4−



2 SO4 Charring of organic compounds: C12 H22 O11 ¾H¾¾ ®12C + 11H2 O

S trong oxidizing agent: Cu + 2H2 SO4 (conc.) → CuSO4 + SO2 + 2H2 O • Uses: Converting calcium phosphate into superphosphate, sulphonation of fatty acids, catalyst in the production of high octane fuels, electrolyte in lead storage batteries as a non-aqueous solvent. 

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Chapter 17_p-Block Elements.indd 416

Dried-up beds of 17 inland lakes and seas containing deposits of NaCl, with smaller amounts of CaCl2, KCl and MgCl2

Sea water

Natural brine Sodium iodate (NaIO3) and sodium periodate (NaIO4)

Chlorine(Cl)

Bromine(Br)

Iodine (I)

Astatine(At)

Fluorspar or 9 fluorite (CaF2) fluoroapatite [3(Ca3(PO4)2∙CaF2]

Fluorine(F)

85

53

35

Occurrence

Atomic number

Elements

1. General Properties



[Xe] 4f14 5d10 6s2 6p5

[Kr] 4d10 5s2 5p5

[Ar] 3d10 4s2 4p5

[Ne] 3s2 3p5

[He] 2s2 2p5

–1, +1 +3, +5, +7,

–1, +1 +3, +4 +5, +6

–1, +1, +3, +4, +5, +6, +7

–1

Electronic Oxidation configuration state

1009

1143

1256

1681

1st Ionization energy (kJ mol−1)

Group 17 Elements – Halogens

–270

–296

–325

–349

–333

Electron gain enthalpy (kJ mol–1)

–274

–339

–370

–513

Hydration enthalpy of X−(kJ mol–1)

+0.3

+0.62

+1.09

+1.40

+2.87

Standard electrode potential E° (V)

b.p. (°C)

114

–7

185

60

–101 –34

–219 –188

m.p. (°C)

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THE p-BLOCK ELEMENTS

417

(a) T  he atomic and ionic radii increase from F to I as the number of quantum shells increase. (b) The atomic radii of halogens are the smallest in the respective periods due to maximum effective attraction of electrons by the nuclear charge. (c) The ionization enthalpies of the halogens decreases as the atoms increase in size. There is little tendency for the atoms to lose electrons and form positive ions. (d) The electron gain enthalpies for all the halogens are negative. (e) The electronegativity of the halogens is very high and it decreases from F to I. Fluorine is the most electronegative element in the group. (f ) The melting and boiling points of the elements increase with increase atomic number. (g) All the elements exist as diatomic molecules, and they are all colored. (h) Bond energy in the X2 molecules decrease as the atoms become larger, other than F2. The bond energy of F2 is abnormally low contributing to its very high reactivity. 2. Chemical Properties (a) Oxidation states (i) Fluorine is always univalent while in Cl, Br and I, the oxidation state may be either (–1) or (+1). (ii) Cl, Br and I also exhibit higher valencies, with oxidation numbers of (+III), (+V) and (+VII). (iii) The oxidation states (+IV) and (+VI) occur in the oxides ClO2, BrO2, Cl2O6 and BrO3. (iv) Chlorine has the highest electron gain enthalpy with a negative sign, so gaseous Cl atoms accept electrons most readily. However, Cl is not the strongest oxidizing agent. (v) Fluorine accepts electrons more readily than chlorine, so fluorine is the strongest oxidizing agent. There are two main reasons for this change of order: • F2 has a low enthalpy of dissociation (arising from the weakness of the F–F bond). • F2 has a high enthalpy of hydration (arising from the smaller size of the F- ion). (b) Anomalous behavior of fluorine: The reasons for the difference of fluorine from remaining group are: (i) The first element is smaller than the rest, and holds its electrons more firmly. (ii) It has no low-lying d orbitals which may be used for bonding. (iii) HF is a liquid, while HCl and HBr are gases. (c) Some important reactions of halogens: Reaction 2F2 + 2H2O → 4H+ + 4F− + O2 2I2 + 2H2O → 4H+ + 4X− + O2 X2 + H2O → H+ + X− + HOX X2 + H2 → 2HX

nX2 + 2M → 2MXn

3X2 + 2P → PX3 5X2 + 2P → PX5

Chapter 17_p-Block Elements.indd 417

Comment Vigorous reaction with F. I reaction in reverse direction. Cl > Br > I (F not at all). All the halogens. Reactivity towards hydrogen decreases down the group. Hydrogen and fluorine react violently. The acidic strength in aqueous solution increases from HF to HI. The stability of halides decreases from HF to HI as the bond dissociation enthalpy decreases. Most metals form halides. F is the most vigorous. The ionic character of metal (monovalent) halides decreases as follows: F > Cl > Br > I. All the halogens form trihalides. As, Sb and Bi also form trihalides. F, Cl and Br form pentahalides. AsF5, SbF5, BiF5, SbCl5.

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Reaction X2 + 2S → S2X2 2Cl2 + S → SCl4 3F2 + S → SF6

Comment Cl and Br Cl only F only

X2 + H2S → 2HX + S

All the halogens oxidize S2– to S.

3X2 + 8NH3 → N2 + 6NH4X

F, Cl and Br

X2 + X′ → 2XX′ X2 + X′X → X′X3

Interhalogen compounds form higher interhalogen compounds.

(d) Reactivity towards oxygen (i) Binary compounds of F and O are fluorides of oxygen rather than oxides of fluorine (eg. OF2 and O2F2). (ii) The other halogens are less electronegative than oxygen and thus form oxides. (iii) The general order of decreasing stability of halogen oxides is I > Cl > Br with higher oxides being more stable than the lower ones. The higher oxidation states are more stable than the lower states. (iv) Most of the chlorine oxides, for example, Cl2O, ClO2, Cl2O6 and Cl2O7 are unstable, are highly reactive and strong oxidizing agents. ClO2 is used in disinfection of water and as a bleaching agent for paper pulp and textiles. (v) The iodine oxides (I2O4, I2O5, I2O7) are the most stable. Bromine oxides (e.g., Br2O, BrO2 and BrO3) are the least stable of halogen oxides but are also strong oxidizing agents. 3. Chlorine (a) Preparation (i) By heating MnO2 with concentrated hydrochloric acid.

4HCl + MnO2 ® MnCl2 + 2H2 O + Cl2

(ii) By heating MnO2 and NaCl with concentrated hydrochloric acid.

4NaCl + MnO2 + 4H2 SO4 → MnCl2 + 4 NaHSO 4 + 2H2 O2



(iii) By the reaction of potassium permanganate with concentrated hydrochloric acid. 16HCl + KMnO4 ® 2 MnCl2 + 8H2 O + 5Cl2 + 2KCl (iv) By the electrolysis of aqueous NaCl solutions where it is liberated at anode in the manufacture of NaOH. 2NaCl + 2H2 O ¾Electrolysis ¾¾¾ ® 2NaOH + Cl2 + 2H2 (v) B  y Deacon’s process where HCl gas is oxidized by atmospheric oxygen in the presence of CuCl2 (catalyst) at 723 K. 2 4HCl + O2 ¾CuCl ¾¾ ® 2Cl2 + 2H2 O (b) Properties: Reaction with Water

Example

Bleaching action

Cl 2 + H2 O ® 2HCl + O

Metals

Cl 2 + 2Na ® 2NaCl 3Cl 2 + 2Fe ® 2FeCl3 6Cl 2 + P4 ® 4PCl3

Chapter 17_p-Block Elements.indd 418

Cl 2 + H2 O ® HCl + HOCl As a result of this reaction, chlorine water loses its yellow color on standing.

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THE p-BLOCK ELEMENTS

Non-metals

10Cl 2 + P4 ® 4PCl5 2Cl 2 + S ® SCl 4 Cl 2 + 2S ® S2 Cl2

Hydrogen and compounds containing hydrogen

Cl 2 + H2 ® 2HCl Cl2 + H2 S ® 2HCl + S

Ammonia

3Cl 2 + 8NH3 ® N 2 + 6NH4 Cl 3Cl 2 (excess) + NH3 ® 3HCl + NCl3

NaOH

Cl 2 + 2NaOH ® NaCl + NaOCl+H2 O 6NaOH + 3Cl2 ® NaClO3 + 5NaCl + 3H2 O

Slaked lime

2Cl 2 + 2Ca(OH)2 → CaCl2 + Ca(OCl)2 + 2H2 O

Carbon monoxide and sulphur dioxide

Cl 2 + CO ® COCl2

419

Cl 2 + SO2 ® SO2 Cl 2

Oxidation of sulphur dioxide Cl 2 + SO2 + 2H2 O ® H2 SO4 + 2HCl Oxidation of sulphite salts

Na 2 SO3 + Cl 2 + H2 O ® Na 2 SO4 + 2HCl 2FeSO4 + H2 SO4 + Cl2 ® Fe 2 (SO4 )3 + 2HCl

Saturated hydrocarbons

Cl 2 + CH4 ® CH3 Cl + HCl Cl 2 + C 2 H4 ® C 2 H4 Cl 2

(d) U  ses: Chlorine is used to make organic chloro compounds, for bleaching and in the manufacture of a variety of organic and inorganic chemicals like 1,2-dichloroethane and DDT, bleaching powder respectively. 4. Hydrogen Chloride (a) Preparation: HCl is made by adding concentrated H2SO4 to rock salt (NaCl). °C NaCl + H2 SO4 ¾150 ¾¾ ® HCl(g ) + NaHSO 4 °C NaCl + NaHSO4 ¾550 ¾¾ ® HCl(g ) + Na 2 SO4

(b) Properties (i) Reaction with water:

K = 107

a HCl(g) + H2 O(l) ¾¾¾ ® H3 O+ (aq) + Cl - (aq)

(ii) Decomposition salts of weaker acids. K 2 CO3 + 2HCl ® 2KCl + H2 O + CO2 KHCO3 + HCl ® KCl + H2 O + CO2 K 2 SO3 + 2HCl ® 2KCl + H2 O + SO2 (iii) Reaction with ammonia

HCl + NH3 ® NH4 Cl

(c) U  ses: The largest use is for “pickling” metals, making metal chlorides, in the manufacture of dyestuffs, and in the sugar industry. It is used in the manufacture of chlorine gas, ammonium chloride and glucose from starch.

Chapter 17_p-Block Elements.indd 419

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OBJECTIVE CHEMISTRY FOR NEET

5. Oxoacids of Halogens: Four series of oxoacids formed by halogens are HOX, HXO2, HXO3 and HXO4. -

X

X

O

O

-

X

O

O OX-

O

X

O

O O

O XO-3

XO-2

-

O

XO-4

6. Interhalogen Compounds: The halogens react with each other to form interhalogen compounds. These are divided into four types XX′, XX′3, XX′5 and XX′7. (X = halogen of larger size, X′ = halogen of smaller size). (a) Preparation: They can all be prepared by direct reaction between the halogens I2 + Cl2 liquid (excess Cl2) → (ICl3)2 (b) Properties: (i) Their melting and boiling points are a little higher than expected, and the melting and boiling points increase as the difference in electronegativity between two atoms increases. (ii) The bonds are essentially covalent because of the small electronegativity difference. (iii) These are diamagnetic in nature. (iv) The interhalogens are generally more reactive than the halogens (except F2). This is because the X   X′ bond in interhalogens is weaker than the X   X bond in the halogens. (v) The reactions of interhalogens are similar to those of the halogens. 7. Pseudohalide Ions and Pseudohalogens: Ions which consists of two or more atoms of which atleast one is nitrogen and have properties similar to those of halide ions are called pseudohalide ions. Some of these pseudohalide ions can be oxidized to form covalent dimmers comparable to halogens (X2). The best known pseudohalide ion is CN–. Some more examples of pseudohalide ions and pseudohalogens are: Pseudohalide ions CN– (Cyanide ion)

Pseudohalogens (CN)2 (Cyanogen)

OCN– (Cyanate ion) SCN– (Thiocyanate ion)

(SCN)2 (Thiocyanogen)

NCN (Cynamide ion) – 2 



Group 18 Elements – Noble Gases

1. General Properties: The elements of Group 18 have been called “the inert gases” or “the rare gases” Elements

Occurrence

Helium (He) Neon (Ne) Argon (Ar) Atmosphere Krypton (Kr) Xenon (Xe) Radon (Rn) Decay of radioactive radium and thorium minerals.

Chapter 17_p-Block Elements.indd 420

Atomic number

Electronic configuration

2 10 18 36

1s2 [He]2s2 2p6 [Ne]3s2 3p6 [Ar]3d10 4s2 4p6

54 86

[Kr]4d10 5s2 5p6 [Xe]4ƒ14 5d10 6s2 6p6

Atomic 1st Ionization radii energies (pm) (kJ mol−1) 120 2372 160 2080 191 1521 200 1351 220

1170 1037

m.p. (°C)

b.p. (°C)

–248.6 –189.4 –157.2

–269.0 –246.0 –186.0 –153.6

–111.8 –71

–108.1 –62

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THE p-BLOCK ELEMENTS

(a) H  elium has two electrons which form a complete shell 1s2. The other noble gases have a closed octet of electrons in their outer shell ns2 np6. This electronic configuration is very stable explaining their inertness. (b) These atoms have an electron affinity of zero and have very high ionization energies. (c) Under normal conditions, the noble gas atoms have little tendency to gain or lose electrons. Thus, they have high electron gain enthalpy, little tendency to form bonds, and so they exist as single atoms. (d) They have high ionization enthalpy which decreases on descending the Group due to increase in atomic size. (e) The atomic radii of the elements are all very large, and increase on descending the group. 2. Chemical Properties (a) Xenon–fluorine and Xenon–oxygen compounds Formula XeF2

Name Xenon difiuoride

Oxidation State (+2)

m.p. (°C) 129

Structure

Preparation

Linear (RnF2 and XeCl2 are similar)

F

400°C, 1 bar

2:1 mixture Xe + F2

1:5 mixture 1:20 mixture

Xe

XeF2

600°C, 7 bar

300°C, 60−70 bar

XeF4 XeF6

F

XeF4

Xenon tetrafluoride

(+4)

117

Square planar (XeCl4 is similar)

F

F Xe

F

XeF6

Xenon hexafluoride

(+6)

49.6

F

Distorted octahedron

F

F F

F

Xe F

F

XeO3

Xenon trioxide

(+6)

Explodes

Pyramidal (tetrahedral with one corner unoccupied)

3XeF4 + 6H2 O ® Xe O

O

1 2Xe + XeO3 + 12HF + 1 O2 2 XeF6 + 6H2 O ® XeO3 + 6HF

O

XeO2F2

(+6)

30.8

Trigonal bipyramid (with one position unoccupied)

F O Xe O F XeO2F2

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OBJECTIVE CHEMISTRY FOR NEET

Formula

Name

XeOF4

XeO4

Xenon tetroxide

Oxidation State (+6)

(+8)

m.p. (°C) –46

–35.9

Structure Square pyramidal (octahedral with one position unoccupied)

O

2XeF6 + SiO2 ® XeOF4 + SiF4

F

XeF6 + H2 O → XeOF4 + 2HF

F

XeF6 + 2H2 O → XeO2 F4 + 4HF

Xe F

F

Tetrahedral

O Xe O

XeO3F2

(+8)

–54.1

Trigonal bipyramid

O F

O

Xe F

Ba2[XeO6]4– Barium perxenate

Preparation

(+8)

dec. > 300

O

O O

Octahedral

(b) Reactions of XeF2, XeF4 and XeF6 Reaction with Hydrogen

Example

Oxidation properties

XeF2 + 2HCl → 2HF + Xe + Cl 2 XeF4 + 4KI → 4KF + Xe + 2I2

XeF2 + H2 → 2HF + Xe XeF4 + 2H2 → 4HF + Xe XeF6 + 3H2 → 6HF + Xe

SO + XeF2 + Ce(SO4 )3 → 2Ce IV (SO4 )2 + Xe + F2 Fluorination properties

XeF4 + 2SF4 → Xe + 2SF6 XeF4 + Pt → Xe + PtF4

Hydrolysis

2XeF2 + 2H2 O ® 2Xe + 4HF + O2

Solved Examples 1. Which of the following has the pp-dp bonding? (1) NO−3 (2) SO2− 3 2− (3) BO3− 3 (4) CO 3

O

S

O-

Solution (2) The sulphite ion exists in crystals and has a pyramidal structure, that is, tetrahedral with one position occupied by a lone pair. There are three s bonds formed between S and O in the usual way. In addition a p bond is formed by the sideways overlap of a p orbital on oxygen with a d orbital on sulphur, giving a pp-dp interaction.

Chapter 17_p-Block Elements.indd 422



O-

2. In NO−3 ion, number of the bond pair and the lone pair of electrons on nitrogen atom are (1) 2, 2 (2) 3, 1 (3) 1, 3 (4) 4, 0

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THE p-BLOCK ELEMENTS Solution

423

6. Which of the following statements is not true?

(4) The structure of nitrate ion is planar triangle. All three oxygen atoms are equivalent. There are four bond pairs and no lone pair of electrons on nitrogen atom.

(1) Among halide ions, iodide is the most powerful reducing agent. (2) Fluorine is the only halogen that does not show a variable oxidation state. (3) HOCl is a stronger acid than HOBr. (4) HF is a stronger acid than HCl.

-

O N O

O

Solution

 

3. Which of the following reactions is not feasible? (1) 2KI + Br2 → 2KBr + I2 (2) 2 KBr + I2 → 2KI + Br2 (3) 2KBr + Cl2 → 2KCl + Br2 (4) 2H2O + 2F2 → 4HF + O2

(4) The acid strength is equal to the sum of various energy terms mentioned below.

Acid strength = enthalpy of dehydration + enthalpy of dissociation + ionization energy of H+ + electron affinity of X− + enthalpy of hydration of H+ and X−



The enthalpy of dissociation of H   F bond is much stronger than the H   Cl, H   Br and H   I bonds. Also, the unexpectedly low value for electron affinity of F− also contributes, and though the enthalpy of hydration of F− is very high, it is not enough to offset these terms. Hence, HF has lower acidity than HCl.

Solution (2) A halogen that is a strong oxidizing agent will displace a halogen that has a lower oxidizing power from one of its compounds. The order of strength as oxidizing agent for halogen is F2 > Cl2 > Br2 > I2. In general, any halogen of low atomic number will oxidize halide ions of higher atomic number. Therefore, iodine cannot displace bromine from its compound.

2− 7. The oxidation stats of sulphur in the anions SO2− 3 , S 2O 4 2− and S 2O6 follow the order

(1) S 2O24− < SO23− < S 2O62− (2) SO23− < S 2O24− < S 2O62−

4. Which of the following statements is true? (1) Silicon exhibits 4 coordination numbers in its compound. (2) Bond energy of F2 is less than Cl2. (3) Mn(III) oxidation state is more stable than Mn(II) in an aqueous state. (4) Elements of 15th group show only +3 and +5 oxidation states.

(3) S 2O24− < S 2O62− < SO23− (4) S 2O62− < S 2O24− < SO23− Solution (1) The oxidation states can be calculated as:

S 2O62− : 2 x + 6( −2) = −2 ⇒ x = 5



S 2O24− : 2 x + 4( −2) = −2 ⇒ x = 3



SO23− : x + 3( −2) = −2 ⇒ x = 4

Hence, the order is S 2O24− F. On the other hand, electron gain enthalpy increases in a period, that is, from left to right as Zeff increase, thus, F > O and Cl > S. Hence, the correct order is O < S < F < Cl.

11. The correct order in which the O—O bond length increases in the following is (1) O3 < H2O2 < O2 (2) O2 < O3 < H2O2 (3) O2 < H2O2 < O3 (4) H2O2 < O2 < O3 Solution (2) Bond length is inversely proportional to the bond order, greater is the bond order lesser is the bond length and vice versa.



For option (1): The atomic radius generally decreases on moving down a group. However, atomic radius of Ga is smaller than that of Al due to ineffective shielding by intervening 3d orbitals.



For option (3): Aluminum halide is trivalent, while thallium halide is monovalent. This is because of the inert pair effect which causes the two s-electrons of the outermost electronic shell to remain non-participative in bonding. The trend increases down the group, so Ga that comes after Al in the group also forms trihalides (e.g., GaF3) but thallium (the last element in the group) has only one p-electron available for bond forming and forms TlCl only.



For option (4): The precipitate of Al(OH)3 formed does not dissolve in excess NH4OH.

13. Which of the following does not exist in free form? (1) BF3 (2) BH3 (3) BCl3 (4) BBr3 Solution

Molecule

Structure

Bond order

O2

O

2

(2) BH3 does not exist in free form because this compound is unstable at room temperature. As a result, it forms higher hydrides such as B2H6, B4H10, B10H14 among others.

Due to resonance the bond order is 1.5

14. In the following sets of reactants which two sets best exhibit the amphoteric character of Al2O3⋅xH2O?

O  +

+

O

O

O3

O

O-

-O

O 

H

H2O2

O

O

1

H 



Solution (2) The elements of Group 13 generally form covalent compounds because the sum of three ionization enthalpies for each element is high. The change from covalent to ionic happens for compounds such as AlCl3 and GaCl3 when the compounds are hydrated and the amount of hydration enthalpy is more than the ionization enthalpy. The compounds thus ionize in solution.

Therefore, the correct order is O2 < O3 < H2O2.

12. Which of the following statements is correct regarding Group 13 elements/compounds? (1) The atomic radius of gallium (Ga) is larger than that of aluminum (Al) because it lies below Al in the group.

Chapter 17_p-Block Elements.indd 424

(I) Al2O3·xH2O (s) and OH−(aq) (II) Al2O3·xH2O (s) and H2O (l) (III) Al2O3·xH2O (s) and H+ (aq) (IV) A12O3·xH2O (s) and NH3 (aq) (1) (I) and (II) (2) (I) and (III) (3) (II) and (IV) (4) (III) and (IV) Solution (2) Al(OH)3 is amphoteric its behavior can be explained when it reacts with acid and base. Al2O3⋅xH2O. exist in hydrated form. The reactions are

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THE p-BLOCK ELEMENTS

  Al2O3·xH2O + H+ → [Al(H2O)6]3+

    Base    Acid



  Al2O3·xH2O + OH– → NaAl(OH)4



      Acid   Base   Sodium aluminate



Thus, (I) and (III) set of reactions explain amphoteric character of Al2O3⋅xH2O.

15. Among the following substituted silanes, the one which will give rise to crosslinked silicone polymer on hydrolysis is (1) R4Si (2) RSiCl3 (3) R2SiCl2 (4) R3SiCl

425



Increasing acid strength: HF < HCl < HBr < HI. Higher bond dissociation energy, lower is the degree of ionization and hence weaker is the acid and the bond dissociation energy is highest for HF.



Increasing order of basic strength: SbH3 < AsH3 < PH3 < NH3. As the size of central atom increases, the electron density on the central metal atom decreases and consequently its tendency to donate a pair of electrons decreases. Hence, the basic strength decreases as we move from NH3 to BiH3.



Increasing ionization enthalpy: B < C < O < N. Nitrogen has half-filled orbitals which have extra stability so removal of electron is difficult as compared to B, C, O.

18. The molecule having smallest bond angle is

Solution

(1) NCl3 (2) AsCl3 (3) SbCl3 (4) PCl3

(2) Hydrolysis of alkyl trichlorosilane RSiCl3 gives complex cross linked polymer.

Solution

Cl R

Si

Cl

H2O

Cl

R

Si

OH OH

Condensation Polymerization

OH

R

Si

Si

O

O

Si

O

Si

O

O

Si

Si

(3) On moving down in a group from nitrogen to antimony (Sb), the size of the central metal atom increases and electronegativity decreases. As a result, of which lone pair–bond pair repulsions increase and bond angle decreases.

R



The order of increase in bond angle is:





n

16. Choose the correct increasing order of dipole moment. (1) H2O < H2S < H2Se < H2Te (2) H2Te < H2Se < H2S < H2O (3) H2Se < H2Te < H2O < H2S (4) H2S < H2O < H2Se < H2Te

NCl3>PCl3>AsCl3>SbCl3

19. The element that has the least tendency to show the inert-pair effect is (1) B (2) P (3) S (4) N

Solution

Solution

(2) Dipole moment depends upon the electronegativity of the atoms in the molecule. Oxygen is the second most electronegative element and the electronegativity decreases on moving down the group. So, the increasing order of dipole moment will be

(4) Nitrogen since it can remove s electron only by removing 3p electron. Therefore, net energy required is very high which is unable to be overcome by lattice enthalpy, which is even less than B.





H2Te < H2Se < H2S < H2O

20. PH3 and NH3 on separate reaction with bleaching powder produce respectively (1) P and N2 (3) PCl3 and N2

17. In which of the following arrangements, the sequence is not strictly according to the property written against it? (1) CO2 < SiO2 < SnO2 < PbO2: increasing oxidizing power. (2) HF < HCl < HBr < HI: increasing acid strength. (3) NH3 < PH3 < AsH3 < SbH3: increasing basic strength. (4) B < C < O < N: increasing first ionization enthalpy.

(2) PCl3 and NCl3 (4) PCl5 and NCl3

Solution (3) Reaction of NH3 with calcium hypochlorite. 4NH3 + 3Ca(OCl)2 → 3CaCl2 + N2 + H2O





Solution



Reaction of PH3 with calcium hypochlorite.

(3) Increasing oxidizing power: CO2 < SiO2 < SnO2 < PbO2. This is because Pb in (+ 4) oxidation state is the strongest oxidizing agent.





   PbO + 4HCl → PbCl + Cl2 + H2O

Chapter 17_p-Block Elements.indd 425

PH3 + Ca(OCl)2 → PCl3

21. (NH4)2CrO7 on heating liberates a gas. The same gas will be obtained by

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OBJECTIVE CHEMISTRY FOR NEET

(1) (2) (3) (4)

(1) Orthorhombic (2) b-monoclinic (3) g-monoclinic (4) Plastic sulphur

heating NH4NO2. heating NH4NO3. treating Mg3N2 with H2O. heating H2O2 on NaNO2.

Solution (1) Orthorhombic sulphur has very less ΔHf so it is very stable.

Solution

25. Which one has the highest boiling point?

(1) The reactions involved are Δ (NH 4 )2 Cr2O7 →  N 2 + Cr2O3 + 4H 2O

NH4NO2 → N2 + 2H2O







Both evolve N2 gas.

22. Which of the following statements regarding sulphur is incorrect? (1) The vapor at 200°C consists mostly of S8 rings. (2) At 600°C the gas mainly consists of S2 molecules. (3) The oxidation state of sulphur is never less than +4 in its compounds. (4) S2 molecule is paramagnetic.

(1) Ne (2) Kr (3) Xe (4) He Solution (3) London dispersion forces increases from He to Xe because the molecular mass increases, so the boiling point also increases from He to Xe. 26. Which of the following statements is correct? (1) H3PO3 is dibasic and reducing. (2) H3PO3 is tribasic and reducing. (3) H3PO3 is tribasic and non-reducing (4) H3PO3 is dibasic and non-reducing.

Solution

Solution

(3) Sulphur can exhibit a minimum oxidation state of −2. The electronic configuration of sulphur is 3s2 3p4.

(1) There are only two  OH groups in orthophosphorous acid, thus is dibasic. The oxidation number of P in this acid is + III. Whereas P may have + 5 oxidation state also. Therefore, H3PO3 can be oxidized which means H3PO3 is a reducing agent.

3s



3p

23. Which of the following is the hybridized state of oxygen atom in ozone molecule? (1) sp (2) sp2 (3) sp3 (4) sp3d Solution (2) Ozone can be represented in terms of the following resonance structures. O

O O



H

So, it can accept two more electrons to attain noble gas configuration. Hence, a minimum of −2 oxidation state is possible. For example, H2S, Na2S.

O

O

O

The structure is described as the central O atom using sp2 hybrid orbitals to bond to the terminal O atoms. The central O atom has one lone pair and the terminal O atoms have two lone pairs. The sigma bonds and lone pairs together account for 14 out of 18 electrons available in the valence shell (six electrons contributed by each O atom). The remaining four electrons are involved in p-bonding using 2pz orbitals on each of the three O atoms.

24. Which of the following allotropic forms of sulphur is thermodynamically most stable?

Chapter 17_p-Block Elements.indd 426

HO

P

OH

O

27. Which of the following isomers of phosphorus are thermodynamically least and most stable? (1) (2) (3) (4)

White (least), Red (most) Yellow (least), Red (most) Red (least), White (most) White (least), Black (most)

Solution (4) Black phosphorous is most stable it has layered structure, whereas white is the least stable and very reactive it catches fire in air (so, it is stored under water). 28. Which of the following silicates is formed when three oxygen atoms of SiO4− 4 tetrahedral units are shared? (1) Pyrosilicates (2) Cyclic silicates (3) Linear chain silicates (4) Sheet silicates Solution (4) When SiO4 4- units share three corners, the structure formed is an infinite two-dimensional sheet of empirical formula (Si 2O5 )n2n−. There are strong bonds within the Si—O sheet, but much weaker forces hold each sheet to the next one.

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THE p-BLOCK ELEMENTS

427

Solution (3) Pseudohalide ions must contain nitrogen atom (at least one) having one negative charge (in general). As SH– does not contain N-atom so it is not a pseudohalide ion. 33. Which of the following pseudohalides does not form dimer like halogen X2? (1) CN– (2) SCN– (3) SeCN– (4) OCN– Solution (4) Some of the pseudohalide ions combine to form dimers comparable with the halogen molecules X2. These include cyanogen (CN)2, thiocyanogen (SCN)2 and selenocyanogen (SeCN)2.

(Si2O5)n2n-sheet

29. Which of the following pairs of species are non-linear?

34. Which gas is evolved when urea is treated with NaOH?

(1) OCN and Br (2) (SCN)2 and I (3) NCN2– and N (4) HN3 and (CN)2 –

– 3  – 3 

− 5 

Solution

(1) Ammonia (2) Nitrogen (3) Laughing gas (4) NO Solution

(2)

(1) The reaction involved is N

S C

S

N

-

I

C and

I I

(SCN)2



NH2CONH2 + 2NaOH → 2NH3 ↑ + Na2CO3

35. O3 reacts with KOH solution to produce I

(1) O2 and KO2. (2) O3 - and K2O. (3) O2 only. (4) O2 and KO3.

I

I5-

30. A + SbF5 → B

  B + tert-butane → [tert-butyl]+ + X− + H2



Then A is (1) HCl (2) HF (3) HBr (4) HI

Solution (2) HF (A) reacts with SbF5 to produce super acid H+ [SbF6]– (B). And this super acid reacts with tertiary hydrogen atom of a hydrocarbon producing H2. 31. Boron nitride is isoelectronic with (1) C2 (2) B2 (3) N2 (4) O2 Solution (1) Boron nitride contains 12 electrons and C2 also contains 12 electrons. 32. Which of the following ions is not a pseudohalide ion? (1) CN– (2) SCN– (3) SH– (4) CNO–

Chapter 17_p-Block Elements.indd 427



Solution (4) The reaction involved is



2KOH + 5O3 → 2KO3 + 5O2 + H2O

36. The variation of the boiling points of the hydrogen halides is in the order





What explains the higher boiling point of hydrogen fluoride?

HF > HI > HBr > HCl.

(1) There is strong hydrogen bonding between HF molecules. (2) The bond energy of HF molecules is greater than in other hydrogen halides. (3) The effect of nuclear shielding is much reduced in fluorine which polarizes the HF molecule. (4) The electronegativity of fluorine is much higher than for other elements in the group. Solution (1) HF forms strong intermolecular hydrogen bonding due to high electronegativity of F. Hence, the boiling point of HF is abnormally high. Boiling points of other

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428

OBJECTIVE CHEMISTRY FOR NEET hydrogen halides gradually increase from HCl to HI due to increase in size of halogen atoms from Cl to I which further increase the magnitude of van der Waals forces.

37. Which is the strongest acid in the following? (1) HClO4 (2) H2SO3 (3) H2SO4 (4) HClO3 Solution (1) HClO4 with highest oxidation number and its conjugate base is resonance stabilized, hence it is most acidic. Cl is more electronegative than S. 38. The correct order of melting point of Group 13 elements is (1) (2) (3) (4)

B < Al < Ga < In < Tl B > Al > Ga > In > Tl B > Al > Tl > In > Ga None of the above

Solution (3) The melting points of Group 13 elements do not show regular trend due to difference in the structures of the elements. Boron has the highest melting point due to an unusual big covalent polymer structure. The melting point decreases from B to Ga and then increases. The melting point of boron is very high because it exists in both solid and liquid states. The low melting point of Ga is due to the fact that it exists as Ga2 molecules and remains as liquid even up to 2000°C. The correct order of melting point is thus, B > Al > Tl > In > Ga.

39. The bond angle is H2S molecule is (1) 109°28′ (2) 90° (3) 92° (4) 105° Solution (3) H2S has bond angle of 92°. The bond angle close to 90° suggests that almost pure p orbitals are used for bonding to hydrogen. In water, the H   O   H bond angle is close to sp3 hybridization (104.28°) due to involvement of lone pairs of electrons in bonding, resulting in distorted tetrahedral geometry. 40. Which of the following has pp– pp bonding? 2− (1) NO2− 3 (2) CO 3 2− 2− (3) BO3 (4) SO3

Solution (4) The bonds between S and O are much shorter than expected for a single bond. A s bond is formed in the usual way and in addition a p-bond is formed by sideways overlap of a p-orbital on the oxygen and a d-orbital on the sulphur due to pp–dp interaction. 41. Which of the following is the strongest reducing agent? (1) Cl2 (2) Cl– (3) Br2 (4) Br– Solution (4) Cl2 is better oxidizing agent than Br2. Thus, Br− is better reducing agent than Cl−. In general, any halogen of low atomic number will oxidize halide ions of higher atomic number.

Practice Exercises Level I Group 13 1. The BCl3 is a planar molecule, whereas NCl3 is pyramidal because (1) N   Cl bond is more covalent than B   Cl bond. (2) B   Cl bond is more polar than N   Cl bond. (3) Nitrogen atom is smaller than boron. (4) BCl3 has no lone pair but NCl3 has one lone pair of electrons. 2. Which of the following has the lowest melting point? (1) B (2) Ga (3) Al (4) Tl 3. The maximum number of atoms is lying in one plan for B2H6 is (1) 4 (2) 5 (3) 6 (4) 8

Chapter 17_p-Block Elements.indd 428

4. In diborane, the two H   B   H angles are nearly (1) 97°, 120° (2) 60°, 120° (3) 120°, 180° (4) 95°, 150° 5. How many bonds between B and N are there in inorganic benzene? (1) 0 (2) 6 (3) 3 (4) 4 6. Which one of the following is the correct statement? (1) Boric acid is a protonic acid. (2) Beryllium exhibits coordination number of six. (3) Chlorides of both beryllium and aluminum have bridged chloride structures in solid phase. (4) B2H6⋅2NH3 is known as “inorganic benzene”. 7. Among the Group 13 elements, which of the following burns in N2 at high temperature forming compound MN?

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THE p-BLOCK ELEMENTS (1) B (2) Al (3) Ga (4) Tl 8. For the construction of a bridge over the connecting channels of the sea, which kind of cement is recommended? (1) Portland cement (2) Slag cement (3) High alumina cement (4) All of these 9. Heating an aqueous solution of aluminum chloride to dryness will give (1) AlCl3 (2) Al2Cl6 (3) Al2O3 (4) Al(OH)Cl2 10. The liquefied metal expanding on solidification is (1) Ga (2) Al (3) Zn (4) Cu 11. Covalent radii for the Group 13 elements follow which order? (1) Tl < In < Ga (2) Ga < Al < B (3) B < Al < Ga (4) B > Al > In 12. Which of the following statements is correct about solid state of boric acid? (1) H-bond is present. (2) Dissolves in water to yield H+. (3) Shape of B is tetrahedral. (4) None of these. 13. Electronic configuration of Al on excluding bonded electron in aluminate ion is (1) [Ne] (2) [Ar] (3) [Ne]3s2 (4) [Ar]4s2 14. Which of the following gives B2H6 on reaction with NaBH4? (1) LiAlH4 (2) NaOH (3) I2 (4) None 15. Which of the following pairs of substances are structurally not similar? (1) (2) (3) (4)

Diamond and silicon carbide Boron nitride and graphite Borazine and benzene Diborane and hydrazine

16. In which of the following elements, +1 oxidation state is more stable than +3? (1) B (2) Al (3) Ga (4) Tl 17. Which of the following is the correct regarding B2H6? (1) (2) (3) (4)

There is direct boron   boron bond. The boron atoms are linked through hydrogen bridges. The structure is similar to C2H6. All the atoms are in one plane.

Chapter 17_p-Block Elements.indd 429

429

18. Which of the following statements incorrect regarding B2H6? (1) Banana bonds are longer but stronger than normal B   H bonds. (2) B2H6 is also known as 3c–2e compound. (3) The hybridization of B in B2H6 is sp3 while that of sp2 in BH3. (4) It cannot be prepared by reacting BF3 with LiBH3 in the presence of dry ether. 19. Which of the following statements is not correct? (1) Aluminum is among the best conductors of electricity. (2) Physical properties of aluminum are characteristic of a true metal. (3) Aluminum dissolves in aqueous sodium hydroxide solution. (4) Aluminum reacts vigorously with hot conc. HNO3, oxides of nitrogen are evolved. 20. Which of the following compounds does not exist? (1) TlF3 (2) TlCl3 (3) TlBr3 (4) TlI3 21. Which of the following Group 13 elements has the highest density? (1) Al (2) Ga (3) In (4) Tl 22. Which of the following compounds is the hardest? (1) Boron nitride (2) Boron carbide (3) Silicon boride (4) Aluminum boride 23. Which of the following is a Lewis acid? (1) AlCl3 (2) MgCl2 (3) CaCl2 (4) BaCl2 24. Which of the following oxides is acidic in nature? (1) B2O3 (2) Al2O3 (3) Ga2O3 (4) In2O3 25. The correct formula of borax is (1) Na2[B4O5(OH)4]·8H2O (2) Na2[B4O6(OH)2]·9H2O (3) Na2[B4O4(OH)6]·7H2O (4) Na2[B4O3(OH)8]·6H2O 26. Alzheimer’s disease is caused by the deposition of _________ metal. (1) Ga (2) Al (3) In (4) Tl 27. The reason for small radius of Ga compared to Al is (1) (2) (3) (4)

poor screening effect of d and f orbitals. decrease in nuclear charge. presence of higher orbitals. higher atomic number.

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OBJECTIVE CHEMISTRY FOR NEET

Group 14 28. Which gas is evolved when PbO2 is treated with concentrated HNO3? (1) NO2 (2) O2 (3) N2 (4) N2O 29. Which of the following exists as covalent crystals in the solid state? (1) Silicon (2) Sulphur (3) Phosphorus (4) Iodine 30. Which of the following is not an example of a threedimensional silicate? (1) Zeolites (2) Ultramarines (3) Feldspars (4) Beryls 31. How many oxygen atoms are shared by silicate units in forming pyrosilicates? (1) Zero (2) 1 (3) 2 (4) 3 32. Statement I: The thermal stability of hydrides of carbon family is in order:



CH4 > SiH4 > GeH4 > SnH4 > PbH4

Statement II: E—H bond dissociation enthalpies of the hydrides of carbon family decrease down group with increasing atomic size. (1) Both statement I and statement II are true and statement II is a correct explanation for statement I. (2) Both statement I and statement II are true, but statement II is not a correct explanation for statement I. (3) Statement I is true, but statement II is false. (4) Both statement I and statement II are false. 33. The most basic oxide of elements in Group 14 of the periodic table is (1) SiO2 (2) GeO (3) SnO2 (4) PbO 34. When tin is treated with concentrated nitric acid, (1) (2) (3) (4)

it is converted into stannous nitrate. it becomes passive. it converted into stannic nitrate. it is converted into metastannic acid.

35. Which of the following is most stable M   F bond, where M is Group 14 element? (1) PbF4 (2) SnF4 (3) CF4 (4) SiF4 36. In which of the following reactions is PbSO4 formed? (1) PbO2 + SO2 (2) PbS + O3 (3) PbS + H2O2 (4) All of the above.

Chapter 17_p-Block Elements.indd 430

37. “Lead pencil” contains (1) PbS. (2) FeS. (3) graphite. (4) Pb. 38. Which of the following has the properties of metals as well as non-metals? (1) C (2) Pb (3) Sn (4) Ge 39. Which of the following oxides has three-dimensional structure? (1) CO (2) CO2 (3) SiO2 (4) SO2 40. Silicon tetrafluoride reacts with HF to form (1) SiF2 (2) SiF6 4(3) SiF6 2- (4) H2[SiF4] 41. In which of the following carbides, the carbon atom is tetrahedrally hybridized? (1) SiC (2) Be2C (3) Al4C3 (4) All of these 42. Which of the following elements has the highest density? (1) Pb (2) C (3) Si (4) Ge 43. The most ionic among the following is (1) SnCl2 (2) SnCl4 (3) SnCl6 (4) CCl4 44. A colorless gas which burns with blue flame and reduces hot copper oxide to copper is (1) CO (2) NO2 (3) SO2 (4) CO2 45. Silicon has a strong tendency to form polymers like silicones. The chain length of silicone polymer can be controlled by adding (1) MeSiCl3 (2) Me2SiCl2 (3) Me3SiCl (4) Me4Si 46. Quartz is extensively used as a piezoelectric material, it contains (1) Pb (2) Si (3) Ti (4) Sn 47. The most commonly used reducing agent is (1) AlCl3 (2) PbCl2 (3) SnCl4 (4) SnCl2 48. Dry ice is (1) solid NH3 (2) solid SO2 (3) solid CO2 (4) solid N2

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THE p-BLOCK ELEMENTS 49. Which of the following statements is correct? (1) Fullerenes have dangling bonds. (2) Fullerenes are ball-like molecules. (3) Graphite is thermodynamically most stable allotrope of carbon. (4) Graphite is slippery and hard and therefore used as a dry lubricant in Machines. 50. Identify the correct resonance structures of carbon dioxide from the ones given below: (1) O   C   O (2) O   C O (3) –O   C   O+ (4) –O   C   O+

Group 15 51. The percentage of p-character in the orbitals forming P   P bonds in P4 is (1) 25 (2) 33 (3) 50 (4) 75 52. Select the true statement for the acids of phosphorus, H3PO2, H3PO3 and H3PO4. (1) (2) (3) (4)

The order of their acidity is H3PO4 > H3PO3 > H3PO2. All of them are reducing in nature. All of them are tribasic acids. The geometry of phosphorus is tetrahedral in all the three.

53. Which of the following is a solid? (1) Nitrogen trioxide (2) Nitrogen pentoxide (3) Dinitrogen tetroxide (4) Nitric oxide 54. The acid strength of the following hydrides increases in the order: (1) SiH4 < PH3 < H2S < HCl (2) H2S < PH3 < HCl < SiH4 (3) PH3 < H2S < HCl < SiH4 (4) H2S < SiH4 < PH3 < HCl 55. N2 forms NCl3 whereas P can form both PCl3 and PCl5. Why? (1) P has d orbitals which can be used for bonding but N2 does not have d orbitals. (2) N atom is larger than P in size. (3) P is more reactive towards Cl than N. (4) The size of N is comparable to Cl while P size is greater than that of Cl. 56. What may be expected to happen when phosphine gas is mixed with chlorine gas? (1) The mixture only cools down. (2) PCl3 and HCl are formed and the mixture warms up. (3) PCl5 and HCl are formed and the mixture cools down. (4) PH3⋅Cl2 is formed with warming up. 57. The number of P   O   P bonds in cyclic metaphosphoric acid is (1) zero. (2) two. (3) three. (4) four.

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431

58. Which of the following statements is not correct regarding PH3 molecule? (1) It is extremely toxic and colorless gas. (2) It has rotten fish smell. (3) It is formed by hydrolysis of Na3P. (4) PH3 is highly soluble in water. 59. Which of the following is tetrabasic acid? (1) (2) (3) (4)

Metaphosphoric acid Orthophosphoric acid Hypophosphoric acid Hypophosphorous acid

60. Out of the following, which does not liberate N2 gas on heating? (1) NaNO2 + NH4Cl (2) Ba(N3)2 (3) O3 + NO (4) (NH4)2 Cr2O7 61. PCl3 on hydrolysis gives fumes of (1) H3PO3 + HCl (2) H3PO4 + HCl (3) H3PO2 and H3PO3 (4) H3PO2 + HCl 62. Zn reacts with dilute HNO3 to give (1) NO2 (2) NO (3) N2O (4) NH3 63. Which of the following allotrope of phosphorous has tetrahedral discrete unit? (1) (2) (3) (4)

Red black P White P Red P Red, black and white P

64. Which of the following tendencies remains unchanged on going down in the nitrogen family (group V A)? (1) (2) (3) (4)

Highest oxidation state Non-metallic character Stability of hydrides Physical state

65. The element which forms oxides in all the oxidation states from +1 to +5 is (1) N (2) P (3) As (4) Sb 66. Each of the following is true for white and red phosphorus except that they (1) (2) (3) (4)

are both soluble in CS2. can be oxidized by heating in air. consists of same kind of atoms. can be converted into one another.

67. White phosphorus has (1) P   P   P bond angle is 75°. (2) four P   P single bonds. (3) polymeric structure. (4) six P   P single bonds.

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68. Which of the following does not show allotropy?

(1) N2O, PbO (2) NO2, PbO (3) NO, PbO (4) NO, PbO2

(1) N (2) P (3) Bi (4) As 78. Which of the following elements does not show allotropy? 69. Molecule with three- electron bond is (1) Cl2 (2) NO (3) H2O (4) Cl2O 70. NO2 is (1) (2) (3) (4)

80. Which of the following statements is wrong?

71. When HNO3 is dropped into the palm and washed with water, it turns into yellow. It shows the presence of (1) NO2 (2) N2O (3) NO (4) N2O5 72. The commercial name for peroxy disulphuric acid is (1) Caro’s acid. (2) Marshall’s acid. (3) sulphuric acid crystals. (4) sulphurous acid 73. Which of the following elements can be involved in pp– dp bonding? (1) Carbon (2) Nitrogen (3) Phosphorous (4) Boron 74. Bond dissociation enthalpy of E   H (E = element) bonds is given below. Which of the compounds will act as strongest reducing agent? Compound



Δdiss (E—H)/kJ mol 389

NH3 –1

PH3

AsH3

SbH3

322

297

255

(1) NH3 (2) PH3 (3) AsH3 (4) SbH3 75. On heating with concentrated NaOH solution in an inert atmosphere of CO2, white phosphorus gives a gas. Which of the following statement is incorrect about the gas? (1) It is highly poisonous and has smell like rotten fish. (2) It’s solution in water decomposes in the presence of light. (3) It is more basic than NH3. (4) It is less basic than NH3. 76. Which of the following acids forms three series of salts? (1) H3PO2 (2) H3BO3 (3) H3PO4 (4) H3PO3 77. On heating lead nitrate forms oxides of nitrogen and lead. The oxides formed are

Chapter 17_p-Block Elements.indd 432

79. Maximum covalency of nitrogen is (1) 3 (2) 5 (3) 4 (4) 6

anhydride of HNO2. anhydride of HNO3. anhydride of H2N2O4. mixed anhydride of HNO2 and HNO3.



(1) Nitrogen (2) Bismuth (3) Antimony (4) Arsenic

(1) Single N   N bond is stronger than the single P   P bond. (2) PH3 can act as a ligand in the formation of coordination compound with transition elements. (3) NO2 is paramagnetic in nature. (4) Covalency of nitrogen in N2O5 is four. 81. Elements of Group 15 form compounds in + 5 oxidation state. However, bismuth forms only one well characterized compound in +5 oxidation state. The compound is (1) Bi2O5 (2) BiF5 (3) BiCl5 (4) Bi2S5 82. On heating ammonium dichromate and barium azide separately we get (1) N2 in both cases. (2) N2 with ammonium dichromate and NO with barium azide. (3) N2O with ammonium dichromate and N2 with barium azide. (4) N2O with ammonium dichromate and NO2 with barium azide. 83. The oxidation state of central atom in the anion of compound NaH2PO2 will be (1) +3 (2) +5 (3) +1 (4) –3 84. In solid state, PCl5 is a (1) covalent solid. (2) octahedral structure. (3) ionic solid with [PCl6]+ octahedral and [PCl4]– tetrahedral. (4) ionic solid with [PCl4]+ tetrahedral and [PCl6]– octahedral. 85. In a cyclotrimetaphosphoric acid molecule, how many single and double bonds are present? (1) (2) (3) (4)

3 double bonds; 9 single bonds. 6 double bonds; 6 single bonds. 3 double bonds; 12 single bonds. Zero double bonds; 12 single bonds.

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THE p-BLOCK ELEMENTS

Group 16

  97. Ozone is

86. Ozone with KI solution produces (1) Cl2 (2) I2 (3) HI (4) IO3 87. Which of the following species does not contain S   S linkage? (1) S2O2 2– (2) S2O3 2– (3) S2O5 2– (4) SO3 2– 88. Hydrolysis of one mole of peroxydisulphuric acid produces (1) two moles of sulphuric acid. (2) two moles of peroxymonosulphuric acid. (3) one mole of sulphuric acid and one mole of peroxymonosulphuric acid. (4) one mole of sulphuric acid, one mole of peroxymonosulphuric acid and one mole of hydrogen peroxide. 89. The number of S   S bonds in sulphur trioxide trimer S3O9 is (1) three. (2) two. (3) one. (4) zero. 90. When S2O8 2– oxidize Fe2+, then the product formed is (1) S 2O62- (2) SO24 (3) S 2O32- (4) SO52− 91. Rhombic sulphur is soluble in

92. Which of the following is stronger reducing agent? (2) SO2 (4) TeO2

93. Caro’s acid is (1) H2SO5 (2) H2SO3 (3) H3S2O5 (4) H2S2O8 94. The most abundant and common oxidation state of sulphur is (1) −2 (2) +4 (3) +2 (4) +6   95. The compounds of ‘S’ obtained by reaction of S and conc. hot KOH when reacted separately with dil. HCl produce (1) (2) (3) (4)

different gases, SO2 and H2S. same gas SO2. S is obtained back. same gas H2S.

  96. Which of the following is thermally the most stable? (1) H2S (2) H2O (3) H2Se (4) H2Te

Chapter 17_p-Block Elements.indd 433

(1) an unstable, dark blue, diamagnetic gas. (2) an unstable, dark blue, paramagnetic gas. (3) a stable, dark blue, paramagnetic gas. (4) found in the upper atmosphere where it absorbs UV radiation.   98. The oxidation states exhibited by oxygen in its compounds are (1) −2 (2) −1 (3) +2 (4) All of these   99. Which of the following are peroxoacids of sulphur? (1) H2S2O6 (2) H2SO3 (3) H2SO5 (4) H2S2O3 100. In the upper layer of atmosphere, ozone is formed by (1) (2) (3) (4)

combination of oxygen. action of electric discharge on oxygen. action of UV rays on oxygen. None of these

101. On addition of conc. H2SO4 to a chloride salt, colorless fumes are evolved but in case of iodide salt, violet fumes come out. This is because (1) H2SO4 reduces HI to I2. (2) HI is of violet color. (3) HI gets oxidized to I2. (4) HI changes to HIO3. 102. Which of the following is not tetrahedral in shape?

(1) CS2. (2) benzene. (3) alcohol. (4) ether.

(1) SO3 (3) SeO3

433

(1) NH +4 (2) SiCl4 (3) SF4 (4) SO2− 4 103. Which of the following element is oxidized by conc. H2SO4 into two gaseous products? (1) Cu (2) S (3) C (4) Zn 104. Which of the following statements is correct? (1) All the elements of Group 16 form monoxides. (2) SeO2 is a volatile solid. (3) TeO2 is a volatile solid. (4) PoO2 is a volatile solid.

Group 17 105. Identify the incorrect statement among the following: (1) Br2 reacts with hot and strong NaOH solution to give NaBr, NaBrO3 and H2O. (2) Ozone reacts with SO2 to give SO3. (3) Silicon reacts with NaOH(aq) in the presence of air to give Na2SiO3 and H2O. (4) Cl2 reacts with excess of NH3 to give N2 and HCl.

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106. The solubility of iodine in water increases in the presence of (1) alcohol. (2) chloroform. (3) sodium hydroxide. (4) potassium iodide.

(1) XeF4 (2) HeF4 (3) SF4 (4) CF4 118. Which element out of He, Ar, Kr and Xe forms least number of compounds?

107. F2 is formed by reacting K2MnF6 with (1) SbF5 (3) KSbF6

(2) MnF3 (4) MnF4

108. Incorrect order of bond energy is

(1) He (2) Ar (3) Kr (4) Xe 119. Who among the following first prepared a stable compound of noble gas? (1) Rutherford (2) Rayleigh (3) Ramsay (4) Neil Barlett

(1) I   Cl < I2 (2) Br   F > F   F (3) Br   F < I   I (4) Br   Cl < Cl   Cl 109. Salicylic acid reacts with ICl vapor to produce (1) (2) (3) (4)

iodinated product. chlorinated product. mixture of both. virtually no reaction.

120. In XeO3 and XeF6 the oxidation state of Xe is (1) +4 (2) +6 (3) +1 (4) +3 121. Amongst the following molecules (I) XeO3 (II) XeOF4 (III) XeF6

110. Which of the following pseudohalides best resembles Cl–, Br– and I–? (1) CN– (2) N −3 (3) ONC– (4) Both N −3 and ONC– 111. The outer configuration of halide ion is (1) ns2 (2) ns2 np5 (3) ns2 np6 (4) ns2 np4 112. In the manufacture of bromine from sea water, the mother liquor containing bromides is treated with (1) CO2 (2) Cl2 (3) I2 (3) SO2 113. In the preparation of chlorine from HCl, MnO2 acts as (1) oxidizing agent. (2) reducing agent. (3) catalytic agent. (4) dehydrating agent. 114. In halogens, covalent bond always has (1) p-bonding. (2) s-bonding. (3) s–p-bonding. (4) None of these. 115. Bromine is obtained on commercial scale from (1) caliche. (2) carnellite. (3) common salt. (4) cryolite.

Group 18 116. Argon is used in arc welding because of its (1) low relativity with metal. (2) ability to lower the melting point of metal (3) flammability. (4) high calorific value. 117. Among the fluorides below, the one which does not exist is

Chapter 17_p-Block Elements.indd 434



those having same number of lone pairs on Xe are (1) (I) and (II) only (2) (I) and (III) only (3) (II) and (III) only (4) (I), (II) and (III)

122. First compound for Xe synthesized was (1) [XeF+] [Xe PtF5]− (2) [XeO2] (3) Xe[PtF6] (4) O2[XeF6] 123. Xenon difluoride is (1) linear. (2) angular. (3) trigonal. (4) pyramidal. 124. XeF4 exists as _______ under ordinary atmospheric conditions. (1) solid (2) liquid (3) gas (4) semi-solid 125. Electron affinity of noble gas element is (1) very high. (2) high. (3) low. (4) zero. 126. The inert gas found most abundantly in the atmosphere is (1) Ar (2) Kr (3) He (4) Xe 127. In XeO3, the Xe is (1) sp3 hybridized. (2) sp3d hybridized. 3 3 (3) sp d hybridized. (4) sp3d2 hybridized. 128. Which of the following substance is a red crystalline solid? (1) XeF2 (2) XeF6 (3) XeF4 (4) Xe[PtF6]

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THE p-BLOCK ELEMENTS

Level II Group 13 1. The compound TlI3 consists of ____________ ions. (1) Tl3+ (2) I(3) I3 - (4) Both (1) and (3) 2. Borax is used for ____________ resistant glazed coating to earthen wares. (1) heat (2) scratch (3) stain (4) all of these 3. Aqueous solution of borax is (1) neutral. (2) acidic. (3) alkaline. (4) strongly acidic. 4. Number of hydroxyl group attached to boron in borax are (1) four. (2) five. (3) six. (4) ten. 5. Anhydrous AlCl3 produces fumes in the air because of (1) oxidation. (2) reduction. (3) dimerization. (4) hydrolysis. 6. The product obtained when one mole of diborane reacts with two moles of NH3 at high temperature is (1) B2H6⋅2NH3 (3) (BN)x

(2) B3N3H6 (4) [BH2(NH3)2]+BH4−

7. Diborane cannot be obtained from (1) Na2B4O7 + HCl (2) NaBH4 + I2 (3) BF3 + LiAlH4 (4) BF3 + NaH 8. The self-protective oxide film on Al can be removed by (1) reacting with Cl2. (2) adding conc. HNO3. (3) amalgamating with Hg. (4) boiling with water. 9. The chemical name of borax is (1) (2) (3) (4)

sodium orthoborate decahydrate. sodium metaborate decahydrate. sodium tetraborate decahydrate. sodium borate decahydrate.

10. The halides of Group 13 elements are (1) Lewis acids. (2) Lewis bases. (3) Bronsted acids. (4) Arrhenius acids. 11. Boron nitride reacts with caustic alkali to give (1) NH3 (2) N2O (3) Na2BO2 (4) NO2 12. The element which exists in liquid state for a wide range of temperature and can be used for measuring high temperature is (1) B (2) Al (3) Ga (4) In

Chapter 17_p-Block Elements.indd 435

435

13. The exhibition of highest co-ordination number depends on the availability of vacant orbitals in the central atom. Which of the following elements is not likely to act as central atom in MF63−? (1) B (2) Al (3) Ga (4) In 14. When borax is heated with ethyl alcohol and concentrated sulphuric acid, vapors of triethylborate are produced. When ignited, these vapors burn with a __________ flame. (1) blue edged (2) green edged (3) red edged (4) brown edged 15. Statement I: Boron always forms covalent bond. Statement II: The small size of B3+ favours formation of covalent bond. (1) Both statement I and statement II are true and statement II is a correct explanation for statement I. (2) Both statement I and statement II are true, but statement II is not a correct explanation for statement I. (3) Statement I is true, but statement II is false. (4) Both statement I and statement II are false.

Group 14 16. The stability of dihalides of Si, Ge, Sn and Pb increases steadily in the sequence: (1) (2) (3) (4)

PbX 2  SnX 2  GeX 2  SiX 2 GeX 2  SiX 2  SnX 2  PbX 2 SiX 2  GeX 2  PbX 2  SnX 2 SiX 2  GeX 2  SnX 2  PbX 2

17. Butter of tin is (1) SnCl2·5H2O (2) SnCl2·2H2O (3) SnCl4·4H2O (4) SnCl4·5H2O 18. The water repellant characteristics of silicone is because (1) Si-atom has no vacant d-orbital available for nucleophillic attack by water molecule. (2) Silicone chain is surrounded by organic side groups and looks like an alkane from outside. (3) The average chain lengths of silicones are very high. (4) Low thermal stability. 19. In silicon dioxide (1) each silicon atom is surrounded by four oxygen atoms and each oxygen atom is bonded to two silicon atoms. (2) each silicon atom is surrounded by two oxygen atoms and each oxygen atom is bonded to two silicon atoms. (3) silicon atom is bonded to two oxygen atoms. (4) there are double bonds between silicon and oxygen atoms.

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436

OBJECTIVE CHEMISTRY FOR NEET

20. Which of the following gives propyne on hydrolysis? (1) Al4C3 (2) Mg2C3 (3) B4C (4) La4C3 21. Which of the following is an organo silicon polymer? (1) Silica (2) Orthosilicic acid (3) Silicon carbide (4) Silicic acid 22. PbCl2 is more ionic than PbO2 because (1) (2) (3) (4)

the radius of Pb2+ is more than that of Pb4+. of inert pair effect. chlorine is more electronegative than oxygen. chlorine atom is smaller than oxygen atom.

23. Which of the halides of carbon is the least stable? (1) Iodide (2) Bromide (3) Chloride (4) Fluoride 24. Which of the following does not exist? (1) [CCl6]2− (2) [SiCl6]2− (3) [GeF6]2− (4) [SnCl6]2− 25. Hydrolysis of dimethyldichlorosilane, (CH3)2SiCl2, followed by condensation polymerization yields straight-chain polymer of (1)

(2)

O

Si

O

O

Si

Si

O

O

O

O

CH3

(3)

O

Si

O

CH3

O

Si

Si

O

CH3

Si

O

CH3 CH3

26. On the basis of structure of graphite, which of the following is correct? (1) It is a paramagnetic substance. (2) It behaves like metallic as well as semiconductor. (3) C   C bond length in graphite is more than the diamond. (4) It is thermodynamically more stable as well as denser than diamond. 27. The most reactive form of carbon is (1) charcoal. (2) fullerene. (3) graphite. (4) soot. 28. The unique ability of carbon to form C   C, C   C, C   S, C   N bonds is due to its ability to form

Chapter 17_p-Block Elements.indd 436

29. Catenation i.e., linking of similar atoms depends on size and electronic configuration of atoms. The tendency of catenation in Group 14 elements follows the order: (1) (2) (3) (4)

C > Si > Ge > Sn C >> Si > Ge ≈ Sn Si > C > Sn > Ge Ge > Sn > Si > C

30. The linear shape of CO2 is due to (1) (2) (3) (4)

sp3 hybridization of carbon. sp hybridization of carbon. pp   dp bonding between carbon and oxygen. sp2 hybridization of carbon.

31. Me3SiCl is used during polymerization of organo silicones because (1) the chain length of organo silicone polymers can be increased by adding Me3SiCl. (2) Me3SiCl blocks the end terminal of silicone polymer. (3) Me3SiCl improves the quality and yield of the polymer. (4) Me3SiCl acts as a catalyst during polymerization.

32. Which of the following statements is wrong?

Si

CH3 CH3

(4)

pp   pp multiple bonds. pp   dp multiple bonds. dp   dp multiple bonds. None of these

Group 15

CH3 O

(1) (2) (3) (4)

(1) Nitrogen cannot form dp   pp bond. (2) Single N   N bond is weaker than the single P   P bond. (3) N2O4 has two resonance structures. (4) The stability of hydrides increases from NH3 to BiH3 in Group 15 of the periodic table. 33. Which one of the following does not have a pyramidal shape? (1) (CH3)3 N (2) (SiH3)3N (3) P(CH3)3 (4) P(SiH3)3 34. In NH3 and PH3, the common is (1) odor. (2) combustility. (3) basic nature. (4) none of these. 35. Choose the correct options from the following orders: (1) (2) (3) (4)

Basicity order: NH3 > NH2   NH2 > NH2OH > NF3 Melting point order: NH3 < SbH3 < AsH3 < PH3 Boiling point order: NH3 > SbH3 > AsH3 > PH3 Thermal stability: NH3 < PH3 < AsH3 < SbH3

36. Which of the following compounds has a P   P bond? (1) H3PO4 (2) H4P2O6 (3) H4P2O7 (4) (HPO3)3

1/5/2018 5:23:59 PM

THE p-BLOCK ELEMENTS 37. Which acid on heating produces phosphine? (1) (2) (3) (4)

38. The number of P   O   P bridges in the structure of phosphorous pentoxide and phosphorous trioxide are respectively (1) 6, 6 (2) 5, 5 (3) 5, 6 (4) 6, 5 39. How many peroxy linkages are present in pyropho­ sphoric acid?

40. Which of the following statements is incorrect? (1) Solid PCl5 exists as tetrahedral [PCl4] and octahedral [PCl6]− ions. (2) Oxides of phosphorus, P2O3 and P2O5 exist as monomers. (3) Solid PCl5 exists as [PCl4]+[PCl6]−. (4) Solid N2O5 exists as NO2+NO3−. +

41. Which of the following is the most acidic? (1) As2O3 (2) P2O3 (3) Sb2O3 (4) Bi2O3 42. Which of the following is correct about the bond angle in the following groups of hydrides? (1) NH3 > SbH3 > AsH3 > PH3 (2) NH3 > PH3 > AsH3 > SbH3 (3) SbH3 > AsH3 > PH3 > NH3 (4) AsH3 > SbH3 > NH3 > PH3 43. How NH3 is generally manufactured for use in fertilizers? (1) By reducing the by-product HNO3. (2) By passing a mixture of N2 and H2 under high pressure and moderate temperature over a catalyst. (3) By passing an electric discharge in the mixture of N2 and H2. (4) 2NH4Cl + Ca(OH)2 → CaCl2 + 2NH3 + 2H2O 44. Which of these statements is not true? (1) NO+ is isoelectronic with O2. (2) B is always covalent in its compounds (3) In aqueous solution, the Tl (I) ion is much more stable than Tl (III) (4) LiAlH4 is a versatile reducing agent in organic synthesis. 45. In the preparation of HNO3, we get NO gas by catalytic oxidation of ammonia. The moles of NO produced by the oxidation of two moles of NH3 will be

Chapter 17_p-Block Elements.indd 437

low oxidation state of phosphorus. presence of two    OH groups and one P   H bond. presence of one    OH group and two P   H bonds. high electron gain enthalpy of phosphorus.

47. In compounds of type ECl3, where E = B, P, As or Bi, the angles Cl   E   Cl for different E are in the order (1) B > P = As = Bi (2) B > P > As > Bi (3) B < P = As = Bi (4) B < P < As < Bi

Group 16 48. Aqueous solution of Na2S2O3 on reaction with Cl2 gives (1) Na2S4O6 (2) NaHSO4 (3) NaCl (4) NaOH

(1) 0 (2) 1 (3) 2 (4) 3

(1) 2 (2) 3 (3) 4 (4) 6

46. Strong reducing behavior of H3PO2 is due to (1) (2) (3) (4)

Phosphoric acid Phosphorous acid Peroxymonophosphoric acid Metaphosphoric acid

437

49. Which of the following is not a peroxy acid? (1) Caro’s acid (2) Marshall’s acid (3) Thiosulphuric acid (4) Sulphurous acid 50. The reagent used to distinguish between H2O2 and O3 is (1) PbS. (2) starch and iodine. (3) KMnO4. (4) bleaching powder. 51. Which of the following has the smallest bond angle? (1) H2S (2) H2O (3) H2Se (4) H2Te 52. SO2 is passed through strongly acidic solution of SnCl2. The precipitate obtained is consisting of S and oxidation state of S in the precipitate is (1) –1 (2) zero. (3) –2 (4) precipitate does not contain S. 53. Ozone can be detected by using (1) Ag (2) AgCl (3) Hg (4) SO2 54. Bleaching action of SO2 is due to the (1) reduction. (2) oxidation. (3) hydrolysis. (4) its acidic nature. 55. Point out in which of the following properties, oxygen differs from the rest of the members of its family? (1) High value of ionization energies. (2) Oxidation states (2, 4, 6) (3) Polymorphism (4) Formation of hydrides 56. Which of the following gases causes tailing of mercury? (1) Ozone (2) Nitrogen (3) Sulphur dioxide (4) Hydrogen sulphide 57. Copper turnings when heated with concentrated sulphuric acid will give

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438

OBJECTIVE CHEMISTRY FOR NEET (1) SO2 (2) SO3 (3) H2S (4) O2

67. Which of the following is formed when ICl3 is treated with water? (1) HIO (2) HIO2 (3) HIO3 (4) HIO4

Group 17 58. Which of the following options are true (T) and which are false (F)? (I) Ionic mobility is the highest for I in water as compared to other halides. (II) Stability order is: Cl3– > Br3– > I3– (III) Reactivity order is: F < Cl< Br < I. (IV) Oxidizing power order is: F2 < Cl2 < Br2 < I2. –

70. In the preparation of compounds of Xe, Bartlett had taken O+2 PtF6− as a base compound. This is because

(1) IF7 (2) (CN)2 (3) ICl2 (4) I -3 60. Which of the following polyhalides is not known to exist? (1) Br3− (2) Cl −3 (3) F3− (4) I −3 61. Which of the following is the least reactive interhalogen compound? (1) BrF3 (2) BrF (3) IF5 (4) ClF3 (1) HCl, HClO (2) HClO (3) HCl (4) HCl, H2SO4 63. Affinity for hydrogen decreases in the group from fluorine to iodine. Which of the halogen acids should have highest bond dissociation enthalpy?

71. Which of the following noble gases is mixed with oxygen in scuba diver’s cylinder? (1) He (2) Ne (3) Ar (4) Kr

(1) XeF2, XeF4, XeF3 (2) XeF2, XeF4, XeF6 (3) XeF2, XeF3, XeF6 (4) XeF2, XeF4, XeF5 73. Which of the following gases is/are called rare gas? (1) Ne (2) He (3) Kr (4) All of these

(1) He (2) Ar (3) Ne (4) Xe

64. White enamel of our teeth is

75. In which of the following noble gases, the van der Waals force is the strongest?

(1) Ca3(PO4)2 (2) CaF2 (3) CaCl2 (4) CaBr2 65. Reduction potentials of some ions are given below. Arrange them in decreasing order of oxidizing power. BrO4−

Reduction potential E o/V E o = 1.19 V E o = 1.65 V E o = 1.74 V (1) ClO−4 > IO−4 > BrO−4 (2) IO−4 > BrO−4 > ClO−4 (3) BrO−4 > IO−4 > ClO−4 (4) BrO−4 > ClO−4 > IO−4 66. Which of the following is formed when HgO is treated with chlorine?

Chapter 17_p-Block Elements.indd 438

both O2 and Xe have same size. both O2 and Xe have same electron gain enthalpy. both O2 and Xe have almost same ionization enthalpy. both Xe and O2 are gases.

74. Which noble gas is most soluble in water?

(1) HF (2) HCl (3) HBr (4) HI

(1) ClO2 (2) Cl2O (3) Cl2O6 (4) Cl2O7

(1) (2) (3) (4)

72. Which of the following series of fluorides is known?

62. Bleaching powder reacts with water to give

IO4−

(1) XeOF4 is formed. (2) XeO2F2 is formed. (3) It is a redox reaction. (4) XeO3 is formed. (1) Kr⋅4H2O (2) Kr⋅6H2O (3) Xe⋅4H2O (4) Xe⋅6H2O

59. Which of the following is a pseudohalogen?

ClO4−

68. Which of the following statement is not true about the hydrolysis of XeF6?

69. The most stable hydrate of noble gases is

(1) TFTF (2) TFFF (3) TFFT (4) FTFT

Ion

Group 18

(1) Kr (2) Xe (3) Ar (4) He 76. Which one of the following reactions of xenon compounds is not feasible? (1) 2XeF2 + 2H2O → 2Xe + 4HF + O2 (2) XeF6 + RbF → Rb[XeF7] (3) XeO3 + 6HF → XeF6 + 3H2O (4) 3XeF4 + 6H2O → 2Xe + XeO3 + 12HF + 1.5O2 77. Which of the following xenon-OXO compounds may not be obtained by hydrolysis of xenon fluorides? (1) XeO2F2 (2) XeOF4 (3) XeO3 (4) XeO4

1/5/2018 5:24:00 PM

THE p-BLOCK ELEMENTS

Previous Years’ NEET Questions

(3) MF > MCl > MBr > MI (4) MF > MCl > MI > MBr

1. Which one of the following anions is present in the chain structure of silicates? 6− (1) SiO4− 4 (2) Si 2O7 2− (3) ( Si 2O5 ) (4) ( SiO2− 3 ) n



(1) Cl2 (2) F2 (3) Br2 (4) I2

n

2. Which one of the following orders correctly represents the increasing acid strengths of the given acids? (1) HOClO3 < HOClO2 < HOClO < HOCl (2) HOCl < HOCIO < HOClO2 < HOClO3 (3) HOClO < HOCl < HOClO3 < HOClO2 (4) HOClO2 < HOClO3 < HOClO < HOCl

9. The stability of +1 oxidation state increasing in the sequence

(AIPMT 2009, RE-AIPMT 2015)

3. Which of the following oxidation states are the most characteristic for lead and tin, respectively? (1) +2, +2 (2) +4, +2 (3) +2, +4 (4) +4, +4

10. Which one of the following molecular hybrids acts as a Lewis acid? (1) CH4 (2) NH3 (3) H2O (4) B2H6

(AIPMT 2007) 4. For the following:

(AIPMT PRE 2010)

11. The tendency of BF3, BCl3 and BBr3 to behave as Lewis acid decreases in the sequence (1) BF3 > BCl3 > BBr3 (3) BBr3 > BCl3 > BF3

(I) I− (II) Cl− (III) Br−

(2) BCl3 > BF3 > BBr3 (4) BBr3 > BF3 > BCl3



The increasing order of nucleophilicity would be (1) Br− < Cl− < I− (2) I− < Br− < Cl− (3) Cl− < Br− < I− (4) I− < Cl− < Br−

5. Which one of the following arrangements does not give the correct picture of the trends indicated against it? (1) F2 > Cl2 > Br2 > I2: Electronegativity. (2) F2 > Cl2 > Br2 > I2: Reactivity in hydrogenation of alkanes. (3) F2 > Cl2 > Br2 > I2: Electron gain enthalpy. (4) F2 > Cl2 > Br2 > I2 : Bond dissociation energy. (AIPMT 2008)

(1) 4 (2) 2 (3) 5 (4) 6 (AIPMT MAINS 2010) 13. Name the type of the structure of silicate in which one oxygen atom of [SiO4]4− is shared. (1) Three dimensional (2) Linear chain silicate (3) Sheet silicate (4) Pyrosilicate

(1) NaHCO3 on heating gives Na2CO3. (2) Pure sodium metal dissolves in liquid ammonia to give blue solution. (3) NaOH reacts with glass to give sodium silicate. (4) Aluminum reacts with excess NaOH to give Al(OH)3.

(1) NO+2 < NO −2 < NO 2 (2) NO−2 < NO +2 < NO 2 (3) NO−2 < NO 2 < NO +2 (4) NO+2 < NO 2 < NO −2

7. In the case of alkali metals, the covalent character decreases in the order

(AIPMT PRE 2011)

14. Which of the following statements is incorrect?

6. The correct order of increasing bond angles in the following triatomic species is

(AIPMT 2008)

(AIPMT PRE 2010)

12. How many bridging oxygen atoms are present in P4O10?

(AIPMT 2007)

Chapter 17_p-Block Elements.indd 439

(AIPMT 2009)

(1) Ga < In < Al < Tl (2) Al < Ga < In < Tl (3) Tl < In < Ga < Al (4) In < Tl < Ga < Al

(AIPMT 2007, NEET- I 2016)

(1) MI > MBr > MCl > MF (2) MCl > MI > MBr > MF

(AIPMT 2009)

8. Among the following, which is the strongest oxidizing agent?

(AIPMT 2007)



439



(AIPMT MAINS 2011)

15. Which of the following oxide is amphoteric? (1) CO2 (2) SnO2 (3) CaO (4) SiO2

(AIPMT MAINS 2011)

1/5/2018 5:24:01 PM

440

OBJECTIVE CHEMISTRY FOR NEET

16. Sulphur trioxide can be obtained by which of the following reaction? D D ¾ (2) H 2SO4 + PCl 5 ¾® ¾ (1) S + H 2SO4 ¾® D D ¾ (4) Fe2(SO 4 )3 ¾® ¾ (3) CaSO 4 + C ¾®

24. Nitrogen dioxide and sulphur dioxide have some properties in common. Which property is shown by one of these compounds, but not by the other? (1) (2) (3) (4)

(AIPMT PRE 2012) 17. Which of the following statements is not valid for oxoacids of phosphorus? (1) Orthophosphoric acid is used in the manufacture of triple superphosphate. (2) Hypophosphorous acid is a diprotic acid. (3) All oxoacids contain tetrahedral four coordinated phosphorus. (4) All oxoacids contain atleast one P   O unit and one P   OH group.

(AIPMT PRE 2012)

18. In which of the following arrangements the given sequence is not strictly according to the property indicated against it? (1) HF < HCl < HBr < HI: increasing acidic strength (2) H2O < H2S < H2Se < H2Te: increasing pKa values (3) NH3 < PH3 < AsH3 < SbH3: increasing acidic character (4) CO2 < SiO2 < SnO2 < PbO2: increasing oxidizing power

(AIPMT MAINS 2012)

(AIPMT 2015) 25. Aqueous solution of which of the following compounds is the best conductor of electric current? (1) (2) (3) (4)

(2) H2SO4 (4) HClO4

20. Which of the following is electron-deficient?

26. Strong reducing behavior of H3PO4 is due to (1) (2) (3) (4)

27. Which one of the following orders is correct for the bond dissociation enthalpy of halogen molecules? (1) I2 > Br2 > Cl2 > F2 (2) Cl2 > Br2 > F2 > I2 (3) Br2 > I2 > F2 > Cl2 (4) F2 > Cl2 > Br2 > I2

21. Which of the following structure is similar to graphite? (1) B2H6 (2) BN (3) B (4) B4C (NEET 2013)



(1) SiO− (2) SiO4− 4 2(3) SiO23 (4) SiO4  (NEET 2013)

23. Acidity of diprotic acids in aqueous solutions increases in the order (1) H2S < H2Se < H2Te (2) H2Se < H2S < H2Te (3) H2Te < H2Te < H2Se (4) H2Se < H2Te < H2S

Chapter 17_p-Block Elements.indd 440

(AIPMT 2014)

Column I

Column II

   (I)  XeF6

(p) Distorted octahedral

 (II)  XeO3

(q) Square planar

(III) XeOF4

(r)  Pyramidal

(IV) XeF4

(s)  Square pyramidal

Code: (I) (1) (p) (2) (p) (3) (s) (4) (s)

22. The basic structural unit of silicates is



(NEET-I 2016)

28. Match the compounds given in Column I with the hybridization and shape given in Column II and mark the correct option.

(NEET 2013)



high oxidation state of phosphorus. presence of two   OH groups and one P   H bond. presence of one   OH group and two P   H bonds. high electron gain enthalpy of phosphorus. (RE-AIPMT 2015)

(2) (CH3)2 (4) (BH3)2



(RE-AIPMT 2015)

(NEET 2013)

(1) PH3 (3) (SiH3)2

Ammonia, NH3 Fructose, C6H12O6 Acetic acid, C2H4O2 Hydrochloric acid, HCl



19. Which of the following is the strongest acid? (1) H2SO3 (3) HClO3

One is a reducing agent, other is not. One is soluble in water, other is not. One is used as a food-preservative, other is not. One constitutes ‘acid-rain’, other does not.



(II) (III) (IV) (r) (s) (q) (q) (s) (r) (r) (p) (q) (p) (q) (r) (NEET-I 2016)

29. Which is the correct statement for the given acids? (1) Phosphinic acid is a monoprotic acid while phosphonic acid is a diprotic acid. (2) Phosphinic acid is a diprotic acid while phosphonic acid is a monprotic acid.

1/5/2018 5:24:01 PM

THE p-BLOCK ELEMENTS (3) Both are diprotic acids. (4) Both are triprotic acids.

(NEET-I 2016)

30. When copper is heated with conc. HNO3 it produces (1) Cu(NO3)2 and NO2. (2) Cu(NO3)2 and NO.



(NEET-I 2016)

31. Boric acid is an acid because its molecule (1) (2) (3) (4)

gives up a proton. accepts OH− from water releasing proton. combines with proton from water molecule. contains replaceable H+ ion.



(NEET-II 2016)

32. AIF3 is soluble in HF only in presence of KF. It is due to the formation of



(1) SO2 (3) P2O3



38. Match the interhalogen compounds of Column I with the geometry in Column II and assign the correct code.

33. Which of the following fluoro compounds is most likely to behave as a Lewis base? (1) PF3 (2) CF4 (3) SiF4 (4) BF3

Column I

Column II

     (I)  XX ¢

(p) T-shape

 (II)  XX ¢3

(q)  Pentagonal bipyramidal

(III)  XX ¢5

(r) Linear

(IV)  XX ¢7

(s)  Square pyramidal (t) Tetrahedral

(NEET-II 2016)

Code: (I) (1) (r) (2) (t) (3) (s) (4) (r)

34. The correct geometry and hybridization for XeF4 are (1) trigonal bipyramidal, sp3d. (2) planar triangle, sp3d3. (3) square planar, sp3d2. (4) octahedral, sp3d2.

(II) (p) (s) (r) (s)

(III) (IV) (s) (q) (r) (q) (q) (p) (p) (q)





(NEET 2017)

(1) Sn2+ is oxidizing while Pb4+ is reducing. (2) Sn2+ and Pb2+ are both oxidizing and reducing. (3) Sn4+ is reducing while Pb4+ is oxidizing. (4) Sn2+ is reducing while Pb4+ is oxidizing. (NEET 2017)

(NEET-II 2016)



(2) NO2 (4) CO2

37. It is because of inability of ns2 electrons of the valence shell to participate in bonding that

(1) K3[AlF6] (2) AlH3 (3) K[AlF3H] (4) K3[AlF3H3]

(NEET-II 2016)

36. Name the gas that can readily decolorize acidified KMnO4 solution.

(3) Cu(NO3)2, NO and NO2. (4) Cu(NO3)2 and N2O.

(3) I +3 has bent geometry. (4) PH5 and BiCl5 do not exist.

441

(NEET-II 2016)

39. In which pair of ions both the species contain S   S bond? (1) S 4O62- , S 2O32- (2) S 2O72- , S 2O82-

35. Among the following, which one is a wrong statement? (1) pp   dp bonds are present in SO2. (2) SeF4 and CH4 have same shape.

(NEET 2017)

(3) S 4O62- , S 2O72- (4) S 4O72- , S 2O32

(NEET 2017)

Answer Key Level I   1. (4)

  2. (2)

 3. (3)

  4. (1)

  5. (3)

  6. (3)

  7. (2)

  8. (3)

  9. (3)

 10. (1)

 11. (3)

 12. (3)

 13. (1)

 14. (2)

 15. (4)

 16. (4)

 17. (2)

 18. (4)

 19. (4)

 20. (4)

 21. (4)

 22. (1)

 23. (1)

 24. (1)

 25. (1)

  26. (2)

 27. (1)

 28. (2)

 29. (1)

 30. (4)

 31. (2)

 32. (1)

 33. (4)

 34. (4)

 35. (4)

 36. (4)

 37. (3)

 38. (4)

 39. (3)

 40. (3)

 41. (4)

 42. (1)

 43. (1)

 44. (1)

 45. (3)

 46. (2)

 47. (4)

 48. (3)

 49. (3)

 50. (4) 

Chapter 17_p-Block Elements.indd 441

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442  51. (4)

OBJECTIVE CHEMISTRY FOR NEET  52. (4)

 53. (2)

 54. (1)

 55. (1)

 56. (2)

 57. (3)

 58. (4)

 59. (3) 

 60. (3)

 61. (1)

 62. (3)

 63. (2)

 64. (1)

 65. (1)

 66. (1)

 67. (4)

 68. (3)

 69. (2)

 70. (4)

 71. (1)

 72. (2)

 73. (3)

 74. (4)

 75. (3)

 76. (3)

 77. (2)

 78. (1)

 79. (3)

 80. (1)

 81. (2)

 82. (1)

 83. (3)

 84. (4)

 85. (1)

 86. (2)

 87. (4)

 88. (1)

 89. (4)

 90. (2)

 91. (1)

 92. (2)

 93. (1)

 94. (4)

 95. (1)

 96. (2)

 97. (1)

 98. (4)

 99. (3)

100. (3)

101. (3)

102. (3)

103. (3)

104. (2)

105. (1)

106. (4)

107. (1)

108. (2)

109. (2)

110. (1)

111. (2)

112. (2)

113. (1)

114. (2)

115. (2)

116. (1)

117. (3)

118. (1)

119. (2)

120. (2)

121. (4)

122. (3)

123. (3)

124. (1)

125. (4)

126. (1)

127. (1)

128. (4)

 1. (4)

 2. (2)

 3. (3)

 4. (1)

 5. (4)

 6. (2)

 7. (1)

 8. (3)

 9. (3)

10. (1)

11. (1)

12. (3)

13. (1)

14. (1)

15. (1)

16. (4)

17. (4)

18. (2)

19. (1)

20. (2)

21. (2)

22. (1)

23. (1)

24. (1)

25. (3)

26. (2)

27. (1)

28. (1)

29. (2)

30. (2)

31. (2)

32. (4)

33. (2)

34. (3)

35. (1)

36. (2)

37. (2)

38. (1)

39. (1)

40. (2)

41. (2)

42. (2)

43. (2)

44. (1)

45. (1)

46. (3)

47. (2)

48. (2)

49. (3)

50. (3)

51. (4)

52. (3)

53. (3)

54. (1)

55. (2)

56. (1)

57. (1)

58. (2)

59. (2)

60. (3)

61. (2)

62. (1)

63. (1)

64. (2)

65. (3)

66. (2)

67. (3)

68. (3)

69. (4)

70. (3)

71. (1)

72. (2)

73. (4)

74. (4)

75. (2)

76. (3)

77. (4)

78. (3)

Level II

Previous Years’ NEET Questions  1. (4)

 2. (2)

 3. (3)

 4. (2)

 5. (4)

 6. (3)

 7. (1)

 8. (4)

 9. (2)

10. (4)

11. (3)

12. (4)

13. (4)

14. (4)

15. (2)

16. (4)

17. (2)

18. (2)

19. (4)

20. (4)

21. (2)

22. (2)

23. (1)

24. (3)

25. (4)

26. (3)

27. (2)

28. (1)

29. (1)

30. (1)

31. (2)

32. (1)

33. (1)

34. (3)

35. (2)

36. (1)

37. (4)

38. (1)

39. (1)

Hints and Explanations Level I

5. (3) The structure of borazine is H

1. (4) BCl3 has no lone pair as B has only three valance electrons thus its shape is trigonal planar; while NCl3 has one lone pair of electron as N has 5 valence electrons, so its shape is pyramidal.

H

H

Cl

N Cl

Cl

Cl

Cl

2. (2) Gallium has an unusual structure. Each metal atom has one close neighbor at a distance of 2.43 Å and six more distant neighbors at distances between 2.70 Å and 2.79 Å. This remarkable structure tends towards discrete diatomic molecules rather than a metallic structure. This accounts for the incredibly low melting point of gallium of 30°C. 4. (1) Due to the presence of bridging hydrogens.

Chapter 17_p-Block Elements.indd 442

B

B N

H N B

H

H Borazine

B Cl

N

6. (3) The structures of AlCl3 and BeCl2 are as follows: Cl Cl



Al

Cl

Cl

Be

Al Cl

Cl

Cl

Cl Cl

Be

Cl

Cl Cl

Be

B3N3 is called inorganic benzene and boric acid is a Lewis acid.

9. (3) Borax is Na2B4O7·10H2O. Its chemical name is sodium tetraborate decahydrate.

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THE p-BLOCK ELEMENTS 14. (2) On heating borax with ethyl alcohol and concentrated sulphuric acid vapors of tri-ethyl borate are produced. When ignited with these vapors burn with a green edged flame.

31. (2) The structure is

-

Na 2B4O7 + H 2SO4 + 5H 2O → Na 2SO4 + 4H 3BO3 -

15. (4) The structures are as follows: H B

H

H

H

H

H

Diborane

N

N

H H

Hydrazine

16. (4) Tl+ shows +1 ionic state due to inert pair effect. The outer shell “s” electrons (ns2) penetrate to (n − 1)d electrons and thus become closer to nucleus and are more effectively pulled towards the nucleus. This results in less availability of ns2 electrons pair for bonding or ns2 electron pair becomes inert. The inert pair effect begins after n ≥ 4 and increases with increasing value of n. The tendency to form M+ ion increases down the group from Ga to Tl.

34. (4) Sn + 4HNO3 ® H 2SnO3 + 4NO 2 + H 2O 37. (3) Lead pencil contains graphite. 38. (4) Germanium acts as metalloid. 39. (3) The structures are as follows: S C

O O

C

O O

O

H B

Si

H

Si

B H

H

18. (4) B2H6 can be prepared by reacting BF3 with LiBH3 in the presence of dry ether.

O

4BF3 + 3LiBH3 → 2B2H6 + 3BF3 + 3LiF

19. (4) Aluminum does not react with hot conc. HNO3 because Al is rendered passive by nitric acid. This is due to oxidation and formation of a thin film of oxide on its surface.

Si

O

O

Si

O

Si

Si O

O O

Si O

O

O

H



Si O

17. (2) The structure is

H

-

Si2O76- ion

32. (1) Down the group the size of atom increases and thus bond length increases and thermal stability decreases.

H

B

- O

-

- Si

H 3BO3 + 3C 2H5OH → B(OC 2H5 )3 + 3H 2O

H

-

-

O

Si

40. (3) SiF4 + 2HF → SiF6 2- + H2↑ 41. (4) In all the given compounds, carbon is tetrahedral sp3 hybridized, due to the presence of four valence electrons. 42. (1) Density increases as Si < C < Ge < Sn < Pb. Density increase down the group but Si has a vacant shell which increases the volume and reduces the density.

20. (4) TlI3 does not exist due to steric hindrance.

43. (1) SnCl2 is the most ionic in the given compounds because Sn has low ionization energy and Cl has high electron gain enthalpy.

21. (4) The density of Group 13 elements increases down the group.

44. (1) CuO + CO → Cu ↑ + CO2

22. (1) Boron nitride is the hardest.

51. (4) Because the hybridization in P4 is sp3 which involves 75% p-character and 25% s-character.

28. (2) PbO2 + HNO3 → Pb(NO3)3 + H2O + 1/2 O2 29. (1) Silicon (Si) exists as covalent solid, sulphur (S8) exists as molecular solid, phosphorus (P4) exists as molecular solid and iodine (I2) exists as molecular solid. 30. (4) Zeolites, Ultramarines and Feldspars are all three dimensional silicates.

Chapter 17_p-Block Elements.indd 443

53. (2) As N2O5 is a colorless solid, whereas all the others exist in solid and liquid form both or as gases. 54. (1) As the polarity of the bond increases with increase in ∆χ (electronegativity difference in a period from left to right with hydrogen), the tendency to lose H+ increases in polar solvent.

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OBJECTIVE CHEMISTRY FOR NEET

55. (1) P has empty d orbitals which can be used for bonding whereas N2 does not have d orbitals.

88. (1) Hydrolysis of peroxydisulphuric acid yields H2SO4.

56. (2) The reaction is

HO

57. (3) Cyclic metaphosphoric acid has 3 P   O   P bonds. O O

O

P

O

O

O

P

64. (1) Highest oxidation state is +5 which remains unchanged. 65. (1) N has the oxidation states from +1 to +5 in the following compounds: N2O, NO, NO2, N2O3, N2O3, N2O5.

98. (4) This is because it has variable valency from −2 to +6. 99. (3) O

60°



69. (2) NO has three-electron bonds. It has bond order 2.5. Two complete bonds and third bond is partial. 70. (4) NO2 is a mixed anhydride of HNO2 and HNO3.

72. (2) Peroxydisulphuric acid, H2S2O8 is commonly known as Marshall’s acid. D ¾ N 2 + 4H 2O + Cr2O3 82. (1) (NH 4 )2 Cr2O7 ¾®

       Ba(N 3 )2 ® Ba + 3N 2 86. (2) The reaction is 2 KI + H 2O + O3 → 2KOH + I 2 + O2 87. (4) S



O -

O -

Chapter 17_p-Block Elements.indd 444

O -

O -

O

OH

HO

O S

O S

O

Peroxomonosulphuric acid also known as Caro’s acid

O

O

S

OH

O (H2S2O8)

Peroxydisulphuric acid also known as Marshall’s acid

104. (2) In Group 16, Se cannot form monoxide and TeO2 and PoO2 are non-volatile solid. 105. (1) Br2 reacts with hot and strong NaOH solution to give NaBr, NaBrO3 and H2O. 106. (4)

I 2 + KI  KI 3

S O

110. (1) The best known pseudohalide is CN– which resembles Cl–, Br– and I– in various respects. It forms dimers, disproportionate in alkaline medium similar to halogens, whereas N -3 and ONC– do not show these characteristics. 111. (2) The outer configuration of halogen atom is ns2np5. After gaining one electron, the configuration of halide ion becomes ns2np6 (stable noble gas configuration). 112. (2) MgBr2 + Cl 2 → MgCl 2 + Br2

O O -

O

S

I 2 + I −  I 3−

2NO2 + H2O → HNO2 + HNO3

S

S

O

O

UV 100. (3) 3O2 ¾rays ¾® 2O 3 ↑

68. (3) Bismuth does not exhibit allotropy. All the other elements of the group exhibit allotropy.

O

O

97. (1) The property of ozone is that it is an unstable, dark blue, diamagnetic gas.

O (H2SO5) P

O

OH

96. (2) As the size of the atom increases, the H   X bond becomes weaker and breaks easily.

HO

P White phosphorus

S

O

S

S 2O23- + H + ® H 2O + SO2 + S ¯

P

S

2 HO

S2− + H+ → H2S



67. (4) The structure of white phosphorus is

P

OH

89. (4) No S   S bond exists in S3O9.



The donating ability of lone pair of P-atom in PH3 is much less because it is present in the almost pure ‘s’ orbital according to Drago’s rule.



S

95. (1) S + KOH → K2S + K2S2O3 + H2O

O

59. (3) Hypophosphoric acid has four acidic hydrogens.



O

94. (4) +6. For example, H2SO4, SO3, etc.

O

58. (4) PH3 + H2O → PH4 + + OH−

O

O

93. (1) H2SO5 is known as Caro’s acid.

O P

S O



PH3 + Cl2 → PCl3 + HCl



O

O

O-

114. (2) In halogen molecules (X   X) covalent bond is always a s-bond.

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THE p-BLOCK ELEMENTS 116. (1) Argon is mainly used to provide an inert atmosphere at high metallurgical extraction, that is, in arc welding of metals or alloys.

15. (1) Due to small size and high charge on B3+, it is highly polarizing and therefore, favors formation of covalent bond.

120. (2) The oxidation state of Xe is +6 in XeO3 and XeF6.

16. (4) Due to inert pair effect, the stability of +2 oxidation state increases as we move down the group. Therefore,

123. (3) XeF2 is linear.

SiX 2  GeX 2  SnX 2  PbX 2



F

18. (2)

R

Xe

Si

F

126. (1) The air contains 0.934% of Ar. 127. (1) In XeO3, the Xe is sp3 hybridized. 128. (4) Xe[PtF6] is in the form of red crystalline solid.

Level II 3. (3) Na 2B4O7 + 7 H 2O → 2NaOH + 4H 3BO3 4. (1) Borax (Na2[B4O5(OH)4]·8H2O) is made up of two triangular and two tetrahedral units. This ion is [B4O5(OH)4]2– and the other water molecules are associated with the metal ions. Thus, four  OH groups are attached to boron in borax. 7. (1) Option (1): Na 2B4O7 + 2HCl + 5H 2O ® 2NaCl + 4H 3BO3 Option (2): 2NaBH4 + I2 → B2H6 + 2NaI + H2







Option (4): 2BF3 + 6NaH → B2H6 + 6NaF

Option (3): 4BF3 + LiAlH4 → 2B2H6 + 3AlF3 + 3LiF

11. (1) BN + 3NaOH ® Na 2BO 3 + NH 3 ↑

Chapter 17_p-Block Elements.indd 445

O

R

Si R

Option (1) is incorrect as Si has vacant 3d orbital.



Option (3) is incorrect as there is no relation between chain length and water repelling characteristics.

19. (1) Si is tetrahedrally surrounded by four O atoms. Each corner is shared with another tetrahedron, thus giving an infinite array. The difference between these structures is the way in which the tetrahedral SiO4 units are arranged. 23. (1) The simple halides of carbon are CF4, CCl4, CBr4 and CI4. All these halides are known to exist. However, the stability of these tetrahalides decreases as the size of the halogen atom increases, that is, CF4 > CCl4 > CBr4 > CI4. This is due to the reason that the bond energies of the carbon–halogen bonds decreases in order: C   F > C   Cl > C   Br > C   I. Hence, iodide is least stable carbon halide. 24. (1) Because carbon forms tetravalency, hence [CCl6]2– does not exist. CH3

25. (3) Cl

Si CH3

CH3 Cl

H2O

HO

Si

CH3 OH + HO

CH3

Si

OH

CH3 -H2O

CH3 HO

Si CH3

H 3BO3 + H 2O ® [B ( OH )4 ]- + H +

10. (1) Since the halides of Group 13 elements accept electron pairs from numerous atoms and ions, they act as Lewis acids.

R

Si



9. (3) It is a weak acid and ionizes mainly as monobasic acid. It does not liberate H+ ion but accepts OH– ion, that is, it behaves as Lewis acid.

O

R

The Si   O   Si skeleton is protected by water repelling alkyl groups that orient themselves towards the surface creating a water proof structure, like an alkane.

Boric acid



Si

R



124. (1) XeF4 is solid under ordinary conditions. 125. (4) This is because they have completely filled orbitals.

O

R



R

CH3 O

Si

OH

CH3

26. (2) Option (1): Graphite is a paramagnetic substance because the fourth valency of each carbon atom remains unsatisfied, that is, the fourth valence electron remains unpaired.

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OBJECTIVE CHEMISTRY FOR NEET



Option (2): It behaves like metallic as well as semiconductor. It conducts electricity due to unpaired electron in the structure.



Option (3): C   C bond length in graphite is less than in the diamond. In graphite C   C covalent distance is 1.42 Å and in diamond the C   C bond distance is 1.54 Å.



Option (4): It is thermodynamically more stable as well as less dense than diamond. Density of graphite is 2.25 g cm–3 and that of diamond is 3.51 g cm–3.



Option (4) is incorrect as the thermal stability depends upon bond energy of M   H bond, which is the highest (shortest bond length) for NH3 and thus it has the highest stability.

36. (2) Hypophosphoric acid contains P   P linkage.

HO

O

O

P

P

HO



OH

OH

27. (1) Charcoal is the most reactive form of carbon because it has highly open structure, giving it high surface area per unit mass. As the surface area increases, the reactivity increases.



Pyrophosphoric acid (H4P2O7) contains four P   OH, two P   O and one P   O   P bonds.



Orthophosphoric acid (H3PO4) contains three P   OH and one P   O bonds.

28. (1) This property of carbon is due to its ability to form pp    pp multiple bonds, because in all the atoms, electrons are in p-orbital.



Metaphosphoric acid (HPO3)3 contains three P   OH, three P   O and three P   O   P bonds.

32. (4) Except statement (4) all the statements are correct.

Stability of hydrides decreases down the group from NH3 to BiH3 as the M   H bond energy decreases. In other words, as we move down the group the size of the central atom increases, and hence its tendency to form covalent bond with small hydrogen atom decreases.

33. (2)

37. (2) Phosphorus acid on heating yield phosphine. Δ 3H 3PO 2 →  PH 3 + 2H 3PO3



39. (1) There is no peroxy linkage present in pyrophosphoric acid. O HO OH

H3Si N

SiH3

SiH3

Due to back bonding between N and Si atom (SiH3)3N become planar.

P

O O

P

OH OH

40. (2) Oxides of phosphorus P2O3 and P2O5 exist as dimers, that is, P2O6 and P4O10. 41. (2) In Group 15, the metallic character increases on descending the group. Thus, N and P are non-metals; As and Sb are metalloids and Bi is a metal. The acidic nature of the oxides thus decreases in the order:



N2O3 > P2O3 > As2O3 > Sb2O3 > Bi2O3

35. (1)

42. (2) The order of bond angles is

Option (1) is correct as all the three bond pairs are attracted towards N-atom and increase the electron density on N-atom in the case of NH3. While in the case of NH2   NH2, bond pair is not attracted towards any N-atom; for NH2OH oxygen withdraws the bond pair towards itself; and finally in NF3, three F-atom having    I effect decrease the electron density on N-atom.





NH3 > PH3 > AsH3 > SbH3





107.48° > 93.36° > 91.48° > 91.18°



The structure of NH3 can either be described as pyramidal or tetrahedral with one position occupied by a lone pair. The regular tetrahedral shape is slightly distorted due lone pair-bond pair interaction and bond angle reduces to 107.48°. The hydrides PH3, AsH3 and SbH3 are expected to have similar structure and bond angles. However, in these the bond pairs of electrons are much further away from the central atom, thus lone pair causes even more distortion, resulting in observed bond angles. The bond angles suggest that the orbitals used for bonding are close to pure p orbitals.



Option (2) is incorrect as due to H-bonding in NH3, the effective molecular mass increases, and overcomes all other molecular masses. Thus, the melting point is highest in NH3.



Option (3) is incorrect as at the boiling point temperature, the extent of H-bonding is less and cannot overcome the molecular weight of SbH3.

Chapter 17_p-Block Elements.indd 446

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THE p-BLOCK ELEMENTS 43. (2) NH3 is manufactured for fertilizers by Haber’s process. In this process, a mixture of N2 and H2 is passed under high pressure and moderate temperature over a catalyst.

N 2(g) + 3H 2(g) ¾¾¾¾¾¾® 2NH 3(g) Fe + Mo 450° C/200- 250 atm

44. (1) The number of electrons in NO+ is 14 while O2 has 16 electrons. Thus, they are not isoelectronic. 5 45. (1) 2NH 3 + O2 ® 2NO + 3H 2O 2 48. (2) Na2S2O3 + 4Cl2 + 5H2O → 2NaHSO4 + 8HCl 49. (3) Thiosulphuric acid, that is, H2S2O3 does not have peroxide linkage O   O. 50. (3) Ozone does not reduce acidified solution of KMnO4. 51. (4) H2Te has the smallest bond angle as Te is very less electronegative therefore, there is very less electron pair repulsion. 52. (3) SO2 + SnCl2 + HCl → SnS2 + SnCl4 + H2O 53. (3) Ozone is detected by using mercury and the reaction involved is 2Hg +O3 ® Hg 2O + O2↑.

The mercurous oxide formed dissolves in mercury and starts sticking of the glass tube (called tailing of mercury).

56. (1) O3 is extremely powerful oxidizing agent. 57. (1) Cu + H 2SO4 ® CuSO4 + 2H 2O + SO2 58. (2) Correct Ionic mobility order: F–(aq) < Cl–(aq) < Br–(aq) < I–(aq).

Correct stability order: Cl3– < Br3– < I3–.



Correct reactivity order: F > Cl > Br > I.



Correct oxidizing powder order: F2 > Cl2 > Br2 > I2.

59. (2) A pseudohalogen is any inorganic compound structurally similar to halogens, but containing at least one non-halogen radical such as cyanide. 60. (3) The Br3- ion is much less stable and less common than I -3. A few unstable Cl -3 compounds are known, and the ion is formed in concentrated solution. No F3compounds are known to exist. 61. (2) Thermal stability decreases as the difference between the electronegativity increases. 62. (1) Bleaching powder dissolves in water to form chloride and hypochlorite ions (HCl + HClO). In hot water it forms chloride and chlorate ions (HCl + HClO3).

Chapter 17_p-Block Elements.indd 447

66. (2) HgO + Cl2 → HgCl2·HgO + Cl2O 67. (3) ICl3 + H2O → ICl + HIO3 + HCl 69. (4) Xe⋅6H2O. Noble gas hydrates are clathrate compounds which is solid crystalline addition compounds having the formula 6H2O:1 gas atom. Due to its large size, Xe is trapped easily and thus has the most stable hydrate. 71. (1) Helium is mixed with oxygen for deep sea divers oxygen cylinder. 72. (2) XeF2 (+2), XeF4 (+4), XeF6 (+6) oxidation states of Xe. 75. (2) The boiling point and melting point of noble gases increases with increase in atomic number. The magnitude of van der Waals forces is the maximum in the case of Xe and so it has the highest boiling point. 76. (3) XeO3 + 6HF → XeF6 + 3H2O

The reaction is not feasible because XeO3 is a powerful oxidizing agent.

1 77. (4)   3 XeF4 + 6H 2O ® 2 Xe + XeO3 + 12HF + O2 2   [XeO6]4− + H2SO4 ® XeO4



Concentrated H2SO4 react with per xenate to give XeO4.

78. (3) The hybridization in XeF2 is sp3d with three lone pairs (linear shape).

Previous Years’ NEET Questions 1. (4) Simple chain silicates or pyroxenes are formed by the sharing the O atoms on two corners of each tetrahedron with other tetrahedra. This gives the formula (SiO 23 )n . (SiO3)n2nChain



 

2. (2) In the series of oxoacids HOCl, HOClO, HOClO2, and HOClO3 an increasing number of oxygen atoms are bonded to the chlorine atom. The more oxygen atoms that are bonded, the more the electrons will be pulled away from the O   H bond, and the more this bond will be weakened. Thus HOClO3 requires the least energy to break the O   H bond and form H+. Hence HOClO3 is the strongest acid. 3. (3) All the Group 14 elements show tetravalency which can be explained on the basis of shifting of one electron

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OBJECTIVE CHEMISTRY FOR NEET of ns to the vacant np orbital. These four orbitals give rise to four sp3 hybrid orbitals. When ns2 electrons of the outermost shell do not participate in bonding it is called inert pair and the effect is called the inert pair effect. Due to this, there is a decrease in stability of the (+IV) oxidation state and an increase in the stability of the (+II) state on descending the group. Sn(+II) exists as simple ions which are strongly reducing but Sn(+IV) is covalent and stable. Pb(+II) is ionic, stable, and more common than Pb(+IV), which is oxidizing.

4. (3) Nucleophilicity is defined as the ability to donate the electron pairs. It increases on going down the periodic table due to increase in size. In aqueous medium, fluoride ion being the smallest is most hydrolyzed and iodide is least. Therefore, the correct order of nucleophilicity is I− > Br− > Cl− > F−. 5. (4) It would be expected that the bond energy in the X2 molecules would decrease as the atoms become larger, since increased size results in less effective overlap of orbitals. Cl2, Br2 and I2 show the expected trend, but the bond energy for F2 is abnormally low which is due to the repulsion between the lone pairs of electrons of two atoms. Hence, the correct order is Cl2 > F2> Br2 > I2.

9. (2) There is an increasing tendency to form univalent compounds among Group 13 elements on descending the group. The atoms in group have an outer electronic configuration of s2p1. Monovalency is explained by the s electrons in the outer shell remaining paired, and not participating in bonding. This is called inert pair effect. 10. (4) B2H6 is an electron deficient compound and hence it acts as a Lewis acid. 11. (3) The valence bond theory predicts the shape of BF3 a planar triangle with hybridization of one s and two p orbitals used for bonding.



6. (3) The bond angles of given triatomic species are: Molecule NO-2  

Structure

Bond angle

N

115°

+ 2

NO  

O  

Bent +

N

O

NO2  

-

O

O 

180°

2s 2p 1s Electronic structure of boron atom - excited state Three singly occupied orbitals form bonds with unpaired electrons from three halogen atoms - shape plane triangle (sp2 hybridization)

In BF3, the empty 2pz atomic orbital of B which is not involved in hybridization is perpendicular to the triangle containing the sp2 hybrid orbitals. This pz orbital may accept an electron pair from a full pz orbital on any one of the three fluorine atoms. Thus a dative p bond is formed, and the B atom attains an octet of electrons, thereby lowering its Lewis acid strength. In case of BBr3 and BCl3, the back bonding is not effective due to large difference in the size of 2pz atomic orbital of boron and 4pz, 3pz atomic orbitals of BBr3 and BCl3 respectively.

12. (4) The structure of P4O10 is

Linear 132°

N O

O P O

O 

Trigonal planar

8. (4) Halogens act as oxidizing agents. Fluorine is the strongest oxidizing agent. There are two main reasons:



• F2 has a low enthalpy of dissociation (arising from the weakness of the F   F bond). • F2 has a high free energy of hydration (arising from the smaller size of the F   F bond).

Chapter 17_p-Block Elements.indd 448

O O

P O

The order of increasing bond angles is NO-2 < NO 2 < NO +2 .

7. (1) According to Fajan’s rule, large negative ions favor covalency. Large ions are highly polarizable, that is easily distorted by the positive ion, because the outermost electrons are shielded from the charge on the nucleus by filled shells of electrons. Therefore, decreasing order of covalent character is MI > MBr > MCl > MF.



O

P O P



O

O



O

 

From the structure, we can see that the numbers of bridging atoms are 6.

13. (4) In case of pyrosilicate, two tehtrahedra units are joined by sharing the O at one corner, thus giving the unit (Si2O7)6−.

-

-



-

Si O

Si2O67 ion

-

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THE p-BLOCK ELEMENTS 14. (4) Aluminum reacts with excess of NaOH to give [Al(OH)4]−. The reaction is

together. The sheets are stacked one on top of the other, giving a layer structure.

2Al(s) + 2NaOH(aq) + 6H 2O(l) ® 2Na +[ Al(OH)4 ]- (aq ) + 3H 2(g )  



SnO2 + 4HCl ® SnCl 4 + 2H 2O  



SnO2 + 2NaOH + 2H 2O ® Na 2[Sn(OH)6 ]  

D Fe2 ( SO 4 )3 ¾® ¾ Fe2O 3 + SO3



B

17. (2) Hypophosphorous acid is a monoprotic acid as it has one replaceable OH (see the structure below).

B

N

N N B



16. (4) The reaction involved is

B

B

15. (2) SnO2 is amphoteric as it reacts with both acids as well as bases to form the corresponding salts. The reactions involved are

B

N

N B

N B

N

C N

C

B

C

N

Boron nitride



C C

B

C C C

C

C C

C C

C

C

C C

C

C

Graphite

22. (2) Silicates are based on (SiO4)2− tetrahedral units. The SiO4 tetrahedra may exist as discrete units or may polymerize into larger units by sharing corners, that is, by sharing O atoms. The structure of silicate is

O O-

P OH H Hypophosphorous acid 

SiO44- ion -O - Si

H



Si

18. (2) In case of Group 16 elements, the stability of hydrides decreases as we move down the group and so the acidic strength increases down the group. More is the pKa value, weaker will be the acid, so, the order of increasing pKa value should be H2O > H2S > H2Se > H2Te. 19. (4) HClO4 is one of the strongest acids known. The anhydrous compound is a powerful oxidizing agent which explodes on contact with organic material and sometimes on its own. 20. (4) In diborane (B2H6) there are 12 valency electrons, three from each B atom and six from the H atoms. The structure of diborane is H

H B

H

H B

H

H







The terminal B—H distances are the same as the bond lengths measured in non-electron-deficient compounds. These are assumed to be normal covalent bonds, with two electrons shared between two atoms. We can describe these bonds as two-center two-electron bonds (2c—2e). Thus the electron deficiency must be associated with the bridge groups.

21. (2) Boron nitride is a white slippery solid. One B atom and one N atom together have the same number of valency electrons as two C atoms. Thus boron nitride has almost the same structure as graphite, with sheets made up of hexagonal rings of alternate B and N atoms joined

Chapter 17_p-Block Elements.indd 449

-





O-

O O-

23. (1) Hydrides of Group 16 elements are all weak acids and there is an increase in acidic strength from H2O to H2Te. This is due to the fact that bond dissociation enthalpy decreases down the group. 24. (3) Sulphur dioxide is used as a food preservative while nitrogen dioxide is not. 25. (4) HCl is a strong acid. It dissociates completely into ions. Thus, the aqueous solution of HCl is a best conductor of electricity. 26. (3) All oxyacid of phosphorus contains P   H bond, therefore, act as reductant. H3PO2, as its structure suggests has one    OH group and two P   H bonds. O P



H

OH H

27. (2) The correct order of bond dissociation enthalpy of halogens is Cl2 > Br2 > F2 > I2. It is expected that the bond energy of X2 molecules (where X is a halogen) would decrease as the atoms become larger, since increased size results in less effective overlap of orbitals Cl2, Br2, I2 show the expected trend, but bond energy of F2 is abnormally low because large electron – electron repulsion between the lone pairs of electrons on two fluorine atoms weaken its bond.

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OBJECTIVE CHEMISTRY FOR NEET

28. (1) XeF6 is sp3d3 hybridized with one lone pair. It has distorted octahedral geometry. F F

O

F

 

Xe O

O

 

XeOF4 is sp3d2 hybridized with one lone pair. It has square pyramidal geometry. F

O Xe

F

F F





F

XeO3 is sp3 hybridized with one lone pair. It has pyramidal geometry.



Xe F



F

32. (1) AIF3 is soluble in HF only in presence of KF due to the formation of K3[AlF6]. AlF3 + 3KF → K3[AlF6]

33. (1) Phosphorous has five outer shell electrons, of which three are used in covalent bonding with fluorine atoms to form PF3 and there is one lone pair. It is due to presence of this lone pair that PF3 acts as Lewis base. CF4 and SiF4 do not have any lone pair of electrons whereas BF3 is electron deficient. 34. (3) XeF4 is a molecule of the type AB4L2, where the central atom (Xe) is surrounded by four other atoms (F) with which it forms bond pairs and two lone pairs. Here the steric number is 6, so hybridization is sp3d2 and expected geometry is octahedral. However, due to presence of two lone pairs, the shape of the molecule is square planar.

  F

XeF4 is sp3d2 hybridized with two lone pairs. It has square planar geometry. Xe F



F

F



F

F

 

29. (1) The structure of phosphinic and phosphonic acids are as follows

F Xe F



35. (2) Option (1): SO2 molecule consists of two sigma bonds and two p bonds. The first p bond is a pp   pp bond and the second p bond involves overlap between 2pz orbital with 3dxz orbital, thus pp-dp bond is present.

Option (2): For CH4, the steric number is 4, so the hybridization is sp3 and the geometry is regular tetrahedron. In SeF4, the steric number is 5 and the molecule is of the type AB4L, so the hybridization is sp3d. The presence of lone pair affects the shape of the molecule and geometry is see-saw.

Phosphinic acid (H3PO2) is a monoprotic acid, while phosphonic (H3PO3) is a diprotic acid.



Option (3): In I3+ , the hybridization is sp3, and with the presence of two lone pairs the shape is bent.

30. (1) When copper reacts with conc. nitric acid, the metal ion coordinates with nitrate ions. The complex gives green color to the solution. The brown gas produced in the reaction is nitrogen dioxide



Option (4): BiCl5 does not exist while BF5 is known to exist.

O



P

OH H Phosphinic acid (Monoprotic) H

O P

H OH Phosphonic acid (Diprotic)   HO

Cu + 4HNO 3(conc.) ® Cu(NO3 )2 + 2NO 2 + 2H 2O   31. (2) Boric acid behaves as weak acid as it partially reacts with water by accepting hydroxyl ion to form [B(OH)4]− and releasing a proton (H3O+). The reaction is



Chapter 17_p-Block Elements.indd 450

B(OH )3 + H 2O  [B(OH )4 ]- + H +

36. (1) Sulphur dioxide (SO2) decolorizes acidified KMnO4 solution. The reaction is +

MnO4- + SO2 ¾H¾ ®

Purple

Mn 2+

Colorless/Light pink

+ SO24- + H 2O

37. (4) Due to inert pair effect, the stability of the lower oxidation state increases on descending the group. Thus Sn2+ is a reducing agent while Pb4+ is oxidizing agent as Pb2+ is stable.

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THE p-BLOCK ELEMENTS 38. (1) The correct shape of interhalogen compounds is as follows: Compounds

Shape

S O-

Linear

XX3 ¢

T-Shape

sp d

XX5 ¢

Square-pyramidal

sp3d2

XX7 ¢

Pentagonal bipyramidal

sp3d3

Chapter 17_p-Block Elements.indd 451

39. (1) The structures are as follows:

Hybridization

XX¢

3

451



S O-

O

OO O

S2O32-

S

S

S

S

O

O-

O S4O26-

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18

General Principles and Processes of Isolation of Elements

Chapter at a Glance 1. Metallurgy deals with the science and technology applied for the extraction of metals economically from their respective ores, their physical and chemical behavior, intermetallic compounds as well as their mixtures like alloys. 2. Mineral are compounds of metals which are naturally available in the earth’s crust and can be obtained by mining. A mineral may consist of one or more metallic compounds, having almost fixed chemical composition. The earthy impurities associated with the metal are known as gangue. But, few metals like Au, Pt, Ag, etc. occur in free-state also. Though Ag (silver) is known occur in combined state also. 3. Ores are the minerals from which a metal can be extracted economically and conveniently. The different types of ores are: oxide ores, oxosalt ore, sulphide ore and halide ore. Some important ores of each type are tabulated as shown below: Oxide ores ZnO: Zincite Fe2O3: Haematite Fe3O4: Magnetite Al2O3⋅2H2O: Bauxite Fe2O3·3H2O: Limonite Cu2O: Cuprite or Ruby copper

Sulphide ores ZnS: Zinc blende or Sphalerite Cu2S: Copper glance or Chalcocite CuFeS2: Copper pyrite or Chalcopyrite

Halide ores Na3AlF6: Cryolite AgCl: Horn silver NaCl: Rock salt CaF2: Fluorite

Oxosalt ores FeCO3: Siderite ZnCO3: Calamine

FeS2: Iron pyrite or Fool’s gold

KCl∙MgCl2∙6H2O: Carnalite

Cu(OH)2⋅2CuCO3: Azurite

Cu(OH)2⋅CuCO3/Cu2(OH)2CO3: Malachite or Basic copper carbonate

Ag2S: Silver glance or Argentite PbS: Galena

4. Extraction of Metal from the Ore (a) Identification of metal deposit in surrounding rocks, stones, gravel and dirt. (b) Concentration of ore or dressing or benefaction of ore: The ores are concentrated by various physical and chemical methods to separate out gangue and other impurities from the metals. Technique Gravity separation or Levigation or Hydraulic washing Magnetic separation

Froth flotation process Leaching

Process The crushed ore is washed by a current of water on a sloping table fitted with a series of corrugated boards known as Wilfley table which is continuously vibrating. The lighter particles move downwards and the heavier particles are left behind the corrugations (as barriers). The magnetic material makes a separate heap because it is held to the roller for a longer time. For example, tin stone (SnO2) is separated from magnetic impurity, wolframite [FeWO4 + MnWO4 (minor)]. The ore is ground as fine powder and mixed with water to form slurry. An oily component (pine oil, eucalyptus oil, coal tar) is added along with sodium ethyl xanthate as collector. Air is bubbled through the mixture and the ore floats to the top with froth and siliceous impurities settle down. It is a chemical process by which the required substance (may be the metal component or impurities) is dissolved out from the ore by using a suitable reagent.

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(c) Extraction of crude metal from concentrated ore (i) Conversion of ore into its oxide: The concentrated ore is converted into corresponding oxide that can be reduced more conveniently to obtain the pure metal. It is generally carried out by two methods: Calcination

Roasting

It is the process in which the concentrated ore is heated to a high temperature (just below its fusion temperature) in the absence of air (or limited supply of air). It is the process in which the concentrated ore is heated to a high temperature (just below its fusion temperature) in presence of excess of air. Roasting is an exothermic process and does not require further heating once the process starts.

(ii) S  melting and reduction (chemical and electrolytic methods): The next step of metal extraction is reduction. The various ways of reduction are summarized below: Method Carbon Reduction or Smelting

Chemical Reduction or Pyrometallurgy

Self-Reduction Thermite Reduction (or Goldschmidt-­Thermite Process) Metal Replacement method (Hydrometallurgy) Electrolytic reduction

Thermal decomposition Hydrometallurgy Cyanide Process

Process It is a process wherein an oxide is added to the concentrated ore to combine with other impurities and form a molten layer that is immiscible with the molten metal. ·  The molten layer formed is called slag. ·  The impurity that is added externally to remove the impurity already present in the ore is known as flux. ·  The choice of flux depends upon the nature of impurity/gangue present within the ore. Flux can be acidic (SiO2) or basic (CaO, MnO). The process involves the reduction of the metal oxide by heating with a reducing agent like carbon. The two limiting factors for its use are: ·  High temperature requirement for the reaction. ·  At high temperatures metal may form carbides. Sulphide ore is partially roasted into its oxide which in turn reacts with unreacted sulphide to produce molten metal. The metal oxides having very high melting point, e.g. the oxides of Cr, Mn, Ti, Mo, Fe, etc., can be reduced by thermite reduction method using Al powder as reducing agent. In this process, the desired metal is extracted out from its solution by the addition of a more electropositive metal. The oxides of strong electropositive metal such as K, Ca, Na, Al. Mg are very stable and it is difficult to reduce them into metallic state by carbon reduction process. Such metals are extracted by passing electricity through their fused chlorides or oxides or hydroxides. The metals which come below Al in the electrochemical series can be extracted by the electrolysis of the aqueous solutions of their salts. This process is applicable for Cu, Zn, Sn, Pb, etc. Some of the metal oxides are thermally unstable and thermal decomposition is utilized to get these metal from their oxides. It is the process of extraction of metal from ores, concentrates and secondary sources such as scrap as ions into the aqueous solution. Gold and silver are purified by, where they are leached into aqueous solution by cyanide ion as Au+ and Ag+ ions.

(iii) Refining methods: Metals obtained from reduction processes are not pure and required further refining. Various methods used are:

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General Principles and Processes of Isolation of Elements

Method Distillation Liquation Poling

Cupellation Electrolytic refining Zone refining

Vapor phase refining

455

Process •  Uses the difference in boiling point of metal and impurity. •  Crude zinc containing Cd, Fe and Pb as impurities is refined by this method. •  Applicable to metals that have low melting points as compared to impurities. •  Used for purification of tin, lead and bismuth. •  Molten metal is stirred with green wood poles. Wood at the high temperature of the molten metals forms hydrocarbons like methane which bring about the reduction of any oxide present in the metal. •  Used for purification of blister copper. Impure metal is heated in a blast of air when impurities get oxidized and are blown away. Used in purification of Cu, Zn, Al, Ag, Au, Al, Pb, Sn and N. •  Based on the principle that impurities have greater solubility in the molten sate than the pure metal. •  Used for metalloids like Si, Ge and Ga. Metal is converted to its volatile compound, the vapors are collected and then decomposed to give pure metal. For example: •  Mond’s process for purification of nickel. - 60° C -180° C Ni(s ) + 4CO(g ) ¾50¾¾¾ ® Ni(CO)4 (g ) ¾150 ¾¾¾ ® Ni(s ) + 4CO(g ) Impure

Volatile

Pure

Re cycled

•  van Arkel process for purification of zirconium, boron and titanium. I2 ( vap ) 1400° C Zr ¾¾ ® ZrI4 or TiI4 ¾Tungsten ¾¾¾¾ ® Zr(s ) or Ti(s ) + 2I2 (g ) or Ti  ¾250 filament °C   Impure

Volatile

Pure

Chromatographic methods Involves separating a mixture of components into individual components by equilibrium distribution between two phases.

Change in Gibbs energy ∆G° (kJ mol−1) of oxygen consumed

5. Thermodynamic Aspect in Reduction by Carbon–Ellingham Diagram: Gibbs energy changes that occur when 1 mol of a common reactant (in this case dioxygen) is used may be plotted graphically against temperature for a number of reactions of metals to their oxides. This graph is called an Ellingham diagram (for oxides). 200

Important observations from Ellingham diagram are:

HgO Ag2O

0 2NiO −200

ZnO

CO2 FeO

Cr2O3 SiO2 TiO2 2MgO Al2O3 CO 2CaO

−400 −600 −800 −1000

·  The graphs for metal to metal oxide all slope upwards, because the Gibbs energy change increases with an increase of temperature. · All the Gibbs energy changes follow a straight line unless the materials melt or vaporize. · When the temperature is raised, a point will be reached where the graph crosses the DG = 0 line. Below this temperature, the Gibbs energy of formation of the oxide is negative, so the oxide is stable. Above this temperature, the Gibbs energy of formation of the oxide is positive, the oxide becomes unstable and should decompose into the metal and dioxygen.

−1200 500

1500 1000 Temperature (°C)

2000

2500

6. Alloys and Amalgams: An alloy is a homogeneous metallic material, which contains two or more metals as a solid solution, and if one of the constituent of the alloy is mercury, it is called amalgam.

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(a) Classification of alloys (i) Ferrous alloys: If the alloy has iron as one of its constituent, it is called a ferrous alloy, e.g., stainless steel (Cr + Fe + Ni), ferrosilicon (Fe + Si), etc. (ii) Non-ferrous alloys: The alloy which does not contain iron is called as non-ferrous alloy, e.g., brass (Cu + Zn), bell metal (Cu + Sn), solder (Sn + Pb), etc. (b) Characteristics of alloys Superior casting

Hardness

Resistance to corrosion

Melting point

Tenacity

(c) Preparation of alloys: The methods utilized for preparation of alloys are described as follows: By fusion

By reduction

By compression

By simultaneous electrodeposition

(d) A  malgam: On treatment of different metals like Sn, Zn, Au, Na, Ag, etc. with mercury produces amalgams. Some uses of amalgam are: (i) Ag-Hg or Au-Hg amalgams are used in filling dental cavities. (ii) Tin amalgam is used for silvering mirrors. (iii) Na-Hg is utilized to have decreased reactivity of Na. 7. Different Types of Furnaces used in Metallurgy Blast furnace Reverberatory furnace Muffle furnace

Electric furnace

Mainly the smelting of the roasted ore is carried out in this furnace (extraction of Fe). Mainly the roasting and calcination is done in this kind of furnace. It is basically a closed chamber which is heated from external heating arrangement so that the material to be heated does not come in direct contact with the fuel. This is used for small scale purposes. This is used where very high temperature is necessary and the high temperature is achieved by an electric arc structured between two graphite electrodes.

8. Extraction of Silver Three processes are commonly used for the extraction of Ag: Cyanide process (Mc-Arthur Forest process), Parke’s process and Pattinson’s process. Cyanide process will be discussed here. (a) Ores: (i) Argentite or silver glance: AgS (ii) Ruby silver: 3Ag2S×Sb2S3 (iii) Stromeyerite or silver copper glance: Ag2S×Cu2S (iv) Horn silver: AgCl (b) Reactions taking place in the process are: + (i) Ag 2 S + 4NaCN(excess )  2[ Ag(CN )2 ] + Na 2 S + 2Na (ii) 4Na 2 S + 5O2 (air ) + 2H2 O ® 2Na 2 SO4 + 4NaOH + 2S (iii) 2Na[Ag(CN )2 ] + Zn ® Na 2 [Zn(CN)4 ] + 2Ag ¯ (iv) Refining of Ag is done by electrolytic method Electrolyte used: AgNO3 solution + 10% HNO3 Cathode used: Pure Ag strip Anode used: Impure Ag slab

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General Principles and Processes of Isolation of Elements

Reactions:

457

At the cathode: Ag + + e - ® Ag At the anode: Ag ® Ag + + e -

9. Extraction of Gold (a) Cynaide process is used for the extraction of gold. Reactions taking place in the process are: 4 Au + 8NaCN + 2H2 O + O2 ® 4Na[ Au(CN )2 ] + 4NaOH 2Na[ Au(CN )2 ] + Zn ® Na 2 [Zn(CN )2 ] + 2 Au ¯ (b) The steps involved in the refining of Au are: (i) Heating with borax: The soluble metaborate of Cu, i.e. Cu(BO2)2 is formed and washed out with water. (ii) Ag dissolves out as Ag2SO4 leaving behind pure Au. 10. Extraction of Aluminium

(a)  Ores: The ore most commonly used for extraction of aluminium is bauxite. Bauxite is of two types: Red bauxite and white bauxite. Hydrated oxides

(i) Bauxite (Al2O3×2H2O)

(ii) Gibbsite (Al2O3×3H2O) (iii) Diaspore (Al2O3×H2O)

Oxide

Corrundum (Al2O3)

Sulphate

Alunite [K2SO4×Al2(SO4)3×4Al(OH)3]

Fluoride

Cryolite (3NaF×AlF3)

Aluminate

Spinel (MgO×Al2O3)

Silicate

Feldspar (K2O×Al2O3×6SiO2)

China clay

Kaolin (Al2O3×2SiO2×2H2O)

(b) Beneficiation of bauxite: Different processes are adopted for beneficiation: (i) Bayer’s process: The reactions involved are: • In the digestion stage, Al2 O3 + 2OH- + 3H2 O ® 2[ Al(OH)4 ]SiO2 + 2NaOH ® NaSiO3 + H2 O 3 H+

3+    • [ Al(OH)4 ]-   Al(OH)3 ¯   Al + 3H2 O OH-

To get Al(OH)3 from [Al(OH)4]-, the pH of the medium is to be reduced by acidification with a weak acid like CO2 gas: CO2 + 2OH- ® CO32 - + H2 O CO2 + H2 O ® H2 CO3 [weak enough to dissolve Al(OH)3 ] [ Al ( OH)4 ]-  Al(OH)3 ¯ +OH(ii)  Hall’s process: This process is adopted for purification of low grade red bauxite. The reactions involved are: • Step 1: Al 2 O3 + Na 2 CO3 ® 2 NaAlO2 + CO2 ↑ SiO2 + Na 2 CO3 ® Na 2 SiO3 + CO2 ↑ Fe 2 O3 + Na 2 CO3 ® 2 NaFeO2 + CO2 ↑ CaO + SiO2 ® CaSiO3 • Step 2: 2NaAlO2 + CO2 + 3H2 O ® 2 Al(OH)3 ¯ + Na 2 CO3 D ® Al2 O3 + 3H2 O • Step 3: 2 Al(OH)3 ¾-¾¾ 3 H2 O

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(iii) Serpeck’s process: It is used for purification of white bauxite. The reactions involved are: • Step 1: ↑

Al 2 O3 + 3C + N 2 2 AlN(s ) + 3CO ↑ SiO2 + 2C Si ↑ + 2CO ↑ ↑

• Step 2: ↑

AlN + NaOH + 3H2 O

Na[ Al(OH)4 ] + NH3 ↑

• Step 3: H O

2   Na[ Al(OH)4 ]  Al(OH)3 ¯ + NaOH

 or [Al(OH)4 ]-   Al(OH)3 ¯ + OH

CO2 + 2OH- ® CO32 - + H2 O (c) Electrolytic reduction of pure Al2O3 Reactions: Al2 O3 ® 2 Al3 + + 3O2 -

At the cathode: Al3 + + 3e - ® Al Since Na and Ca are more electropositive compared to Al, only Al3+ gets deposited at the cathode. At the anode: 2O2 - ® O2 + 4e (d) Electrorefining of aluminium: Impure aluminium mixed with copper melt is taken in an iron tank with graphite lining. The layer of pure Al acts as the cathode and graphite rods at the top are essential for electrical connection. The electrolyte is molten mixture of cryolite, BaF2 saturated with Al2O3. The reactions involved in the process are: At the anode: Al ® Al3 + + 3e At the cathode: Al3 + + 3e - ® Al 11. Extraction of Copper (a) Ores: The chief ore used for extraction of copper is copper pyrite (Cu2S·Fe2S3). Other ores are:  (i) Chalcocite or copper glance: Cu2S (ii) Malachite: Cu(OH)2×CuCO3 (iii) Azurite:Cu(OH)2×2CuCO3 (b) At the roasting stage: Cu 2 S × Fe 2 S3 + O2 ® Cu 2 S + 2FeS + SO2 ↑ Cu 2 S × Fe 2 S3 + 4O2 ® Cu 2 S + 2FeO + 3SO2 ↑ Since, iron is more electropositive as compared to copper, its sulphide is preferentially oxidized and Cu2S remains unaffected. 3 Cu 2 S + O2 → Cu 2 S + SO2 ↑ 2 Cu 2 O + FeS → Cu 2 S + FeO (c) During the smelting step: Coke is used here as fuel to maintain the temperature in such way that the mixture is in the molten state. 3 FeS + O2 → FeO + SO2 ↑ 2 Cu 2 O + FeS ® Cu 2 S + FeO FeO + SiO2 ® FeSiO3 (Slag) (d) In the Bessemer converter: The reactions involved are 3 FeS + O2 → FeO + SO2 ↑ 2

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SiO2 + FeO ® FeSiO3 (Slag) 3 Cu 2 S + O2 → Cu 2 O + SO2 ↑ 2 Cu 2 S + 2Cu 2 O → 6Cu( l ) + SO2 ↑ Cu 2 S + 2O2 → Cu 2 SO4 Cu 2 S + Cu 2 SO4 → 4Cu + 2SO2 ↑ The molten Cu obtained is poured into large container and allowed to cool and during cooling the dissolved SO2 comes up to the surface and forms blister. It is known as blister copper. 12. Extraction of Zinc (a) Ores: The chief ore used for extraction of zinc is zinc blende (ZnS). Other ores are: (i) Zincite : ZnO (ii) Franklinite: ZnO×Fe2O3 (iii) Calamine: ZnCO3 (iv) Willemite: ZnSiO3 (v) Electric calamine: ZnSiO3×ZnO×H2O (b) At froth floatation stage: This is done in two steps to separate out PbS and ZnS depending upon their different floating characteristics. (c) At the roasting stage: 3 Above ZnS + O2 ¾850 ¾¾ ® ZnO + SO2  ↑ °C 2 Below ZnS + 2O2 ¾850 ¾¾ ® ZnSO 4 °C During ZnSO 4 + 4C ¾carbon ¾¾¾¾ ® ZnS + 4CO ↑ reduction

(d) At the smelting stage:

ZnO + C ® Zn + CO 2ZnO + C  2Zn + CO2 (This reaction is reversible) CO2 + C ® 2CO

®

(e) At the electrorefining stage: For electrorefining of Zn (crude), Al sheet is used as cathode instead of pure Zn strip. This is because the electrolyte used is ZnSO4 + H2SO4(dil.), and in dil. H2SO4 Zn gets dissolved while Al does not. Zn + H2 SO4 (dil .)® ZnSO 4 + H2 Al + H2 SO4 (dil .)® No reaction 2 Al + 6H2 SO4 (conc.) ® Al 2 (SO4 )3 + 3SO2 + H2 O Reaction during electrorefining: ZnSO 4 ® Zn 2 + + SO24 At the cathode: Zn 2 + + 2e- ® Zn ..

®

  At the anode : OH.. ® O H + e 4 O H ® 2 H2 O + O 2

13. Extraction of Iron (a) Ore: The chief ore used for extraction of iron is haematite (Fe2O3). Other common ores are:   (i) Haematite: Magnetite: Fe3O4  (ii) Brown haematite or Limonite: Fe2O3·3H2O (iii) Siderite or spathic iron ore: FeCO3  (iv) Iron pyrite: FeS

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OBJECTIVE CHEMISTRY FOR NEET

(b) During roasting Fe3 O4 → FeO + Fe 2 O3 FeCO3 → FeO + CO2 ↑ 2FeO + 1 O2 → Fe 2 O3 2 Fe 2 O3 .3H2 O → Fe 2 O3 + 3H2 O ↑ (c) During smelting  (i) At 600–900°C,

Fe 2 O3

( Partially reduced )

↓ +3CO → 2Fe + 3CO2 ↑

 (ii) At 900–1000°C, CaCO3 → CaO + CO2 ↑ CO2 + C → 2CO ↑ (iii) At 1000–1300°C, Fe 2 O3 + 3C → 2Fe + 3CO ↑ CaO + SiO2 ® CaSiO3 (slag )  (iv) At 1500°C (i.e. at the hearth): The coke powder crossing the line of tuyers, does not have the scope of burning anymore and reacts with MnO2 and SiO2 to produce impurities like Mn and SiO as follows: MnO2 + 2C ® Mn + 2CO SiO2 + 2C ® Si + 2CO Ca3(PO4)2 present in the lime stone reacts with SiO2 to produce slag and P2O5 is reduced by coke to produce P4 as impurity. Ca 3 (PO4 )2 + 3SiO2 → 3CaSiO3 + 2P2 O5 2P2 O5 + 10C → P4 + 10CO ↑  (v) Finally the cast iron produced consists of impurities like Mn, Si, P, C and S. 14. Steel (a) Preparation: Steel is made by removing most of the C and other impurities from pig iron. Composition of various steels depending upon percentage of carbon is %C 0.15 – 0.3 0.3 – 0.6 0.6 – 0.8 0.8 – 1.4

Name Mild steel Medium steel High carbon steel Tool steel

(b) Properties: Different elements present in steel provide different properties: Element P above 0.05% Mn Cr and Ni N (above 0.01%) C

Properties Imparts low tensile strength and cold brittleness. Imparts high hardness and increases tensile strength, e.g., rail road contains 13% Mn. Imparts stainless characteristic producing impervious coating of their oxides on the surface. Makes steel brittle as well as makes weldability poor. Improves the hardness and strength.

(c) Various processes used for preparation of good quality steel are:  (i) Pudding process

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(ii) Bessemer and Thomas process (iii) Siemens open hearth process  (iv) Basic oxygen process (also called as L.D. process)   All these processes are developed to economize the production of steel from iron. (d) Some heat treatment processes related to steel are: annealing, quenching or hardening and tempering.

Solved Examples 1. The method of zone refining of metals is based on the principle of (1) greater mobility of the pure metal than that of the impurity. (2) higher melting point of the impurity than that of the pure metal. (3) greater noble character of the solid metal than that of the impurity. (4) greater solubility of the impurity in the molten state than in the solid.

(1) Ag (2) Ca (3) Cu (4) Cr Solution (2) The reactions at cathode for the electrolysis of aqueous salt solution of Ag, Ca, Cu and Cr salts are as follows: Ag + + 2e - → Ag 1 H 2O + e - → H 2 + OH 2 Cu 2+ + 2e - → Cu

Solution

Cr 3+ + 3e - → Cr

(4) Metals like Si, Ge and Ga of high purity are purified by zone refining method. It is based upon fractional crystallization as the impurity prefers to stay in the melt and on solidification only the pure metal solidifies on the top surface of the melt. 2. Al2O3 can be converted to anhydrous AlCl3 by heating (1) Al2O3 with HCl gas. (2) Al2O3 with NaCl in solid state. (3) a mixture of Al2O3 and carbon in dry Cl2 gas. (4) Al2O3 with Cl2 gas. Solution (3) Al2O3 can be converted to anhydrous AlCl3 by heating a mixture of Al2O3 with carbon and dry chlorine gas. The reaction is



Al2O3 + 3C + 3Cl2 → 2AlCl3 + 3CO

3. During the concentration of sulphide ores by froth floatation process, the separation of sphalerite and galena is achieved by which of the following substances used as depressant? (1) Potassium xanthate (2) Sodium cyanide (3) Copper sulphate (4) Pine oil Solution (2) Collector, that is, xanthate acts only on solid particles through adsorption. The two ores show different reactivity towards NaCN: ZnS + 4NaCN → Na 2S +

Na 2 Zn(CN )4

(Soluble complex)

PbS + NaCN → No reaction 4. The metal that cannot be obtained by electrolysis of an aqueous solution of its salts is

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As Ca lies in the bottom of electrochemical series, this implies very high negative reduction potential of calcium ions. Water has less negative reduction potential, so, water is reduced instead of calcium from its aqueous solution. Thus, Ca cannot be obtained by electrolysis of its aqueous salt solution, H2 is evolved instead at the cathode.

5. The reduction of a metal oxide easier if the (1) (2) (3) (4)

metal exists in solid form. metal exists in liquid form. oxide is basic. oxide is acidic.

Solution (2) In the liquid state, the entropy of the system is higher than that in the solid state. The value of ΔG° becomes more negative (i.e., ΔG = ΔH - TΔS) and the reduction becomes more spontaneous. 6. Which of the following is true for a smelting process? (1) Gangue + flux ® Slag (2) Flux + slag ® Gangue (3) Gangue + slag ® Flux (4) None of these. Solution (1) The impurity that is added externally to remove the impurity already present in the ore is known as flux. For example, if the sulphide ore of copper contains iron as impurity, it is mixed with silica before heating. 2CuFeS2(s) + 4O 2(g) → Cu 2S(s) + 2FeO(s) + 3SO2(g) Cu 2S(s) + FeO(s) + SiO2 → FeSiO3 + Cu 2S Gangue

Flux

Fusible Slag

7. Which method of purification is represented by the following equation?

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OBJECTIVE CHEMISTRY FOR NEET K K Ti(s) + 2I 2(g ) ¾523 ¾¾ ® TiI 4(g ) ¾1700 ¾¾ ® Ti(s) + 2I 2(g )

13. Which of the following statement is correct? (1) Slag is carefully chosen to combine with the slag present in the ore to produce easily fusible gangue to carry away the impurities. (2) Gangues are carefully chosen to combine with the slag present in the ore to produce easily fusible flux to carry away the impurities. (3) Gangues are carefully chosen to combine with flux present in the ore to produce easily fusible slag to carry away the impurities. (4) Fluxes are carefully chosen to combine with the gangue present in the ore to produce easily fusible slag to carry away the impurities.

(1) Zone refining (2) Cupellation (3) Poling (4) van Arkel Solution (4) Ti, Zr , Hf metals are purified by van Arkel method. In this process, crude Ti is heated with I2 first at 523 K in an evacuated vessel. It forms TiI4 which is a volatile compound and gets decomposed to Ti on heating over tungsten filament at 1700 K. K K Ti(s) + 2I 2(g ) 523  → TiI 4 1700  → Ti(s) + 2I 2(g )

8. Commercially, aluminum is obtained from purified aluminum oxide by (1) (2) (3) (4)

electrolysis of aqueous Al2(SO4)3. electrolysis of aqueous KAl(SO4)2. electrolysis of a fused mixture of Al2O3 and Na3AlF6. reduction with coke at high temperatures.

Solution (3) Aluminium is obtained by electrolytic reduction of purified alumina mixed with cryolite (Na3AlF6) and fluorspar (CaF2) in an iron tank with carbon lining. The process is called Hall-Heroult process. 9. The form of iron obtained from blast furnace is (1) steel. (2) cast iron. (3) pig iron. (4) wrought iron. Solution (3) The molten iron obtained from blast furnace run into moulds made of sand and allowed to solidfy into ingots called pigs. Pig iron is hard and brittle form. 10. Cryolite is an ore of (1) iron. (2) silver. (3) zinc. (4) aluminium. Solution (4) Cryolite (AlF3×3HF) is an ore of aluminium. 11. Which one of the following ores is known as malachite? (1) Cu2O (2) Cu2S (3) CuFeS2 (4) Cu(OH)2×CuCO3 Solution (4) Malachite is copper carbonate hydroxide mineral Cu(OH)2×CuCO3. 12. The salt which is least likely to be found in mineral is (1) chloride. (2) sulphate. (3) sulphide. (4) nitrate. Solution (4) Nitrates: Chile salt petre (NaNO3), Nitre or Indian salt petre: KNO3. These are the only two ores obtained as nitrates.

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Solution (4) Fluxes are the substances that combine with the gangue particles to form an easily fusible material called slag. The slag thus formed is insoluble in the molten metal and being lighter floats on the surface of the metal from which it can be removed from time to time.

Gangue cannot be chosen as they are the impurities already present in the metal ore.



Slag cannot be chosen as it is formed when fluxes react with the gangue particles only.

14. Which of the following is used for the separation of ZnS and PbS during the froth floatation process? (1) KCl (2) KCN (3) NH4NO3 (4) None of these Solution (2) KCN complexes with ZnS to form K2[Zn(CN)4] and this complex forms a layer on the surface of ZnS and hence prevents ZnS from forming the froth. 15. Calcination is used in metallurgy for removal of (1) water and sulphide. (2) CO2 and H2S. (3) H2O and H2S. (4) water and CO2. Solution (4) It is a process of heating the concentrated ore in limited supply of air. It removes H2O and CO2. 16. Lead is mainly extracted by (1) (2) (3) (4)

carbon reduction method. self-reduction method. the method of electrolysis. leaching with aqueous solution of NaCN followed by reduction.

Solution (2) Pb is extracted from are galena (PbS). PbS is partially oxidized by heating and blowing air through it. After some time, air is turned off and heating is continued, the mixture undergoes self-reduction. in Heat in 3PbS Heat  → PbS + 2PbO absence  → 3Pb( l ) + SO 2 (g ) air of air

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General Principles and Processes of Isolation of Elements 17. Heating mixture of Cu2O and Cu2S will give (1) Cu + SO2 (2) Cu + SO3 (3) CuO + CuS (4) Cu2SO3 Solution (1) The reaction is 2Cu 2O + Cu 2S ® 6Cu + SO 2 18. One of the following metals forms a volatile compound and this property is taken advantage for its extraction. This metal is

463

serves as an added impurity and lowers the melting point of the mixture to about 950°C present in these freshly cut poles reduce CuO to Cu. 22. Which metal is separated from silver in cupellation? (1) Zn (2) Ti (3) Pb (4) Sn Solution

Solution

(3) Cupellation is the process by which impure sample of metal (say Pb in Ag) is fused in a bone ash crucible (cupel) on the hearth of furnace in a blast of air. The impurity (Pb) present is oxidized and blown away with air. Some PbO is absorbed by cupel.

(4) Nickel forms volatile nickel carbonyl [Ni(CO)4] when reacted with carbon monoxide. Ni(CO)4 is then heated to 230°C, it decomposes to give pure metal.

23. van Arkel method of purification of metals involves converting the metal to a

(1) cobalt. (2) iron. (3) tungsten. (4) nickel.

19. For metal oxides, when ΔG° is plotted against temperature, there is a point where the graph crosses ΔG° = zero line. Above this temperature, (1) (2) (3) (4)

the oxide is stable. the oxide is unstable. the oxide melts. the oxide is reduced.

Solution (2) Standard Gibbs energy values of metal oxides show a gradual increase with the increase of temperature because the entropy decreases due to disappearance of one mole of oxygen. Hence ΔG° becomes less negative and metal oxides are less stable at high temperatures. 20. Chemical reduction is not suitable for converting (1) zinc oxide into zinc. (2) cuprite into copper. (3) bauxite into aluminium. (4) haematite into iron. Solution (3) Chemical reduction is not suitable for converting bauxite into aluminium because Al is a highly reactive metal and reduction can be done only by electrolytic method; whereas other metals such as Zn, Cu and Fe are less reactive metals so CO can be used as a reducing agent. 21. Electrolytic reduction of alumina to aluminium by Hall– Heroult process is carried out (1) in the presence of NaCl. (2) in the presence of fluorite. (3) in the presence of cryolite which forms a melt with lower melting temperature. (4) in the presence of cryolite which forms a melt with higher melting temperature. Solution (3) This is because cryolite improves the electrical conductivity of the cell as Al2O3 is a poor conductor. The cryolite

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(1) (2) (3) (4)

volatile stable compound. non-volatile stable compound. volatile unstable compound. non-volatile unstable compound.

Solution (1) In van arkel method, the crude metal (to be purified) is heated in an evacuated vessel with iodine at 870 K. due to which it forms a volatile tetraiodide complex. This complex can be then decomposed by heating over tungsten filament to give pure metal. So, the complex formed should be a volatile stable compound. For example, Tungsten

Zr + 2I 2 → ZrI 4 filament  → Zr + 2I 2

Crude

Volatile

Pure

24. Heat treatment alters the properties of steel due to (1) (2) (3) (4)

changes in lattice structure. changes in residual energy. partial rusting. chemical reaction on heating.

Solution (1) In heat treatment, the lattice structure of steel changes because of differential rates of cooling. 25. The method of concentrating the ore which makes use of the difference in density between ore and impurities is called (1) leaching. (2) liquefaction. (3) levigation. (4) magnetic separation. Solution (3) In the levigation process, lighter earthy particles are freed from the heavier particles by washing with water. 26. In the extraction of iron, Fe2O3 is reduced by (1) carbon dioxide. (2) aluminium.

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OBJECTIVE CHEMISTRY FOR NEET (3) carbon and carbon monoxide. (4) electrolytic reduction.

Solution (3) The reactions involved are K Fe2O3 + 3C 823  → 2Fe + 3CO

Fe2O 3 + CO → 2FeO + CO 2 27. Aluminium is the most abundant in the earth’s crust, yet it is obtained from bauxite because (1) (2) (3) (4)

bauxite contains maximum aluminium. bauxite is available in large quantity. bauxite is less impure. aluminium can be easily extracted from bauxite.

Solution (1) The earth’s crust contains aluminium in the form of two well-known minerals – bauxite (Al2O3×2H2O) and China clay (Al2O3×2SiO2×2H2O). But the extraction of aluminium is cheaper and easy from bauxite ore. 28. Which of the following chemical is used as a depressant in separating ZnS from PbS in froth floatation process? (1) NaCl (2) NaCN (3) ZnSO2 (4) BaCl2 Solution (2) Depressants depress the floating property of one of the components of the ore and thus help in the separation of different minerals present in the same ore. NaCN and KCN are the examples of depressants.

Practice Exercises Level I

(3) to derive benefits from the ore. (4) to heat the ore in presence of oxygen.

Occurrence of Metals

9. Which one of the following ores is best concentrated by froth floatation method?

1. Calamine is an ore of (1) aluminium. (2) copper. (3) iron. (4) zinc. 2. Which of the following is not an ore? (1) Malachite (2) Calamine (3) Sand (4) Cerussite 3. Which ore contains both iron and copper? (1) Chalcopyrite (2) Malachite (3) Chalcocite (4) Cuprite 4. Which of the following does not contain Mg? (1) Magnetite (2) Magnesite (3) Asbestos (4) Carnallite 5. Cassiterite is an ore of (1) Mn (2) Ni (3) Sb (4) Sn 6. In bauxite AlOx(OH)3–2x the value of x is (1) 0 < x < 1 (3) x = 0

(2) x = 1 (4) 1 < x < 0

7. Which of the following is a carbonate ore? (1) Pyrolusite (2) Malachite (3) Diaspore (4) Cassiterite

Concentration of Ore 8. Benefaction means (1) to treat the metal oxide with carbon monoxide. (2) to concentrate or remove impurities from an ore.

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(1) Magnetite (2) Malachite (3) Galena (4) Cassiterite 10. The substance which is mixed with the ore for removal of impurities is termed as (1) slag. (2) catalyst. (3) gangue. (4) flux. 11. Name the process for the concentration of the ore if the impurities present in the ore are lighter. (1) Magnetic separation (2) Froth floatation (3) Leaching (4) Hydraulic washing 12. The ore ZnS is concentrated by (1) froth floatation. (2) leaching. (3) magnetic separation. (4) washing with water. 13. Which of the following ores are concentrated by chemical leaching method? (1) Galena (2) Copper pyrite (3) Cinnabar (4) Argentite 14. Which one of the following is not a basic flux? (1) CaCO3 (2) SiO2 (3) CaO (4) MgO 15. Flux added in the extraction of iron is (1) feldspar. (2) flint. (3) silica. (4) limestone. 16. Gravity separation process is used for the concentration of

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General Principles and Processes of Isolation of Elements (1) calamine. (2) haematite. (3) chalcopyrite. (4) bauxite.

Extraction of Crude Metal from Concentrated Ore 17. In the isolation of metals, calcinations process usually results in (1) metal carbonate. (2) metal oxide. (3) metal sulphide. (4) metal hydroxide. 18. Roasting, a process used in metallurgy is (1) conversion of metal carbonate compounds to metal oxides by heating in air. (2) conversion of metal sulphides to metal oxides by ­heating in air. (3) conversion of metal carbonate compounds to metal sulfides by heating with sulfur. (4) conversion of metal sulphides to metal carbonate compounds by heating with carbon in the presence of air. 19. The reaction 2ZnS + 3O2 ® 2ZnO + 2SO2 in the metallurgical process of zinc is called (1) roasting. (2) smelting. (3) cupellation. (4) calcination. 20. What reduces haematite in the blast furnace? (1) C (2) CO (3) CO2 (4) SO2

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26. When limestone is heated strongly, it gives off CO2. In metallurgy, this process is known as (1) smelting. (2) calcination. (3) roasting. (4) ore dressing. 27. In blister copper, the blisters are formed due to passing out of the following gas. (1) Nitrogen (2) CO (3) CO2 (4) SO2

Thermodynamic Principles of Metallurgy 28. Which of the following metals can be obtained by thermal decomposition of their oxides at attainable temperatures? (1) Silver (2) Mercury (3) Gold (4) All of these 29. From the Ellingham graphs of carbon, which of the following statements is false? (1) CO2 is more stable than CO at less than 983 K. (2) CO reduces Fe2O3 to Fe at less than 983 K. (3) CO is less stable than CO2 at more than 983 K. (4) CO reduces Fe2O3 to Fe in the reduction zone of blast furnace. 30. Ellingham diagrams can be drawn for which of the following? (1) Sulphides (2) Oxides (3) Halides (4) All of these

21. The metal extracted by cyanide process is

31. Identify M and N in the following reaction. controlled (1) Ag (2) Cu ¾¾¾ ® M + SO2 Copper glance ¾heating (3) Fe (4) Na

22. The metallurgical process in which metal is obtained in fused state is called (1) calcination. (2) smelting. (3) froth floatation. (4) roasting. 23. Calcium is obtained by the (1) (2) (3) (4)

roasting of limestone. electrolysis of solution of calcium chloride in H2O. electrolysis of molten anhydrous calcium chloride. reduction of calcium chloride with carbon.

24. In the extraction of copper from its sulphide ore, the metal is formed by the reduction of Cu2O with (1) FeS (2) CO (3) Cu2S (4) SO2 25. Main function of roasting is (1) oxidation. (2) reduction. (3) slag formation. (4) to remove volatile substance.

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M N → Cu + SO2 ↑ (1) (2) (3) (4)

M = Cu 2O; N = Self-reduction M = Cu 2O + Cu 2S; N = only heating M = Cu 2O ; N = carbon reduction M = Cu 2O ; N = Electrolytic reduction

32. Select the true statement. (1) (2) (3) (4)

Below 710°C, C is better reducing agent than CO. Below 710°C, CO is better reducing agent than C. Below 710°C, CO is an oxidizing agent. Below 710°C, CO2 is a reducing agent.

33. Poling process is used for (1) (2) (3) (4)

removal of Cu from Cu2O. removal of Al from Al2O3. removal of Fe from Fe2O3. all of these.

34. Refractory materials are generally used in furnaces because (1) they can withstand high temperature. (2) they are chemically inert.

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OBJECTIVE CHEMISTRY FOR NEET (3) they do not require replacement. (4) they possess great structural strength.

Electrochemical Principles of Metallurgy 35. Before introducing FeO in blast furnace, it is converted to Fe2O3 by roasting so that (1) it may not be removed as slag with silica. (2) it may not evaporate in the furnace. (3) presence of it may increase the melting point of charge. (4) None of these. 36. The metal extracted by electrolysis of its fused salt is (1) copper. (2) sodium. (3) iron. (4) lead. 37. In the metallurgy of aluminium, (1) Al3+ is oxidized to Al (s). (2) graphite anode is oxidized to carbon monoxide and carbon dioxide. (3) oxidation state of oxygen changes in the reaction at anode. (4) oxidation state of oxygen changes in the overall reaction involved in the process 38. In the extraction of chlorine by electrolysis of brine (1) (2) (3) (4)

oxidation of Cl– ion to chlorine gas occurs. reduction of Cl– ion to chlorine gas occurs. For overall reaction ΔG° has negative value. a displacement reaction takes place.

Refining of Metals 39. In order to refine “blister copper”, it is melted in a furnace and is stirred with green logs of wood. The purpose is (1) (2) (3) (4)

to expel the dissolved gases in blister copper. to bring the impurities to surface and oxidize them. to increase the carbon content of copper. to reduce the metallic oxide impurities with hydrocarbon gases liberated from the wood.

40. Read the following statements: (I) (II) (III) (IV)

Al has greater affinity than Fe, for oxygen. Cast iron has impurity of zinc and lead. Refining of nickel is done by vapour phase refining. In cyanide process, oxygen and zinc dust are used as oxidizing agent and reducing agent, respectively.

Choose the correct set of statements. (1) I, III (2) II, III, IV (3) I, III, IV (4) I, IV

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41. For obtaining high purity metals, which of the following methods of refining is preferred? (1) Poling (2) Liquation (3) Zone refining (4) Electrolytic method 42. Electrolytic refining is used to purify which of the following metals? (1) Cu and Zn (2) Ge and Si (3) Zr and Ti (4) Zn and Hg 43. Which of the following activities is not related to poling? (1) Metallic impurities having higher oxidation potential than the metal to be refined are oxidized first. (2) Non-volatile oxides are removed in the form of scum. (3) Metallic impurities having lower oxidation potential than the metal to be refined settle down at the bottom of the furnace. (4) Green poles of wood are used as stirrers. 44. The process of liquation for refining of the metal is applicable to (1) Bi (2) Pb (3) Sn (4) All of these 45. To which of the following metals, the process of distillation is not applicable for refining? (1) W (2) Hg (3) Zn (4) All of these 46. In electrolytic refining, the impure metal is made is used to make (1) anode. (2) cathode. (3) electrolytic bath. (4) None of these.

Alloys and Amalgams 47. Which of the following is ferrous alloy? (1) Magnalium (2) Solder (3) Invar (4) Type metal 48. Which of the following is not an alloy? (1) Steel (2) Copper (3) Brass (4) Bronze 49. Brass is an alloy of (1) copper and tin. (2) copper and nickel. (3) copper and aluminium. (4) copper and zinc. 50. Amalgams are (1) (2) (3) (4)

alloys of silver. alloys of mercury. alloys in liquid state. highly coloured alloys.

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Level II

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(2) an acidic flux is needed. (3) both (1) and (2). (4) neither of them is needed.

Occurrence of Metals 1. Which of the following statements is correct? (1) Anthracite and chalcocite are both ores of copper. (2) Anthracite and chalcocite are both sulphide ores. (3) Both German silver and horn silver have zero percent silver content. (4) Malachite and azurite are both basic copper carbonate. 2. Which of the following metals are found in native state? (1) Ag, Pb, Fe (2) Ag, Pt, Mg (3) Au, Pt, Ag (4) Cu, Al, Zn 3. Cast iron is brittle because it contains (1) < 2% carbon. (2) > 6% carbon. (3) > 5% lead. (4) < 6% carbon. 4. Which of the following is not an ore of iron? (1) Dolomite (2) Magnetite (3) Siderite (4) Haematite 5. A number of elements are available in earth’s crust but most abundant elements are (1) Al and Fe (2) Al and Cu (3) Fe and Cu (4) Cu and Ag

Concentration of Ore 6. Match the ores with the preferred methods of ore dressing:

  (I) PbS

(p) Hydraulic washing



 (II) Fe2O3

(q) Froth floatation



(III) Bauxite

(r) Magnetic separation



(IV) FeCr2O4

(s) Leaching

(1) I ® (p); II ® (p, r); III ® (r); IV ® (s) (2) I ® (p); II ® (r); III ® (p, r); IV ® (q) (3) I ® (q); II ® (q, r); III ® (s); IV ® (r) (4) I ® (s); II ® (r); III ® (p, q); IV ® (p) 7. Froth floatation method is successful in separating impurities from ores because (1) the pure ore is soluble in water containing additives like pine oil, cresylic acid, etc. (2) the pure ore is lighter than water containing additives like pine oil, cresylic acid, etc. (3) the impurities are soluble in water containing additives like pine oil, cresylic acid, etc. (4) the pure ore is not easily wetted by water as by pine oil, cresylic acid, etc. 8. When a metal is to be extracted from its ore, if the gangue associated with the ore is silica, then (1) a basic flux is needed.

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9. Bauxite ore is concentrated by (1) (2) (3) (4)

electromagnetic separation. chemical separation. froth floatation. hydraulic washing.

10. 4M(s) + 8CN-(aq) + 2H2O(aq) + O2(g) ® 4[M(CN)2]−(aq) + 4OH−(aq). Here, (1) O2 acts as oxidizing agent and CN− acts as complex forming agent. (2) O2 acts as complex forming agent and CN− acts as oxidizing agent. (3) O2 and CN− both act as oxidizing agents. (4) O2 and CN− both act as complex forming agents. 11. In the froth floatation process for the purification of minerals, the particles float because (1) (2) (3) (4)

they are light. they are insoluble. their surface is preferentially wetted with oil. they bear an electrostatic charge.

12. Purpose of smelting of an ore is (1) (2) (3) (4)

to oxidize it. to remove vaporization impurities. to reduce it. to obtain an alloy.

13. General method for the extraction of metal from oxide ore is (1) (2) (3) (4)

carbon reduction. reduction by aluminium. reduction by hydrogen. electrolytic reduction.

14. Which one of the following is correct? (1) (2) (3) (4)

A mineral cannot be ore. An ore cannot be a mineral. All minerals are ores. All ores are minerals.

Extraction of Crude Metal from Concentrated Ore 15. The process of ore dressing is carried out to (1) (2) (3) (4)

remove the siliceous materials. add flux to the mineral. convert the ore to the oxide. remove the poisonous impurities.

16. Which process of reduction of mineral to the metal is suited for the extraction of copper from its ores with low copper content? (1) Metal displacement

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468

OBJECTIVE CHEMISTRY FOR NEET (2) Auto reduction (3) Chemical reduction (4) Electrolytic reduction

17. The overall floating of the ore occurs due to (1) absorption process between ore particle and collector. (2) only adsorption between ore particles and collector. (3) adsorption and lyophilic–lyophobic aptitude of collector. (4) only lyophilic–lyophobic aptitude of collector. 18. Extraction of gold and silver involves leaching the metal with CN– ion. The metal is recovered by (1) displacement of metal by some other metal from the complex ion. (2) roasting of metal complex. (3) calcination followed by roasting. (4) thermal decomposition of metal complex. 19. Which of the following reactions is an example of autoreduction? (1) Fe3O4 + 4CO ® 3Fe + 4CO2 (2) Cu2O + C ® 2Cu + CO (3) Cu2+ (aq) + Fe (s) ® Cu (s) + Fe2+ (aq) 1 1 (4) Cu 2O + Cu 2S ® 3Cu + SO2 2 2 20. Pb and Sn are extracted from their chief ore, respectively, by (1) (2) (3) (4)

electrolysis and self-reduction. self-reduction and electrolysis. carbon reduction and self-reduction. self-reduction and carbon reduction.

21. Select the incorrect statement. (1) In the Bayer’s Al2O3 goes in to solution as soluble Al(OH)3 remain insoluble while other basic oxides as TiO2 and Fe2O3. (2) Extraction of zinc from zinc blende is achieved by roasting followed by reduction with carbon. (3) The methods chiefly used for the extraction of lead and tin are respectively carbon reduction and electrolytic reduction. (4) Extractive metallurgy of magnesium involves fused salt electrolysis. 22. Matte contains mainly (1) Cu2S and FeS (2) Cu2S (3) CuS and Fe2S3 (4) Fe

(3) 2Cu2O + Cu2S → 6Cu + SO2↑ (4) 2Cu2S + 3O2 → 2Cu2O + 2SO2↑ 24. The furnace which gives the highest temperature is (1) blast furnace. (2) reverberatory furnace. (3) electrical furnace. (4) muffle furnace. 25. Which of the following factors is of no significance for roasting sulphide ores to the oxides and not subjecting the sulphide ores to carbon reduction directly? (1) Metal sulphides are thermodynamically more stable than CS2. (2) CO2 is thermodynamically more stable than CS2. (3) Metal sulphides are less stable than the corresponding oxides. (4) CO2 is more volatile than CS2. 26. Alumino-thermite process is commonly used for extraction of metals, whose oxides are (1) (2) (3) (4)

strongly basic. strongly acidic. fusible. not easily reduced by carbon.

27. Which of the following characteristics of steel are developed by the tempering process? (1) (2) (3) (4)

Steel becomes hard and brittle. Steel becomes soft. Steel remains hard but brittleness disappears. Only the surface layer becomes hard.

28. Statement I: Reduction of ZnO with carbon is done at 1100°C. Statement II: DG° is negative at this temperature thus, process is spontaneous. (1) If both statement I and statement II are true and statement II is a correct explanation for statement I. (2) If both statement I and statement II are true, but statement II is not a correct explanation for statement I. (3) If statement I is true, but statement II is false. (4) If both statement I and statement II are false. 29. In the modern blast furnaces, the charge consists of a mixture of (1) (2) (3) (4)

iron pyrites + bituminous coal. hydrated iron oxide + limestone + coke. calcined iron oxides + limestone + coke. calcined iron oxides + lime + anthracite coal.

Thermodynamic Principles of Metallurgy

Electrochemical Principles of Metallurgy

23. In the extraction of copper, the reaction which takes place in Bessemer converter is

30. Which of the following processes is used in extraction of magnesium?

(1) 2FeS + 3O2 → 2FeO + SO2↑ (2) 2CuFeS2 + O2 → Cu2S + FeS + SO2↑

Chapter 18_General Principles and Processes of Isolation of Elements.indd 468

(1) Fused salt electrolysis (2) Self-reduction

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General Principles and Processes of Isolation of Elements (3) Thermite reduction (4) Aqueous solutions extraction 31. Brine is electrolysed by using inert electrodes. The reaction at anode is 1 o = 1.36 V (1) Cl – (aq ) → Cl 2 (g ) + e – ; E Cell 2 o = 1.23 V (2) 2H 2O(l) → O 2(g ) + 4H + + e - ; E Cell o = 2.71 V (3) Na + (aq) + e – → Na(s); E Cell 1 + o (4) H (aq ) + e → H 2 (g ) ; E Cell = 0.00 V 2

Refining of Metals 32. There are different categories of lead pencil like 4B, 3B, 2B, HB, H, 2H, 3H and 4H (where B indicates softness and H indicates hardness). Which of the following statements is correct if all these varieties of lead pencil are made of different kinds of lead? (1) (2) (3) (4)

Impurity content increases from 4B to 4H. Impurity content decreases from 4B to 4H. Impurities have no impact on hardness. All are incorrect.

33. In the refining process of nickel by Mond’s process, the metal is obtained by (1) (2) (3) (4)

electrochemical reduction. thermal decomposition. chemical reduction by aluminium. reduction by carbon.

34. Which of the following is true for vapor phase refining?

469

39. By adding chromium to steel which of the following property is enhanced? (1) Resistance to corrosion. (2) Electrical characteristics. (3) Magnetic property. (4) Ductility. 40. Duralumin is an alloy of (1) (2) (3) (4)

aluminium and copper. aluminium and iron. aluminium and carbon. aluminium and mercury.

Previous Years’ NEET Questions 1. Which of the following statements, about the advantage of roasting of sulphide ore before reduction is not true? (1) Roasting of the sulphide to the oxide is thermodynamically feasible. (2) Carbon and hydrogen are suitable reducing agents for metal sulphides. (3) The ΔfGq of the sulphide is greater than those for CS2 and H2S. (4) The ΔfGq is negative for roasting for sulphide ore to oxide. (AIPMT 2007) 2. Sulphide ores of metals are usually concentrated by the froth flotation process. Which one of the following sulphide ores offers an exception and is concentrated by chemical leaching?

(1) The metal should form a volatile compound. (1) Sphalerite (2) Argentite (2) The metal should form a non-volatile compound. (3) Galena (4) Copper pyrite (3) The impurities in the metal should form a non-­ volatile compound. (AIPMT 2007) (4) The impurities in the metal should form a volatile 3. Match List-I (substance) with List-II (processes) compound. employed in the manufacture of the substances and 5. Elemental silicon to be used as a semiconductor is puri3 select the correct option. fied by (1) heating under vacuum. (2) floatation. (3) zone refining. (4) electrolysis. 36. Purification of aluminium by electrolytic refining is known as (1) Hall’s process. (2) Baeyer’s process. (3) Hoop’s process. (4) Serpeck’s process.

Alloys and Amalgams 37. Monel metal is an alloy of (1) Zu-Sn. (2) Cu-Sn. (3) Mg-Al. (4) Cu-Ni. 38. Which of the following is used for making mirrors? (1) Zn-Hg (2) Pure Sn (3) Sn-Hg (4) None of these

Chapter 18_General Principles and Processes of Isolation of Elements.indd 469

List-I Substances

List-II Processes

  (I)  Sulphuric acid

(p)  Haber process

 (II) Steel

(q)  Bessemer process

(III)  Sodium hydroxide

(r)  Leblanc process

(IV) Ammonia

(s)  Contact process

Options:  

(I)

(II)

(III)

(IV)

(1)

(p)

(q)

(r)

(s)

(2)

(s)

(r)

(q)

(p)

(3)

(s)

(q)

(r)

(p)

(4)

(p)

(s)

(q)

(r)

(AIPMT MAINS 2010)

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470

OBJECTIVE CHEMISTRY FOR NEET

4. Which of the following pairs of metals is purified by van Arkel method?

10. Aluminium is extracted from alumina (Al2O3) by electrolysis of a molten mixture of

(1) Ni and Fe (2) Ga and In (3) Zr and Ti (4) Ag and Au (AIPMT PRE 2011) 5. Which of the following elements is present as the impurity to the maximum extent in the pig iron? (1) Phosphorus (2) Manganese (3) Carbon (4) Silicon (AIPMT PRE 2011) 6. The following reactions take place in the blast furnace for the preparation of impure iron. Identify the reaction pertaining to the formation of the slag (1) 2C(s) + O2(g)→ 2CO(g) (2) Fe2O3(s) + 3CO(g)→2Fe(l) + 3CO2(g) (3) CaCO3(s) → CaO(s) + CO2(g) (4) CaO(s) + SiO2(s) → CaSiO3(s)

(AIPMT MAINS 2011)

(1) Al2O3 + Na3AlF6 + CaF2 (3) Al2O3 + HF + NaAlF4



List-II

Substances

Composition

  (I) Plaster of Paris

(p) CaSO 4 × 2H 2O

(II) Epsomite

(q) CaSO 4 × 1 2 H 2O

(III) Kieserite

(r) MgSO4×7H2O

(IV) Gypsum

(s) MgSO4×H2O

(NEET 2013)

12. ‘Metals are usually not found as nitrates in their ores’. Out of the following two (I and II) reasons which is/are true for the above observation?

 (I) Metal nitrates are highly unstable.



(II) Metal nitrates are highly soluble in water. (1) (2) (3) (4)

I and II are false. I is false but II is true. I is true but II is false. I and II are true. (AIPMT 2015)

13. Which of the following processes does not involve oxidation of iron?



(1) Decolorization of blue CuSO4 solution by iron. (2) Formation of Fe(CO)5 from Fe. (3) Liberation of H2 from steam by iron at high temperature. (4) Rusting of iron sheets. (AIPMT 2015)

14. In the extraction of copper from its sulphide ore, the metal is finally obtained by the reduction of cuprous oxide with (1) copper(I) sulphide. (2) sulphur dioxide. (3) iron(II) sulphide. (4) carbon monoxide.

(AIPMT MAINS 2011)

8. Which one of the following is a mineral of iron? (1) Pyrolusite (2) Magnetite (3) Malachite (4) Cassiterite (AIPMT PRE 2012)

9. Identify the alloy containing a non-metal as a constituent in it. (1) Bell metal (2) Bronze (3) Invar (4) Steel

(1) H2S (2) SO2 (3) CO2 (4) SO3



List-I

Code: (I) (II) (III) (IV) (1) (s) (r) (q) (p) (2) (r) (s) (p) (q) (3) (q) (r) (s) (p) (4) (p) (q) (r) (s)



(AIPMT PRE 2012)

11. Roasting of sulphides gives the gas X as a by-product. This is a colorless gas with choking smell of burnt sulphur and causes great damage to respiratory organs as a result of acid rain. Its aqueous solution is acidic acts as a reducing agent and its acid has never been isolated. The gas X is

7. Match List-I with List-II for the composition of substances and select the correct answer using the code given below the lists:



(2) Al2O3 + KF + Na3AlF6 (4) Al2O2 + CaF2 + NaAlF4

(AIPMT PRE 2012)

Chapter 18_General Principles and Processes of Isolation of Elements.indd 470



(RE-AIPMT 2015)

15. Match items of Column I with the items of Column II and assign the correct code Column I (I) Cyanide process

Column II (p) Ultrapure Ge

(II) Froth floatation process

(q) Dressing of ZnS

(III) Electrolytic reduction

(r)  Extraction of Al

(IV) Zone refining

(s)  Extraction of Au

 

(t)  Purification of Ni

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General Principles and Processes of Isolation of Elements Code: (I) (II) (III) (IV) (1) (s) (q) (r) (p) (2) (q) (r) (p) (t) (3) (p) (q) (r) (s) (4) (r) (s) (t) (p)

471

16. Extraction of gold and silver involves leaching with CN– ions. Silver is later recovered by (1) Distillation (2) Zone refining (3) Displacement with Zn (4) Liquation

(NEET 2017)

(NEET 2016)

Answer Key Level I  1. (4)

 2. (3)

 3. (1)

 4. (1)

 5. (4)

 6. (1)

 7. (2)

 8. (2)

 9. (3)

10. (4)

11. (4)

12. (1)

13. (4)

14. (2)

15. (4)

16. (2)

17. (1)

18. (2)

19. (1)

20. (2)

21. (1)

22. (2)

23. (3)

24. (3)

25. (4)

26. (2)

27. (4)

28. (4)

29. (3)

30. (4)

31. (2)

32. (2)

33. (1)

34. (1)

35. (1)

36. (2)

37. (2)

38. (3)

39. (4)

40. (3)

41. (3)

42. (1)

43. (3)

44. (4)

45. (1)

46. (1)

47. (3)

48. (2)

49. (4)

50. (2)

 1. (4)

 2. (3)

 3. (1)

 4. (1)

 5. (1)

 6. (3)

 7. (4)

 8. (1)

 9. (2)

10. (1)

11. (3)

12. (3)

13. (1)

14. (4)

15. (1)

16. (2)

17. (3)

18. (1)

19. (4)

20. (4)

21. (3)

22. (1)

23. (3)

24. (3)

25. (3)

26. (4)

27. (3)

28. (1)

29. (3)

30. (1)

31. (1)

32. (1)

33. (2)

34. (1)

35. (3)

36. (3)

37. (4)

38. (3)

39. (1)

40. (1)

 7. (3)

 8. (2)

 9. (4)

10. (1)

Level II

Previous Years’ NEET Questions  1. (2)

 2. (2)

 3. (3)

 4. (3)

 5. (3)

 6. (4)

11. (2)

12. (2)

13. (2)

14. (1)

15. (1)

16. (3)

Hints and Explanations Level I

10. (4) Flux helps in removal of impurities from the ore.

1. (4) Calamine (ZnCO3) is an ore of zinc.

11. (4) This technique utilizes the density difference between the ore and the impurity to concentrate the ore.

2. (3) Sand is not an ore. It is SiO2. 3. (1) Chalcopyrite (CuFeS2) is a copper iron sulphide mineral. 4. (1) Magnetite is an ore of iron.

Magnetite is Fe3O4 Siderite is Fe(CO)5 Haematite is Fe2O3⋅2H2O

12. (1) Sulphide ores are normally concentrated by froth floatation process. 14. (2) SiO2 reacts with basic fluxes such as CaCO3, CaO and MgO to give fusible matter called slag. SiO2 itself is acidic in nature. (It acts as an acidic flux).

7. (2) Malachite, CuCO3 ⋅Cu(OH)2

15. (4) Limestone is added which decomposes into CaO and CO2. The lime thus produced acts as a flux and combines with silica (present as an impurity) to produce slag.

8. (2) The concentration (dressing or benefaction) of ore is done by separating unwanted materials (sand, clays, etc.).

17. (1) Calcination is used generally for metal carbonates.

9. (3) Galena is a sulphide ore and sulphide ores are concentrated by froth floatation method.

18. (2) It is an exothermic process in which the concentrated ore (mostly sulphide ore) is heated to a high temperature

Chapter 18_General Principles and Processes of Isolation of Elements.indd 471

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OBJECTIVE CHEMISTRY FOR NEET (just below its fusion temperature) in presence of excess of air to convert it into its oxide form.

19. (1) The process of oxidizing sulphide ores is known as roasting. In roasting, concentrated ore is roasted in the presence of excess of air at about 1200 K to convert ZnS into ZnO. 20. (2) The reactions involved are Fe2O3 + CO ® 2FeO + CO2 FeO + CO ® Fe + CO2 21. (1) Metals like gold and silver are only extracted by cyanide process or Macarthur process. 22. (2) Smelting is the process in which the metal is extracted from its ore in a fused state. 25. (4) While roasting, the impurities present are removed as their volatile oxides. 26. (2) Calcination is the process in which the concentrated ore is heated strongly in the presence of a limited supply of air. D CaCO3 →  CaO + CO 2

45. (1) The boiling point of W is very high, so it cannot be purified by distillation. 46. (1) Impure metal undergoes oxidation at anode. 47. (3) Invar is an alloy of nickel and iron. 50. (2) Amalgams are the alloys which contain Hg as one of the component.

Level II 1. (4) Malachite: Cu(OH)2·CuCO3.

Azurite: Cu(OH)2·2CuCO3. Anthracite is a variety of coal. Chalcocite is the Cu2S.

2. (3) Au, Pt and Ag metals are found in native state. The elements are said to be in native state if they are found in their elementary form. Generally, less reactive elements are found in native state. 3. (1) Cast iron normally contains 1.5% carbon. 4. (1) Dolomite is CaCO3·MgCO3

28. (4) Silver, mercury and gold can be obtained by thermal decomposition of their oxides at attainable temperature.

6. (3) Hydraulic washing is mainly applicable to oxide ores, such as Fe2O3, FeCr2O4.

30. (4) Ellingham diagram is used to evaluate the case of reduction of metal oxides, sulphides and halides.



31. (2) Due to partial roasting the (M) will be (Cu2O + Cu2S) and this mixture on heating (N) only undergoes selfreduction. Self-reduction is the process, not a condition of the reaction. 32. (2) At 600°C, ferric oxide is reduced by carbon monoxide to give iron. Fe2O3(s) + 3CO ® 2Fe(s) + 3 CO2(g ) 34. (1) In furnaces heating can be done to much higher temperatures in a controlled way. 36. (2) Na is extracted by the electrolysis of fused NaCl. 39. (4) Blister copper is purified by heating it strongly in a reverberatory furnace in presence of excess of air. Some of the Cu changes to CuO. This is reduced back to Cu by stirring the molten metal with green poles of wood.



Froth floatation is applicable to sulphide ores, such as galena (PbS). In bauxite ore, Al2O3 is leached out as sodium aluminate. Magnetic separation is used to separate non-magnetic impurities from magnetic ores, or vice versa, such as Fe2O3.

7. (4) This is because the pure ore is not easily wetted by water as by pine oil, cresylic acid, etc. 8. (1) As SiO2 is acidic, basic flux is needed. 9. (2) Bauxite ore is concentrated by chemical separation. 11. (3) The pine oil gets absorbed at the surface of sulphide ore particles which become water repellent and comes out along with froth when air is blown through the mixture. 12. (3) Smelting is the process of reducing the calcined or roasted ores. 20. (4) Pb is extracted by self-reduction, that is,

41. (3) High purity elements like Ge, Si, B, Ga, In, etc. are purified by zone refining method. 44. (4) Liquation process can be used for metals with low melting points, so that the metal is separated from high melting impurities.

Chapter 18_General Principles and Processes of Isolation of Elements.indd 472

2PbS + 3O2 ® 2PbO + 2SO2 2PbO + PbS ® 3Pb + SO 2

Sn is extracted by reduction with either C or CO. SnO2 + C ® Sn + CO

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General Principles and Processes of Isolation of Elements



Change in free energy ∆G°, kJ mol−1

21. (3) Lead (self-reduction) 2PbO + PbS ® 3Pb + SO2 Tin (carbon reduction) SnO2 + 2C ® Sn + 2CO

23. (3) 2Cu2O + Cu2S → 6Cu + SO2 24. (3) An electrical furnace has a heat producing equipment which can be heated to produce high temperature. The temperatures attained in other furnaces depend on heats of combustion of fuels used. 25. (3) As DG is positive for reduction of metal sulphides by carbon, whereas it is negative for reduction of metal oxides by carbon. So, CO2 is thermodynamically more stable than CS2, thus metal sulphides are more stable than metal oxides. 26. (4) Metal oxides like Cr2O3, Mn3O4 cannot be reduced effectively with carbon or CO. In such cases, Al is used as a reducing agent. 28. (1) All three oxidation curves for the carbon system lie above that for oxidation of zinc, until a temperature of approximately 1000°C is reached. At this point C is thermodynamicallv capable of reducing ZnO to Zn. Since this temperature is greater than the boiling point of Zn (907°C), it will be formed as a vapor. The overall equation for reduction is



ZnO(s) + C (s) ® Zn(g) + CO(g)

29. (3) The added limestone serves as a flux and coke serves as fuel.



Ni

Impure nickel

- 350 K 450- 470 K + 4CO 330  → Ni(CO)4  → Ni + CO Thermal decomposition

Pure nickel

34. (1) In this refining process, the metal is converted to its volatile compound and the vapors are collected and decomposed to give pure metal. The volatile compound should be easily decomposable, so that the recovery is easy. 37. (4) Monel metal is an alloy of Cu, Ni, Mn or Fe. 38. (3) Tin-mercury mirrors were made by coating glass with tin amalgam.

Previous Years’ NEET Questions 1. (2) Many metals occur as sulphide ores. Though carbon is a good reducing agent for oxides, it is a poor reducing agent for sulphides. The reason why carbon reduces so many oxides at elevated temperatures is that the ΔG°/T line for CO has a negative slope.

Chapter 18_General Principles and Processes of Isolation of Elements.indd 473

2

−400

→2

CO

C + O2 → CO2

−600 −800 −1000 500 710 1000 1500 Temperature, °C

2500

2000

There is no compound CS analogous to CO with a steep negative ΔG°/T line. Thus sulphides are normally roasted in air to form oxides before reducing with carbon. 2MS + 3O2 ® 2MO + 2SO2

2. (2) Leaching is the process by which the required substance (may be the metal component or impurities) is dissolved out from the ore by using a suitable reagent. For example, Ag2S is dissolved out from silver glance or argentite (Ag2S + impurity) by using NaCN solution in the presence of air. Ag 2S + 4NaCN ® 2[Ag(CN)2 ]- + 4 Na + + S 2 3. (3)

Option (I) → (s): H2SO4 is the most important acid used in the chemical industry. By far the most important ­commercial process for its manufacture is the contact process.



Option (II) → (q): Bessemer process is used in the production of steel from iron.



Option (III) → (r): In Leblanc process, the chemicals used are H2SO4, NaCl, CaCO3 and C, and the products are NaOH and Cl2.



Option (IV) → (p): Most of ammonia is manufactured synthetically from H2 and N2 by the Haber-Bosch process.

30. (1) Electrolysis of fused MgCl2 takes place. 33. (2) The reaction involved is

−200 2C +O

473

4. (3) van Arkel method is used to get ultrapure metals, such as Zr and Ti. The reaction involved in the purification of Zr and Ti is 1400° C 2 ( vap ) Zr or Ti I250  → ZrI 4 or TiI 4 Tungsten  → Zr(s)or Ti(s) + 2I 2(g )    filament °C      

Impure

Volatile

Pure

5. (3) Pig iron has very high carbon content (3.5 – 4.5%) which makes it brittle. 6. (4) Slag is obtained when flux reacts with the acidic or basic impurity present in the oxide ore. CaO(s)

( Basic flux )

+

SiO 2 (s)

( Acidic flux )

→ CaSiO3(s)   (Slag )

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OBJECTIVE CHEMISTRY FOR NEET

7. (3) The composition of the given substances is as follows: Substance

Composition

Plaster of Paris

CaSO 4 × H 2O

Epsomite

MgSO 4 × 7H 2O

Kieserite

MgSO 4 × H 2O

Gypsum

CaSO 4 × 2H 2O

1 2



Option (3): Iron reacts with steam to give a mixture of ferric oxide and hydrogen. 3Fe + 4H 2O → Fe3O4 + 4H 2 (steam)



Option (4): Corrosion of iron is an oxidation process. 3 2Fe(s) + O 2(g ) + xH 2O( l ) ® Fe2O 3 × xH 2O(s) 2

14. (1) This is self-reduction process. The reaction is

8. (2) Magnetite, that is, Fe3O4 is the mineral ore of iron.



9. (4) Steel always contains some percent of carbon in it.

15. (1)

Metal sulphide + O2 → Metal oxide + SO2 ↑

16. (3) Process used for extraction of silver is cyanide process. Reactions taking place at the different stages in the cyanide process are as follows:

SO2 is a colorless gas, which has choking smell, and is very soluble in water.

Ag 2S + 4NaCN(excess)  2[ Ag(CN )2 ] + Na 2S

12. (2) Metals are usually not found as nitrates in their ores because they are highly soluble in water.



Option (1): This is a displacement reaction, involves the oxidation of iron. Fe + CuSO4 ® FeSO4 + Cu



4Na 2S + 5O 2(air ) + 2H 2O ® 2Na 2SO4 + 4NaOH + 2S

13. (2)

Option (2): It does not involve oxidation of iron. Fe + 5 CO ® Fe(CO)5

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Cu2S + 2Cu2O → 6Cu + SO2 ↑

(I) → (s): Cyanide process is used for the extraction of Au in hydrometallurgy. (II) → (q): Froth flotation process is used for the dressing of sulphide ores of Zn. (III) → (r): Electrolytic reduction is used for the extraction of Al. (IV) → (p): Zone refining process is used for obtaining ultrapure Ge.

10. (1) In electrolytic reduction of Al2O3, that is, molten alumina (20%) mixed, with cryolite (Na3AlF6) and fluorspar (CaF2) is taken in an iron tank with carbon lining that acts as the cathode. They increase the conductivity and also decrease the melting point of the melt. 11. (2) Roasting is a process mainly applicable for sulphide ores to get corresponding metal oxides with the evolution of sulphur dioxide gas (X). The reaction is



In the step involving precipitation of Ag, a little excess of Zn powder is added by which Na[Ag(CN)2] becomes the limiting reagent, otherwise the loss of Ag will be more. Here Zn is chosen because it is more electropositive as compared to Ag and the replacement reaction occurs very easily. 2Na[ Ag(CN )2 ]  Zn  Na 2[Zn(CN )4 ]  2 Ag 

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19

The d- and f-Block Elements

Chapter at a Glance 1. The Transition Elements or d-block Elements The d-block elements are four series of elements formed by filling the 3d, 4d, 5d and 6d shells: 3d: 21Sc to 30Zn 4d: 39Y to 48Cd 5d: 57La, 72Hf to 80Hg 6d: 89Ac (Incomplete series) The 3d, 4d and 5d series have 10 elements each, whereas 6d is incomplete. Electronic configuration  (i) The electronic configuration of d-block elements can be represented as (n− 1) d1–10ns1–2. Half-filled and completely filled orbitals are extra stable. (ii) Zn, Cd and Hg are not considered as transition metals as their orbitals are completely filled in the ground state as well as in their common oxidation states. (iii) Transition elements with similar d n configuration, where n = 1 to 10, show similarities in electronic and magnetic properties. These elements show greater horizontal similarity (within a row) in properties as compared to the main group elements.  (iv) All transition elements with partly filled d-orbitals show certain characteristic properties such as variable oxidation state and formation of complexes with ligands. They form many colored and paramagnetic compounds. (b) Melting and boiling points (i) The melting and boiling points of the transition elements are generally very high. The high melting point is due to the higher number of electrons of the (n − 1)d orbitals along with the ns electrons involved in the interatomic metallic bonding. Bonding gets stronger with increase in the number of valence electrons. More the number of unpaired electrons involved in the metallic bonding, higher will be the melting point. (ii) Across a row, the melting point of transition elements increases up to d5 and then fall regularly as the atomic number increases. Mn and Tc, however, are exceptions. (c) Enthalpies of atomization: Transition metals have high enthalpies of atomization and these increase with increasing number of electrons up to d 5 and then decrease. This is due to increased interatomic interactions with increasing number of electrons. (d) Variation in atomic and ionic sizes of transition metals  (i) The atomic radii of the elements decrease from left to right across a row in the transition series, until near the end when the size increases slightly. (ii) On passing from left to right, extra protons are placed in the nucleus and extra orbital electrons are added. The orbital electrons shield the nuclear charge incompletely due to poor shielding by d electrons. Thus the nuclear charge attracts all of the electrons more strongly and a size contraction occurs. The shielding effect of electrons decreases in the order s >p > d > f. The contraction in size from one element to another is fairly small. (iii) The elements in the first group in the d-block (Group 3) show the expected increase in size, Sc → Y → La. However, in the subsequent Groups (3–12) there is an increase in radius of 10 → 20 pm between the first and second member, but hardly any increase between the second and the third elements.  (iv) There is a gradual decrease in size of the 14 lanthanide elements from cerium to lutetium and this is known as the lanthanoid contraction. Lanthanoid contraction almost exactly cancels the normal size increase on (a)

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descending a group of transition elements. Therefore, the second and third row transition elements have similar radii. Thus, the differences in properties between the first row and second row elements are much greater than the differences between the second and third row elements. (e) Ionization enthalpies (i) Ionization enthalpy increases across a period in the transition series. (ii) A small change in the values of the first ionization enthalpy is observed. The increase in second and third ionization enthalpies is much higher for the successive members of the first transition series. (f ) Oxidation states (i) The common oxidation state is +2. The maximum oxidation state is the number of valance electrons. (ii) Lower oxidation states are ionic whereas higher oxidation states are covalent. (iii) Among first five elements, the correlation between electronic structure and minimum and maximum oxidation states in simple compounds is complete. (iv) In the last five elements, the tendency for all the d electrons to participate in bonding decreases. Thus, Fe has a maximum oxidation state of (+6). However, the second and third elements in this group attain a maximum oxidation state of (+8). (g) Trends in standard electrode potentials (i) For M2+/M: The standard electrode potential for M2+ ions (E °) in solution for first row transition elements becomes less negative as we move across the series. This is due to a successive increase in the sum of first and second ionization enthalpies. (ii) For M3+/M2+: The standard electrode potential for M3+/M2+ ions (E °) in solution for first row transition elements do not depict any regular trend. (h) Some important trends and irregularities in halides and oxides are listed as follows: (i)  In halides: The characteristic feature of fluorides is that they are stable in higher oxidation states and are unstable in low oxidation states. Trifluorides (MF3) are generally ionic, whereas chlorides, bromides and iodides (where known) show considerable covalent character. (ii)  In oxides: Highest oxidation state of 7 is attained by Mn2O7. The ability of oxygen to form multiple bonds with metals explains the ability of oxygen to stabilize higher oxidation states. (i) Chemical reactivity and E ° values: The first row transition metals are generally reactive and oxidized to M2+ in presence of non-oxidizing acids (H+) with the exception of copper. The tendency to form M2+ ions reduces as we move down the series. Mn3+ and Co3+ are the strongest oxidizing agents in aqueous solution, and Ti2+, V2+ and Cr2+ are strongest reducing agents. (j) Magnetic properties: Many compounds of the transition elements are paramagnetic because of the presence of unpaired electrons or partially filled electron shells. (i) If the magnetic moment is measured, the number of unpaired electrons can be calculated. Magnetic moment (spin only), ms  n(n  2) BM, where n is the number of unpaired electrons. (ii) Certain elements, such as Fe, Cu and Ni, are easily magnetized therefore act as ferromagnetic substances. (iii) Elements like Zn, Cd and Hg and certain transition metal ions, like Zn2+, Cd2+, Hg2+, Sc3+, Ti4+ are diamagnetic due to the absence of unpaired electron. (k) Color: Most of the transition metal compounds (ionic and covalent) are colored due to d-d transitions resulting from excitation of electrons from lower energy to higher energy d-orbitals. (i) In a free isolated gaseous ion, the five d orbitals are degenerate. In real-life situations, the ion will be surrounded by solvent molecules or other ligands or ions, which affect the energy of some d orbitals more than others. Thus, the d orbitals are no longer degenerate and thus it is possible to promote electrons from one d level to another d level of higher energy in transition element ions with a partly filled d shell. This corresponds to a fairly small energy difference, and so light is absorbed in the visible region. (ii) In transition metal ions which have completely filled d orbitals, the electrons cannot undergo d-d transitions, hence these are colorless. (iii) KMnO4 is colored due to the formation of charge transfer spectra.

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(l) Complex formation: Transition metals have tendency to form coordination compounds (complexes) with Lewis bases (ligands) which can donate an electron pair. The reason transition elements form complexes is that they have small, highly charged ions and vacant low energy orbitals to accept lone pairs of electrons donated by other groups or ligands. (m) Catalytic properties: The catalytic properties of transition metals are attributed to their ability to exist in multiple oxidation states and form complexes. Variable oxidation states make them more efficient catalysts. Iron, titanium vanadium, nickel, platinum and palladium are examples of transition metal catalysts. (n) Formation of interstitial compounds: The transition metals form some interstitial compounds when small atoms such as hydrogen, carbon, boron or nitrogen get entrapped inside voids (interstitial spaces) in the crystal metal lattice. Some examples include titanium carbide, titanium nitride, Fe3H, Mn4H, VH0.56, TiH1.7, VSe0.98 and Fe0.94O. 2. Some Important Compounds of Transition Elements (a) Oxides and oxoanions of metals  (i) The transition metals (except scandium) react with oxygen to form ionic oxides. The ionic character ­decreases as the oxidation number increases. The higher oxides show acidic character. For e.g., formation of HMnO4 from Mn2O7 and H2CrO4 and H2Cr2O7 from CrO3. (ii) A change from basic to amphoteric character is shown in vanadium and chromium. (iii) In vanadium, V2O5 is acidic and reacts with alkalis to form VO3 4+ salts; it also behaves as base and forms VO4 + salts.V2O3 is basic and V2O4 is less basic. (b) Potassium dichromate (K2Cr2O7) (i) Preparation: • From chromite ore: 1000 1300C 4FeCr2 O4  8Na 2 CO3  7O2   8Na 2 CrO4  2Fe 2 O3  8CO2  Red hot in presence of air

Chromite ore

Obtained solid after cooling

Dissolve in water and filter

Filterate Na2CrO4

Residue Fe2O3

2Na 2 CrO4  H2 SO4  Na 2 SO4  Na 2 Cr2 O7  H2 O • Potassium dichromate can be obtained by double displacement reaction using hot concentrated ­Na2Cr2O7 solution and potassium chloride. Double Na 2 Cr2 O7  2KCl   K 2 Cr2 O7  2NaCl decomposition

(ii) Properties: • It decomposes on heating to give Cr2O3, K2CrO4 and O2. • Its acidic solution contains CrO2− 4 .

− Cr2 O72− , in which two tetra• On acidifying, chromates CrO2− 4 form HCrO 4 and orange–red dichromates hedral units join together by sharing the oxygen atom at one corner with bond angle of 126°.

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−

Cr

−

Chromate

−

Cr

Cr

−

Dichromate

• Chromate (Cr2 O72−) and dichromate (Cr2 O72−) exist in equilibrium over a wide range of pH from 2-6. • They are a strong oxidizing agent. • The oxidation state of chromium in chromate and dichromate is the same. • Its equivalent mass in acidic medium is 49 (M/6). • It is used in volumetric analysis as a standard solution. (c) Potassium permanganate (KMnO4): (i) Preparation • KMnO4 is prepared from MnO2 in the presence of alkali when dark green K2MnO4 is formed, which further disproportionate to form KMnO4. Fuse in NaOH

MnO2 + KNO2  K 2 MnO4 + NO Electrolytic oxidation

K 2 MnO4 + H2 O  KMnO4 + KOH + H2 • KMnO4 is manufactured on a large scale by the alkaline oxidative fusion of MnO2 followed by electrolytic oxidation in alkaline medium: Fuse with KOH

Electrolytic oxidation

MnO2   MnO24   in  MnO4 Oxidize with air or KNO3 alkaline solution Manganate

Permanganate

• KMnO4 is prepared in the laboratory by the oxidation of Mn(II) salt using peroxodisulphate. 2Mn2+ + 5S2 O82  + 8H2 O  2MnO4 + 10SO24  + 16H+ (ii) Properties • KMnO4 is a dark purple–black solid. MnO4 − ion is deep purple colored due to charge transfer spectra. • KMnO4 is only partially soluble in water and decomposes on heating at 200°C. Addition of little water to this residue, followed by filtration, gives green filtrate and black residue. ∆ 2KMnO4   K 2 MnO4  MnO2  O2 200C

Green



Black

• KMnO4 is used as an oxidizing agent. It should be stored in dark bottles and standardized just before use because it is not a primary standard and also it is slowly reduced to MnO2 in presence of acid or light. • Its equivalent mass in acidic medium is M/5; basic medium M/1; strong basic medium M/3 and neutral medium M/5. The oxidation reactions are: In alkaline medium: 2MnO4  2OH  2 MnO24   H2 O +[O] 2MnO4  2H2 O  2MnO2  4OH + 2[O]



Some examples, 2KMnO4  H2 O  KI  2MnO2  2KOH+KIO3 2KMnO4  3HCO2 K  2 MnO2  2KHCO3 +2 K 2 CO3  H2 O 2KMnO4  3H2 O2  2KOH  2MnO2 + 2H2 O+ 3O2

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In acidic medium: 2MnO4  6H  5SO32   2Mn 2+  5SO24   6H2 O

3. The Inner Transition Elements (f-Block) or The Lanthanides Elements The 14 elements between La and Hf in which the 4f orbitals are successively filled are called lanthanides because of their chemical resemblance to lanthanum. These are placed at the bottom of the periodic table. Together with actinides, they are called f-block elements. (a) Electronic configurations: The general electronic configuration is (n – 2)f 1–14 (n – 1)d 0–1 ns2. (b) Atomic and ionic radii (i) The atomic radius decreases gradually from Ce (187 pm) to Lu (171 pm) by nearly 20 pm. This contraction in radius is due to lanthanide contraction. But this regular decrease in radius is not followed by Eu (At. no. 63) and Yb (At. no. 70). (ii) Ionic radii depend on how many electrons are removed. The properties of an ion depend on its size and its charge. The Ln3+ lanthanide ions change by only a small amount from one element to the next and their charge is the same, and so their chemical properties are very similar. (c) Oxidation states: The very common and stable oxidation state for lanthanides is +3. Hence they are placed in group III B in the periodic table. Oxidation states (+2) and (+4) occur, particularly when they lead to the following: (i) A noble gas configuration, example, Ce4+ ( f 0); (ii) A half-filled f shell, example, Eu2+ and Tb4+ ( f 7); (iii) A completely filled f  level, example, Yb2+ ( f 14); The 4f electrons do not take part in bonding. They are neither removed to produce ions nor do they take any significant part in crystal field stabilization of complexes. This is because; in the antipenultimate shell 4f electrons are very effectively shielded from their chemical environment outside the atom by the 5s and 5p electrons. (d) Metallic properties: They exhibit metallic structure and are good conductors of heat and electricity. The metals are all soft and silvery white. (e) Melting and boiling points: The hardness, melting points and boiling points of the elements all increase from Ce to Lu. This is because the attraction between the atoms increases as the size decreases. (f ) Color: Elements with (n) f electrons often have a similar color to those with (14 – n) f electrons. However, the elements in other valency states do not all have colors similar to their isoelectronic 3+ counterparts. (g) Magnetic nature (i) Lu3+ has f 14configuration. These have no unpaired electrons, and thus, they are diamagnetic. All other f states contain unpaired electrons and are, therefore, paramagnetic. (ii) From Ce3+ to Gd3+, the paramagnetic behavior increases and then decreases and finally becomes diamagnetic into Lu3+. Ce4+, Yb2+ and Lu3+ are diamagnetic. (h) Ionization enthalpy (i) The sum of the first three ionization enthalpies varies with minima at La3+, Gd3+ and Lu3+ which are associated with attaining an empty, half-full or full f shell thereby, giving stability. (ii) Maxima occur at Eu3+ and Yb3+ associated with breaking a half-filled or fully filled shell. The first and second ionization enthalpies of lanthanides are around 600 kJ mol−1 and 1200 kJ mol−1, respectively. (i) Some other properties (i) Ionization energies are fairly low and comparable to alkaline earth metals. Hence, they are good reducing agent. (ii) Complex formation tendencies are less and Lu3+ forms several complexes due to its smallest size. (iii) Basic nature gradually decreases from Ce(OH)3 to Lu(OH)3 due to higher polarization caused by gradual decrease in size of M3+ ion. The hydroxides Ln(OH)3 is less basic than Ca(OH)2 and more basic than Al(OH)3. (iv) They form carbide like MC2 type which on hydrolysis produces C2H2.

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4. The Actinides The actinides are a second inner transition series, beginning with thorium and ending with lawrencium. Since these man-made elements have atomic numbers higher than 92U they are sometimes called the transuranium elements. (a) Electronic configurations: The electronic structures of the actinides do not follow the simple pattern found in the lanthanides. The participation of the 5f orbitals explains the higher oxidation states shown by the earlier actinide elements. (b) Ionic sizes: The size of the ions decreases regularly along the series, because the extra charge on the nucleus is poorly shielded by the f electrons. This results in an “actinide contraction”. However, the contraction from element to element in the actinide series is more because of the poor shielding by 5f electrons. (c) Oxidation states (i) Variable oxidation states exist because 5f, 6d and 7s levels are of comparable energies. The actinides in general, have an oxidation state of +3, like the lanthanides. However, this is not always the most stable oxidation state in the actinides. (ii) +3 is not the most stable oxidation state for the first four elements Th, Pa, U and Np. The most stable oxidation states for the first four elements are Th (+4), Pa (+5) and U (+6). Though Np (+7) exists, it is oxidizing and the most stable state is (+5). (iii) The +3 state is the most stable state for the later elements. Their properties are similar to those of the lanthanides. (d) Magnetic properties: Ions are paramagnetic or diamagnetic that depends upon number of 5f electrons which can be obtained from their electronic configuration. (e) Color: In general, actinide ions are colored unless they are diamagnetic. (f ) Melting points and boiling points: There is no regular trend with increase in atomic number. (g) Ionization enthalpies: The ionization enthalpy values are lower than those of early lanthanides. This is because 5f electrons are more effectively shielded from the nuclear charge than the 4f electrons of the corresponding lanthanides. The outer electrons are thus loosely held and are available for bonding in the actinides. 5. Comparison Between Lanthanides and Actinides (a) Similarities:  (i) Both are electropositive and strongly reducing in nature. (ii) Both are having stable oxidation state of +3. (iii) Both kinds of ions are colored due to f-f transition.  (iv) Both series show the gradual decrease in ionic radius. (b) Differences: Lanthanides Other than +3 oxidation state, +2, and +4 are available for some elements. All are non-radioactive except Pm. Lanthanides contraction is less than actinide contraction. Lanthanides have fewer tendencies towards complex formation. Oxides and hydroxide are relatively less basic. Densities are less.

Actinides Other than +3 oxidation states +4, +5, +6, +7 are available for some elements. All are radioactive. Actinide contraction is more than lanthanide contraction. Actinides have relatively stronger tendency towards complex formation. Oxides and hydroxide of actinides are more basic. Densities are relatively more.

6. Applications of d-and f-block Elements (a) d-Block elements (i) Titanium is lighter than other elements and resistant to corrosion. So it is used in turbine engines and industrial chemical, aircraft and marine equipment.

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(ii) V  anadium is commercially used in alloy steels and cast iron, to which it lends ductility and shock ­resistance. (iii) For construction materials, iron and steel is of prime importance. Chromium, being resistant to c­ orrosion is used as an electroplated protective coating. (iv) The oxides of Ti and Mn are used in pigment industry and in dry battery cells. (v) Metals such as silver, gold, copper are used for making coins. (vi) Nickel, being resistant to attack by air or water at ordinary temperatures is electroplated as a protective coating. (vii) AgBr finds application in photographic industries. (viii) V2O5 is used as a catalyst for oxidation of sulphur dioxide in manufacture of sulphuric acid. Raney nickel is used for catalytic reductions. Nickel compounds are also used in the polymerization of alkynes. Hydrated manganese dioxide is used in organic chemistry for the oxidation of alcohols and other compounds. Iron catalysts are used in the manufacture of ammonia. Copper compounds catalyze a number of organic reactions. TiCl4 with Al(CH3)3 are catalysts in polythene manufacture. (b) f-block elements (i) Misch metal (50% Ce, 40% La, 7% Fe and 3% other metals) is added to steel to improve its strength and workability. (ii) CeO2 is used to polish glass. (iii) Thorium is used as fuel rods in nuclear reactors and in treatment of cancer. (iv) Uranium is used as nuclear fuel. The salts are used in medicines and to impart green color to gases.

Solved Examples 1. General electronic configuration of lanthanides is (1) (n − 2)f 1−14(n − 1)d 0−1ns2 (2) (n − 2)f 10−14(n − 1)d 0−1ns2 (3) (n − 2)f 0−14(n − 1)d10ns2 (4) (n − 2)d 0−1(n − 1)f 1−14ns2 Solution (1) The general electronic configuration for lanthanides is [Xe]4f 1−145d 0−16s2.

Solution (3) Among the given elements, Manganese shows the maximum number of oxidation states as all of the s and d electrons are being used for bonding. Mn has oxidation states (+II), (+III), (+IV), (+V), (+VI) and (+VII). 4. Zn gives H2 gas with H2SO4 and HCl but not with HNO3 because (1) Zn acts as an oxidizing agent when reacts with HNO3. (2) HNO3 is weaker acid than H2SO4 and HCl. (3) in electrochemical series Zn is above hydrogen. (4) NO3− is reduced in preference to hydronium ion.

2. An atom has electronic configuration 1s2 2s2 2p6 3s2 3p6 3d3 4s2, in which group it will be placed? (1) Fifth (2) Fifteenth (3) Second (4) Third Solution (1) The general valence shell electronic configuration for d-block elements is ns0−2(n–1)d1−10 . From the electronic configuration 1s2 2s2 2p6 3p6 3d3 4s2, we can see that the last electron enters into the d-orbital, therefore, the element belongs to d-block.

Group number of d-block element = Total number of valence shell electrons = (3 + 2) = 5

3. Which of the following shows maximum number of oxidation states? (1) Cr (2) Fe (3) Mn (4) V

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Solution (3) An element can displace all elements that lie below it in the series, from their salt solutions. For example, zinc has lower reduction potential than hydrogen, therefore, it displaces hydrogen from diluted HCl and H2SO4. HNO3 is a strong oxidizing agent, so, hydrogen obtained in the reaction with zinc is converted to water.



4Zn + 10HNO3 → 4Zn (NO3)2 + NH4NO3 + 3H2O

5. Which one of the following characteristics of the transition metals is associated with their catalytic activity? (1) (2) (3) (4)

High enthalpy of atomization. Paramagnetic behavior. Color of hydrated ions. Variable oxidation states.

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Solution

Solution

(4) The catalytic property of transition metals is due to their tendency to form reaction intermediates with suitable reactants. These intermediates give reaction paths of lower activation energy and increase the rate of the reaction. These reaction intermediates readily decompose yielding the products and regenerate metals form these reaction intermediates due to the presence of vacant orbitals or their tendency to form variable oxidation states.

(2) Lanthanoids are 14 elements between lanthanum (La) and Hf in which 4f orbitals are successively filled; starting from Ce (At. no. 58) to Lu (At. no. 71).

6. The basic character of the transition metal monoxides follows the order

(4) Among the given elements, iron belongs to the transition elements. One of the most striking features of the transition elements is that the element usually exists in several different oxidation states. For example, iron exists in +3 (FeCl3) and +2 (FeCl2) oxidation states.

(1) (2) (3) (4)

VO > CrO > TiO > FeO CrO > VO > FeO > TiO TiO > FeO > VO > CrO TiO > VO > CrO > FeO

(1) Zn (2) K (3) Ca (4) Fe Solution

10. Among the following series of transition metal ions, the one where all metal ions have 3d2 electronic configuration is (At. no. of Ti = 22; V = 23; Cr = 24; Mn = 25)

(At. no. Ti = 22, V = 23, Cr = 24, Fe = 26)

Solution (4) With the decrease in size of metal atom from Ti to Fe, the basic character of their monoxide decreases. Therefore, the order is TiO > VO > CrO > FeO. 7. The correct order of ionic radii of Y3+, La3+, Eu3+ and Lu3+ is (1) Y3+ < La3+ < Eu3+ < Lu3+ (2) Y3+ < Lu3+ < Eu3+ < La3+ (3) Lu3+ < Eu3+ < La3+ < Y3+ (4) La3+ < Eu3+ < Lu3+ < Y3+

9. Among K, Ca, Fe and Zn, the element which can form more than one binary compound with chlorine is

(1) Ti+, V4+, Cr6+, Mn7+ (3) Ti2+, V3+, Cr4+, Mn5+ Solution

(3) The electronic configuration of the given ions is Ti2+: [Ar]3d24s0, V3+: [Ar]3d24s0, Cr4+: [Ar]3d24s0, Mn5+: [Ar]3d24s0. 11. The aqueous solution containing which one of the following ions will be colorless? (1) Fe2+ (3) Ti3+

(At. no. Y = 39, La = 57, Eu = 63, Lu = 71)

Solution



(2) The placement of ions in the periodic table can be shown as:

Solution

Period Y 90

6

La 103.2

Eu 94.7

Lu 86.1

Lanthanoid contraction

Therefore, correct order of ionic radii is Y3+ < Lu3+ < Eu3+ < La3+ 8. Lanthanoids are (1) 14 elements in the seventh period (At. no. = 90 - 103) that are filling 5f sublevel. (2) 14 elements in the sixth period (At. no. = 58 - 71) that are filling 4f sublevel. (3) 14 elements in the seventh period (At. no. = 58 - 71) that are filling 4f sublevel. (4) 14 elements in the sixth period (At. no. = 90 - 103) that are filling 4f sublevel.

Chapter 19_d and f-Block Elements.indd 482

(2) Mn2+ (4) Sc3+

(At. no. Sc = 21, Fe = 26, Ti = 22, Mn = 25)

(4) Color in transition metal ion arises from incomplete d shells. Among the given ions, Sc(III) has an empty d shell; hence d−d spectra is impossible, therefore, it is colorless.

Elements

5

(2) Ti4+, V3+, Cr2+, Mn3+ (4) Ti3+, V2+, Cr3+, Mn4+

12. Four successive members of the first-row transition elements are listed below with their atomic numbers. Which one of them is expected to have the highest third ionization enthalpy? (1) Vanadium (Z = 23) (2) Manganese (Z = 25) (3) Chromium (Z = 24) (4) Iron (Z = 26) Solution (2) The electronic configuration of the species is as follows: Elements

Electronic configuration of M

Electronic configuration of M2+

V

3d 34s2

3d 3

Mn

3d 54s2

3d 5

Cr

3d 54s1

3d 4

Fe

3d 64s2

3d 6

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THE d- AND f-BLOCK ELEMENTS

The third ionization energy will be highest for Mn as it has half-filled electronic configuration.

13. The main reason for larger number of oxidation states exhibited by the actinides than the corresponding lanthanides is (1) lesser energy difference between 5f and 6d orbitals than between 4f and 5d orbitals. (2) more energy difference between 5f and 6d orbitals than between 4f and 5d orbitals. (3) greater reactive nature of the actinides than the lanthanides. (4) larger atomic size of actinides than the lanthanides.

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Option (4): Once d 5 configuration is exceeded then tendency to involve 3d electrons decreases because pairing of electron takes place. 16. Knowing that the chemistry of lanthanoids (Ln) is dominated by its +3 oxidation state, which of the following statements is incorrect? (1) Because of the large size of the Ln(III) ions the bonding in its compounds is predominantly ionic in character. (2) The ionic sizes of Ln(III) decrease in general with increasing atomic number. (3) Ln(III) compounds are generally colorless. (4) Ln(III) hydroxides are mainly basic in character.

Solution (1) The main reason for larger number of oxidation states exhibited by the actinides than the corresponding lanthanides is lesser energy difference between 5f and 6d orbitals than between 4f and 5d orbitals.

Solution

Option (1): L anthanoids(III) ions form ionic compounds due to its large size.

14. Identify the incorrect statement among the following.



Option (2): T  he decrease in size of atoms and ions with increase in atomic number is known as lanthanide contraction.



Option (3): M  ost lanthanoids are silvery white metals and most of trivalent metal ions are colored due to partially filled f-orbitals which permit f–f transition.



Option (4): L  anthanoids form hydroxides of the type M(OH)3 which are basic in nature.

(1) 4f and 5f orbitals are equally shielded. (2) d block elements show irregular and erratic chemical properties among themselves. (3) La and Lu have partially filled d orbitals and no other partially filled orbitals. (4) The chemistry of various lanthanides is very similar. Solution (1) 4f and 5f belongs to different energy levels, hence shielding effect is not the same for both of them. Shielding effect of 4f is more than 5f. Also, 5f is less deeply buried than 4f.

(3) The correct statements are:

17. Iron exhibits +2 and +3 oxidation states. Which of the following about iron is incorrect? (1) Ferrous oxide is more basic in nature than the ferric oxide. (2) Ferrous compounds are relatively more ionic than corresponding ferric compounds. (3) Ferrous compounds are less volatile than the corresponding ferric compounds. (4) Ferrous compounds are more easily hydrolyzed than corresponding ferric compounds.

15. In context with the transition elements, which of the following statements is incorrect? (1) In addition to the normal oxidation states, the zero oxidation state is also shown by these elements in complexes. (2) In the highest oxidation states, the transition metal show basic character and form cationic complexes. (3) In the highest oxidation states of the first five transition elements (Sc to Mn), all the 4s and 3d electrons are used for bonding. (4) Once the d5 configuration is exceeded, the tendency to involve all the 3d electrons in bonding decreases. Solution (2) The correct statements are: Option (1): Transition metals show zero oxidation state, example, Cr(CO)6 Option (2): They show acidic behavior in higher oxidation state. Option (3): Some of them use their 4s and 3d orbitals for bonding for higher oxidation state.

Chapter 19_d and f-Block Elements.indd 483

Solution (4) The correct statements are:

Option (1): FeO > Fe2O3 (Metals are more basic in lower oxidation on state).



Option (2): FeO < Fe2O3(With increase in oxidation state covalent character increases).



Option (3): FeO > Fe2O3(Covalent compounds are more volatile).



Option (4): F  eO is ferrous oxide in which the oxidation number of Fe is +2, it is more basic, more ionic and less volatile than ferric oxide (Fe in +3 oxidation state).

18. Given o o o E Cr = −0.74 V ; E MnO = 1.51 V, E Cr = 1.33 V ; E Clo /Cl − = 1.36 V 3+ − /Cr /Mn 2 + O2 − /Cr 3+ 4

2

7

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OBJECTIVE CHEMISTRY FOR NEET

Based on the data given above, strongest oxidizing agent will be (1) Cr3+ (2) Mn2+ (3) MnO 4− (4) Cl−

Solution (3) As per the data given, MnO 4− is the strongest oxidizing agent as it has maximum standard reduction potential (SRP) value.

The values of SRP are

  E Cro

3+

(1) Blue (2) Green (3) Red (4) Orange Solution (1) The complimentary color of orange is blue and the observed color of K2Cr2O7 is orange. 23. Which series of reactions correctly represents chemical reactions related to iron and its compound? H 2 SO4 2 SO 4 , O 2  → FeSO 4 H → Fe2(SO4 )3 Heat  → Fe (1) Fe dil.

o o = −0.74 V , E MnO = 1.51 V , E Cr 3+ = 1.33 V , 4 /Mn 2 O7 /Cr = 1.36 V

/Cr

E Clo /Cl −



22. K2Cr2O7 absorbs which of the following color?

H 2 SO4 2 , heat → FeO dil.  → FeSO4 Heat  → Fe (2) Fe O air 2 , heat → FeCl 3 Heat,  → FeCl 2 Zn  → Fe (3) Fe Cl

On the basis of given data, correct order of oxidizing behavior is MnO 4− > Cl > Cr2O72− > Cr 3+.

19. Chloro compound of vanadium has only spin magnetic moment of 1.73 BM. This vanadium chloride has the formula (At. no. of V = 23) (1) VCl2 (2) VCl4 (3) VCl3 (4) VCl5 Solution

600 ° C 700 ° C 2 , heat → Fe3O4 CO,  → FeO CO,  → Fe (4) Fe O

Solution (4) The reaction in option (4) is correct.

In option (1), Fe2(SO4)3 produces oxide(s) of iron, not Fe.



In option (2), FeSO4 cannot be converted to Fe on heating, in this reaction, oxide(s) will be formed.



In option (3), FeCl3 cannot be reduced to FeCl2 in the presence of air.

(2) m = n(n + 2) BM

ms = 1.73 BM is obtained when n = 1 thus one unpaired electron must be present in chloride of vanadium.

24. Lanthanides contraction is caused due to (1) the appreciable shielding on outer electrons by 4f electrons from the nuclear charge. (2) the appreciable shielding on outer electrons by 5d electrons from the nuclear charge. (3) the same effective nuclear charge from Ce to Lu. (4) the imperfect shielding on outer electrons by 4f electrons from the nuclear charge.

V = 3d34s2

For one unpaired electron, vanadium must be present in +4 (VCl4) oxidation state. V4+ = 3d1

20. Zn shows yellow color on heating due to (1) (2) (3) (4)

d-d transition. charge transfer spectra. high polarization caused by Zn2+ ions. presence of F-center.

Solution (4) On heating, O2- ions are escaped from the system in the form of O2 leaving behind electrons d in its place and these free electrons are called F-centers. 21. The oxidation state of chromium in the final product formed by the reaction between KI and acidified potassium dichromate solution is (1) +4 (2) +6 (3) +2 (4) +3 Solution (4) The reaction involved is   Cr2O72- + 14H+ + 6I - ® 2Cr 3+ + 7H 2O + 3I 2

Solution (4) In the lanthanides as we move from one element to another, the nuclear charge increases by one unit and one electron is added. The 4f electrons shield each other from the nuclear charge quite poorly because of the very diffused shapes of f-orbitals. 25. Which of the following cations will have the same color in aqueous solution with the color of aqueous solution of Pm3+ (atomic number = 61)? (1) Eu3+ (63) (2) Tb3+ (65) (3) Yb3+ (70) (4) Ho3+ (67) Solution (4) The ions having nf n and nf 14−n configuration will have similar color (expected) because they have the same number of unpaired electrons.

Pm3+: [Xe]4f 4, that is, ↑ ↑ ↑ ↑ electrons.



The electronic configurations of other ions are as follows:

Reduction from +6 to +3

Chapter 19_d and f-Block Elements.indd 484

four unpaired

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THE d- AND f-BLOCK ELEMENTS



Ho3+: [Xe]4f 10, that is, unpaired electrons.



Eu3+: [Xe]4f 6, that is, ↑ ↑ ↑ ↑ ↑ ↑ electrons.



Yb3+: [Xe]4f 13, that is, ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ ↑ unpaired electrons.



Tb3+: [Xe]4f 8, that is, ↑↓ ↑ ↑ ↑ ↑ ↑ ↑ electrons.

↑↓ ↑↓ ↑↓ ↑ ↑ ↑ ↑

four

28. The spin only magnetic moment of Cr3+ is (atomic number of Cr is 24) (1) 1.73 BM (2) 2.83 BM (3) 3.87 BM (4) 4.90 BM

six unpaired

Solution one

six unpaired

(3) The electronic configuration of Cr (Z = 24): [Ar]3d54s1

Therefore, the electronic configuration of Cr3+ is [Ar]3d3



The number of unpaired electrons is 3, so the spin only magnetic moment is

26. Which of the following relationship is not correct regarding their atomic radius? (1) Ag ≈ Au (2) Pd > Ni (3) W >> Mo (4) La > Y Solution (3) Due to lanthanide contraction, the radius of elements of first transition series is less than that of radius of corresponding second transition series elements but that of third transition series elements is nearly same as the second transition series, that is





The exception to this behavior is observed in case of Sc < Y < La. Lanthanide contraction is not observed in case of in La because in its electronic configuration 4f 14 is not present. Hence,

Radius First transition series. < Radius second transition series and Radius second transition series. ≈ Radius third transition series



Option (1) is correct: rAg = 1.34 Å and rAu = 1.34 Å.



Option (2) is correct: rPd = 1.28 Å and rNi = 1.15 Å.



Option (3) i s incorrect: rW = 1.30 Å and rMo = 1.29 Å, that is, both are almost the same.



Option (4) is correct: rLa = 1.69 Å and rY = 1.62 Å.

27. Which of the following cations are paramagnetic? (1) Yb2+ (2) Lr3+ (3) Th4+ (4) None of these

m = n(n + 2) = 3( 3 + 2) = 15 = 3.87 BM

29. Ionization enthalpy values of lanthanides are quite comparable to which of the following alkaline earth elements? (1) Barium (2) Strontium (3) Calcium (4) Radium Solution (3) The value of the first ionization enthalpy, that is, 588 kJ mol−1 and second ionization enthalpy, that is, 1155 kJ mol−1 are comparable to those of alkaline earth metals particularly Ca. 30. Which of the following configuration will have the same color as that of a lanthanide ion having configuration of 4f n? (1) 5f n (2) 5f 14 – n (3) 4f 14 – n (4) 4f 14 – (n + 1) Solution (3) Lanthanide ions having mf n and mf 14–n configuration have the same number of unpaired electrons and hence they have same color. Thus, option (3) is correct. Other options are automatically incorrect. 31. Which among the following first row transition metals shows the highest oxidation state in its compounds? (1) Cr (2) Mn (3) Fe (4) V

[Atomic numbers are Yb = 70, Lr = 103, Th = 90]

Solution (4) The electronic configuration of above cations is as follows:

Yb2+: [Xe]4f 14 is diamagnetic.



Lr3+: [Rn]5f 14 is diamagnetic.



Th4+: [Rn] is diamagnetic.



Hence, none of them is paramagnetic.

Chapter 19_d and f-Block Elements.indd 485

485

Solution (2) Among the given first row transition metals, Mn shows the highest oxidation state of +7. The other possible oxidation states are: (+2), (+3), (+4), (+5), (+6).

The possible oxidation states for other metals are:



For Cr: (+1), (+2), (+3), (+4), (+5), (+6)



For Fe: (+2), (+3), (+4), (+5), (+6)



For V: (+2), (+3), (+4), (+5)

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Practice Exercises Level I General Properties of the Transition Elements (d-Block) 1. Electronic configuration of a transition element X in +3 oxidation state is [Ar] 3d5. What is its atomic number? (1) 25 (2) 26 (3) 27 (4) 24 2. The number of d-electrons retained in Fe2+ (At. no. of Fe = 26) ion is (1) 3 (2) 4 (3) 5 (4) 6 3. The correct order of E Mo 2+ /M values with negative sign for the four successive elements Cr, Mn, Fe and Co is (1) Mn > Cr > Fe > Co (2) Cr > Mn > Fe > Co (3) Cr > Fe > Mn > Co (4) Fe > Mn > Cr > Co 4. Which of the following compounds is expected to be colored? (1) Ag2SO4 (2) CrF2 (3) MgF2 (4) CuCl 5. Although Zirconium belongs to 4d transition series and Hafnium to 5d transition series even then they show similar physical and chemical properties because (1) (2) (3) (4)

both belong to d-block. both have same number of electrons. both have similar atomic radius. both belong to the same group of the periodic table.

6. Select the most stable ion from the following: (1) Mn2+ (2) Fe2+ (3) Cr2+ (4) All are equally stable 7. Zinc carbonate is precipitated from zinc sulphate solution by the addition of (1) Na2CO3 (2) CaCO3 (3) MgCO3 (4) NaHCO3 8. Which of the following ion has smallest radii? (1) Mn2+ (2) Ni2+ (3) Ti2+ (4) V2+ 9. Copper becomes green when exposed to moist air for a long period. This is due to (1) the formation of a layer of cupric oxide on the surface of copper. (2) the formation of a layer of basic carbonate of copper on the surface of copper. (3) the formation of a layer of cupric hydroxide on the surface of copper. (4) the formation of basic copper sulphate layer on the surface of the metal.

Chapter 19_d and f-Block Elements.indd 486

10. Calomel (Hg2Cl2) on reaction with ammonium hydroxide gives (1) HgNH2 Cl (2) NH2   Hg   Hg   Cl (3) Hg2O (4) HgO 11. The oxide formed by manganese in which of following oxidation state is unstable? (1) +6 (2) +7 (3) +3 (4) +4 12. Which of the following ions have the same magnetic moment? (1) Ca2+ and Ti2+ (2) Fe2+ and Cr2+ (3) Mn2+ and Cr2+ (4) Cu2+ and Ti2+ 13. Which of the following metals have more than one oxidation state? (1) Zn (2) Mn (3) Sr (4) Ca 14. Which of the following ions shows the highest spin magnetic moment? (1) Mn3+ (2) Cr3+ (3) Ti4+ (4) Ni2+ 15. The elements with maximum and minimum melting points in the second transition series respectively are (1) Cr and Zn. (2) Cr and Cd. (3) Cr and Hg. (4) Mo and Cd. 16. Which one of the following statements is wrong? (1) HgCl2 dissolves in hot water. (2) HgCl2 gives HgS when treated with sulphuric acid. (3) HgCl2 gives yellow precipitate with NaOH. (4) HgCl2 gives white precipitate with ammonium hydroxide. 17. Ammonium nitrate is formed when Zn is treated with (1) (2) (3) (4)

very dilute nitric acid. dilute nitric acid. concentrated nitric acid. dilute hydrochloric acid.

18. In the highest oxidation states of the first 5 elements of 3d series, the electrons of which orbitals are used for bonding? (1) s (2) p (3) d (4) Both (1) and (3) 19. The color of zinc sulphide is (1) yellow. (2) white. (3) brown. (4) black.

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THE d- AND f-BLOCK ELEMENTS 20. Nitrous oxide is evolved when Zn is treated with (1) (2) (3) (4)

very dilute nitric acid. dilute nitric acid. concentrated nitric acid. dilute hydrochloric acid.

(1) Cu(OH)2 (2) Fe(OH)3 (3) Cr(OH)3 (4) Zn(OH)2 22. The magnetic nature of elements depends on the presence of unpaired electrons. Identify the configuration of transition element, which shows highest magnetic moment. (1) 3d7 (2) 3d5 (3) 3d8 (4) 3d2 23. Which of the following is amphoteric oxide?

Mn2O7, CrO3, Cr2O3, CrO, V2O5, V2O4

(1) V2O5, Cr2O3 (3) CrO, V2O5

(2) Mn2O7, CrO3 (4) V2O5, V2O4

Some Important Compounds of Transition Elements 24. When acidified K2Cr2O7 solution is added to Sn2+ salts then Sn2+ changes to (1) Sn (2) Sn3+ (3) Sn4+ (4) Sn+ 25. On addition of small amount of KMnO4 to concentrated H2SO4, a green oily compound is obtained which is highly explosive in nature. Identify the compound from the following. (1) Mn2O7 (2) MnO2 (3) MnSO4 (4) Mn2O3 26. What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid? (1) Cr3+ and Cr2O72− are formed. (2) Cr2O72− and H2O are formed. (3) CrO2− is reduced to +3 state of Cr. 4 (4) CrO2− is oxidized to +7 state of Cr. 4 solvent  → X + Y , X and 27. Acidified chromic acid + H 2O2 Organic Y are

(1) CrO5 and H2O (2) Cr2O3 and H2O (3) CrO2 and H2O (4) CrO and H2O 28. The purple color of KMnO4 will not be discharged by which of the following reagents? (1) MnSO4 + Zn2+ (2) H2O2/H+ (3) KNO3 (4) K2S/H+ 29. Which of the following reagents does not convert copper sulphate to cuprous state?

Chapter 19_d and f-Block Elements.indd 487

(1) KCN (2) KCNS/SO2 (3) KI (4) K4[Fe(CN)6] 30. Blood-red colored solution is produced when ferric chloride solution is treated with

21. The hydroxide which is soluble in excess of NaOH solution is



487

(1) KSCN (2) KCN (3) K4Fe(CN)6 (4) K3Fe(CN)6 31. In Cr2O72− every Cr is linked to (1) two O atoms. (2) three O atoms. (3) four O atoms. (4) five O atoms. 32. When KMnO4 solution is added to oxalic acid solution, the decolorization is slow in the beginning but becomes instantaneous after some time because (1) CO2 is formed as the product. (2) reaction is exothermic. (3) MnO 4− catalyzes the reaction. (4) Mn2+ acts as autocatalyst. 33. KMnO4 acts as an oxidizing agent in acidic medium. The number of moles of KMnO4 that will be needed to react with one mole of sulphide ions in acidic solution is (1) 2/5 (2) 3/5 (3) 4/5 (4) 1/5 34. KMnO4 acts as an oxidizing agent in alkaline medium. When alkaline KMnO4 is treated with KI, iodide ion is oxidized to ____________. (1) I2 (2) IO– (3) IO3− (4) IO4−

General Properties of the Inner Transition Elements (f-Block) 35. There are 14 elements in actinide series. Which of the following elements does not belong to this series? (1) U (2) Np (3) Tm (4) Fm 36. The outer electron configuration of Gd (atomic number 64) is (1) 4f 8 5d0 6s2 (3) 4f 7 5d1 6s2

(2) 4f 4 5d4 6s2 (4) 4f 3 5d5 6s2

37. Which of the following ions is diamagnetic in nature? (1) La3+ (2) Lu3+ (3) Gd3+ (4) Both La3+ and Lu3+ 38. How many electrons are there in ground state of Pu in 5f orbitals? (1) 3 (2) 4 (3) 5 (4) 6 39. The most common oxidation state for actinides is (1) +2 (2) +3 (3) +4 (4) +5

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OBJECTIVE CHEMISTRY FOR NEET

40. Which of the following statements is true for f-block elements? (1) They can have electrons from f 0 to f 14. (2) Group number is 3 in the periodic table. (3) With the increase in number of f-electrons, the radius decreases due to poor shielding effect of f-electrons. (4) All of these. 41. Ce (Z = 58) and Yb (Z = 70) exhibit stable +4 and +2 oxidation states, respectively. This is because (1) Ce4+ and Yb2+ acquire f 7 configurations. (2) Ce4+ and Yb2+ acquire f 0 configurations. (3) Ce4+ and Yb2+ acquire f 0 and f 14 configurations. (4) Ce4+ and Yb2+ acquire f 7 and f 14 configurations. 42. The oxidation state not observed for lanthanides is (1) +2 (2) +3 (3) +4 (4) +5 43. In the context of lanthanides, which of the following statement is not correct? (1) All the members exhibit +3 oxidation state. (2) Because of similar properties, the separation of lanthanides is not easy. (3) Availability of 4f electrons results in the formation of compounds in +4 state for all the members of the series. (4) There is a gradual decrease in the radii of the members with increasing atomic number in the series. 44. Cerium (Z = 58) is an important member of the lanthanides. Which of the following statements about cerium is incorrect? (1) The common oxidation states of cerium are +3 and +4. (2) Cerium (IV) acts as an oxidizing agent. (3) The +4 oxidation state of cerium is not known in solutions. (4) The +3 oxidation state of cerium is more stable than the +4 oxidation state. 45. Which of the following actinide does not have any electron in 6d orbital in its ground state? (1) Lr (2) Cm (3) Pa (4) Cf 46. Due to lanthanide contraction (1) Fe, Co, Ni have equal size. (2) Zr and Hf have equal size. (3) all f-block ions have equal size. (4) all isoelectronic ions have equal size. 47. Which of the following properties varies between lanthanoids and actinoids? (1) Highest oxidation state. (2) Radioactive nature. (3) Basicity of hydroxides. (4) All of these.

Chapter 19_d and f-Block Elements.indd 488

48. Which of the following salts are insoluble for the actinides? (1) Sulphates (2) Perchlorates (3) Fluorides (4) Nitrates 49. For which of the following actinides, the +3 oxidation state is the most stable? (1) Am (2) Th (3) Pa (4) U 50. Which of the following is diamagnetic in +3 oxidation state? (1) U (2) Th (3) Ac (4) Both (2) and (3)

Some Applications of d- and f-Block Elements 51. Which of the following element is used in the treatment of cancer? (1) Uranium (2) Thorium (3) Cerium (4) Plutonium

Level II General Properties of the Transition Elements (d-Block) 1. Which of the following atomic numbers do not correspond to an exceptional electronic configuration? (1) 29 (2) 46 (3) 78 (4) 43 2. Amongst the following, identify the species with an atom in + 6 oxidation state. (1) [MnO4]− (2) [Cr(CN)6]3− (3) Cr2O3 (4) CrO2Cl2 3. Four successive members of the first row transition elements are listed below with atomic numbers. Which one of them is expected to have the highest E Mo 3+ /M2+ value? (1) Mn (Z = 25) (2) Fe (Z = 26) (3) Co (Z = 27) (4) Cr (Z = 24) 4. The known oxidation states for both V and Co are (1) +2, +3, +4, + 5 (2) +2, +3, +4, + 5, +6 (3) +2, +3, + 4 (4) + 2, + 3 5. The electronic configuration of Cu(II) is 3d9, whereas that of Cu(I) is 3d10. Which of the following is correct? (1) Cu(II) is more stable. (2) Cu(II) is less stable. (3) Cu(I) and Cu(II) are equally stable. (4) Stability of Cu(I) and Cu(II) depends on nature of copper salts.

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THE d- AND f-BLOCK ELEMENTS 6. How many d-block elements have the ability to evolve hydrogen gas from 2% nitric acid? (1) 1 (2) 2 (3) 3 (4) Many

(1) (n − 1)d ns (2) (n − 1)d ns (3) (n − 1)d3ns2 (4) (n − 1)d5ns1 2

5

2

8. The atomic numbers of vanadium (V), chromium (Cr), manganese (Mn) and iron (Fe) are, respectively, 23, 24, 25 and 26. Which one of these may be expected to have the highest second ionization enthalpy? (1) V (2) Cr (3) Mn (4) Fe

16. Which of the following pairs of transition metal ions are the stronger oxidizing agents in aqueous solutions? (1) V2+ and Cr2+ (2) Ti2+ and Cr2+ (3) Mn3+ and Cr3+ (4) V2+ and Fe2+ 17. CuSO4(aq) + 4NH3 → X, then X is (1) [Cu(NH3)4]2+ (2) Dimagnetic (3) Colored (4) Has magnetic moment 1.73 BM 18. Which of the following ions shows the lowest spin magnetic moment?

9. Metallic radii of some transition elements are given below. Which of these elements will have highest density?

Element

Fe

Co

Ni

Cu



Metallic radii/pm

126

125

125

128

(1) Fe (2) Ni (3) Co (4) Cu 10. Maximum magnetic moment is shown by (1) d5 (2) d6 (3) d7 (4) d8 11. White colored thiosulphate which changes to black color automatically belongs to the following d-block metal. (1) Copper (2) Manganese (3) Chromium (4) Silver 12. Which of the following metal ion is expected to be colored? (1) Zn2+ (2) Ti3+ (3) Sc3+ (4) Ti4+ 13. Which of the following statements is not correct? (1) Copper liberates hydrogen from acids. (2) In its higher oxidation states, manganese forms stable compounds with oxygen and fluorine. (3) Mn3+ and Co3+ are oxidizing agents in aqueous solution. (4) Ti2+ and Cr2+ are reducing agents in aqueous solution. 14. Generally transition elements form colored salts due to the presence of unpaired electrons. Which of the following compounds will be colored in solid state? (1) Ag2SO4 (2) CuF2 (3) ZnF2 (4) Cu2Cl2

Chapter 19_d and f-Block Elements.indd 489

15. The following elements belong to second transition series: (1) V, La, Hf, Ta (2) Nb, Mo, Tc, Ru (3) Ir, Os, Re, Ta (4) Re, Os, Ir, Pr

7. Of the following outer electronic configurations of atoms, the highest oxidation state is achieved by which one of them? 8

489

(1) Mn3+ (2) Cr3+ (3) Ti4+ (4) Ni2+ 19. Which of the following metal ions is expected to be colored? (1) Zn2+ (2) Ti3+ (3) Sc3+ (4) Ti4+ 20. If NaOH is added to an aqueous solution of zinc ions, a white precipitate appears and on adding excess NaOH, the precipitate dissolves. In this solution zinc exists in the (1) (2) (3) (4)

cationic part. anionic part. both in (1) and (2). there is no zinc in solution.

21. Ag+ ion is isoelectronic with (1) Cu+ (2) Cd2+ (3) Zn2+ (4) Pd2+

Some Important Compounds of Transition Elements 22. Which of the following is not formed when H2S reacts with acidic K2Cr2O7 solution? (1) CrSO4 (2) Cr2(SO4)3 (3) K2SO4 (4) S 23. The color of KMnO4 is due to (1) d–d transition. (2) L → M charge transfer transition. (3) σ–σ* transition. (4) M → L charge transfer transition. 24. Which of the following reaction(s) can be used for the complete conversion of K2MnO4 to KMnO4? (1) 3K 2MnO4 + 2H 2O  2KMnO 4 + MnO 2 + 4KOH (2) 2K 2MnO 4 +Cl 2 → 2KMnO 4 +2KCl

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490

OBJECTIVE CHEMISTRY FOR NEET (3) 2K 2MnO 4 +O3 +H 2O → 2KMnO 4 +2KOH+ O 2 (4) Both (2) and (3)

25. KMnO4 + HCl → H2O + X(g), X is a (acidified) (1) red liquid. (2) violet gas. (3) greenish yellow. (4) yellow brown gas. 26. The product of oxidation of I− with MnO 4− in acidic medium is (1) IO3− (2) I2 (3) IO− (4) IO4− 27. How many gases are evolved when potassium permanganate and potassium dichromate are heated, respectively? (1) 1,2 (2) 2,1 (3) 1,1 (4) 2,2 28. Which of the following reactions are disproportionation reactions? (I) Cu+ → Cu2+ + Cu (II) 3MnO 4− + 4H + → 2MnO 4− + MnO 2 + 2H 2O (III) 2KMnO4 → K2MnO4 + MnO2 + O2 (IV) 2MnO4− + 2Mn 2+ + 4H 2O → 5MnO 2 + 4H+ (1) I, II (2) I, II, III (3) II, III, IV (4) I, IV 29. Why is HCl not used to make the medium acidic in oxidation reactions of KMnO4 in acidic medium? (1) Both HCl and KMnO4 act as oxidizing agents. (2) KMnO4 oxidises HCl into Cl2 which is also an oxidizing agent. (3) KMnO4 is a weaker oxidizing agent than HCl. (4) KMnO4 acts as a reducing agent in the presence of HCl.

General Properties of the Inner Transition Elements (f-Block) 30. A reduction in atomic size with increase in atomic number is a characteristic of elements of (1) high atomic masses. (2) d-block. (3) f-block. (4) radioactive series. 31. Which of the following salts are insoluble for the lanthanides? (1) Sulphates (2) Perchlorates (3) Fluorides (4) Nitrates 32. Which of the following lanthanides forms an exceptional oxide? (1) Lu (2) Ce (3) Gd (4) Sm 33. Which of the following is a synthetic element? (1) U (2) Pa (3) Fm (4) Th

Chapter 19_d and f-Block Elements.indd 490

34. Which of the following is diamagnetic in +4 oxidation state? (1) Th (2) Pa (3) Pu (4) All of these 35. Which of the following lanthanide ions is colorless? (1) Pr3+ (2) Nd3+ (3) Yb3+ (4) Sm3+

Some Applications of d- and f-Block Elements 36. Which of the following sets is of coinage metals? (1) Cu, Ag, Au (2) Zn, Cd, Hg (3) Au, Ag, Zn (4) Li, Na, K

Previous Years’ NEET Questions 1. Identify the incorrect statement among the following: (1) Shielding power of 4f electrons is quite weak. (2) There is a decrease in the radii of the atoms or ions as one proceeds from La to Lu. (3) Lanthanide contraction is the accumulation of successive shrinkages. (4) As a result of lanthanide contraction, the properties of 4d series of the transition elements have no similarities with the 5d series of elements. (AIPMT 2007) 2. The correct order of decreasing second ionization enthalpy of Ti (22), V (23), Cr (24) and Mn (25) is (1) Ti > V > Cr > Mn (2) Cr > Mn > V > Ti (3) V > Mn > Cr > Ti (4) Mn > Cr > Ti > V (AIPMT 2008) 3. Out of TiF62− , CoF63− , CuCl 2 and NiCl 2− 4 (At. no. Ti = 22, Co = 27, Cu = 29, Ni = 28) the colorless species are (1) CoF63− and NiCl 24− (2) TiF62− and CoF63− (3) Cu 2Cl 2 and NiCl 42− (4) TiF62− and Cu 2Cl 2 (AIPMT 2009) 4. Which of the following pairs has the same size? (1) Zr 4+ , Ti 4+ (2) Zr 4+ , Hf 4+ (3) Zn 2+ , Hf 4+ (4) Fe2+ , Ni 2+ (AIPMT PRE 2010) 5. Which of the following ions will exhibit color in aqueous solutions? (1) Ti 3+ (Z = 22) (2) Lu 3+ (Z = 71) (3) Sc3+ (Z = 21) (4) La 3(Z = 57 ) (AIPMT PRE 2010)

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THE d- AND f-BLOCK ELEMENTS 6. Which one of the following ions has electronic configuration [ Ar ] 3d 6 ?

13. Which of the following statements about the interstitial compounds is incorrect?

(1) Mn 3+ (2) Fe3+ (3) Co3+ (4) Ni 3+

(1) (2) (3) (4)

(AIPMT PRE 2010)

They retain metallic conductivity. They are chemically reactive. They are much harder than the pure metal. They have higher melting points than the pure metal.

7. Which of the following oxidation states is the most common among the lanthanides?

(NEET 2013) 14. Magnetic moment 2.83 BM is given by which of the following ions? (At. No. Ti = 22, Cr = 24, Mn = 25, Ni = 28)

(1) 2 (2) 5 (3) 3 (4) 4

(1) Ti3+ (2) Ni2+ (3) Cr3+ (4) Mn2+

(AIPMT MAINS 2010) 8. Identify the alloy containing a non-metal as a constituent in it.



(1) (2) (3) (4)

(AIPMT PRE 2012) 9. Which one of the following does not correctly represent the correct order of the property indicated against it? (1) Ti < V < Cr < Mn: increasing number of oxidation states (2) Ti3+ < V3+ < Cr3+ < Mn3+: increasing magnetic moment (3) Ti < V < Cr < Mn: increasing melting points (4) Ti < V < Mn < Cr: increasing 2nd ionization enthalpy (AIPMT MAINS 2012) 10. Four successive members of the first series of the transition metals are listed below. For which one of them the standard potential (E Mo 2+ /M) value has a positive sign? (1) Co (Z = 27) (2) Ni (Z = 28) (3) Cu (Z = 29) (4) Fe (Z = 26) (AIPMT MAINS 2012) 11. The catalytic activity of transition metals and their compounds is ascribed mainly to their magnetic behavior. their unfilled d-orbitals. their ability to adopt variable oxidation state. their chemical reactivity. (AIPMT MAINS 2012) 12. KMnO4 can be prepared from K2MnO4 as per the reaction: 3MnO 24− + 2H 2O → 2MnO 4− + MnO2 + 4OH −



The reaction can go to completion by removing OH− ions by adding (1) HCl (2) KOH (3) CO2 (4) SO2



Chapter 19_d and f-Block Elements.indd 491

(AIPMT 2014)

15. Reason of lanthanide contraction is

(1) Invar (2) Steel (3) Bell metal (4) Bronze

(1) (2) (3) (4)

491

(NEET 2013)



negligible screening effect of f-orbitals. increasing nuclear charge. decreasing nuclear charge. decreasing screening effect. (AIPMT 2014)

16. Magnetic moment 2.84 BM is given by (At. no. Ni = 28, Ti = 22, Cr = 24, Co = 27) (1) Ti3+ (2) Cr2+ (3) Co2+ (4) Ni2+

(AIPMT 2015)

17. Because of lanthanide contraction, which of the following pairs of elements have nearly same atomic radii? (Numbers in the parenthesis are atomic numbers.) (1) Zr (40) and Nb (41) (2) Zr (40) and Hf (72) (3) Zr (40) and Ta (73) (4) Ti (22) and Zr (40)

(AIPMT 2015)

18. Which one of the following statements is correct when SO2 is passed through acidified K2Cr2O7 solution? (1) The solution turns blue. (2) The solution is decolorized. (3) SO2 is reduced. (4) Green Cr2(SO4 )3 is formed. (NEET-I 2016) 19. Which one of the following statements related to lanthanons is incorrect? (1) The basicity decreases as the ionic radius decreases from Pr to Lu. (2) All the lanthanons are much more reactive than ­aluminum. (3) Ce (+4) solutions are widely used as oxidizing agent in volumetric analysis. (4) Europium shows +2 oxidation state. (NEET-II 2016)

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OBJECTIVE CHEMISTRY FOR NEET

20. HgCl2 and I2 both when dissolved in water containing I− ions the pair of species formed is (2) HgI 24− , I 3− (4) HgI 2 , I 3-

(1) HgI2, I− (3) Hg2I2, I−

(1) actinide contraction. (2) 5f, 6d and 7s levels having comparable energies. (3) 4f and 5d levels being close in energies. (4) the radioactive nature of actinides. (NEET 2017)

(NEET 2017) 21. The reason for greater range of oxidation states in actinides is attributed to

Answer Key Level I  1. (2)

 2. (4)

 3. (1)

 4. (2)

 5. (3)

 6. (1)

 7. (4)

 8. (2)

 9. (2)

10. (1)

11. (1)

12. (2)

13. (2)

14. (1)

15. (4)

16. (2)

17. (1)

18. (4)

19. (2)

20. (2)

21. (4)

22. (2)

23. (1)

24. (3)

25. (1)

26. (2)

27. (1)

28. (3)

29. (4)

30. (1)

31. (3)

32. (4)

33. (1)

34. (3)

35. (3)

36. (3)

37. (4)

38. (4)

39. (2)

40. (4)

41. (3)

42. (4)

43. (3)

44. (3)

45. (4)

46. (2)

47. (4)

48. (3)

49. (1)

50. (3)

 1. (4)

 2. (4)

 3. (3)

 4. (3)

 5. (1)

 6. (1)

 7. (2)

 8. (2)

 9. (4)

10. (1)

11. (4)

12. (2)

13. (1)

14. (2)

15. (1)

16. (2)

17. (1)

18. (3)

19. (2)

20. (2)

27. (3)

28. (1)

29. (2)

30. (3)

51. (2)

Level II

21. (2)

22. (1)

23. (2)

24. (4)

25. (1)

26. (2)

31. (3)

32. (2)

33. (3)

34. (1)

35. (3)

36. (1)

Previous Years’ NEET Questions  1. (4)

 2. (3)

 3. (4)

 4. (2)

 5. (1)

 6. (3)

 7. (3)

 8. (2)

 9. (3)

10. (3)

11. (3)

12. (3)

13. (2)

14. (2)

15. (1)

16. (4)

17. (2)

18. (4)

19. (2)

20. (2)

21. (2)

Hints and Explanations Level I 2. (4) Configuration of Fe is 3d6 4s2 and that of Fe2+ is 3d6. 3. (1) This is in accordance with the reactivity series. 4. (2) CrF2 is colored as Cr2+ contains two unpaired electrons. 6. (1) Mn2+ is 3d5 which is a half-filled stable configuration. 7. (4) The reaction involved is ZnSO 4 + NaHCO3 → ZnCO 3 + Na 2SO4 + H 2

Chapter 19_d and f-Block Elements.indd 492

8. (2) The ionic radius of transition metals decreases with increase in atomic number but the decrease is small after midway, that is, after Mn because after that the increase in d electrons (pairing starts) counterbalances the increase nuclear charge. 9. (2) Copper very slowly gets oxidized in the presence of moist air and gives a green coating of basic copper carbonate. CuCO3⋅Cu (OH)2. 10. (1) The reaction involved is Hg 2Cl 2 + 2NH 4OH → Hg + Hg(NH 2 )Cl + NH 4Cl + 2H 2O

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THE d- AND f-BLOCK ELEMENTS

493

11. (1) The oxidation state of +6 is rarely shown by Mn, as it is unstable and readily converts to Mn7+.



12. (2) Both Fe2+ and Cr2+ has four unpaired electrons so the value of magnetic moment will be the same, that is, n(n + 2) = 4( 4 + 2) = 4.90 BM.

38. (4) Pu (Z = 94); [Rn] 5f 6 6d0 7s2

13. (2) Transition elements show variable oxidation states, for example, Mn.

42. (4) The energy gap between 4f and 5d subshells is very large, so they do not show higher oxidation states, that is, +5.

14. (1) Mn3+ = 3d4 with 4 unpaired electrons. 17. (1) Zn + HNO3( very dil.) → NH 4NO 3 18. (4) Both s and d electrons are used for bonding.

Both La3+ and Lu3+ do not have unpaired electrons and therefore are diamagnetic in nature.

41. (3) Both acquire stable configurations, that is, f 0 and f 14.

43. (3) The general oxidation state of lanthanides is +3, only few elements exhibit +4 oxidation state. They show limited number of oxidation states because the energy gap between 4f and 5d subshells is large.

19. (2) Since it has completely filled d orbital, so contains no unpaired electrons.

45. (4) Cf is [Rn] 5f10 6d0 7s2

20. (2) 4Zn + 10HNO3(dil.) → 4Zn(NO3 )2 + N 2O + 5H 2O

48. (3) Fluorides of actinides are high melting solids and are insoluble.

21. (4) Zn(OH )2 + 2NaOH → Na 2[ Zn(OH )4 ]

49. (1) Am = [Rn] 5f 7 6d0 7s2 and Am3+ = 5f 6.

26. (2) The reaction involved is

50. (3)Ac3+ has 5f 6 configuration.

Soluble

2CrO 24− + 2H + →  Cr2O72− + 2H 2O

Level II

28. (3) Mn2+ (in presence of Zn2+), H2O2 and S2− are reducing agents and reduce Mn7+ by which color is discharged. While in KNO3, nitrogen is in +5 oxidation state and cannot be further oxidized and hence, cannot act as reducing agent.





where the oxidation state of Cu in the product is +2 (cupric state).

30. (1) The reaction involved is Fe3+ + 3CNS − → Fe(CNS)3

Blood red ppt.

31. (3) The structure is as follows: O

O

Cr

O−

O

O

O

36. (3) Gd is gadolinium and it is a lanthanide. The electronic configuration is [Xe]4f 75d16s2 37. (4) La3+ = [Xe]4f 0



Lu3+ = [Xe]4f 14





Gd3+ = [Xe] 4f 7

Chapter 19_d and f-Block Elements.indd 493

x + 2 × (−2) + 2 × (−1) = 0 ⇒ x = −6

3. (3) We know that

CuSO4 + K4Fe(CN)6 → Cu2Fe(CN)6 + K2SO4



Cr

2. (4) Let the oxidation state of Cr in CrO2Cl2 molecule be x.

29. (4) The reaction involved is

−O

1. (4) Tc (At. no. 43) is has 5s1 4d6 configuration.

o o E Mn = 1.57 V, E Fe = 0.77 V 3+ 3+ /Mn 2 + /Fe2 + o o E Co = 1.97 V and E Cr = −0.41 V. 3+ 3+ /Co2 + /Cr 2 +

6. (1) Mn reacts with HNO3 to evolve H2 gas. 7. (2) (n − 1)d5ns2. This is because it can lose upto 7 electrons. 8. (2) Considering the outermost configuration of all, V = 3d3 4s2; Cr = 3d5 4s1; Mn = 3d5 4s2; Fe = 3d6 4s2. In case of chromium, elimination of one electron from 4s orbital will result in a stable configuration, that is, 3d5. So, it has high second ionization enthalpy. 10. (1) For d5 configuration, the number of unpaired electrons is 5, so the magnetic moment is m = n(n + 2) = 5(5 + 2) = 35 BM

11. (4) White colored Ag2S2O3 absorbs moisture and converts into black color. Ag 2S 2O3 + H 2O → Ag 2S + H 2SO4 White

Black ppt.

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494

OBJECTIVE CHEMISTRY FOR NEET

12. (2) Paramagnetic substance (Ti3+, that is, [Ar]3d1) is colored in general. 18. (3) Mn3+ is 3d4 (four unpaired electrons, high spin); Cr3+ is 3d3 (three unpaired electrons, high spin); Ti4+ is 3d0 (no unpaired electrons, lowest spin) Ni2+ (two unpaired electrons, high spin). 19. (2) Substances with unpaired electrons generally formed colored ions. Of the given options:

Zn2+:[Ar]3d10 is colorless



Ti3+ :[Ar]3d1 is colored (purple)



Sc3+: [Ar] is colorless



Ti4+: [Ar] is colorless

20. (2) The reaction involved is Zn 2+ + 2OH − → Zn(OH)2



If excess of NaOH is used then Zn(OH)2 + 2OH- → [Zn(OH)4]2-

21. (2) Ag+ and Cd2+ both have 46 electrons with electron configuring 4d10. 22. (1) The reaction of acidic K2Cr2O7 with H2S is

K2Cr2O7 + H2SO4 + H2S → Cr2 (SO4) + K2SO4 + S + H2O



K2Cr2O7 is oxidizing agent and oxidizes S2− to sulphur.

23. (2) Color of KMnO4 is due to charge transfer from O2− to Mn. 24. (4) Option (1) is incorrect because from three moles of KMnO4 only two moles K MnO4 are formed, that is, conversion ratio is 66%, when the reaction is irreversible. Since the reaction is reversible, the conversion ratio is even less than 66%. About one-third of Mn is lost as MnO2.

In options (2) and (3), the conversion ratio is almost 100%.

26. (2) MnO 4− + H 2O + I − → Mn 2+ + 8H 2O + I 2 ¾¾ ® K 2MnO 4(s) + MnO2(s) + O2 ↑ 27. (3) 2KMnO4 ¾Heat K 2Cr2O7 Heat  → 4K 2CrO4(s) + 2Cr2O3(s) + 3O2 ↑ 30. (3) Because of lanthanide contraction in f-block elements. 31. (3) Fluorides of lanthanides are insoluble, while with heavier lanthanides they become sparingly soluble due to the formation of complex. 32. (2) Ce is present in +4 oxidation state in Ce2O3 and CeO2 oxides. 33. (3) Fm is a synthetic element which is made artificially by neutron bombardment of lighter elements.

Chapter 19_d and f-Block Elements.indd 494

34. (1) Th4+ = 5f 0 (diamagnetic), Pa4+ is 5f 1 and Pu4+ is 5f 4. 35. (3) Color depends on number of unpaired f electrons. Yb3+ has 4f13 configuration. 36. (1) Coinage metals are those metallic elements that were previously used as components in alloys used to make coins.

Previous Years’ NEET Questions 1. (4) There is a steady decrease in the radii as the atomic number of the lanthanide elements increase. For every additional proton added in nucleus, the corresponding electron goes to 4f subshell. The shape of f-orbitals is very much diffused and they have poor shielding effect. The effective nuclear charge increases which cause the contraction in the size of electron charge cloud. This contraction in size is quite regular and known as lanthanide contraction. Since the change in the ionic radii in the lanthanide series is very small, thus their chemical properties are similar. 2. (2) For d-block elements, there is slight increase in ionization energy due to increase in Zeff values along the period but it is not prominent. In case of the second ionization enthalpy, the electronic configuration of Cr (d5) is half-filled, hence more stable than that of Mn (d5s1). Therefore, the correct order is Cr > Mn > V > Ti. 3. (4) In these compounds, TiF62− and Cu 2Cl 2 , it is not possible to promote an electron as Ti4+ has d0 configuration and Cu+ has d10. Hence d-d spectra are not possible. 4. (2) Zr 4+ ( 4d series) and Hf 4+ (5d series) have almost same size due to lanthanide contraction. 5. (1) Ti 3+ ( Z = 22) = 1s 2 2s 2 2p 6 3s 2 3p 6 4s1 3d 0

The presence of an unpaired electron in the s orbital of Ti3+ causes it to exhibit color in aqueous solution.

6. (3) The electronic configuration of Co3+ is [ Ar ] 3d 6 . 7. (3) The very common and stable oxidation state for lanthanides is +3 which can also be explained easily from their electronic configurations, that is, two electrons from 6s orbital and one electron from 5d orbital are lost easily from all elements except Eu and Yb. 8. (2) Steel always contains some percent of carbon in it. 9. (3) The trend in melting points of the transition elements can be explained on the basis that more the number of unpaired electrons involved in the metallic bonding, higher will be the melting point. Therefore, the correct increasing order of melting point is Ti < V < Mn < Cr.

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THE d- AND f-BLOCK ELEMENTS o = −0.28 V ; 10. (3) The E Mo 2+ /M values are as follows E Co 2+ /Co o o o E Ni 2+ /Ni = −0.25 V ; E Cu 2+ /Cu = +0.34 V and E Fe2+ /Fe = −0.44 V .

11. (3) The catalytic activity of transition metals is attributed to their ability to adopt multiple oxidation states and to form complexes. For example, V2O5 used in contact process and finely divided iron used in Haber’s process. 12. (3) The reaction in the presence of CO2 is 3 K 2MnO 4 + 4CO2 + 2H 2O → 2 KMnO4 + MnO 2 + 4KHCO 3

13. (2) These are formed mostly by transition elements, and some of the lanthanides and actinides. The Cr, Mn, Fe, Co and Ni groups form a large number of carbides with a wide range of stoichiometries. They are typically infusible or are very high melting, and are very hard. Interstitial carbides are generally unreactive. They do not react with H2O like ionic carbides.

17. (2) Zr (40) and Hf (72) will have nearly same atomic radii because of lanthanide contraction. As after lanthanum electron filling takes place in f sub-shell. Electrons ­present in f sub-shell didn’t do good shielding due to which with the increasing atomic number or increasing effective nuclear charge size gets constricted and size of Zr and Hf becomes almost equal. 18. (4) When SO2 is passed through acidified K2Cr2O7, the following reaction takes place K 2Cr2O7 + 3SO2 + H 2SO4 → K 2SO4 + Cr2(SO4 )3 + 3H 2O   (green)

19. (2) For option (1): In lanthanides, the basic character gradually decreases from Ce(OH)3 to Lu(OH)3 due to high polarization caused by gradual decrease in size of M3+ ions.

For option (2): Lanthanons are less reactive than aluminum due to their high ionization potential. This is a consequence of lanthanide contraction which results from outer orbital electrons being incompletely shielded by extra nuclear charge.



For option (3): Ce4+ is a good oxidizing agent and easily converted to Ce3+ (most stable oxidation state).



For option (4): The electronic configuration of Eu(63) is [Xe] 4f 75d0 6s2, the stability of oxidation state +2 for Eu can be explained on the basis of extra stability associated with the half-filled f-orbitals.

14. (2) The magnetic moment can be calculated as Ion

Number of unpaired Dipole moment electrons (n)

3+

Ti

3d ; n = 1

m = n(n + 2) = 3 = 1.713 BM  

Ni2+

3d8; n = 2

m = n(n + 2) = 2 2 = 2.8 BM  

Cr3+

3d3; n = 3

m = n(n + 2) = 15 = 3.8 BM  

Mn

3d ; n = 5

m = n(n + 2) = 35 = 5.9 BM  

1

2+

5

495

20. (2) The reactions are as follows: 15. (1) The atomic radius decreases gradually from Ce (187 pm) to Lu (171 pm) by nearly 20 pm. This contraction in radius is due to lanthanide contraction which results because the extra orbital electrons are incompletely shielded by the extra nuclear charge. 16. (4) We have, magnetic moment =

n(n + 2) BM.



where n is number of unpaired electrons.



2.84 =



Ni (3d ) contain two unpaired electrons. 2+

n(n + 2) ⇒ n = 2 8

Chapter 19_d and f-Block Elements.indd 495

Hg 2+ KI →

HgI 2

→ 

Scarlet red ppt. −

I 2 + I (excess) →

[HgI 4 ]2− Colorless soluble complex

I 3−

Dark brown complex

21. (2) In actinides, the 5f orbitals extend in space beyond 6s and 6p orbitals and participate in bonding as energy difference between 5f, 6d and 7s levels become comparable.



2Na[Ag(CN)2] + Zn → Na2[Zn(CN)4] + 2Ag ↓

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Chapter 19_d and f-Block Elements.indd 496

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497

20

Coordination Compounds

Chapter at a Glance 1. Coordination chemistry is the branch of chemistry which distinctively deals with the study of coordination compounds for their chemical, structural, magnetic and spectral properties. 2. Coordination compounds (also called complex compounds or simply complexes) are a special class of compounds that consist of a central metal atom or ion, which is surrounded by oppositely charged ions or neutral molecules in more than its normal valence. 3. Werner’s Theory: The main postulate of the theory is that the metal ion encapsulated inside the coordination entity exhibits two types of valencies: primary valency and secondary valency. (a)  The primary valency is ionizable as well as non-directional. It corresponds to the oxidation state in the modernday terminology. It corresponds to the number of charges present on the complex ion. (b)  The secondary valency is directional, non-ionizable and corresponds to coordination number in the modernday terminology. It equals the total number of ligands coordinately bonded to the central metal ion inside the coordination sphere. (c)  The metal complex first satisfies its secondary valency and then its primary valency. (d)  Every metal ion has fixed number of secondary valencies, which are directed towards fixed positions in space around the central metal atom, resulting in definite geometry of the complex. (e)  The coordination compounds are represented as [Ni(NH3)6]Cl2 where the entity in the square bracket is the coordination complex and the ions outside the bracket are called counter ions. 4. Difference between Double Salt and Complexes: When two or more stable compounds join together in stoichiometric amounts, addition compounds are formed, which are of two types. Double salts Lose their identity in solution. Show properties of individual ions in solution. Example, Mohr salt [(NH4)Fe(SO4)2·6H2O]. and potassium alum K 2 SO4 ⋅ Al 2 (SO4 )3 ⋅ 24H2 O

Complexes Retain their identity in solution. Exist as distinct identities as solid and in solution. Example, cuproammonium ion [Cu(H2O)2(NH3)4]2+ and the ferrocyanide ion [Fe(CN)6]4-

5. Important Terms (a) C  oordination entity: It is the central metal atom or ion which is bonded to a definite number of ions or molecules which is fixed. (b) Central atom/ion: It is the central cation that is surrounded and coordinately bonded to one or more neutral molecules or negatively charged ions in a definite geometric arrangement. (c) Coordination number: The total number of ligands coordinately bonded to the central metal atom or ion in the primary coordination sphere represents the coordination number of that complex. (d) Coordination sphere: This is represented by the central metal ion and the ligands coordinately bonded to it. It is that part of the complex which remains as the single entity, that is, it does not lose its identity and is non-ionizable. (e) Donor atom: The coordinating atom of the ligand which is actually donating electron pair to the central metal ion is called a donor atom. (f ) Oxidation number of central atom: When all the ligands are removed along with the electron pairs that are shared with the central atom, the charge that the central atom would carry is called the oxidation number. It is represented in roman numeral in parenthesis after the name of the central atom.

Chapter 20_Coordination Compounds.indd 497

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OBJECTIVE CHEMISTRY FOR NEET

(g) H  omoleptic and heteroleptic complexes: When the central metal ion is bound to only one kind of donor group, the complex is called homoleptic. The complex in which the metal atoms are bonded to more than one type of donor group is called heteroleptic species. 6. Ligands: It can be an atom, ion or molecule that binds to a central metal ion to form a coordination complex. There are several ways to classify the ligands and these are discussed as follows:

Based upon denticity of the ligand

Based upon the charges

Classification Type Neutral ligands

Example H2 O, NO, CO, C 6 H6 , etc. +

Positive ligands

 NO and NH 2 - NH3

Monodentate: Only one donation is accepted from the ligand. Bidentate: Two donations are accepted from the ligand.

H2 O, NO, CO, NH3, CO32 -, Cl - , etc.

+

en: ethylenediamine    Oxalate ion CH2

CH2

NH2

Polydentate: These are ligands that bind to the metal ion through multiple bonding sites and include tridentate, tetradentate, pentadentate, hexadentate ligands.

CH2

COO−

NH2

COO− dien: diethylenetriamine trien: triethylenetetraamine CH2

H2N

CH2

CH2

CH2

NH

H2N

CH2

NH

CH2

CH2

H 2N

NH2

Flexidentate ligand: A ligand which shows variable denticity is called a flexidentate ligand.

CH2

NH

Tridentate

CH2

Tetradentate O

O O

O

O and

S

S O

O

Bidentate ion

Other Classification

CH2

O

Monodentate ion

Ambidentate ligands: These are ligands that (i)  NO - : Nitro or Nitro-N (bonding occurs through N). 2 have more than one kind of donor sites but at (ii) (ONO-)/ NO - : Nitrito or Nitro-O (bonding occurs 2 a time only one kind of donor site is utilized. through O). 2+ 2+ Chelate effect: It is the enhanced stability OH2 OH2 CH2 NH2 CH NH 2 2 OH2 NH2 CH2 of complex on coordination with polydenNi Ni CH2 NH2 CH2 NH2 tate ligand due to formation of ring- like OH2 NH2 CH2 OH2 OH2 structures or chelates. Chelation results in formation of ring structure which are Chelate with one Chelate with two kinetically as well as thermodynamically ethylenediamine ligands ethylenediamine ligands very stable.

CH2

NH2

CH2

NH2

CH2 CH2 NH2 NH2 Ni NH2 NH2 CH2

2+

CH2 Chelate with three ethylenediamine ligands

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7. Effective Atomic Number (EAN) (a) I t is the total number of electrons neighboring the nucleus of a metal atom in a coordination complex. It is composed of the metal atom’s electrons and the bonding electrons from the surrounding ligands. EAN of a central atom in a complex = Atomic number of the central atom (Z)   – (oxidation number of the central atom with sign)   + (number of electrons collected from the ligands). (b) The following points need to be noted with regards to EAN. (i) All donations contribute two electrons, while NO is considered as 3 electron donor. (ii) For p - donors, number of p -electrons involved in donation from a particular ligand are to be considered. (c) Sidgwick EAN rule (i) Sidgwick suggested that electron pairs from ligands were added to the central metal atom until the central atom was surrounded by the same number of electrons as the next noble gas. (ii) The metal carbonyls exhibit a strong tendency to achieve Sidgwick EAN values.    The number of Co molecule attached in mononuclear carbonyls can be predicted. Some important observations based on this are described as follows: · Th  e number of Co molecule attached in mononuclear carbonyls can be predicted. For example, in Fe(CO)x : x = 5, Ni(CO) y : y = 4, and Cr(CO)z : z = 6. · [ Mn(CO)6 ]0 can act as a reducing agent. As a reducing agent, the complex will lose electron to attain noble gas configuration and hence obey Sidgwick EAN rule. e →[ Mn(CO)6 ]+ [ Mn(CO)6 ] - EAN = 37 EAN = 36

· [ V(CO)6 ]0 can act as oxidizing agent. As an oxidizing agent, the complex will gain an electron to attain noble gas configuration and hence obey Sidgwick EAN rule. e →[ V(CO)6 ] [ V(CO)6 ]0 + EAN = 35 EAN = 36 0 · [Mn(CO)5 ] undergoes dimerization. By dimerization, the complex can attain noble gas configuration and hence obey Sidgwick EAN rule.

→[Mn 2 (CO)10 ] 2[Mn(CO)5 ]0  EAN = 36 EAN = 35 8. Nomenclature of Coordination Compounds: The rules for systematic naming of coordination compounds are discussed as follows: (a) T  he name of the cationic part is written first followed by the anionic part, for example, diamminesilver(I) chloride, [Ag(NH3)2]Cl (b) The name of the ligands is listed before the name(s) of the central atom(s), for example, hexaamminecopper(III) chloride, [Co(NH3)6]Cl3 (c) Prefixes are used to designate the number of each type of ligand in the complex ion; for example, di-, tri- and tetra-. If the ligand already contains a prefix (e.g., ethylenediamine) or if it is a polydentate ligand then Greek prefixes bis-, tris-, tetrakis-, pentakis-, etc. are used instead. Example: tris(bipyridine)iron(II), [Fe(NH4C5-C5H4N)3]2+ (d) Ligand names are listed in the alphabetical order (multiplicative prefixes indicating the number of ligands are not considered in determining that order), for example, tetraamminedichlorocobalt(III), [Co(NH3)4Cl2]+ (e) For naming the central metal if the complex ion is a cation, the metal is named same as the element, for example, tetraammineplatinum(II), [Pt(NH3)4]2+. If the complex ion is an anion, the name of the metal ends with the suffix –ate, for example, tetrachloroplatinate(II), [PtCl4]2-

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(f ) A  nionic ligands should end in “-o”. Thus, anions that end in “-ide” (e.g., chloride), “-ate” (e.g., sulphate, nitrate), and “-ite” (e.g., nitrite) should be changed into -ido, -ato and -ito, respectively. (g) For neutral ligands, the common name of the molecule is used, for example, H2NCH2CH2NH2 (ethylenediamine). Important exceptions include water called as “aqua”, ammonia called as “ammine”, carbon monoxide called as “carbonyl ”, N2 called as “dinitrogen” and O2 called as “dioxygen”. x (h) For the p - donors, the prefix like ‘h ’ is to be used, where h indicates p- electron donation and x is known as hapticity of the ligand, i.e., the number of atoms involved in the p - donation. For example, 5 (i) p - C5 H5 : h - cyclopentadienyl or pentahaptocyclopentadienyl 6 (ii) p - C 6 H6 : h - benzene or hexahaptobenzene. 3 (iii) p - C3 H5 : h - allyl or trihaptoallyl. (i) For bridging ligands between two metal ions as shown below the prefix m is used. H3N H3N

NH3

H2 N

Co NH3

NH3

4+ NH3

Co O H

NH3

NH3

μ-amido-μ-hydroxobis tetraminecobalt(IV)

9. Isomerism (a) G  eometric isomerism: This type of isomerism occurs in di-substituted (heteroleptic) complexes with coordination numbers 4 and 6 having square planar and octahedral geometries, respectively. This isomerism can be of two types: cis: When the two identical ligands are adjacent to each other. trans: When the two identical ligands are on the opposite side. (i) Geometrical isomerism in complexes with coordination number 4: Square planar complexes of the type MA2X2, [M(AB)2] and MABX2 can exist, where A, X are neutral or anionic ligands and AB is an unsymmetrical bidentate ligand. · The complexes with formulae MA2X2, can have two geometrical isomers. X

X

A

X

M A

M A

cis

X

trans

A

· For complex with the type [ M(AB)2 ], two geometrical isomers are possible. A

A

A

B

M B

M B

B

A

cis

trans

· For the complexes of the type [ M(AB)X 2 ], two geometrical isomers are possible. B

X

A

M A

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X M

X

B

X

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(ii) G  eometrical isomerism in complexes with coordination number 6: In an octahedral complex shown as follows, positions 1–6, 2–4 and 3–5 are trans, while positions 1–2, 2–3, 2–5, 6–3, 6–4, 1–3, 1–4 and 1–5 are cis to each other. 1

5

2 M

4

3 6

· Complexes of the following types can show geometrical isomerism: MA4X2, MA2X4, MA4X4, M(AA)2X2 and M(AA)2XY, where A and B are neutral or anionic ligands and (AA) is a symmetrical bidentate, ligand. · Complexes of type M(ABCDEF) can also exhibit geometrical isomerism. These isomers may be written by fixing a ligand at one position and then placing the other ligands trans to it. · Complexes having unsymmetrical bidentate ligands also exhibit geometrical isomerism. (b) Optical isomerism: It is ability of certain compounds to rotate the plane of polarized light in different directions. The compounds are optically active and are called dextrorotatory and laevorotatory complexes, depending on if they rotate plane of light in the right or left direction, respectively. Coordination complexes of the following type show optical isomerism: (i) M(AA)3 where (AA) is a symmetrical bidentate ligand like ethylenediamine. (ii) M(AA)2X2 and M(AA)2XY, where AA is a symmetrical bidentate ligand and X and Y are neutral or anionic ligands. Examples: [Co(en)2Cl2], [RhCl2(en)2]. (iii) [M(AA)X2Y2] having only one symmetrical bidentate ligand also shows optical isomerism. Example: [CoCl2(en)(NH3)2] (c) Linkage isomerism: This type of isomerism arises due to presence of ligands with two different donor atoms, which may thus attach to the central metal atom through either of the two atoms. This is exhibited by the complexes having ambidentate ligands. H3N H3N

2+

NH3

H3N

ONO

Co

NH3

and

2+

NH3 Co

H3N

NH3

NO2 NH3

NH3

Red Nitritopentamminecobalt(III) ion

Yellow Nitropentamminecobalt(III) ion

Nitrito and nitro complexes (d) C  oordination isomerism: This type of isomerism occurs in those complexes which have both anionic as well as cationic entities and there is a difference in the distribution of ligands within these entities. Both these complexes share the same molecular formula, but they differ in the cationic and anionic entities. Example: [Pt(NH3)4]2+[PtCl4]2–; [PtCl(NH3)3]1+[PtCl3(NH3)]-1 (e) Ionization isomerism: This type of isomerism occurs in complexes which have the same molecular formula but give different ions in solution on ionization. Example: complex having the formula Co(NH3)5BrSO4 can have two possible structures: [Co( NH3 )5 Br ]SO4

Pentamminebromocobalt(III) sulphate (Violet - red)

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[Co( NH3 )5 SO4 ]Br

Pentamminesulphatecobalt(III) bromide (Red)

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(f) H  ydrate or Solvate isomerism: This isomerism is very similar to ionization isomerism, but the difference is that in this case the solvent molecule can act as the ligand. For example, CrCl3·6H2O can have any of the following structures: [Cr(H2O)6]Cl3, [CrCl(H2O)5]Cl2·H2O, [CrCl2(H2O)4]Cl·2H2O, [CrCl3(H2O)3]·3H2O 10. Valence Bond Theory: The main postulates of the theory are: (a) C  oordination compounds contain complex ions, in which ligands form coordinate bonds to the metal. (b) Ligand must have lone pair of electrons, and the metal must have an empty orbital of suitable energy available for bonding. (c) The atomic orbitals made available by the central metal ion are a mixture of s, p, d orbitals, all of which are of different energies and orientation. These hybridize to give orbitals that are equivalent in energy and symmetry, which then form bonds with the ligands. (d) The d orbitals that are involved in the process of hybridization can be either from the inner d orbital (n − 1) or the outer d orbital (n). The complexes thus formed are referred to as inner orbital and outer orbital complexes, respectively. (e) If the complex contains unpaired electrons, it is paramagnetic in nature; whereas, if it does not contain unpaired electrons, it is diamagnetic in nature. 11. Crystal Field Theory (a) I n this theory, the interaction between the ligand and the central metal ion is treated as a purely electrostatic interaction, as opposed to the valence bond theory where the interaction is treated as covalent in nature. (b) Crystal field splitting in octahedral complexes: When the six ligands approach the central metal cation the lobes of d z 2 and d x 2 - y 2 orbitals (eg set of orbitals) experience greater repulsion exerted by the electron clouds of the ligands than that experienced by the electrons in the dxy, dyz and dzx orbitals (t2g set of orbitals). Hence, the orbitals d x 2 - y 2 and d z 2 are raised in energy, whereas the orbitals dxy, dyz and dzx are lowered in energy relative to the excited d levels. eg +0.6∆o ∆o

Energy Average energy of metal ion in spherical field

Average energy level (Bari center)

−0.4∆o t2g Metal ion in octahedral field

(i) T  he separation of five d orbitals into t2g and eg sets of different energies is known as crystal field splitting. (ii) The energy difference between eg and t2g sets is denoted by Do (the subscript o stands for octahedral) and is called the crystal field stabilization energy (CFSE). The magnitude of Do depends on three factors. · Nature of ligands. · Charge on the metal ion. · Whether the metal is in the first, second or third row of transition elements. (iii) The energy of the two eg orbitals increases by 0.6Do (3/5Do) and that of t2g orbitals is lowered by 0.4Do (2/5Do). (iv) CFSE and electronic arrangements in octahedral complexes

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3.87 4.90 5.92 4.90 3.87 2.83 1.73 0.00

-1.2 -1.2 + 0.6 = -0.6 -1.2 + 1.2 = -0.0 -1.6 + 1.2 = -0.4 -2.0 + 1.2 = -0.8 -2.4 + 1.2 = -1.2 -2.4 + 1.8 = -0.6 -2.4 + 2.4 = -0.0

d 3

4

d 5

d 6

7

d 8

d 9

10

d 

d 

d 

2.83

-0.8

CFSE Do

d 2

eg -0.4

t2g

Spin only magnetic moment lo(D) 1.73

d 1

Number of d electrons

Arrangement in weak ligand field t2g

eg

-2.4 + 2.4 = -0.0

-2.4 + 1.8 = -0.6

-2.4 + 1.2 = -1.2

-2.4 + 0.6 = -1.8

-2.4

-2.0

-1.6

-1.2

-0.8

-0.4

CFSE Do

Arrangement in strong ligand field

0.00

1.73

2.83

1.73

0.00

1.73

2.83

3.87

2.83

Spin only magnetic moment lo(D) 1.73

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(c) Jahn–Teller theorem (i) It states that ‘Any non-linear molecular system in a degenerate electronic state will be unstable, and will undergo some sort of distortion (Jahn-Teller distortion) to lower its symmetry and remove the degeneracy.’ (ii) More simply, molecules or complexes (of any shape except linear), which have an unequally filled set of orbitals (either t2g or eg), will be distorted. In octahedral complexes distortions from the t2g level are too small to be detected. (iii) The two eg orbitals d x 2 - y 2 and d z 2 are normally degenerate. However, if they are asymmetrically filled then this degeneracy is destroyed, and the two orbitals are no longer equal in energy. If the d z 2 orbital contains one more electron than the d x 2 - y 2 orbital then the ligands approaching along +z and –z will encounter greater repulsion than the other four ligands. The repulsion and distortion result in elongation of the octahedron along the z axis. This is called tetragonal distortion. (iv) Asymmetrical electronic arrangements Electronic configuration

eg

t2g

Nature of ligand field

Examples

d 4

Weak field (high-spin complex)

Cr(+II), Mn(+III)

d 7

Strong field (low-spin complex)

Co(+II), Ni(+III)

d 9

Either strong or weak

Cu(+II)

(d) C  rystal field splitting in square planar complexes: Square planar complexes are formed by d 8 ions with strong field ligands, for example [NiII(CN)4]2–. The crystal field splitting Δo is larger for second and third row transition elements, and for more highly charged species. All the complexes of Pt(+II) and Au(+III) are square planar – including those with weak field ligands such as halide ions. d 8 arrangement in very strong octahedral field is shown below. Tetragonal distortion splits (a) the eg level; and (b) also splits the t2g level. dx2−y2 ↑↓ dz2 eg

dx2−y2 ↑↓ dz2 eg

Energy ↑↓ dxy ↑↓ ↑↓ ↑↓ t2g (a)

↑↓ ↑↓ dxy and dyz t2g (b)

 The amount of tetragonal distortion that occurs depends on the particular metal ion and ligands. Sometimes the tetragonal distortion may become so large that the d z 2 orbital is lower in energy than the dxy orbital as shown in below figure. dx2 −y2 eg Energy

dxy dz2 t2g

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dxy and dyz

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(e) Crystal field splitting in tetrahedral complexes: (i) The three t2g (dxy , dyz and dzx ) orbitals are of higher energy and two eg (d z 2, d x 2 - y 2) orbitals are of lower energy. t2g +0.4∆t Energy

∆1 Average energy of metal ion in spherical f ield

Average energy level (Bari centre)

+0.6∆t eg Metal ion in tetrahedral field

(ii) The crystal field splitting is denoted by ∆t (the subscript “t” indicating tetrahedral complex). (iii) In the tetrahedral field, since the d orbitals are not interacting directly with the ligand field, the splitting of d orbitals is less than that in the octahedral complexes. Dt < D o 4 Dt = D o 9 12. Color and Magnetic Properties (a) C  olor: Crystal field theory explains the color in coordination compounds and attributes it to the d−d transition of the electron between the split t2g and eg levels. (i) Most of the transition metal complexes are colored in their solution or solid state. (ii) When light falls on transition metal complexes the electrons in the lower energy level jump to the higher energy level. The absorbed light is actually that portion of light which is sufficient to excite the electrons from lower energy level to the higher energy level. The portion of light reflected back is responsible for the color of the complex. (b) Magnetic properties: These properties are useful for identifying and characterizing the compounds. (i) The magnetic moment of the coordination compounds can be obtained by measuring the magnetic susceptibility. (ii) The magnetic nature of coordination compounds is determined by the number of unpaired electrons present. For example: · [Mn(CN)6]3– has magnetic moment corresponding to two unpaired electrons, while [MnCl6]3– shows paramagnetic character corresponding to four unpaired electrons. · [Fe(CN)6]3– has magnetic moment corresponding to single unpaired electron, while [FeF6]3– shows paramagnetic character corresponding to five unpaired electrons. · [CoF6]3– is paramagnetic with four unpaired electrons while [Co(C2O4)3]3– is diamagnetic. 13. Bonding in Metal Carbonyls (a) T  he simplest carbonyls of transition metals are of the type M(CO)x which have simple well-defined structures. For example, V(CO)6 and Cr(CO)6 are octahedral, Fe(CO)5 is trigonal bipyramidal and Ni(CO)4 is tetrahedral. (b) Polynuclear carbonyls can be homonuclear, such as Fe3(CO)12 or heteronuclear, such as MnRe(CO)10. In these compounds, there are not only linear M   C   M groups but also either M   M bonds or M   M bonds with bridging carbonyls. (c) The bonding in linear M   C   O groups can be explained as: (i) First, there is a dative overlap of the filled carbonyl carbon s-orbital with the vacant metal orbital. (ii) The second dative overlap is of the filled dp- metal orbital with an empty antibonding pp orbital of the carbonyl group.

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(iii) T  his bonding mechanism is synergic, that is, the effects of s-bond formation strengthen the p bonding and vice versa. p∗

C

M

O

M

C

O

p∗

   B  y valence bond or molecular orbital theory, it is well understood that the bond order of C   O bond + decreases and C   O bond length must increase due to synergic effect. Similarly, C N and N O are isoelectronic with CO, hence back donation takes place here also in p * orbitals and same conclusion can be drawn for the bond order and bond lengths. · In case of PR 3, the back donation may be depicted as: Vaccant 3d orbital accepts the back donation here. PR3

M

· In case of C 2 H4 , the back donation may be depicted using the example of Zeise’s salt. H CI

CI

H

p∗ C

Pt CI p∗

C H H

   H  ere back donation is accepted in p * orbital of C   C bond. Hence, bond order of C   C bond decreases and bond length increases compared to free C 2 H4 molecule. 14. Stability of Coordination Complexes (a) I n general, the coordination complexes are highly stable and the stability arises from the interaction between the metal ion and the ligands. (b) For the general reaction for the formation of the coordination complex, Ma + + nLb - → [ ML n ]

c+

  w  here a+, b– and c+ are the respective charges on the metal, ligand and complex. The stability constant is represented as c+ ( ML n )] [ K = [ Ma + ][Lb - ]n (c) I f the reaction for complex formation proceeds in steps, them the overall stability constant can be related to the stepwise stability constants by the relation: b n = K 1 × K 2 × K 3 ... K n 15. Applications of Coordination Compounds (a) Analytical chemistry (i)  Qualitative and Quantitative analysis: The reactions of metals to form colored coordination complexes are used for detection of these metal ions using classical or instrumental methods such as gravimetry or colorimetry.

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(ii)  Volumetric analysis: Hardness of water can be estimated by titration with EDTA. The metal ions causing hardness, that is, Ca2+ and Mg2+ form stable complexes with EDTA. (b) Metal extraction and purification: Extraction of metals, such as silver and gold is carried out by forming their water soluble cyanide complexes with the ore. (c) Catalysis: Coordination compounds are used as catalysts in important commercial processes. For example, the Zeigler–Natta catalyst (TiCl4 and trialkyl aluminium) is used as a catalyst in the formation of polyethylene. (d) Electroplating: Coordination compounds of gold, silver and copper are used as components in the baths used for electroplating articles of other metals with these metals. (e) Biological importance: Some important biological compounds are coordination complexes. For example, chlorophyll is a complex of Mg2+. This green pigment plays a vital role in photosynthesis in plants. Similarly, hemoglobin, the red pigment present in blood, is a coordination complex of Fe2+ and vitamin B12, an essential nutrient, is complex compound of Co3+. (f ) Medicinal uses: Complexing or chelating agents are used in treating metal poisoning, wherein, the coordination complex is formed between toxic metal in excess metal and the complexing agent. Some examples are: (i) EDTA is used in lead poisoning. EDTA, when injected intravenously into the bloodstream, traps lead forming a compound that is flushed out of the body with the urine. (ii) Chelation therapy can also be used for other heavy metal poisonings with mercury, arsenic, aluminum, chromium, cobalt, manganese, nickel, selenium, zinc, tin and thallium.

Solved Examples 1. Atomic number of Cr and Fe are respectively 24 and 26, which of the following is paramagnetic?

3d Fe

(1) [Cr(CO)6] (2) [Fe(CO)5] (3) [Fe(CN)6]4− (4) [Cr(NH3)6]3+

×× ××

[Fe(CN)4]2−

Solution (4) Option (1): In [Cr(CO)6], the electronic configuration of Cr is 3d54s1. 3d

4s

×× ××

[Cr(CO)6]

××

Option (4): In [Cr(NH3)6]3+, the electronic configuration of Cr3+ is 3d3 3d

4s

4p

××

×× ×× ××

Cr3+



Option (2): In [Fe(CO)5], the electronic configuration of Fe is 3d64s2. 4s

4p

××

×× ×× ××

Fe (Ground state) ××

 



3

CO is a strong field ligand due to which pairing of electrons takes place. Hence, the complex is diamagnetic.

[Fe(CO)5]

×× ×× ××

Cyanide is also a strong field ligand, therefore, electrons get paired up to give a diamagnetic complex.

×× ×× ××



3d

××



d sp 2

4p

d2sp3

4p

Cr (Ground state)

4s

2+

dsp3

[Cr(NH3)6]3+

×× ××

d2sp3



 

Ammonia is a weak field ligand, therefore, pairing of electrons will not take place resulting in the formation of high spin or paramagnetic complex.

2. The number of unpaired electrons in the complex ion [CoF6]3− is (Z of Co = 27) (1) 2 (2) 3 (3) 4 (4) Zero



The electrons pair up because of strong field ligand, carbonyl. Hence, the complex is diamagnetic.

Solution



Option (3): In [Fe(CN)6]4−, the electronic configuration of Fe2+ is 3d6

(3) The electronic configuration of Co3+ is 3d64s0. The unpaired electrons can be calculated as Co3+ (3d64s0):

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OBJECTIVE CHEMISTRY FOR NEET 3d



4s

Solution

4p

F− is a weak field ligand, therefore, pairing of electrons will not take place. 3d

4s

4p

××

(4) The compounds containing metal-carbon double bond are called p-bonded organometallics. Since there is no C C double bond in (CH3)4Sn, therefore, it does not form p-bonded organometallic compounds.

4d

×× ×× ××

×× ××

5. According to IUPAC nomenclature, sodium nitroprusside is named as

sp3d 2



(1) (2) (3) (4)

Therefore, the complex has four unpaired electrons

3. Which one of the following octahedral complexes will not show geometric isomerism? (A and B are monodentate ligands) (1) [MA2B4] (2) [MA3B3] (3) [MA4B2] (4) [MA5B]

Solution (3) The formula of sodium nitroprusside is Na2[Fe(CN)5NO]. Its IUPAC name sodium pentacyanonitrosylferrate(II). 6. CN− is a strong field ligand. This is due to the fact that

Solution

(1) (2) (3) (4)

(4) All the complexes except [MA5B] will show geometrical isomerism.

Option (1): [MA2B4] A

B

B

B

B

(2) CN− is a strong ligand as it not only donates the lone pair of electrons to the central atom but also accept the electron cloud from the central atom in its low-lying vacant orbitals. This kind of back donation is known as synergic effect or synergic bonding.

M B

B

B

A

A Trans

B Cis

 

Option (2): [MA3B3] A B

B A

A

A

B

M B

A

B

B

Facial isomer

Meridional isomer



Option (3): [MA4B2] A

A A

B A Cis

4. Among the following, which is not the p-bonded organometallic compound? (1) K[PtCl3(h 2 ‑ C2H4)] (2) Fe(h 5 − C5H5)2 (3) Cr(h 6 − C6H6)2 (4) (CH3)4Sn

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(1) In [Mn (H2O)6]2+, Mn is present in +2 oxidation state. The electronic configuration of Mn2+ is [Ar] 3d54s0. 3d

4s

3d

4s ×

4p

Mn2+

B M

A B Trans

Solution

A

M A

(1) five. (2) two. (3) four. (4) three.

A



A

7. Considering H2O as a weak field ligand, the number of unpaired electrons in [Mn(H2O)6]2+ will be (At. no. of Mn = 25)

M

B

it is a pseudohalide. it can accept electrons from metal species. it forms high spin complexes with metal species. it carries negative charge.

Solution

A

M



Sodium nitroferricyanide Sodium nitroferrocyanide Sodium pentacyanonitrosyl ferrate(II) Sodium pentacyanonitrosyl ferrate(III)

[Mn(H2O)6]2+ Unpaired electrons



×

4p × ×

4d ×

×

sp3d2

H2O is a weak field ligand, thus, no pairing of electrons takes place in 3d orbital. Therefore, number of unpaired electrons is five.

8. Among [Ni(CO)4], [Ni(CN)4]2−, [NiCl4]2− species, the hybridiza­tion states at the Ni atom are, respectively (At. no. of Ni = 28)

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Coordination Compounds (1) sp3, dsp2, sp3 (2) sp3, sp3, dsp2 (3) dsp2, sp3, sp3 (4) sp3, dsp2, dsp2

Cl Cl

(1) The hybridization of Ni in the given complexes is as follows:

3d

4s

4p

××

×× ×× ××

H3N

CO being a strong field ligand will cause the pairing of electrons.



[Ni(CN)4]2−: The arrangement of electrons in Ni2+(3d8) will be 4s

3d

4p

××

××

×× ××

[NiCl4]2−: The arrangement of electrons in Ni2+ (3d8) will be 4p

××

×× ×× ××

(4) All the species shown below contain metal-carbon bond.



CH2 CH3 MgBr Grignard reagent

2

Pt Cl

Cl

CH

2

Ni CO CO CO

While in Al(OC2H5)3, Al is attached to oxygen. So, there is no metal-carbon bond present in this species. CH3

AI H3C

 

Cl− is a weak field ligand, hence, pairing of electrons will not take place.

9. Which of the following coordination compounds would exhibit optical isomerism? (1) Diamminedichloroplatinum (II) (2) trans-dicyanobis(ethylenediamine)chromium(III) chloride (3) tris-(ethylenediamine)cobalt(III) bromide (4) Pentaamminenitrocobalt (III) iodide Solution (3) Compounds that do not possess any symmetry elements shows optical isomerism. In the following compounds, plane of symmetry exists; hence, they do not exhibit optical isomerism.

Chapter 20_Coordination Compounds.indd 509

CO

CH

Cl

O

sp3 (Tetrahedral)



Br−

Co

Ni2+ [NiCl4]2−

+

10. Which of the following does not have a metal-carbon bond?

CN− ion is a strong field ligand; hence pairing of electrons takes place.

4s

Due to lack of any symmetry element in the below compound, it is optically active. Therefore, optical isomerism exists.

Solution

dsp hybridization (Square planar)  

3d

l−

NH3

(1) C2H2MgBr (2) K[Pt(C2H4)Cl3] (3) Ni(CO)4 (4) Al(OC2H5)3

2



NO2

en

Ni2+ [Ni(CN)4]2−

Co

en



en Cl−

+ NH3

en

sp3 hybridization (Tetrahedral shape)  

+

CN

NH3

H3N



[Ni(CO)4]

Cr

NH3

[Ni(CO)4]: The arrangement of electrons in Ni0 (3d8 4s2) will be Ni



en

Pt

Solution



CN

NH3

O

O

CH3

11. Which one of the following pairs represents stereo­ isomerism? (1) (2) (3) (4)

Linkage isomerism and geometrical isomerism. Chain isomerism and rotational isomerism. Optical isomerism and geometrical isomerism. Structural isomerism and geometrical isomerism.

Solution (3) Two molecules are described as stereoisomers of each other if they are made of the same atoms, connected in the same sequence, but the atoms are positioned differently in space. The difference between two stereoisomers can only be seen when the three dimensional arrangement of the molecules is considered. Stereoisomers can be subdivided into geometrical isomerism and optical isomerism.

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OBJECTIVE CHEMISTRY FOR NEET

12. Which one of the following is an inner orbital complex as well as diamagnetic in behavior? [At. no. of Zn = 30; Cr = 24; Co = 27; Ni = 28] (1) [Zn(NH3)6]2+ (3) [Cr(NH3)6]3+

cis-[Pt(NH3)2Cl2] and trans-[Pt(NH3)2Cl2] plane of symmetry exists, therefore they are not optical isomers. H3N

(2) [Ni(NH3)6]2+ (4) [Co(NH3)6]3+

(4) Option (1): In [Zn(NH3)6]2+, the electronic configuration of Zn2+ is 3d10. 4s

4p

××

×× ×× ××

×× ××

 



This is an outer orbital complex and diamagnetic in nature.



Option (2): In [Ni(NH3)6] , the electronic configuration of Ni2+ is 3d8. 4p

××

×× ×× ××

Cl en

×× ××

This is an outer orbital complex and paramagnetic in nature.



Option (3): In [Cr(NH3)6]3+, the electronic configuration of Cr3+ is 3d3. 4p

××

×× ×× ××

d2sp3

Since low energy inner d orbitals are used this is called an inner orbital complex and paramagnetic due to the presence of three unpaired electrons.



Option (4): In [Co(NH3)6]3+, the electronic configuration of Co3+ is 3d6. 4s

4d

××

×× ×× ××

d2sp3



en

Cl

Cl

Cl

Cl

Co

en

Co en

en

cis-[Co(en)2Cl2]

(1) Linkage isomerism, ionization isomerism and optical isomerism. (2) Linkage isomerism, ionization isomerism and geometrical isomerism. (3) Ionization isomerism, geometrical isomerism and optical isomerism. (4) Linkage isomerism, geometrical isomerism and optical isomerism. Solution

(a) Ionization isomerism: It is due to the exchange of groups between the complex ion and the ions outside it. The isomers are [Co(NH3)4(NO2)2]Cl and [Co(NH3)4(NO2)Cl]NO2. (b) Linkage isomerism: It arises due to the fact that in the NO2- ion, either N or O atoms can act as the electron pair donors. Two different complexes are [Co(NH3)4(NO2)2Cl] and [Co(NH3)4(NO2)(ONO)]Cl (c) Geometrical isomerism: The geometrical isomers are

 

This is an inner orbital complex and diamagnetic in nature.

13. Which one of the following is expected to exhibit optical isomerism (en = ethylenediamine)? (1) cis-[Pt(NH3)2Cl2] (2) cis-[Co(en)2Cl2] (3) trans-[Co(en)2Cl2] (4) trans-[Pt(NH3)2Cl2] Solution

O2N

(2) A molecule is said to exhibit optical isomerism if it cannot be superimposed on its mirror image and there is lack of any symmetry elements. In both the complexes,

O2N

Chapter 20_Coordination Compounds.indd 510

Co

(2) The isomerism exhibited by the complex [Co(NH3)4 (NO2)2]Cl is as follows:

Co3+ ×× ××

en

 



[Co(NH3)6]3+

en

14. [Co(NH3)4(NO2)2]Cl exhibits

4s

3d

en

Cl



×× ××

Co

Complex cis-[Co(en)2Cl2] is optically active as it has nonsuperimposable mirror image.

sp3d2

3d

Cl

Cl



4d

trans-[Pt(NH3)2Cl2]

The complex trans-[Co(en)2Cl2] have superimposable mirror image as shown below.

2+

4s

NH3

Cl

cis-[Pt(NH3)2Cl2]



sp d

Cl

H3N

4d

3 2

3d

Pt

Pt

Solution

3d

Cl

H3N

Cl

 

NH3 Co NH3 cis

NH3

H3N

NH3

NO2

NH3 Co NH3 Trans

NO2 NH3

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Coordination Compounds 15. [Cr(H2O)6]Cl3 (At. no. Cr = 24) has a magnetic moment of 3.83 BM. The correct distribution of 3d electrons in chromium in the complex is

18. Which of the following ligands is tridentate type? (1) oxalate (ox2–) (2) NOS– (3) iminodiacetate [imda2–; (CH2COO-)2NH] (4) propylenediamine (pn)

1 1 (1) 3dx12 - y 2 , 3dz12 , 3dxz (2) 3dzy , 3d 1x 2 - y 2 , 3d 1yz 1 1 (3) 3dxy , 3d 1yz , 3dxz



(4) 3dxy1 , 3d 1yz , 3dz12

Solution

Solution (3) The ligands can be represented as

(3) The electronic configuration of Cr3+ in complex [Cr(H2O)6]Cl3 is 3d3. Therefore, the distribution of electrons will be ( 3dxy )1( 3d yz )1( 3dxz )1. 3d

pn: propylenediamine

ox2− oxalate

CH3

− O

O C

eg

CH2

NH2

NH2

Bidentate

C

Bidentate

O −

O

CH

t2g

 



CH2 O

16. Which of the following metal carbonyl may act as either oxidizing or reducing agents having EAN of 36? (1) [Co(CO)4] (2) [Mn(CO)6] (3) [Cr(CO)6] (4) None of these. Solution (4) To achieve the Sidgwick EAN value, that is, 36, 54 and 86, the metal carbonyl may act as oxidizing agent when its EAN is less than Sidgwick EAN value. It may act as reducing agent when its EAN value is greater than Sidgwick EAN value and cannot act as either of the two if its EAN value is equal to any one of Sidgwick EAN value. 17. Which of the following statements is/are correct for [Cr(NH3)5 Br] SO4 and [Cr(NH3)5SO4] Br? (I) Coordination number of the central atom is 6 for both. (II) In both the cases, the anionic ligand satisfies the primary valency in equal manner. (III) Electrical conductivities of both the complexes are equal. (1) (I), (II) (2) (I), (III) (3) (II), (III) (4) (I) only Solution (4) Statement I is correct. Here SO2acts as monodentate 4 ligand.

Statement II is incorrect. Br- satisfies only one primary 3+ valency while SO24 satisfies two primary valency of Cr .



Statement III is incorrect. Since the charges of both anions are different (though number of ions produced are same), the electrical conductivities are different.

Chapter 20_Coordination Compounds.indd 511

NOS− O

imda2− iminodiacetate

The lobes of eg orbitals (dx 2 - y 2 and dz 2) point along the axes x, y and z. The lobes of t2g orbitals (dxy, dyz and dxz) point in between the axes.

NH

CH2

N C

C O−

O

O−

S Monodentate

Tridentate

19. A solution containing 2.675 g of CoCl3⋅6NH3 (molar mass = 267.5 g mol-1) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of AgNO3 to give 4.78 g of AgCl (molar mass = 143.5 g mol-1). The formula of the complex is (atomic mass of Ag = 108 u). (1) [CoCl(NH3)6]Cl2 (2) [Co(NH3)6]Cl3 (3) [CoCl2(NH3)4]Cl (4) [CoCl3(NH3)3] Solution (2) The reaction is CoCl 3 ⋅ 6NH 3 → xCl – + AgNO3 → xAgCl n( AgCl ) = xn(CoCl 3 ⋅ 6NH 3 ) 4.78  2.675  = x  267.5  143.5

On solving, we get x = 3. Therefore, the complex is [Co(NH3)6]Cl3.

20. The IUPAC name for the complex [Co(NO2)(NH3)5]Cl2 is (1) nitrito-N-pentaamminecobalt (III) chloride (2) nitrito-N-pentaamminecobalt (II) chloride (3) pentaammine nitrito-N-cobalt (II) chloride (4) pentaammine nitrito-N-cobalt (III) chloride Solution (4) In naming the coordination entity, the ligands are named first in alphabetical order in preference. Cobalt is attached through nitrogen of nitrito group, therefore, nitrito-N-cobalt and oxidation state of cobalt is x – 5(0)– 1 – 2 = 0 ⇒ x = +3.

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21. The possible number of stereoisomers for the formula n± [Ma2b3 ] is (1) 5 (2) 6 (3) 4 (4) 9

(1) Six (2) Three (3) One (4) Two Solution

Solution (2) For coordination number 5, two geometries are possible, that is trigonal bipyramidal and square pyramidal. TBP (I)

(3) EDTA is a hexadentate ligand and in an octahedral complex the coordination number is 6. Therefore, only one EDTA molecule is required to make an octahedral complex. 24. Which of the following facts about the complex [Cr(NH3)6]Cl3 is wrong?

C.N = 5

(1) The complex is paramagnetic. (2) The complex is an outer orbital complex. (3) The complex gives white precipitate with silver nitrate solution. (4) The complex involves d2sp3 hybridization and is octahedral in shape.

Square pyramidal (II)



For (I): a b

b b

M b

a a

M

b

M

b a



b

a

Solution

a b

b

For (II): a

b

b

a

a

M

b b

M

b

b

a

a

b

(2) In [Cr(NH3)6]Cl3, the electronic configuration of Cr3+ is 3d3 and the hybridization is d2sp3, thus forming an inner orbital octahedral complex. The compound is paramagnetic as there are three unpaired electrons. This compound gives white precipitate on reaction with AgNO3 due to the reaction with Cl- ions. 3d

M b

a

Cr(III) b 3d

22. An octahedral complex with molecular composition M·5NH3·Cl·SO4 has two isomers, A and B. The solution of A gives a white precipitate with AgNO3 solution and the solution of B gives white precipitate with BaCl2 solution. The type of isomerism exhibited by the complex is (1) (2) (3) (4)

linkage isomerism. ionization isomerism. coordinate isomerism. geometrical isomerism.

[Cr(NH3)6]

××

4p ××

××

××

25. The correct statement about the magnetic properties of [Fe(CN)6]3- and [FeF6]3- is (Z = 26). (1) both are paramagnetic. (2) both are diamagnetic. (3) [Fe(CN)6]3- is diamagnetic, [FeF6]3- is paramagnetic. (4) [Fe(CN)6]3- is paramagnetic, [FeF6]3- is diamagnetic.

(2) A and B are two isomer of M·5NH3·Cl·SO4. 3 M ⋅ 5NH 3 ⋅ Cl ⋅ SO4 AgNO  → white ppt.

A 2 M ⋅ 5NH 3 ⋅ Cl ⋅ SO4 BaCl  → white ppt.

B Ag+ ion gives white precipitate with both Cl- and SO24 but Ba2+ ion gives white precipitate only with SO2Hence in 4 complex B, SO24 must present outside the complex sphere.

Solution (1) In [Fe (CN)6]3- and [FeF6]3- the electronic configuration of Fe3+ is 3d5 Fe3+ : 3d5



24



Structure of B = [M(NH3)5Cl] SO



Structure of A = [M(NH3)5SO4]Cl

23. How many EDTA (ethylenediaminetetraacetic acid) molecules are required to make an octahedral complex with a Ca2+ ion?

Chapter 20_Coordination Compounds.indd 512

××

d2sp3 hybridization Six electron pairs donated by six NH3 molecules

Solution



4s ××

3+

4s0

4p

In strong field ligand, the hybridization of Fe3+ is d2sp3 due to pairing of electrons. [Fe(CN)6]3−: d2sp3



In weak field ligand, the hybridization of Fe3+ is sp3d2.

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Coordination Compounds

the region of red, green, yellow and blue, respectively. The increasing order of ligand strength of the four ligands is

[FeF6]3−: sp3d2



In both the complexes unpaired electrons are present. Thus, both complexes are paramagnetic.

26. Nickel (Z = 28) combines with a uninegative monodentate ligand to form a diamagnetic complex [NiL4]2-. The hybridization involved and the numbers of unpaired electrons present in the complex are respectively: (1) sp3, two (2) dsp2, zero (3) dsp2, one (4) sp3, zero

(2) In [NiL4]2- complex ligand is uninegative monodentate hence the oxidation state of Ni is +2. Since complex is diamagnetic hence all the electrons should be paired.

Ground state Ni2+: 3d [NiL4] : 2−

××

(1) L4 < L3 < L2 < L1 (2) L1 < L3 < L2 < L4 (3) L3 < L2 < L4 < L1 (4) L1 < L2 < L4 < L3 Solution (2) Strong field ligands cause higher splitting in the d orbitals. This means that the absorption of energy for d–d transition is more, and so transmitted energy is more. In case of L1, L2, L3 and L4, as the energy decreases in the order Violet > Indigo > Blue > Green > Yellow > Orange > Red

Solution



513

4s

4p

××

×× ××

dsp2 hybridization

27. The octahedral complex of a metal ion M3+ with four mono­ dentate ligands L1, L2, L3 and L4 absorb wavelengths in



so, the order ligand strength is L1 < L3 < L2 < L4.

28. Which of the following complex has lowest IR stretching vibration frequency for C–O bond? (1) [Fe(CO)6]2+ (2) [Mn(CO)6]+ (3) [V(CO)6]- (4) [Ti(CO)6]2Solution (4) Metal carrying maximum negative charge causes maximum extent of synergic bonding. Hence the bond order of C—O bond in case of (4) is decreased in highest extent and hence the IR stretching vibration frequency for C—O bond will be lowest. The Synergic effect order is Fe2+ < Mn+ < V- < Ti2-

Practice Exercises Level I Werner’s Theory of Coordination Compounds 1. One mole of the complex compound Co(NH3)Cl3 gives three moles of ions on dissolutions in water. One mole of the same complex reacts with two moles of AgNO3 solution to yield two moles of AgCl(s). The structure of the complex is (1) [Co(NH3)5Cl]Cl2 (2) [Co(NH3)3Cl3]×2NH3 (3) [Co(NH3)4Cl2]Cl×NH3 (4) [Co(NH3)4Cl]Cl2×NH3 2. According to Werner’s theory, the primary valencies of the central metal atom (1) (2) (3) (4)

are satisfied by negative ions. are satisfied by negative ions or neutral molecule. decide the geometry of the complex. are equal to its coordination number.

3. KCl×MgCl2×6H2O is a (1) mixed salt. (2) double salt. (3) basic salt. (4) complex salt. 4. The compound [CoCl3I(C5H5N)2]Br will show the chemical test for which of the following ions?

Chapter 20_Coordination Compounds.indd 513

(1) Br - (2) Cl(3) I- (4) Both (1) and (2). 5. When 0.1 mol of CoCl3(NH3)5 is treated with excess of AgNO3, 0.2 mol of AgCl are obtained. The conductivity of solution will correspond to (1) 1:3 electrolyte. (2) 1:2 electrolyte. (3) 1:1 electrolyte. (4) 3:1 electrolyte.

Definition of Some Important Terms Pertaining to Coordination Compounds 6. The coordination number of central metal atom in a complex is determined by (1) the number of ligands around a metal ion bonded by sigma bonds. (2) the number of only anionic ligands bonded to the metal ion. (3) the number of ligands around a metal ion bonded by sigma and p-bonds both. (4) the number of ligands around a metal ion bonded by p-bonds.

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7. Which of the following compounds has the minimum coordination number? (1) Co2(CO)8 (3) [Co(NH3)5Cl]Cl2

(2) Mn2(CO)10 (4) [Co(NH3)4Cl2]

17. Which of the following is not a bidentate ligand? (1) Oxalate (2) Diethylenetriamine (3) DMG (4) 1,10-Phenanthroline 18. What is not true about a ligand?

8. Which of the following complex has highest EAN values? (1) [Fe(CO)5] (2) [Ti(s–C2H5)2(p–C5H5)2] (3) [Ni(en)3]Cl2 (4) [Mn(C2O4)3]3–

(1) (2) (3) (4)

19. Which of the following complexes obeys the EAN rule?

9. What is the coordination number of nickel in [Ni(NO2)2(C2O4)2]4-? (1) 4 (2) 5 (3) 6 (4) 8

(1) [Pt(NH3)4]2+ (2) [Ni(NH3)6]2+ (3) [Cr(NH3)6]3+ (4) [Fe(CN)6]4 20. Which of the following is non-ionizable?

10. In all the following complexes, the coordination number of iron is six. In which of them is the oxidation state of iron the lowest? (1) K4[Fe(CN)6] (2) Fe(CO)6 (3) Fe[(EDTA)]- (4) [Fe(CN)6]3 11. The coordination number of the central atom in the complex [Co(NH3)4SO4]NO3 is (1) four (2) five (3) six (4) seven 12. In the complex [Pt(py)4][PtCl4], the oxidation number of Pt atoms in the former and latter parts of the complex are, respectively, (1) 0 and 0 (2) +4 and +2 (3) +2 and +2 (4) 0 and +4

(1) [Co(NH3)3Cl3] (2) [Co(NH3)4Cl2]Cl (3) [Co(NH3)5Cl]Cl2 (4) [Co(NH3)6]Cl2 21. Which of the following species is not expected to be a ligand? (1) NO (2) NH +4 (3) NH2CH2CH2NH2 (4) CO 22. The coordination number and the oxidation state of the element ‘E’ in the complex [E(en)2(C2O4)]NO2 (where (en) is ethylenediamine) are, respectively, (1) 6 and 2 (2) 4 and 2 (3) 4 and 3 (4) 6 and 3

Nomenclature of Coordination Compounds 23. The ligand called p acid is

13. What is the coordination number of chromium in [Cr(NH3)2(H2O)2]Cl3? (1) 3 (2) 4 (3) 5 (4) 6

(1) CO (2) NH3 (3) C 2O42- (4) Ethylenediamine 24. Which of the following name formula combinations is not correct? Formula

Name

(1)

K2[Pt(CN)4]

Potassium tetracyanoplatinate(II)

(2)

[Mn(CN)5]2-

Pentacyanomagnate(II) ion

(3)

K[Cr(NH3)2Cl4]

Potassium diammine tetrachlorochromate(III)

(4)

[Co(NH3)4(H2O)I] Tetraammine aquaiodo cobalt(III) SO4 sulphate

14. Which of the following complexes obeys the EAN rule? (1) [Pt(NH3)4] (2) [Ni(NH3)6] (3) [Cr(NH3)6]3+ (4) [Fe(CN)6]42+

2+

15. In which of the following compounds, the metal is in the lowest oxidation state? (1) [Co(NH3)5Br]2SO4 (2) Fe3[Fe(CN)6]2 (3) Mn2(CO)10 (4) K[PtCl3(C2H4)] 16. Which of the following statements is correct for the complex [Fe(H2O)5NO]SO4? (1) The EAN value of Fe in this complex depends on the charge of NO ligand. (2) The EAN value of Fe in this complex does not depend on the charge of NO ligand. (3) The hybridization of the central atom is d2sp3. (4) It is paramagnetic with μ = 173. BM.

Chapter 20_Coordination Compounds.indd 514

It can act as Lewis base. It can be monodentate or multidentate. A monodentate ligand cannot be chelating ligand. A multidentate ligand cannot cause chelation.

25. What is the correct name for the following complex? [Co(NH2)2(NH2CH3)2]Cl (1) (2) (3) (4)

Diamidodimethylaminecobalt(III) chloride. bis(Methylamine) diamidocobalt(III) chloride. Diamidobis(methylamine)cobalt(III) chloride. Diaminedimethylaminecobalt(III) chloride.

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Coordination Compounds 26. Choose the correct name for [Zn(NCS)4](NO3)2. (1) (2) (3) (4)

Tetrathiocyanato zincate(II) nitrate. Tetrathiocyanato zinc(II) nitrate Tetraisothiocyanato zinc(II) nitrate Tetrathiocyanate-N-zincate(II) nitrate

27. The IUPAC name of the complex having formula [(CO)3Fe(CO)3Fe(CO)3] is (1) monocarbonylferrate(0). (2) tricarbonyliron(0)-m-tricarbonyliron(0). (3) tri-m-carbonylbis-{tricarbonyliron(0)}. (4) hexacarbonyl-m-tricarbonyliron(III). 28 The complex Hg[Co(CNS)4] is correctly named as (1) (2) (3) (4)

mercury tetrathiocyanatocobaltate(II). mercury cobalttetrasulphocyano(II). mercury tetrasulphocyanidecobaltate(II). mercury sulphocyanatocobalt(II).

29. IUPAC name of [Pt(NH3)2Cl(NO2)] is (1) Platinum diaminechloronitrite. (2) Chloronitrito-N-ammineplatinum(II). (3) Diamminechloridonitrito-N-platinum(II). (4) Diamminechloronitrito-N-platinate(II).

Isomerism in Coordination Compounds 30. Which of the following ligands can show linkage isomerism? (1) NO (2) NH 3 (3) NO-3 (4) None of these 31. In which of the following pairs both the complexes show optical isomerism? (1) cis-[Cr(C2O4)2Cl2]3−, cis-[Co(NH3)4Cl2] (2) Co(en)3]Cl3, cis-[Co(en)2Cl2]Cl (3) [PtCl(dien)]Cl, [NiCl2Br2]2− (4) [Co(NO3)3(NH3)3], cis-[Pt(en)2Cl2] 32. Which of the following complex species is not expected to exhibit optical isomerism? (1) [Co(en )2 Cl 2 ]+ (2) [Co(NH 3 )3 Cl 3 ] (3) [Co(en )(NH 3 )2 Cl 2 ]+ (4) [Co(en )3 ]3+ 33. Which of the following pairs represents linkage isomers? (1) [Cu (NH3)4] [PtCl4] and [Pt(NH3)4] [CuCl4] (2) [Pd(PPh3)2(NCS)2] and [Pd(PPh3)2(SCN)2] (3) [CO (NH3)5NO3]SO4 and [CO(NH3)5SO4]NO3 (4) [PtCl2(NH3)4]Br2 and [PtBr2(NH3)4] Cl2 34. The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]+ is (py = pyridine) (1) 3 (2) 4 (3) 6 (4) 2 35. Which of the following compounds shows optical isomerism? (1) [Cu(NH3)3]2+ (2) [ZnCl4]2(3) [Cr(C2O4)]3- (4) [Co(CN)6]3-

Chapter 20_Coordination Compounds.indd 515

515

36. Find the total number of cis-isomers with respect to ligands a and b for [Mabcdef]n± type of complex among its all possible stereoisomers. (1) 6 (2) 24 (3) 8 (4) 22 37. Find the kind of isomerism is present between these two complexes.

  [Co(NH3)4(H2O)Cl]Br2 and [Co(NH3)4Br2]Cl . H2O (1) Hydrate isomerism (2) Ionization isomerism (3) Both (1) and (2) (4) Cannot be predicted

38. The isomers [(C6H5)3P2Pd(SCN)2] and [(C6H5)3P2Pd(NCS)2] show which isomerism? (1) (2) (3) (4)

Linkage isomerism Coordination isomerism Ionization isomerism Geometrical isomerism

39. Which one of the following square planar complexes will form geometrical isomers? (1) MA4 (2) MA3B (3) MA2B2 (4) MAB3 40. The total number of possible isomers for the complex compound [CuII(NH3)4] [PtIICl4] are (1) 3 (2) 4 (3 5 (4) 6 41. The ligand ac ac has __________ donor sites. (1) 4 (2) 3 (3) 2 (4) 5 42. The compounds [Co(SO4)(NH3)5]Br and [Co(SO4)(NH3)5] Cl represent (1) (2) (3) (4)

Linkage isomerism. Ionization isomerism. Coordination isomerism. No isomerism.

43. Indicate the complex ion which shows geometrical isomerism. (1) [Cr(H2O)4Cl2]+ (2) [Pt(NH3)3 Cl] (3) [Co(NH3)6]3+ (4) [Co(CN)5(NC)]3–

Bonding in Coordination Compounds 44. Amongst the following, the diamagnetic species is (1) [Ni(CN)4]2− (2) [NiCl4]2− (3) [CoCl4]2− (4) [CoF6]2− 45. The wavelength of light absorbed is highest in (1) [Co(NH3)5Cl]2+ (3) [Co(NH3)6]3+

(2) [Co(NH3)5H2O]3+ (4) [Co(en)3]3+

46. In [Ag(CN)2]−, the number of p bonds is (1) 2 (2) 3 (3) 4 (4) 6

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47. Ammonia forms the complex ion [Cu(NH3)4]2+ with copper ions in alkaline solutions but not in acidic solutions. What is the reason for it? (1) In acidic solutions hydration protects copper ions. (2) In acidic solutions protons coordinate with ammonia molecules forming NH +4 ions and NH3 molecules are not available (3) In alkaline solutions insoluble Cu(OH)2 is precipitated which is soluble in excess of any alkali (4) Copper hydroxide is an amphoteric substance. 48. An octahedral complex of Co3+ is diamagnetic. The hybridization involved in the formation of the complex is (1) sp d (2) dsp (3) d2sp3 (4) dsp3d 3

2

2

49. When spherically symmetrical field of ligands surrounds the central metal ion, then which of the following options is correct for the change in energy order of d-orbitals of the central metal ion? (1) t2g > eg (2) eg > t2g (3) eg = t2g (4) Cannot be predicted 50. Which of the following patterns is correct for [Fe(H 2O)6 ]3+ and [Fe(CN)6 ]3- respectively? [Fe(H2O)6]3+

[Fe(CN)6]3− eg

eg

(1) E

t2g eg

(2) E

t2g

eg

(3) E

t2g eg

t2g

Both (I) and (II) are paramagnetic. Both (I) and (II) are diamagnetic. I is paramagnetic while II is diamagnetic. I is diamagnetic while II is paramagnetic

53. Which of the following complex has same magnetic moment as that of [Cu(NH3)4]2+? (1) [Cr(NH3)6]3+ (2) [Mn(H2O)6]2+ (3) [NiCl4]2– (4) [Ti(H2O)6]3+ 54. An octahedral complex is formed when hybrid orbitals of the following type are involved (1) sp3 (2) dsp2 (3) sp3d2 (4) sp3d 55. Which of the following is colorless? (1) Ni(CO)4 (2) Fe(CO)5 (3) V(CO)5 (4) All of these. 56. In which of the following complexes, the nickel metal is in the highest oxidation state? (1) Ni(CO)4 (2) K2NiF6 (3) [Ni(NH3)6](BF4)2 (4) K4[Ni(CN)6] 57. The stabilization of coordination compounds due to chelation is called the chelate effect. Which of the following is the most stable complex species?

58. The CFSE for octahedral [CoCl6]4– is 18,000 cm–1. The CFSE for tetrahedral [CoCl4]2– will be (1) 18,000 cm–1 (2) 16,000 cm–1 (3) 8,000 cm–1 (4) 20,000 cm–1 59. The V – C distances in V(CO)6 and [ V(CO)6 ]- are respectively (in pm). (1) 200, 200 (2) 193, 200 (3) 200, 193 (4) None of these

eg t2g

eg t2g

E

(1) (2) (3) (4)

(1) [Fe(CO)5] (2) [Fe(CN)6]3– (3) [Fe(C2O4)3]3– (4) [Fe(H2O)6]3+

t2g eg

(4)

52. Which of the following statements is true for the complexes: (I) [CoF6 ]3- and (II) [NiF6 ]2- ?

t2g

60. Which one of the following complexes in an outer orbital complex? (1) [Fe(CN)6]4– (2) [Ni(NH3)6]2+ (3) [Co(NH3)6]3+ (4) [Mn(CN)6]4–

Bonding in Metal Carbonyls 61. The oxidation number of cobalt in K[Co(CO)4] is (1) +1. (2) +3. (3) –1. (4) –3.

51. K3CoF6 is high-spin complex. What is the hybrid state of Co atom in this complex? (1) sp3d (2) sp3d2 (3) d2sp3 (4) dsp2

Chapter 20_Coordination Compounds.indd 516

62. In which of the following compounds, C   O bond length is the maximum? (1) [Fe(CO)5]− (2) [Ni(CO)4] (3) [Cr(CO)6] (4) [Mn(CO)5]+

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517

Coordination Compounds 63. Complex in which central atom contains minimum electrons in valence shell is (1) [Co(CO)4]+ (2) [Mn(CO)5]− (3) V(CO)6 (4) Cr(CO)6

Definition of Some Important Terms Pertaining to Coordination Compounds 5. Find the ligand having the highest denticity from the following options.

Importance and Applications of Coordination Compounds 64. Coordination compounds have great importance in biological systems. In this context which of the following statements is incorrect? (1) Chlorophylls are green pigments in plants and contain calcium. (2) Carboxypeptidase–A is an enzyme and contains zinc. (3) Cyanocobalamin is B12 and contains cobalt. (4) Haemoglobin is the red pigment of blood and contains iron. 65. Which of the following is not an example of organometallic compound?

O

C O

CH2

NH2

NH2

CH2 CH2

(2)

N

O

C

NH2

O O

O

C

(3)

CH2

NH

O

(4) NH2

CH2

C

O

6. Which of the following ligand is tridentate type? (1) (2) (3) (4)

Butane-1,2-diamine (bn) Propane-1,3-diamine (tn) Diethylamine (dien) Triethyltetraamine (trien)

(1) [Ni(h 5− C5H5)2]+ (2) [V(h 6 − C7H8)(h 7 − C7H7)]+ (3) [Fe(CO)2(NO)2]0 (4) [Fe(CN)6]3-

Level II Werner’s Theory of Coordination Compounds 1. Consider the coordination compound, [Co(NH3)6]Cl3. In the formation of this complex, the species which acts as the Lewis acid is -

(1) [Co(NH3)6] (2) Cl (3) Co3+ (4) NH3 2. Which of the following pair contains complex salt and double salt, respectively? (1) FeSO4, K4[Fe(CN)6] (2) [Cu(NH3)4]SO4, FeSO4⋅7H2O (3) [Cu(NH3)4]SO4, K2SO4⋅Al2(SO4)3⋅24H2O (4) MgSO4⋅7H2O, CuSO4 3. When 1 mol CrCl3⋅6H2O is treated with excess of AgNO3, 3 mol of AgCl are obtained. The formula of the complex is (1) [CrCl3 (H2O)3]⋅3H2O (2) [CrCl2(H2O)4]Cl⋅2H2O (3) [CrCl(H2O)5]Cl2⋅H2O (4) [Cr(H2O)6]Cl3 4. The complex [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br can be distinguished by (1) BaCl2 (2) AgNO3 (3) PbCl2 (4) All of these.

Chapter 20_Coordination Compounds.indd 517

(1)

NH2

CH

7. Which of the following complexes contains a cationic ligand?

(1) Trimethylboron (2) Trimethylaluminium (3) Trimethoxytitanium chloride (4) Tetracarbonylnickel

3+

CH3

8. Effective atomic number (EAN) of Fe(CO)5 is (1) 35 (2) 34 (3) 36 (4) 37 9. Which of the following set of ligands contains anionic ligands? (1) (2) (3) (4)

Aqua, oxalato, chloro Imido, nitrito, sulphido Aqua, nitrito, sulphido Nitronium, hydrazinium, sulphato

10. A group of atoms can function as a ligand only when (1) (2) (3) (4)

it is a small molecule. it is capable of acting as donor of electron pair. it is a negatively charged ion. it is a positively charged ion.

11. Ligand en is an example of a (1) monodentate ligand. (2) bidentate ligand. (3) tridentate ligand. (4) hexadentate ligand. 12. Ferrocene is (1) Fe(h 5 − C5H5)2 (2) Fe(h 2 − C5H5)2 5 (3) Cr(h  − C5H5)5 (4) Os(h 5 − C5H5)2

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OBJECTIVE CHEMISTRY FOR NEET

13. A chelating agent has two or more than two donor atoms to bind to a single metal ion. Which of the following is not a chelating agent?

(2) In meridional isomers, same ligands are present at 90° and 180° angles. (3) In facial isomers, same ligands are present only at 90° angle. (4) Both isomers (facial and meridional) are optically active.

(1) Thiosulphato (2) Oxalato (3) Glycinato (4) Ethane-1,2-diamine

Nomenclature of Coordination Compounds 14. Which among the following will be named as dibro­ midobis(ethylene diamine)chromium(III) bromide?

22. The sum of stereoisomers of complex-A, complex-B and complex-C in following reaction is ___________. NH 3 2(pyridine) [PtCl 4 ]2- ¾+¾¾¾ ® [Complex - A] ¾+-¾¾ ® -2 Cl Cl -

(1) [Cr(en)3]Br3 (2) [Cr(en)2Br2]Br (3) [Cr(en)Br4]- (4) [Cr(en)Br2]Br

-

+ Br [Complex - B] ¾-¾¾¾ ® [Complex -C] (pyridine)

15. The IUPAC name of the coordination compound K3[Fe(CN)6] is (1) (2) (3) (4)

Potassium hexacyanoferrate(II) Potassium hexacyanoferrate(III) Potassium hexacyanoiron(II) Tripotassium hexacyanoiron(II)

(1) 7 (2) 5 (3) 6 (4) 9 23. A similarity between optical and geometrical isomerism is that (1) each gives equal number of isomers for a given compound. (2) If in a compound one is present then so is other. (3) both are included in stereoisomerism. (4) they have no similarity.

16. The correct IUPAC name of [Pt(NH3)2Cl2] is (1) Diamminedichloridoplatinum(II) (2) Diamminedichloridoplatinum(IV) (3) Diamminedichloridoplatinum(0) (4) Dichloridodiammineplatinum(IV) 17. The correct IUPAC name of [Fe(C5H5)2] is (1) cyclopentadienyliron(II). (2) bis(cyclopentadienyl)iron(II). (3) dicyclopentadienylferrate(II). (4) ferrocene. 18. The correct IUPAC name of [Mn3(CO)12] is (1) magneticdodecylcarbonyl(0). (2) dodecacarbonylmanganate(0). (3) dodecarbonylmangemic(II). (4) dodecacarbonyltrimanganese(0).

Isomerism in Coordination Compounds 19. Which of the following compounds shows both linkage and ionization isomerism? (1) [Co(NH3)5NO2](NO3) (2) [Co(NO2)(py)2(NH3)2]NO3 (3) [Co(NO2)(NH3)5]Cl2 (4) [Co(NO2)Br(NH3)4]Cl 20. Which of the following has an optical isomer? (1) [Co(H2O)4(en)]3+ (2) [Co(en)2(NH3)2]3+ (3) [Co(NH3)3Cl]+ (4) [Co(en)(NH3)2]2+ 21. Ma3b3 complex has two geometrical forms: facial and meridional. Then which of the following statement is incorrect? (1) In facial isomers, three same ligands occupy adjacent positions on octahedron face.

Chapter 20_Coordination Compounds.indd 518

24. What kind of isomerism exists between [Cr(H2O)6]Cl3 (violet) and [Cr(H2O)5Cl]Cl2⋅H2O (greyish-green)? (1) (2) (3) (4)

Linkage isomerism Solvate isomerism Ionization isomerism Coordination isomerism

25. For hydrate and linkage isomerism, the following respective ligands are responsible: (1) OH-, CN- (2) H2O, NH3 (3) H2O, NO-2 (4) OH-, NO-

Bonding in Coordination Compounds 26. The correct order of magnetic moments (spin only values in BM) among is (1) [MnCl4]2– > [CoCl4]2– > [Fe(CN)6]4– (2) [Fe(CN)6]4– > [CoCl4]2– > [MnCl4]2– (3) [Fe(CN)6]4– > [MnCl4]2– > [CoCl4]2– (4) [MnCl4]2– > [Fe(CN)6]4– > [CoCl4]2–

(Atomic numbers: Mn = 25; Fe = 26, Co = 27)

27. Which one of the following cyano complexes would exhibit the lowest value of paramagnetic behavior? (1) [Cr(CN)6]3– (2) [Mn(CN)6]3– (3) [Fe(CN)6]3– (4) [Co(CN)6]3–

(Atomic numbers: Cr = 24, Mn = 25, Fe = 26, Co = 27)

28. Electronic configuration of Co in an octahedral complex 6 0 is t2g  eg with a ligand L, then L is

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Coordination Compounds (1) SCN− (3) F−

(2) Oxalate (4) Cl−

29. Which of the following is diamagnetic? (1) [Cu(NH3)4] 2+ (3) [PtCl4]2−

(2) [NiCl4]2− (4) [Cu(H2O)4]2+

30. An aqueous solution of CoCl2 on addition of excess of concentrated HCl turns blue due to formation of (1) [Co(H2O)4Cl2] (2) [Co(H2O)2Cl4]2− (3) [CoCl4]2− (4) [Co(H2O)2Cl2] 31. How many EDTA (ethylenediaminetetraacetic acid) molecules are required to make an octahedral complex with a Ca2+ ion? (1) Six (2) Three (3) One (4) Two 32. Nickel (Z = 28) combines with a uninegative monodentate ligand X- to form a paramagnetic complex [NiX4]2–. The number of unpaired electron(s) in the nickel and geometry of this complex ion are, respectively (1) one, tetrahedral. (2) two, tetrahedral. (3) one, square planar. (4) two, square planar. 33. The magnetic moment (spin only) of [NiCl4]2– is (1) 5.46 BM (2) 2.83 BM (3) 1.41 BM (4) 1.82 BM 34. Which one of the following complexes will most likely absorb visible light? (Atomic number: Sc = 21, Ti = 22, V = 23, Zn = 30) (1) [Sc(H2O)6]3+ (2) [Ti(NH3)6]4+ (3) [V(NH3)6]3 + (4) [Zn(NH3)6]2+ 35. The splitting of the d - orbitals of transition metals by a particular complex geometry gives rise to the pattern, dxy dyz dxz Energy



dx2 − y2

dz2

Which one of the following geometries produces the above splitting pattern? (1) Octahedral (2) Planar triangular (3) Square planar (4) Tetrahedral

36. Find the total number of unpaired electrons in t2g orbital of [Ni(en)3]2+. (1) 2 (2) 0 (3) 4 (4) 3 37. Which of the following shall form an octahedral complex? (1) d 4 (low spin) (2) d 8 (high spin) (3) d 6 (low spin) (4) All of these 38. The color of the coordination compounds depends on the crystal field splitting. What will be the correct order

Chapter 20_Coordination Compounds.indd 519

519

of absorption of wavelength of light in the visible region, for the complexes, [Co(NH3)6]3+, [Co(CN)6]3–, [Co(H2O)6]3+ (1) [Co(CN)6]3– > [Co(NH3)6]3+ > [Co(H2O)6]3+ (2) [Co(NH3)6]3+ > [Co(H2O)6]3+ > [Co(CN)6]3– (3) [Co(H2O)6]3+ > [Co(NH3)6]3+ > [Co(CN)6]3– (4) [Co(CN)6]3– > [Co(NH3)6]3+ > [Co(H2O)6] 39. Which of the following system has the maximum number of the unpaired electrons in an inner octahedral complex? (1) d 4 (2) d 9 (3) d 7 (4) d 5 40. Which one of the following has a square planar geometry (At. no. of Co = 27, Ni = 28, Fe = 26, Pt = 78)? (1) [PtCl4]2- (3) [FeCl4]2-

(2) [CoCl4]2(4) [NiCl4]2-

41. Among the following species the one which causes the highest CFSE, Do as a ligand is: (1) CN- (2) NH3 (3) F- (4) CO 42. Comment on the magnetic behavior of K2[NiF6] and K4[NiF6]. (1) (2) (3) (4)

Paramagnetic and paramagnetic. Diamagnetic and paramagnetic. Diamagnetic and diamagnetic. Paramagnetic and diamagnetic.

43. Choose the correct order for D o for the following complexes. (I) [Co(H 2O)6 ]2+ (III) [Fe(H 2O)6 ]3+

 (II)   [Co(H 2O)6 ]3+ (IV)   [Fe(CN )6 ]3-

(1) I < II < III < IV (2) I < III < II < IV (3) I < II = III < IV (4) I < II < IV < III

Bonding in Metal Carbonyls 44. In Fe(CO)5, the Fe   C bond possesses (1) p-character only. (2) both s and p characters. (3) ionic character. (4) s-character only. 45. Select the true statement from the following for metal carbonyls? (1) The p back bonding strengthens M   C bond order as well as CO bond order. (2) The p back bonding weakens M   C bond order as well as CO bond order. (3) The p back bonding weakens M   C bond order but strengthens CO bond order. (4) The p back bonding strengthens M   C bond order and weakens CO bond order.

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520

OBJECTIVE CHEMISTRY FOR NEET

Stability of Coordination Compounds

(1) [Co(en)3 ]3+ (2) [Co(en)2Cl 2 ]+ (3) [Co(NH 3 )3Cl 3 ]0 (4) [Co(en )Cl 2(NH 3 )2 ]+

46. Which of the following complexes formed by Cu2+ ions is most stable? (1) Cu2+ + 4NH3 → [Cu(NH3)4]2+, log K = 11.6 (2) Cu2+ + 4CN– → [Cu(CN)4]2–, log K = 27.3 (3) Cu2+ + 2en → [Cu(en)2]2+, log K = 15.4 (4) Cu2+ + 4H2O → [Cu(H2O)4]2+, log K = 8.9

(AIPMT 2009) 7. Crystal field stabilization energy for high spin d4 octahedral complex is (1) -1.6 D o (2) -1.2 D o (3) -0.6 D o (4) -0.8 D o

Previous Years’ NEET Questions 1. Which of the following will give a pair of enantiomorphs? (en = NH2CH2 CH2NH2) (1) [Pt(NH3)4][PtCl6] (2) [Co(NH3)4Cl2]NO2

(AIPMT PRE 2010) 8. The existence of two different colored complexes with the composition of [Co(NH3)4Cl2]+ is due to

(3) [Cr(NH3)6][Co(CN)6] (4) [Co(en)2Cl2]Cl

(1) (2) (3) (4)

(AIPMT 2007)

2. The d electron configuration of Cr2+, Mn2+, Fe2+ and Ni2+ are 3d4, 3d5, 3d6 and 3d8, respectively. Which one of the following aqua complex will exhibit the minimum paramagnetic behavior? (At. no. Cr = 24, Mn = 25, Fe = 26, Ni = 28)



(1) [Ni(H2O)6]2+ (3) [Cr(NH3)6]3+

(3) [Fe(H2O)6]2+ (4) [Ni(H2O)6]2+ (AIPMT 2007)

3. Which of the following complexes exhibits the highest paramagnetic behavior? (Z of Ti = 22; V = 23; Fe = 26; Co = 27) (1) [Co(ox)2(OH)2]

(2) [Ti(NH3)6]

(3) [V (gly)2(OH)2(NH3)2]+

(4) [Fe(en) (bpy) (NH3)2]2+

−

(where gly = glycine, en = enthylenediamine and bpy = bipyridyl moieties)



(AIPMT 2008)

4. In which of the following coordination entities the magnitude of Do (CFSE in the octahedral field) will be maximum (At. No. Co = 27)? 3-

3+

(1) [Co(C 2O4 )3 ] (2) [Co(H 2O)6 ] (3) [Co(NH 3 )6 ]3+ (4) [Co(CN)6 ]3

(1) [Zn(NH3)6]2+ (3) [Ti(en)2(NH3)2]4+

(2) [Sc(H2O)3(NH3)3]3+ (4) [Cr(NH3)6)]3+

(1) [Pt (NH3)2Cl2] (2) [Ni(NH3)2Cl2] (3) [Ni(en)3]2+ (4) [Ni(NH3)4(H2O)2]2+

Chapter 20_Coordination Compounds.indd 520

(AIPMT MAINS 2010)

11. The d-electron configurations of Cr2+, Mn2+, Fe2+ and Co2+ are d4, d 5, d 6 and d 7 respectively. Which one of the following will exhibit minimum paramagnetic behavior? (1) [Mn(H2O)6]2+ (3) [Co(H2O)6]2+

(2) [Fe(H2O)6]2+ (4) [Cr(H2O6)]2+



(AIPMT PRE 2011)

12. The complex [Pt(Py)(NH3)BrCl] will have how many geometrical isomers? (1) 3 (2) 4 (3) 0 (4) 2

(AIPMT PRE 2011)

13. Of the following complex ions, which is diamagnetic in nature? (1) [NiCl4]2− (3) [CuCl4]2−

(AIPMT 2009)

6. Which of the following cannot show optical isomerism? (en = ethylenediamine)

(AIPMT PRE 2010)

10. Which one of the following complexes is not expected to exhibit isomerism?

(AIPMT 2008)

5. Which of the following complex ions is expected to absorb visible light?

(2) [Ni(CN)4]2− (4) [Fe(H2O)6]2+



3+



(AIPMT PRE 2010)

9. Which of the following complex ion is not expected to absorb visible light?

(1) [Cr(H2O)6]2+ (2) [Mn(H2O)6]2+

ionization isomerism. linkage isomerism. geometrical isomerism. coordination isomerism.



(2) [Ni(CN)4]2− (4) [CoF6]3− (AIPMT PRE 2011)

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Coordination Compounds 14. The complex [Co(NH3)6] [Cr(CN)6] and [Cr(NH3)6] [Co(CN)6] are the examples of which type of isomerism?

21. Which of the following complexes is used as an anti­ cancer agent?

(1) Linkage isomerism (2) Ionization isomerism (3) Coordination isomerism (4) Geometric isomerism

(AIPMT PRE 2011)

15. Which of the following carbonyls will have the strongest C   O bond? (2) [Mn(CO)6 ]+ (4) [ V(CO)6 ]-

(1) Fe(CO)5 (3) Cr(CO)6

(1) mer-[Co(NH3)3Cl3] (2) cis-[PtCl2(NH3)2] (3) cis-K2[PtCl2Br2] (4) Na2CoCl4

(1) [Zn(NH3)6] (3) [Cr(NH3)6]3+

2+



(1) CoCl3×4NH3 (3) CoCl3×6NH3

(1) [Co(CN)6]3− has four unpaired electrons and will be in a low-spin configuration. (2) [Co(CN)6]3− has four unpaired electrons and will be in a high-spin configuration. (3) [Co(CN)6]3− has no unpaired electrons and will be in a high-spin configuration. (4) [Co(CN)6]3− has no unpaired electrons and will be in a low-spin configuration.

(2) [Ti(NH3)6] (4) [Co(NH3)6]3+ 3+

(2) [Co(NH3)6]3+ (4) [Zn(NH3)6]2+ (AIPMT PRE 2012)

18. Low spin complex of d6 cation in an octahedral field will have which of the following energy? (Δo = Crystal field splitting energy in an octahedral field, P = Electron pairing energy) -2 -2 D o + 2P (2) Do + P 5 5 -12 -12 D o + P (4) D o + 3P (3) 5 5

(1)



(AIPMT 2015) 24. Number of possible isomers for the complex [Co(en)2Cl2]Cl will be (en = ethylenediamine) (1) 3 (2) 4 (3) 2 (4) 1

(1) (2) (3) (4)

20. Among the following complexes, the one which shows zero crystal field stabilization energy (CFSE) is



Chapter 20_Coordination Compounds.indd 521

(RE AIPMT 2015)

26. The hybridization involved in complex [Ni(CN)4]2− is (Z of Ni = 28) (1) d2sp2 (3) dsp2

(NEET 2013)

(1) [Mn(H2O)6]3+ (3) [Co(H2O)6]2+

Tricyanoferrate (III) ion. Hexacyanidoferrate (III) ion. Hexacyanoiron (III) ion. Hexacyanitoferrate (III) ion.



(1) 0.01 (2) 0.001 (3) 0.002 (4) 0.003

(RE AIPMT 2015)

25. The name of complex ion [Fe(CN)6]3− is

(AIPMT MAINS 2012)

19. An excess of AgNO3 is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromium (III) chloride. The number of molecules of AgCl precipitated would be

(AIPMT 2015)

23. Which of the following statements about [Co(CN)6]3- is true?

17. Which one of the following is an outer orbital complex and exhibits paramagnetic behavior?



(2) Fe(H2O)6]3+ (4) [Co(H2O)6]3+ (AIPMT 2014)

(2) CoCl3×5NH3 (4) CoCl3×3NH3



(AIPMT MAINS 2011)

(1) [Cr(NH3)6]3+ (3) [Ni(NH3)6]2+

(AIPMT 2014)

22. Cobalt (III) chloride forms several octahedral complexes with ammonia. Which of the following will not give test of chloride ions with silver nitrate at 25°C?

(AIPMT MAINS 2011)

16. Which of the following complex compounds will exhibit highest paramagnetic behavior? (At. No. Ti = 22, Cr = 24, Co = 27, Zn = 30)

521



(2) d2sp3 (4) sp3 (RE AIPMT 2015)

27. Which of the following has longest C   O bond length? (Free C   O bond length is CO is 1.128 Å). (1) Ni(CO)4 (2) [Co(CO)6]– (3) [Fe(CO)4]2− (4) [Mn(CO)6]+ (NEET-I 2016)

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OBJECTIVE CHEMISTRY FOR NEET

28. The correct increasing order of trans-effect of the following species is

(2) It is d2sp3 hybridized and octahedral. (3) It is dsp2 hybridized and square planar. (4) It is sp3d2 hybridized and octahedral.

(1) CN - > C6 H5 - > Br - > NH 3 (2) Br - > CN - > NH 3 > C6 H5(3) CN - > Br - > C6 H5- > NH 3

(NEET 2017)

(4) NH 3 > CN - > Br - > C6 H5(NEET-II 2016) 29. Jahn–Teller effect is not observed in high spin complexes of

32. The correct order of the stoichiometries of AgCl formed when AgNO3 in excess is treated with the complexes: CoCl3·6NH3, CoCl3·5NH3, CoCl3·4NH3 respectively is (1) 3AgCl, 1AgCl, 2AgCl (2) 3AgCl, 2AgCl, 1AgCl (3) 2AgCl, 3AgCl, 1AgCl (4) 1AgCl, 3AgCl, 2AgCl

(1) d8 (2) d4 (3) d9 (4) d7 (NEET-II 2016) 30. An example of a sigma bonded organometallic compound is (1) grignard’s reagent. (2) ferrocene. (3) cobaltocene. (4) ruthenocene. (NEET 2017) 31. Pick out the correct statement with respect to [Mn(CN )6 ]3-.



(NEET 2017)

33. Correct increasing order for the wavelengths of absorption in the visible region for the complexes of Co3+ is (1) [Co(H2O)6]3+, [Co(en)3]3+, [Co(NH3)6]3+ (2) [Co(H2O)6]3+, [Co(NH3)6]3+, [Co(en)3]3+ (3) [Co(NH3)6]3+, [Co(en)3]3+, [Co(H2O)6]3+ (4) [Co(en)3]3+, [Co(NH3)6]3+, [Co(H2O)6]3+

(1) It is sp3 hybridized and tetrahedral.

(NEET 2017)

Answer Key Level I  1. (1)

 2. (1)

 3. (2)

 4. (1)

 5. (2)

 6. (1)

 7. (1)

 8. (3)

 9. (3)

10. (2)

11. (3)

12. (3)

13. (2)

14. (4)

15. (3)

16. (2)

17. (2)

18. (4)

19. (4)

20. (1)

21. (2)

22. (4)

23. (1)

24. (2)

25. (3)

26. (3)

27. (3)

28. (1)

29. (3)

30. (1)

31. (2)

32. (2)

33. (2)

34. (1)

35. (3)

36. (2)

37. (3)

38. (1)

39. (3)

40. (4)

41. (2)

42. (4)

43. (1)

44. (1)

45. (1)

46. (3)

47. (2)

48. (3)

49. (3)

50. (1)

51. (2)

52. (1)

53. (4)

54. (3)

55. (1)

56. (2)

57. (3)

58. (3)

59. (3)

60. (2)

61. (3)

62. (1)

63. (3)

64. (1)

65. (3)

 2. (3)

 3. (4)

 4. (4)

 5. (3)

 6. (3)

 7. (2)

 8. (3)

 9. (2)

10. (2)

Level II  1. (3) 11. (2)

12. (1)

13. (1)

14. (2)

15. (2)

16. (1)

17. (2)

18. (4)

19. (4)

20. (2)

21. (4)

22. (1)

23. (3)

24. (2)

25. (3)

26. (1)

27. (4)

28. (2)

29. (3)

30. (3)

31. (3)

32. (2)

33. (2)

34. (3)

35. (4)

36. (2)

37. (4)

38. (3)

39. (1)

40. (1)

41. (4)

42. (2)

43. (2)

44. (2)

45. (4)

46. (2)

Previous Years’ NEET Questions  1. (4)

 2. (4)

 3. (1)

 4. (4)

 5. (4)

 6. (3)

 7. (3)

 8. (3)

 9. (2)

10. (2)

11. (3)

12. (1)

13. (2)

14. (3)

15. (2)

16. (3)

17. (3)

18. (4)

19. (2)

20. (2)

21. (2)

22. (4)

23. (4)

24. (1)

25. (2)

26. (3)

27. (3)

28. (1)

29. (1)

30. (1)

31. (2)

32. (2)

33. (4)

Chapter 20_Coordination Compounds.indd 522

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Coordination Compounds

523

Hints and Explanations Level I 1. (1) This is because the group which is outside the square bracket (coordination sphere) is precipitated out.

[Co(NH 3 )5Cl]Cl 2 + AgNO3 → [Co(NH 3 )5Cl]2+ + AgCl ↓

2. (1) Primary valencies are ionizable. If the complex ion exhibits positive charge, then primary valency corresponds to the number of charges present and is balanced by the same number of negative ions. Hence, primary valency can also be defined by the number of anions neutralizing the charge on the complex. 3. (2) KCl×MgCl2×6H2O is a double salt as it loses its entity and ionizes as K+ + 2Mg2+ + 3Cl- + 6H2O

12. (3) In both the complexes the oxidation number of Pt atom [Pt (py)4]2+[PtCl4]2- is +2. 13. (2) Coordination number = Total number of ligands. Both NH3 and H2O are unidentate ligands. 14. (4) EAN = Atomic number - Oxidation state + 2 × Coordination number

EAN of any molecule should be equal to the atomic number of a noble gas, only then that molecule will obey EAN rule.



[Pt(NH3)4]2+: EAN = 78 - 2 + 2 × 4 = 84 (does not obey EAN rule)



[Ni(NH3)6]2+: EAN = 28 - 2 + 2 × 6 = 38 (does not obey EAN rule)



[Cr(NH3)6]3+: EAN = 24 - 3 + 2 × 6 = 33 (does not obey EAN rule)



[Fe(CN)6]4-: EAN = 26 - 2 + 2 × 6 = 36 (obeys EAN rule) EAN = Atomic number of Kr.

4. (1) The compound ionizes as follows: [CoCl 3 I(C6 H5N )2 ]Br → [CoCl 3 I(C6 H5N )2 ]+ + Br

Therefore, it will give positive test for Br-.

6. (1) The number of ligands around a metal ion bonded by sigma bonds. For example, the coordination number of Pt in [PtCl4]2- is four. 8. (3) EAN of all above complexes are as follows:

15. (3) In metal carbonyls, CO is a neutral ligand so Mn has zero oxidation state.



Option (1): [Fe(CO)5]:    26 – 0 + 2 × 5 = 36

17. (2) Diethylenetriamine is a tridentate ligand, and rest all are bidentate ligand.



Option (2): [Ti(s-C2H5)2(p-C5H5)2]: 22 – 0 + 2 × 6 = 34



Option (3): [Ni(en)3]Cl2:     28 – 2 + 2 × 6 = 38

18. (4) Multidentate ligands can cause chelation as EDTA.



Option (4): [Mn(C2O4)3]3–:      25 – 3 + 2 × 6 = 34

9. (3) There are two bidentate ligands (C 2O42- ) and two monodentate ligands ( NO-2 ), so the coordination number is six. 10. (2) In K4[Fe(CN)6]: oxidation state is +2.

In Fe(CO)6: oxidation state is 0.



In Fe[(EDTA)]-: oxidation state is +3.



In [Fe(CN)6]3-: oxidation state is +3.



The carbonyl ligand (CO) is a neutral molecule, so iron is in the oxidation state zero.

11. (3) The total number of ligands coordinately bonded to the central metal atom or ion in the primary coordination sphere represents the coordination number of the complex. In the given complex, there are four NH3 ligands coordinately bonded to central atom and SO2acts as a 4 bidentate ligand, so the coordination number is six.

Chapter 20_Coordination Compounds.indd 523

19. (4) Oxidation state of Fe in [Fe(CN )6 ]4- = + 2

Number of electrons in Fe (II) of the complex = 26 - 2 = 24



Number of electrons donated by six CN- ligands = 6 × 2 = 12



So, EAN of Fe(II) in [Fe(CN )6 ]4- = 24 + 12 = 36



Since 36 is equal to the nearest noble gas (Kr), the [Fe(CN )6 ]4- follows the EAN rule.

20. (1) Since there is no ion outside the coordination sphere so the compound will not ionize. 22. (4) Ethylenediammine is a neutral ligand, C 2O42- has a (–2) charge and is bidentate. Therefore, coordination number of E is 6 and its oxidation state is +3. +

en E en

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524

OBJECTIVE CHEMISTRY FOR NEET

24. (2) The correct name for the complex is [Mn(CN)5]2- = Pentacyanomagnate(III) ion.

36. (2) Total possible stereoisomer for [Mabcdef]n± type complex = 30

25. (3) (NH -2 ) is named as ‘amido’, so (NH -2 )2 is named diamido.

MeNH2 is named methylamine. For (MeNH2)2 – ligand, bis convention is followed to avoid confusion from ligand Me2NH– whose name is dimethylamine. Thus the correct name of the complex is diamidobis(methylamine) cobalt(III) chloride.

a

a

M

M

c



b For trans

d /e/f For cis

b/ c/ d/ e/ f

26. (3) [Zn(NCS)4]2+ is a cationic complex. Hence, ‘ate’ cannot be added at the end of metal ions name, hence (1) and (4) are incorrect.



27. (3) Carbonyl is acting as a bridging ligand here.

37. (3) (a) The number of water of crystallization is different, that is, zero and one hence (1) is correct.

28. (1)Hg[Co(CNS)4] is mercury tetrathiocyanatocobaltate(II) with oxidation state on Co as x - 4 = +2 ⇒ x = -2.



32. (2) [Co(NH3)3Cl3] exists in two forms (facial and meridonial). Cl

NH3

NH3

H3N

NH3

Cl

Co Cl

Cl Facial



NH3 Co

Cl

39. (3) The type MA4, MA3B and MAB3 do not show geometrical isomerism.

Meridional

Both of these forms are achiral. Hence [Co(NH3)3Cl3] does not show optical isomerism.

33. (2) Linkage isomerism occurs in those compounds which contain ambidentate ligand, that is, SCN-.

47. (2) This is because NH3 is not available as ligand. NH 3 + H +  NH 4+ 48. (3) The electronic configuration of Co3+ is

34. (1) Type Mabcd has three geometrical isomers. Cl

+

NH3

Cl

(b) Ion produced in solution is different. Hence they are ionization isomer of each other and (2) is also correct.

38. (1) SCN- and NCS- are ambidentate ligands which can cause linkage isomerism.

Cl

NH3

Total trans isomers with respect to a and b = 6 (all are optically active). Hence, total cis isomers with respect to a and b = 30 - 6 = 24. (All are optically active)

Co3+: NH3

+ Co3+ complex :

×× ××

××

×× ×× ××

Pt

Pt

d2sp3 NH2OH

Py

HOH2N

Cl

Py

Py +

Pt HOH2N

ox

3−

ox

3−

ox

Cr

Cr

ox

ox

Non superimposable mirror image

Chapter 20_Coordination Compounds.indd 524

Thus, the hybridization of Co3+ complex is d2sp3. All the electrons are paired in Co3+ complex which is octahedral and diamagnetic.

49. (3) If a symmetrical field of negative charges surrounds a metal ion, the d-orbitals remain degenerate.

NH3

35. (3) The optical isomers of [Cr(C2O4)]3- are as follows: ox



50. (1) Since CN- is a stronger ligand than H2O, therefore, CFT splitting in [Fe(CN)6]3- will be more than [Fe(H2O)6]3+. 51. (2) [CoF6]3- is sp3d2 as F- is a weak field ligand. 52. (1) Both the complexes are paramagnetic as F- is a weak field ligand and both of them are high spin complexes.

1/4/2018 5:23:05 PM

Coordination Compounds

Level II

53. (4) Complex

Shape

Configuration

No. of unpaired electron

[Cu(NH3)4]2+

Square planar

d9

1

[Cr(NH3)6]3+

Octahedral

d3

3

[Mn(H2O)6]2+

Octahedral

d5 with weak field ligand

5

[NiCl4]2–

Tetrahedral

d8 with weak field ligand

2

[Ti(H2O)6]3+

Octahedral

d1 with weak field ligand

1

1. (3) [Co(NH3)6]Cl3 → [Co(NH3)6]3+ + 3Cl-

55. (1) Ni(CO)4 is sp3 hybridized, tetrahedral complex having no unpaired electron, therefore, colorless.

Ni(CO)4 has Ni = 0 (oxidation state)



K4[Ni(CN)6] has Ni = +2 (oxidation state)



[Ni(NH3)6](BF4)2 has Ni = +2 (oxidation state)

59. (3) The extent of synergic bonding is more in [V(CO)6]– compared to that in [V(CO)6], and increases the partial double bond character of V   C bond. 60. (2) In [Ni(NH3)6]2+, the electronic configuration of Ni2+ is 3d8: [Ni(NH3)6]2+ 3d

4s

4p

××

×× ×× ××

4d



NH3 being a weak field ligand does not cause pairing of electrons. So, it is an outer orbital complex.

61. (3) In K[Co(CO)4] as the oxidation number of K is +1, so the overall oxidation number on the square bracket should be -1 and as CO is a neutral ligand, the oxidation state of cobalt is -1. 64. (1) Chlorophylls are green pigments in plants and contain calcium. This is because chlorophyll contains magnesium and because oxidation potential of cerium(IV) is negative in aqueous solutions. 65. (3) When carbon is directly linked with an electropositive metal, it results in organometallic compound.

Chapter 20_Coordination Compounds.indd 525

NH3

NH3

NH3 act as Lewis base and Co3+ ion act as Lewis acid in this complex.

2. (3) Double salts are those which lose their entity in water while complex salts do not lose their entity in water.

Pair (1) represents double salt as both of them will lose + their entity and will ionize as Fe2+, SO24 , K , CN ions.



Pair (2) represents complex salts, that is, [Cu(NH3)4SO4] and [Fe(H2O)6]SO4



Pair (3) represents a complex salt and a double salt, that is, K2SO4×Al2(SO4)3×24H2O will ionize as 2K+ + 2Al3+ + 4 SO24



Pair (4) represents both double salts and will ionize as 2+ Mg2+, SO24 , Cu

4. (4) On adding BaCl2 in the first complex, BaSO4 (white precipitate) will be formed because SO4 will ionize to 2+ SO2ions. On adding AgNO3 in 4 and combine with Ba the second complex, precipitate of AgBr will be formed as Br will ionize as Br- and combine with Ag+ ions.

×× ××

sp3d2 hybridization Six electron pairs donated by six NH3 molecules

NH3 Co

57. (2) K2[NiF6] has Ni = +4 (oxidation state)

3+

NH3 H3N

H3N

54. (3) Octahedral complex has either d2sp3 or sp3d2 hybridization.



525

Similarly, on adding PbCl2 to the first complex, precipitate of PbSO4 will be formed.

5. (3) Denticity is the number of donor groups from a given ligand attached to the same central atom. The denticity of compound (1) = 2; (2) = 3; (3) = 4; (4) = 2. 6. (3)

Option (1) and (2): Butane-1,2-diamine (bn) and Propane-1,3-diamine (tn) are bidentate.



Option (3) Diethylamine (dien) is tridentate.



Option (4) Triethyltetraamine (trien) is tetradentate.

7. (2) h7 – C7H7 is the tropylium cation, which is a cationic ligand and is resonance stabilized as follows:

+ +

+

p -electron cloud is being donated here

1/4/2018 5:23:06 PM

526

OBJECTIVE CHEMISTRY FOR NEET

8. (3) We first find the oxidation number and the coordination number of central atom to determine the EAN.

H3N

In [Fe(CO)5], Fe has zero oxidation number and its coordination number is 5.

H3N

EAN = Atomic number - Oxidation state + 2 × Coordination number = 26 - 0 + 2 × 5 = 36 2

-

24

9. (2) Imido NH , nitrito ONO , sulphido SO . 10. (2) Ligand should be capable of donating lone pair of electrons only then a coordinate bond will be formed. It is not necessary for the ligand to be small in size because large ligands such as EDTA do form stable coordination compounds. Presence of positive and negative charges is also not the criteria for a molecule to behave as ligand, since neutral ligands such as NH3 also form stable coordination compounds.

en

Co

Co en

en

NH3 NH3

Mirror

23. (3) Both optical and geometrical isomerism are included in stereoisomerism because in both the isomerisms, the relative arrangement of atoms in space are differently oriented. For example, cis and trans. 25. (3) For linkage isomerism, the ligand should be an ambidentate ligand. For example, NO-2 and for hydrate isomerism, H2O molecules should be present as ligands. 26. (1) The order is [MnCl4]2- > [CoCl4]2- > [Fe(CN)6]4

Mn: 3d54s2

11. (2) Because it has two electron pairs. 12. (1) Ferrocene is bis(h5-cyclopentadienyl) iron

en

[MnCl 4 ]2- : ↑ ↑ ↑ ↑ ↑

Hence, maximum magnetic moment.



Co: 3d74s2 [CoCl 4 ]2- : ↑↓ ↑↓ ↑ ↑ ↑

Fe



Fe: 3d64s2 [Fe(CN)6 ]4- : ↑↓ ↑↓ ↑↓

14. (2) According to the rules, dibromido is Br2, bis(ethylenediamine) is CH2NH2CH2NH2 or en and bromide is outside the square bracket. Let oxidation state on Cr be x, then we have



Pairing takes place because CN– is a strong field ligand.

27. (4) Cr(III): Configuration is d3 ↑ ↑ ↑ 3 unpaired electrons.

.

Thus, there are



Mn(III): Configuration is d4 ↑↓ ↑ ↑ 2 unpaired electrons.

.

Thus, there are

15. (2) Since CN is a negatively charged ligand, iron is named as ferrate. The oxidation state of iron is



Fe(III): Configuration is d5 ↑↓ ↑↓ ↑ unpaired electron.

.

Thus, there is 1

x - 6 = -3 ⇒ x = +3.



. Co(III): Configuration is d6 ↑↓ ↑↓ ↑↓ Thus, there are no unpaired electrons and is least paramagnetic.

x + 0( 2) + ( -1)2 = +1 ⇒ x = +3

17. (2) Fe(C5H5)2 is bis(cyclopentadienyl)iron(II) as C5H5 is −

. 20. (2) Complex compound which do not have plane of symmetry or center of symmetry shows optical isomerism. Complex having general formula [M(AA)2X2] does not possess point of symmetry hence they show optical isomerism. It is an octahedral complex of the type [M(AA)2X2]

Chapter 20_Coordination Compounds.indd 526

31. (3) EDTA is a hexadentate ligand and in an octahedral complex the coordination number is 6. Therefore, only one EDTA molecule is required to make an octahedral complex. 32. (2) The electronic configuration of Ni(28) is 3s2, 3p6, 3d8, 4s2 and Ni2+ is 3s2, 3p6, 3d8. For a paramagnetic complex no pairing of 3d electrons should occur. Therefore, hybridization involved is sp3, that is, tetrahedral shape.

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527

Coordination Compounds 3d

4s

36. (2) The electronic configuration for Ni2+ is [Ar] 3d8 and en is a strong field ligand.

4p

Ni (II)

eg

X− is a weak field ligand, so 3d

4s

4p

××

[NiX4]2−

××

××

××

sp3 hybridization Four electron pairs donated by four X− ions

33. (2) In [NiCl4]2−, the electronic configuration of Ni2+ is 3d8.Cl− ion is a weak field ligand, so it does not lead to the pairing of unpaired 3d electrons and undergoes sp3 hybridization. 3d

4s

4p

××

[NiCl4]2−

××

××

××

sp3 hybridization Four electron pairs donated by four Cl− ions



Due to the presence of two unpaired electrons, the magnetic moment is

t2g



For a d8 system, irrespective of the nature of the ligand, the number of electron(s) in eg and t2g set will remain the same. Hence, there are zero unpaired electrons in t2g○ orbitals. Automatically, the other options are incorrect.

37. (4) The d4 (low spin) ↑↓ ↑↓ and d6 (low spin) ↑↓ ↑↓ ↑↓ will have two vacant d orbitals required for d2sp3 configuration and will form an octahedral complex. The d8 (high spin) ↑↓ ↑↓ ↑↓ ↑ ↑ does not have vacant inner d orbitals but can use outer d orbitals to form octahedral complex. 4 0. (1) In [CoCl4]2−, [FeCl4]2− and [NiCl4]2−, the electronic configurations of Co2+ is 3d7, Fe2+ is 3d6, Ni2+ is 3d8. In all these cases, Cl− being a weak field ligand does not cause pairing, so hybridization is sp3 and the geometry of the complex is tetrahedral.

m = n(n + 2) = 2( 2 + 2) = 2.82 BM

3d

34. (3) Complex compound with unpaired electron in its d orbital may undergo d-d transition and absorb light of visible region.

[V(NH3)6]3+



The electronic configuration of V = 3d3 4s2



The electronic configuration of V3+ = 3d2 4s0



Here d – d transition is possible thus e– may absorb light of visible region.

D0

t2g

35. (4) In tetrahedral complexes, the ligands t2g (dxy, dyz and dxz) orbitals point closer to the direction of the approaching ligands than the eg ( dx 2 - y 2 and dz 2 ) orbitals. Thus the three t2g orbitals are of higher energy and two eg orbitals are of lower energy.

××

4p ××

××

××

sp3 hybridization Four electron pairs donated by four Cl− ions



eg

Chapter 20_Coordination Compounds.indd 527

[MCl4]2−

4s

All complexes of Pt are square planar irrespective of the fact whether the ligand is of weak or strong field.

41. (4) On the basis of spectrochemical CO>CN - >NH -3 >F series CO is found to be the strongest ligand among given ligands. It is a s donor as well as a p acceptor ligand. 42. (2) In K2[NiF6]: Ni4+- d6 , F–-weak field ligand

Due to +4 (high charge) on Ni, Δ0 > P thus all the electrons in d orbitals gets paired up and the compound becomes diamagnetic.



In K4[NiF6]: Ni2+ - d8 , F– - weak field ligand. This complex has 2 unpaired electrons thus is paramagnetic.

43. (2) (i) Do increase with increase of charge. (ii) Among II and III, ligands are same, charge of the cation same but Zeff of Co3+ is greater than that of Fe3+. (iii) In IV, CN - is a strong field ligand, causes higher value of Do.

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528

OBJECTIVE CHEMISTRY FOR NEET

44. (2) The electronic configuration of Fe is 3d6 4s2. CO being a strong field ligand causes pairing of electrons. 3d

4s xx

[Fe(CO)5]

3. (1) The arrangement of d electrons in the given octahedral complexes is shown below:

4p

xx

xx

xx

dsp3 hybridization Five electron pairs donated by four CO molecules



Trigonal bipyramidal structure arises due to (i) the donation of lone pair of electrons of carbon of CO into empty orbital of metal atom forms a s bond. (ii) p orbital overlap involving donation of electrons from filled metal d-orbitals into vacant antibonding p molecular orbitals of CO forms a p bond. This is also called back bonding.

45. (4) The carbonyl ligands have vacant p orbitals in addition to lone pairs. These vacant orbitals accept electron density from filled metal orbitals to form a type of p bond that strengthens the s bonding arising from lone pair donation. Since the electrons fill the antibonding orbitals of CO, the C  O bond is weakened.

Previous Years’ NEET Questions 1. (4) If a molecule is asymmetric, it cannot be superimposed on its mirror image. The molecule and its mirror image have the type of symmetry as shown by the left and right hands and are called an enantiomorphic pair. The two forms are optical isomers. They are called either dextro or laevo (often shortened to d or l), depending on the direction in which they rotate the plane of polarized light in a polarimeter. [Co(en)2Cl2]Cl exhibits enantiomorphs. Their structures are

N

N

Cl

Cl

Co N

N

N

Cl

N

↑

[V(gly)2(OH)2(NH3)2]+(+1, d2)

↑ ↑

[Fe(en)(bpy)(NH3)2]2+(+2, d6)

↑↓ ↑↓ ↑↓

[Co(ox)2(OH)2]-(+5, d4)

↑ ↑ ↑

eg

↑

4. (4) The difference in energy between the two d levels is known as crystal field splitting energy (∆o). It depends upon the nature of the ligand, the charge on the metal ion and whether the metal is in the first, second or third row of transition elements. As all the given complexes contain the same metal ion, that is, Co3+, therefore, ∆o would depend upon the ligands which are attached. The crystal field splitting energy produced by CN− is highest among the given ligands. Hence, the order is [Co(C 2O4 )3 ]3- < [Co(H 2O)6 ]3+ < [Co(NH 3 )6 ]3+ < [Co(CN)6 ]3-. 5. (4) The electronic configuration of the metal ions in the given complexes is Zn2+: d10; Sc3+: d0; Ti4+: d0; Cr3+: d3. In the complexes, [ Zn(NH 3 )6 ]2+ , [Sc(H 2O)3(NH 3 )3 ]3+ and [ Ti(en)2(NH 3 )2 ]4+ it is not possible to promote electrons within d level as Zn2+ has a d10 configuration and the d level is full. Sc3+ and Ti4+ has d0 configuration and the d level is empty; hence d-d spectra are impossible and they should be rather colorless. Cr has three unpaired electrons. The d-d transition absorbs light of wavelength that falls in the visible region. 6. (3) The complex [Co(NH 3 )3Cl 3 ]0 has plane of symmetry hence it cannot show optical isomerism.

N

Plane of symmetry

Cl Cl N Cl

In [Cr(H2O)6]2+, Cr3+(3d4) has four unpaired electrons.



In [Mn(H2O)6]2+, Mn2+(3d5) has five unpaired electrons.



In [Fe(H2O)6]2+, Fe2+(3d6) has four unpaired electrons.



In [Ni(H2O)6]2+, Ni2+(3d8) has two unpaired electrons. Since Ni2+ has the least number of unpaired electrons, therefore, it is least paramagnetic.

NH3 NH3

2. (4)

NH3 Co

 

Chapter 20_Coordination Compounds.indd 528

t2g

[Ti(NH3)6]3+(+3, d1)

Co Cl

Arrangement of electrons

Complex

xx

7. (3) The electronic arrangement for high spin d4 octahedral complex is





t 2g

eg

↑ ↑ ↑

↑

 

CFSE = (3 × −4∆o) + (0.6∆o) = −0.6∆o

1/4/2018 5:23:22 PM

Coordination Compounds 8. (3) The geometrical isomers of complex [Co(NH 3 )4Cl 2 ]+ are as follows: Cl Cl

NH3 Co NH3

+ NH3 NH3

H3N H3N

Cis

+

Cl Co Cl Trans

NH3 NH3

 

9. (2) The presence of a strong field ligand CN− in [Ni(CN)4]2− causes pairing of electrons in Ni2+. Hence, there are no unpaired electrons to absorb light in the visible range. 10. (2) [Ni(NH3)2Cl2] has tetrahedral geometry and thus does not exhibit isomerism due to presence of symmetry elements. 11. (3) Magnetic moment depends upon the number of unpaired electrons.

15. (2) When CO binds to a metal center, a s  bond is formed via overlap between a filled non-bonding orbital containing the lone pair of electrons on CO and empty s-type orbital on the metal. The metal responds by pushing electron density away from filled d orbital to overlapping p * orbital on CO (known as back bonding). Overall this kind of bonding is known as synergic, because as more electron density is donated to the metal via s  bond, more can be back donated via the p. Both strengthen the metal-­carbon bond. Since, [Mn(CO)6 ]+ has lack of electrons on the metal atom, therefore, the M   CO bond becomes weak, consequently, C   O bond is strongest. 16. (4) Complexes containing unpaired electrons exhibit paramagnetic behavior.

ms = n(n + 2)  

Ions

Electronic configuration

Cr2+

d4

↑ ↑ ↑ ↑

Mn2+

d5

↑ ↑ ↑ ↑ ↑  

Fe

d

↑↓ ↑ ↑ ↑ ↑

Co2+

d7

↑↓ ↑↓ ↑ ↑ ↑



6

Option (1): In [Zn(NH3)6]2+, the arrangement of electrons for Zn2+ (d10) is  

The number of unpaired electrons in the given cations is as follows:

2+

529

No. of unpaired electrons



Since, it doesn’t contain any unpaired electrons, therefore, is not paramagnetic.



Option (2): In [Ti(NH3)6]3+, the arrangement of electrons for Ti3+ (d1) is

 

 

It contains one unpaired electrons, therefore is paramagnetic.



Option (3): In [Cr(NH3)6]3+, the arrangement of electrons for Cr3+ (d3) is

   

 

Since, complex Co2+ has the minimum number of unpaired electrons. Therefore, its magnetic moment is minimum.



12. (1) The complex is a square planar of the type [Mabcd]. It can have three possible geometrical isomers.

Since, it contains maximum unpaired electrons among the given complexes, therefore, exhibit highest paramagnetic behavior.



Option (4): In [Co(NH3)6]3+, due to the presence of strong field ligand NH3, the pairing of electrons takes place. The arrangement of electrons for Cr3+ (d6) is

Cl

Py

Cl

Br

Cl

Py

Br

NH3 NH3

Py

NH3

Br

   

13. (2) In the complex [Ni(CN )4 ]2- , the number of unpaired electrons is 0; therefore, it is diamagnetic in nature. 14. (3) When both the positive and negative ions are complex ions, isomerism may be caused by the interchange of ligands between the anion and the cation. Therefore, the complex [Co(NH3)6] [Cr(CN)6] and [Cr(NH3)6] [Co(CN)6] exhibit coordination isomerism.

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Due to lack of any unpaired electron, the complex is diamagnetic.

17. (3) In [Ni(NH3)6]2+, the oxidation state of Ni is +2 and electronic configuration is 3d8. NH3 causes pairing of electrons leaving only one 3d orbital empty. Thus, it cannot undergo d2sp3 hybridization. Therefore, Ni(II) undergoes sp3d2 hybridization instead of d2sp3 and forms an outer orbital complex. Since there are two unpaired electrons, the complex is paramagnetic.

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530

OBJECTIVE CHEMISTRY FOR NEET 3d

4s

4p

4d

Ni(II) (sp3d2 hybridized) sp3d2 3d

4s

4p

××

[Ni(NH3)6]2+

××

××

4d ××

××

××

Six electron pairs donated by six NH3 ligands

18. (4) When a low spin complex is formed (pairing takes place), the d-orbital occupation is t 26g eg0 . Thus, the crystal field stabilization energy = −0.4 × 6 Δo + 3P = −(12/5) Δo + 3P.

(Δo = Crystal field splitting energy in an octahedral field, P = Electron pairing energy)

19. (2) The reaction involved is

[Cr(H 2O)4 Cl 2 ]Cl + AgNO3 → [Cr(H 2O)4 Cl 2 ]NO3 + AgCl ↓ n( Complex solution ) = M × V = 0.01 × (100 /1000) = 0.001 mol



23. (4) The electronic configuration of Co3+ is 3d 64s0. CN− is a strong field ligand due to which pairing of electrons takes place that leads to the formation of low spin configuration. 24. (1) For the complex of type [M(AA)2a2], the total number of stereoisomers are three, the number of enantiomer pairs are one and geometric isomers are two. 25. (2) The name of the complex ion is hexacyanidoferrate(III) ion. 26. (3) In [Ni(CN)4]2−, oxidation state of Ni is +2.

3d

The complex [Cr(H2O)4Cl2]Cl has one ionizable Cl−, therefore, 0.001 mol will precipitate the same amount of AgCl.

20. (2) The crystal field stabilizations energy (CFSE) can be calculated as Complex

Therefore, the electronic configuration of Ni2+ = [Ar]183d84s0



4s

4p

Since, CN− is a strong field ligand due to which all the unpaired electrons are paired up. 3d

4s

4p

Electronic Crystal field stabilizations configuration energy (CFSE)

4 3 1 [Mn(H 2O)6 ]3+   3d (t2g  eg )

(3 × −0.4 ∆o) + (0.6 ∆o) = −0.6 ∆o

5 3 2 [Fe(H 2O)6 ]3+   3d (t2g  eg )

(3 × −0.4 ∆o) + (2 × 0.6 ∆o) = 0 ∆o

7 5 2 [Co(H 2O)6 ]2+   3d (t2g  eg )

(5 × −0.4 ∆o) + (2 × 0.6 ∆o) = −0.8 ∆o

6 4 2 [Co(H 2O)6 ]3+   3d (t2g  eg )

(4 × −0.4 ∆o) + (2 × 0.6∆o) = −0.4 ∆o

21. (2) Cis-platin is one of the most widely used anticancer agents. The structure is Cl

NH3 Pt



Cl

NH3

22. (4) [Co(NH3)3Cl3] does not give positive test with silver nitrate solution at 25°C.

Chapter 20_Coordination Compounds.indd 530

dsp2

27. (3) The metal carbon bond in metal carbonyls possesses both s and p character. On donation of a pair of electrons from filled metal orbitals (dp) to vacant antibonding orbital (pp) of CO, a p bond is formed between metal and carbon.

More is the negative charge on the metal atom, more easily the transfer of electron from the metal to p* orbital of CO would occur. Thus, as M   C bond order increases, C   O bond order decreases.

28. (1) Trans effect is the property of a ligand to direct substitution at the position trans to itself in substitution reactions in square planar complexes. For example, in the following complex, the substitution of ligand X by nucleophile Y is directed by ligand T at the position trans to it.

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Coordination Compounds X

L M T



Y

L

Y −X



M T

L

L

30. (1) Compounds that contain carbon–metal bonds are called organometallic compounds. This bond can be either a direct carbon to metal bond (s bond or sigma bond) or a metal complex bond (p bond or pi bond). Among the given compounds, Grignard reagent is an example of sigma bonded organometallic compound. 31. (2) Mn atom has the outer electronic structure 3d54s2. Thus a Mn3+ ion will have the structure 3d4, and the electrons will be arranged as: 3d

4s

4p

CN– is a strong field ligand; therefore, pairing of electrons takes place which leads to d2sp3 hybridization with octahedral shape. [Mn(CN)6]3−:

This effect arises due to kinetic (stabilization of transition state) and thermodynamic (weakening of M   L bond at the trans position) factors. The order of trans effect: CN - > C6 H5 - > Br - > NH 3

29. (1) According to Jahn–Teller effect, molecules or complexes of any shape (other than linear), which have unequally filled set of orbitals (either t2g or eg) will be distorted. The asymmetrical electronic arrangements are observed in d4 (high spin), d7 (low spin) and d9 (high and low spin) complexes. The d8 electronic configuration is symmetrical, hence no distortion is observed.

531

××

××

××

××

××

××

d2sp3 hybridization (Octahedral shape)

32. (2) The reactions involved are as follows: [Co(NH 3 )6 ]Cl 3 + 3 AgNO3 → [Co(NH 3 )6 ](NO 3 )3 + 3 AgCl ↓ [Co(NH 3 )5 ]Cl 2 + 2 AgNO3 → [Co(NH 3 )5 Cl](NO3 )2 + 2 AgCl ↓ [Co(NH 3 )4 Cl 2 ]Cl + AgNO3 → [Co((NH 3 )4 Cl 2 ](NO3 ) + AgCl ↓ 33. (4) Energy absorbed is inversely proportional to the 1 absorbed wavelength, that is, E ∝ . l

Presence of strong field ligand increases the energy gap Δo between the t2g and eg levels, therefore, decreases the wavelength of light absorbed. The order of ligand field strength is H2O < NH3 < en. Hence, increasing order for the wavelength of absorption in the visible region is [Co(en)3 ]3+ < [Co(NH 3 )6 ]3+ < [Co(H 2O)6 ]3+ .

4d

Mn : 3+

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21

Environmental Chemistry

Chapter at a Glance 1. Environment The term environment generally refers to natural surroundings of an organism and how the physical, chemical and biological factors influence its survival through various natural processes. The earth’s environment is constituted by the following. (a) Atmosphere: It is a gaseous mixture of air that surrounds the earth. The different layers of atmosphere are: (i) Troposphere: It is the lowest region of atmosphere extending from the earth’s surface to the lower boundary of stratosphere. It contains water vapors and is greatly affected by air pollution. (ii) Stratosphere: The layer of the earth’s atmosphere above troposphere and below mesosphere is called stratosphere. Ozone layer is present in this region. (iii) Mesosphere: It is the region of the earth’s atmosphere above stratosphere and below the thermosphere. It is the coldest region with temperature between –2 to 92°C. (iv) Thermosphere: The upper region of atmosphere above the mesosphere is called thermosphere. It is the hottest region with temperature up to 2000°C. (v) Exosphere: It is the upper most region of atmosphere. It contains atomic and ionic O2, H2 and He. 2. Environmental Pollution It may be described as the degradation of the natural environment with harmful contaminants (pollutants) into the air, water or soil mainly arising from by natural or human activities. (a) Pollution can be of two types: (i) Natural pollution: This is caused by natural sources, such as volcanic eruptions, forest fires, etc. (ii) Man-made pollution: These results from human activity, such as deforestation, use of pesticides, industrial effluents, etc. (b) Pollutant: Any substances produced either by natural sources or human activity which causes adverse effect on environment is called a pollutant. Environmental pollutants can be classified on the following basis. (i) On the basis of degradation: • Biodegradable: These are capable of being destroyed by biological or microbial action. For example, domestic waste. • Non-biodegradable: These are not normally acted upon by microbes. These undergo biological magnification. These are further divided into wastes (e.g., glass, plastic) and poisons (e.g., heavy metals, pesticides). (ii) On the basis of occurrence in nature • Primary pollutants: Present in the same form, as added by man, for example, DDT, pesticides. • Secondary pollutants: Formed by the reaction of primary pollutants with sunlight, for example, HNO3, H2SO4, PAN. 3. Tropospheric Pollution The release of contaminants in the air affects mainly the troposphere (layer closest to the earth) and stratosphere. Tropospheric pollution is caused by gaseous pollutants and particulate matter.

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OBJECTIVE CHEMISTRY FOR NEET

(a) Air pollution: The release of undesirable materials into the atmosphere either by natural phenomena or humanactivity that adversely affect the quality of air causes atmospheric pollution. It occurs when the concentration of normal components of air or a new chemical substance released or formed in air, builds up to undesirable levels/ proportions. The major air pollutants are gases such as oxides of carbon, sulphur, nitrogen, hydrocarbons and particulate pollutants like, smoke, dust, fumes and smog. (i) The types of air pollutants, their sources of generation and harmful effects caused by them are as follows: Pollutant

Types

Generation

Effects

Oxides of carbon

Carbon monoxide Released by incomplete (CO) combustion of fossil and hydrocarbon fuels

Toxic and lowers the dissolution of oxygen in blood by forming carboxyhaemoglobin which is more stable than oxygen-haemoglobin complex. It causes drowsiness, affects vision and causes cardiovascular diseases.

Carbon dioxide (CO2)

Released by burning of fossil fuels, vehicular emission and exhaust from industrial flue gases

Enhanced concentration can lead to greenhouse effect causing global warming. Hinders respiratory process.

Oxides of sulphur

Suphur dioxide Released by volcanic (SO2) and sulphur eruptions, burning of fossil trioxide (SO3) fuels, vehicular emissions and industrial processes

Causes acid rain and smog. SO2 is harmful (even at 5 ppm) for plants, animals and causes diseases of lungs in humans. SO3 is an irritant and harmful even at 1 ppm, causing discomfort to elderly and falling-off of flowers. Taj Mahal is reported to be affected by SO2.

Oxides of nitrogen

Primarily include NO, NO2 and N2O

Formed in the atmosphere or released by burning of fuels, production of explosives, fertilizers and nitric acid

Cause formation of smog and acid rain. NO2 causes chlorosis in plants and chronic lung condition. Both gases are corrosive to metals and harmful to vegetation.

Hydrocarbons

Incomplete combustion of fuels in automobiles and evaporation of fuels

Carcinogenic at levels between 500 to 1000 ppm.

Chlorofluorocarbons (CFCs)

Refrigerants

Smoke

Incomplete combustion of fossil fuels, garbage, dry leaves, and from cement, iron and steel works. (ii) Smog: It is a mixture of smoke (compound of tiny particles of carbon, ash and air, etc.) and fog in suspended droplet form. Smog is of two types: classical (London) smog and photochemical smog.

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Environmental Chemistry

Classical (London) smog

Photochemical smog

Formed when fog produced in chilly and damp areas gets mixed with SO2 and SO3 formed by burning of coal.

Contains oxidized hydrocarbons, ozone and NOx.

Has sulphuric acid as aerosol.

Occurs in warm, dry and sunny climate in areas with dense populations. Brown in color. Oxidizing in nature.

Causes bronchial irritation and acid rain. Reducing in nature. Occurs in cold climates.

535

The photochemical smog gets its name as it is formed by following photochemical reactions:   NO + hv → NO + O  O + O2 → O3 RH + O → RO ⋅



RO ⋅ + O2 → RO⋅3

RO×3 + NO ® RO×2 + NO2 ® Peroxy acetyl nitrate (PAN) (b) Greenhouse effect and Global warming: Certain gases called greenhouse gases (carbon dioxide, methane, ozone, chlorofluorocarbon compounds and water vapor) in atmosphere absorb the heat given out by the earth and radiate it back to its surface; thus warming it. With increase in emission of greenhouse gases, there has been a slow but continuous increase in the average temperature of air and water sources near the surface of the earth and this phenomenon is known as global warming. The amount of warming which a greenhouse gas imparts to a square meter of earth is called radiative force. (c) Acid rain: The pH of normal rain water is 5.6 due to dissolution of CO2 from the atmosphere. H2 O + CO2  H2 CO3



H2 CO3  H+ + HCO3−  When pH decreases below 5, it is called acid rain. Thus usually occurs due to presence of gaseous oxides of nitrogen and sulphur which lead to the formation of corresponding acids. NO + O3 → NO2 + O2 NO2 + O2 → NO3 + O2 NO2 + NO3 → N 2 O5 N 2 O5 + H2 O → 2HNO3 1 Soot particles SO2 + O2 + H2 O → H2 SO4 2 Acid rain causes damage to marble, aquatic life and respiratory diseases.

4. Stratospheric Pollution (a) Ozone layer: Ozone present in the stratosphere forms a protective layer by filtering harmful ultraviolet rays from the sun. Ozone layer absorbs nearly all solar photons in the range 180 to 360 nm. The overall amount of ozone in the stratosphere is maintained by balance between photochemical production and recombination. O2 (g ) ¾hv ¾® O(g)+O(g) O(g)+O2 (g) ® O3 (g ) ¾® O2 (g) + O(g) O3 (g ) ¾hv O(g) + O3 (g ) ® 2O2 (g)

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OBJECTIVE CHEMISTRY FOR NEET

(b) Ozone depletion: When the chlorofluorocarbons (Freon’s) emitted on the earth reach the stratosphere, the following reactions take place on absorption of UV rays. One molecule of these chlorofluorocarbons has the same impact as 25000 molecules of carbon dioxide. The free chlorine radicals react with ozone, leading to its depletion. Thus, there is a steady decline in total volume of stratospheric ozone. CF2 Cl 2 (g ) hv → Cl ⋅ + ⋅ CClF2 (g) CFCl3 (g ) hv → Cl ⋅ + ⋅ CCl2 F(g) Cl ⋅ (g ) + O3 (g ) hv → ClO ⋅ (g) + O2 (g) ClO ⋅ (g) + O(g) → Cl ⋅ (g ) + O2 (g) The free Cl radical is regenerated and continues to react with ozone. CH4 and NO2 act as natural sinks for chlorine monoxide and chlorine free radicals and prevent depletion of ozone. ⋅ ClO(g ) + NO2 (g ) →⋅ ClONO2 (g) ⋅ Cl(g) + CH4 (g) →⋅ CH3 (g ) + HCl(g) (c) Ozone hole: This refers to the larger seasonal decrease in stratospheric ozone over the earth’s polar regions (Antarctica). Ozone depletion is prevented in winter months by polar clouds but occurs in spring/summer in presence of sunlight due to photolysis. 5. Water Pollution The contamination of water by foreign substances which would constitute a health hazard and make it unfit for all purposes is known as water pollution. The pollutants may be physical, biological or chemical substances. Water pollutants can be categorized on the basis of the source they are emitted from as: • Point sources: These are distinct and defined, like pipes from industrial and municipal wastes. These can be controlled by on-site treatment. • Non-point sources: These are diffused and intermittent, such as run-off water from streets or fields. These are difficult to monitor and control. (a) Major pollutants (i) Pathogens: Disease causing microorganisms that leads to waterborne diseases. (ii) Organic waste: This includes dead organic matter which is decayed by bacteria by using oxygen dissolved in water. (iii) Chemicals: These include inorganic insecticides like sodium chlorate and sodium arsenates which are toxic to mammals and pesticides like BHC, DDT, 2-4 D, etc. The other chemical pollutants are acids, oil discharge, organic chemicals, polychlorinated biphenyls, synthetic detergents, etc. Chemical pollutants enter the food chain and the concentration of a substance (like DDT) increases in living tissues as it moves through a food web (bio accumulation). (b) Sewage treatment: The following steps are involved in this: (i) Preliminary process: This involves removal of large suspended matter, such as, gravel, etc. (ii) Settling process: The residual water is allowed to stand in tanks and the oil and grease that floats on the top is removed. (iii) Secondary treatment: This involves aerobic chemical oxidation forming secondary sludge. (iv) Tertiary treatment: It involves treatment for removal of phosphates. (c) Aerobic and anaerobic oxidation: The oxidation of organic compounds present in the sewage in the presence of large amount of free or dissolved oxygen by aerobic bacteria is known as aerobic oxidation. As soon as we find that the oxygen level in any water form is less than certain value it is known as stale. The anaerobic bacteria bring about the purification by producing and releasing H2S, NH3, CH4, etc. The optimum value of dissolved oxygen must be 4–6 ppm. (d) Biological oxygen demand (BOD): It is defined as the amount of free oxygen required for oxidation of organic matter under aerobic condition at 20°C for 5 days. Its unit is ppm. Bacterial respiration and organic decomposition can use up dissolved oxygen, depriving fish and invertebrates of available oxygen in the water (eutrophication).

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Environmental Chemistry

537

(e) Chemical oxygen demand (COD): It is the actual measurement of all types of oxidizable impurities present in sewage. COD values are always higher than BOD. It is expressed in terms of ppm and is detected using potassium dichromate titration. (f ) Eutrophication: It is a process by which a body of water develops high concentration of nutrients (nitrogen and phosphorous) which causes an increase in growth of aquatic plants and blue-green algae on the surface of water. This reduces light availability and hence photosynthesis is leading to decay of plant life. The resulting increase in BOD affects other organisms like fish. (g) Control of water pollution (i) Recycling of waste. (ii) Use of chemicals to reduce lead poisoning. (iii) By using adsorption, ion exchange, reverse osmosis processes. (iv) By waste water reclamation. (h) International standards for drinking water (i) Fluorides: up to 1.5 ppm. (ii) Lead : up to 50 ppb. (iii) Metals: Fe: 0.2 ppm, Al: 0.2 ppm, Mn: 0.005 ppm, Zn: 5 ppm, Cu: 3 ppm and Cd: 0.005 ppm. (iv) Sulphates: up to 500 ppm are harmless. (v) Nitrates: up to 50 ppm. 6. Soil and Land Pollution Any substance which when added to the soil, alters the productivity of the soil is a pollutant and causes soil pollution. (a) Sources (i) Use of large amount of fertilizer, pesticides, etc. (ii) Farm wastes. (iii) Soil conditioners containing toxic metals like lead, mercury, arsenic, etc. (iv) Domestic refuge and industrial waste. (v) Radioactive substance waste disposal. (b) Control (i) Use of manures, must be encouraged. (ii) Use of biofertilizers. (iii) Developing proper sewage system. (iv) Recycling. 7. Strategies to Control Environmental Pollution (a) Waste management: Integrated waste management (IWM) includes reuse, source reduction, recycling, composting, landfill and incineration. The three Rs of IWM are reduce, reuse and recycle. (b) Collection and disposal: Inadequate waste management practices like poorly controlled open dumps and illegal roadside dumping can spoil scenic resources, pollute soil and water resources and produce health hazards. At the disposal site, garbage is sorted out and separated into biodegradable and non-biodegradable materials. Non-­biodegradable materials such as plastic, glass, metal scraps etc. are sent for recycling. Biodegradable wastes are deposited in landfills and are converted into compost. Composting is a biochemical process in which organic materials such as kitchen scraps decompose to a rich, soil-like material. The process involves rapid partial decomposition of moist solid organic waste by aerobic organisms. 8. Green Chemistry (a) G  reen chemistry is an area in which various processes are carried out in a way that there is no adverse impact on the environment. (b)  It promotes the use of techniques that avoid use of toxic solvents and reagents and harmful reaction conditions.

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OBJECTIVE CHEMISTRY FOR NEET

(i) Dry cleaning of clothes • A non-polar solvent, tetrachloroethylene, is used to dissolve non-polar compounds (dirt and grease). This solvent is inflammable and was considered an ideal choice for dry cleaning. However, it is found to contaminate ground water and is a potential carcinogen. • Liquefied carbon dioxide with a suitable detergent is now being used as an alternate dry cleaning agent to avoid contamination of ground water. (ii) Bleaching of paper • The use of chlorine gas in bleaching paper is being replaced by hydrogen peroxide in the presence of a catalyst. (iii) Green solvents • It involves replacement of volatile organic solvents with ionic liquids. • These are salts made up of complex ionic components, as a result of which they are liquid at room temperature. Being ionic in nature, the components are held together strongly by electrostatic forces, so unlike conventional solvents, they are not lost to the environment. (iv) Synthesis of chemicals • Chemicals are now being synthesized on a commercial scale using environment-friendly solvents, reagents and reactions conditions. For example, oxidation of ethene to ethanal is carried out in a single step using an ionic catalyst in aqueous medium. CH2   CH2 + O2 → CH3CHO

Solved Examples 1. Identify the wrong statements in the following:

(2) The problem of ozone depletion is more serious at poles because ice crystals in the clouds over poles act as catalyst for photochemical reactions involving the decomposition of ozone by Cl• and ClO• radicals. (3) Freons, chlorofluorocarbons, are inert chemically, they do not react with ozone in stratosphere. (4) Oxides of nitrogen also do not react with ozone in stratosphere.

(1) Chlorofluorocarbons are responsible for ozone layer depletion. (2) Greenhouse effect is responsible for global warming. (3) Ozone layer does not permit infrared radiation from the sun to reach the earth. (4) Acid rains are mostly because of oxides of nitrogen and sulphur. Solution

Solution

(3) Ozone layer does allow infrared radiation from sun to reach earth, whereas it does not allow to pass UV radiation.

(2)

2. The gas leaked from a storage tank of the Union Carbide plant in Bhopal gas tragedy was (1) methylamine. (3) phosgene.

(2) ammonia. (4) methyl isocynate.

Cl• + O3 ® ClO + O2 ClO + O → Cl + O2

Reaction without polar stratospheric clouds



CFCl3 + UV light ® Cl• + CFCl2 Cl• + O3 ® ClO + O2

Solution (4) The gas leaked from a storage tank of the Union Carbide plant in Bhopal gas tragedy was methyl isocyanate (C2H3NCO). 3. Which of the following statements about the depletion of ozone layer is correct? (1) The problem of ozone depletion is less serious at poles because NO2 solidifies and is not available for consuming ClO• radicals.

Chapter 21_Environmental Chemistry.indd 538

CFCl 3 + UV light → CFCl 2 + Cl

  ClO + NO2 ® ClONO2 (Reservoirs)

With polar stratospheric clouds ClONO2 + Clouds + HCl (Moisture)

Cl2

Cl

.

O3

.

O3

+ Cl

. ClO +

.

ClO

Cl2O2

O2

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Environmental Chemistry

The main reason for ozone depletion in polar region is that ice crystals catalyze ozone depletion reaction.

4. Global warming is due to increase of (1) (2) (3) (4)

methane and nitrous oxide in atmosphere. methane and CO2 in atmosphere. methane and O3 in atmosphere. methane and CO in atmosphere.

Solution (2) The increasing percentage of CO2 and CH4 in the atmosphere is the main cause of global warming. CO2 is most prevalent greenhouse gas and CH4 is second most prevalent greenhouse gas. 5. CO is more toxic than CO2 because (1) it affects the nervous system. (2) it damages lungs. (3) it reduces the oxygen carrying capacity of haemoglobin. (4) it forms acid with water. Solution (3) CO is more toxic than CO2 because it reduces the oxygen carrying capacity of haemoglobin. 6. Which of the following statement is true about ozone layer? (1) It is harmful because ozone is dangerous to living organism (2) It is beneficial because oxidation reaction can proceed faster in the presence of ozone (3) It is beneficial because ozone cuts-off the ultraviolet radiation of the sun (4) It is harmful because ozone cuts out the important radiations of the sun which are vital for photosynthesis.

(3) Ozone layer in the upper atmosphere absorbs UV radiations and is beneficial to human life. 7. About 20 km above the earth there is an ozone layer. Which one of the following statements about ozone and ozone layer is true? (1) Ozone layer is beneficial to us because ozone cuts out the ultraviolet radiation of the sun. (2) The conversion of ozone to oxygen is an endothermic reaction. (3) Ozone has a triatomic linear molecule. (4) Ozone layer is harmful to us because it blocks radiations that are useful for photosynthesis. Solution

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8. Taj Mahal is badly affected by acid rain. This is because (1) (2) (3) (4)

Taj Mahal is made of marble. acid rain attacks the steel structure of Taj Mahal. acid rain has a pH above 5.6. the paint on Taj Mahal is soluble in acids.

Solution (1) Taj Mahal is made of marble (CaCO3) which reacts with H2SO4 of the acid rain causing pitting, decolorization and lusterless. 9. Photochemical smog consists of excessive amount of X, in addition to aldehydes, ketones, peroxy acetyl nitrile (PAN), and so forth. X is (1) CH4 (3) CO2

(2) CO (4) O3

Solution (4) Photochemical smog consists of excessive amount of ozone (O3) in addition to nitrogen oxides (NOx), peroxyacetyl nitrate (PAN) and volatile organic compounds (VOCs) that are produced by reaction of sunlight. 10. The smog is essentially caused by the presence of (1) O2 and O3       (2) O3 and N2 (3) oxides of sulphur and nitrogen.   (4) NO2 and N2. Solution (3) Classical smog contains oxides of sulphur, while photochemical smog contains oxides of nitrogen. 11. Addition of phosphate fertilizers to water bodies causes (1) (2) (3) (4)

enhanced growth of algae. increase in amount of dissolved oxygen in water. deposition of calcium phosphate. increase in fish population.

Solution

Solution

(3) The structure of ozone is O structure.

539

O

O. It has a linear

(1) Addition of nutrients such as nitrogen and phosphorus (in the form of nitrates and phosphates) cause an increase in the growth of aquatic plants and production of blue green bacteria and algae. This process is known as eutrophication. 12. Eutrophication is caused by (1) (2) (3) (4)

excess growth of algae. increase in rate of photosynthesis. decrease in biological oxygen demand. all of these.

Solution (1) Excessive growth of algae indicates eutrophication in which there is an increase in the amount of nutrients present in the water especially nitrogen and phosphorus. These algae form surface mats and reduce the light availability to the algae below the surface which further

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540

OBJECTIVE CHEMISTRY FOR NEET reduces photosynthesis. Besides, the decomposition of dead algae also increases the biological oxygen demand and reduces the oxygen content of water.

13. The term “biomagnification” refers to (1) growth of organism due to food consumption. (2) increase in population size. (3) increase in the concentration of non-degradable pollutants as they pass through food chain. (4) decrease in population size. Solution (3) Biomagnification is the accumulation or increase in the concentration of a substance in living tissue as it moves through a food web (also known as bioaccumulation).



Eutrophication is carried out by presence of excess nitrogen and phosphorus in run-off waters from agricultural fields.

16. What is DDT among the following? (1) (2) (3) (4)

Greenhouse gas A fertilizer Biodegradable pollutant Non-biodegradable pollutant

Solution (4) DDT is a well-known insecticide. It is a non-biodegradable pollutant. DDT is not rapidly metabolized by animals, it affects the reproductive system of animals.

The structure of DDT is Cl

Cl

Cl

14. Which of the following is the best measure for pollution in water? (1) COD (2) Dissolved oxygen (3) BOD (4) None of these

Cl 1, 1, 1 - Trichloro-2,2-bis (4 - chlorophenyl)ethane

Solution (3) Dissolved oxygen depends upon various factors, such as temperature of the water, the amount of oxygen used up by organisms, and the amount replenished by photosynthesizing plants, aeration, etc. COD refers to chemical oxygen demand, and is less specific as it measures both organic and oxidizable inorganic material that can be chemically oxidized, rather than just levels of biologically active organic matter. Biochemical oxygen demand (BOD) is a more specific measure as it measures the amount of the dissolved oxygen that would be needed by the microorganisms to oxidize organic waste present in sewage water.

Cl

17. Eutrophication can be observed in (1) saline soil. (2) desert. (3) lake. (4) agricultural fields. Solution (3) Eutrophication can be observed in lake. The addition of phosphorus to water, in the form of the phosphate anion PO3− 4 , encourages the formation of algae, which reduces the dissolved oxygen concentration of water. This process is known as eutrophication. It is normally seen in lakes.

15. The industrial waste pipeline that is discharged into a water body is (1) a non-point source of pollution. (2) a point source of pollution. (3) called acid mine drainage. (4) responsible for eutrophication.

18. Domestic waste mostly constitutes

Solution

Solution

(2) Point sources of pollution are distinct and defined; their origin can be identified and corrective measures for control can be taken. Therefore, water being discharged through industrial pipeline is a point source that can be treated and the pollutants can be regulated before it is discharged into the water body.

Non-point sources of water pollution are diffused and intermittent. They cannot be identified easily, and hence cannot have specific control measures.



Acid mine drainage refers to water with high concentration of sulphuric acid that drains from mines, mostly coal mines.

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(1) non-biodegradable pollutants. (2) biodegradable pollutants. (3) effluents. (4) air pollutants.

(2) Domestic waste mostly constitutes biodegradable pollutants. 19. Modes of controlling pollution in large cities include (1) less use of insecticides. (2) proper disposal of organic wastes, sewage and industrial effluents. (3) shifting of factories out of the residential area. (4) all the above. Solution (4) All the three (1), (2) and (3) modes can prevent pollution in large cities.

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Practice Exercises Level I

10. PAN and ozone are

Atmospheric Pollution 1. The presence of undesirable substance in the lowest layer of atmosphere is (1) (2) (3) (4)

air pollution. tropospheric pollution. photochemical pollution. greenhouse pollution.

(1) Ferrocene (2) Fullerenes (3) Freons (4) Polyhalogens 3. The CO content in the atmosphere can be minimized by (1) Development of more efficient internal combustion engine. (2) Use of natural gas. (3) Use of liquefied natural gas. (4) All of these. 4. Which of the following is correct? CO is equally fatal as CO2 CO is more fatal than CO2 CO is less fatal than CO2 CO is not fatal at all.

5. Which of the following is a living component of atmosphere? (1) Lithosphere (2) Biosphere (3) Hydrosphere (4) Troposphere 6. Lung diseases are four times more in urban areas than rural areas. This is due to the presence of (1) SO2 (2) CO2 (3) N2 (4) water vapor. 7. Depletion of ozone layer allows UV rays to the earth surface that causes which of the following problems? (1) (2) (3 (4)

Skin cancer Deterioration of immune systems Retardation of plant growth All of these

8. Formula of PAN is (1) CH3CONO2 (3) CH3COOONO2

(2) CH3COONO2 (4) CH3OCOONO2

9. Pick out the correct match. (1) (2) (3) (4)

London smog : Brown in color Los Angeles smog : Brown in color Los Angeles smog : Greyish in color London smog : Blue in color

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primary pollutants. secondary pollutants. responsible for respiratory problems. particulate pollutants.

11. London smog consists of

2. Which of the following is responsible for depletion of the ozone layer in the upper strata of the atmosphere?

(1) (2) (3) (4)

(1) (2) (3) (4)

(1) Carbon soot arising from unburnt coal. (2) Sulphur dioxide arising from oxidation of sulphur present in coal. (3) Both (1) and (2). (4) Neither (1) nor (2). 12. Which layer of the atmosphere consists of ozone layer? (1) Troposphere (2) Stratosphere (3) Mesosphere (4) Exosphere 13. Which of the following is the result of global warming? (1) (2) (3) (4)

Change in rainfall patterns. Rise in sea land. Water evaporation form water bodies. All of these.

14. Choose the correct statement(s) among the following. (1) NO is more harmful than NO2. (2) Acid rain contains mainly HNO3. (3) CO2 is not responsible for greenhouse effect. (4) CO2 can absorb infrared radiations but does not allow them to pass through.

Water Pollution 15. What is the minimum value of BOD in polluted water? (1) 10 ppm (2) 10 ppb (3) 17 ppm (4) 17 ppb 16. Organic wastes are (1) (2) (3) (4)

non-biodegradable wastes. biodegradable wastes. toxic heavy metals. sediments.

17. Fluorosis, a bone disease, is caused by the presence of (1) (2) (3) (4)

pesticides in water. fluorides in water. carbon monoxide in air. sulphur dioxide in air.

18. The permissible limit of pesticides in drinking water is (1) 0.1 ppm. (2) 0 1 ppm. (3) 10 ppm. (4) 0.005 ppm. 19. Which of the following is not a chemical pollutant? (1) Cadmium (2) Mercury (3) Nickel (4) Caesium

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OBJECTIVE CHEMISTRY FOR NEET

20. The main source(s) of soil pollution is (are) (1) radioactive pollutants. (2) acid rain. (3) soil erosion. (4) All of these. 21. Most abundant water pollutant is (1) detergent. (3) pesticide.

(2) industrial waste. (4) oil spill.

22. Which of the following are common water pollutants? (1) (2) (3) (4)

Oxygen demanding wastes Fertilizers Sewage All of these

23. Water is often treated with chlorine to (1) (2) (3) (4)

increase oxygen content. remove insoluble impurities. cause sedimentation. kill germs.

Soil Pollution 24. Which of the following is not a soil pollutant? (1) Polythene bags (2) Pesticides (3) Detergents (4) Nitrate and phosphate fertilizers 25. Negative soil pollution is (1) reduction in soil productivity due to erosion and overuse. (2) reduction in soil productivity due to addition of pesticides and industrial wastes. (3) converting fertile land into barren land by dumping ash, sludge and garbage. (4) None of these. 26. Chief source of water and soil pollution is/are (1) mining. (2) thermal power plant. (3) agro-industry. (4) all of these.

Strategies to Control Environmental Pollution and Green Chemistry 27. Green chemistry involves (1) production of chemicals of our daily use from greenhouse gases. (2) such chemical processes in which green plants are used. (3) those reactions which are of biological origin. (4) use of non-toxic reagents and solvents to produce environment-friendly products. 28. Proper management of disposal of household and industrial wastes can be done by (1) recycling the waste material to give useful products again.

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(2) burning and incineration of combustible waste. (3) sewage treatment. (4) all of these. 29. The pollution in your city may be controlled to great extent, by which of the following methods? (1) By proper sewage and proper exit of chemicals from factories. (2) By wide roads and factories away from the city. (3) By cleaning city and scanty use of pesticides. (4) All of these. 30. Which of the following is not a part of green chemistry? (1) Photochemistry (2) Sonochemistry (3) Nuclear chemistry (4) Biochemistry

Previous Years’ NEET Questions 1. Green chemistry means such reactions which (1) study of the reactions in plants. (2) produce color during reactions. (3) reduce the use and production of hazardous chemicals. (4) are related to the depletion of ozone layer. (AIPMT 2008) 2. Which one of the following statements is not true? (1) Oxides of sulphur, nitrogen and carbon are the most widespread air pollutant. (2) pH of drinking water should be between 5.5–9.5. (3) Concentration of DO below 6 ppm is good for the growth of fish. (4) Clean water would have a BOD value of less than 5 ppm. (AIPMT PRE 2011) 3. Which one of the following statements regarding photochemical smog is not correct? (1) Carbon monoxide does not play any role in photochemical smog formation. (2) Photochemical smog is an oxidizing agent in character. (3) Photochemical smog is formed through photochemical reaction involving solar energy. (4) Photochemical smog does not cause irritation in eyes and throat. (AIPMT PRE 2012) 4. Which one of the following is not a common component of photochemical smog? (1) Ozone (2) Acrolein (3) Peroxyacetyl nitrate (4) Chloroflurocarbons (AIPMT 2014)

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543

Answer Key Level I  1. (2)

 2. (3)

 3. (4)

 4. (2)

 5. (2)

 6. (1)

 7.  (4)

 8. (3)

 9. (2)

10.  (2)

11.  (3)

12.  (2)

13.  (4)

14.  (4)

15.  (3)

16.  (2)

17.  (2)

18.  (4)

19.  (4)

20.  (4)

21.  (2)

22.  (4)

23.  (4)

24.  (3)

25.  (1)

26.  (4)

27.  (4)

28.  (4)

29.  (4)

30.  (3)

Previous Years’ NEET Questions  1. (3)

 2. (3)

 3. (4)

 4. (4)

Hints and Explanations Level I 1. (2) The presence of undesirable substances in the

22. (4) Oxygen demanding waste, synthetic organic compounds, inorganic chemicals and minerals, fertilizers and sewage are common water pollutants.

lowest layer of atmosphere is known as troposphere pollution.

23. (4) Chlorine is used to kill germs as it is strong oxidizing agent.

4. (2) CO is more fatal than CO2 as it combines with haemoglobin and reduces the oxygen transport efficiency.

30. (3) Green chemistry may be defined as the program of developing new chemical products and chemical processes or making improvements in the already existing compounds and processes so as to make them less harmful to human health and environment. It includes photochemistry (chemical reactions taking place in the presence of light), biochemistry (chemical processes in living organisms), and sonochemistry (effect of sonic waves and wave properties on chemical systems).

6. (1) This is due to the presence of SO2 gas. A concentration of 0.5 ppm of SO2 gas for full day or 0.2 ppm lasting for 3 days has been reported to damage the lungs and increase the rate of mortality. 10. (2) Ozone and PAN are secondary pollutants. 14. (4) NO is less toxic than NO2 and cannot enter the blood stream, whereas NO2 causes respiratory diseases and is fatal at high concentration (100 ppm).

Acid rain contains both HNO3 and H2SO4.

15. (3) 17 ppm is the maximum value of BOD for polluted water and clean water must have 5 ppm or below. 16. (2) Organic wastes are biodegradable wastes that pollute water as a result of run-off. 18. (4) The permissible limit of pesticides in water is 0.005 ppm. 19. (4) Chemical pollutants are water soluble chemicals like heavy metals such as Cd, Hg, Ni, etc. 20. (4) The main sources of soil pollution are industrial wastes, urban and domestic wastes, agrochemicals, soil erosion, radioactive pollution and acid rain.

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Previous Years’ NEET Questions 1. (3) Green chemistry is used in the environmental field for methods that prevent the control of hazardous effect of chemicals. 2. (3) The consumption of dissolved oxygen (DO) in water is by respiration of plants and by oxidation of organic matter by microorganisms. When the BOD is high, the DO content of the water may become too low to support life in water. The aquatic fauna is affected when DO falls below 6 ppm. 3. (4) Photochemical smog causes irritation in eyes and throat. 4. (4) Chlorofluorocarbons are responsible for depletion of ozone layer rest all of them are components of photochemical smog.

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ORGANIC CHEMISTRY

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22

Organic Chemistry – Some Basic Principles and Techniques

Chapter at a Glance General Concepts of Organic Chemistry 1. Introduction (a) O  rganic chemistry is branch of chemistry dealing with carbon and its compounds, with the exception of CO, CO2, HCO3 -, CO3 2- and inorganic cyanides. (b) Carbon forms covalent bonds with other carbon atoms and produces straight or branched chain molecules. It may also share two or three electrons with another carbon atom as in ethylene (H2C   CH2) or acetylene (CH CH) and gives rise to a double or triple bonds, respectively. 2. Hybridization (a) T  he hybridization in alkanes, (e.g., ethane) is sp3 (tetrahedral), the hybridization in alkenes (e.g., ethene) is sp2 (trigonal) and the hybridization in alkynes, (e.g., ethyne) is sp (linear). (b) The C   C, C   C, and C C bond lengths are 1.54 Å, 1.34 Å and 1.20 Å, respectively. Hybridization and bond angles are; sp3, 109° 28′; sp2, 120° and sp, 180°, respectively. Percentage s-character is 25% in sp3, 33.3% in sp2 and 50% in sp orbitals. (c) A single bond between carbon and carbon (as in CH4, C2H6) is always a sigma bond. A double bond is composed of one sigma and one pi (π) bond (as in C2H4). A triple bond between carbon-carbon comprises of single sigma and two pi bonds. The shorter is the bond, greater is its strength. Thus, bond formed by sp hybridization as in acetylene is the strongest, because it has maximum bond energy. The bond formed by sp3 hybridized carbon as in methane is the weakest, as it has minimum bond energy. (d)  Bond energy is directly proportional to the bond strength. The greater the bond energy, greater is the strength of the bond. 3. Structural Representation of Organic Compounds The structural formulas can be represented by Lewis dot structure, dash structure, condensed structure or bond-line structure. H H H H C C C H H O H H Dot structure

H

H

H

H

C

C

C

H H H Propane

H

H

H

H

H

C

C

C

H

H

Propene

Dash structure CH3CHCH3

CH3CH(OH)CH3

OH Condensed structure

CH3CHClCH2CH3 Cl Bond line structure

4. Classification of Organic Compounds (a) Organic compounds can be classified based on open chain and cyclic structures as follows.

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Organic compounds

Acyclic (Open-chain or aliphatic)

Cyclic (ciosed-chain or ring compound)

Carboxylic or homocyclic

H2C

CH2 C H2

Cyclopropane

Alicyclic

Aromatic

CH2

H2C H2C

Heterocyclic

Nonaromatic

CH2

C H2 Cyclopentane

Aromatic 4 5 3 4 6

2 5

N

N H Pyrrole

Pyridine

3 4

3 4 2 5

O

2

Furan

3

5

S

2

Thiophene

(b) A carbon atom can be classified on the basis of number of carbon atoms to which it is attached. Organic compounds Primary 1°

Secondary 2°

Quartenary 4°

Tertiary 3°

Carbon is directly Carbon is directly Carbon is directly attached to one attached to two attached to three carbon atom carbon atoms carbon atoms H

C C 1°

C H 1°

C

C

2°

C

C

C C

C 3°

C C

C C 4° C

5. Nomenclature of Organic Compounds (a)  The systematic and internationally accepted method of naming organic compounds is IUPAC (International Union of Pure and Applied Chemistry) system. However, common or trivial names are also used along with abbreviations. (b)  The name of an organic compound consists of five parts: (i) Prefix: Shows number of carbon atoms in the carbon chain. Root word 1 2 3 4 5 6 7 8 9

Number of carbon atoms methethpropbutpenthexheptoctnon-

Root word 10 11 12 13 14 18 19 20

Number of carbon atoms decundecdodectridectetradecoctadec nonadeceicos-

    Prefixes are of two types: • Primary: Bi, tri, tetra, etc. • Secondary:    X: halo (X = F, Cl, Br, I);    N2: diazo;    N   O: nitroso;    NO2: nitro; SO3H: sulpho

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(ii) Suffix: These are of two types: • Primary suffix (infix): Shows the nature of carbon-carbon bonds present in the parent chain. Primary suffix -an -en-yn

Nature of carbon–carbon bonds in the parent chain All single bonds One or more double bonds One or more triple bonds

• Secondary suffix: Shows the class of the compound Secondary suffix -e -ol -al -one -oic acid -amine -nitrile Sulphonic acid

Class of compounds Hydrocarbon Alcohol Aldehyde Ketone Carboxylic acid Amine Cyanide Sulphanic acid

(c) The name must have Secondary prefix + Primary prefix + Root + Primary suffix + Secondary suffix (d) IUPAC rules for naming organic compounds containing one or more functional groups are as follows. (i) Select the longest continuous carbon chain in the molecule. (ii)  The numbering of the carbon chain is done in a manner such that the carbon bearing the functional group gets the lowest number. (iii)  If two or more same functional groups are present, then they are named using prefixes di, tri, tetra, etc. (iv)  In compounds containing two or more different functional groups (polyfunctional compounds), one group is regarded as the principal functional group and the other is treated as a substituent.      The reactivity series for selection of principal functional group is as follows: —COOH > SO3H > —COOR > —COCl > —CONH2 > CN > NC > CHO > C   O > OH > —NH2 > —OR > Alkene > Alkyne > NO2 > Alkyl (v)  If functional group, multiple bonds and side chains are all present in the organic molecule, then the order of preference is Functional group > Double bond > Triple bond > Side chain. (e) Naming of saturated hydrocarbons (Alkanes): In the word root we add the word ane. For example, if the carbon chain contains six carbons, it is hex + ane = hexane. (i) There are a few common names for branched chain alkanes. CH3 CH3

CH

CH2

CH3

CH

CH3

CH2

CH3

CH3

Isopropyl

Isobutyl

CH CH3

Sec-butyl

CH3

C CH3

Tertiary butyl

(ii) In case of cyclic hydrocarbons, we prefix the word cyclo.

Cyclohexane

Cyclobutane

(f ) Naming of unsaturated hydrocarbons (Alkenes/Alkynes) (i)  For naming alkenes, the root word is selected and word ene added. For example,

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CH2

CH3 CH CH2 Prop-1-ene

CH CH CH2 Buta-1,3-diene

(ii)  For naming alkynes, the root is selected and the word yne added to it. For example, CH3

CH2 C But-1-yne

CH

(iii) If double and triple bond both are present together, then there are two possibilities: • If both are terminal, we prefer the alkene, for example, 1

2

CH2

3

CH

4

C

CH

But-1-en-3-yne

• If triple bond is terminal and double bond is non-terminal, we prefer triple bond, for example: 1

2

3

C

CH

4

5

CH2 CH CH Hex-4-en-1-yne

6

CH3

(g) Naming of alkyl halides: Halogen is treated as a substituent of the alkane chain. For example, 3

2

1

1

CH3CH2CH2Cl 1-Chloropropane

CHCl3 Trichloromethane

(h) Naming of alcohols: These are named after parent alkanes by replacing suffix ane with ol. For example, 4

3

2

1

CH3CH2CH(OH)CH3 Butan-2-ol

(i) Naming of ethers: The smallest alkane chain is treated as alkoxy substituent. For example, CH3 O CH2CH2CH3 Methoxy propane

(j) Naming of aldehydes and ketones: In aldehydes and ketones, the suffix ane is replaced with al and one, respectively. For example CH3CH2CHO Propanal

CH3CH2COCH3 Butan-2-one

  In case both aldehyde and ketone are present in a molecule, preference is given to aldehydes and ketone is treated as a prefix keto or oxo. For example, 4

3

CH3

C

2

1

CHO

CH2

O 3-Oxobutanal

(k) Naming of carboxylic acids: These are named by replacing ane with oic acid. For example, 1 COOH 4

3

2

1

CH3CH2CH2COOH

2 COOH

Butanoic acid

Ethan-1,2-dioic acid (oxalic acid)

(l) Naming of esters: These are named by taking the right side chain as parent carbon chain and in the left hand chain, ane is replaced by oate. For example, O CH3 CH2COOCH3 Methyl propanoate

C

O

CH2

CH3

Ethyl benzoate

(m) Naming of amines: In primary amine, the e in the parent alkane chain is replaced with amine. For example, CH3CH2NH2 Ethanamine

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In secondary and tertiary amines, the smallest chain is considered as substituent. For example, CH3 CH3 NH C2H5 N-Methyl ethanamine

CH3

N

CH2

CH3

N,N-Dimethyl ethanamine

6. Dipole Moment (a) If electronegativity differences exist between two bonded atoms, and they are not large, the electrons are not shared equally and a polar covalent bond is the result. (b) Dipole moment is given by µ = e × d 7. Isomerism It is the phenomenon in which two or more than two organic compounds have the same general molecular formula but different structural or spatial arrangement of atoms. These may or may not have the same physical and chemical properties. Isomerism is mainly classified into two types: Isomerism

Structural or constitutional

Stereoisomerism

(a) Structural or constitutional isomers: These are compounds with same molecular formula but different structural arrangement of atoms. These are further classified as: Structural isomerism

Chain

Position

Functional

Metamerism

(i) Chain isomers: These are compounds with different chain lengths. For example, C6H14 can be: CH3 CH3

CH2

CH2

CH2

1

CH3

2

3

CH

CH3

Hexane

5

4

CH2

1

CH3

CH2

CH3

Cl

2

C

3

CH2

4

CH3

CH3

2-Methyl pentane

2,2-Dimethyl butane

(ii) Position isomers: These are compounds with different position of substituent or functional group. For example, C4H9Cl can be: 4

CH3

3

CH2

2

CH2

1

CH2

4

Cl

3

CH3

CH2

2

CH

1

CH3

Cl 2-Chlorobutane

1-Chlorobutane

(iii) Functional isomers: These are organic compounds with different functional groups. For example, C3H8O can be: 3

2

1

CH3CH2CH2OH Propan-1-ol

CH3

O

C3H5

Methoxy ethane

(iv) Metamers: These arise due to different alkyl groups on either side of the same bivalent functional group in a molecule. For example, C4H10O can be: CH3

O

C3H7

Methoxy propane

CH3CH2 O CH2CH3 Ethoxy ethane

(b) Stereoisomerism: Isomers that have the same molecular formula and substituent but differ in the arrangement of those substituent in three dimensional spaces.

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(i) Geometrical isomerism • Geometric isomers are those which have the same structural formulas but differ in the arrangement of groups at a single atom, at double bonds, or in rings. • When two similar groups lie on the same side of a double bond, the arrangement is called cis and when two similar groups lie on the opposite side of a double bond, the arrangement is called a trans. CH3

CH3 C

CH3

C

H

H C

H

H

Cis-2-Butene

C CH3

Trans-2-Butene

E and Z notation for geometrical isomers The Cahn-Ingold and Prlog sequence rule are used for naming geometric isomers. • Assign priority to the groups attached to the two double bonded carbons. • If groups of higher priority are on same side of double bond, the isomer is Z isomer (zusammen = together); if groups of higher priority are on opposite side of double bond, the isomer is E isomer (entgegen = across). • Rules for assigning priority are based on the highest atomic number of the atom directly bonded to the carbon. 1

H

high priority

17

Cl

1

H

6

CH3

high priority H3C 6

high H3C Cl high 17 6 priority priority (Z )-2-Chloro-2-butene

CH3 6

(E )-2-Chloro-2-butene

If carbon is bonded to same atoms, rules are applied at first point of difference. • If carbon is bonded to isotopes of same element, isotope with higher mass number is given higher priority. • If carbon is bonded to double or triple bond, it is considered equivalent to the same number of single bonded atoms. Physical properties of geometrical isomers • The boiling point of the cis isomer is higher than that of the trans isomer. The boiling point of isomeric compounds depends upon the dipole–dipole interactions. Since cis isomers usually have a higher value of dipole moment than the corresponding trans isomer, therefore their boiling point is more. • Cis isomers have higher dipole moments than trans isomers which may have zero dipole moments because dipole moment of similar groups on opposite sides cancel out each other. For example, Cl

H C C H

Cl m =1.85 D (Cis)

Cl

H C C

H Cl m=0D (Trans)

(ii) Optical isomerism      These are substances that can rotate the plane of plane-polarized light and generally contain a carbon atom attached to four different groups (chiral center). The stereoisomers that can exhibit optical isomerism can be either enantiomers or diastereomers. • Enantiomers: These are optical isomers that are non-superimposable mirror images of each other. • Diastereomers: They are one of a set of stereoisomers that are not enantiomers i.e. stereoisomers that are not mirror images of one another and are non-superimposable on one another. Stereoisomers with two or more stereocenters can be diastereomers. • Mesomers: These are optically inactive substances having two or more chiral centers and a plane of symmetry, which bisects the compound into two equal parts that are mirror images of each other.

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The number of optically active forms of a compound depends on the number of chiral carbon atoms. If there are n chiral carbons, the number of optically active form will be 2n, provided the chiral atoms are not identically substituted. There are compounds which show optical activity even though they do not possess any assymetric carbon. Compounds in which the central carbon atom is hybridized in a manner such that the substituents at one end are in a plane perpendicular to the substituents at the other end, can also exist as non-superimposable mirror images and exhibit optical activity, such as 1,3-diphenyl propadiene. C6H5 H

C

C

C

H

H

C6H5

H5C6

C

C

C

C6H5 H

(c) Nomenclature of chiral compounds (i) D and L nomenclature • The main carbon chain is oriented vertically (Fischer projection) with the lowest numbered carbon at the top. • Next, arrange the structure around the particular chiral carbon whose configuration is to be assigned such the horizontal bonds to that carbon extend towards observer and the vertical bonds extend away from observer. • The relative positions of the substituents on the horizontal bonds at the chiral centers are examined. If the main substituent is on the left of the main chain, the L configuration is assigned; if this substituent is on the right, the D configuration is assigned. Main substitutent on left CHO

Main substitutent on left

CHO

CHO

H = HO

HO

CH2OH

H

OH

H

CH2OH

=

CH2OH

L

CHO

H

OH CH2OH

D

(ii) (R) and (S) configurations are assigned on the basis of the following procedure.        The Cahn-Ingold-Prelog priority rules are used for naming chirality centers. • Each of the four groups attached to the stereogenic carbon is assigned a priority or preference a, b, c, or d. Priority is first assigned on the basis of the atomic number of the atom that is directly attached to the stereogenic carbon. The group with the lowest atomic number is given the lowest priority, d; the group with next higher atomic number is given the next higher priority, c; and so on.       We can illustrate the application of the rule with the 2-butanol enantiomer, I: CH3 (a) HO

C CH2

(b or c) H (d) (b or c)

CH3

• We now rotate the formula (or model) so that the group with lowest priority (4) is directed away from us. Then we trace a path from a to b to c. If, as we do this, the direction of our finger (or pencil) is clockwise, the enantiomer is designated (R). If the direction is counterclockwise, the enantiomer is designated (S). On this basis the 2-butanol enantiomer I is (R)-2-butanol. (c) CH3 CH3 HO

C CH2 CH3

H

(d) H

OH

(a) Viewer

CH3CH2 (b) Arrows are clockwise (R)-2-Butanol

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8. Fission Organic molecules are mainly composed of covalent bonds and the organic reactions proceed with the cleavage of these bonds. Fission

Homolytic • Produces intermediates called free radicals, which have a single odd unpaired electron.

Heterolytic

• Planar species, sp2 • Stability of alkyl free radicals follows the order: 3° > 2° 1° > vinyl

Carboanion • sp3 hybridized with pyramidal structure. • Stability order is 1° > 2° >3°.

Carbocation • sp2 hybridized with planar structure. • Stability order is 3° > 2° >1° > −CH3.

9. Electron Displacement Effects in Organic Reactions The bonding pairs of the covalent bond in organic compounds undergo electronic displacements on their own or under the influence of attacking reagents. (a) Attacking species

Nucleophiles • Nucleus loving species. • Negative ion or any neutral molecule. • Examples: NH3, H2 O, CN− and RCOO−

Electrophiles • Electrons seeking or electron loving species. • BF3, ZnCl2, NO+2 and Ag+.

(b) Inductive (I) effect (i)  The process of electron shift along the chain of atoms due to the presence of polar covalent bond in it is called inductive effect. (ii) When the hetero atom or group is such that is it attracts the electron pair towards itself, it is said to exert  -I effect or electron withdrawing inductive effect. When the hetero atom or substituent group of atoms pushes the electrons away from itself it exerts +I effect or electron releasing inductive effect. Electron releasing groups such as alkyl groups decrease the acidity, while electron withdrawing groups such as Cl, Br, OH, CN, etc. increase the acidity. (c) Resonance (R) effect (i)  The phenomenon due to which a compound is said to be a hybrid of various cannonical forms is called resonance. The effect caused by resonance in a molecule is called resonance (R) or mesomeric (M) effect. (ii) It is a permanent effect and it operates via the p-orbitals in organic compounds containing at least one double bond attached in configuration with other double bond or a lone pair (conjugated systems). (iii) Mesomeric effect is of two types, +M (positive mesomeric) and -M (negative mesomeric) effect.

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(iv) W  hen the substituent is an electron withdrawing group,  -M effect is observed. When the substituent is electron releasing, +M effect is observed. For example,    NO2,    CN groups show -M effect and  OH,   NH2 groups show +M effect. CH2

CH

C

O

+

CH2

CH

O− (−M effect)

C

H

CH2

CH

−

Cl

CH2

CH

Cl+ (+M effect)

H

(d) Hyperconjugation (i) This effect is a stabilizing interaction that results from interaction of sigma bond with adjacent empty or partially filled p-orbitals. This interaction results in extended molecular orbitals which increases the stability of the system. It involves conjugation of s electrons of a single C   H bond and the unshared p orbital or the p electrons of the adjacent multiple bond. (ii) Hyperconjugation explains the stability of alkenes. More the number of a -hydrogens in an alkene, more will be its stability. Thus, the order of stability some alkenes are as follows: CH3

C

CH3

C

CH3

>

CH3

CH3

C

C

CH3

(12a -hydrogens)

CH3

>

H

CH3

C

CH3

(9a -hydrogens)

CH2

(6a -hydrogens)

10. Types of Organic Reactions (a) Substitution reactions: In these reactions, one group is replaced by another. These are characteristic reactions of saturated compounds such as alkanes and alkyl halides and of aromatic compounds. For example, CH3Cl + OH- → CH3OH + Cl(b) Elimination reactions: In these reactions, one molecule loses the elements of another small molecule. These reactions are used for preparing compounds with double and triple bonds. (i) An elimination reaction in which both the groups and atoms are removed from the same carbon of the molecule is called a-elimination reaction. This reaction is mainly given by gem dihalides and gem trihalides having at least one a-hydrogen. (ii) An elimination reaction in which functional group (i.e., leaving group) is removed from a-carbon and other group (generally a hydrogen atom) from the b-carbon is called b-elimination reaction. For example, H H

b

H a

C

C

H

Br

H

KOH (−HBr)

H

H C

C

H

H

(c) Addition reactions: Reactions in which the components of a species A–B are added to adjacent atoms across a carbon–carbon multiple bonds. An addition reaction is the reverse of an elimination reaction. HCl + CH2

CH2 → CH3CH2Cl

(d) Rearrangement reactions: In these reactions, a molecule undergoes a reorganization of its constituent parts. For example, H

H

H3C

C

H

C

H3C

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C

Acid catalyst

H3C

H3C

CH3

C

C CH3

CH3

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Purification and Characterization of Organic Compounds 1. Th  e purification of organic compounds depends upon the nature of the compound and the kind of impurity present in it. Melting and boiling points, or chromatographic and spectroscopic techniques are used to determine the purity of the compounds. 2. Purification of Solids The steps in techniques used for purification of solids are described as follows: Crystallization

Fractional Crystallization

Sublimation

Saturated solution of impure substance is prepared in hot solvent

Solution

Substance

Heated animal charcoal adsorbs impurities Least soluble

Pure substance

Most soluble

Sublimation

No Sublimation

Pure

Impure

or by mixing different solvents

Seeding Crystal formation

3. Purification of Liquids Distillation is used to separate substances that have different boiling points or to separate volatile substances from non-volatile substances. Various distillation techniques used are as follows.

(a) Simple distillation Mixture Heated Vaporization

No-Vaporization

Pure substance

Impure

(b) Fractional distillation: The process of fractional distillation is used when the boiling points of the components to be separated differ by less than 30–40°C and a fairly complete separation is desired. Liquid

High boiling point

Low boiling point

Heat Separated as pure

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Pure

Impure

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(c) Distillation under pressure: Liquids which boil at very high temperatures or which decompose below their boiling points or undergo chemical changes on heating are separated by carrying out distillation at reduced pressure. (d) Steam distillation: The liquids insoluble in water, steam volatile in nature, having high molecular mass and high vapor pressure are purified by steam distillation. For example, ortho-hydroxyacetophenone and para-­ hydroxyacetophenone. 4. Differential Extraction (a)  Solvent extraction is frequently used in the laboratory to separate out the desired compound out of the mixture or from impurities depending upon their differential solubility in various solvents. (b)  The solubility characteristics of a given compound governs its distribution between two phases of immiscible solvents. Thus, an organic compound can be separated from an aqueous mixture by shaking with an organic solvent in a separating funnel. 5. Chromatographic Methods (a) C  hromatography is defined as a method of separating a mixture of components into individual components by equilibrium distribution between two phases, one mobile and one stationary. It is based on the principle of adsorption. Adsorption

Partition

Paper

Stationary phase → Solid Mobile phase → Liquid/Gas

Fixed phase → Liquid supported Mobile → Liquid/Gas

Whatmann paper

(b)  The various methods of chromatography are categorized on the basis of the phases involved as: (i) Solid–liquid chromatography. (ii) Liquid–liquid chromatography (paper). (iii) Vapor phase chromatography (gas–liquid or simply gas chromatography). Different solid phases have different retardation factor (Rf  ) value. Rf =

Distance moved by substance from base line ( X ) Distance moved by solvent from base line

Qualitative Analysis of Organic Compounds 1. Detection of Carbon and Hydrogen The presence of carbon and hydrogen is detected by Liebig method. C (organic) + 2 CuO Organic compound

CO2 + Cu Ca(OH)2

CuO

2H + CuO

CuSO4 · 5H2O Blue color

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∆

∆

CaCO3 + H2O (Lime water test) H2O + Cu

CuSO4 (colorless) Anhydrous

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OBJECTIVE CHEMISTRY FOR NEET

2. Detection of Other Elements (a) Lassaigne’s test N Organic compound

Na + C + N 2Na + S       

Na + X

∆ ∆ ∆

Na

Plunge in water/oil

S

Fused

X

Lassainge’s extract or Sodium extract

NaCN Na2S NaX (where X = Cl, Br and I)

(b) Detection of nitrogen

Lassaigne extract + Ferrous sulphate (Freshly prepared) Heated

Cooled

Dil. H2SO4

Green or Prussian blue confirms Presence of nitrogen

6CN- + Fe2+

[Fe(CN)6]4-

   O  n addition of sulphuric acid, some part of Fe2+ is oxidized to Fe3+ which further reacts with sodium hexacynoferrate. 3[Fe(CN )6 ]4 - + 4Fe3 + → Fe 4 [Fe(CN )6 ]3 Prussianblue

  If sulphur is present with nitrogen, red color appears due to formation of [Fe(CNS)3]. Hydrazine (NH2NH2) and diazonium salt do not give this test. (c) Detection of sulphur CH3COOH Lassaigne extract ( →[Fe(CN )5 NOS]4 CH COO ) Pb 3

S2 - + [Fe(CN)5 NO]2 - →

2

Nitroprusside (violet color)

[Fe(CN)5 NOS]4 -

Violet color confirms the presence of sulphur

(d) Detection of halogens HNO

3 Lassaigne extract AgNO  → AgX (where X = Cl, Br , I)



3

ppt

        The color of the precipitate is indicative of the halogen present, AgCl is white, AgBr is yellow and AgI is bright yellow. Beilstein test is also used to detect halogens. In this test, a copper wire is heated in Bunsen burner flame until it ceases to impart any green/blue color to the flame. It is then dipped in organic compound and heated again in the flame. Appearance of green/bluish-green flame due to cupric halides confirms the presence of halogens. (e) Detection of phosphorous ∆ Organic compound Na  → Phosphate O 2

Phosphate

2

Boil HNO3 solution Ammonium  → Yellow coloration confirms molybdate

Na 3 PO4 + 3HNO3 → H3 PO4 + 3NaNO3 H3 PO4 + 12( NH4 )2 MoO4 + 21HNO3 → ( NH4 )3 PO4 ⋅ 12 MoO3

the presence of phosphorus

Ammonium phosphomolybdate (Yellow ppt.)

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+ 21 NH4 NO3 + 12 H2 O

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559

(f ) Detection of oxygen: There is no direct test for detection of oxygen. It is generally present in the form of functional groups like    OH,    CO,    COOH,    NO2. Quantitative Estimation of Elements 1. Estimation of Carbon and Hydrogen: The amount of carbon and hydrogen in an organic sample is determined by heating a known amount of sample in a combustion tube called Liebig’s tube based on the following equation.

(y)

C x H y + (x + y )O2 → x CO2 + 2 H2 O % of Carbon =

12 × Weight of carbon dioxide 44 × Weight of organic compound

% of Hydrogen =

2 × Weight of H2 O formed 18 × Weight of organic compound

× 100 × 100

2. Estimation of Nitrogen (a) Dumas method: A known amount of organic compound containing nitrogen is placed in a combustion tube with copper oxide and heated in an atmosphere of nitrogen. y  2x + y   2x + y  C x H y N z +  2  CuO → xCO2 + 2 H2 O + 2z N 2 +  2  Cu     28 × Volume of nitrogen evolved × 100 % Nitrogen = 22400 × Weight of inorganic compound

()

()

(b) Kjeldahl’s method: The organic compound containing nitrogen is heated with concentrated sulphuric acid in the presence of CuSO4. Ammonium sulphate formed is heated with sodium hydroxide to liberate ammonia which is neutralized by sulphuric acid. Organic compound + H2 SO4 → (NH4 )2 SO4 (NH4 )2 SO4 + 2NaOH → Na 2 SO4 + 2NH3 + 2H2 O 2NH3 + H2 SO4 → (NH4 )2 SO4 1.4 ´ N ´ V % of Nitrogen = ´ 100 Weight of organic compound      where N is Normality of acid and V is the volume of acid in ml used to neutralize ammonia. 3. Estimation of Sulphur Oxidising agent H2SO4 + BaCl2

Organic compound  → BaSO4 % Sulphur =

32 ´ Weight of BaSO4 233 ´ Weight of organic compound

´ 100

4. Estimation of Halogens (Carius Method) HNO

3 Organic compound AgNO  → AgX ( where X = Cl,Br,I) 3

% Cl =

ppt

35.5 ´ Weight of AgCl 143.5 ´ Weight of organic compound

% Br =

80 ´ Weight of AgBr 188 ´ Weight of organic compound

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´ 100 ´ 100

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OBJECTIVE CHEMISTRY FOR NEET

% I=

127 × Weight of AgI 255 × Weight of organic compound

× 100

5. Estimation of Phosphorous Fuming → H3 PO4 (a) Organic compound  HNO 3

Magnesia mixture

H3 PO4  → MgNH4 PO4 (MgSO4 + NH4Cl + NH4 OH) Ignition

MgNH4 PO4  → % of Phosphorus =

Mg 2 P2 O7

Magnesium pyrophosphate

62 ´ Weight of Mg 2 P2 O7 222 ´ Weight of organic compound

´ 100

(b)  The compound is heated with sodium peroxide which oxidizes the phosphorus present in the compound to phosphate and reagent (NH4)2MoO4 (ammonium molybdate) is added in the presence HNO3 to produce a canary yellow precipitate of ammonium phosphomolybdate. Na 3 PO4 + 3HNO3 → H3 PO4 + 3NaNO3 H3 PO4 + 12( NH4 )2 MoO4 + 21 HNO3 → ( NH4 )3 ⋅ 12MoO3 + 21 NH4 NO3 + 12H2 O Canary yellow ppt.

6. Estimation of Oxygen 100 − [Sum of percentages of all the other elements present].

Solved Examples 1. What alkyl groups make up the following ketone? Ph O

(1) Phenyl, pentyl (2) Hexyl, phenyl (3) Benzyl, hexyl (4) Benzyl, heptyl Solution

of one s orbital and one p orbital), those of the sp2 orbitals of ethene have 33.3% s character, while those of the sp3 orbitals of ethane have only 25% s character. Therefore, the correct order is sp > sp2 > sp3. 3. Which one the following does not have sp2 hybridized carbon? (1) Acetone (2) Acetamide (3) Acetonitrile (4) Acetic acid

(3) O Hexyl

Ph

sp 3

sp

(3) Acetonitrile ( CH 3 Benzyl

2. The correct order regarding the electronegativity of hybrid orbitals of carbon is (1) sp > sp2 < sp3 (3) sp < sp2 > sp3

Solution

(2) sp > sp2 > sp3 (4) sp < sp2 < sp3

Solution (2) Electronegativity measures an atom’s ability to hold bonding electrons close to its nucleus, and having electrons closer to the nucleus makes it more stable. Greater the percentage of s character, greater is the electronegativity. The sp orbitals of the C   H bonds of ethyne have 50% s character (because they arise from the combination

Chapter 22_Organic Chemistry - Some Basic Basic Principles and Techniques.indd 560

C

N) has no double bonds.

4. In allene (C3H4), the type(s) of hybridization of the carbon atoms is (are) (1) sp and sp3 (3) only sp2

(2) sp2 and sp (4) sp2 and sp3

Solution (2) In allene molecule (C3H4), the central carbon atom is sp hybridized and other terminal carbon atoms are sp2 hybridized. H C H

p p C C s s

H H

sp2 sp sp2

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Organic Chemistry – Some Basic Principles and Techniques 5. The C4-C5 carbon-carbon bond in the following molecule results from the overlap of which orbitals (in the order C4-C5)?

(1) 2-bromo-6-chlorocyclohex-1-ene. (2) 6-bromo-2-chlorocyclohexene. (3) 3-bromo-1-chlorocyclohexene. (4) 1-bromo-3-chlorocyclohexene.

O 7

6

5

4

3

Solution

1

2

(3)

(1) sp-sp2 (2) sp-sp3 (3) sp2-sp2 (4) sp3-sp2

Cl 1

6

2 3

5

Solution

4

(4) C4-C5 bond made by sp3-sp2 orbital overlapping.

sp2 o 7

6

5

4

3

1

2

561

Br

According to the order of precedence of groups the substituents are ordered alphabetically and the position is mentioned by the first carbon of the double bond.

9. The IUPAC name of the following compound is

sp3

6. The shape of a carboanion is (1) linear. (2) planar. (3) pyramidal. (4) tetrahedral.

(1) 3-ethyl-4, 4-dimethylheptane. (2) 1,1-diethyl-2,2-dimethylpentane. (3) 4,4-dimethyl-5,5-diethylpentane. (4) 5,5-diethyl-4,4-diemthylpentane.

Solution (3) The carbon atom bearing the negative charge in the cabanion is in sp3 hybridized state. It is bonded to three other atoms and the unshared electron pair occupies the apex of the tetrahedron. Thus, the shape of carboanion is pyramidal.

Solution (1) Naming of given organic compound is done by following step wise process: (i) Selection of longest chain (ii) Numbering is done from that end having lowest set of locants for substituents. (iii) Writing the name by using the below formula.

R R C− R sp3 hybridized carbon

7. Which of the following molecules has two sigma (s) and two pi (p) bonds? (1) C2H4 (3) C2H2Cl2



Name of substituent with position in alphabetical order + Root word + Suffix (nature of functional group) 6

(2) N2F2 (4) HCN

7

5

4

2 3

1

Solution 3-Ethyl-4, 4-dimethylheptane

(4) The number of s and p bonds in the given molecules are: C

5s and 1p

C H

H

N

N

3s and 1p

C

N

2s and 2p

F

10. The correct decreasing order of priority for the functional groups of organic compounds in the IUPAC system of nomenclature is (1) (2) (3) (4)

CI

CI C H

F

H

H

5s and 1p

C

H

H

8. The IUPAC name of the compound shown below is

       

 COOH,    SO3H,    SO3H,    COOH,    CHO,    COOH,    CONH2,    CHO,  

 CONH2,    CHO  CONH2,    CHO  SO3H,    CONH2  SO3H,    COOH

Solution

Cl

(1) According to latest IUPAC system priority order of various functional group is Br

Chapter 22_Organic Chemistry - Some Basic Basic Principles and Techniques.indd 561





   COOH > SO3H >   CONH2> CHO

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OBJECTIVE CHEMISTRY FOR NEET

11. The IUPAC name of neopentane is

P (CH3)2

(1) 2-methylbutane. (2) 2,2-dimethylpropane. (3) 2-methylpropane. (2) 2,2-dimethylbutane.

+

Solution

N H

(2) The structure is of neopentane is CH3 3

2

P (CH3)2

1

C

H3C

+

CH3



(1)

(2) N+ H

The IUPAC name is 2,2-dimethylpropane.

(1) diketones. (2) carboxylic acids. (3) diols. (4) dialdehydes.

(3)

P (CH3)2 +

(4) N H

Solution (2) This is a general formula for carboxylic acids. For example, for n = 2. CH3COOH −

N H

+ P (CH ) 3 2

12. The general formula CnH2nO2 could be for open chain

13. CH2

P (CH3)2

CH3

C

CH3

O

CH2

C

CH3

− O

and

Solution (3)

+

Solution (1) Structure CH3 Ph-H2IC and II represents H structures C resonating Ph-Hthe 2 C can C as one structure be converted CintoC another one by H H CH3 H changing only the (cis) positions of the electrons. (trans) C

CH3

CH2

C

N

H

H

+

P (CH3)2

CH3 N

(II)

14. Which molecule would you expect to have no dipole moment (i.e., µ = 0)? (2)

N

O-

O (I)

(1) CHF3

+ P (CH3)2

P (CH3)2

are

(1) resonating structures. (2) tautomers. (3) geometrical isomers. (4) optical isomers.

CH2

N H

F

H

H

F

(3) NF3

(4)

F

F

H

H

Solution (2) In the following molecule, net dipole moment cancels out. F

H C

H

C F

15. Which of the following species is a resonance form of the species in the box?

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H

16. Geometrical isomers differ in (1) (2) (3) (4)

position of the functional group. position of atoms. spatial arrangement of atoms. length of the carbon chain.

Solution (3) Geometrical isomers are those isomers that have same molecular formulas but different spatial arrangement of atoms. 17. Which of the following pairs of compounds are positional isomers?

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Organic Chemistry – Some Basic Principles and Techniques (1) CH3

CH2

CH2

CH3

CHO

CH2

CH2

and CH3

(2) CH3

CH2

CH2

CH2

CH2

C

C

CH

CH3

CH

CH2

CH3 and

H3C

C

CH2

C

CH2

CH3

C

CH2

CH2

CH3 and

CH

CH3

CH2

Ph-H2C

CHO

(3) Compounds that differ in the position of the substituent or the functional group on the parent carbon chain are known as the positional isomers.

C

CH2

C

CH2

+

(1) CH3CH2

CH2

CH3 are positional isomers.

18. The alkene that exhibits geometrical isomerism is (1) propene. (2) 2-methylpropene. (3) 2-butene. (4) 2-methyl-2- butene. Solution

H3C

C

C cis

H CH3

H H3C

C

Ph-H2C

H

C

H

(cis)

H

C

CH3

(trans)

(I)

(II)+

(2) CH3

C

CH3 H

trans

19. The number of structural isomers for C6H14 is (1) 3 (2) 4 (3) 5 (4) 6 Solution (3) Since the structural isomers refers to the compounds that have same molecular formula but different arrangement of atoms or groups within the molecule. Based in this, following are the structural isomers for C6H14:

Chapter 22_Organic Chemistry - Some Basic Basic Principles and Techniques.indd 563

(4) CH3

C+

+

CH CH3

CH3

Solution (3) Among the given species, tertiary carbocation is the most stable, and the methyl cation is the least stable. The overall order of stability is as follows

(3) 2-Butene exhibits geometrical isomerism. In rest of the compounds, two identical groups are attached to the doubly bonded carbon atom, so their geometrical isomerism is not possible in them. H

CH3

CH3

CH3 and

(3) CH3

C

C

-

O CH3

2,3-Dimethyl butane

21. WhichCHamongst following stable car­ CHthe CH2 isCtheCHmost C 3 3 2 bocation? O O-

O

CH2

CH3

CH3 CH3

H

Thus, CH3

CH

(4) 1-Phenyl-2-butene

Solution



CH

Solution

O CH3

H3C

CH3

(1) 3-Phenyl-1-butene (2) 2-Phenyl-1-butene (3) 1,1-Diphenyl-1-propane (4) 1-Phenyl-2-butene

O

(4) CH3

CH2

20. Which of the following compounds will exhibit geometrical isomerism?

O CH3

3-Methyl pentane

2,2-Dimethyl butane

CH3 and

C

CH3

CH3

CHO

CH2

CH2

CH3

CH3

CH2

CH

CH2

2-Methyl pentane

CH3

(3) CH3

CH2

CH3

CH3

O CH3

CH2

CH2

CH2

n-Hexane H3C CH3

O H3C

CH2

CH3 CH3

C+ CH3



>

CH3

+

CH

>

CH3

+

CH2

>

+

CH3

CH3

This order of stability of carbocations can be explained on the basis of hyperconjugation which involves conjugation between s-bond of C   H and positive charge on carbon. More the number of a-hydrogens, more will be the stability of carbocation.

22. Which one of the following pair represents stereo­ isomerism? (1) (2) (3) (4)

Linkage isomerism and geometrical isomerism. Chain isomerism and rotational isomerism. Optical isomerism and geometrical isomerism. Structural isomerism and geometrical isomerism.

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Solution (3) Two molecules are described as stereoisomers of each other if they are made of the same atoms, connected in the same sequence, but the atoms are positioned differently in space. The difference between two stereoisomers can only be seen when the three dimensional arrangement of the molecules is considered. Stereoisomers can be subdivided into geometrical isomerism and optical isomerism. Br

H3C

(3) 3-Bromopentane is an achiral compound. H H3C

H2C

∗

C

CH2

(2)

CHO

Cl

HO

H SH

(2) R (4) Z

CH2OH

Solution

(3)

(2) Each of four groups attached to the stereogenic carbon is assigned a priority which is first done on the basis of atomic number. Therefore, the priority of atoms are Br > Cl > CH3 > H. Br1 H4 3

H2N

(4)

NH2

H

COOH H2N

H Ph

Ph

H H

Solution

C

(1) The compound has a chiral carbon atom.

Cl2

H3C

CH3

Br

(1)

H

(1) E (3) S



Solution

26. Which of the following molecules is expected to rotate the plane of plane-polarized light?

23. The chirality of the following compound is

C

(3) 3-Bromopentane (4) 2-Hydroxypropanoic acid

The molecule is designated as (R) due to its clockwise rotation.

(A carbon attached

CHO

to four different HO

24. The optically inactive compound from the following is (1) 2-chloropropanal. (2) 2-chloropentane. (3) 2-chlorobutane. (4) 2-chloro -2-methylbutane.

Chiral carbon

Option (1):

*

substituents)

H

CH2OH Plane of symmetry

Option (2):

(which divides the

Solution

molecule into two

(4) Among the given compounds, 2-chloro-2-methylbutane does not contain any chiral center hence is optically inactive.

CH *

CHO

CH3

2-Chloropropanal

CH2

* CH

CH2

CH3

Cl 2-Chloropentane

CH2

* CH

CH3

Cl 2-Chlorobutane

CH3

CH2

C

(1) 2-Butanol (2) 2,3-Dibromopentane

Chapter 22_Organic Chemistry - Some Basic Basic Principles and Techniques.indd 564

NH2

H

H Ph

Plane of symmetry Achiral

CH3

Cl 2-Chloro-2-methylbutane

25. Which of the following is not chiral?

H2N

Ph

CH3 CH3

Achiral

Option (3):

Cl CH3

equal halves) SH

Option (4):

COOH H2N

H

Two identical groups attached Achiral

H

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Organic Chemistry – Some Basic Principles and Techniques 27. The number of stereoisomers possible for a compound of the following molecular formula CH3   CH   CH   CH(OH)   CH3 is

31. The order of stability of the following carbocations is +

CH2

(1) 3 (2) 2 (3) 4 (4) 6

H2C

Solution (3) The compound has one chiral center. Therefore, 2n = 21 = 2 optically active compounds are possible. Also, two geometrical isomers (cis and trans) are possible. So, total of four isomers are possible. H CH3

C

H

H

C

C*

OH

29. The correct order of stability of the following free radicals is

CH2

+

CH2 ;

(II)

(III)

Solution (3) Stability is directly proportional to extent of delocalization of positive charge. Thus, order of stability is CH+2

> CH2

(1) chemically reactive. (2) chemically inactive. (3) anions. (4) cations.

R − H → R + H ( where R is an alkyl group)

CH2; H3C

(1) II > III > I (2) I > II > III (3) III > I > II (4) III > II > I

28. Due to the presence of an unpaired electron, free radicals are

(1) Free radicals are highly reactive due to the presence of unpaired electron.

+

CH

(I)

CH3

Solution

565

CH

+

CH2

> CH3

+

CH2

CH2

32. Which of the following can be applied to explain relative order of stability of carbocations? (1) Resonance (2) Inductive effect (3) Hyperconjugation (4) All of these Solution (4) Stability of carbocations can be explained by resonance, inductive effect and hyperconjugation effect. (1) Resonance: For example consider allyl carbocation

(I)

(II)

+

CH2

(III)

(IV)

(1) III > I > II > IV (2) I > II > IV > III (3) IV > II > I > III (4) I > III > II > IV Solution (1) The free radicals are electron-deficient species, whose stability is enhanced by electron donating groups. Therefore in (III), allylic free radical is most stable since electron density can be transferred by both inductive as well as resonance effects. As (I), (II) and (IV) are 3°, 2° and 1° free radicals, respectively, the correct order of stability is III > I > II > IV. 30. Which of the following is not isomeric with diethyl ether? (1) n-Propyl methyl ether (2) 1-Butanol (3) 2-Methyl-2-propanol (4) Butanone Solution (4) For compounds to be isomeric, they should have same molecular formula. Hence, diethyl ether (C4H10O) and butanone (C4H8O) are not isomers.

Chapter 22_Organic Chemistry - Some Basic Basic Principles and Techniques.indd 565

CH

CH2

CH2

CH

+

CH2

(2) Inductive effect: The order of stability of carbocation can also be explained on the basis of +ve inductive effect of alkyl group. It is clear that lesser the positive formal charge on the carbon atom of the carbocation due to +IE of alkyl group, greater is the stability of carbocation. (3) Hyperconjugation: The stability of carbocations can be explained on the basis of number of hyperconjugating structures. As the number of hyperconjugating structure increases, the stability of carbocation increases. 33. In the anion HCOO-, the two carbon–oxygen bonds are found to be of equal length. What is the reason for it? (1) Electronic orbitals of carbon atom are hybridized. (2) The C   O bond is weaker than the C   O bond. (3) The anion HCOO– has two equivalent resonating structures. (4) The anion is obtained by removal of a proton from the acid molecule. Solution (3) The anion HCOO-  can be represented as following resonating structures.

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OBJECTIVE CHEMISTRY FOR NEET

H

C

Solution

O-

O O-

H

C

( 4) Lassaigne’s test results

O

(1) Aniline

(r) Acidic solution of FeSO4 (2) Benzene sulphonic Acid (p) Red color with FeCl3 (3) Thiourea (q) sodium nitroprusside

34. Addition of acids to alkenes is an example of (1) +I effect. (2) -I effect. (3) +E effect. (4) -E effect. Solution (3) If the electrons of the p bond are transferred to that atom of the double bond to which the reagent gets finally attached, the effect is called +E effect. For example, addition of acids to alkenes.

38. The suitable technique of separation of the components from a mixture of calcium sulphate and camphor is (1) sublimation. (2) distillation. (3) crystallization. (4) chromatography. Solution

C

C

+

H

+

+

C

C H

35. Liebig’s test is used to estimate

(1) Camphor sublimes on heating and can be collected as the sublimate. 39. The presence of halogen in an organic compound is detected by (1) iodoform test. (2) silver nitrate test. (3) Beilstein’s test. (4) Millon’s test.

(1) H (2) C (3) Both (1) and (2) (4) N Solution

Solution

(3) In Liebig’s test, the amount of carbon and hydrogen present in the organic compound is determined by heating a known amount of sample in a combustion tube (also known as Liebig’s combustion tube) in the presence of excess oxygen and copper(II) oxide. Carbon is oxidized to carbon dioxide and hydrogen to water.

(3) In Beilstein test, a (clean) copper wire is heated in the non-luminous flame of the Bunsen burner until it ceases to impart any green or bluish green color to the flame. The heated end of the wire is dipped into the organic compound and is again introduced into the Bunsen flame. The appearance of a bluish green or green flame due to the formation of volatile cupric halides shows the presence of halogens in the given organic compound.

36. In Kjeldahl’s method, CuSO4 acts as (1) oxidizing agent. (2) reducing agent. (3) hydrolyzing agent. (4) catalytic agent. Solution (4) In Kjeldahl’s method, the organic compound containing nitrogen is heated with concentrated sulphuric acid in the presence of CuSO4, which converts the nitrogen in the compound to ammonium sulphate. 37. Match the organic compounds in Column-I with the Lassaigne’s test results in Column-II appropriately: Column-I Column-II (1) Aniline (p) Red color with FeCl3 (2) Benzene sulphonic acid (q) Violet color with sodium nitroprusside (3) Thiourea (r)  Blue color with hot and acidic solution of FeSO4 (1) (1) – (q); (2) – (p); (3) – (r) (2) (1) – (r); (2) – (q); (3) – (p) (3) (1) – (q); (2) – (r); (3) – (p) (4) (1) – (r); (2) – (p); (3) – (q)

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40. In Kjeldahl’s method, 29.5 mg of an organic compound containing nitrogen was digested and the evolved ammonia was absorbed in 20 mL of 0.1 M HCl solution. The excess of the acid required 15 mL of 0.1 M NaOH solution for complete neutralization. The percentage of nitrogen in the compound is (1) 29.5 (2) 59.0 (3) 47.4 (4) 23.7 Solution (4) Weight of substance taken = 29.5 × 10−3 g

Volume of acid taken = 20 mL of 0.1 N HCl



Now, 15 mL of 0.1 N NaOH = 15 mL of 0.1 N HCl



Therefore, volume of acid used = 20 − 15 = 5 mL of 0.1 N HCl % Nitrogen = =

1.4 × Volume × Normality × 10 0 Weight of substance taken 1.4 × 5 × 10 -3 L × 1 g L-1 × 100 = 23.73% 29.5 × 10 -3 g

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Organic Chemistry – Some Basic Principles and Techniques 41. In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The percentage of bromine in the compound is (At. mass of Ag = 108; Br = 80) (1) 36 (2) 48 (3) 60 (4) 24

42. 1.4 g of an organic compound was digested according to Kjeldahl’s method and the ammonia evolved was absorbed in 60 mL of M/10 H2SO4 solution. The excess sulphuric acid required 20 mL of M/10 NaOH solution for neutralization. The percentage of nitrogen in the compound is (1) 3 (2) 5 (3) 10 (4) 24

Solution (4) Percentage of Br Atomic mass of Br × Weight of AgBr × 100 = Molecular mass of AgBr × Weight of organic substance 80 × 141 × 100 = 24% = 188 × 250

Solution (3) % of Nitrogen =

1.4 × m.eq of H 2SO4 used to neutralize NH 3 mass of the compound

1 1  1.4 ×  60 × × 2 - 20 ×    10 10 = 10% = 1.4

Practice Exercises Level I Hybridization, Shapes of Simple Molecules and Nomenclature 1. What alkyl groups make up the following ether?

(1) (2) (3) (4)

sec-Butyl, ethyl, propyl iso-Butyl, isopropyl, ethyl sec-Butyl, ethyl, isopropyl Butyl, ethyl, propyl

5. Which compound(s) contain(s) tertiary carbon atom(s)?

O

F

(1) Ethyl and phenyl (2) Propyl and benzyl (3) Ethyl and benzyl (4) Propyl and phenyl

(I)

(II)

(III)

Br OH

2. The correct IUPAC name of the following compound is

OH

(IV)

(V)

(1) I, II, III (2) I (3) II, III (4) I, IV (1) 4-methyl-3-ethylhexane (2) 3-ethyl-4-methylhexane (3) 3, 4-ethylmethylhexane (2) 4-ethyl-3-methylhexane 3. Among the following molecules, the correct order of C–C bond length is (1) C2H6 > C6H6 > C2H2 (2) C2H6 > C6H6 > C2H4 > C2H2 (C6H6 is benzene) (3) C2H4 > C2H6 > C2H2 > C6H6 (4) C2H6 > C2H4 > C2H2 > C6H6 4. What alkyl groups make up the following 3o amine?

N

Chapter 22_Organic Chemistry - Some Basic Basic Principles and Techniques.indd 567

6. The correct IUPAC name of the following compound is

(1) 1-Cyclopropylcyclobutane (2) 1,1′-Dicyclobutane (3) 1-Cyclobutane-1-cyclopropane (4) All of these 7. What functional group is present in the following compound? HO

(1) 1o alcohol (2) ether (3) 2o alcohol (4) 3o alcohol

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OBJECTIVE CHEMISTRY FOR NEET

(I)

(II)

(III)

(IV)

8. What is the correct IUPAC name for the following structure?

(1) 2−Ethylheptane (2) 2−Methylhexane (3) 6−Methylhexane (4) 2−Methylpentane 9. What is the hybridization of the N atom in the following molecule? N

(1) s (3) sp

(1) I and II (2) I and III (3) I, II, and III (4) I, II, III, and IV 14. For a molecule to possess a dipole moment, which following condition is necessary but not sufficient? (1) (2) (3) (4)

(2) p (4) sp2

Three or more atoms in the molecule. Presence of one or more polar bonds. A non-linear structure. Presence of oxygen or fluorine.

15. Which molecule does not have a dipole moment?

10. Which functional group is not present in the following molecule?

F

F

(1)

OH

F O OH

O

F

(2)

F

(3)

(4)

F

F

16. Which of the following represent pairs of constitutional isomers?

O

(1)

OH

(1) 1° alcohol (2) Ketone (3) 3° alcohol (4) Carboxylic acid 11. How many sigma 1s-2sp3 bonds are there in ethane?

O

(2)

(1) 7 (2) 6 (3) 5 (4) 3 12. Which molecule has the shortest carbon-carbon single bond?

(2)

(3)

(4)

O CH3

Br

H

and

CH3

H

Br

H CH2

(3)

H Br

and

H

H CH3

(4) All of these pairs

17. Which compound contains a nitrogen atom with a formal positive charge? H N

Isomerism and Dipole Moment 13. Which of the following structures represent compounds that are constitutional isomers of each other?

and

CH3

Br

(1)

O

(I)

N N

(II)

(III)

(1) I (2) II (3) III (4) More than one of the above 18. Of the following solvents which one does not have a zero dipole moment?

(I)

(II)

Chapter 22_Organic Chemistry - Some Basic Basic Principles and Techniques.indd 568

(1) Pentane (2) Cyclohexane (3) Diethyl ether (4) Cyclopentane

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569

Organic Chemistry – Some Basic Principles and Techniques 19. Out of the following, the alkene that exhibits optical isomerism is (1) 2-methyl-2-pentene. (2) 3- methyl-2-pentene. (3) 4-methyl-1-pentene. (4) 3-methyl-1-pentene.

27. Among the following four structures I to IV, it is true that, CH3 C2H5

H

(3)

CHCH3

Br

and

C2H5

CH3 C2H5

CH

C3H7

(1) (2) (3) (4)

(IV)

all four are chiral compounds. only I and II are chiral compounds. only III is a chiral compound. only II and IV are chiral compounds.

28. Heterolytic fission of a covalent bond in organic molecules gives

H

Br

H

CH

Reaction Intermediates

Br

H

C

C

(II)

(III)

and

Br

+

CH3

H

21. Which of the following represent a pair of constitutional isomers?

CH2 and CH2

CH3

H

(1) Methyl fluoride (2) Methyl iodide (3) Methyl chloride (4) Methyl bromide

(2) CH3CH

C3H7

(I)

20. Which of the following has the greatest dipole moment?

(1)

CH

O

H

(1) free radicals. (2) cations and anions. (3) only cations. (4) only anions.

(4) More than one of these

- -

-

-

9. Arrange the carbanions, (CH3)3C, CCl3, (CH3)2CH, C6H5CH2 2 (CH3)3C, CCl3, (CH3)2CH, C6H5CH2 in order of their decreasing stability: (1) HCOOC2H5 and CH3COOCH3 (1) C6H5CH2 > CCl3 > (CH3)3C > (CH3)2 CH (2) HCOOC2H5 and C3H7COOH (3) CH3COOCH3 and C3H7COOH (2) (CH3)2CH > CCl3 > C6H5CH2 > (CH3)3C (4) C3H7OH and CH3COOCH3 (3) CCl3 > C6H5CH2 > (CH3)2 CH > (CH3)3C (4) (CH3)3C > (CH3)2CH > C6H5CH2 > CCl3 23. Which of following compound can show geometrical isomerism? 30. The heterolysis of O forms a (1) CH 2 CHCl (2) CH 3CH 2 CH 2Cl (1) carbanion. (2) carbocation. (3) CH CCl (4) (Cl)(Br)C CH(I) (3) free radical. (4) cleavage not possible. 24. Which of the following will have mesoisomer also? 22. Isomers of propionic acid are

- -

-

(1) 2-Chlorobutane (2) 2-Hydroxypropanoic acid (3) 2,3-dichloropentane (4) 2,3-dichlorobutane 25. Which of the following compound is not chiral? (1) 1-Chloropentane (2) 3-Chloro-2-methylpentane (3) 1-Chloro-2-methylpentane (4) 2-Chloropentane 26. Which types of isomerism is shown by 2,3-dichlorobutane? (1) Conformational (2) Optical (3) Geometrical (4) Structural

Chapter 22_Organic Chemistry - Some Basic Basic Principles and Techniques.indd 569

Electrophiles, Nucleophiles, Inductive Effect, Resonance and Hyperconjugation 31. The inductive effect

(1) decreases with increase of distance. (2) increases with increase of distance. (3) indicates the transfer of p pair of electron from less electronegative atom to more electronegative atom in a molecule. (4) shows the transfer of lone pair of electrons. 32. Electrophilic reagents are (1) electron pair donors. (2) Lewis acids. (3) odd electron molecules. (4) None of these.

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OBJECTIVE CHEMISTRY FOR NEET

33. Which of these shows incorrect representation of inductive effect? (1)

(2)

CH3

(1) CH3CH +

(4) CH3CH2

39. Which of the following pairs are not resonance structures?

OH

C

(1) H3C (2) O

O C

N

O

+

and H3C

O

+

and O

O

C

CH3

(3)

(4) CH3

O C

O

C

CH2

CH3

O

(3) H3C

O

N

O

−

-

-

S

S

-

36. A nucleophilic reagent must necessarily have (1) an overall positive charge. (2) an overall negative charge. (3) an unpaired electron. (4) a species with complete octet and lone pair of electrons.

(I)

(II) HS

S -

(III)

(IV)

(1) I (2) II and III (3) III and IV (4) I and IV 41. Which of the following species is not a resonance form of the following species? O +

37. Which of the following reactions involves a nucleophile?

+

O

N

S

(1) delocalization of s-electrons into an adjacent p- bond. (2) delocalization of lone pair electrons into an adjacent double bond. (3) delocalization of p-electrons into an adjacent bond. (4) All of these are correct.

(III) C6H6 + CH3CO

−

(4) Each of these pairs represents resonance structures.

35. Hyperconjugation involves

-

O

−

O

(1) delocalization of p-electrons along a conjugated system. (2) delocalization of n-electrons along a conjugated system. (3) delocalization of s-electrons into an adjacent p-bond. (4) All of these are correct.

(II) CH3COCH3 + CN

O +

and H3C

N

40. Which of the following species is/are not a resonance form(s) of the anionic species in the box?

34. Resonance effect involves

(I) CH3COOH + HO-

CH2

+

(3) (CH3)2 CH2

O

+

(2) (CH3)3C

CH2

CH3COO- + H2O O(CH3)2C CN C6H5COCH3

(1) O +

(1) I and II (2) I and III (3) III only (4) II and III 38. Which of the following molecule/ion cannot be stabilized by hyperconjugation?

Chapter 22_Organic Chemistry - Some Basic Basic Principles and Techniques.indd 570

+

(2) O

(3) O

+

(4)

+

O

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Organic Chemistry – Some Basic Principles and Techniques

Purification and Characterization of Organic Compounds 42. A mixture of camphor and benzoic acid can be separated by (1) chemical method. (2) sublimation. (3) fractional distillation. (4) extraction with a solvent. 43. An organic substance can be separated from its aqueous solution by (1) distillation. (2) steam distillation. (3) solvent extraction. (4) fractional distillation. 44. A liquid decomposes at its normal pressure. It can be purified by (1) sublimation. (2) vacuum distillation. (3) fractional distillation. (4) steam distillation. 45. Chromatography is a valuable method for the separation, isolation, purification and identification of the constituents of a mixture and it is based on general principles of (1) phase rule. (2) phase distribution. (3) interphase separation. (4) phase contact. 46. Two solids which are soluble in the same liquid to different extents may be separated by (1) crystallization. (2) sublimation. (3) evaporation. (4) fractional crystallization. 47. In steam distillation, the vapor pressure of the volatile organic compound is (1) (2) (3) (4)

equal to atmospheric pressure. less than atmospheric pressure. more than atmospheric pressure. None of the above.

Qualitative Analysis of Organic Compounds 48. In Carius tube, the compound ClCH2COOH was heated with fuming HNO3 and AgNO3. After filtration and washing, a white precipitate was formed. The precipitate is (1) AgCl (2) AgNO3 (3) Ag2SO4 (4) ClCH2COOAg 49. In the Lassaigne’s test, sulphur present in the organic compound first changes into (1) CS2 (2) Na2S (3) Na2SO4 (4) Na2SO3

571

Quantitative Estimation of Elements 51. The empirical formula of compound is CH2. The mass of one mole of the compound is 42 g. Therefore, its structural formula is (1) CH3CH2CH3 (2) CH3   CH   CH2 (3) CH2   CH   CH   CH2 (4) CH3   C CH 52. Empirical formula of a hydrocarbon containing 80% carbon and 20% hydrogen is (1) CH (2) CH2 (3) CH3 (4) CH4 53. The percentage of N2 in urea is about (1) 18.05% (2) 28.29% (3) 46.66% (4) 85.56% 54. A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g of CO2, The empirical formula of the hydrocarbon is (1) C3H4 (2) C6H5 (3) C7H8 (4) C2H4 55. In sulphur estimation 0.2 g of an organic compound gave 0.6 g of BaSO4. The percent of S is (1) 41.20 (2) 4.120 (3) 0.4120 (4) 24.68 56. An organic compound is found to have the formula C5H10ONCl. The percentage of nitrogen present in it is (1) 21.36% (2) 10.3% (3) 44.05% (4) 20.6%

Level II Reaction Intermediates 1. Which of the following would yield a carbon radical on heating? (1) CH2N2 (2) Et4Pb (3) PhCOO   OOCPh (4) Me3CO   COMe3 2. In which of the following pairs A is more stable than B? A

B

Ph3C

(CH3)3C

(1)

(2)

50. In Kjeldahl’s method, nitrogen present is estimated as (1) N2 (2) NH3 (3) NO2 (4) NH3

(3) (4)

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OBJECTIVE CHEMISTRY FOR NEET

Isomerism and Dipole Moment

H3C C

3. Five alcohols can be drawn for formula C4H10O. How many of these are optically active?

H3C CH3

Cl

C

CH3

H3C

Cl

C

C Cl

C

Cl

CH3

(II)

(III)

(1) I > II > III (2) III > II > I (3) III > I > II (4) II > I > III

Electrophiles, Nucleophiles, Inductive Effect, Resonance and Hyperconjugation 9.

H3C

(I) CH2

CH

CH

CH

CH

CH2

(II) CH2

CH

CH

CH

CH

CH2

CH

CH

CH

CH

CH2

+



Cl

CH3

(I)

4. How many chiral carbons are there in the following compound?

CH3

H3C

C

H3hC

(1) 1 (2) 2 (3) 3 (4) 4

Cl

+

-

-

HO



(III) CH2

(1) 6 (2) 7 (3) 8 (4) 9



Among these three canonical structures (through more are possible) what would be their relative contribution in the hybrid?

5. The stereochemical relationship between the following molecules is I

Br

(1) I > II > III (2) III > II > I (3) I > III > II (4) III > I > II 10. The total number of contributing structures for hyperconjugation in propene is

Br

I

(1) identical. (2) diastereomers. (3) enantiomers. (4) constitutional isomers.

11. Which of the following statements is correct?

6. Indicate the relationship of the pair of molecules shown below. H

H

I

(1) 3 (2) 2 (3) 5 (4) 4

12. Hyperconjugation explains the stability of

I

(1) Constitutional isomers (2) Enantiomers (3) Diastereomers (4) Different molecules 7. What kind of isomerism is possible in the following organic compound? H3C

(1) The C   Cl bond in chlorobenzene is shorter than in chloroethane. (2) The C   Cl bond in chlorobenzene has some double bond character. (3) It is difficult to replace chlorine from chlorobenzene than from benzoyl chloride. (4) All of these

CH3 CH3

(1) Optical (2) Geometrical (3) Both (1) and (2) (4) None of these 8. Arrange the following in decreasing order of dipole moments:

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(1) carbocations. (2) free radicals. (3) Both (1) and (2) (4) None of these

Purification and Characterization of Organic Compounds 13. Which of the following methods is not used for purification of an organic compound? (1) Simple crystallization (2) Vaporization (3) Sublimation (4) Steam distillation 14. Two volatile liquids A and B differ in their boiling points by 15 K. The process that can be used to separate them is

1/4/2018 5:24:52 PM

Organic Chemistry – Some Basic Principles and Techniques (1) fractional distillation. (2) steam distillation. (3) fractional crystallization. (4) simple distillation. 15. Which of the following is not commonly used solvent for crystallization?

Previous Years’ NEET Questions 1. CH3   CHCl   CH2   CH3 has a chiral center. Which one of the following represents its R configurations? CH3

(1) Water (2) Alcohol (3) Phenol (4) Acetone

(1) H

16. The organic mixture is dissolved in a solvent in which two components have different solubility. The process is (1) (2) (3) (4)

sublimation. distillation. fractional crystallization. simple crystallization.

(3) H

C

(2) H3C

C

Cl

C2H5

H

C2H5

C2H5

C

(4) Cl

CH3

Cl

C

CH3

H

(AIPMT 2007)

2. How many stereoisomers’ does the molecule have? CH3CH   CHCH2 CHBrCH3 (1) 2 (2) 4 (3) 6 (4) 8

17. Sodium fusion extract of an organic compound gives a blood red coloration with few drops of FeCl3 solution. This indicates the presence of

(AIPMT 2008) 3. Base strength of the following compounds is in the order of CH-2

(I) H3C

(II) H2C CH(III) H C C

18. Lassaigne’s test for the detection of nitrogen will fail in case of

(1) I > II > III (2) II > I > III (3) III > II > I (4) I > III > II

(1) NH2CONH2 (2) NH2CONH NH2 ⋅ HCl (3) NH2NH2⋅ HCl (4) C6H5NH2 NH2⋅ 2HCl

Quantitative Estimation of Elements

C2H5 Cl

Qualitative Analysis of Organic Compounds

(1) nitrogen. (2) sulphur. (3) both nitrogen and sulphur. (4) both sulphur and cholrine.

573

(AIPMT 2008) 4. The stability of carbanions in the following is in the order of:

19. The silver salt of a monobasic acid on ignition gave 60% of Ag. The molecular weight of the acid is

(I) RC

-

C-

(II)

CH-

(IV) R3C

(1) 37 (2) 33 (3) 73 (4) 77 (III) R2C

20. If a compound on analysis was found to contain C = 18.5%, H = 1.55%, Cl = 55.04% and O = 24.81%, then its empirical formula is (1) CHClO (2) CH2ClO (3) C2H2OCl (4) ClCH2O 21. 5/19 g of an organic compound gave 22.4 cm3 of moist nitrogen measured at 280 K and 732.7 mm pressure. The percentage of nitrogen in the substance is approximately (Aqueous tension at 280 K = 12.7 mm.) (1) 9.8 (2) 19.6 (3) 4.9 (4) 9.0 22. On complete combustion, 0.3 g of an organic compound gave 0.2 g of CO2 and 0.1 g of water. What is the percentage of carbon in the compound? (1) 18.18 (2) 21.95 (3) 28.36 (4) 46.24

Chapter 22_Organic Chemistry - Some Basic Basic Principles and Techniques.indd 573

CH2-

(1) I > III > II > IV (2) I > II > III > IV (3) II > III > IV > I (4) IV > II > III > I (AIPMT 2008) 5. In the hydrocarbon CH3 6



CH 5

CH 4

CH2 3

C 2

CH 1

The states of hybridization of carbons 1, 3 and 5 are in the following sequence: (1) sp 3, sp 2, sp (2) sp 2, sp, sp 3 3 2 (3) sp, sp , sp (4) sp, sp 2, sp 3 (AIPMT 2008)

6. The state of hybridization of C2, C3, C5 and C6 of the hydrocarbon

1/4/2018 5:24:53 PM

574

OBJECTIVE CHEMISTRY FOR NEET CH3 7 5 6 CH3 C CH

4 CH

CH3 3 CH

2 C

(3) helps in the precipitation of AgCl. (4) increases the solubility product of AgCl.

1 CH

(AIPMT PRE 2011)

CH3



13. The IUPAC name of the following compound is

is in the following sequence: (1) sp, sp2, sp3 and sp2 (3) sp3, sp2, sp2 and sp

(2) sp, sp3, sp2 and sp3 (4) sp, sp2, sp2 and sp3



Cl

(AIPMT 2009) 7. Which of the following compounds will exhibit cis-trans (geometrical) isomerism? (1) 2-Butenol (2) 2-Butene (3) Butanol (4) 2-Butyne

CH2CH3

CH3

(1) (2) (3) (4)

I

cis-2-chloro-3-iodo-2-pentene. trans-2-chloro-3-iodo-2-pentene. cis-3-iodo-4-chIoro-3-pentene. trans-3-iodo-4-chloro-3-pentene.

(AIPMT 2009)

(AIPMT MAINS 2011)

8. The correct order of increasing bond length of C   H, C   O, C   C and C   C is

14. Which nomenclature is not according to IUPAC system?

(1) C  (2) C  (3) C  (4) C 

 H < C   H < C   C < C   O < C 

 O < C   C < C   C < C   H < C 

 C < C   O < C   O < C   C < C 

 O  C  H  C

(1) Br

CH2

(2) CH3

9. In Dumas method for estimation of nitrogen, 0.25 g of an organic compound gave 40 mL of nitrogen collected at 300 K temperature and 725 mm pressure. If the aqueous tension at 300 K is 25 mm, the percentage of nitrogen in the compound is (1) 18.20 (2) 16.76 (3) 15.76 (4) 17.36 (AIPMT 2015) 10. Considering the state of hybridization of carbon atoms, find out the molecule among the following which is linear?

 CH2   CH2   CH3  CH   CH   CH3  C C   CH3  CH   CH2   C CH

CH2, 1-Bromo-prop-2-ene

CH3 CH2

C

CH2

Br

(AIPMT 2011)

(1) CH3  (2) CH3  (3) CH3  (4) CH2 

CH

(3) CH3

CH

CH

CHCH3, CH3 4-Bromo-2,4-di-methylhexane

CH2

CH3, 2-Methyl-3-phenylpentane

CH2

CH2 COOH, 5-Oxohexanoic acid

CH3

(4) CH3

C

CH2

O

(AIPMT PRE 2012)

15. Which of the following acids does not exhibit optical isomerism? (1) Maleic acid (2) a-amino acids (3) Lactic acid (4) Tartaric acid (AIPMT PRE 2012) 16. Structure of the compound whose IUPAC name is 3-ethyl-2-hydroxy-4-methylhex-3-en-5-ynoic acid is OH

(AIPMT PRE 2011)

11. In Dumas’ method of estimation of nitrogen 0.35 g of an organic compound gave 55 mL of nitrogen collected at 300 K temperature and 715 mm pressure. The percentage composition of nitrogen in the compound would be (Aqueous tension at 300 K = 15 mm.)

OH COOH

(1)

COOH

(2)

OH

(3)

COOH

COOH

(4) OH

(1) 14.45 (2) 15.45 (3) 16.45 (4) 17.45 (AIPMT PRE 2011) 12. The Lassaigne’s extract is boiled with conc. HNO3 while testing for halogens. By doing so it, 3

(1) increase the concentration of NO ions. (2) decomposes Na2S and NaCN, if formed.

Chapter 22_Organic Chemistry - Some Basic Basic Principles and Techniques.indd 574

(NEET 2013) 17. Which of the following organic compounds has same hybridization as its combustion product (CO2)? (1) Ethane (2) Ethyne (3) Ethene (4) Ethanol (AIPMT 2014)

1/4/2018 5:24:54 PM

575

Organic Chemistry – Some Basic Principles and Techniques 18. The total number of p-bond electrons in the following structure is H

H3C

H

24. Which of the following biphenyls is optically active? O2N

H

(1)

CH3

H3C H

H2C

Br

(2) I

CH3

CH3

(3) (AIPMT 2015)

(4) I

CH3

19. Consider the following compounds: CH3 CH3

C

(NEET I 2016)

Ph CH

Ph

C

Ph

CH3

CH3

(I)

I

I

I

(1) 8 (2) 12 (3) 16 (4) 4

Br

(II)

25. In which of the following molecules, all atoms are coplanar? CH3

(1)

(III)

(2)

CN C

C CN

CH3

Hyperconjugation occurs in (1) II only. (2) III only. (3) I and III. (4) I only.

(3)

(4)

(AIPMT 2015) 20. In which of the following compounds, the C   Cl bond ionization shall give most stable carbonium ion? (1)

(2)

H3C

C

H3C

(3)

H

CH

Cl

(NEET II 2016)

Cl

26. In pyrrole

CH3

H O2NH2C

C H

Cl

(4) H3C

H C

5 Cl

3 2 N1 H

H 3C

(AIPMT 2015) 21. The number of structural isomers possible from the molecular formula C3H9N is (1) 2 (2) 3 (3) 4 (4) 5



the electron density is maximum on (1) 3 and 4 (2) 2 and 4 (3) 2 and 5 (4) 2 and 3 (NEET II 2016)

(RE AIPMT 2015) 22. Two possible stereostructures of CH3CHOH·COOH, which are optically active, are called (1) enantiomers. (2) mesomers. (3) diastereomers. (4) atropisomers.

23. The pair of electron in the given carbanion, CH 3C present in which of the following orbitals?

C is

(2) sp (4) sp3 (NEET I 2016)

Chapter 22_Organic Chemistry - Some Basic Basic Principles and Techniques.indd 575

27. The most suitable method of separation of 1:1 mixture of ortho and para-nitrophenols is (1) chromatography. (2) crystallization. (3) steam distillation. (4) sublimation. (NEET 2017)

(RE AIPMT 2015)

(1) sp2 (3) 2p

4

28. The correct statement regarding electrophile is (1) electrophile is negatively charged species and can form a bond by accepting a pair of electrons from another electrophile. (2) electrophiles are generally neutral species and can form a bond by accepting a pair of electrons from a nucleophile.

1/4/2018 5:24:55 PM

576

OBJECTIVE CHEMISTRY FOR NEET (3) electrophile can be either neutral or positively charged species and can form a bond by accepting a pair of electrons from a nucleophile.

(4) electrophile is a negatively charged species and can form a bond by accepting a pair of electrons from a nucleophile. (NEET 2017)

Answer Key Level I  1. (3)

 2. (2)

 3. (2)

 4. (1)

 5. (4)

 6. (1)

 7. (4)

 8. (2)

 9. (4)

10.  (4)

11.  (2)

12.  (4)

13.  (3)

14.  (2)

15.  (4)

16.  (4)

17.  (4)

18.  (3)

19.  (4)

20.  (1)

21.  (4)

22.  (1)

23.  (4)

24.  (4)

25.  (1)

26.  (2)

27.  (2)

28.  (2)

29.  (3)

30.  (1)

31.  (1)

32.  (2)

33.  (4)

34.  (1)

35.  (1)

36.  (4)

37.  (1)

38.  (4)

39.  (3)

40.  (4)

41.  (1)

42.  (2)

43.  (3)

44.  (2)

45.  (3)

46.  (4)

47.  (2)

48.  (1)

49.  (2)

50.  (2)

51.  (2)

52.  (3)

53.  (3)

54.  (3)

55.  (1)

56.  (2)

 1. (2)

 2. (4)

 3. (2)

 4. (3)

 5. (1)

 6. (3)

 7. (3)

 8. (1)

 9. (3)

10.  (4)

11.  (4)

12.  (3)

13.  (2)

14.  (1)

15.  (3)

16.  (3)

17.  (3)

18.  (3)

19.  (3)

20.  (1)

21.  (1)

22.  (1)

Level II

Previous Years’ NEET Questions  1. (4)

 2. (2)

 3. (1)

 4. (2)

 5. (3)

 6. (2)

 7. (2)

 8. (2)

 9. (2)

10.  (3)

11.  (3)

12.  (2)

13.  (2)

14.  (1)

15.  (1)

16.  (2)

17.  (2)

18.  (1)

19.  (2)

20.  (1)

21.  (3)

22.  (1)

23.  (2)

24.  (2)

25.  (4)

26.  (3)

27.  (3)

28.  (3)

Hints and Explanations Level I

2. (2) The IUPAC name of the compound is 3-ethyl-4-­ methyl hexane

1. (3)

2 1

O

Ethyl

Benzyl

Chapter 22_Organic Chemistry - Some Basic Basic Principles and Techniques.indd 576

3

4

6 5

3. (2) C2H6 (1.54 Å) > C6H6 (1.39 Å) > C2H4 (1.34 Å) > C2H2 (1.20 Å)

1/4/2018 5:24:56 PM

Organic Chemistry – Some Basic Principles and Techniques

As the s-character increases in hybridized orbitals, the tendency of overlapping becomes more, giving rise to smaller bond length and higher bond energy.

577

13. (3)

4. (1)

C7H16

C7H16

C7H16

C7H14

(III)

(IV)

(I)

Propyl

(II)

N Ethyl

sec-butyl

15. (4) F

5. (4) 4ç

4ç

3ç

m=0

F

(I)

(II) 3ç

2ç

Br

F

(III) 17. (4) OH

N +

OH

(IV)

(V)

N

(I)

8. (2) The IUPAC name of the compound is 2−methylhexane. 5 6

3

1

4



2

OH

OH

11. (2) There are total six 1s-2sp3 bonds present in ethane.

sp3-s

C

C

sp3-sp3

CH3CHCH2CH

CH2

CH3 4-Methyl-1-pentene (No chiral carbon)

C

*

CH2CH3

CH3CH2CH

C

CH3

CH3

CH3

2-Methyl-2-pentene (No chiral carbon)

20. (1) CH3F has the greatest dipole due to the highest electronegativity of F which results in greater charge generation.

alkene

H

CH

3-Methyl-1-pentene (One chiral carbon)

1° alcohol

O

H H

CH2CH3

H

3° alcohol

O

C

CH3 3-Methyl-2-pentene (No chiral carbon)

H2C O

(III)

So, (II) and (III) have positive charge on nitrogen atom.

CH3CH

Two s bonds + 1 lone pair sp2 hybridization

10. (4) —COOH is carboxylic acid which is absent in the following molecule Ketone

(II)

19. (4) The alkene that has chiral center exhibits optical isomerism.

9. (4) N

N +

H

H H H

sp3-s

Chapter 22_Organic Chemistry - Some Basic Basic Principles and Techniques.indd 577

21. (4) For constitutional isomers molecular formula should be same.

Option (1): Molecular formula is not the same.

C3H8

C3H6

1/4/2018 5:24:58 PM

578

OBJECTIVE CHEMISTRY FOR NEET

Option (2): Compounds are same.

CH2-



Option (3): Molecular formula is the same.

22. (1) Carboxylic acids (RCOOH) and esters (RCOOR′) are functional isomers. O H

O

CH3CH2

CH3

C

O

24. (4) 2,3-Dichlorobutane has a plane of symmetry that divides the molecule into halves that are mirror images of each other. CH3 Cl

H

Cl

25. (1) 1-Chloropentane. This is because it does not have a chiral or asymmetric carbon atom. 26. (2) Optical isomerism is shown by 2,3-dichlorobutane. CH3

CH -

CH3

C-

CH3

+I Effect

OH +

O H

H

Heterolysis

O + H2O -

H

Carbanion

31. (1) The inductive effect decreases with increases of distance. C4

ddd + C3

dd + C2

d+ C1

d− Cl

32. (2) Electron deficient species that attack a negative center or a carbanion (C-) are known as electrophiles. These accept electron pairs in sharing being electron pair acceptors, and thus act as Lewis acids.

Plane of symmetry

CH3

CH3

CH3

30. (1) The heterolysis takes place as follows:

CH3

23. (4) Geometrical isomerism is not observed in a molecule if either of the doubly bounded atoms has two similar groups.

H

CCl3

Resonance

O

C

CH3

CH3

CH3

H

Cl

H

Cl

Cl

H

H

Cl

Cl

H

H

Cl

33. (4) Inductive effect may be defined as the induction of polarity in an otherwise covalent bond, due to incomplete shifting of the electron pair between the two atoms that have different electronegativity. It is a permanent effect.

It is represented by the symbol →



It propagates through the carbon chain. CH3

C

CH2

CH3

O CH3

CH3

CH3

27. (2) Because in structure I and II, there is asymmetric center present, that is, carbon atom having four different substituent attached. 28. (2) Heterolytic fission of covalent bond in organic molecules results in the formation of a carbocation, in which the carbon atom bears a positive charge and has six electrons in the valence shell. R

R R

C R

X

Heterolytic fission

R

+

C +X

-

R Carbocation

29. (3) In CCl 3- , Cl-  being electronegative stabilizes the carbanion, in C6 H5CH 2- , the presence of phenyl group stabilizes it due to resonance and in case of secondary and tertiary, the order of reactivity is 2° > 3°.

Chapter 22_Organic Chemistry - Some Basic Basic Principles and Techniques.indd 578



CH3 Crepresentation CH2 CH CHof 33 inductive effect This is the incorrect

correct representation O is CH3

C

CH2

CH3

O

35. (1) Hyperconjugation involves delocalization of s electrons through overlapping of p-orbitals of a double bond with s-orbital of the adjacent single bond. 37. (1) The reactions I and II involve nucleophiles, OH-  being the nucleophile in the (I) and CN-  being the ambident nucleophile in II. 38. (4) Hyperconjugation is not possible in cation (CH3)2C   CH2+  as it does not have C   H bond directly attached to C+ carbon atom. 39. (3) In resonance, position of atom does not change, thus, compounds in option (3) are not the correct representation of resonating structures.

1/4/2018 5:25:01 PM

Organic Chemistry – Some Basic Principles and Techniques 40. (4) The resonance structure are as follows: S

S

53. (3) The structure of urea is NH2CONH2. Its molecular mass is 60. Therefore, percentage of N2 in urea is S

-



-

(II)

28 × 100 = 46.66% 60

N2 =

54. (3) The equation is (III)

(

CxH y + x + Weight

41. (1)

Moles + O

+ O

O

579

+

)

y O → xCO2 + 4 2 3.08 g 0.07

x 0.07 x 7 = ⇒ = y / 2 0.04 y 8



Therefore, we have



Hence, empirical formula is C7H8.

55. (1) % of sulphur =

+ O

y HO 2 2 0.72 g 0.04

32 × 0.6 × 100 = 41.20% 233 × 0.20

56. (2) Molecular mass of C5H10ONCl = 135.5 g 42. (2) Sublimation process can be used for separation of camphor and benzoic acid. This technique is used to separate volatile substances like camphor from nonvolatile impurities like benzoic acid. 43. (3) Solvent extraction is the process of removing an organic substance from its solution by using another immiscible solvent such as ether, chloroform, CCl4, etc.



Percentage of nitrogen =

14 × 100 = 10.3% 135.5

Level II 1. (2) Tetraethyl lead decomposed by homolytic cleavage at 140°C to give lead and ethyl radical.

.

∆ Pb(CH 2CH 3 )4 →  Pb + 4C H 2CH 3

44. (2) Vacuum distillation is also known as distillation under reduced pressure.

2. (4) Stability of radicals can be compared on the basis of electronic effect and angle strain.

46. (4) The solid with lesser solubility crystalizes out first followed by the other.



In option (1): B is more stable than A due to resonance.


H2C

=

12 × 0.2 × 100 = 18.18 44 × 0.3

CH3

CH2- > HC

C-

4. (2) The greater the percentage of s-character of a carbon, greater is its electronegativity and hence more easily it can hold electrons and greater is the stability (as lower charge density means higher stability). Electron releasing group decreases the stability of carbanion. The order is

22. (1) The percentage of carbon can be calculated as 12 × Weight of carbon dioxide %Carbon = × 100 44 × Weight of organic compound

CH2 > H3C

Relative basicity of carbanions

21. (1) Pressure of dry nitrogen = (732.7 −12.7) mm = 720 mm

CH3

Br

20. (1)



581

R

C

-

C- > sp

CH- > R3C

> R2C

sp (+l effect)

sp

2

CH2 sp3 (+l effect)

2

5. (3) The hybridization is 6

Previous Years’ NEET Questions

5

CH3

CH2 sp

4

CH

3

CH 3 2 sp

2

1

C

CH sp

6. (2) In the compound, C2 = sp; C3 = sp ; C5 = sp2 and C6 = sp3. 3

1. (4) The R configuration is shown by the compound as shown below. 2

C2H5

H

R 1

7. (2) For geometrical isomerism, the two atoms or groups attached to the double-bonded carbon should be different.

3

Cl

CH3

H

Chapter 22_Organic Chemistry - Some Basic Basic Principles and Techniques.indd 581

CH3

CH3

CH3

H

C H

4

C

cis-2-butene

C

H

C CH3

trans-2-butene

1/4/2018 5:25:08 PM

582

OBJECTIVE CHEMISTRY FOR NEET

8. (2) C   H has the lowest bond length due to the smallest size of hydrogen followed by C   C bond length. In C   O, because of greater electronegativity of oxygen atom, the bond strength of C   O is stronger than that of C   C, consequently, bond length of C   O is shorter than that of C   C. Hence the correct order is C - H < C = C < C - O < C - C.

13. (2) The IUPAC name of the given compound is

9. (2) Given: Mass of organic compound = 0.25 g; Volume of moist nitrogen (V1) = 40 mL at 725 mm; Pressure of Nitrogen (p1) = 725 – 25 = 700 mm

14. (1) In option (1), double bond should be given preference over Br group. So, the IUPAC will be 3-bromoprop-1-ene.



We know that

p1V1 p2V2 = T1 T2 p1V1 T2 × T1 p2 700 × 40 × 273 = 300 × 760 7644000 = = 33.52 mL 228000

Cl

22,400 mL of nitrogen at STP weighs = 28 g



Therefore, 33.52 mL of nitrogen at STP weighs =



15. (1) The mirror image of maleic acid is exactly the same, so it exhibits no optical isomerism. H

COOH

H

CH3

C

C

sp

sp

CH3

6





700 × 55 = 0.002 mol 760 × 1000 × 0.0821 × 300

Weight of N 2 = 28 × 0.002 = 0.0575 g

Therefore,

17. (2) The combustion reaction is 5 C 2H 2 + O2 → 2CO2 + H 2O 2

The hybridization of both CO2 and CH

Chapter 22_Organic Chemistry - Some Basic Basic Principles and Techniques.indd 582

CH is sp.

18. (1) There are four p-bonds in the given structure as shown below. H

CH3 CH3

p



H

H

p

p

p CH2

H

CH3 CH3

Each p-bond contains two electrons; therefore, total number of p-bond electrons is eight.

19. (2) When s electrons of a C   H bond are in conjugation with an adjacent p bond, then this type of conjugation is called hyperconjugation. 20. (1) The order of stability of carbonium ions are as follows:

0.0575 × 100 % of N 2 = = 16.43% 0.35

12. (2) In the presence of nitrogen or sulphur, the sodium fusion extract is boiled with concentrated nitric acid first to decompose sodium cyanide or sulphide formed during Lassaigne’s test which would otherwise interfere with silver nitrate test.

4

3-Ethyl-2-hydroxy-4-methylhex-3-en-5-ynoic acid

We know, pV = nRT. Substituting the values, we get n(of N 2 ) =

1 COOH

5

= 715 - 15 = 700mm Hg



OH 2

3

11. (3) We have, pN 2 = pressure - aqueous tension

pN 2 = 700/760 atm

COOH Maleic acid

16. (2)

0.0418 × 100 = 16.76 % 0.25 10. (3) sp hybridized carbon atoms consist of linear arrangements of atoms. The bond angle between them is 180°.

C

C

28 × 33.52 938.56 = = 0.0418 g 22400 22400

Hence, % of nitrogen in organic compound =

I

trans-2-Chloro-3-iodo-2-pentene

V2 =



C

1 2 CH3

4 5 CH2CH3

3 C

H3C H3C



+ C > CH3

H3C + CH2 > H3C

H + CH > O2NH2C

C+

Tertiary carbocation is most stable followed secondary carbocation whose stability is equivalent to benzylic carbocation. Primary carbocation is least stable.

1/4/2018 5:25:11 PM

583

Organic Chemistry – Some Basic Principles and Techniques 21. (3) There are four structural isomers possible for molecule C3H9N. The structures are CH3

CH2

CH3

CH

CH2

NH2

CH3

and

CH3

CH2

CH3

N

NH2

NH

CH3,

CH3

CH3

25. (4) The coplanarity of atoms in the given compounds can be determined on the basis of the hybridizations of the carbon atoms. In option (1), cyclohexane, all carbons are sp3 hybridized, so not planar. The compound in option (2) has carbons in sp2 and sp3 hybridization, so non­-planar. In option (3), all carbons are again sp3 hybridized. In option (4), carbon atoms of biphenyl are sp2 hybridized so coplanar.

22. (1) There is one chiral center in the molecule. Thus, two optically active enantiomers are possible. The structures are COOH

H

COOH

OH

HO

CH3

H

26. (3) In pyrrole, the four p-electrons are contributed by the carbon atoms of the pyrrole ring and two by sp2 hybridized nitrogen to complete the aromatic sextet. The possible resonance structures are as follows: 4

CH3

3

24. (2) Substituted biphenyl compounds can show optical activity even in the absence of chiral carbon because of restricted rotation about carbon-carbon single bond. Steric hindrance arises due to presence of bulkier groups at ortho-positions of benzene rings, the biphenyl system becomes non-planar and hence optically active. Br

Br

I

I

Restricted rotation around bond



Biphenyl systems in option (2) are optically active because (i) the ortho positions in both the benzene rings are occupied by groups, thus restricting the free rotation around the single bond. (ii) neither of the rings have plane of symmetry and center of symmetry.

Chapter 22_Organic Chemistry - Some Basic Basic Principles and Techniques.indd 583

-

2

5

23. (2) In the carbanion, the carbon has one σ bond, two p bonds and one lone pair. Therefore, carbon is sp hybridized.

-

N H

N+ H

N+ H

(I)

(II)

(III)

-

N+ H

(V)

-

N+ H

(IV)

From the resonance structures, we can see that the maximum electron density is at position (2) and (5) in the ring as resonating structures (III) and (IV) are more stable than (II) and (V) so are the major contributors.

27. (3) The ortho and para isomers can be separated by steam distillation. Steam distillation is used for purifying substances that are steam volatile and immiscible with water. Ortho-nitrophenol is steam volatile due to intramolecular hydrogen bonding while para-nitrophenol is less volatile due to intermolecular hydrogen bonding which cause association of molecules. 28. (3) Electrophiles are reagents that seek electrons so as to achieve a stable shell of electrons like that of a noble gas. They can be neutral (BF3 and AlCl3) or positively charged (CH +3 ) species and can form a bond by accepting a pair of electrons from a nucleophile.

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23

Aliphatic Hydrocarbons

Chapter at a Glance Alkanes 1. These are open chain compounds with general molecular formula CnH2n+2. These are also known as paraffins. 2. Isomerism: Alkanes can exhibit only chain isomerism. For example, butane, 2-methyl propane, 2,3-dimethyl hexane and 3,3-dimethyl hexane are chain isomers.

CH3 CH3CH2CH2CH3 Butane

CH3CHCH3 2-Methylpropane

2,3-Dimethylhexane

3,3-Dimethylhexane

3. Methods of Preparation (a) From hydrogenation R

Ni / C2H5OH

CH2+ H2

CH

RCH2CH3

(25°C, 50 atm)

Alkene R

Alkane Pt Solvent, pressure

CH + 2H2

C

RCH2CH3

Alkyne

Alkane

(b) From alkyl halides R

Zn HX

X

LiAlH4 [H]

RX CH3CH2Br + H2

R

H + ZnX2 RH

Pd

CH3CH3 + HBr

or Raney nickel

CH3CH2I + HI

Red P

CH3CH3 + l2

(c) Wurtz reaction R

X + 2Na + X

R

Dry ether

R + 2NaX

R

(d) From Grignard reagents R’MgX + ROH

R’H

R’MgX + RCOOH

R’H

(e) Corey–House synthesis RMgX + RLi

Chapter 23_Aliphatic hydrocarbon.indd 585

CuX R’X

R

R’

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586

OBJECTIVE CHEMISTRY FOR NEET

(f ) Decarboxylation of carboxylic acids RCOOH

NaOH CaO

RH + CO2

4. Conformations of Alkanes (a) A  ny three dimensional arrangement of atoms that results from rotation about carbon–carbon single bond are called conformations and each possible structure is called conformer. Only sigma bond can undergo this rotation. The different conformations of the molecules: • have the same molecular formula. • have the same atomic connections. • have different three-dimensional shapes. • are interchanged by the rotation of single bonds. (b) The conformers can be represented by two types of structures, Newman projections and sawhorse formula.

Newman projection formula

Sawhorse formula

(c)  There are infinite numbers of structures possible for conformers but two main cases are described with the help of conformations for ethane molecule. (i) Eclipsed conformation: In this, the hydrogen atoms on the back carbon are eclipsed by the hydrogen atoms on the front carbon. The dihedral angle of hydrogen atoms is 0°. φ = 0° H

H

H H

H H

(ii) Staggered conformation: In this, the hydrogen atoms on the two carbon atoms are farthest apart. The dihedral angle between the hydrogen atoms is 60°. H H

H φ − 180°

H

H H

(iii) The intermediate positions of hydrogens on the two carbons are called skew conformations.

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Aliphatic Hydrocarbons

587

(d) Relative stability of conformations of ethane HH H H

H H

Potential energy

Eclipsed

12 kJ mol−1

H H H

H H

H

H

H

H H H Staggered

H Staggered Rotation

5. Physical Properties (a)  For unbranched chains, at room temperature C1 to C4 of the homologous series are gases, C5 to C17 are liquids and C18 onwards are solids. (b)  Alkanes and cycloalkanes are almost totally insoluble in water due to their low polarity and inability to form intermolecular hydrogen bonding. (c)  The boiling points of the unbranched alkanes show a regular increase with increasing molecular weight. Branching of the chain, however, lowers the boiling point due to reduced surface area and strength of van der Waals’ forces operating between them. 6. Chemical Properties (a) Halogenation R

H + X2

R

X + HX (X = F, Cl, Br, I)

 he reaction follows free radical mechanism and the reaction produces a mixture of haloalkane and hydrogen T halide. The order of reactivity is F2 > Cl2 > Br2 and I2 does not react. (b) Nitration

R

H + HO

NO2

∆

R

NO2 + H2O

Nitroalkane

(c) Sulphonation R

H + HO

SO3H

SO3

R

SO3H + H2O

(d) Combustion: General reaction for alkanes is CnH2n+2 + 3n + 1 O2 2

nCO2 + (n + 1)H2O

(e) Pyrolysis: It is also known as cracking in which carbon-carbon bonds and carbon-hydrogen bonds of higher alkanes are broken by the action of heat to yield lower alkanes and alkenes. Cracking may be carried out by heating (thermal cracking) or use of catalyst (catalytic cracking). The process is used in petroleum industry. C16H34 Alkane

Chapter 23_Aliphatic hydrocarbon.indd 587

Heat Catalyst

C8H18 + C8H16 Alkane

Alkene

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588

OBJECTIVE CHEMISTRY FOR NEET

Alkenes 1. Alkenes are hydrocarbons that contain at least one carbon- carbon double bond. The common name for alkenes is olefins and the general molecular formula is CnH2n. 2. Isomerism (a)  Alkenes can exhibit structural (chain, position and cyclic or ring-chain) isomerism and geometrical isomerism. For example, Isomerism

Position isomers

Chain isomers CH3

CH3 CH2

CHCH2CH2CH3 CH2CH 1-Pentene

CHCH2CH3

CH2

2-Pentene

CCH2CH3

CH2C

Ring-chain (cyclic) isomers CH3 CHCH3

CH2

CHCHCH3

CH3

2-Methyl-1-butene 2-Methyl-2-butene 3-Methyl-1-butene

CH

CH2

Propene

Cyclopropane

(b) Geometrical isomerism: Alkenes of the type abC=Cab, abC=Cde and abC=Cad can show geometrical isomerism. The geometrical isomers are named as cis and trans. H

H

H C

C

H3C

C

H 3C

CH3

cis-2-Butene



CH3 C

H

trans-2-Butene

 or the other two types E and Z system of nomenclature is used. The two groups attached on each carbon are F ­assigned priority (based on Cahn-Ingol-prelog system). If the two groups of higher priority are on the same side of double bond, the alkane is designated Z and if on opposite side, it is named as E. For example C

Br

Cl

Cl

Cl

C

C

Br H (E)-1-Bromo-1,2-dichloroethane

C

Cl H (Z)-1-Bromo-1,2-dichloroethane

3. Stability (a) Cis isomers are generally less stable than trans isomers due to strain caused by crowding of two alkyl groups. (b)  Greater the number of attached alkyl groups, that is more highly substituted the carbon atoms of the double bond, the greater is the alkene’s stability (quantitatively determined based on heat of hydrogenation). Relative stability order is R

R C

>

C

R

R

R

Tetrasubstituted

R C

R

R

C

H C

H

R

>

C

H

R

H

Trisubstituted

C H

R

C

R C

R

H

Disubstituted

>

C H

H

R C H

H >

C H

Monosubstituted

H C

H

C H

Unsubstituted

4. Methods of Preparation (a) From alkyl halides RCH2CH2X

Chapter 23_Aliphatic hydrocarbon.indd 588

alc. KOH

RCH

CH2 + HX

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Aliphatic Hydrocarbons

589

(i)  The reaction is called dehydrohalogenation, as a hydrogen halide molecule is eliminated. These reactions are also referred to as a-eliminations as the hydrogen is eliminated from b-carbon. Also, called 1,2-elimination reactions. (ii)  The reaction follows Saytzeffs’ rule which states that in dehydrohalogenation, the formation of alkene that has greater number of alkyl groups attached to the doubly bonded carbon atom is preferred. Thus the ease of dehydrohalogenation of alkyl halides is 3° > 2° > 1°. (b) From vicinal dihalides CH2Br + Zn

CH3 CHBr

CH3CH

CH2 + ZnBr2

(c) From alcohols RCH2 CH2 OH

conc. H2SO4 Heat (443 K)

CH2 + H2O

RCH

(d) From alkynes: Different catalysts can be used for hydrogenation to give cis- or trans-alkenes. (i) Syn–addition to give cis-alkenes using Lindlar catalyst and nickel boride. R

C

C

R

H2, Pd/CaCO3 (Lindlar’s catalyst)

R

R C

Quinoline

C

H

CH3CH2C

CCH2CH3

H2/Ni2B

H

CH3CH2

3-Hexyne

CH2CH3 C

C

H H (Z)-3-Hexene (cis-3-hexene) (97%)

(ii) Anti-addition to give trans-alkenes metal dissolved in liquid NH3 or alkylamines.

CH3(CH2)2

C

C

(CH2)2CH3

4-Octyne

1. Li, C2H5NH2, −78°C 2. NH4Cl

CH3(CH2)2

H C

H

C (CH2)2CH3

(E)-4-Octene (trans-4-Octene (52%))

5. Physical Properties (a) These are similar to those of alkanes with the same carbon skeleton. (b)  These are weakly polar due to the presence of the p-electrons of the double bond. So they have higher dipole moment than corresponding alkanes. (c) Alkenes dissolve in non-polar solvents or solvents with low polarity and are very slightly soluble in water. (d)  The boiling point increases with the size and unbranched alkenes have higher boiling point than the branched chain compounds. 6. Chemical Properties (a) A  lkenes are more reactive than the corresponding alkanes due to the presence of carbon-carbon double bond and the most characteristic reaction is addition to carbon-carbon double bond. The important reactions of alkenes are summarized as follows.

Chapter 23_Aliphatic hydrocarbon.indd 589

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590

OBJECTIVE CHEMISTRY FOR NEET

X RCH X

X

CH

CH2

RCH2

CH2

Br

=

g ro ani xid c es

(X = Cl, Br)

X2

r) ,B

pe

Cl

n

R

HX

(X

Init iato r lym eriz atio n)

(Po

+

OH

e ridin , py OsO 4

H+

R

CH

CH3 (2ç Alcohol)

3 , fu min g HS 2 O

O /H 2

6

RCH2CH2OSO3H

)

CO2 + H2O

H (B 2

RCHO or RCOR

4

BH 3

OH OH (Diol)

OH

H2O

CH2

SO

/Zn O3

CH2

CH

Com bus

RCH

R

H2

OH

KMnO4/H

Nl 573 K

CH2

tion

RCH

(Primary halide) (Anti-Markovnikov’s reaction)

Or

CH2

CH

HB ro nl

y

R

CH3

(Secondary halide) (Markovnikov reaction)

(RCH2CH2)3B

RCH2CH2BH2

H2O2

R

CH2

CH2

OH

Hydroboration-oxidation

RCH2CH3

Alkynes 1. Hydrocarbon molecules containing at least one carbon-carbon triple bonds are called alkynes. These are also called ethynes or acetylenes. The general molecular formula for alkynes is CnH2n−2. 2. Isomerism: Alkynes can exhibit position (for alkynes with three or more carbon atoms), and chain isomerism (for alkynes with five or more carbon atoms). They can also exhibit ring chain isomerism. Isomerism

Position isomers CH3 CH2

C

CH CH3

But-1-yne

Chain isomers C

C

CH3

CH3

But-2-yne

CH2

C

CH2

Ring-chain (cyclic) isomers

CH CH3

Pent-1-yne CH3 CH

C

C

CH2 CH3

CH3 CH2 C

Pent-2-yne CH

C

1-Butyne

CH3

CH Cyclobutene

Methyl cyclopropene

CH3 3-Methyl but-1-yne

3. Methods of Preparation (a) Dehydrogenation of vicinal or geminal halides H

X

C

C

H

X

H or

H

H

C

C

X

X

alc. KOH

CH

CH NaNH2

C

C

(b) From alkyl substitution CH

Chapter 23_Aliphatic hydrocarbon.indd 590

CH + Na

Liq. NH3

CH

C

Na

RX

CH

C

R

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Aliphatic Hydrocarbons

591

(c) From calcium carbide CaC2 + CO

CaO + 3C

Ca(OH)2 + CH

CaC2 + H2O

CH

(d) From electrolysis: This process is called Kolbe’s synthesis. CHCOOK CHCOOK

+ 2H2O

CH

Electrolysis

CH

+ 2CO2 + H2 + KOH

(e) From tetrahalides Br Br R

C

Methanol Heat

R + 2Zn

C

CR + 2ZnBr2

RC

Br Br

(f ) From carbon and hydrogen 2C + H2

Electric arc

HC

CH

Heat

HC

CH + 6AgCl

(g) From haloforms CHCl3 + 6Ag + CHCl3

4. Physical Properties (a) The properties of alkynes are similar to those of corresponding alkanes. (b) Alkynes upto three carbon atoms are gases at room temperature. (c)  These are relatively non-polar and dissolve in non-polar solvents or solvents with low polarity. These are very slightly soluble in water. 5. Chemical Properties (a) Acidic character: The hydrogen bonded to the carbon of a terminal alkyne is much more acidic than those bonded to carbons of an alkene or alkane because the sp hybridized orbitals of carbon atoms in alkynes exhibit the highest electronegativity due to the maximum percentage (50%) of s character. H

C

C

pKa = 25

H

H H

C

C

pKa = 44

H H

H

H

H

C

C

H

H H pKa = 50

(b) Addition reactions: The triple bond is very reactive due to the presence of loosely held p-electrons and undergo electrophilic addition reactions by addition of two molecules of reagents (e.g., hydrogen, halogens, hydrogen halides, etc.).

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592

OBJECTIVE CHEMISTRY FOR NEET

CH + R¢OH

RC CH3

RCH2

KOH ∆

RCH

CH

C

O

O

R’O KO H H

RC

H2SO4

OR’ RCH

OH R’CO 2+ Hg HBr

CH

R

H

R

Br2 CCl4

C

Br

H

R

Br

Br

C

C

C

H H

H2O2

Br Br Tetrabromoalkane

Dibromoalkene

CH3

H C

Br

C Br

R

C

Br

2 /C

Cl

4

1. O 3

OCOR’

HBr

Br O s 2 n/H lysi 2. Z zono O

R

CH2

C

CH

Br

C

CH

BH R2

R

H2O

CHOH

RCH2CHO

23K

2H 2

HC N Ba( CN )2

RCH

RCH

Ni 5

CN

NaOH

BR2 CH3COOH NaOH

RCH2CHO

R

CH

CH2

(c) Linear polymerization: Ethyne undergoes polymerization to yield polyacetylene. H C

C

H

H C H

C

H C

C

H

H C H

C

H C H

C

H C H

C n

(d) Combustion  CnH2n−2 + 3n −1 O2  2 

nCO2 + (n −1)H2O

Solved Examples 1. IUPAC name of the following compound is CH2   CH   CH2   CH2   C 

 CH

(1) 1,5-hexenyne. (2) 1-hexene-5-yne. (3) 1-hexyne-5-ene. (4) 1,5-hexynene.

1



2

CH

3

CH2

4

CH2

5

C

6

CH2

The IUPAC name of the compound is 1-hexene-5-yne as double bond gets priority over triple bond.

Chapter 23_Aliphatic hydrocarbon.indd 592

is

(1) 6-ethyl-3,4-dimethylheptane. (2) 2-ethyl-4,5-dimethylheptane. (3) 3,4,6-trimethyloctane. (4) 3,5,6-trimethyloctane.

Solution (2) CH2

2. The IUPAC name for

Solution (3) The IUPAC name of the compound is 3,4,6-trimethyloctane.

1/4/2018 5:25:10 PM

Aliphatic Hydrocarbons

2 1

3

CH3

8

7

(1) CH3 4

6

5

593

CH2

CH2

CH

CH

CH2CH3

CH2CH3 3-Methyl-4-ethylheptane

3. A correct IUPAC name for the following compound is

(2) CH3

CH

CH

OH

CH3

CH3

3-Methyl-2-butanol

(3) CH3

CH2

C

CH

CH3

CH2 CH3 2-Ethyl-3-methyl-but-l-ene OH

(4) CH3

Cl

C

C

CH(CH3)2

4-Methyl-2-pentyne

(1) 4-propyl-5-chloro-3-heptanol. (2) 4-propyl-3-chloro-5-heptanol. (3) 4-(1-chloropropyl)-3-heptanol. (4) 5-chloro-4-propyl-3-heptanol.

Solution (1) According to the IUPAC rule, if two different substituents are present on the ring, they are listed in alphabetical order. Therefore, the correct IUPAC name of the below compound is 4-ethyl-3-methylheptane.

Solution (4) The IUPAC name of the compound is 5-chloro-4-­ propyl-3-heptanol.

CH3 1

2

3

4

5

6

7

CH3CH2CHCHCH2CH2CH3 CH2CH3 2 1

3

5

4

OH

6

6. How many constitutional isomers are possible for the formula C6H14?

7

Cl

(1) 2 (2) 3 (3) 4 (4) 5

4. IUPAC name of the compound given below is

Solution

CH3 H3C

(4) There are five constitutional isomers possible for C6H14.

CH3 CH3

(1) 4-ethyl-3-methyloctane. (2) 3-methyl-4-ethyloctane. (3) 2,3-diethylheptane. (4) 5-ethyl-6-methyloctane.

(1)

(2)

(3)

Solution

(1) The IUPAC name of the compound is 4-ethyl-3-­ methyloctane.

1

(5)

7. Isopentyl is the common name for which alkyl group? CH3 H

CH3

(4)

2

CH2

3

C

4

C

5

CH2

6

CH2

7

CH2

8

CH3

(1) CH3CH2CH2CH CH3

C2H5

5. IUPAC names of some compounds are given. Which one is incorrect?

Chapter 23_Aliphatic hydrocarbon.indd 593

(3) CH3CHCH2CH2 CH3

(2) CH3CH2CHCH2 CH3

(4) CH3CH2CH CH2CH3

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594

OBJECTIVE CHEMISTRY FOR NEET

Solution

Solution

(3) The isopentyl group is CH3CHCH2CH2

(2) The reaction involved is the synthesis of 2-bromohexane. Br

CH3 HBr

8. The most stable conformation of 1,2-dibromoethane is (1)

(2)

H Br Br

2-Bromohexane

H H

H

Br Br

H H

H

10. Reaction of HBr with propene in the presence of peroxide gives (1) 3-bromopropane. (2) allyl bromide. (3) n-propyl bromide. (4) isopropyl bromide.

H

(3)

(4)

H

H

H

Br

H

Br

Br

H

H

Br H

H

Solution (3) The anti-conformation does not have torsional strain and has no steric hindrance because the groups are staggered and the bromine groups are far apart. Hence, it is the most stable. The bromine groups in the gauche conformations are close enough to each other for the dispersion forces between them to be repulsive; the electron clouds of the two groups are so close that they repel each other.

Solution (3) Anti-Markovnikov addition of hydrogen bromide to alkenes occurs when peroxides are present in the reaction mixture. The reaction takes place through a free radical mechanism. CH3

H H

Br

H

(1)

H Gauche

H

H

Br

H

Br

Br

H

H

Br H Gauche

H Anti

9. Which of the following would serve as the best synthesis of 2-bromohexane? (1) CH2

CHCH2CH2CH2CH3 + HBr

Peroxides Heat

(2) CH2

CHCH2CH2CH2CH3 + HBr

Heat

(3) CH3CH

CHCH2CH2CH3 + HBr

Heat

(4) CH3CH

CHCH2CH2CH3 + HBr

Peroxides Heat

Chapter 23_Aliphatic hydrocarbon.indd 594

CH2 CH2 Br n-Propyl bromide

CH2CH3

(2)

(3)

CH2CH3

CH2CH3 CH2CH3

(4)

CH2CH3

Br Br Eclipsed

H

CH3

CH2CH3

H H H H

HBr Peroxides

11. Which one of the following alkenes has the maximum heat of hydrogenation?

Eclipsed conformation has the greatest energy of all because of the added large repulsive dispersion forces between the eclipsed bromine groups. Br

CH CH2 Propene

CH2 CH2CH3

Solution (4) All alkenes contain single double bond, so one equivalent of hydrogen can be added to each alkene. After hydrogenation, almost all products are equally stable but reactants are not equally stable. The hydrogenation reaction is exothermic, so more energy will release from the most unstable alkene.

The stability of alkenes can be explained by hyperconjugation which is directly proportional to number of alpha hydrogen atoms. The alpha hydrogen atoms in (1), (2), (3) and (4) options are 5, 8, 4 and 3, respectively. Hence, the alkene given in option (4) is the least stable for which heat of hydrogenation will be the maximum.

Quick Tip: Just look for the least stable alkene by counting the least number of alpha hydrogens. 12. How many moles of hydrogen (H2) will react with cis2,3,3-trimethylhepta-1,5-diene? (1) 0 (2) 1 (3) 2 (4) 3

1/4/2018 5:25:13 PM

595

Aliphatic Hydrocarbons Solution

Solution CH3

CH3

(3) Compound

reacts with 2 mol of CH3

hydrogen due to the presence of two double bonds.

(2) Hydrogen Hb would be abstracted first as this leads to the formation of tert-radical.

16.

Which alkene is the most stable?

(1)

(2)

(3)

(4)

13. What product is formed when 3-vinylcyclo-2-penten-1-­ol is reacted with Pd/C and H2? (1) 3-Vinylcyclopentanol (2) 3-Ethylcyclo-2-penten-1-ol (3) 4-Ethylcyclopentanol (4) 3-Ethylcyclopentanol

Solution

Solution

(1) The greater the number of attached alkyl groups (i.e., the more highly substituted the carbon atoms of the double bond), the greater is the alkene’s stability.

(4) The reaction is OH

OH

17. Which product(s) would be produced by acid-catalyzed

Pd/C; H2

dehydration of 2-methyl-2-pentanol?

3-Vinylcyclopent-2-en-1-ol

14.

3-Ethylcyclopentanol

(1)

Which of the following reactions would have the smallest energy of activation?

(1)

+ Br

+ HBr

(2)

+ Br

+ HBr

(3)

+ Br

+ HBr

(4)

+ Br

+ HBr

(3) Reaction (3) involves the smallest energy of activation as this leads to the formation of most stable 3° radical. 15. Which hydrogen would be abstracted first when monobrominating with Br2 and light?

and

(3)

(4) Solution (2)

Solution

Hc

and

(2)

H+/∆ E1

OH

+

18. Which one of the following alcohols would dehydrate most rapidly when treated with sulphuric acid? (1)

Hc

(2) OH

OH

Ha

Hd Hd

Hb

Hd

Ha

(3)

Ha

OH

He He

He

(1) Ha (2) Hb (3) Hc (4) Hd

Chapter 23_Aliphatic hydrocarbon.indd 595

(4) OH

Solution

(2) Rate of dehydration is directly proportional to the stability of the carbocation. Compound (2) will dehydrate

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596

OBJECTIVE CHEMISTRY FOR NEET most rapidly as it involves the formation of most stable tert-carbocation.

19. An alkene with the molecular formula C10H18 is treated

Solution (2) The steps involved in the acid catalyzed dehydration are as follows:

with ozone and then with zinc and acetic acid. The only product isolated from these reactions is

H+

RDS

OH

+

Alkyl shift

O

O

What is the structure of the alkene?

(1)

(2)

(3)

(4)

21.

+

H2O

+



+

OH2

Markovnikov addition of HI to 2-methyl-2-butene involves (1) (2) (3) (4)

initial attack by an iodide ion. initial attack by an iodine atom. isomerization of 2-iodo-2-methylbutene. formation of a carbocation at C-2.

Solution (4) 1

2

4

3

H

+

1

2

4

3

+

I

-

I

Solution

22. What is the major product for the following reaction?

(3) O3/Zn/H2O O

HBr

Reductive Ozonolysis

O

(2) Br

(1) Br

20.

Which mechanistic step in the acid-catalyzed dehydration of 3,3-dimethyl-2-butanol is the rate determining step?

(4)

(3)

(1)

Br

Br + H3O+

+ H2O + OH2

OH

Solution (4) H+

(2)

Rearrangement

+

+ H 2O +

+ OH2

+

-

Br

Br

(3) +

23.

+

Which of the following reactions would yield 2-pentyne?

(4) +

+ H2O

+ H3O+

(3)

(1 mol)

Br CH3OH

Br

Chapter 23_Aliphatic hydrocarbon.indd 596

I

NaNH2

(1)

(2)

NaNH2

Br

(1 mol)

(4)

HA Heat

OH

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597

Aliphatic Hydrocarbons Solution (2) NaNH2 Acid base reaction

-

C

CH3CH2Br

24. What is the major product for the following reaction sequence?

H

(2)

(3)

(4)

N H

SN2

Pent-2-yne

H

(1)

Solution (2) CH

CH

NaNH2, NH3

CH

-

C

CH3CH2CH2Br

CH

1a. NaNH2, NH3(liq.) 1b. CH3CH2CH2Br

C

CH2CH2CH3

H2 (excess) Pd/C

2. H2 (excess), Pd/C

Pentane

Practice Exercises Level I

   (I) 3-Ethyl-2,4,5-trimethyloctane (II) Isobutylcyclohexane (III) 4-Ethyl-2,2-dimethylheptane (IV) 4-Ethyl-2-methyloctane

Alkanes Nomenclature and Isomerism

(1) I, IV (2) II, III (3) I, II (4) III, IV 1. An IUPAC name for

is

5. Which of the following properties are not identical for constitutional isomers?    (I) Molecular formula (II) Molecular weight (III) Order of attachment of atoms (IV) Physical properties

(1) 3-isobutyl-2,4-dimethylhexane. (2) 3-sec-butyl-2,5-dimethylhexane. (3) 4-sec-butyl-2,5-dimethylhexane. (4) 4-isopropyl-2,5-dimethylheptane. 2. Which is the IUPAC name for the following alkane?

(1) 2-Ethyl-3-methylpentane (2) 3,4-Dimethylhexane (3) 2,3-Diethylbutane (4) 3-Methyl-4-ethylpentane

3. An IUPAC name for

(II)

(III)

(IV)

(1) I and IV (2) II and III (3) I, II, and IV (4) I, II, and III is

of the following are constitutional isomers of 4-isopropyloctane?

Chapter 23_Aliphatic hydrocarbon.indd 597

6. Which of the following molecules are constitutional isomers?

(I)

(1) 5-methyl-4-(1-methylpropyl)hexane. (2) 2-methyl-3-(1-methylpropyl)hexane. (3) 2-methyl-3-(2-methylpropyl)hexane. (4) 3-methyl-4-(1-methylethyl)heptane.

4. Which

(1) I, IV (2) II, III (3) I, II (4) III, IV

Methods of Preparation 7. Which of the following reaction would not produce saturated hydrocarbon as a major product? (1) CH3CH2Br + Na (2) CH3COCH3 (3) CH3CHO

ether

Zn(Hg)/HCl, D

N2H4/KOH

(4) CH3COOCH3

LiAlH4

1/4/2018 5:25:27 PM

598

OBJECTIVE CHEMISTRY FOR NEET

Conformations

(1) the low energy of activation for the chain-propagating steps. (2) the large negative overall DH ° for the reaction. (3) the energy of activation of the chain-initiating step so fluorine is quite unreactive. (4) (1) and (2).

8. Which conformation of 2-methylbutane is least stable? CH3H

(1) H

CH3

CH3

(2)

H CH3 CH3

(3) H

CH3

H

H CH3

H

14.

CH3 CH3 CH3

CH3

(4)

H

H

H

CH3 H

H

(1) Tertiary butyl chloride (2) Neopentane (3) Isohexane (4) Neohexane 15. The major product obtained in the photocatalyzed bromination of 2-methylbutane is

Physical Properties 9. Which compound has the lowest boiling point? (2) CH3

(1) CH3CH2CH2CH2CH3

C

CHCH2CH3

CH3

CH3

(3) CH3

Which branched chain isomer of the hydrocarbon with molecular mass 72 u gives only one isomer of monosubstituted alkyl halide?

CH3

(1) l-bromo-2-methylbutane (2) l-bromo-3-methylbutane (3) 2-bromo-3-methylbutane (4) 2-bromo-2-methylbutane 16. Which of the following reactions should have the smallest energy of activation?

(4)

CH3

(1) CH4 + Cl

CH3 + HCl

(2) CH3CH3 + Cl

CH3CH2 + HCl

10. Which compound has the highest boiling point? (1) CH3CH2CH2CH2CH2CH2CH3

(2) CH3

C

CHCH2CH2CH3

CH3

CH2CH3

(3) CH3

(3)

CH3

(4)

(4)

CH2CH3

11. Which are the stoichiometric coefficients that complete the following equation? + xO2

yCO2 + zH2O

1 (1) x = 9 , y = 6, z = 7 (2) x = 18, y = 12, z = 14 2 (3) x = 1, y = 1, z = 2 (4) x = 6 , y = 6, z = 4 12. Which hydrogen would be abstracted first when monobrominating with Br2 and light? He

Hd

CH3 CH3CHCH3

+ HCl

CH3 CH3CCH3

+ Cl

CH3

+ HCl

CH3

(1) (2) (3) (4)

A new free radical is formed. The process is endothermic. Eact = 0. DH° is positive.

Alkenes Nomenclature and Structure 18. What is the IUPAC name for the following structure? CH3 CH3

C

CH

CH2

CH3 Ha Ha

Hc Hc Hb Hb

(1) Ha (2) Hb (3) Hc (4) Hd 13. Free radical fluorination of an alkane is not typically conducted because of fluorine’s high reactivity due to

Chapter 23_Aliphatic hydrocarbon.indd 598

CH3CHCH2

+ Cl

17. Which is true for a chain-terminating step?

Chemical Reactions

He He

CH3CHCH3

(1) 3,3-Dimethyl-4-pentene (2) 3-Methyl-3-ethyl-1-butene (3) Isopropylpentene (4) 3,3-Dimethyl-1-butene

19. The IUPAC name of hydrocarbon with seven carbon atoms containing a neopentyl and a vinyl group is

1/4/2018 5:25:31 PM

599

Aliphatic Hydrocarbons (1) 2,2-dimethyl-4-pentene. (2) 4,4-dimethylpentene. (3) isopropyl-2-butene. (4) 2,2-dimethyl-3-pentene.

(I)

(II)

20. Name the following compound: HO

(III)

(IV)

(1) II, IV, III, I (2) IV, II, III, I (3) III, IV, II, I (4) I, II, III, IV

(1) (E)-4-Phenyl-4-methylbut-3-en-1-ol (2) (E)-4-Phenylpent-3-en-1-ol (3) (Z)-4-Phenylpent-3-en-1-ol (4) (Z)-4-Phenyl-4-methylbut-3-en-1-ol

Methods of Preparation 26. Which alcohol would be most easily dehydrated?

21. Which

of the following cycloalkenes shows cis-transisomerism?

(1)

(2)

(1) Cyclobutene (2) Cyclopentene (3) Cyclohexene (4) Cyclooctene

OH

OH

22. Which is the correct structure for vinylcyclobutane? (1)

(2) (3)

(4) OH

CH2

(3)

OH

(4)

Chemical Reactions CH2

CH

CH2

23. Which of the following alkenes show cis-trans-isomerism? (I) 2,5-Dimethyl-3-hexene (II) 2,4-Dimethyl-2-pentene (III) 2,4-Dimethyl-1-pentene (IV) 4,5-Dimethyl-2-hexene (1) I and II (2) II and III (3) I and IV (4) II and IV HO Cl

I

Br H2N

Br

(I)

(II)

Cl

O

IV

II

28. The oxymercuration-demercuration reaction produces a  ___  addition. (1) markovnikov (2) anti-Markovnikov (3) syn addition (4) anti-addition

Cl

(III)

III

I

(1) I (2) II (3) III (4) IV

24. Which of the following alkenes have E configuration? I

27. Using Markovnikov’s rule, predict the position of the Cl atom in the major product from the reaction of 1-­methylcyclohexene with HCl.

(IV)

(1) I and II (2) II and IV (3) II and III (4) I and IV

29. Arrange these carbocations in order of increasing stability (least to most). + +

+

Physical Properties 25. Arrange the following molecules in order of increasing boiling point (lowest first).

Chapter 23_Aliphatic hydrocarbon.indd 599

(I)

(II)

(III)

+

(IV)

1/4/2018 5:25:34 PM

600

OBJECTIVE CHEMISTRY FOR NEET (1) I, II, III, IV (2) III, II, I, IV (3) II, IV, I, III (4) III, I, II, IV

(1) (CH3)3C   CH2 > H3CHC   CH2 > H2C   CH2 (2) H3CHC   CH2 > (CH3)3C   CH2 > H2C   CH2

30. Which is the intermediate formed in the reaction of pro-

(3) H2C   CH2 > (CH3)3C   CH2 > H3CHC   CH2

pene with HBr? Br

(1)

(2)

+

(4)

+

Br

+

(3)

+

(4) H2C   CH2 > H3CHC   CH2 > (CH3)3C   CH2 38. An unsaturated hydrocarbon upon ozonolysis gives one mole each of formaldehyde, acetaldehyde and methylglyoxal (CH3COCHO). The structure of the hydrocarbon is (1) CH2   CH   CH2   CH   CH (2) CH2   CH   C(CH3)   CH   CH3 (3) (CH3)2C   CH   CH3 (4) CH3   CH   C(CH3)   CH3

31. Carbocations are frequent intermediates in acidic reactions of alkenes, alcohols, etc. What do carbocations usually do? They may (1) (2) (3) (4)

rearrange to a more stable carbocation. lose a proton to form an alkene. combine with a nucleophile. do all of the above.

39. Which of the following will react with sodium metal?

(1) Ethene (2) Propyne (3) But-2-yne (4) Ethane

32. The hydroboration-oxidation reaction produces a  ___  40. What is the product of the following reaction? addition. 1. O3, CH2, Cl2, -78°C

(1) Markovnikov (2) anti-Markovnikov (3) syn addition (4) (2) and (3) 33. HBr reacts with CH2   CH   OCH3 under anhydrous conditions at room temperature to give (1) CH3CHO and CH3Br (2) BrCH2CHO and CH3 (3) BrCH2   CH2   OCH3 (4) H3C   CHBr   OCH3

2. Zn/AcOH

+

(1)

34. Which one of the following alkenes will react faster with

O OH

H

+ O

O

(3)

R

R

R

R

R

R

R

H

(2)

(4)

R

R

H

H

R

H

R

H

[R = Alkyl substituent]

35. In the hydroboration-oxidation reaction of propene with diborane, H2O2 and NaOH, the organic compound formed is (1) CH3CH2OH (2) CH3CHOHCH3 (3) CH3CH2CH2OH (4) (CH3)3COH

O

one of the steps is endothermic in HCl and HI. both HCl and HI are strong acids. HCl is oxidizing and HI is reducing. all the steps are exothermic in HCl and HI.

37. The correct order of reactivity of the following alkenes to acid-catalyzed hydration is

Chapter 23_Aliphatic hydrocarbon.indd 600

O

41. What is the product of the following reaction? Br2

Br

Br

(1)

(2) Br

Br Br

OH

(3)

(4) Br

Br

42. Ozonolysis of compound Z yields the products shown below. What is the structure of Z? O

36. In the presence of peroxide, HCl and HI do not give antiMarkovnikov’s rule addition to alkenes because (1) (2) (3) (4)

OH

+

(4)

H2 under catalytic hydrogenation conditions? (1)

+ CO2

(2)

O

O

(3)

H

H

H

Z

2. Zn, HOAc

O

O

2HCH + CH3 CCH2CH

(2)

(1)

(3)

1. O3

O

O

(4)

H

1/4/2018 5:25:35 PM

Aliphatic Hydrocarbons 43. Which alkene would you expect to be most reactive toward acid-catalyzed hydration?

50. The major result of treating 1-butyne with 6 M aqueous NaOH would be

(1) 1 Pentene (2) trans-2-Pentene (3) cis-2-Pentene (4) 2-Methyl-1butene

(1) (2) (3) (4)

44. What is the major product of the following reaction sequence?



(3)

(2)

(1) CH3CH2CH2COOH (2) CH3CH2COOH (3) CH3CH2COOH + CO2 (4) CH3CH2COOH + HCOOH

Br

(4)

Br

52. What product is formed when 2,3,3-trimethylhept-1-en6-yne is reacted with Pd/C and H2?

Br

(1) 2,3,3-Trimethylheptan-6-yne (2) 2,3,3-Trimethylheptan-1-ene (3) 5,5,6-Trimethylheptan-6-yne (4) 2,3,3-Trimethylheptane

Alkynes Chemical Reactions

45. Which statement is/are true about acetylide anions? (1) They do not alkylate with secondary alkyl halides. (2) Primary alkyl halides are best suited for alkylation. (3) In the presence of tertiary alkyl halides, the acetylide anion acts as base to give an elimination product. (4) All of the statements are true. 46. Order the four compounds according to increasing acidity: H CH3

H

(I)

(II)

the production of the sodium acetylide. the production of an alkene. the production of an enol. nothing, as the alkyne would not react.

51. l-Butyne reacts with cold alkaline KMnO4 to produce

HBr

(1) Br

NH2

(III)

(IV)

Level II Alkanes Chemical Reactions 1. How many chiral compounds are possible on monochlorination of 2-methylbutane? (1) 8 (2) 2 (3) 4 (4) 6

H

(1) I < II < III < IV (2) II < I < III < IV (3) II < I < IV < III (4) II < IV < III < I

2. The reaction of 2,2-dimethylbutane with chlorine would yield how many monochloro derivatives? (not including stereoisomers)

47. Which of the following reactions will yield 2,2-dibromopropane? (1) CH3   C   CH + 2HBr (2) CH3CH   CHBr + HBr (3) CH   CH + 2HBr

(1) 1 (2) 2 (3) 3 (4) 4

3.

What is the product for the following reaction sequence? 1. Br2, hv 2. H2O

(4) CH3   CH   CH2 + HBr OH

48. The hydrocarbon which can react with sodium in liquid ammonia is

(1)

(1) CH3CH2CH2C   CCH2CH2CH3 (2) CH3CH2C   CH (3) CH3CH   CHCH3 (4) CH3CH2C   CCH2CH3 49. The treatment of CH3MgX with CH3C   C   H produces (1) CH3   CH   CH2 (2) CH3C   C   CH3 (3) CH3CH   CHCH3 (4) CH4

Chapter 23_Aliphatic hydrocarbon.indd 601

601

(2) OH

(3)

(4)

4. For the following reaction C5H12 + Cl2 (A)



hn

C5H11Cl (B)

Na/ether

(C)

the structures of (A), (B) and (C) are

1/4/2018 5:25:38 PM

602

OBJECTIVE CHEMISTRY FOR NEET (1)

CH3 CH3

C

CH3

CH3

CH3

C

CH3

CH3

CH3

CH3 CH2

C

8. The gas liberated by the electrolysis of dipotassium succinate solution is

CH3

CH2

C

CH3

CH2Cl

(1) ethane. (2) ethyne. (3) ethene. (4) propene. 9. Which of the following conditions/reagents would you employ to obtain the best yields in the following reaction?

CH3

CH3 Br

(2)

CH3 CH3

CH

CH3 CH2CH3

CH3

C Cl

CH3 CH3 CH3

CH2

C

C

(1) NaOH(aq), heat (2) C2H5ONa/C2H5OH, heat (3) (CH3)3COK/(CH3)3COH, heat (4) H2SO4, heat

CH2CH3

CH2CH3

10. What is the reagent needed for the following reaction?

CH3 CH3

OH

(3) Both (1) and (2) (4) None of these 5. Mono-bromination of the alkane, Br2 with light) would be

(1) (2) (3) (4)

, (using

Br

(1)

(2)

Br

Concentrated sulphuric acid and high heat. 85% phosphoric acid and high heat. Dilute sulphuric acid and heat. Both (1) and (2).

Chemical Reactions 11. In the following sequence of reactions, the alkene forms the compound B.

(3)

(4) Br Br

6. Which reaction would you expect to have the smallest energy of activation?







(1) CH3· + CH3· (2) CH4 + F· (3) CH4 + I· (4) CH4 + Br·

    ∆H°(kJ mol ) -1

CH3CH3 −378 CH3· + HF −130 CH3· + HI +142 CH3· + HBr +104

CH3 CH3

CH3

(I)

C2H5CH2 C

CH2

CH3

(II)

C2H5CH2 C

Chapter 23_Aliphatic hydrocarbon.indd 602

OCH3

CH3

(III)

(1) I and II (2) I and III (3) III only (4) All of these



The compound B is

CHCH 3

O3

A

H2 O Zn

B

(1) CH3CH2CHO (2) CH3COCH3 (3) CH3CH2COCH3 (4) CH3CHO 12. One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44 u. The alkene is

13. Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the presence of

7. 2-Chloro-2-methylpentane on reaction with sodium methoxide in methanol yields:

C



(1) ethane. (2) propene. (3) 1-butene. (4) 2-butene.

Alkenes Methods of Preparation

C2H5CH2

CH 3CH



(1) (2) (3) (4)

a vinyl group. an isopropyl group. an acetylenic triple bond. two ethylenic double bonds.

14. 1-Butene reacts with the following reagents given in the options. Out of these, which will react through free radical mechanism? (1) HCl + peroxide (2) (i) BH3, (ii) H2O2, OH− (3) HBr (4) HBr + peroxide

1/4/2018 5:25:39 PM

Aliphatic Hydrocarbons 15. 3-Phenylpropene on reaction with HBr gives (as a major product)

603

19. What is the major product for the following reaction?

(1) C6H5CH2CH(Br)CH3 (2) C6H5CH(Br)CH2CH3 (3) C6H5CH2CH2CH2Br. (4) C6H5CH(Br)CH   CH2

HCl

(2)

(1)

16. An unknown compound, A, has the molecular formula C7H12. On oxidation with hot aqueous potassium permanganate, A yields CH3CH2COOH and (CH3)2CHCOOH. Which of the following structures best represents A?

Cl



(1)

(2)

(3)

(4)

Cl

(3)

Cl

(4) Cl

Alkynes Methods of Preparation

17. Compound C has the molecular formula C7H12. On catalytic hydrogenation, 1 mol of C absorbs 1 mol of hydrogen and yields a compound with the molecular formula C7H14. On ozonolysis and subsequent treatment with zinc and acetic acid, C yields only:

20. What is the major product of the following reaction sequence? 1. Br2, CCl4 2a. NaNH2 (excess), NH3 + 2b. H3O

O

(1)

(2)

(3)

(4)

O



The structure of C is

Br

(1)

(2)

Chemical Reactions 21. 2-Hexyne gives trans-2-hexene on treatment with

(3)

(1) Pt/H2 (2) Li/NH3 (3) Pd/BaSO4 (4) LiAlH4

(4)

18. An alkene adds hydrogen in the presence of a catalyst

22. The reagent needed for converting Ph

to give 3,4-dimethylhexane. Ozonolysis of the alkene followed by treatment with zinc and acetic acid gives a single organic product. The structure of the alkene is

CH3

CCH3

(1)

(cis or trans)

CH3

H

Chapter 23_Aliphatic hydrocarbon.indd 603

Ph

Cl

H3C

(4) H3C

Cl

OH Cl Cl

Cl

CH2

CH3

C

Cl

O H3C

CH3

(2)

O Cl2HC

(3)

CCH2CHCH2 CH3

(4) CH3CH2 CCH CH2 CH3

H C

(1) cat. Hydrogenation (2) H2/Lindlar cat. (3) Li/NH3 (4) LiAlH4

CH2 CH3

(3) CH2

Ph

23. Addition of hypochlorous acid to propyne gives

CH3

(2) CH3CH2C

C

is

CHCH2CH3 (cis or trans)

C

C

H



CH3

(1) CH3 CH

Ph

OH

24. Products of the following reaction are



CH3C

CCH2CH3

1. O3 2. Hydrolysis

1/4/2018 5:25:41 PM

604

OBJECTIVE CHEMISTRY FOR NEET (1) CH3COOH + CH3COCH3 (2) CH3COOH + CH3CH2COOH (3) CH3CHO + CH3CH2CHO (4) CH3COOH + CO2

(3) hydrolysis. (4) oxidation. 

(AIPMT PRE 2010)

6. In the following, the most stable conformation of n-butane is

Previous Years’ NEET Questions

CH3

(1)

1. Predict the product C obtained in the following reaction of butyne-1. CH3

CH2

C

CH + HCl

HI

B

C

H

H

H

H

l

(1) CH3CH2

C

CH2

(2) CH3

CH3

CH Cl

l C

CH2

CH2CH2l

H

(4) CH3

CH3 H

(3)

CH2



CH

H H

CH2Cl

(AIPMT 2007)

2. Which of the compounds with molecular formula C5H10 yields acetone on ozonolysis?

 CH2 + HBr

(1)

CH

CH3

Br

CH3

(2) CH3

C

CH2CH3

CH2

CH2Br

CH3

Br

(3) CH3

CH

(4)

CH3

CH

Br

CH3



CH3

(AIPMT 2008)

4. The IUPAC name of the compound having the formula CH   C   CH   CH2 is (1) 1-butene-3-yne. (2) 3-butene-1-yne. (3) 1-butyn-3-ene. (4) but-1-yne-3-ene. 

(AIPMT 2009)

5. Liquid hydrocarbons can be converted to a mixture of gaseous hydrocarbons by (1) cracking. (2) distillation under reduced pressure.

Chapter 23_Aliphatic hydrocarbon.indd 604

(2)

OH H

HO H

HH

(4)

OH

HH OH

H

OH

H

H

H

H

H

CH

CH3

H H

OH OH

(3)

CH

H

H3C H

H H

A

A (predominantly) is (1) CH3

CH3

7. Which of the following conformers for ethylene glycol is most stable?

CH3



H

(AIPMT PRE 2010)

(AIPMT 2007) CH

CH3



(1) 2-Methyl-1-butene (2) 2-Methyl-2-butene (3) 3-Methyl-l-butene (4) Cyclopentane

CH

HH

(4)

l

Cl

3. H3C

H H

CH3

Cl

(3) CH3

CH3 CH3

(2)

H

H OH



(AIPMT MAINS 2010)

8. The IUPAC name of the compound

CH3CH   CHC 

 CH is

(1) Pent-3-en-1-yne (2) Pent-2-en-4-yne (3) Pent-1-yn-3-ene (4) Pent-4-yn-2-ene (AIPMT MAINS 2010) 9. In the following reactions, CH3

(I) CH3

CH

CH

CH3

OH

(II) A

HBr, dark in absence of peroxide

H+/heat

B A + (Minor (Major product) product)

D C + (Minor (Major product) product)

1/4/2018 5:25:43 PM

605

Aliphatic Hydrocarbons

13. The structure of isobutyl group in an organic compound is

The major product (A) and (C) are respectively CH3

(1)

CH2

C

CH3 CH3 and CH3

CH2

C

(1) CH2

(2)

CH2

C

(2) CH3

CH3 CH3 and CH2

CH2

CH2

CH3

C

CH3 and CH3

C

CH2

C

CH

CH3 and CH3

CH

CH2

C CH3

CH3



Br CH3

CH3

(4)

(4) CH3 CH2

CH2

CH3

CH3

CH3 CH

CH2

CH2

CH3

Br

(3)

CH2

CH

(3) CH3

CH

CH3

CH

CH3

CH3

Br CH3

CH3

(NEET 2013)

14. Given: CH

CH3

H3C

CH3 H3C

CH2 H2C

CH2

Br



(AIPMT PRE 2011)

10. The correct IUPAC name of the compound is

CH3

CH3

CH2

(I)

(II)

(III)

The enthalpy of the hydrogenation of these compounds will be in the order as: (1) III > II > I (2) II > III > I (3) II > I > III (4) I > II > III

(1) (2) (3) (4)



3-(1-ethyl propyl) hex-1-ene. 4-ethyl-3-propyl hex-1-ene. 3-ethyl-4-ethenyl heptane. 3-ethyl-4-propyl hex-5-ene.

(AIPMT 2015)

15. The reaction of C6H5CH   CHCH3 with HBr produces (1) C6H5CH2 CH CH3

(2) C6H5CH2CH2CH2Br

Br

(AIPMT PRE 2011) (3)

11. In the following reaction

CH

(4) C6H5CH CH2 CH3

CHCH3

Br

CH3 H3C

C

CH

CH2

H2O/H+

CH3



A B + Major product Minor product



The major product is (1) H3C

C

CH

CH3

OH CH3

(3)

C

CH

CH3 OH



CH2

C

OH

CH3

(4)

CH3 H3C

CH3

(2)

CH3

CH3

Br

CH2

CH3

(AIPMT 2015)

16. A single compound of the structure is obtainable from ozonolysis of which of the following cyclic compounds? CH3 OHC

CH3 H3C

C CH3

CH2

CH2

(1)

OH

(AIPMT PRE 2012) (3)

Chapter 23_Aliphatic hydrocarbon.indd 605

(2)

O

H3C

CH3

CH3

(4) H C 3 CH3

(1) NaNH2 (2) HCl (3) O2 (4) Br2 (AIPMT MAINS 2012)

C C C H C H2 H2

H3C

12. Which of the following reagents will be able to distinguish between 1-butyne and 2-butyne?



H3C

CH3

CH3



(AIPMT 2015)

1/4/2018 5:25:44 PM

606

OBJECTIVE CHEMISTRY FOR NEET

17. In the reaction with HCl, an alkene reacts in accordance with the Markovnikov’s rule, to give a product 1-chloro1-methylcyclohexane. The possible alkene is

(3) The eclipsed conformation of ethane is more stable than staggered conformation even though the eclipsed conformation has torsional strain. (4) The staggered conformation of ethane is more stable than eclipsed conformation, because staggered conformation has no torsional strain.

CH3

CH2

(2)

(1)

 CH3

(NEET I 2016)

21. In the reaction H

(3)

(4)

Both (1) and (2)





18. 2,3-Dimethyl-2-butene can be prepared by heating which of the following compounds with a strong acid? CH

CH2

CH2

CH

CH2

(3) (CH3)2CH

CH

CH

CH2

CH

19. Which of the following is not the product of dehydration

(1)

(3)

X

1. NaNH2/liq. NH3 2. CH3CH2Br

(NEET I 2016)

(NEET II 2016)

23. With respect to the conformers of ethane, which of the following statements is true? (1) Bond angle changes but bond length remains same. (2) Both bond angle and bond length change. (3) Both bond angles and bond length remains same. (4) Bond angle remains same but bond length changes.

?

(2)



(NEET 2017)

24. Predict the correct intermediate and product in the following reaction:

(4)

H3C

C

CH

H2O, H2SO4 HgSO4

Intermediate (A)

Product (B)

(1)

A: H3C

C

CH2

B: H3C

C

20. The correct statement regarding the comparison of staggered and eclipsed conformations of ethane is

(2)

A: H3C

OH C CH3

B: H3C

SO4 C CH

(1) The staggered conformation of ethane is less stable than eclipsed conformation, because staggered conformation has torsional strain. (2) The eclipsed conformation of ethane is more stable than staggered conformation, because eclipsed conformation has no torsional strain.

(3)

A: H3C

O C

CH2

B: H3C

C

(4)

A: H3C

CH2

B: H3C



(RE AIPMT 2015)

Chapter 23_Aliphatic hydrocarbon.indd 606

Y,

X and Y are

 (RE AIPMT 2015)

OH

2. CH3CH2Br

(1) C2H2 (2) C4H10 (3) C2H4 (4) C3H6

CH2



of

1. NaNH2/liq. NH3

22. The compound that will react most readily with gaseous bromine has the formula

CH3

(4) (CH3)3C

H



CH3

(2) (CH3)2CH

C

(1) X = 1-Butyne; Y = 3-Hexyne (2) X = 2-Butyne; Y = 3-Hexyne (3) X = 2-Butyne; Y = 2-Hexyne (4) X = 1-Butyne; Y = 2-Hexyne

(RE AIPMT 2015)

(1) (CH3)2C

C

OH C SO4



CH2

CH3

O C

CH3

O

(NEET 2017)

1/4/2018 5:25:46 PM

607

Aliphatic Hydrocarbons

Answer Key Level I 1.  (4)

 2. (2)

 3. (4)

 4. (4)

 5. (4)

 6. (4)

 7. (4)

 8. (4)

 9. (3)

10.  (1)

11.  (1)

12.  (4)

13.  (4)

14.  (2)

15.  (4)

16.  (4)

17.  (3)

18.  (4)

19.  (2)

20.  (2)

21.  (4)

22.  (4)

23.  (3)

24.  (3)

25.  (3)

26.  (2)

27.  (3)

28.  (1)

29.  (2)

30.  (4)

31.  (4)

32.  (4)

33.  (4)

34.  (2)

35.  (3)

36.  (1)

37.  (1)

38.  (2)

39.  (2)

40.  (3)

41.  (4)

42.  (2)

43.  (4)

44.  (3)

45.  (4)

46.  (4)

47.  (1)

48.  (2)

49.  (4)

50.  (4)

51.  (3)

52.  (4)

Level II 1.  (2)

 2. (3)

 3. (1)

 4. (1)

 5. (3)

 6. (1)

 7. (4)

 8. (3)

 9. (3)

10.  (2)

11.  (4)

12.  (4)

13.  (1)

14.  (4)

15.  (2)

16.  (2)

17.  (3)

18.  (2)

19.  (4)

20.  (2)

21.  (2)

22.  (3)

23.  (3)

24.  (2)

Previous Years’ NEET Questions 1.  (1)

 2. (2)

 3. (3)

 4. (1)

 5. (1)

 6. (1)

 7. (3)

 8. (1)

 9. (3)

10.  (2)

11.  (1)

12.  (1)

13.  (1)

14.  (1)

15.  (4)

16.  (4)

17.  (3)

18.  (4)

19.  (4)

20.  (4)

21.  (1)

22.  (4)

23.  (3)

24.  (3)

Hints and Explanations Level I

15. (4) The reaction is

7. (4) Major products of the following reactions are as follows: ether

CH3CH2Br + Na

Zn(Hg)/HCl, D

CH3COCH3 CH3CHO

N2H4/KOH LiAlH4

CH3COOCH3

CH2

CH3

CH3CH2CH3

CH3CH3 CH3CH2OH

CH3

CH3 CH3

CH3 Neopentane All hydrogens are equivalent

Cl2/hv

CH3

C

CH2Cl

CH3 Monosubstituted alkyl halide

Molecular mass = C5H12 = 5 × 12 + 12 × 1 = 72 u

Chapter 23_Aliphatic hydrocarbon.indd 607

CH

Br2

CH3

hv

C

CH2

CH3

Br 2-Bromo-2-methylbutane (major)

14. (2) The reaction involved is C

CH3

CH3

CH3CH2CH2CH3

12. (4) Hydrogen Hd would be abstracted first as it will lead to the formation of 3° radical.

CH3

CH3

16. (4) Reaction of option (4) has smallest energy of activation as it involves the formation of most stable 3° radical as the product. 17. (3) In the chain terminating step, all radicals combine with each other to form product. Hence, activation energy for this step is zero. 19. (2) The hydrocarbon is 4,4-dimethylpentene. H3C H3C

CH3

26. (2) Rate of dehydration is directly proportional to the stability of the carbocation formed. As compound in

1/4/2018 5:25:48 PM

608

OBJECTIVE CHEMISTRY FOR NEET option (2) involves the formation of 3° carbocation, hence would be dehydrated most easily.

40. (3) Ozonolysis of a trisubstituted alkene produces an aldehyde and a ketone.

H2C

CH

HBr

OCH3

+

H2C

CH

1. O3, CH2Cl2

Zn/AcOH

O O (Ozonide)

−78°C

OCH3

Br

H

H

O

33. (4) The reaction follows Markonikov’s rule.

O + O

Br H3C

CH

H OCH3

34. (2) Greater the number of alkyl groups attached to the doubly bonded carbon atoms, the more stable is the alkR

R

H

H

ene. Therefore, alkene

41. (4) When a dihalide is added to a double bond, the double bond is replaced by the two halogens, which add anti to each other. Br

reacts faster with H2 in the

Br2

presence of a catalyst as it is least stable and less sterically hindered.

Br O

35. (3) CH3



CH

1. B2H6

CH2

CH3

2. H2O2/OH -

CH2

CH2

1. O3

42. (2)

OH

44.

2. Zn, HOAc

(3)

The product formed is opposite to Markonikov’s regioselectivity.

H+

+

36. (1) Peroxides do not give anti-Markonikov’s addition because one of the steps is endothermic in HCl and HI. 37. (1) Acid catalyzed hydration is the addition of water molecule in the presence of acid. It occurs through carbocation formation in the first step which is also slow step, so, more stable carbocation formation indicates faster reaction. H2C

CH2

H3CHC

CH2

(H3C)2 C

H

H

47. (1) The reaction is CH3

C

CH + HBr

(Markovnikov’s addition)

CH2

H

C(CH3)

Br CH3

(CH3)2 CCH3 (3° carbocation, most stable)

CH

CH3

O3/Zn

H2C

O + O

CH3

48. (2) Terminal alkynes are acidic hence they will react with Na in presence of liquid ammonia. CH

C

O

Methyl glyoxal CH

CH3

CH3

+ H3 C C C Na + 1 H2 2 Sodium propynide

CH2

C

CH + Na

NH 3

CH3

CH2

−

C Na+

C

49. (4) Grignard reagents react with active H-compounds and liberate alkanes them. CH3

MgX + CH3

C

CH

CH4↑ + CH3

↑

39. (2) Terminal alkynes react with sodium metal to form sodium alkynides due to the presence of acidic hydrogen. −

Na

C

Br 2,2-Dibromopropane

Acetaldehyde

H

CH2

Br

+

+

+ O

Chapter 23_Aliphatic hydrocarbon.indd 608

C

+

Formaldehyde

C C Propyne

CH3

CH3 C HCH3 (2° carbocation, moderately stable)

CH3

H3C

Br −

H2 CCH3 (1° carbocation, least stable) +

Thus, the order of reactivity is (CH3)3C   CH2 > H3CHC   CH2 > H2C   CH2

CH

+ Br

(Markovnikov’s HBr addition)

38. (2) The reaction is H2C

Alkyl shift

+

+

CHO

2HCHO +

C

−

C MgX

+

Thus, CH3MgX acts as a base to abstract acidic hydrogen.

50. (4) The terminal alkynes are weakly acidic, the pKa value is around 25. NaOH is not as much as basic that it can deprotonate alkynes. Hence, there are no major changes occurring in this reaction.

1/4/2018 5:25:52 PM

609

Aliphatic Hydrocarbons 51. (3) The reaction is H 3C

CH2

4. (1) C

CH

H 3C

CH3

cold KMnO4(alk.)

CH2

CH3

COOH + CO2

C

CH3 Cl CH3 hv2

CH3 Neopentane (A)

52. (4)

CH3

C

CH2CI

CH3 Neopentyl chloride (B)

Pd/C; H2

Na/ether Wurtz reaction

CH3

2,3,3, Trimethylhept-1-en-6-yne

CH3

2,3,3-Trimethylheptane

C

CH3 CH2

CH2

CH3

Level II

C

CH3

CH3 (C)

5. (3)

1. (2) The reaction involved is 1

CH3

2

CH

3

4

CH2

CH3

Br2/hv

Br

CH3

7. (4)

2-Methylbutane

Cl

Cl2/hv or heat

CH3 Cl CH3

CH

CH2

CH2Cl + CH3

(Achiral)

(Achiral)

CH2

CH3 + CH3

(Chiral)

* CH

CH3ONa CH3OH

CH

+ CH2

CH3

C

CH

CH3 C CH2

CH2

CH3 (I)

CH2

CH3 (II)

CH2

C

CH2

CH3 (III)

CH3



2. (3)

In case of products I and II, b -elimination takes place as methoxide acts as a strong base in presence of methanol. Product (III) is also possible because the reaction is carried out between haloalkane and sodium methoxide (Williamson’s synthesis).

8. (3)

Cl

+ Cl

CH2

COOK

CH2

COO-

CH2

COOK

CH2

COO-

+ 2K+

H 2O  OH − + H +

Cl

3. (1) Br Br2/hv

H2O Solvolysis

+



1 At the cathode: H + + e − → H 2 2



At the anode: O

Rearrange

CH2

−

COO

CH2

C

O

CH2

CH2

COO−

CH2

C

O

CH2

O H2O

OH

Chapter 23_Aliphatic hydrocarbon.indd 609

CH3

+ CH3

Out of all the four isomers formed only two compounds are optically active.

+

CH3

CH3 OCH3

(Chiral)

Cl2/hv

CH2

CH3

CH3 Cl

CH2Cl



CH2

CH3

+ H3C

CH2

CH3

CH

CH3

* CH

C

Nu −

+

(g) + 2CO2(g)

Ethene

9. (3) For elimination, strong base is required. But more hindered base will produce less substituted alkene.

1/4/2018 5:25:54 PM

610

OBJECTIVE CHEMISTRY FOR NEET

Br

RO

+

t-BuOK t-BuOH, 75°C

2-Methyl-1-butene (72.5%) (less substituted)

R



Secondary alcohols can be dehydrated using 85% phosphoric acid at 165–170°C.

H

An alkoxy radical

R

Br

H + Br

O

A bromine radical

10. (2) The temperature and concentration of acid required to dehydrate an alcohol depend on the structure of the alcohol substrate. Primary alcohols are the most difficult to dehydrate. They require concentrated sulphuric acid and a temperature of 180°C in order to dehydrate

+

O

2RO

∆

An alkyl peroxide 2-Methyl-2-butene (27.5%) (more substituted)



light or

OR

Br

+ CH2

CHCH2CH3

CH2CHCH2CH3 Br

CH2CHCH2CH3

+ H

Br

Br

CH2

CHCH2CH3

Br

H

15. (2) H

OH CH2

85% H3PO4 165−170°C

CH

CH2

+ CH

CH

Br

11. (4) This is an example of reductive ozonolysis. CH

CH

O3

CH3

CH

O H3C

CH

CH

O

O

H CH3

C

(2 mol) (B)

H

KMnO4, H+/∆ Oxidative ozonolysis

(A)

CH3

Zn/H2O

C

CH3

CH3

CH2 O

H O

H

H

C

C

18. (2) CH3

Option (1): H2/Ni

H

3,4-Dimethyl hexane

O3

CH2

Zn

O

H2O

O

Zn/H2O

HCHO + HCHO

14. (4) Electrophilic addition reactions are the characteristic reactions of alkenes and these reactions are ionic in nature, so, the reaction is completed by electrophiles and nucleophiles. However, in the case of presence of peroxide, reaction takes place through free radical formation. This behavior is observed only in the case of addition of HBr.

Chapter 23_Aliphatic hydrocarbon.indd 610

COOH

CH3

O + O3

+

O3/Zn/H2O

13. (1) Vinyl group gives formaldehyde on ozonolysis which confirms the presence of vinylic group — CH   CH2 in compound.

CH2

COOH

O

2CH3CHO Molecular mass (24 + 16 + 4 = 44 u)

2-Butene

CH

CH3

17. (3) O3

H O+O

CH2

Br-

16. (2)

C

12. (4) The reaction is CH CH 2-Butene (Symmetrical alkene)

+ CH

CH3

Major product

O

CH3

CH2

CH3 (Ozonide)

(A) H2O/Zn

H3C

CH3

(2°)

HBr

H3C

+ Br

+ O (Two products)

Option (2):

H2/Ni

(cis or trans)

3,4-dimethyl hexane O3 Zn/H2O

O (Single product)

1/4/2018 5:25:57 PM

611

Aliphatic Hydrocarbons 19. (4)

O H

CH3C

+

C CH2CH3

CH3

Hydride shift

+

C

C

O

O

CH2CH3

2. Hydrolysis

Cl

20.

1. O3

+

CH3COOH + CH3CH2COOH

Previous Years’ NEET Questions (2) Br 1. Br2/CCl4

1. (1) The reaction will take place according to Markovnik-

Br 2a. NaNH2/NH3

ov’s rule. It states that in the addition of HX to an alkene, the hydrogen atom adds to the carbon atom of the double bond that already has the greater number of hydrogen atoms. The reaction is CH3

Na +

−

C

CH + HCl

CH3

CH2

C

CH2

Cl (B)

+

2b. H3O

CH2

Hl

l CH3

CH2

C

21. (2) An anti-addition of hydrogen atoms to the triple bond occurs when alkynes are reduced with lithium or sodium metal in ammonia at low temperature. H C

CH3(CH2)2 C

CH3

2-Hexyne

Li/NH3

CH3 C

Cl (C)

2. (2) Ozonolysis of 2-methyl-2-butene, gives acetone and acetaldehyde. The reaction involved is

C H

CH3CH2CH2

CH3

trans-2-Hexane



C

CH

CH3

O3

CH3

CH3

CH3

In this reaction, trans product is also formed because of an equilibrium between cis and trans alkenyl radicals. The trans radical is formed because it is more stable.

C

C

Ph

Ph C−

− C

NH3

C

H3C

C

CH

OH Cl

H3C

C

CH3

O

OH Cl Unstable

O

C CH3 + CH3 C H Acetone Acetaldehyde

3. (3) 3-Methylbutene on reaction with hydrogen bromide yields 2-bromo-2-methylbutane via formation of more stable 3° carbocation through hydride shift.

CH

-H2O

Cl

H3C C CH Cl 1,1-Dichloropropan-2-one

24. (2) Treating alkynes with ozone followed by hydrolysis leads to cleavage at the carbon-carbon triple bond. The products are carboxylic acids.

Chapter 23_Aliphatic hydrocarbon.indd 611

O

Ph

OH Cl HOCl

CH3

C

23. (3) The reaction is C CH + HOCl Propyne

O

Zn/H2O

H

H

Ph

CH3

H C

O Ph

Ph

O C

2-Methyl-2-butene

22. (3) The reaction is Li/NH3

CH3

CH3 CH3

CH

CH3

CH2

CH H

CH3

Br

CH

+

CH3

CH

CH3 CH3

Hydride shift

2° Carbocation (Less stable)

3-Methyl butene

C

CH2

CH3

Br (A) 2-Bromo-2-methylbutane

Br −

CH3

+

C

CH2

CH3

CH3 3° Carbocation (More stable)

1/4/2018 5:25:59 PM

612

OBJECTIVE CHEMISTRY FOR NEET

4. (1) In numbering the carbon atoms, double bond is preferred over the triple bond. Therefore, the IUPAC name of 1

2 CH

the compound CH2

3 C

4 CH is 1-butene-3-yne.

11. (1) The reaction involved is CH3 CH3

5. (1) In case of cracking higher hydrocarbons are broken into smaller alkanes. A process used in the petroleum industry for breaking down the molecules of larger alkanes into mixture of gaseous hydrocarbons. Cracking may be accomplished with heat (thermal cracking), or with a catalyst (catalytic cracking).

C

CH3 C

+ CH2 H

due to hydrogen bonding interactions. H H

3 CH

2 C

CH3

-H+

C

CH

CH3

CH3 OH (Minor product) (B)

1,2-methyl shift

CH3 CH3

+

C

CH

OH CH3 CH3

H2O

CH3

-H+

C

CH3

CH

CH3

CH3 (Major product) (A)

(3º carbocation) More stable

CH3

8. (1) The correct name of the compound 5 4 CH3 CH

CH3

+

12. (1) 1-Butyne reacts with sodium amide and liberates ammonia gas and sodium butynide precipitate as it contains acidic hydrogen, while 2-butyne does not react with sodium amide due to the lack of acidic hydrogen.

OH

H

CH

(2º carbocation) less stable

7. (3) Among the given conformations, the most stable is H

C CH3

CH3

6. (1) Anti-conformation of butane is the most stable conformation as the torsional and steric strains are minimum because both the bulky CH3 groups are 180° apart. This conformation corresponds to minimum energy and is hence most stable. OH

CH3

CH3 H2O

1 CH is pent-3-en-1-yne.

CH2

C

CH + NaNH2

CH3

1-Butyne CH3

9. (3) The reactions involved are

CH2

C− Na++ NH3

C

Sodium butynide

CH3 + NaNH2 C C 2-Butyne

No reaction

CH3 CH3

CH

CH

CH3

C

CH2 CH3 + CH3

C CH (Major) (A) HBr

CH3 CH3

Heat

OH

CH3

CH2

13. (1) The isobutyl group is

H+

CH3

C CH2 (Minor) (B)

CH3

CH3 CH3

CH3

CH

CH

Br

Br

(C)

(D)

CH3

10. (2) This is the longest chain including functional group,

3

15. (4) The reaction takes place via the formation of benzyl intermediate which is stable due to the dispersal of positive charge in the ring. The reaction is Br CH

1 4

CH2

14. (1) The quantity of heat evolved when one mole of unsaturated compound is hydrogenated is called heat of hydrogenation. The differences in heat of hydrogenation arise due to difference in stability. Greater the stability of alkene, lesser is the heat of hydrogenation and greater the number of alkyl group attached to the doubly bonded carbon atom, more stable the alkene. Hence, the correct order of enthalpy of hydrogenation is I < II < III.

along with lowest sum rule. 2

CH

5 6

CH

CH

CH3

CH2

CH3

+ HBr

4-Ethyl-3-propyl-hex-1-ene

Chapter 23_Aliphatic hydrocarbon.indd 612

1/4/2018 5:26:01 PM

613

Aliphatic Hydrocarbons 16. (4) The reaction is

20. (4) Staggered conformation of ethane is more stable.

CH3 1. O3, CH2Cl2 −78°C

H H

O

CH3 OHC

CH CH2

HC

CH3

17. (3) The first step involves the formation of a carbocation, next step is the attack of nucleophile. +

CH3

CH3

CH2

CH3

CH3 Cl

+

Cl−

H+

H3C

C

CH3 CH

CH2

H+

CH3

CH3

CH3

CH3

C

C

CH3

C

−H+

+

CH3

CH3

CH3 CH3 3° Carbocation (More stable)

2,3-Dimethyl-2-butene

+

H+

−+

HC

CNa

CH3

CH2

H Staggered form (More stable)

H3C

H3C

CH2

Br

CH2

Br



CH3

CH CH2 Propene

CH3

CH2 CH2 Butane

C CH2 (X) 1-Butyne

H 3C

CH2

C

CH3 NaNH2 liq. NH3

CNa

CH2 CH3

CH

CH2

CH3

CH

CH2

CH3

Gaseous bromine will react faster with propene because the resulting allylic free radical is more stable due to conjugation.

23. (3) Any three-dimensional arrangement of atoms that results from rotation about a single bond is called a conformation. In conformation of ethane, there is no change in bond length and bond angle. Change occurs only in dihedral or torsional angle.

O

OH CH3C +

Propyne

CH + H2O

H2SO4 HgSO4

CH3C

CH2

Propen-2-oI (an enol) (A)

Chapter 23_Aliphatic hydrocarbon.indd 613

HC

22. (4) Given that the reaction involves gaseous bromine, suggests that the reaction is taking place at higher temperature, as bromine is a liquid at room temperature conditions. Thus the bromination reaction would take place by free radical mechanism which is possible with butane (C4H10) and propene (C3H6). The reaction that results in more stable free radical will take place faster.

-H+

+

H

H

24. (3) In the presence of concentrated sulphuric acid and mercuric salts, alkynes undergo the addition of water in a reaction that follows Markovnikov’s rule; that is, it corresponds to the addition of H to the less substituted carbon of the triple bond and –OH to the more substituted carbon.

19. (4) The possible products are

OH

CH3

Alkyl shift

CH

C

CH3 CH3

+ CH

NaNH2 liq. NH3

CH2

18. (4) On heating 3,3-dimethyl-2-butene in the presence of strong acid gives 2,3-dimethyl-2-butene. CH3

CH

C C (Y) 3-Hexyne

H3C

Cl

Cl−

+

H

21. (1) The reaction is

O

C

CH2

H

Eclipsed form (Torsional strain)

2. Zn/HOAC

H

H

H

O O

CH3

H

HH

CH3

CH3

+−

CH3

CH3CCH3 Propanone (acetone) (B)

1/4/2018 5:26:04 PM

Chapter 23_Aliphatic hydrocarbon.indd 614

1/4/2018 5:26:04 PM

24

Aromatic Hydrocarbons

Chapter at a Glance 1. Aromatic compounds are a class of hydrocarbons, so called because many of them possess a fragrant smell. These are carbocylic compounds containing at least one benzene ring within their structure. Benzene and its derivatives belong to a class of compounds called arenes. 2. Structure and Hybridization   The bond angles of the carbon atoms in the benzene ring are all 120° which suggests that the carbon atoms are sp2   hybridized and the molecule forms a planar six-membered ring from these carbon atoms. Each carbon is sp2 hybridized. H

H

H

H

H

H

  All aromatic compounds have planar geometry and there is a continuous delocalization of p-electrons. 3. Nomenclature of Benzene Derivatives (a) Monosubstituted benzenes (i) Benzene is the parent name and the substituent is simply indicated by a prefix. For example, fluorobenzene, chlorobenzene etc. F

Cl

Br

NO2

Fluorobenzene

Chlorobenzene

Bromobenzene

Nitrobenzene

(ii) The substituent and the benzene ring together form a commonly accepted base name. For example, toluene, phenol, aniline, anisole. H

CH3

Toluene

O

Phenol

H

N

H

Aniline

(b) Disubstituted benzenes (i) Positions are indicated by numbering or the prefixes ortho- (1, 2 relation), meta-(1, 3 relation), and para(1, 4 relation).

Chapter 24_Aromatic Hydrocarbon.indd 615

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OBJECTIVE CHEMISTRY FOR NEET

Br

Br Br

Br

Br

1,2-Dibromobenzene (o-dibromobenzene) ortho

1,3-Dibromobenzene (m-dibromobenzene) meta

Br 1,4-Dibromobenzene (p-dibromobenzene) para

(ii) The dimethyl benzenes are often called xylenes. CH3

CH3

CH3

CH3

H3C

CH3

1,2-Dimethylbenzene 1,3-Dimethylbenzene 1,4-Dimethylbenzene (o-xylene) (p-xylene) (m-xylene)

(c) Polysubstituted benzenes (i) Groups are numbered so as to give the lowest possible numbers to the substituents, with the substituents listed in alphabetical order. Cl 1

6

6

2

5

2

Br

5

3 4 Br 1,2,4-Tribromobenzene (not 1,3,4-tribromobenzene)

3

4

Br 1

1-Chloro-3-ethylbenzene

(ii) When a substituent is one that together with the benzene ring gives a new base name, that substituent is assumed to be in position 1 and the new parent name is used. O2N 3 4

2

5 NO2

O 1

6 5

OH

4

6

3,5-Dinitrobenzoic acid

F

1 SO3H 2

3

F

2,4-Difluorobenzenesulphonic acid

(iii) When the C6H5— group is named as a substituent, it is called a phenyl (Ph) group.

C6H5 Butylbenzene

(Z)-2-Phenyl-2-butene

2-Phenylheptane

4. Resonance and Stability   Resonance theory accounts for the much greater stability of benzene when compared to the hypothetical 1,3,5-­cyclohexatriene. According to resonance theory whenever equivalent resonance structures can be drawn for a molecule, the molecule (or hybrid) is much more stable than any of the resonance structures would be individually if they could exist. All aromatic compounds show resonance.

Chapter 24_Aromatic Hydrocarbon.indd 616

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Aromatic Hydrocarbons

617

5. Aromaticity The three general requirements for a compound to be aromatic are: (a) The compound must be cyclic. (b) The compound must contain a cyclic cloud of delocalized p electrons above and below the plane of the molecule. The delocalization of p electrons over the ring must results in the lowering of the electronic energy. (c) The compound must follow Hückel’s Rule (delocalized electrons cloud must contain a total of (4n+2) p electrons, where n is an integer equal to 0, 1, 2, 3, ... and number of p electrons are 2, 6, 10, 14, ... Some examples of aromatic compounds based on Hückel’s rule are:

-

+

Naphthalene (n = 2, 10p electrons)

Cyclopentadienyl Cycloheptatrienyl anion cation (n = 1, 6p electrons)

Anthracene

Phenanthrene

(n = 3, 14p electrons)

6. Methods of Preparation (a) General methods COONa

Decarboxylation (−CO2)

OH

NaOH/CaO ∆

Polymerization

Reduction

CH

CH

Zn H3PO2, H2O

dil. HCl 150°C

Hydrolysis

+

Reduction

N2Cl−

SO3H

(b) Wurtz–Fittig reaction

X

R

+ Na + RX

Chapter 24_Aromatic Hydrocarbon.indd 617

Dry ether

+ NaX

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OBJECTIVE CHEMISTRY FOR NEET

7. Physical Properties (a) These have characteristic fragrance. (b) They are non-polar, insoluble in water but soluble in organic solvents. (c)  Alkyl benzenes are non-polar and have lower boiling points than benzene with polar substituents like phenol, aniline, etc. (d)  Benzene with no substituents has higher melting point than many substituted benzenes due to increase in van der Waals’ forces of attraction. 8. Chemical Properties (a) Electrophilic aromatic substitutions: The most characteristic reactions of benzenoid arenes are the sub+ stitution reactions with electrophilic reagents. The electrophiles are either a positive ion (E ) or some other electron­-deficient species with large partial positive charge. Important electrophilic substitution reactions are:

ion inat

Br2, AlBr3

Cl2, AlCl3

HNO3, H2SO4

Nit rat ion

r

Chlo

Desulphonation

ation

Bromin

Frie del– Craf ts

Su lph on ati on Dilute H2SO4

Friedel– Crafts a cyla

tion

alky

O

latio

n

Cl

CH3Cl, AlCl3

Fuming H2SO4

AlCl3 O

Br

Cl

Reduction

NO2

CH3

SO3H

Excess NBS

Fe or Zn, HCl

Benzylic bromination

NH2

1. KMnO4 NaOH, heat 2. H3O+ Oxidation

Zn(Hg)/HCl heat (Clemmensen reduction)

O

CBr3

OH

(b) Addition reactions: These include hydrogenation and chlorination. + 3H2

Ni

Cyclohexane Cl +

3Cl2

Cl

Cl

Cl

Cl

UV 500 K

Cl

Chapter 24_Aromatic Hydrocarbon.indd 618

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Aromatic Hydrocarbons

619

(c) Oxidation reactions: These yield different products with different oxidizing reagents. (i) With oxygen • Complete combustion 12CO2 + 6H2O

2C6H6 + 15O2

12CO2 + 6H2O

2C6H6 + 15O2

• Catalytic oxidation

V2O5, 773 K

C6H6 + 9 O2 2 C6H6 + 9 O2 2

V2O5, 773 K

HO2CCHCHCO2H + 2CO2 + H2O Maleic acid HO2CCHCHCO2H + 2CO2 + H2O Maleic acid

(ii) With potassium permanganate CH2CH2CH3

COOH + 2CO2 + 3H2O

Hot KMnO4/OH−

Benzoic acid

Propylbenzene

(iii) With ozone

CH HC

CH

CH CH Benzene

HC

+ 3O3

Benzene triozonide

CHO + 3H2O2 3 CHO Glyoxal

H2 O

9. Directive Influence of Functional Groups

Meta-directing Chapter 24_Aromatic Hydrocarbon.indd 619

Strongly activating

NH2

NHR

NH2

OH

O

O

O

NHCAr

OCR

OCAr

O

Moderately activating

NHCR

Weakly activating

R

Weakly deactivating

F

Cl

Br

l

OR

Moderately deactivating

O

O

O

O

O

O

CH

CR

COH

COR

CNH2

SOH

Strongly deactivating

NO2

NH3+

CF3

CCl3

O

Ortho-para directing

(a)  In mono-substituted benzenes, certain substituents direct a second substituent preferentially to the ortho and para positions (ortho-para directing); other substituents direct it preferentially to a meta position (meta directing). (b)  In mono-substituted benzenes, certain substituents cause the rate of a second substitution to be greater than that of benzene itself (activating groups), whereas other substituents cause the rate of a second substitution to be lower than that of benzene (deactivating groups).

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OBJECTIVE CHEMISTRY FOR NEET

Solved Examples 4

1. Which structures have the correct IUPAC names?

3

Cl

Br

Br para-Bromo-meta-chlorotoluene

ortho-Bromotoluene

(I)

(II) O

NO2

Br 2

5 Br 6

1 NH2

3. Phenol is the name commonly assigned to (1) hydroxybenzene. (2) aminobenzene. (3) methylbenzene. (4) ethylbenzene. Solution (1) Phenol is also known as hydroxybenzene. OH

Cl

meta-Chloroaniline

para-Ethylanisole

(III)

(IV)

(1) I, II (3) I, IV

(2) III, IV (4) II, III



Solution (3) The correct IUPAC names are as follows: O Br

para-Ethylanisole

2. The correct name for the compound shown below is Br Br

Solution

5. Which of the following statements is correct? (1) (2) (3) (4)

Toluene has higher boiling point than benzene. Benzene has higher melting point than toluene. Both are insoluble in water. All of these.

Solution NH2

(1) 3,4-dibromoaniline. (2) 2,4-dibromoaniline. (3) 2,5-dibromoaniline. (4) 3,6-dibromoaniline. Solution NH2 is a main functional group, so numbering starts from that carbon to which NH2 is attached. The numbering done in such a way that substituent gets the lowest position.

Chapter 24_Aromatic Hydrocarbon.indd 620

(1) Benzene tends to undergo substitution rather than addition reactions, even though it has a high index of hydrogen deficiency. (2) All of the hydrogen atoms of benzene are equivalent. (3) The carbon-carbon bonds of benzene are alternately short and long around the ring. (4) Only one o-dichlorobenzene has ever been found.

(3) The carbon–carbon bonds of benzene are all of the same length (1.39 Å), a value in between that of a carbon–­ carbon single bond and a carbon–carbon double bond, due to resonance.

ortho-Bromotoluene

(3)

4. Which of the following is not true of benzene?

(4) Toluene is slightly more polar than benzene due to electron releasing CH3 group, hence has higher boiling point whereas benzene is more symmetrical than toluene and thus, fits better into crystal lattice resulting in higher melting point. Both cannot form hydrogen bonds with water, therefore, are insoluble in water. 6. Of Hückel’s requirements for aromatic character, only this one is waived in the case of certain compounds considered to be aromatic.

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621

Aromatic Hydrocarbons (1) (2) (3) (4)

The ring system must be planar. The system must be monocyclic. There must be (4n + 2) p electrons. The Hückel number of electrons must be completely delocalized.

9. How many dichloronaphthalenes are possible? (1) 7 (3) 9



(2) 8 (4) 10

Solution (1) The possible dichloronaphthalenes are as follows:

Solution

Cl Cl

(1)

Cl

Cl

Cl

Cl

Anthracene (n = 3, 14 π electrons) Cl

7. Which of the following statements regarding the cyclopentadienyl cation is correct? (1) (2) (3) (4)

(3) Cl

(2) Cl

Naphthalene (n = 2, 10 π electrons)

Cl

Cl

Cl

(2) Polycyclic compounds may be aromatic if they fulfill the conditions of Hückel’s rule, that is, (4n+2)p electrons rule. For example, naphthalene (bicyclic) and anthracene (tricyclic) are aromatic compounds.

(4)

It is aromatic. It is not aromatic. It obeys Hückel’s rule. It undergoes reactions characteristic of benzene.

(5)

(6) Cl

Cl

(7)

Solution (2) Cyclopentadienyl cation does not follow Hückel’s rule, that is, (4n+2)p electrons rule, hence, it is not aromatic.

10. Why would 1,3-cyclohexadiene undergo dehydrogenation readily? (1) It is easily reduced. (2) Hydrogen is a small molecule. (3) 1,3-Cyclohexadiene has no resonance energy. (4) It would gain considerable stability by becoming benzene.

+

4p electrons

8. Which of the following would you expect to be ­aromatic? (1)

(2)

(3)

(4)

Solution (4)

+

H

H

-

H

Dehydrogenation −H2

Solution

H

(2)

11. Which reagent(s) would you use to carry out the following transformation?

sp3 hybridized carbon +

H H 6p electrons (Not aromatic)

Benzene

6p electrons (Aromatic)



Ethylbenzene

2- and 4-Chloro-1-ethylbenzene

(1) Cl2, light, and heat (2) Cl2, FeCl3 (3) SOCl2 (4) C2H5Cl, AlCl3 Solution

−

8p electrons (Not aromatic)

Chapter 24_Aromatic Hydrocarbon.indd 621

7p electrons (Not aromatic)

(2) It is an example of halogenation of benzene ring which takes place in the presence of halogen and Lewis acid.

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622

OBJECTIVE CHEMISTRY FOR NEET Solution

Cl

(2) The reactions involved are as follows:

Cl2, FeCl3

+

(Electrophilic substitution)

Br Br

2-Chloro-1- ethyl benzene

Cl 4-Chloro-1- ethyl benzene

Br

O

Br

Br

+ NO2

+ NO2

12. Arrange the following compounds in order of decreasing reactivity in electrophilic substitution: O

Br

Br

Br

Br

+

+

Br

Br

Br NO2 Br

Br

(I)

(II)

(III)

(IV)

(V)

(2) Rate of electrophilic substitution is directly proportional to the presence of electron donating group and inversely proportional to the presence electron withdrawing group.

Three products

15. What would you expect to be the major product obtained from the following reaction? O

O CH3

OCH3

-R

+R

NO2

CH3

-R



Br2, FeBr3

Br

O

-I O

(1) Br

(2) (4)

NO2 C N

Solution (1)

NHCOCH3 (amide group) is a +R or electron donating group, hence, it is ortho-para directing and meta deactivating.

14. Which dibromobenzene can, in theory, yield three mononitro derivatives? (1) o-Dibromobenzene  (2) m-Dibromobenzene (3) p-Dibromobenzene  (4) All of these

Chapter 24_Aromatic Hydrocarbon.indd 622

O Br

O

(3)

(4)

O

O

13. Which of the following is not a meta-directing substituent when present on the benzene ring? NHCOCH3 N(CH3)3+

O

(2)

O

+I

Hence, the correct order is II > V > III > I > IV.

(1) (3)

Br

Br

Br

Solution

O2N

NO2

+ NO2

(1) V > II > I > III > IV (2) II > V > III > I > IV (3) IV > I > III > V > II (4) III > II > I > IV > V

Br

NO2

+ NO2

NO2

NO2 Br

Br

O Br

Solution (4) Electrophile will attack on activated ring and at ortho and para position. O



O

Ph C activator O Ph O C is an as well as ortho-para director is a deactivator as well as Ph O while C Ph C O O

meta director.

O

1/4/2018 5:25:59 PM

Aromatic Hydrocarbons

18. Which reagent(s) would serve as the basis for a simple chemical test that would distinguish between ethylbenzene and vinylcyclohexane?

O Br2 / FeBr3

O Br

O

(1) H2CrO4 (2) LAH (3) NaBH4, H2O (4) O3/Zn, H+

O +

O

Br

O

(Minor)

Solution (4)

(Major) O3/Zn /Zn O 3

16. Which of the following compounds would be most reactive toward ring nitration?

No reaction reaction No

+

+ H H

CHO CHO

O CF3

O

(1)

(2)

(3)

OH

(4)

O3/Zn /Zn O 3

(4) −OH is the strongest activating group among the given substituents; therefore, it will activate the ring most towards the nitration. 17. What would you expect to be the major product obtained from the monobromination of m-dichlorobenzene? Br

HCHO HCHO

(1) NaOH in H2O (2) Br2 in CCl4 (3) AgNO3 in C2H5OH (4) NaHSO3 in H2O Solution (2) Br2/CCl4

Cl

Cl

+ +

19. Which reagent(s) would serve as the basis for a simple chemical test that would distinguish between benzene and cyclohexene?

Solution

(1)

623

No reaction

(2)

(Brown color)

Br

Cl

Br

Br2/CCl4

Cl

Cl

Br (colorless)

(3)

(4) Equal amount of (1) and (2) Br

Cl

20. Which of the following reactions could be used to synthesize tert-butylbenzene? (1) C6H6 + CH2

Solution (2) Cl

(2) C6H6 + (CH3)3 COH

Cl

Cl

(3) C6H6 + (CH3)3 CCl

Br Br2, FeBr3

Cl

+ Cl

Cl Br

Minor Major (More steric hinderance) (Less steric hinderance)

Chapter 24_Aromatic Hydrocarbon.indd 623

C(CH3)2

H2SO4

H2SO4

AlCl3

(4) All of the above Solution (4) Following are the of Friedel–Crafts alkylation reactions taking place.

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624

OBJECTIVE CHEMISTRY FOR NEET CH3 H3C

(CH3)2C

+

+

H

CH2

(CH3)2C

C

CH3

CH3 H3C

(CH3)3C

OH

H

+

+

(CH3)3C

C

CH3 CH3

+

OH2

(CH3)3C CH3 CH3 CH3

C (CH3)3C

Cl

AlCl3

(CH3)3C

+

+

−

Cl

−AlCl4

AlCl3

+

(CH3)3C

21. Which of these liquids would be unsuitable as an inert solvent for a Friedel–Crafts reaction?

23. What would be the product of the following reaction sequence?

(1) Chlorobenzene  (2)  Nitrobenzene (3) Acetophenone    (4)  (Trifluoromethyl) benzene

OH

1. SOCl2 2. C6H6 , AlCl3 3. Zn(Hg), HCl, heat

O

Solution (1) Strong electron withdrawing group attached to benzene ring does not give Friedel–Crafts reaction, hence, can be used as a solvent. Therefore, chlorobenzene is not suitable to use as an inert solvent. 22. The reaction of benzene with

O

O OH

(2)

(1)

in the pres-

OH

Cl

ence of anhydrous aluminum chloride produces principally which of these?

(3)

(4)

Solution (1)

(4)

(2)

OH O

(3)

Cl SOCl2 Darzen process

(4)

O

AlCl3, Friedel–Craft acylation

Solution

O

(4) It is an example of Friedel–Crafts reaction. Zn(Hg), HCl/∆

CH2Cl

AlCl3 −

+

CH2

+ AlCl4

Less stable

+ Rearrange

More stable

CH2

Minor

Chapter 24_Aromatic Hydrocarbon.indd 624

Clemmensen reduction

Major

24. Which of these is the rate-determining step in the nitration of benzene? (1) Protonation of nitric acid by sulfuric acid. (2) Protonation of sulfuric acid by nitric acid. (3) Loss of a water molecule by the protonated species to produce the nitronium ion. (4) Addition of the nitronium to benzene to produce the arenium ion.

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625

Aromatic Hydrocarbons Solution

NH2

(4) In aromatic electrophilic substitution, attack of electrophile on benzene ring is rate determining step of the reaction. H E

Br

(2) Cl

H2N

E

Cl NH2

Cl Br

Br

RDS

Br

Br

(1)

+

+

Br

Br

(3)

(4)

(Arenium ion) NH2

25. What is the expected product of the following reaction?

Cl

Solution

NH2

(4) It is an example of electrophilic substitution reaction.

Br2, H2O

NH2

NH2 Br

Br Br2, H2O

Cl Cl

Cl

Practice Exercises Level I

(1) I: 3, II: 4, III: 4 (2) I: 2, II: 3, III: 4 (3) I: 2, II: 4, III: 4 (4) I: 3, II: 3, III: 4

Nomenclature

5. How many different dibromophenols are possible?

1. Anisole is the name commonly assigned to (1) hydroxybenzene. (2) aminobenzene. (3) methylbenzene. (4) methoxybenzene. 2. Aniline is the name commonly assigned to (1) hydroxybenzene. (2) aminobenzene. (3) methylbenzene. (4) ethylbenzene. 3. What is the correct name for the following? NH2

(1) 8 (2) 7

(4) 5

6. Which of the following has the highest melting point? (1) o-Xylene (2) m-Xylene (3) p-Xylene (4) Toluene

7. To be aromatic, a compound must be ___ , ___ , contain alternating double and single bonds and have a(n) ___ of electrons. (1) (2) (3) (4)

Br

(3) 6

noncyclic, coplanar, even number cyclic, symmetric, odd number cyclic, planar, Hückel (4n+2) p cyclic, aliphatic, Hückel (4n+2) p

8. Number of bonds in benzene is (1) o-Bromoaniline (2) 2-Bromoaniline (3) 1-Amino-2-bromobenzene (4) All of the above

(1) 6s and 3p. (2) 12s and 3p. (3) 3s and 12p. (4) 6s and 6p.

9. We now know that the two Kekule structures for benzene are related in the following way:

Structure and Physical Properties 4. How many Kekule structures are possible for benzene, naphthalene, and anthracene?

(I)

Chapter 24_Aromatic Hydrocarbon.indd 625

(II)

(III)

(1) They are each equally correct as a structure for benzene. (2) Benzene is sometimes one structure and sometimes the other. (3) The two structures are in a state of rapid equilibrium. (4) Neither of the two structures adequately describes benzene; benzene is a resonance hybrid of the two.

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OBJECTIVE CHEMISTRY FOR NEET

10. The carbon-carbon bonds in benzene are (1) of equal length and are shorter than the double bond of ethene. (2) of equal length and are intermediate between a double bond and a single bond. (3) of unequal length and are alternately short and long around the ring. (4) due only to p-orbital overlap. 11. In which of the following compounds would the longest carbon-carbon bond(s) be found? (1) 2-Bromobenzaldehyde (2) Vinylbenzene (3) 1,3,5-Heptatriene (4) 2,4,6-Octatriene 12. Which of the following is slightly soluble in water? (1) 1,4-Cyclohexadiene (2) 1,3-Cyclohexadiene (3) Benzene (4) All are soluble in water

19. Benzene adds on ozone to form a (1) mono-ozonide. (2) diozonoide. (3) triozonide. (4) polyozonide. 20. In the Friedel–Crafts acylation, the electrophile is +

-

(1) C6H5  (2) AlCl4  + (3) CH3CO+ (4) C6H5CH2  21. The number of disubstituted products of benzene is (1) 2. (2) 3.

(3) 4.

14. Which of these would you expect to have significant resonance stabilization energy?

23. Which of the following will undergo faster dehydrobromination? (1)

Br

(3)

Br

(2)

Br

(4)

Br

(2) N

N

24. Benzyl chloride (C6H5CH2Cl) can be prepared from toluene by chlorination with

H

(3)

(4) All of the above

Chemical Reactions 15. In Friedel–Crafts synthesis of toluene, reactants in addition to anhydrous AlCl3 are

(1) SO2Cl2 (2) SOCl2 (3) Cl2 (4) NaOCl 25. In which of the following polysubstitution takes place? COCH3

(1)

+ CH3COCI

(1) C6H6 + CH4 (2) C6H6 + CH3Cl (3) C6H5Cl + CH3Cl (4) C6H5Cl + CH4 16. This molecule does not normally participate as a reactant in a Friedel–Crafts reaction

CH3

(2)

+ CH3CI

(1) Benzene (2) Chlorobenzene (3) Nitrobenzene (4) Toluene 17. In electrophilic aromatic substitution, the attacking species (the electrophile) necessarily is a (1) (2) (3) (4)

(4) 5.

(1) benzene. (2) nitrobenzene. (3) toluene. (4) chlorobenzene.

It is aromatic. It is not aromatic. It obeys Hückel’s rule. It undergoes reactions characteristic of benzene.

(1)

(1) cold and dilute (2) hot and dilute (3) hot and concentrated (4) mixed with HNO3

22. Among the following, the compound that can be most readily sulphonated is

13. Which of the following statements regarding the cycloheptatrienyl anion is correct? (1) (2) (3) (4)

18. Benzene reacts with sulphuric acid to form benzenesulphonic acid, only when the sulphuric acid is

neutral species. positively charged species. Lewis acid. All of these.

Chapter 24_Aromatic Hydrocarbon.indd 626

NO2

(3)

+ HNO3

H2SO4

SO3H

(4)

+ H2SO4

1/4/2018 5:26:05 PM

(1)

CH3

(2)

26. Which of the following is not an ortho-para director in electrophilic aromatic substitution? (1) (3)

(2) (4)

 CF3  CH3

(1) An oxygen atom is directly attached to the aromatic ring. (2) The atom attached to the aromatic ring possesses an unshared pair of electrons. (3) The group has the ability to delocalize the positive charge of the arenium ion. (4) The atom directly attached to the aromatic ring is more electronegative than carbon.

(3)

29. Using anhydrous AlCl3 as catalyst, which one of the following reactions produces ethylbenzene (PhEt)? (1) CH3   CH   CH2 + C6H6

(2) H2C   CH2 + C6H6

(3) H3C   CH3 + C6H6

(4) H3C   CH2OH + C6H6

30. Which is the electrophile in the following reaction?

+

+

(3)

(2)

CH2NO2

+ AgNO2

CH3CH(OH)CN

Level II Structure and Physical Properties 1. Which of these is an aromatic molecule? O

(1)

(2)

(3)

(4)

CH3

(4)

(1) (2) (3) (4)

+

(1) Benzene (2) Toluene (3) m-Xylene (4) p-Xylene

(2)

+ CH Cl 3

(3)

CH2Cl

(4) CH3CHO + HCN

Chapter 24_Aromatic Hydrocarbon.indd 627

Boiling

Anhyd. AlCl3

+ AgNO2

H

(3) 5

CH2Cl

CH3

(4) 7

4. Recalling that benzene has a resonance energy of 152 kJ mol1 and naphthalene has a resonance energy of 255 kJ mol-1, predict the positions which would be occupied by bromine when phenanthrene (below) undergoes addition of Br2. 3

5

4

6 7

1

8 10

(1) 1, 2 (2) 1, 4

32. Which one of the following is a free-radical substitution reaction?

+ CI2

CH3

It is aromatic. It is not aromatic. It obeys Hückel’s rule. It undergoes reactions characteristic of benzene.

2

CH3

N

3. How many equivalent resonance structures can be written for the cycloheptatrienyl cation?

31. Which of the following compounds would you expect to be most reactive toward ring nitration?

(1)

B

O

N

(1) 3 (2) 4

H3PO4

+

627

2. Which of the following statements regarding the cyclopropenyl anion is correct?

(1) III > II > I (2) II > III > I (3) I < II > III (4) I > II > III

(1)

CH2Cl

(4) CH3CHO + HCN

28. The correct order of reactivity towards the electrophilic substitution of the compounds aniline (I), benzene (II) and nitrobenzene (III) is

+

CH3 + CH Cl Anhyd. AlCl3 Aromatic Hydrocarbons 3

 OCH3  F

27. What is a feature found in all ortho-para directing groups?

CH2Cl

Boiling

+ CI2

9

(3) 3, 4

(4) 9, 10

5. Which of these is the single best representation for naphthalene? (1)

(2)

(3)

(4)

CH2NO2

CH3CH(OH)CN

1/4/2018 5:26:07 PM

628

OBJECTIVE CHEMISTRY FOR NEET

Chemical Reactions 6. Which of the following structures contribute to the resonance hybrid of the intermediate formed when nitrobenzene undergoes meta-chlorination? O

O

CI

O

N

H

(2)

Cl NO2

Cl

CI

(3)

O

(4) Equal amount of (I) and (II) Cl

O 2N

+

(3)

(2)

Cl

+

H +

NO2

(1)

N

(1)

Cl

Cl

O

O

N

10. What would you expect to be the major product(s) obtained from the mononitration of meta-dichloro­ benzene?

(4) All of these

H CI

11. What would you expect to be the major product obtained from the following reaction?

7. The direct iodination of benzene is not possible because (1) (2) (3) (4)

1. Equiv. Br2 FeBr3

iodine is an oxidizing agent. resulting C6H5I is reduced to C6H6 by HI. HI is unstable. the ring gets deactivated.

Br

Br

8. Which of the compounds listed below would you expect to give the greatest amount of meta-product when subjected to ring bromination? HO

OH

(1)

(3)

O

(2)

(2)

(1)

Br

12. In the following reaction, the structure of the major product ‘X’ is OCH3

NH2

(3)

O

(4)

N H

9. Which is the best sequence of reactions for the following transformation? O

(1)

NO2

(1) (i) HNO3, H2SO4; (ii) CH3MgBr, Et2O; (iii) H3O+, heat (2) (i) CH3MgBr, Et2O; (ii) H3O+, heat; (iii) HNO3, H2SO4 (3) (i) HNO3, H2SO4; (ii) NaBH4, H2O; (iii) H3O+, heat (4) (i)HNO3, H2SO4; (ii) LiAlH4, H2O; (iii) H3O+, heat

Conc. HNO3 Conc. H2SO4

NO2

N H

X

O2N

O

H

Chapter 24_Aromatic Hydrocarbon.indd 628

(4) Br

(2)

O N H

O

O

(3)

(4)

N H

O2N

N H

NO2

13. Which of the following reactions would produce isopropylbenzene?

1/4/2018 5:26:09 PM

Cl

COOH

(1)

(2) OH

Aromatic Hydrocarbons NO 2

NHCOCH3

Cl

Cl

(1) Benzene AlCl3

Cl

(3) (2) Benzene

(4)

HF

CH3

OH

(3) Benzene

18. How might the following synthesis be carried out?

H2SO4

(4) All of these



Several steps

14. Which would be the product of the following reaction sequence? O OH

629

Cl

(1) C6H6

1. SOCl2 2. Benzene, AlCl3 3. Zn(Hg), HCl

(2) C6H6

Cl2

CH3CH2Cl AlCl3

product

FeCl3

Cl2

CH3CH2Cl

FeCl3

AlCl3

product

O

(1)

(2)

SO2

CO2

(3) C6H6

CH3CCl

Cl2

Zn(Hg)

AlCl3

FeCl3

HCl

(4) C6H6

CH3CCl

Zn(Hg)

Cl2

AlCl3

HCl

FeCl3

O

O

(3)

(4)

CH2

15. Which of these compounds gives essentially a single product on electrophilic substitution of a third group?

O

OH

16. The major product(s) of the following reaction,

NO2

Excess NBS

(1) (i) HNO3, H2SO4; (ii) CH3MgBr, Et2O; (iii) H2O, NH4Cl

ROOR

(2) (i) CH3MgBr, Et2O; (ii) H3O+, heat; (iii) HNO3, H2SO4

Br Br

(1)

Br

Br Br

(4)

(3)

(3) (i) HNO3, H2SO4; (ii) NaBH4, H2O; (iii) H3O+, heat (4) (i) HNO3, H2SO4; (ii) LiAlH4, H2O; (iii) H3O+, heat

(2) Br

20. Toluene is subjected to the action of the following reagents in the order given: (i) KMnO4, OH−, heat; then H3O+ (ii) HNO3, H2SO4 (iii) Br2, FeBr3. What is the final product of this sequence?

17. Each of the five disubstituted benzenes shown below is nitrated. In which of these cases does the arrow not indicate the chief position of nitration. Cl

(2) Br

Br

NO2

NO2 Br

CO2H NO2

(3)

(2) NO2

(4) Br

NO2

NHCOCH3

Cl

(3)

(1)

COOH

OH

CO2H

CO2H

Br

(1)

product

19. Which is the best sequence of reactions for the following transformation?

(1) p-Chlorotoluene (2) m-Ethylanisole (3) 1-Bromo-2-chlorobenzene (4) m-Xylene

Br

product

Cl

Chapter 24_Aromatic Hydrocarbon.indd 629

(4) 1/4/2018 5:26:11 PM

630

OBJECTIVE CHEMISTRY FOR NEET

Previous Years’ NEET Questions

CH3

1. The order of decreasing reactivity towards an electrophilic reagent for the following: (I) Benzene (III) Chlorobenzene

CH3

(1)

(2)

 (II) Toluene (IV) Phenol

CH2OH

OCH3

CH3

CH3

(4)

(3)

would be

(AIPMT 2007) 2. Which one of the following is most reactive towards electrophilic attack? Cl

NHCOCH3

OH

(1) IV > II > I > III (2) I > II > III > IV (3) II > IV > I > III (4) IV > III > II > I

(AIPMT PRE 2010, 2011) 7. Among the following compounds, the one that is most reactive towards electrophilic nitration is (1) benzoic acid. (2) nitrobenzene. (3) toluene. (4) benzene.

CH2OH

(1)

(AIPMT PRE 2012)

(2)

8. Nitrobenzene on reaction with conc. HNO3/H2SO4 at 80–100°C forms which one of the following products? Cl

(3)

Cl NO2

(4)

(1) (2) (3) (4)

OH

1, 2-Dinitrobenzene 1, 3-Dinitrobenzene 1, 4-Dinitrobenzene 1, 2, 4-Trinitrobenzene (NEET 2013)

(AIPMT 2008) 3. Benzene reacts with CH3Cl in the presence of anhydrous AlCl3 to form (1) xylene. (2) toluene. (3) chlorobenzene. (4) benzylchloride.

9. Some meta-directing substituents in aromatic substitution are given. Which one is most deactivating? (1) (3)

C N (2) COOH (4)

SO3H  NO2 (NEET 2013)

(AIPMT 2009) 4. Nitrobenzene can be prepared from benzene by using a mixture of conc. HNO3 and conc. H2SO4. In the mixture, nitric acid acts as a/an (1) catalyst. (2) reducing agent. (3) acid. (4) base. (AIPMT 2009) 5. The reaction of toluene with Cl2 exists in presence of FeCl3 gives X and reaction in presence of light gives Y. Thus, X and Y are (1) X = m-chlorotoluene, Y = p- chlorotoluene

10. Which of the following compounds will not undergo Friedel–Craft’s reaction easily? (1) Cumene (2) Xylene (3) Nitrobenzene (4) Toluene (NEET 2013) 11. The radical

+

CH2 is aromatic because it has

(1) 6p-orbitals and 6 unpaired electrons. (2) 7p-orbitals and 6 unpaired electrons. (3) 7p-orbitals and 7 unpaired electrons. (4) 6p-orbitals and 7 unpaired electrons.

(2) X = o- and p-chlorotoluene, Y = trichlorome­ thylbenzene (3) X = benzyl chloride, Y = m-chlorotoluene (4) X = benzyl chloride, Y = o-chlorotoluene. (AIPMT PRE 2010) 6. Which one of the following is most reactive towards electrophilic reagent?

Chapter 24_Aromatic Hydrocarbon.indd 630

(NEET 2013)

12. What products are formed when the following compound is treated with Br2 in the presence of FeBr3? CH3

CH3

1/4/2018 5:26:18 PM

631

Aromatic Hydrocarbons (1) faster. (2) slower. (3) unchanged. (4) doubled.

CH3

CH3 Br

(1)

and

(NEET I 2016) CH3

CH3

15. Which of the following can be used as the halide component for Friedel–Crafts reaction?

Br

CH3

CH3

Br

and

(2)

(1) Bromobenzene (2) Chloroethene (3) Isopropyl chloride (4) Chlorobenzene

Br

(NEET II 2016)

CH3

CH3

16. In the given reaction

CH3

CH3 Br

and

(3)

CH3

CH3



Br

CH3

HF

+

CH3

the product P is (1)

0°C

P

(2)

F

and

(4) CH3 Br

CH3

Br

(AIPMT 2014)

(3)

(4)

F

13. The oxidation of benzene by V2O5 in the presence of air produces (1) benzoic acid. (2) benzaldehyde. (3) benzoic anhydride. (4) maleic anhydride. (RE AIPMT 2015)

(NEET II 2016)

14. Consider the nitration of benzene using mixed conc. H2SO4 and HNO3. If a large amount of KHSO4 is added to the mixture, the rate of nitration will be

Answer Key Level I  1. (4)

 2. (2)

 3. (4)

 4. (2)

 5. (3)

 6. (3)

 7. (3)

 8. (2)

 9. (4)

10.  (2)

11.  (4)

12.  (3)

13.  (2)

14.  (4)

15.  (2)

16.  (3)

17.  (4)

18.  (3)

19.  (3)

20.  (3)

21.  (2)

22.  (3)

23. (4)

24.  (3)

25. (2)

26.  (1)

27.  (3)

28. (4)

29.  (2)

30.  (2)

31.  (3)

32.  (1)

 1. (1)

 2. (2)

 3. (4)

 4. (4)

 5. (1)

 6. (4)

  7. (2)

 8. (2)

 9. (1)

10. (2)

11. (3)

12. (2)

13. (4)

14. (3)

15. (4)

16. (3)

17. (1)

18. (4)

19. (3)

20. (1)

Level II

Chapter 24_Aromatic Hydrocarbon.indd 631

1/4/2018 5:26:19 PM

632

OBJECTIVE CHEMISTRY FOR NEET

Previous Years’ NEET Questions  1. (1)

 2. (4)

 3. (2)

 4. (4)

 5 (2)

 6. (3)

 7. (3)

 8. (2)

 9. (4)

10.  (3)

11.  (1)

12.  (3)

13.  (4)

14.  (2)

15.  (3)

16.  (2)

Hints and Explanations 23. (4)

Level I

Br

1. (4) All compounds are planar, containing 6p electrons, that is, all are aromatic compounds, so all have significant resonance stabilization energy. 2. (1) NHCOCH3 is ortho, para directing while all other groups are meta directing. 12. (3) The delocalized electron in benzene is more easily polarized, resulting in greater van der Waals attractive forces in water. 18. (3) Sulphonation of benzene is usually carried out with fuming sulphuric acid at room temperature. Sulphonation also takes place in concentrated H2SO4 but more slowly. Thus, we must use hot and concentrated H2SO4.

H H

−HBr

Cl2

24. (3) C6H5CH3

C6H5CH2Cl + HCl

25. (2) Friedel–Crafts alkylation reaction forms polysubstitution products. Due to activating effect of an alkyl group connected to an aromatic ring, the monoalkylated reaction product is more reactive towards electrophilic substitution than the original starting ­ material. 29. (2) This is a Friedel−Crafts alkylation reaction. CH2

19. (3) Benzene forms triozonide as it has three p bonds. O

O

+ 3O3

CH CH

CH

Benzene

O

+ H2C

O

O O O O

CH

O

CH

O

O

CH

O

Benzene triozonide

CH2

CH3

AlCl3 (anhyd)

Benzene

Ethylbenzene

Level II 3. (4) + +

20. (3) O

O CH3

C

Cl + AICI3

CH3

+

+

C + AICI4−

Y

(II)

(III)

Cycloheptatrienyl cation

21. (2) Benzene yields three isomeric disubstituted products: Y

+

(I)

Y

(IV) +

+

Y +

(VIII)

Y Y

22. (3) Toluene is readily sulphonated because it contains methyl group, which activates the ring.

Chapter 24_Aromatic Hydrocarbon.indd 632



+

(VII)

(VI)

(V)

Structures (I) and (VII) are same, hence, there are seven possible resonating structures for cycloheptatrienyl cation.

1/4/2018 5:26:20 PM

CH3

+I group, activating or o/p directing group

633

Aromatic Hydrocarbons Activating ring

6. (4)

CH3

NO2

NO2 Cl

NO2

Br2 / FeBr3 Br +

+

+

Cl H

(I) +

Cl H

(II)

CH3

CH3 Br +

NO2

NO2

Br +

Cl

−H+

(III)

Cl H

(I), (II) and (III) all are resonating structure of intermediate so they all contribute towards resonance hybrid.

12. (2) The lone pair of electrons on the nitrogen atom activates the ring i.e. increases the electron density at ortho and para positons, so the substitution will occur at ortho or para (preferably) positions. O

8. (2) COOH is an electron withdrawing group (−R), hence it is a meta directing group while remaining groups are ortho-para directing groups.

N

C

H Conc. HNO3 Conc. H2SO4

9. (1) −+ OMgBr CHO

CHO HNO3

CH

O NH

O2N

CH3

C

(X)

CH3MgBr, Et2O

H2SO4

NO2

16. (3) NBS and ROOR is a reagent for allylic or benzylic halogenation.

NO2

Br

+

H3 O

CH

CH2

OH CH

Br CH3

∆

ROOR

NO2

17. (3)

NO2

Cl

10. (2) Halogen is a ortho-para directing group. Cl

CH3

Excess NBS

Cl Cl

Cl

Cl

Cl

+ NO2

NO2

HNO3 / H2SO4

+

∆

Cl

Minor Cl

Cl Less stable (minor) ⇓ due to steric hinderance

Cl

Cl

NO2 More stable (major)

NO2 Major

18. (4) COCH3 CH3COCl AlCl3 (Friedal−Craft acylation)

11. (3) CH3

Zn(Hg)/HCl Clemmensen reduction

+I group, activating or o/p directing group

CH2CH3

Cl

Cl2 /FeCl3

+ (Minor)

Activating ring CH3

Halogenation

Cl (Major)

Br2 / FeBr3 Br + Chapter 24_Aromatic Hydrocarbon.indd 633

1/4/2018 5:26:22 PM

634

OBJECTIVE CHEMISTRY FOR NEET

19. (1) The reactions involved are as follows: OH

O O CH3

HNO3 / H2SO4

CH3 MgBr NH4Cl

NO2

NO2 −

O

O

H3O +

CH3 MgBr

HNO3

∆

H2SO4

O

O

NO2

OH ∆

LiAlH4, H2O

HNO3 / H2SO4

NO2

NO2

NO2 NaBH4, H2O

OH ∆

NO2

NO2

COO

CH3

HNO3 / H2SO4

−

+

KMnO4 /OH

activating group, that is, causes the ring to be more reactive, while CH2OH is weakly activating. Cl is weakly deactivating group whereas NO2 group is strongly deactivating.

COOH

−

20. (1)

H3O

∆

Nitration

E COOH

COOH

+ E+

Br

NO2

Arenium ion complex

NO2

Previous Years’ NEET Questions

3. (2) This is an example of Friedel–Crafts alkylation. Step 1: CH3

1. (1) The most characteristic reactions of benzenoid arenes are the substitution reactions that occur when they react with electrophilic reagents. The order of groups to activate the benzene ring towards electrophilic substitution reaction is OH > CH3> H > Cl. Therefore, correct order of reactivity towards electrophile is C6H5OH > C6H5CH3> C6H6> C6H5Cl. 2. (4) Any such group which donate electrons towards the ring (via +I, +H or +R effect) tends to stabilize the arenium ion complex, and hence increases the reactivity of the ring towards electrophilic attack. OH is strong

Chapter 24_Aromatic Hydrocarbon.indd 634

−H+

+

Br2 / FeBr3 Halogenation

E

H

Cl

Cl Cl + Al Cl Cl

CH3

+

Cl

Al

Cl +

Cl

CH3 + Cl

Al

Cl

Cl

Cl

Step 2: + +

CH3

+ CH3 H Step 3: Cl

+

CH3 + Cl H

Al

Cl

CH3 + HCl + AlCl3

Cl

1/4/2018 5:26:23 PM

635

Aromatic Hydrocarbons 4. (4) The steps involved in the nitration of benzene are as follows:

CH3

Step 1: In this step, nitric acid accepts a proton from the stronger acid, sulphuric acid. H

O HO3SO

H+H

+

O

H

N

O

(HNO3) Base

(H2SO4) Acid



7. (3) The order of reactivity is

+

+

−

−

O

H

O

O

+

O

N O

H2O + N

−

+

O Nitronium ion

5. (2) The formation of first product is through electrophilic aromatic substitution. CH3

CH3

>

Toluene

+ HSO4

Step 2: Now that it is protonated, nitric acid can dissociate to form a nitronium ion. H

>



Benzene

NO2

NO2 HNO3 H2SO4

CH3

CI (X) p-Chlorotoluene



o-Chlorotoluene

CCl3

hn

Trichloromethyl benzene (Y)

6. (3) Electron releasing groups increase the reactivity of aromatic ring towards electrophilic aromatic substitution by +I and +R effect. All alkyl groups are activating groups. The hydroxyl group OH is strong activating group while −NHCOCH3 , OCH3, CH2OH are mod-

+

+

NO2

NO2 NO2 (1%)

(6%)

9. (4) The nitro group ( NO2) is a very strong deactivating group. The carboxyl group ( COOH), the sulpho group ( SO3H), and the cyanide group ( CN) are also deactivating group. 10. (3) Friedel–Craft’s reaction is an electrophilic substitution reaction. The presence of nitrobenzene ( NO2) which is an electron withdrawing, deactivating group, deactivates the benzene ring from electrophilic substitution reactions.

12. (3) m-xylene reacts readily with bromine in the presence of Lewis acid FeBr3 to yield 1-bromo-m-xylene as the major and 2-bromo-m-xylene as the minor product (due to steric hindrance). CH3

is most reactive OH

towards electrophilic substitution reaction.

CH3

CH3 Br

CH3

Chapter 24_Aromatic Hydrocarbon.indd 635

NO2

11. (1) Hückel’s rule states that planar monocyclic rings with 2, 6, 10, 14…delocalized electrons should be aromatic. Since the compound has 6p orbitals and 6 unpaired electrons contribute to its aromaticity.

Cl2

erately activating. Hence,

Nitrobenzene

NO2

(93%)

The formation of second product is due to free radical substitution reaction that takes place in presence of light. CH3

Benzoic acid

8. (2) Nitrobenzene undergoes nitration at a rate only 10−4 times that of benzene. The nitro group is a meta director. When nitrobenzene is nitrated with nitric and sulphuric acids, 93% of the substitution occurs at the meta position.

+

FeCI3

>

Electron donating groups like CH3 increases the electron density on the ring and makes it highly activating towards electrophilic substitution while electron withdrawing groups like COOH and NO2 decreases the electron density and makes the ring deactivated towards electrophilic substitution.

CI CI2

NO2

O

N

O

COOH

Br

Br2/FeBr3

CH3

+ CH3 1-Bromo-m-xylene (major)

CH3 2-Bromo-m-xylene (minor)

1/4/2018 5:26:25 PM

636

OBJECTIVE CHEMISTRY FOR NEET

13. (4) The oxidation of benzene by V2O5 in the presence of air produces maleic anhydride. The reaction is

for Friedel–Crafts reaction because the lone pair on halogen is delocalized with the p-bond(s) of the parent molecule and hence not available for abstraction by AlCl3 to form AlCl 4 . Only isopropyl chloride is a suitable reagent.

O O

V2O5

CH3 CH3

CI +

O

14. (2) The nitration of benzene proceeds through electrophilic aromatic substitution reaction.



Anhyd. AICI3

CH3

+ HSO4− + H2O

Presence of excess of KHSO4, increases the concentration of HSO4 – ions, thus shifting the equilibrium backwards. This decreases the formation of nitronium (NO2 +) ions, hence lowering the rate of reaction.

15. (3) From the given option, chlorobenzene, bromobenzene and chloroethene cannot be used as halide component

Chapter 24_Aromatic Hydrocarbon.indd 636

CH 3 + AICI4−

16. (2) The reaction involves generation of a carbocation in the first step followed by electrophilic aromatic substitution.

NO2 + H2SO4 + HNO3

CH

HF

+

Carbocation + +

Electrophilic aromatic substitution Friedel−Crafts reaction

1/4/2018 5:26:26 PM

25

Organic Compounds Containing Halogens

Chapter at a Glance Alkyl Halides 1. General Formula of alkyl halides is CnH2n + 1X, where X = F, Cl, Br, I. They are also known as haloalkanes. 2. Classification (a) Based on number of halogen atoms: On the basis of number of halogen atoms present, halogen derivatives are classified as mono-, di-, tri- and tetra-halogen derivatives. Monohalogen derivatives are called alkyl halides. (b) Based on nature of the carbon atom: The carbon atom to which the halogen atom is attached may be sp3, sp2 or sp hybridized.  When halogen is attached to sp3 hybridized carbon, these are categorized as alkyl, allylic and benzylic halides. These are further classified as primary (e.g., 1-cholorobutane), secondary (e.g., 2-cholorobutane) or tertiary (e.g., 2-methyl 2-choloropropane) based on the nature of halogen-bearing carbon atom. 1° Carbon

H

H

H

C

C

H

H

CI

H

2° Carbon H

H

Cl

C

C

C

H

H

H

H

H3C

C

CH2X

3° Alkyl chloride

R

C

C

C X

C

2° Allylic (R = Alkyl)

Ar

C

C

X

3° Allylic

H

1° Benzylic (Ar = Aromatic)

R’

R

C

1° Allylic

CH2X

CI

CH3

H

Ar

C

2° Alkyl chloride

1° Alkyl chloride

C

3° Carbon CH3

R X

R 2° Benzylic

Ar

C

X

R’ 3° Benzylic

3. Methods of Preparation (a) From alcohols: Alkyl halides can be obtained from alcohols by replacement of hydroxide with halide using a number of reagents. ROH reactivity order: CH3OH < 1° < 2° < 3° < benzylic and allylic. (i) Halogen acids ROH + HX + ZnCl2 RX + H2O Lucas Reagent       Order of reactivity of halogen acids: HI > HBr > HCl > HF

Chapter 25_Organic Compounds Containing Halogens.indd 637

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638

OBJECTIVE CHEMISTRY FOR NEET

(ii) Phosphorous halides ROH + PCl5 → RCl + POCl3 + HCl 3ROH + PCl3 → 3RCl + H3 PO3 (iii) Thionyl chloride: The reaction is called Darzen’s procedure and is one of the best methods for preparing alkyl halides because both the byproducts formed (SO2 and HCl) are gaseous and escape easily. ROH + SOCl2 ¾Pyridine ¾¾® RCl + HCl ↑ + SO2 ↑



(iv) Sodium halide in presence of sulphuric acid. 2 SO4 ROH + NaX ¾H¾¾ ® R ¾ X + NaHSO4 + H2 O Reflux

(b) From alkanes: Alkanes react with X2 (Cl2 or Br2) in the presence of light by free radical mechanism. R ¾ H + Cl 2 ¾Light ¾¾ ® R ¾ Cl + HCl   Order of reactivity: Allylic H > 3° > 2° > 1° > CH4 (c) From alkenes (i) Addition of hydrogen halides to alkenes C

C

+ HX

C

C

H

X

   The order of reactivity of hydrogen halides is HI > HBr > HCl > HF. In case of unsymmetrical alkenes, the addition proceeds in accordance with the Markovnikov’s rule. However, in presence of organic peroxide, anti-Markovnikov addition product is obtained. C

C

C

+ HBr

C

C

C

Br 2-Bromopropane (Markovnikov product) C

C

+ HBr

C

Organic

C

peroxide

C

C

Br

1-Bromopropane (Anti-markovnikov product)

(ii) Addition of halogens: Addition of halogens to alkenes is an electrophilic addition reaction. When an alkene is allowed to react with halogen (X = Cl or Br) in the presence of solvent like CCl4 or CH2Cl2, it forms vicinal dihalide. CCl4 CH2 CH2 + X2 XCH2 CH2 X (d) From alkynes: Addition of halogen to alkyne is also an electrophilic addition reaction. When an alkyne is allowed to react with halogen in the presence of CH2Cl2, it will form trans-dihalide. C

Chapter 25_Organic Compounds Containing Halogens.indd 638

C

+ X2

CH2Cl2

H X

X C

C

trans

H

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Organic Compounds Containing Halogens



639

(e) Halogen exchange (i) Finkelstein reaction: This is one of the halogen replacement reaction, used to prepare iodo-alkanes from RCl and RBr by reacting them with NaI in the presence of acetone. It is one of the methods for preparation of iodoalkanes. RX + NaI ¾Acetone ¾¾® RI + NaX (ii) Swarts reaction: This is the only method to prepare RF by halide exchange by treating RX (X = Cl or Br) with metallic fluorides like AgF, SbF3 and Hg2F2. R   X + AgF → RF + AgX (iii) Hunsdiecker reaction: When silver salt of carboxylic acid is allowed to react with Cl2 and Br2 in the presence of carbon tetrachloride, it forms alkyl halide. 4 RCOOAg + Br2 CCl  → RBr + CO2 ↑ + AgBr ↓



    The yield of halides is in the order 1° > 2° > 3°. 4. Physical Properties (a)  Lower members up to three carbon atom chain, are gaseous at room temperature. Next thirteen members are liquids and higher are solids. (b)  Haloalkanes, though polar, are not soluble in water, as they do not form intermolecular hydrogen bonding. (c)  The boiling point increases with increase in the size of carbon chain (surface and molecular mass increases) and decreases with branching of carbon chain. For example, consider the increasing order of boiling point in the following alkyl halides: CH3Cl < CH3CH2Cl < CH3CH2CH2Cl < CH3CH2CH2CH2Cl CH3 CH3CH2CH2CH2CH2Cl > CH3

CH

CH2

CH2

Cl > CH3

C

CH2

Cl

CH3

CH3

(i)  The boiling points are higher than the corresponding hydrocarbons because of strong dipole-dipole and van der Waals forces of attraction. (ii)  The boiling point order with respect to halogen atom is R   I > R   Br > R   Cl > R   F (d)  Bond strength of haloalkanes decreases as the size of the halogen atom increases. Thus, the order of bond strength is CH3   F > CH3   Cl > CH3   Br > CH3   I (e) Dipole moment decreases as the electronegativity of the halogen atom decreases. 5. Chemical Properties The chemical reactions of haloalkanes can be classified into three main categories: (i) nucleophilic substitution reactions, (ii) elimination reactions and (iii) reactions with metals. (a) Nucleophilic substitution reactions (i) General reaction Nu:− + R

X

Nucleophile Alkyl halide (Substrate)

R

Nu + X

−

Product Halide ion

Mechanism: Nucleophilic substitution reactions proceed through two type of mechanisms − unimolecular (SN1) or bimolecular (SN2) depending on the structure of the alkyl halide, nature of nucleophile and the solvent used.

Chapter 25_Organic Compounds Containing Halogens.indd 639

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OBJECTIVE CHEMISTRY FOR NEET

· Substitution nucleophilic unimolecular (SN1 type): The substitution reaction proceeds in two steps. CH3 CH3

C

X

CH3

CH3



CH3

+

C

CH3

CH3

C

Nu + Cl−

CH3

Planar carbocation

It is a first order reaction, so the rate depends only on the initial concentration of the alkyl halide and is given by r = k[RX]. The order of reactivity of alkyl halides towards SN1 mechanism based on the stability of carbocation formed is 3° > 2° > 1°. Polar solvents, low concentration of nucleophiles and weak nucleophiles favor SN1 mechanism. In SN1 reaction, partial racemization occurs due to the possibility of frontal as well as backside attack on planar carbocation. R1

+

R2 R1 R3

R2

Nu− Frontal Backside R3 attack attack Planar carbocation Nu−



CH3 +Nu−(fast) Step 2

C

C

Nu + Nu

C

R2 R1 R3

Enantiomers (Racemization)

· S  ubstitution nucleophilic unimolecular (SN2 type): This substitution reaction proceeds in two steps and is a second order reaction because the rate depends on the initial concentration of alkyl halide and nucleophile both. The rate is given by r = k[RX][Nu]. During SN2 reaction, inversion of configuration occurs (Walden inversion) that is, starting with dextrorotatory halide, laevorotatory halide is obtained. Primary halides undergo substitution reactions by SN2 mechanism. H Nu + C H

H

H Cl CH3

Nu

C H

Cl

CH3

C

Nu + Cl−

H

CH3

Transition state



For substrates, the order of reactivity in SN2 mechanism is CH3   X > R   CH2   X > R2   CH   X Tertiary halides do not react by SN2 mechanism. The effect of leaving group is the same in both SN1 and SN2 reactions, that is R   I > R   Br > R   Cl

Factor favoring SN1 vs. SN2 reactions Factor

SN1

SN2

Substrate

3° (requires formation of a relatively stable carbocation). Weak Lewis base, neutral molecule, nucleophile may be the solvent (solvolysis). Polar protic (e.g., alcohols, water).

Methyl > 1° > 2° (requires unhindered substrate).

Nucleophile Solvent Leaving group

Strong Lewis base, rate favored by high concentration of nucleophile. Polar aprotic (e.g., DMF, DMSO).

I > Br > Cl > F for both SN1 and SN2 (the weaker the base after the group departs, the better the leaving group)

(ii) Elimination reactions: In the presence of a strong base, HX from adjacent carbon atoms in alkyl halides is eliminated and the reaction is known as dehydrohalogenation. These are also called b-eliminations (or 1,2-eliminations) because hydrogen is removed from b-carbon atom.

Chapter 25_Organic Compounds Containing Halogens.indd 640

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Organic Compounds Containing Halogens

CH3

CH2

CH

CH3

alc. KOH

CH3

CH

CH

641

CH3

Cl

· E  2 elimination: For example, consider the following reaction. It is first order in each reactant and second order overall.

C2H5O- + CH3CHBrCH3 → CH2   CHCH3 + C2H5OH + BrRate = k[CH3CHBrCH3][C2H5O-]



· E  1 elimination: The reaction proceeds with the formation of common carbocation intermediate as in SN1 reaction, so both are competing reactions. Subsequent substitution or elimination in the next step takes place depending on the nature of solvent. When the solvent molecule acts a nucleophile, substitution occurs and when it acts as a base to remove b-hydrogen, elimination occurs. (iii) Reactions with metals Dry ether

· R ¾ X + Mg ¾ ¾¾®

RMgX

Grignard reagent

· Wurtz’s reaction: R ¾ X + 2Na + X ¾ R ® R ¾ R + 2NaX (Alkane)

· Frankland reaction: R ¾ X + Zn + X ¾ R ® R ¾ R + ZnX 2 Ether ® RLi + LiX · R ¾ X + 2Li ¾¾¾

(Alkane)

(iv)  Reaction with silver nitrate is used to detect which halogen is present in the haloalkane. The order of reactivity of halides for the same alkyl group is I > Br > Cl and the order for alkyl group for the same halogen is 3° > 2° > benzylic > 1° > vinylic. 6. Some Important Polyhalogen Compounds Compound

Uses

Harmful effects

Dichloromethane (CH2Cl2)

·  Solvent for paint removal. ·  Aerosol propellant. ·  Manufacture of drugs. · Metal cleaning and finishing solvent.

·  Affects human CNS. ·  Lower levels-impaired hearing and vision. · Higher levels-dizziness, nausea, tingling and numbness. ·  Skin contact- causes intense. ·  Burning and mild redness. ·  Eye contact- burning of cornea.

Trichloromethane (Chloroform, CHCl3)

·  Solvent for fats, alkaloids. · Production of the freon refrigerants. ·  General anaesthetic. ·  Antiseptic

·  Inhaling vapors depresses the CNS. ·  Mild exposure-dizziness, fatigue, and headache. · Chronic exposure-damage to the liver, kidney; skin sores.

Triiodomethane (CHI3) Carbon Tetrachloride (CCl4)

·  Manufacture of refrigerants. · Common effects-dizziness, light headedness, nausea and vomiting. ·  Propellants for aerosol cans. · Synthesis of chlorofluorocarbons. ·  Severe effects – coma, unconsciousness or death. ·  Eye irritation. ·  Solvent use. · Depletion of O3 layer-increased UV exposure, increased skin cancer, eye disorders, and disruption of the immune system.

Freons (chloroflurocarbon, · Aerosol propellants, refrigeraCFCs) tion and air conditioning. p,p’-Dichlorodiphenyl·  Insecticide trichloroethane (DDT)

Chapter 25_Organic Compounds Containing Halogens.indd 641

·  Disturb ozone balance in atmosphere. ·  Insect resistance to DDT. ·  Toxicity to fish. · Incomplete metabolism, deposition and storage in fatty tissues.

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642

OBJECTIVE CHEMISTRY FOR NEET

Aryl Halides Compounds that have halogen atoms attached to a benzene ring are called aryl halides or haloarenes. 1. Nomenclature of Aryl Halides (a)  In naming the aryl halides, we prefix halo and name it as halobenzene. This naming convention is accepted both as IUPAC and common names. For example, CI

Br

Cholrobenzene



Bromobenzene

I n the case of dihalogen compounds, the prefixes ortho-, meta- or para- are used in common names and numbers 1,2; 1,3 and 1,4 in IUPAC names. Cl

Cl

Cl

Cl

Cl Cl 1,2-Dichlorobenzene (o-Dichlorobenzene)



1,3-Dichlorobenzene (m-Dichlorobenzene)

1,4-Dichlorobenzene (p-Dichlorobenzene)

In case of polyhalogen compounds, the position of halogen substituents is indicated by numbers. Br

Cl

Br

Br

Br

1,3,5-Tribromobenzene

1-Bromo-3-chlorobenzene

(b)  In case of presence of other substituents on the ring, the numbering of the ring begins at the halogen substituted carbon and proceeds in the direction of next substituted carbon that possesses the least number. CI

CI CH3

C2H5 1-Chloro-2-methylbenzene

1-Chloro-4-ethylbenzene

2. Physical Properties (a) Aryl halides are aromatic in nature. (b)  The halogen atom is attached to sp2 hybridized carbon atom of the aromatic ring by single sigma bond, but due to resonance the carbon−halogen bond length in aryl halide is shorter than in alkyl halide. (c)  Only chlorobenzene, bromobenzene and iodobenzene are liquids. Boiling point increases as the size of halogen atom increases. CI

Br

< Boiling point

Chapter 25_Organic Compounds Containing Halogens.indd 642

132°C

I

< 156°C

189°C

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Organic Compounds Containing Halogens

643

 he boiling point increases because the van der Waals forces of attraction increase as the number of electrons in T the molecule increases. (d)  There are permanent dipole-dipole attractions involved in chloro- and bromo-benzene but these are very poor in iodobenzene (because iodine has the same electronegativity as carbon). (e)  They are insoluble in the water and denser than water and form a separate layer because they unable to form intermolecular hydrogen bonding with water.

3. Methods of Preparation (a) Halogenation of benzene: Benzene reacts with chlorine or bromine in the presence of Lewis acid to form chloro- or bromo-benzene derivatives by electrophilic substitution reaction. + Cl2

Cl

FeCl3

+ HCl

25˚C

Chlorobenzene (90%)

(b) From diazonium salts NH2

X

N2+Cl−

N2+Cl− NaNO2 + HCl

(Sandmeyer reaction)

CuX

0−5˚C

Diazonium salt N2+Cl−

(X = Cl, Br)

X Cu/HX

N2+Cl−

I

+ N2 (Gattermann reaction)

∆

KI ∆

(X = Cl, Br) N2+Cl−

F HBF4 ∆

(Balz-Schiemann reaction)

4. Chemical Properties (a) Electrophilic aromatic substitution Aryl halides undergo electrophilic aromatic substitution though they are less reactive than benzene, due to the presence of halogen atom which is a weak deactivator. This weak deactivation is due to the electronegativity of the halogen making the intermediate cations less stable than those are produced when benzene undergoes substitution. X

X

X

E E+

+

+ H+

E



 alide ions direct subsequent substitution to ortho and para-positions. (Normally deactivating groups are metaH directing and activating groups are usually ortho and para-directing.) The directing effect is due to the resonance stabilization of the cationic intermediates, derived from ortho- and para-attack and not meta-attack.

Chapter 25_Organic Compounds Containing Halogens.indd 643

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OBJECTIVE CHEMISTRY FOR NEET

(i) Halogenation

X

X

X Cl

FeCl3 (Anhyd.)

+

Cl2

Cl

(ii) Nitration

X

X

X NO2

HNO3

+

conc. H2SO4

NO2

(iii) Sulphonation

X

X

X SO3H

conc. H2SO4

+

SO3 Fuming

SO3H

(iv) Friedel–Crafts alkylation X

X

X R

+ RX

AICI3

+

(Anhyd.)

R

(v) Friedel–Crafts acylation X

X

X COCH3

+ CH3COCI

AICI3

+

(Anhyd.)

COCH3

(b) Nucleophilic aromatic substitution Nucleophilic substitution reactions can occur when strong electron withdrawing groups are present ortho or para to the halogen. The reaction involves addition- elimination mechanism with the formation of a carbanion called a Meisenheimer complex. The process is called nucleophilic aromatic substitution (SNAr). (i) Replacement by hydroxyl group CI

OH 1. NaOH, 623 K, 300 atm 2. H+

X

MgX

+ Mg

Et2O Diethyl ether or tetrahydrofuran

X = Cl, Br, I

Chapter 25_Organic Compounds Containing Halogens.indd 644

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Organic Compounds Containing Halogens

645

(ii) Halide reactivity order is: I > Br > Cl. (iii) Reactions with other nucleophiles CI

CI

CN CuCN

OR

+ NaOR

300°C, high pressure, pyridine

High pressure 300°C, Cu

Alkoxy benzene

Cyanobenzene CI

CI

SH

+ KSH(aq)

High pressure

+ Zn

300°C, Cu salt

Ni Al

+ HCI

Thiobenzene

Benzene

(iv) Benzyne mechanism CH3

CH3

CH3

CH3 NH2

CI KNH2

+

NH3

NH2

Benzyne intermediate

(c) Reactions with metals (i) Wurtz–Fittig reaction

X R + 2Na + RX

Ether

+ NaX

(ii) Fittig reaction Dry ether

Cl + 2Na + Cl

+ 2NaCl

Chlorobenzene (two molecules)

Diphenyl or Biphenyl

(iii) Ullmann reaction ∆

l + 2Cu + l

+ 2Cul

(iv) Preparation of DDT Cl

CCl3CHO + 2

Cl conc. H2SO4 ∆ (Number of steps)

CCl3CH Cl DDT p,p-Dichloro diphenyl trichloro ethane or 1,1,1-Trichloro-2,2-bis(p-chlorophenyl) ethane

Chapter 25_Organic Compounds Containing Halogens.indd 645

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OBJECTIVE CHEMISTRY FOR NEET

Solved Examples 1. The carbon atom to which the halogen is attached can be (1) sp2 hybridized. (2) sp3 hybridized. (3) sp hybridized. (4) both (1) and (2).

3



S

O

2. For the compounds CH3Cl, CH3Br, CH3I and CH3F, the correct order of increasing C   H halogen bond length is (1) CH3F < CH3Cl < CH3Br < CH3I (2) CH3F < CH3Br < CH3Cl < CH3I (3) CH3F < CH3I < CH3Br < CH3Cl (4) CH3Cl < CH3Br < CH3F < CH3I

O

-O 1

Solution (4) In alkyl halides, allylic halides and benzylic halides the halogen atom is bonded to an alkyl group or sp3 hybridized carbon atom. In vinylic halides and aryl halides, the halogen atom is bonded to the sp2 hybridized carbon next to doubly bonded carbon.

O

4 -

O

2O-

O

(1) 1 (2) 2 (3) 3 (4) 4 Solution (4) The strongest nucleophilic site is 4 as the negative charge on C is localized unlike on the other sites. 5. Identify the product C in the following reaction. CH3

(1)

CH

CH

Cl

Cl

CH3

CCl2

CH3 CH2

alc. KOH

A

Hg2+/H+

(2) CH3

CH3

Solution (1) Bond length of carbon halogen bond depends on the size of halogen atom. Larger the size of halogen atom, larger is the bond length.

F < Cl < Br < I    Atomic size



H3C   F < CH3   Cl < CH3   Br < CH3   I  Bond length

(3)

CH3

(4) CH3

CHCl2

CH

CH

Cl

Cl

CH2

CH3

CH

CH2

Cl

Cl

(1) The reaction involved is CH3

CH3   C   CH + 2HBr CH3CH CHBr + HBr CH   CH + 2HBr CH3   CH CH2 + HBr

CH

CH

Cl

Cl

CH3 alc. KOH CH3

C

(A)

C

CH3

Hg2+/H+

O CH3

CCl2

CH2

CH3

PCl5

CH3

(C)

Solution

C

CH3

CH2

(B)

6. The major product formed when 1,1,1-trichloropropane is treated with aqueous potassium hydroxide is

(1) CH3

CH2

C

Solution

3. Which of the following reactions will yield 2,2-­ dibromopropane? (1) (2) (3) (4)

CH2

PCl5

B

C

CH + HBr

(Markovnikov’s addition)

CH3

C

CH2

(1) propyne. (2) 1-propanol. (3) 2-propanol. (4) propanoic acid.

Br (Markovnikov’s addition)

Solution

HBr

(4)

Br CH3

C

Cl Cl

CH3

C

OH CH2

CH3 + KOH(aq)

HO

CH3

−H2O

O

2,2-Dibromopropane

Chapter 25_Organic Compounds Containing Halogens.indd 646

CH2

OH

Cl

Br

4. Which one is the strongest nucleophilic site in the following species?

C

HO

C

CH2

CH3

Propanoic acid

1/4/2018 5:26:57 PM

Organic Compounds Containing Halogens 7. The synthesis of alkyl fluorides is best accomplished by (1) (2) (3) (4)

Nu -

Sandmeyer’s reaction. Finkelstein reaction. Swarts reaction. Free radical fluorination.

Nucleophile



Solution (3) Alkyl fluorides are best prepared by Swarts reaction.

R   X + AgF

∆

+

R   F + AgX

8. The major organic compound formed by the reaction of 1,1,1-trichloroethane with silver powder is (1) acetylene (2) ethene (3) 2-butyne (4) 2-butene

Leaving group

R X Heterolysis occurs here

Among the given compounds, vinyl chloride is a least reactive in a nucleophilic substitution reaction due to the formation of partial double bond character between C and Cl because of resonance; this makes the C   Cl bond more stable. R

R

X

-

Cl3 + 6Ag + Cl3

C

CH3

H3C

C

C

CH3

(1) NaI (2) NaOEt/EtOH (3) NaOH/H2O (4) NaOH/H2O-EtOH

2-Butyne

9. Which of the following is least reactive in a nucleophilic substitution reaction? (1) CH3  (3) CH2 

X+

Br

(3) The reaction involved is C

Nu R + X-

1 0. Which one of the following reagents is not suitable for the elimination reaction?

Solution H3C

647

(2) CH3CH2Cl (4) (CH3)3C   Cl

CHCl CHCH2Cl

Solution (1) In nucleophilic substitution reactions, the carbon-halogen bond of the substrate undergoes heterolysis, and the unshared electron pair of the nucleophile is used to form a new bond to the carbon atom.

1 1. In the reaction (CH3)2CH   CH   CH2

HCl

Solution (1) Treatment of a primary alkyl halide with an alkali metal halide (e.g. KF, NaI) leads to replacement of the halogen. Thus, on reaction of 1-bromobutane with NaI, substitution takes place and 1-iodobutane is formed. Br



I

Nal

In reaction with other reagents, elimination product (alkene) is obtained.

the product obtained is

(1) (CH3)2CCl   CH2 CH3



(2) (CH3)2CH   CH

(3) a mixture of (1) and (2)



Cl (4)  none of these.

 CH3

Solution (3) The reaction involved is Cl H3C

CH

CH

CH2

H

+

H3C

CH

+ CH

CH3

CH3

CH3

H3C

CH

shift

H3C

CH2

CH3 CI

Cl−

Hydride

+ C

CH

CH3

Cl−

H3C

C

CH2

CH3

CH3

More stable CH3

CH3

Chapter 25_Organic Compounds Containing Halogens.indd 647

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OBJECTIVE CHEMISTRY FOR NEET

12. Which of the following is the correct order of decreasing SN2 reactivity? (1) RCH 2 X > R 3CX > R 2CHX

(1) carbine. (2) carbocation. (3) free radical. (4) carbanion.

(2) RCH 2 X > R 2CHX > R 3CX (3) R 3CX > R 2CHX > RCH 2 X

Solution

(4) R 2CHX > R 3CX > RCH 2 X



15. A solution of (–)-1-chloro-1-phenylethane in toluene racemizes slowly in the presence of a small amount of SbCl5, due to the formation of

(2) When a chiral substrate undergoes SN1 reaction due to formation of carbocation intermediate it forms racemized products.

(where X = halogen)

Solution (2) SN2 reactivity depends on the steric hindrance because it involves the attack of nucleophile from the back side. So, more is the steric hindrance, less is the SN2 reactivity.

CH

13. The organic chloro compound, which shows complete stereochemical inversion during a SN2 reaction, is

H

CH3

SbCl5

[Ph

Cl

C

(1) (C2H5)2CHCl (2) (CH3)3CCl (3) (CH3)2CHCl (4) CH3Cl

CH CH3]+[SbCl6]− Planar structure

+

CH3 + SbCl5

C

Cl

H Planar carbocation

Solution (4) In SN2 reaction nucleophile attack from backside hence stereochemical inversion is observed which is called Walden inversion. Steric hindrance is the major factor considered for a substrate undergoing SN2 reaction. In CH3Cl, the hindrance being minimum favors SN2 to a maximum extent. H H H

C

Cl + OH−

HO

C

H H H

14. Which of the following reactions would give the best yield? O

Heat

(1) CH3OH +

ONa

(3) CH3ONa + (4)

+ CH3Br

CH3

(1) (2) (3) (4)

m-chlorotoluene. benzoyl chloride. benzyl chloride. o- and p-chlorotoluene.

CH3

CH3

Br

Alc. KOH Heat

CH3

Cl2 FeCl3

O

(2) The product can be formed with best yield if SN2 reaction conditions are used, otherwise elimination product is also formed in significant amount. For SN2 mechanism, primary alkyl halides are best suited. Hence, the reaction in option (2) will take place with SN2. In other options, reaction conditions also initiate elimination reaction, so, mixture of products is obtained in other cases.

Chapter 25_Organic Compounds Containing Halogens.indd 648

CH3 Cl

CH3

Solution



16. The reaction of toluene with Cl2 in presence of FeCl3 gives predominantly

O

+ CH3OH Br

Enantiomeric pair

(4) The reaction involved is electrophilic addition reaction:

O

(2)

R+S

Solution

CH3

Br

CH3

Toluene (o, p-directing in nature)

+

o-Chlorotoluene (Minor)

Cl p-Chlorotoluene (Major)

17. When the alkyl bromides (listed here) were subjected to hydrolysis in a mixture of ethanol and water (80% EtOH/20% H2O) at 55°C, the rates of the reaction showed the order: (I) CH3Br (III) (CH3)2CHBr

 (II) CH3CH2Br (IV) (CH3)3CBr

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649

Organic Compounds Containing Halogens (1) I > II > III > IV (2) IV > III > II > I (3) IV > I > II > III (4) IV > III > I > II

Cl NO2

Solution

NO2

(3) Two different mechanisms are involved in the hydrolysis of these compounds. (CH3)3CBr reacts by SN1 mechanism and apparently this reaction takes place faster. The other three alkyl halides react by SN2 mechanism, and their reactions are slower because the nucleophile H2O is weak. The reaction rates are affected by the steric hindrance and thus their order of reactivity is I > II > III > IV.

CF3

NH3

CF3

Cl

(III)

NO2

(IV) 

NO2



(2) Aromatic nucleophilic substitution in haloarenes can occur when strong electron withdrawing groups are present at ortho or para position to the halogen atom attached to the nucleus. (II) is the most reactive due to presence of two –NO2 at ortho and para positions, and (IV ) is the least reactive due to lack of resonance activation by nitro group from meta position. (I) is more reactive than (III) due to less steric hindrance and greater electrophilicity at a-carbon of p-fluoronitrobenzene.

NaNH2

(3)

Cl

Solution Cl

NH2

NO2

 (II) 



(1) I < IV < III < II (2) IV < III < I < II (3) I < IV < II < III (4) IV < I < II < III

1 8. The major product of the reaction

(1)

(I) F

Cl

(2)

CF3

(4)

CF3

Cl

NH2 NH2

2 0. The compound C is C7H8

NH2

Solution (4) In this reaction, substitution takes place through benzyne formation. Since this benzyne is asymmetric and also bears strong electron withdrawing group, so, further reaction is very specific and only meta product is formed.

3Cl2/∆

A

Br2/Fe

B

Zn/HCl

C

(1) o-bromotoluene. (2) m-bromotoluene. (3) p-bromotoluene. (4) 3-bromo-2,4,6-trichlorotoluene. Solution (2) The reaction involved is

CF3

CH3

NH2

3Cl2/∆

CCl3

CCl3 Br2/Fe

−

CF3

Br −

NH2

Less stable carbanion

Zn/HCl

CF3

CF3 −

NH3

NH2

CH3 − + NH2 NH2

19. Arrange the following in increasing order of reactivity towards aromatic nucleophilic substitution reaction.

Chapter 25_Organic Compounds Containing Halogens.indd 649

Br m-Bromotoluene

1/5/2018 5:27:38 PM

HBr

Br

650

(1)

OBJECTIVE CHEMISTRY FOR NEET

Br

(2)

Br

Practice Exercises (3)

Level I

(4)

Alkyl Halides

Br

1. Which of the following structures have correct common names? CH3

(I) CH2

CH

CH2

F

(II)

CH

CH3

Allyl fluoride

Isobutyl bromide

(III) CHCl3

(IV) CH3

CH

CH2

CH3

(1) I, II (2) III, IV (3) I, III (4) II, IV 2. Which of the following compounds are primary haloalkanes? CH3

CH3

I

H

C

C

H

CH3

CH3

(II) CH3

C

CH3

CH2

CH3

CBr3

(IV) FCH2

CH2

CH CH3

(1) I, II (2) III, IV (3) II, III (4) I, IV 3. Which of the following compounds has the highest boiling point? (1) CH3CH2CH2Cl (2) CH3CH2CH2CH2Cl (3) CH3CH(CH3)CH2Cl (4) (CH3)3CCl 4. Which of the following is the best nucleophile? (1) C- (3) O-

(2) N(4) F-

5. The best leaving group produces a _________ base that is a(n) _________ anion. (1) weak, stable (2) strong, stable (3) weak, unstable (4) strong, unstable 6. The product obtained for the following reaction is HBr

Br

(1)

(3)

Br

(2)

Br

(1) R  (2) R  (3) R  (4) R 

 F > R   I > R   I > R   F > R 

 Cl > R   Br > R   Cl > R   I > R 

 Br > R   Cl > R   Br > R   Br > R 

 I  F  F  Cl

9. SN2 reactions proceed with a(n) ________ of configuration. (1) inversion (2) retention (3) racemization (4) None of these 10. Which statements apply to an SN1 reaction?

CI

CH3

(III)

unimolecular, 2nd order unimolecular, 1st order bimolecular, 2nd order bimolecular, 1st order

8. The reactivity order of halides for dehydrohalogenation is

Isopropyl iodide

Methylene chloride

7. SN2 reactions are ________ and have ________ kinetics. (1) (2) (3) (4)

CH2 Br

I

(I)

Br

(I) The rate limiting step of the reaction involves the alkyl halide and the nucleophile. (II) The order of reactivity is methyl > 1° > 2° > 3°. (III) The rate limiting step of the reaction involves only the alkyl halide. (IV) There is an intermediate carbocation. (1) I, II (2) III, IV (3) I, IV (4) II, III 11. In an SN1 reaction, if the [nucleophile] doubles, the reaction rate ________. In an SN1 reaction, if the [substrate] doubles, the reaction rate ________. (1) (2) (3) (4)

doubles, doubles halves, halves doubles, halves will not be effected, doubles

12. In SN2 reactions, the order of reactivity of the halides CH3X, C2H5X, n-C3H7X, n-C4H9X is (1) CH3X > C2H5X > n-C3H7X > n-C4H9X (2) C2H5X > n-C3H7X > n-C4H9X > CH3X (3) n-C4H9X > n-C3H7X > n-C2H5X > CH3X (4) None of these. 13. SN1 reactions run best in __________ solvents and SN2 reactions run best in _________ solvents. (1) (2) (3) (4)

polar protic, polar protic polar protic, polar aprotic polar aprotic, polar aprotic polar aprotic, polar protic

14. Carbon tetrachloride does not have a dipole moment due to (4)

Chapter 25_Organic Compounds Containing Br Halogens.indd 650

Br 1/4/2018 5:27:03 PM

Organic Compounds Containing Halogens (1) (2) (3) (4)

its regular tetrahedral structure. its planar structure. the similar electron affinities of carbon and chlorine. the similar size of the carbon and chlorine atoms.

15. __________ increases the likelihood of elimination occurring instead of substitution. (1) (2) (3) (4)

Br2 alc. KOH KCN 21. In the reaction CH 3CH 2I   X   Y   Z, the product Z is

(1) CH3CH2CN (2) CH2BrCH2CN (3) CNCH2CH2CN (4) BrCH   CHCN 22. Which of the following statements is correct? (1) The addition of HBr in the presence of peroxide to propene gives 2-bromo­propane. (2) The addition of HBr to propene gives 1-bromo­ propane. (3) The addition of HCl to vinyl chloride gives ethylidene chloride. (4) The addition of HCl to vinyl chloride gives ethylene chloride.

increased pressure increased temperature decreased pressure decreased temperature

16. The ease of dehydrohalogenation with alcoholic KOH in case of (I) chloroethane, (II) 2-chloropropane, (III) 2-chloro-2-methylpropane is of the order: (1) III > II > I (2) I > II > III (3) II > I > III (4) I > III > II

23. When hydrochloric acid gas is treated with propene in presence of benzoyl peroxide, it gives (1) 2-chloropropane. (2) allyl chloride. (3) no reaction. (4) n-propyl chloride.

17. SN2 reactions proceed fastest with ________ substrates. (1) primary (2) secondary (3) tertiary (4) none of these

651

24. What is the major product of the following reaction?

18. Vicinal and geminal dihalides can be distinguished by NaOMe/MeOH

(1) KOH (aq.) (2) KOH (alc.) (3) Zn dust (4) None

Br

(1)

19. An alkyl bromide produces a single alkene when it reacts with sodium ethoxide and ethanol. This alkene on hydrogenation produces 2-methylbutane. What is the identity of the alkyl halide? (1) 1-Bromo-2,2-dimethylpropane (2) 1-Bromobutane (3) 1-Bromo-2-methylbutane (4) 2-Bromo-2-methylbutane



CH2

(1)

CI

(2)

CH3

CI CH2CI

(3)

(4)

  2H2O + RX → ROH + H3O+ + X(1) X- (2) H3O+ (3) ROH (4) H2O

26. Which nucleophilic substitution reaction is not likely to occur? (1) I- + CH3CH2Cl → CH3CH2I + Cl(2) I- + CH3CH2Br → CH3CH2I + Br(3) I- + CH3CH2OH → CH3CH2I + OH(4) CH3O- + CH3CH2Br → CH3CH2OCH3 + Br-

E2

CH3

Na

OMe

(3)

25. Identify the nucleophile in the following reaction:

20. Which of the following alkyl halides yields the product shown as the only possible product of an E2 reaction?

?

(2)

(4)

CH3

Cl

Chapter 25_Organic Compounds Containing Halogens.indd 651

27. Chloroform, when kept open, is oxidized to (1) O2 (2) COCl2 (3) O2, C2 (4) none of these. 28. Which of the following halogen-exchange reaction will occur? (1) R–I + NaCl (2) R–F + KCl (3) R–Cl + NaI (4) CH3–F + AgBr

1/4/2018 5:27:04 PM

652

OBJECTIVE CHEMISTRY FOR NEET

29. The major product formed in the following reaction is CH3 CH3

C

CH2l

(1) Allyl bromide (2) Vinyl bromide (3) tert-Butyl bromide (4) Propyl bromide

CH3OH

H H

CH3

(1) CH3

(3)

C

CH2OCH3

H



(2) CH3

CH3

C

C

CH2CH3

OCH3 CH3

(4) CH3

CH3

C

CH3

OCH3

CH2



30. When ethyl iodide is heated with silver nitrate, the product obtained is (1) C2H5Ag (2) Ag—O—NO2 (3) C2H5NO2 (4) C2H5I—NO2 31. (CH3)3CMgBr on reaction with D2O produces (1) (CH3)3CD (2) (CH3)3COD (3) (CD3)3CD (4) (CD3)3OD 32. By analyzing the starting material and the product(s), the following reaction is an example of what type of mechanism? I

36. In terms of reactivity towards nucleophiles, bromobenzene is most similar to which of these?

HO

(1) SN1 (2) SN2 (3) E1 (4) E2 33. Elimination reactions are favored over nucleophilic substitution reactions (1) at high temperatures. (2) when 3° alkyl halides are used as substrates. (3) when nucleophiles are used which are strong bases and the substrate is a 2° alkyl halide. (4) in all of these cases.

37. Monochlorination of ethyl benzene in the presence of light and heat yields which of the following as the major product? (1) m-Chlorotoluene (2) A mixture of o- and p-chlorotoluene (3) 1-Chloro-2-phenylethane (4) 1-Chloro-1-phenylethane 38. Which of the following will give yellow precipitate on shaking with an aqueous solution of NaOH followed by acidification with diluted HNO3 and addition of AgNO3 solution? (1)

(2)

C2H5I

(3)

CH2I

I

CH2CI

(4) CH3

Br

CI

39. Which of the following is not an example of Wurtz−Fittig reaction? (1) C6H5CH2Cl + CH3Cl

Na/Ether

(2) C6H5CH2Cl + C6H5Cl

Na/Ether

(3) C6H5Cl + CH3Cl

Na/Ether

(4) None of these. 40. Which of the following reactions is most suitable for the preparation of n-propylbenzene? (1) Friedel–Crafts alkylation (2) Wurtz reaction (3) Wurtz–Fittig reaction (4) Grignard reaction 41. The structure of the major product formed in the following reaction is

Aryl Halides 34. The IUPAC name of the compound shown below is

Cl

CI

NaCN DMF

I Br

(1) 2-bromo-6-chlorocyclohex-1-ene. (2) 6-bromo-2-chlorocyclohexene. (3) 3-bromo-1-chlorocyclohexene. (4) 1-bromo-3-chlorocyclohexene. 35. Which of the following possesses highest melting point? (1) Chlorobenzene (2) o-Dichlorobenzene (3) m-Dichlorobenzene (4) p-Dichlorobenzene

Chapter 25_Organic Compounds Containing Halogens.indd 652

CN

(1)

(2)

CN NC

CN Cl

(3)

CN

I

(4)



CN

I

1/4/2018 5:27:07 PM

Organic Compounds Containing Halogens 42. Which of the following would be most likely to undergo a nucleophilic substitution reaction with aqueous sodium hydroxide by an addition-elimination mechanism? Br

(1)

(2)

Br

(3)

Br

(4)

NO2

NO2

O

CHO

(1) (2) (3) (4)

of its covalent bond. of its low boiling point. of its high melting point. it gives incombustible vapors.

(1) insolubility. (2) instability. (3) inductive effect. (4) steric hindrance.

KNH2/NH3

(II)

OH

CHO

(III)

1. Which of the following compounds are secondary haloalkanes?

3. Tertiary alkyl halides are practically inert to SN2 mecha­ nism because of

Cl

NH2

Alkyl Halides

2. CCl4 is used as a fire extinguisher because

43. What is the product of the following reaction?

(I)

Level II

(I) Isobutyl bromide (II)   2-Iodobutane (III) Isopropyl fluoride (IV)   Neopentyl chloride (1) I, II (2) III, IV (3) II, III (4) I, IV

Br

O

CHO

4. Which of the following halides is prepared from the following reaction?

HBr

(IV)

NH2

653

NH2

CHO

Br

(1)

Br

(2)

(1) I (2) II (3) I and IV (4) II and III 44. What is the major product expected in the following reaction?

(3)

Br

(4) Br

CH3 CI

5. In an SN2 reaction, if the [nucleophile] doubles, the reaction rate ________. In an SN2 reaction, if the [substrate] doubles, the reaction rate ________.

NaNH2 NH3

(1)

(2)

CH3

CH3

(1) doubles, doubles (2) halves, halves (3) doubles, halves (4) halves, doubles

NH2

6. Br has a low reactivity in CH2   CH   Br because NH2 CH3

(3) Na

NH2

(4) None of these.

NH2

Chapter 25_Organic Compounds Containing Halogens.indd 653

(1) (2) (3) (4)

the C   Br bond has a partial triple bond character. of the +M effect of bromine. Br is electronegative. None of the above.

7. The SN2 mechanism is characterized by the presence of a(n)________. The SN1 mechanism is characterized by the formation of a(n) _________ called a(n) ________.

1/4/2018 5:27:08 PM

654

OBJECTIVE CHEMISTRY FOR NEET (1) (2) (3) (4)

14. The intermediate during the addition of HCl to propene in presence of peroxide is + · (1) CH 3 C HCH 2Cl (2) CH 3 C HCH 3 + · (3) CH 3CH 2 C H 2 (4) CH 3CH 2 C H 2

intermediate, transition state, carbocation transition state, intermediate, carbanion intermediate, intermediate, carbanion transition state, intermediate, carbocation

8. Phosgene is a common name for (1) (2) (3) (4)

phosphonyl chloride. thionyl chloride. carbon dioxide and phosphine. carbonyl chloride.

15. Which of the following is an SN2 reaction? (1) (CH3)3CBr + KOH (2) CH3CH2Br + CH3CH2ONa

9. SN1 reactions produce ________ products. (3) CH3

10. In an SN2 reaction, reaction rates are affected by the __________ of the substrate. In an SN1 reaction, reaction rates are determined by the __________ of the carbocation. stability, substrate steric hindrance, stability substrate, concentration stability, steric hindrance

CH2

CH3 Br + KOH

C

C

OH + KBr

CH3

(4) All of these. 16. Under SN1 conditions, the hydrolysis of neopentyl bromide in the presence of methanol, CH3 CH3

C

CH2Br

CH3 Cl



A (Major)

Zn Ether

gives the major product: CH3

CH3

(1) CH3

Cl

The product A is

(1)

CH3CH2

CH3

11. In the given reaction



CH3CH2OCH2CH3 + NaBr

CH3

(1) inverted (2) identical (3) racemic (4) None of these

(1) (2) (3) (4)

(CH3)3COH + KBr

C

CH3

(2) CH3

CH2CH3

C

CH3

(3) CH3

(2) CH3

C

CH2OH

CH3 CH

CH3

(4) None of these.

CH3

CH3

17. Which reactions proceed according to an E2 mechanism?

(3)

(4)

(I)

CH3

CH3

T = 78°C Ethanol

Br

12. What is the total number of products obtained on monochlorination of 2,2,5-trimethyl hexane in the presence of light and chlorine?

(II)

Br

T = 78°C Ethanol

(1) 5 (2) 7 (3) 8 (4) 10 13. The order of reactivity of the following halide towards a SN1 reaction is Cl

Cl

Cl

(III)

−

+

O K , T = 56°C

Br

(IV)

Acetone

−

+

O K , T = 50°C Dimethyl formamide

(I)

(II)

(1) II > III > I (2) III > II > I (3) III > I > II (4) I > III > II

Chapter 25_Organic Compounds Containing Halogens.indd 654

(III)

Br

(1) I and II (2) II and III (3) III and IV (4) I and IV

1/4/2018 5:27:10 PM

655

Organic Compounds Containing Halogens 18. SN1 reactions of the following type:



Nu:- + R   X → R   Nu + :X-







are favored (1) by the use of tertiary substrates (as opposed to primary or secondary substrates). (2) by increasing the concentration of the nucleophile. (3) by increasing the polarity of the solvent. (4) by use of a strong base.

19. What is the major product of the following reaction?

Assuming no other changes, what effect on the rate would result from simultaneously doubling the concentrations of both butyl bromide and OH- ion? (1) (2) (3) (4)

No effect. It would double the rate. It would triple the rate. It would increase the rate four times.

Aryl Halides 23. The reagent B in the following reaction is Br

Br

KOt-Bu

NaNO2/HCl

∆ (Heat)

Br

0−4°C NH2

KOt-Bu = Potassium tert-butoxide

(1) OtBu

24. The reactivity of the compounds (I) CH3Br, (II) C6H5CH2Br, (III) CH3Cl, (IV) p-CH3OC6H4Br decreases as

K

(1) I > II > III > IV (2) IV > II > I > III (3) IV > III > I > II (4) II) > I > III > IV

(4)

20. Which of the following compounds can be prepared by free radical halogenation without complication by formation of isomeric by-products? (1)

Br

25. Benzyl chloride (C6H5CH2Cl) can be prepared from toluene by chlorination with which of the following (I) SO2Cl2, (II) SOCl2, (III) Cl2/hn, (IV) NaOCl? (1) I, II and III (2) I and III (3) II, III and IV (4) All of these.

(2)

26. Replacement of Cl of chlorobenzene to give phenol requires drastic conditions but chlorine of 2,4-dinitrochlorobenzene is readily replaced. This is because

Br

(3)

(4)

Cl

Br

(1) NO2 makes the ring electron rich at ortho and para positions. (2) NO2 withdraws electrons from meta position. (3) NO2 donates electrons at meta position. (4) NO2 withdraws electrons from ortho/para position.

21. What is the major product of the following reaction? OK ∆

27. What product(s) would you expect from the following reaction?

NaBr + MeOH + ?

Br

Br

(1)

Br

(1) Cu powder and HBr (2) Cuprous bromide (3) Either (1) or (2) (4) Neither (1) nor (2)

(2)

(3)

B

A

1. NaOH, NaHCO3,

(2)

O2N

NO2

H2O, heat

?

2. H3O+

Br O

(3)

(4)

HO

NaO

O2N

NO2 O2N

(I) 22. Consider the SN2 reaction of butyl bromide with OH ion.

CH3CH2CH2CH2Br + OH- → CH3CH2CH2CH2OH + Br -

Chapter 25_Organic Compounds Containing Halogens.indd 655

-

NO2

(II)

HO

NH2

(III)

(1) I (2) II (3) III (4) Substantial amounts of I and II

1/4/2018 5:27:11 PM

656

OBJECTIVE CHEMISTRY FOR NEET

28. What is the major contributing structure to the cation intermediate of the bromination of anisole? O

(1)

32. What is the product of the following reaction? Cl

O

(2)

1. NaOH, NaHCO3, heat 2. H3O+

+

+ H

Br

H

O

(3)

CHO

Br

(1)

(2)

Cl

Cl CH

+O

(4)

+ CHO

CHO H

Br

H

Br

(3)

29. Fluorobenzene (C6H5F) can be synthesized in the laboratory (1) by heating phenol with HF and KF. (2) from aniline by diazotization followed by heating the diazonium salt with HBF4. (3) by direct fluorination of benzene with F2 gas. (4) by reacting 1-bromobenzene with NaF solution. 30. Chlorobenzene reacts with Mg in dry ether to give a compound which further reacts with ethanol to yield (1) phenol. (2) benzene. (3) ethyl benzene. (4) phenyl ether. 31. What are A and B in the following reaction? Br

OH

OH

A

Cl

2. CH3CHO 3. aq.NH4Cl

Previous Years’ NEET Questions 1. In an SN2 substitution, reaction of the type DMF R  Br +Cl    R  Cl + Br 



Which one of the following has the highest relative rate?

(1)

(2) CH3

CH2Br

CH2

(3) CH3

CH

CH2Br

(4) CH3

CH

(AIPMT 2008)

Br

Br Cl

2. Trichloroacetaldehyde, CCl3CHO reacts with chlorobenzene in presence of sulphuric acid and produces (1) Cl

(2)

CH2Br

CH3

CH3

and

Cl

CH2Br

CH3

B

CHOHCH3

MgCl

CHO

CHO

(1) CH3 1. Mg/Et2O

(4)

and MgBr

CH

Cl

CCl3

CHOHCH3

Cl MgCl

MgCl

(3)

and Br

CHOHCH3

(2) Cl

(4) None of these

C

Cl

H Cl

(3) Cl

C

Cl

CH3Cl

Chapter 25_Organic Compounds Containing Halogens.indd 656

OH

1/4/2018 5:27:12 PM

Cl

(2) Cl

C

(2) Cl

Organic Compounds Containing Halogens

H

657

NO2

(3)

Cl

(3) Cl

Cl

(1)

Cl

C

(4)

Cl

Cl

Cl

CH3Cl CH3

OH

(4) Cl

C

(AIPMT MAINS 2011)

Cl

6. Consider the reaction:

CH

(AIPMT 2009)

3. The correct order of increasing reactivity of C   X bond towards nucleophile in the following compounds is X

OCH3

X

(I)

(CH3)2CH

CH2Br

(II)

(CH3)2CH

CH2

C2H5OH

Br

(CH3)2CH

C2H5O−

(CH3)2CH

CH2OC2H5 + HBr CH2OC2H5 + Br−

The mechanisms of reactions (I) and (II) are, respectively,

NO2

(1) SN2 and SN1 (2) SN1 and SN2 (3) SN1 and SN1 (4) SN2 and SN2 NO2

(I)

(CH3)3C

(II)

X

(CH3)2CH

(III)

(AIPMT MAINS 2011)

X

(IV)

7. In an SN1 reaction on chiral centers, there is

(1) II < III < I < IV (2) IV < III < I < II (3) III < II < I < IV (4) I < II < IV < III

(1) 100% retention. (2) 100% inversion. (3) 100% racemization. (4) inversion more than retention leading to partial racemization. (RE AIPMT 2015)

(AIPMT PRE 2010) 4. Which one is a nucleophilic substitution reaction among the following? (1)

CH3CHO + HCN

(2)

CH3

CH

CH3CHOHCN H+

CH2 + H2O

8. Which of the following reaction(s) can be used for the preparation of alkyl halides?

CH3

CH

CH3

(I) CH3CH2OH + HCl (II) CH3CH2OH + HCl

OH

(3)

RCHO + R’MgX

R

CH

(III) (CH3)3COH + HCl

R’

(IV) (CH3)3COH + HCl

OH

CH3

CH2

CH

CH2Br + NH3

(RE AIPMT 2015) 9. For the following reactions.

CH3 CH3

CH2

CH

I CH3CH2CH2Br + KOH

CH2NH2

II H3C

5. Which of the following compounds undergoes nucleophilic substitution reaction most easily? Cl

(2) Cl

(3)

Cl

(4)

CH3 Chapter 25_Organic Compounds Containing Halogens.indd 657

III

CH3CH   CH2 + KBr + H2O H3C

CH3

+ KBr

OH Br + Br2 Br

NO2

CH3 + KOH Br

(AIPMT PRE 2011)

(1)

anhy. ZnCl2

(1) (IV) only (2) (III) and (IV) only (3) (I), (III) and (IV) only (4) (I) and (II) only

CH3

(4)

anhy. ZnCl2

Which of the following statements is correct? (1) I and II are elimination reactions and III is addition reaction.

Cl

OCH3 1/4/2018 5:27:14 PM

OCH3

658

NaNH2

OBJECTIVE CHEMISTRY FOR NEET

A

Br

(2) I is elimination, II is substitution and III is addition reaction. (3) I is elimination, II and III are substitution reactions. (4) I is substitution, II and III are addition reactions.

OCH3 NH2

(1)

and elimination addition reaction

(NEET I 2016) 10. Which of the following compounds shall not produce propene by reaction with HBr followed by elimination or direct only elimination reaction? (1) H3C

CH2

CH2OH (2) H2C

(3)

CH2

CH2Br

H3C

C

(4) H2C

OCH3 Br

(2)

and cine substitution reaction

O OCH3

CH2 C H2

(3)

and cine substitution reaction

(NEET II 2016) OCH3

11. Identify A and predict the type of reaction. OCH3

(4) NaNH2

NH2

A

and substitution reaction

Br

(NEET 2017)

OCH3 NH2

Answer Key (1)

and elimination addition reaction

Level I  1. (1)

OCH3

 2. (2)

11. (4) (2) 21. (3)

22. (3)

12. (1)

31. (1) 41. (4)

Br

 3. (2)

 4. (1)

 5. (1)

 6. (1)

 7. (3)

 8. (2)

 9. (1)

10. (2)

13. (2)

14. (1)

15. (2)

16. (1)

17. (1)

18. (1)

19. (3)

20. (3)

23. (1)

24. (1)

25. (4)

26. (3)

27. (2)

28. (3)

29. (4)

30. (3)

32. (1)

33. (4)

34. (3)

35. (4)

36. (2)

37. (4)

38. (3)

39. (2)

40. (3)

42. (4)

43. (1)

44. (1)

and cine substitution reaction

OCH3

(3) II Level

and cine substitution reaction

 1. (3)

 2. OCH(4)

 3. (4)

 4. (3)

 5. (1)

 6. (2)

 7. (4)

 8. (4)

 9. (3)

10. (2)

11. (1)

12. (3)

13. (3)

14. (2)

15. (2)

16. (4)

17. (3)

18. (1)

19. (3)

20. (3)

23. (3)

24. (4)

25. (2)

26. (4)

27. (1)

28. (4)

29. (2)

30. (2)

 5. (2)

 6. (2)

 7. (4)

8. (3)

 9. (2)

10. (2)

21. (2) (4) 31. (2)

3

22. (4) 32. (3) NH 2

and substitution reaction

Previous Years’ NEET Questions  1. (1)

 2. (1)

 3. (4)

 4. (4)

11. (1)

Chapter 25_Organic Compounds Containing Halogens.indd 658

1/4/2018 5:27:15 PM

659

Organic Compounds Containing Halogens

Hints and Explanations Level I 3. (2) For a given halogen, boiling point increases with addition of each carbon. For the same number of carbon atoms, straight chain has a higher boiling point than branched chain. 4. (1) In the periodic table, basicity increases as you go up the chart in a group or as you go left across a period. Carbon is the furthest left across the second period. It is the strongest base and the most unstable base. This makes it the best nucleophile. 5. (1) The more stable the anion produced by the leaving group, the more likely the leaving group is to break off the substrate. That makes it a good leaving group. Weak bases are the most stable anions. If an anion is unstable it wants to react and that makes it a strong base. 8. (2) Better the leaving group ability, greater is the reactivity of that alkyl halide RI > RBr > RCl > RF. 11. (4) In an SN1 reaction, the reaction rate is affected by only the substrate. 12. (1) Less hindered is the substrate, greater is its reactivity for SN2 reaction. 14. (1) CCl4 has a net dipole moment equal to zero, as it is symmetrical molecule. 15. (2) Increasing the temperature is one way to favor an elimination over a substitution. 16. (1) Ease of dehydrohalogenation with alcoholic KOH by (E2) mechanism is 3° RX > 2° RX > 1°RX. 17. (1) Primary substrates have the least steric hindrance and their reaction rates are the fastest.

23. (1) When propene is treated with HCl in the presence of peroxide it undergoes Markovnikov addition reaction to give 2-chloropropane. 24. (1) The reaction below proceeds with SN2 mechanism. NaOMe/MeOH ∆ (Heat)

Br

27. (2) Chloroform is generally stored in completely filled dark-brown or blue bottles. This is because in presence of sunlight and air, it is slowly oxidized to form carbonyl chloride (phosgene) which is extremely poisonous. CHCl 3 +

CH3

CH2

CH

CH2Br

29. (4) The reaction is an example of nucleophilic substitution reaction. The mechanism followed is SN1. Methanol is a protic solvent that is able to solvate the carbocation. Ionization of carbon-iodide bond leads to formation of primary cation which undergoes hydride shift to form a more stable tertiary carbocation. 30. (3) Ethyl nitrate is prepared by the reaction of ethyl iodide and silver nitrate. 3 C 2H5I AgNO  → C 2H5NO2 Ether

31. (1) The reaction involved is CH3 CH3

−

C

CH3 +

MgBr + D2O

NaOEt Ethanol

CH3

CH2

C

CH2

CH3

21. (3) The reaction is as follows: alc. KOH

H2C

CH2

Br2

H2C Br

CH2

KCN

Br

22. (3) CH2   CHCl + HCl → CH3   CHCl2



Vinyl chloride    Ethylidene chloride

Chapter 25_Organic Compounds Containing Halogens.indd 659

CH3

CH3

H2C CN

CH2 CN

C

D + Mg(OD)Br

CH3

34. (3) 3-Bromo-1-chlorocylohexene Cl 6

CH3

CH3CH2l

1 O2 → COCl 2 +HCl 2

acetone 28. (3) R  Cl  NaI   RI  NaCl

18. (1) Vicinal and geminal dihalides can be distinguished by KOH (aq.). 19. (3) The reaction is as follows:

OMe

5

2

4

3

Br

35. (4) p-Dichlorobenzene has a symmetrical structure and therefore, its molecules can easily pack closely in the crystal lattice. As a result, intermolecular forces of attraction are stronger and therefore, greater energy is required to break its lattice and it melts at higher temperature. 37. (4) The reaction proceeds by free radical benzylic substitution:

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660

OBJECTIVE CHEMISTRY FOR NEET CH Cl2, hn

6. (2) Carbon–halogen has a double bond character because of mesomeric effect of bromine as depicted by the following structures

CH3

CI

3 8. (3) Halide groups attached to benzylic carbon are easily replaced. CH2l AgNO3

Cl

Agl Yellow ppt.

This is not Wurtz–Fittig reaction, as one of the halides is aryl halide and the other is alkyl halide in Wurtz–Fittig reaction.

4 0. (3) The preparation of n-propylbenzene can be carried out by Wurtz–Fittig reaction as follows:

Acetone C6 H5CH 2Br  NaI  CH 3CH 2Br   C6 H5CH 2CH 2CH 3

4 1. (4) Benzylic carbon is a much better substrate, when compared with phenylic carbon for nucleophilic substitution reactions. Cl

CN

NH2

CH3

NH3

CH3

CH2Cl ∆

NH2

CH3 CH3

CH3 +

CH3C

Level II

CH3 +

CH3

CH3CCH2CH2CHCH2Cl + HCl

Cl

CH3

1 3. (3) Greater the stability of carbocation higher will be the tendency of the alkyl halide to undergo SN1 reaction. +

Chapter 25_Organic Compounds Containing Halogens.indd 660

+

+ its enantiomer

CH3

2. (4) CCl4 gives incombustible vapors that reduce the spreading of fire, and eventually extinguish it. 5. (1) In an SN2 reaction, the reaction rate is affected by both the nucleophile and the substrate.

CH3 * CHCH2CHCH3

CH3 Cl

CH3CCH2CH2CCH3 CH3

CH3

CH3CCH2CH2CHCH3

CH3 Cl

CH3

CH3

CH3

1 2. (3) In this chlorination, there are five different possibilities to produce five constitutional isomers, out of which two contain chiral carbon centers. Since no stereospecific reagent is used in this reaction, optical isomers are also produced and racemic mixture is produced in the case of enantiomers. Hence, a total of seven chlorination products are obtained.

+ its enantiomer

−

− HCl

Benzyne

CH3

Cl

* CH3CCH2CHCHCH3

CH3 NH2−

Br

Zn Ether

CH3

Cl NaNH2

+

CH

Cl

CH3

4 4. (1) The reaction involved is CH3

CH2

CH3CCH2CH2CHCH3 + Cl2

I

4 3. (1) Benzyne elimination-addition reaction.

CH3

−

Br :

1 0. (2) In an SN2 reaction, the reaction rate is affected by how bulky and hindered the substrate is for back-side attack. In an SN1 reaction, the reaction rate is most affected by the stability of a carbocation to form.

CH3

NaCN DMF

I

CH

1 1. (1) The reaction is

Na/ether  C6 H5CH 2C6 H5 3 9. (2) C6 H5CH 2Cl  C6 H5Cl 



CH2



:

CH2CH3

+ >

Allylic carbocation

+ >

2°

1°

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661

Organic Compounds Containing Halogens 14. (2) The reaction involved is CH3

CH

CH2

OK

HCI (C6H5COO)2O

CH3CH(Cl)CH3

H+

CH3



∆

Br

Major

+

CH

Cl−

CH3

15. (2) The alkoxide ion reacts with the alky halide by SN2 mechanism. 16. (4) The reaction involved is CH3 CH3

C

CH3 CH2Br

H

+

CH3

CH3

C

NaBr + MeOH +

23. (3) The reaction represents the method of preparation of aryl halides from diazonium salts. When reagent B is Cu powder and HBr, the reaction is called Gattermann reaction and when reagent B is cuprous bromide, the reaction is called Sandmeyer reaction. 24. (4) PhCH2Br > CH3Br > CH3Cl > p-CH3OC6H4Br.

+

CH2

25. (2) It can be prepared by free radical substitution reaction on toluene using SO2Cl2 or Cl2/hn.

CH3

27. (1)

CH3

C

Br

CH3

CH3 CH2

CH3

CH3

C +

CH2

OH NO2

CH3

NO2 NaOH

OCH3

SNAr mechanism

NO2

19. (3)

NO2

KOt-Bu Br



NaHCO3

∆ (Heat) OH

Elimination reaction occurs with primary substrate and potassium tert-butoxide.

O−Na+ NO2

NO2 H 3O

+

20. (3) Option (3): In cyclopentane, all hydrogen atoms are equivalent, so only one product is produced on chlorination.

NO2

NO2

Option (1): Although bromination is more selective than chlorination for substitution of particular hydrogen atom, but 1,4-dimethylcyclohexane contains 1°, 2° and 3° hydrogen atoms out of which 2° and 3° hydrogen atom would participate in the reaction and mixture of isomers is produced.

29. (2) The reaction is known as Balz–Schiemann reaction.



Option (4): n-pentane also contains two types of 2° hydrogen atoms, so, a mixture of at least two isomeric forms is produced.

Benzene diazonium chloride



Option (2): In the case of bromination of 2-methylhexane, mixture of structural as well as optical isomers is produced since it contains 2° and 3° hydrogen atoms for bromination.



21. (2) According to Hofmann’s rule, dehydrohalogenation with a bulky base such as potassium tert-butoxide favors the formation of the less substituted alkene. Thus, this reaction gives an elimination product that follows Hofmann’s rule.

Chapter 25_Organic Compounds Containing Halogens.indd 661

+ −

+ −

N2Cl

N2BF4 + HBF4 Fluoroboric acid

273–278 K

+ HCI Benzene diazonium fluoroborate ∆

F + N2 + BF3 Fluorobenzene

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662

OBJECTIVE CHEMISTRY FOR NEET

31. (3) The reaction is

Cl OMgBr

Br

MgBr 1. Mg/Et2O

2. CH3

Cl

CH3

O C

H

C

compound NO2 will undergo nucleophilic substitution reaction most easily.

H

6. (2) Ethoxide anion is a good nucleophile for a substitution reaction with 1-bromo-2-methyl propane. It reacts rapidly by an SN2 mechanism to form the given product. Ethanol, 3. aq. NH4Cl on the other hand is a solvent and is a poor nucleophile, so, the reaction takes place by SN1 mechanism.

Cl

Cl

(A)

CH3

CH

(B)

OH

7. (4) SN1 reaction proceeds through the formation of an intermediate carbocation and the carbocation, because of its trigonal planar configuration, is achiral. It reacts with nucleophile at equal rates from either side that lead to the formation of racemic mixture with some amount of the formation of isomer corresponds to inversion.

Cl

32. (3) Nucleophilic aromatic substitution reaction.

Previous Years’ NEET Questions 1. (1) For SN2 reactions, lesser is the steric hindrance on the carbon undergoing the attack, faster is the rate of reaction. CH3CH2Br is sterically least hindered and hence reacts fastest. 2. (1) The reaction involved is H CCl3

CH3

Cl

O

Cl -H2O

H

8. (3) These reactions involve substitution of alcohol hydroxyl group. Because chloride ion is a weaker nucleophile than bromide or iodide ions, hydrogen chloride does not react with primary or secondary alcohols unless zinc chloride or similar Lewis acid is added to the reaction mixture as well. Zinc chloride, a good Lewis acid, forms a complex with the alcohol through association with an unshared pair of electrons on the oxygen atom. This provides a better leaving group for the reaction than H2O.

CCl3

CH

O

R

H

Cl

Cl

CI

(DDT)

3. (4) Aryl halides are less reactive than alkyl halides towards nucleophilic substitution. Among aryl halides, the one having strong electron withdrawing group attached to it is more reactive. 4. (4) In nucleophilic substitution reactions the carbonhalogen bond of the substrate undergoes heterolysis, and the unshared electron pair of the nucleophile is used to form a new bond to the carbon atom. 5. (2) Nucleophilic substitution reactions of aryl halides occur readily when an electronic factor makes the aryl carbon bonded to the halogen susceptible to nucleophilic attack. Presence of strong electron withdrawing group (−NO2) at ortho or para to the halogen atom facilitates the attack of nucleophile readily. Therefore,

Chapter 25_Organic Compounds Containing Halogens.indd 662

O + ZnCl2

R

+

-+R

O

+ ZnCl 2

H ZnCl2

H [Zn(OH)Cl2] - + H +

R + [Zn(OH)Cl2]

CI

ZnCl2 + H2O

9. (2) (I) is elimination reaction. It involves breaking of two s bonds and formation of one p bond.

CH3CH2CH2Br + KOH → CH3CH   CH2 + KBr + H2O



(II) is substitution reaction. It involves replacement of Br– with OH–. H3C

CH3

H3C + KOH

Br



CH3 + KBr OH

(III) is addition reaction. It involves breaking of one p bond and formation of two s bonds.

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Organic Compounds Containing Halogens

11. (1) This is an example of elimination addition reaction.

Br + Br

H - NH2

10. (2) The reactions with the given compounds are as follows: CH2

CH2

CH3

CH2

CH2

Br

Elimination

CH3

CH

CH2

CH2

OH

HBr

Elimination

C

CH3

CH2

O

HBr

CH2

H2C Br

C

-

CH2

CH

CH2

Elimination

No propene formation CH3

Chapter 25_Organic Compounds Containing Halogens.indd 663

CH

CH2

OCH3



NH2

NH3

NH2

OCH3

NH2 NH3 +

OH

Br

Br

Br CH2

-

OCH3 H3C

OCH3

-Br-

C H2 CH3

OCH3

OCH3

Br

HBr

663

-

More stable carbanion (The negative charge is farther to electron donating group)

NH2 Less stable carbanion

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Chapter 25_Organic Compounds Containing Halogens.indd 664

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26

Alcohols, Phenols and Ethers

Chapter at a Glance Alcohols 1. Alcohols are compounds whose molecules have one or more hydroxyl group attached to a c­ arbon atom. They are represented by ROH and have the general formula CnH2n+2O. 2. Nomenclature (a)  In common functional class nomenclature, alcohols are called alkyl alcohols such as methyl alcohol, ethyl alcohol, and so on. (b)  Alcohols containing two hydroxyl groups are commonly called glycols and in the IUPAC substitutive system, they are called diols. CH2CH2CH2 OH

OH

Trimethylene glycol Propan-1,3-diol (1°)

(c) Rules for naming alcohols (i)  Select the longest continuous carbon chain to which the hydroxyl is directly attached. Change the name of the alkane corresponding to this chain by dropping the final e and adding the suffix -ol. 3

2

5

1

4

3

2

1

CH3CHCH2CH2CH2OH

CH3CH2CH2OH

CH3

1-Propanol or Propan-1-ol (1°)

4-Methyl pentan-1-ol (1°)

(ii) When more than the functional groups are present, number the longest continuous carbon chain so as to give the carbon atom bearing the hydroxyl group the lower number. Indicate the position of the hydroxyl group by using this number as a locant; indicate the positions of other substituents (as prefixes) by using the numbers corresponding to their positions along the carbon chain as locants. CH3 3

2

1

1

CICH2CH2CH2OH

2

3

4

5

CH3CHCH2CCH3 OH

3-Chloro-propan-1-ol (1°)

CH3

4,4-Dimethyl-pentan-2-ol (1°)

3. Methods of Preparation (a) By substitution on alkyl halides H KO

R

X Mo

)

(aq

AgOH Ag2O

istu

re

R R R

OH OH OH

(b) From alkenes: The preparation of alcohols from alkenes proceeds by three methods.

Chapter 26_Alcohols Phenols and Ethers.indd 665

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OBJECTIVE CHEMISTRY FOR NEET

(i) Acid-catalyzed hydration: Alkenes undergo electrophilic addition of water in the presence of an acid catalyst to yield alcohols. These reactions are usually regioselective and the addition of water to the double bond follows Markovnikov’s rule. CH3

CH3

+

CH2 + HOH

C CH3

H3O

CH3

25°C

C

CH3

OH 2-Methyl propan-2-ol (tert-Butyl alcohol)

2-Methyl prop-1-ene (Isobutylene)

(ii) Hydroboration-oxidation: An alkene reacts with BH3:THF or diborane to produce an alkylborane. Oxidation and hydrolysis of the alkylborane with hydrogen peroxide and base yields an alcohol. 3 CH3CH

CH2

Propene

BH3 : THF Hydroboration

(CH3CH2CH2)3B Tripropylborane

H2O2/OH−

3 CH3CH2CH2OH

Oxidation

1-Propanol (1°)

Hydroboration–oxidation takes place with syn stereochemistry, as well as anti-Markovnikov regiochemistry. (iii) Oxymercuration-demercuration: Alkenes react with mercuric acetate in a mixture of water and tetrahydrofuran (THF) to produce (hydroxyalkyl) mercury compounds. These can be reduced to alcohols with sodium borohydride and water. R

H C

H

HO

+

C H

H

1. Hg(OAc)2/THF−Η2Ο 2. NaBH4 , OH−

R

H

H

C

C

HO

H

H

The net orientation of the addition of the elements of water, H– and –OH, is in accordance with Markovnikov’s rule. (c) From carbonyl compounds (i) Reduction of aldehydes and ketones RCHO

RCHO

[H] LiAlH4

[H] LiAlH4

RCH2OH Primary alcohol R

CH

R

OH Secondary alcohol

Aldehydes and ketones are also reduced to alcohols by hydrogen and a metal catalyst, by sodium in alcohol, and by sodium borohydride (NaBH4). O CH3CH2CH2CH Butanal

NaBH4 or C2H5OH/Na

CH3CH2CH2CH2OH 1-Butanol (85%)

(ii) Reaction with Grignard reagent: Formaldehyde on treatment with Grignard reagent followed by hydrolysis yields primary alcohol.

Chapter 26_Alcohols Phenols and Ethers.indd 666

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Alcohols, Phenols and Ethers

H C

H

H2O

RMgX

+

O

Formaldehyde

667

RCH2OH Primary alcohol

Grignard reagent

Aldehydes other than formaldehyde on treatment with Grignard reagent followed by hydrolysis yields secondary alcohol. R

C

H

O + R’MgX

R

H2O

H C

R’ OH Seconday alcohol

Aldehyde other than formaldehyde

Ketone on treatment with Grignard reagent followed by hydrolyses yield tertiary alcohols. R1 C

O + R’MgX

R1

H2O

C R2

R2

OH R’

Tertiary alcohol

Ketone

(d) From carboxylic acids and esters (i) Reduction of carboxylic acids RCOOH

LiAIH4

RCH2OH Primary alcohol

(ii) Bouveault-Blanc reduction: In this reaction, the esters are reduced to primary alcohols in the presence of absolute alcohol and sodium metal. Na

RCOOR’

Ethanol

RCH2OH’ + R’OH Primary alcohol

(iii) Reaction with Grignard reagent: Esters react with two molar equivalents of Grignard reagent to form tertiary alcohols.

CH3CH2MgBr + Ethylmagnesium bromide

CH3

CH3 C

C2H5O

O

CH3CH2

Et2O

C

OMgBr

CH3

C

O

CH3CH2

OC2H5

Ethyl acetate CH3 CH3CH2MgBr

C2H5OMgBr

CH3CH2

C

CH3 CH2CH3

H2O NH4CI

CH3CH2CCH2CH3 OH 3-Methyl-3-pentanol

OMgBr

(e) Hydrolysis of esters RCOOR’

Chapter 26_Alcohols Phenols and Ethers.indd 667

H

+

H2O

RCOOH + R’OH

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OBJECTIVE CHEMISTRY FOR NEET

(f ) Hydrolysis of ethers 2 SO4 ROR + H2 O H → 2ROH

(g) From primary amines 2 / HCl RNH2 + HONO ¾NaNO ¾¾¾ ¾ ® 2ROH + N 2 + H2 O

Nitrrous acid

4. Physical Properties (a) Initial members of homologous series of alcohols are colorless liquid and higher members are solids. (b)  Alcohols are soluble in water as they form intermolecular hydrogen bonding. Methanol, ethanol, propyl alcohols and tert-butyl alcohol are completely miscible with water. However, with the increase of carbon chain, the solubility of alcohol in water decreases as hydrophobic part. (c)  Alcohols have much higher boiling points than that of ethers or hydrocarbons of comparable molecular weights because molecules associate with each other through intermolecular hydrogen bonding. Boling point order is 1° > 2° > 3° (due to carbon chain). H3C H

O

H O

H

O CH3

CH3

(d)  Molecules that have symmetry generally have abnormally high melting points. For example, tert-butyl alcohol has a much higher melting point than the corresponding isomeric alcohols, due to its geometry. 5. Chemical Properties (a) Acid character: The oxygen atom of the alcohol polarizes both C   O bond and O   H bond. This imparts partial positive charge to hydrogen and alcohols thus act as weak acid (weaker than water). Alcohols react with sodium or any other active metal to liberate hydrogen gas. ROH + Na ® RONa + H2 ↑

Any compound which reacts with active metal like sodium and liberates hydrogen is acidic. (b) Reaction with H2SO4 (conc.): At different temperatures different products are obtained. RCH2CH2HSO3 SO 4

H2

3K H2SO4

38

RCH2CH2OH

413 K

H2SO4 443 K

RCH2CH2OCH2CH2R Ether (symmetrical) RCH CH2 (3° > 2° > 1°) Alkene

(c) Reaction with halogen acids R   OH + HX ® RX + H2O (X = Cl, Br, F)



The order of reactivity is: HF > HBr > HCl (i) Lucas test (Lucas reagent: HCl + anhyd. ZnCl2) 1° Alcohol + Lucas reagent → precipitate formed after long time or may not form. 2° Alcohol + Lucas reagent → precipitate formed after 5−10 minutes. 3° Alcohol + Lucas reagent → precipitate formed immediately.

Chapter 26_Alcohols Phenols and Ethers.indd 668

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Alcohols, Phenols and Ethers

669

(d) Reaction with phosphorous halides PCI5

R

RCI + POCI3 + HCI

PX3

OH

RX + H3PO3

X = CI, Br

P/I2/I−

R

I + H3PO3

(e) Esterification: Alcohols react with carboxylic acids, acid chlorides and acid anhydrides to form esters. O

O

C

R

OH

+ R‘

HA

OH

R

C

OR‘

+ H2O

(f ) Reaction with thionyl chloride ROH + SOCI2

RCI + SO2

+ HCI

(g) Dehydrogenation RCH2OH (1°) R

CH

Cu

RCHO (Aldehyde)

300°C Cu

R

OH (2°)

(CH3)3CHO

R

300°C

C

R (Ketone)

O CH3

Cu 300°C

C

CH3

(3°)

CH2 (Alkene)

(h) Oxidation of primary and secondary alcohols (i) Primary alcohols: Potassium dichromate can be used to oxidize primary alcohols to produce aldehydes and carboxylic acids. O

O R

CH2OH

[O]

[O]

C R

1° Alcohol

H

C R

Aldehyde

OH

Carboxylic acid

It is difficult to stop the oxidation at aldehyde stage, therefore different reaction conditions are used to exclusively obtain carbonyl compound or carboxylic acid. Use of pyridinium chlorochromate (PCC) stops the oxidation of primary alcohols at the aldehyde stage. RCH2OH

PCC

RCHO

On oxidation of primary alcohols with potassium permanganate, carboxylic acids are obtained. R

CH2OH + KMnO4

OH− H 2O Heat

RCOO- K+ + MnO2 H3O+

RCO2H (Carboxylic acid)

Chapter 26_Alcohols Phenols and Ethers.indd 669

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OBJECTIVE CHEMISTRY FOR NEET

(ii) Secondary alcohols: Secondary alcohols are oxidized to ketones and the reaction usually stops at the ketone stage because further oxidation requires breaking of carbon-carbon bond. Various oxidizing agents based on Cr(IV) are used. The use of CrO4 in aqueous acetone is called oxidation by Jones reagent. R

CH

R

CrO4/Acetone

R

OH

C

R

O

6. Distinction between 1°, 2° and 3° Alcohols (a) Iodoform test: Iodoform test is given when compound has free CH3

group.

C O

CH3CH2OH + 4I2 + 6NaOH

CHI3 + HCOONa + 5Nal + 5H2O

Ethanol

Iodoform yellow ppŁ.

CH3OH + I2 + NaOH

No reaction

(b) Victor Meyer Test Primary OH

CH2

RCH2

P + I2

RCH2

I

CH2

AgNO2

RCH2

NO2

CH2

conc. HNO3

R

CH3

P + I2

R2

OH

I

C

OH

P + I2

R

R

AgNO2

CH3

CH

Tertiary R

NO2

Alkali soln.

Red color

NOH Nitrolic acid

Secondary RCH

C

R2

CH3

conc. HNO3

R

I

AgNO2

R

R

C

C

NO2

Alkali soln.

Blue

N O Pseudo nitrol

NO2 R

R C

CH

NO2

conc. HNO3

No reaction

Alkali soln.

Colorless

R

Phenols 1. Compounds that have a hydroxyl group directly bonded to a benzene ring are called phenols. OH Phenol

OH

H3C 4-Methylphenol (a phenol)

2. Nomenclature (a)  Phenol is the common as well as accepted IUPAC name for hydroxybenzene. It is also used as the parent name when substituents are attached to the ring. Phenol is also known as carbolic acid. CI

Br NO2 OH

OH 4-Chlorophenol (p-chlorophenol)

Chapter 26_Alcohols Phenols and Ethers.indd 670

2-Nitrophenol (o-nitrophenol)

OH 3-Bromophenol (m-bromophenol)

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Alcohols, Phenols and Ethers

671

(b) The methylphenols are commonly called cresols. CH3

CH3

CH3

OH

OH 2-Methylphenol (o-cresol)

OH 4-Methylphenol (p-cresol)

3-Methylphenol (m-cresol)

(c) The benzene-diols have common names, catechol, resorcinol and hydroquinone for the o-, m- and p-substituents. 3. Methods of Preparation (a) Decarboxylation OH

OH COOH NaOH CaO

Salicylic acid

Phenol

(b) From haloarenes (Dow’s process) CI

ONa H+

700 K 320 atm

+ NaOH

OH

Acidif ication

Sodium phenoxide

Phenol

(c) Oxidation of benzene OH 1 O2 — 2 V 2O 5

conc. H2SO4

OH

SO3H 1. NaOH 2. H+

Benzene sulphonic acid

Phenol

(d) From diazonium salts N2+Cl-

OH H2O

+ N2

H+

+ HCI

warm Benzene diazonium salt

Chapter 26_Alcohols Phenols and Ethers.indd 671

Phenol

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OBJECTIVE CHEMISTRY FOR NEET

(e) From cumene (Industrial process) CH3 CH3

CH

CH3

CH3

C

O

OH

OH

[O]

H3O+

+ CH3COCH3

50°−90°C

Isopropyl benzene (Cumene)

Phenol

4. Physical Properties (a)  Due to the presence of hydroxyl group, phenols, like alcohols, can form strong intermolecular hydrogen bonds, which cause the molecules to be bonded and have higher boiling point. (b) Phenol is a colorless liquid or crystalline solid. But if kept in open for a long time it turns colored. (c)  Phenol is soluble in water; however phenols are generally less soluble than corresponding alcohols water due to hydrophobic benzene part. 5. Chemical Properties (a) Acidity of phenols Phenols are much stronger acids than structurally similar alcohols. The pKa value of alcohols is higher than for phenols. (i)  The higher electronegativity of sp2 hybridized carbon decreases the electron density on oxygen making O   H bond more polar. (ii) The phenoxide ion formed is stabilized due to delocalization of negative charge on resonating structures. +

O

−

+

+

O

−

O

O

Ο−

− −

(b) Electrophilic substitution reactions The presence of a hydroxyl group on the ring activates the ring towards electrophilic substitution and directs the substituents to ortho- and para-positions. (i) Bromination Monobromination is achieved when the reaction is carried out in presence of CS2, whereas bromine water produces white precipitate of 2,4,6-tribromophenol. OH

OH

OH + Br2

CS2

Br

o-Bromophenol 40% OH Br

+

Br p-Bromophenol 60%

Br

Br2-water

Br 2,4,6,-Tribromophenol (white ppt.)

(White precipitate obtained in confirmatory test for phenol.)

Chapter 26_Alcohols Phenols and Ethers.indd 672

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Alcohols, Phenols and Ethers

673

(ii) Sulphonation OH

OH

+ H2SO4 (conc.)

OH SO3H

SO3 Fuming

+

SO3H

(iii) Nitration OH

OH

OH NO2

+ HNO3 (conc.)

+

o-Nitrophenol 40%

NO2 p-Nitrophenol

OH HNO3 (conc.)

6%

O2N

NO2

H2SO4

NO2 2,4,6,-Trinitrophenol (Picric acid)

(c) Other Reactions of Phenols (i) Reimer–Tiemann reaction ONa

OH

ONa CHCI2

CH

NaOH

CHCI3

OH

aq. NaOH 310 K

NaOH

OH

CCI4

CHO

OH

OH

H+, −H2O

COOH Salicyaldehyde Salicylic acid

(ii) Kolbe–Schmidt reaction OH

OH

ONa

COOH NaOH

CO2, H

+

High temp. and pressure

Salicylic acid

Chapter 26_Alcohols Phenols and Ethers.indd 673

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OBJECTIVE CHEMISTRY FOR NEET

(iii) Esterification: Phenols react with carboxylic acid anhydrides and acid chlorides to form esters. OH

(RCO)2O

O

Base

O

O

CR + R

CO−

O OH

RCOCI Base

O

R + CI−

C

(iv) Reaction with zinc dust OH

+ Zn

Heating

+ ZnO Benzene

(v) Fries rearrangement OH

OCOCH3

OH

OH

AlCl3

+ CH3COOH

COCH3

+

∆

COCH3 (60%)

(40%)

(vi) Reactions of salicylic acid OH

OH

OH COOH

CCI4, KOH H+

Phenol

OCOC6H5

∆ C6H5OH

Salicylic acid

Salol

OH COOCH3

OCOCH3 COOH CH3OH

CH3COCI

H2SO4

or CH3COOH

Aspirin

Oil of winter green

6. Tests for Phenol (a) (b) (c) (d)

It turns blue litmus paper red. It liberates hydrogen gas when reacted with active metal like Na. It reacts with sodium hydroxide to form sodium phenoxide salt. It reacts with neutral FeCl3 to give colored compound. C 6 H5OH + FeCl3 (alkaline ) → (C 6 H5O)3 Fe ( Violet )

Violet

(e) When reacted with Br2 water, it forms white precipitate of 2,4,6-tribromophenol.

Chapter 26_Alcohols Phenols and Ethers.indd 674

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Alcohols, Phenols and Ethers

675

Ethers 1. Ethers are obtained when the hydrogen atom of a hydrocarbon is replaced by an alkoxy or aryloxy group. The general molecular formula is CnH2n+2O, so these are structural isomers (functional) of alcohols. 2. The general representation is ROR′, where the groups R and R′ may be different or same and can be derived from saturated, unsaturated or aromatic hydrocarbons. The oxygen atom is sandwiched between two alkyl groups (same or different), two aryl groups (same or different) or one alkyl and one aryl group. R O R Ar O Ar Symmetrical

R

O

R′ R O Unsymmetrical

Ar

3. Methods of Preparation (a) Intermolecular dehydration of alcohols 2R

OH

H2SO4

ROR

413 K

Dehydration of ether takes place at a lower temperature than dehydration to alkenes. H2SO4 180˚C

CH2

CH2

Ethene

CH3CH2OH H2SO4 140˚C

CH3CH2OCH2CH3 Diethyl ether

(b) From alkyl halides 2R

X

Ag2O (dry)

R

O

R

(c) Williamson synthesis: Symmetrical and unsymmetrical ethers can be obtained from nucleophilic substitution reaction of sodium alkoxide with an alkyl halide or alkyl suphonate. If tertiary alkyl halide is used in this reaction, it undergoes elimination to give alkene. This is the limitation of the method. Phenols can also be converted to ethers through Williamson synthesis. OR

ONa

+ RX

+ NaX

4. Chemical Properties (a) Reaction with halogen acids O

OH

R

+ HX

+ RX

CH3 CH3

CH3 O

C

CH3 + HI

CH3OH + CH3

C

I

CH3

CH3

(b) Reaction with halogens RCH2OCH2CH3

CI2 Excess

R

CHOCH2CH3 CI

Chapter 26_Alcohols Phenols and Ethers.indd 675

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OBJECTIVE CHEMISTRY FOR NEET

(c) Reaction with PCl5 R

∆

R + PCl5

O

RCl + POCl3

(d) Reaction with CO R

High temp.

R + CO

O

RCOOH

High pressure

(e) Electrophilic substitution OCH3

OCH3 Br OCH3

+ Br 2 CS 2

(40%)

Conc. HNO3

Br (60%) OCH3

OCH3

H4SO4

NO2 + (40%)

OCH3

NO2 (60%)

OCH3

COCH3

OCH3

CH3COCI anhyd. AICI3 Friedel-Crafts acylation

(40%)

COCH3 (60%)

OCH3

CH3

OCH3

CH3CI anhyd. AICI3 Friedel-Crafts alkylation

(40%)

CH3 (60%)

Solved Examples 1. The gas evolved on heating, CH3MgBr in methanol is (1) methane. (2) ethane. (3) propane. (4) HBr. Solution (1) Methyl magnesium bromide (Grignard reagent) on reaction with methanol evolves methane gas. CH 3MgBr  CH 3OH  CH 4  Mg(OCH 3 )Br 2. Anti-Markovnikov hydration of the carbon-carbon double bond occurs when an alkene reacts with:

Chapter 26_Alcohols Phenols and Ethers.indd 676

(1) BH3:THF; then H2O2/OH(2) BH3:THF; then CH3COOH (3) Hg(OAc)2, THF, H2O; then NaBH4, OH(4) Hg(OAc)2, THF, CH3OH; then NaBH4, OHSolution (1) Anti-Markovnikov hydration of the carbon-carbon double bond occurs when an alkene reacts with BH3·THF or diborane to produce an alkylborane. Oxidation and hydrolysis of the alkylborane with hydrogen peroxide and base yield an alcohol.

1/4/2018 5:27:16 PM

Alcohols, Phenols and Ethers 3. Which statement is true concerning the formation of alcohols by the hydroboration-oxidation sequence? (1) Overall, the process results in syn-addition and antiMarkovnikov orientation. (2) Overall, the process results in anti-addition and anti-Markovnikov orientation. (3) Overall, the process results in syn-addition and Markovnikov orientation. (4) Overall, the process results in anti-addition and Markovnikov orientation.

677

+

3O C6 H6 MgBr + CH 3OH ¾H¾¾ ® C6 H6 + Mg(OCH 3 )Br

6. n-Propyl alcohol and isopropyl alcohol can be chemically distinguished by which reagent? (1) PCl5 (2) Reduction (3) Oxidation with potassium dichromate (4) Ozonolysis Solution

(3) n-Propyl alcohol and isopropyl alcohol yield different products on reaction with potassium dichromate. The (1) The addition of    H and    OH occurs with anti-­ reactions are Markovnikov regioselectivity and syn stereoselectivity. [O] Solution

CH3CH2CH2OH

4. Oxymercuration-demercuration of 3-methylcyclopentene produces which of this/these product(s)?

n-propyl alcohol CH3

CH2OH

CH3

HO (1)

CH3 OH

(3)

Propionic acid [O]

CH3

CH

K2Cr2O7

CH3

CH3

7. The increasing order of reactivity of following alcohols towards HCl is:

CH3 OH

(4) Both

C

O Acetone

OH Isopropyl alcohol

(2)

CH3

CH3CH2COOH

K2Cr2O7

and

(I)

F

CH3

F

CH3

(II)

OH

OH

OH H3C

H3C (III)

Solution (4) Alkenes react with mercuric acetate in a mixture of water and tetrahydrofuran (THF) to produce (hydroxyalkyl) mercury compounds. These can be reduced to alcohols with sodium borohydride and water. The overall alkene hydration is not stereoselective because even though the oxymercuration step occurs with anti-addition, the demercuration step is not stereoselective (radicals are thought to be involved), and hence a mixture of syn and anti-products results. CH3

CH3 1. Hg(OAc)2, H2O THF 2. NaBH4, OH−

(2) Reactivity of alcohols towards HCl is decided by formation of more stable carbocation. Greater the stability of carbocation, more will be the reactivity.

Among the given four alcohols, (IV) makes the most stable carbocation as it is benzylic carbocation which is resonance stabilized. Alcohols (I), (II) and (III), all form secondary carbocations, of which that formed by (III) is most stable due to positive inductive effect of two methyl groups. The carbocation formed by (II) is more stable than that from (I) due to lower negative inductive effect of fluoro group.



From the above observations the correct order of formation of stable carbocation is the same as the reactivity towards HCl. Thus, the order is

OH

(1) (2) (3) (4)

a mixture of anisole and Mg(OH)Br a mixture of benzene and Mg(OMe)Br a mixture of toluene and Mg(OH)Br a mixture of phenol and Mg(Me)Br

Ph OH

Solution (2) Alcohols react with Grignard reagent to form hydrocarbons. The reaction is

Chapter 26_Alcohols Phenols and Ethers.indd 677

OH

Solution

+

5. Phenyl magnesium bromide reacts with methanol to give

Ph

(1) II < I < III < IV (2) I < II < III < IV (3) IV < III < II < I (4) I < III < II < IV

CH3 OH

(IV) OH



>

CH3

H3C

OH

>

F CH3 HO

>

F

CH3 OH

So, I < II < III < IV

1/4/2018 5:27:17 PM

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OBJECTIVE CHEMISTRY FOR NEET

8. In the following sequence of reactions, P + I2

CH3CH2OH

Mg

A

B

Ether

10. The major product of the reaction

HCHO

H2O

C

NaNO2/H2SO4

D NH2 OH

the compound D is (1) butanal. (2) n-butyl alcohol. (3) n-propyl alcohol. (4) propanal.

(2)

(1)

Solution

(3)

(3) CH3CH2OH

P + I2

(4)

N H

CH3CH2I

Solution

(A) Mg

CH3CH2I

O

H OH

(2) The complete reaction is

CH3CH2MgI Grignard reagent (B)

Dry ether

(A) H

H

OH

N2 OH

CH3CH2CH2OMgl CH3

(C) +

H2O/H

CH3CH2CH2OMgl

+ +

NH2OH

O

C

CH3CH2Mgl Grignard reagent (B)

NaNO2/H2SO4

+

O

CH3CH2CH2OH + Mg(OH)l

O +

OH

H

Propanol (C)

11. From amongst the following alcohols, the one that would react fastest with conc. HCl and anhydrous ZnCl2 is

(D)

9. The main product of the following reaction is

(1) 1-butanol. (2) 2-butanol. (3) 2-methylpropan-2-ol. (4) 2-methylpropanol.

C6 H5CH 2CH(OH )CH(CH 3 )2 ¾¾¾¾¾ ®? conc. H 2 SO4

(1) H5C6

H C

(2) C6H5CH2

C

H

CH(CH3)2

(3) C6H5

C

CH(CH3)2

C

H

C

C

(4) H5C6CH2CH2

C

H3C

CH3

OH

CH3

conc. H2SO4

CH3

C6H5CH2

CH +

CH

CH3

(1) (2) (3) (4)

CH3

C6H5

CH

CH

CH

CH3

CH3

C6H5

C

CH(CH3)2 H

trans isomer (More stable) (Major)

Chapter 26_Alcohols Phenols and Ethers.indd 678

ZnCI2 −OH−

H3C

C +

+ CI−

CH3

H3C

3° carbocation

C

CH3

CI

C6H5

C

C

CH(CH3)2 H

H cis isomer (Less stable) (Minor)

Tertiary alcohol by SN1 Secondary alcohol by SN2 Tertiary alcohol by SN2 Secondary alcohol by SN1

Solution (1) Lucas reagent reacts with alcohol via SN1 mechanism via formation of carbocation intermediate.

(Conjugated system) C

conc. HCI

12. An unknown alcohol is treated with the Lucas reagent to determine whether the alcohol is primary, secondary or tertiary. Which alcohol reacts fastest and by what mechanism?

Loss of proton

H

CH3

CH3 CH3

C OH

CH2

(1) CH

(3) The reactivity of carbocation is 3° > 2° > 1°.

H3C

Solution

C6H5CH2CH

Solution

CH3

H

H

CH3

-

2 R - OH + HCl ¾ZnCl ¾¾ ® R + ¾Cl¾® R - Cl



The reaction proceeds through SN1 mechanism with tertiary alcohol and the rate of reaction is directly proportional to stability of carbocation formation. As tertiary carbocation formed by tertiary alcohol is highly stable, hence the reaction is the fastest.

1/4/2018 5:27:19 PM

679

Alcohols, Phenols and Ethers 13. In the Victor-Meyer’s test, the color given by 1°, 2° and 3° alcohols are respectively (1) red, colorless, blue. (2) red, blue, colorless. (3) colorless, red, blue. (4) red, blue, violet. Solution (2) CH3

CH2

OH

(1°)

P+I2

CH3

CH2

CH3

NaOH

Red color

AgNO2

I

CH3 C N

CH2

Solution (2) This reaction is known as Reimer–Tiemann reaction which involves the treatment of phenol with chloroform in presence of aqueous sodium or potassium hydroxide at 340 K followed by hydrolysis of the resulting product gives 2-hydroxybenzaldehyde (salicylaldehyde). OH

NO2

NO2

ONa

ONa CHCI2

HNO2

+ CHCl3

OH

NaOH, 340 K

CH(OH)2

2NaOH 2NaCI

−NaCI, −H2O

Nitrolic acid CH3 CH3

CH

CH3

CH3 OH

P+I2

CH3

CH

I

(2°) NaOH

Blue color

AgNO2

CH3 CH3

CH

CH3

ONa NO2

NO2

CH3

CH3

C

OH

P+I2

HNO2

CH3

CH3

C

16. Which one of the following compounds will not be soluble in sodium bicarbonate?

CH3 I

AgNO2

CH3

C

HNO2 NaOH

No color or colorless

No reaction

14. Which of the following is true about C   O bond lengths? X

CH3 Y O

O H

(1) 2,4,6-Trinitrophenol (2) Benzoic acid (3) ortho-Nitrophenol (4) Benzene sulphonic acid

NO2

CH3

CH3

Solution (3) Compounds which react with NaHCO3 are soluble in it. Compounds with low acidic strength do not react with NaHCO3 hence insoluble. OH NO2

H

(1) X = Y (2) X > Y (3) X < Y (4) Cannot be predicted Solution

− H



H

15. When phenol is treated with CHCl3 and NaOH, the product formed is (1) benzaldehyde. (2) salicylaldehyde. (3) salicylic acid. (4) benzoic acid.

Chapter 26_Alcohols Phenols and Ethers.indd 679

No reaction

(1) Boiling point of o-nitrophenol is lower than p-­nitrophenol. (2) Nitration of phenol with nitrating mixture gives p-nitrophenol as the major product. (3) p-nitrophenol is a stronger acid than o-nitrophenol. (4) m-nitrophenol is the weakest acid among them.

+ O

It has a partial double bond character, while CH3   O   H has no resonance. Therefore, it is a pure single bond, and hence Y > X.

NaHCO3

17. Which of following statements is not true about o-, m-, and p-nitrophenol?

(3) The resonance in the first structure is shown below O

CHO dil. HCI −NaCI

−H2O

C

N O Pseudo nitrol

CH3

OH CHO

Solution (2) The acidic strength order: p-nitrophenol > o-nitrophenol > m-nitrophenol.

Due to intramolecular hydrogen bonding in o-nitrophenol, its boiling point is less than p-nitrophenol.



The nitration of phenol produces o-nitrophenol as the major product, due to hydrogen bonding.

1/4/2018 5:27:21 PM

680

OBJECTIVE CHEMISTRY FOR NEET

1 8. The structure of the compound that gives a tribromo derivative on treatment with bromine water is CH3

20. Phenol, when it first reacts with concentrated sulphuric acid and then with concentrated nitric acid, gives (1) 2,4,6-trinitrobenzene. (2) o-nitrophenol. (3) p-nitrophenol. (4) nitrobenzene.

CH2OH

(1)

Solution

(2)

(2)    OH group is ortho and para directing, but    SO3H group being bigger preferably occupies para position.

OH CH3

CH3

OH

OH

(3)

OH

(4)

Conc. H2SO4

OH NO2

Conc. HNO3

OH

(1) Both the groups CH3 and OH are ortho-para directing in nature; therefore, Br attaches at 2, 4 and 6 positions in option (1). CH3

21. Phenol is heated with a solution of mixture of KBr and KBrO3. The major product obtained in the above reaction is (1) 3-bromophenol. (2) 4-bromophenol. (3) 2,4,6-tribromophenol. (4) 2-bromophenol.

CH3 Br

Br Br2/H2O

Solution (3)

OH

OH Br

OH

OH

1 9. The electrophile involved in the reaction is

Br

OH

Br

KBr. KBrO3

O Na −

+

+ CHCI3 + NaOH

Phenol

CHO

(1) (2) (3) (4)

o-Nitrophenol

SO3H

Solution

Br 2,4,6-Tribromophenol

+

dichloromethyl cation (C HCl 2 ). dichlorocarbene (:CCl2). trichloromethyl+anion (CCl 3 ). formyl cation (CHO).

22. Sodium phenoxide when heated with CO2 under pressure at 125°C yields a product which on acetylation produces C.

Solution

ONa

(2) The reaction is named as Reimer–Tiemann reaction.

+ CO2

Fast

125° 5 atm

CHCl 3 + NaOH  Na + CCl -3 + H 2O CCl -3 ¾Slow ¾® : CCl 2 + Cl



O− CI

C

C

CI

NaOH

OCOCH3

CI

O

H

CI −

O−

O H

COOH

(1)

H

Chapter 26_Alcohols Phenols and Ethers.indd 680

COCH3

(2)

COCH3 O−

O C

C

OH

H OH

H+ AC2O

The major product C is

Mechanism OH

B

CI

C NaOH

CI CI H

OH

(3)

OCOCH3 COOCH3

(4) COOH

1/5/2018 12:58:04 PM

Alcohols, Phenols and Ethers Solution

HO

COOH

(1) The reaction is ONa

681

OH

+2 OH

COOH Phthalic acid

OH COONa

CO2

COOH

H3O+

Resorcinol H2SO4

125˚C, 5 atm

O

HO

O

Salicylic acid

(B)

COOH

(CH3CO)2O

OCOCH3 Fluorescein

COOH

25. The major organic product in the reaction, Aspirin (C)







is

CH3   O 

(1) CH3OH + (CH3)2CHI (2) CH2OCH(CH3)2

23. The following reaction is known as

(3) CH3OC(CH3)2

OH + HCI + HCN

(1) (2) (3) (4)

Solution CHO

Perkin reaction. Gattermann-Koch Formylation. Kolbe’s reaction. Gattermann reaction.

(4) The reaction involved is CH3

CH3 CH3

O

CH

CH3

H−I

CH3

+

O

CH

CH3

H

Solution

I−

(4)

SN2

CH3I + (CH3)2CHOH

OH + HCI + HCN

anhy. ZnCl2

OH CHO



(4) CH3I + (CH3)2CHOH

I

OH anhyd. Zncl2

 CH(CH3)2 + HI → Product

Due to formation of hydrogen bond ortho substituted product is major. This reaction is called Gattermann reaction.

24. Phthalic acid reacts with resorcinol in the presence of concentrated H2SO4 to give (1) phenolphthalein. (2) alizarin. (3) coumarin. (4) fluorescein. Solution (4) Phthalic acid reacts with resorcinol in the presence of conc. H2SO4 to give fluorescein.

Chapter 26_Alcohols Phenols and Ethers.indd 681

26. Which one of the following reactions will produce tertbutyl methyl ether in high yield? (1) tert-butyl chloride + Sodium methoxide (2) tert-butanol + methanol in presence of H2SO4 (3) tert-butyl bromide + bromomethane in presence of NaOH (4) Sodium tert-butoxide + bromomethane Solution (4) Williamson’s synthesis is an important method used to synthesize unsymmetrical ethers. It involves an SN2 reaction of a sodium alkoxide with an alkyl halide (also, alkyl sulphate or alkyl sulphonate). Due to SN2 mechanism, best results are obtained when the alkyl halide is primary (or methyl) that is most unhindered. If the substrate (alkyl halide) is tertiary, elimination reaction becomes predominant.

1/4/2018 5:27:25 PM

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OBJECTIVE CHEMISTRY FOR NEET

27. The product(s) of the following reaction 1 equiv. HI Heat

O

(1) (3)

28. The product of the following reaction is

(1)

(2) I

O

(4)

OH

O

(2) (3) Br

OH

I

I

Excess HBr Heat

O

OH and Br

(4)

O

Solution

OH

Br Br

Br

O

(3) When one side of the epoxide is a tertiary position, the reaction is observed to occur at more substituted, tertiary site. 1 equiv. HI

O

I

Solution (3) Due to presence of HBr in excess, addition of two mole equivalents is possible.

OH

Heat

O

Excess HBr Heat

Br

OH

+ Br

Br

Practice Exercises Level I

(1) I, III, II, IV (2) III, I, IV, II (3) I, IV, II, III (4) IV, I, III, II

Alcohols 1. Which is the IUPAC name for the following structure?

6. Which among the following compounds is most acidic in character? (1) HOH (3) CH3

OH

(1) Cyclohexenol (2) 3-Cyclohexen-1-ol (3) 1-Cyclohexen-4-ol (4) 4-Cyclohexenol

(1) CH3CH2CH2CH3 (2) CH3CH2OCH3 (3) CH3CH2CH2OH (4) HOCH2CH2OH

(1) methanol. (2) ethanol. (3) acetone. (4) benzene.

+

5. Arrange the compounds in the order of increasing solubility in water (least soluble first).

(III)

OH HO

Chapter 26_Alcohols Phenols and Ethers.indd 682

(IV)

CH3

8. Which compound would have the highest boiling point?

4. Wood spirit is known as

(II)

(4) H3C

(1) by removing the water in the using concentrated sulphuric acid. (2) by removing the water in it using phosphorus pentoxide. (3) by distilling with the appropriate amount of benzene. (4) by distilling over plenty of quick lime.

3. Which of the following is a dihydric alcohol?

OH

OH

7. Absolute alcohol can be obtained from rectified spirit

(1) CnH2nO2 (2) CnH2nO (3) CnH2n+1O (4) CnH2n+2O

(I)

CH2

OH OH CH3

2. The general molecular formula, which represents the homologous series of alkanols, is

(1) Glycerol (2) Ethylene glycol (3) Catechol (4) Resorcinol

(2) CH3

3O 9. C 4H8O2 + CH 3MgBr ¾H¾¾ ® C 4H10O (Ester A) (2 parts) Alcohol(B)



OH

OH

Alcohol B reacts faster with Lucas reagent. Hence A and B are (1) (2) (3) (4)

CH3   COO   C2H5, (CH3)3COH HCOOC3H7, (CH3)2CHOH CH3   COO   C2H5, (CH3)2CHOH HCOOC3H7, (CH3)3COH

1/4/2018 5:27:27 PM

683

Alcohols, Phenols and Ethers 10. The conversion of CH3 H3C

CH

(3) CH3CH2CH2COH

CH3 CH

CH3

to

H3C

CH

CH

CH3

(4) CH3CH2CH2COCH3

15. Which alcohol would undergo acid-catalyzed dehydration most rapidly?

Br

OH



O

O

is best achieved through use of which of these reagents in a low temperature reaction? (1) conc. HBr (2) Br2 (3) NaBr, H2SO4 (4) PBr3

(1) 3,3-Dimethyl-1-butanol (2) 2,2-Dimethyl-1-butanol (3) 3,3-Dimethyl-2-butanol (4) 2-Methyl-2-butanol 16. Which of the following is used as antifreeze?

11. The    OH group of an alcohol or the carboxylic acid can be replaced by    Cl, using (1) hypochlorous acid. (2) chlorine. (3) hydrochloric acid. (4) phosphorous pentachloride.

(1) Glycerol (2) Ethyl alcohol (3) Water (4) Methanol 17. What is the product of the following reaction? OH O

12. Which is the product of the following reaction?

NaBH4/H2O

O

(1)

(2)

PCC

OH

dioxane

O

OH

O

OH

(3) O

(1)

(2)

HO

OH

(4)

H

OH

OH

OH

O

O OH

18. In the following sequence of reactions, Z is

HO

Z O H

(3)

H

(4)

O

1. PCl5

X

2. Alc. KOH

1. conc. H2SO4 2. H2O, boil

Z

(1) CH3CH2CH2OH

(2) CH3CHOHCH3

(3) CH3CH2CH2CH2OH

(4) (CH3)3CCH2OH

19. What is the product of the following reaction?

13. Which reaction can accomplish the following transformation in good yield? PCC, CH2Cl2, 25˚C

? OH

OH

OH

(1)

(1) H+/H2O (2) Oxymercuration/demercuation (3) Hydroboration/oxidation (4) Reaction with NaOH

O

14. Which of these compounds will not be reduced by LiAlH4? H (1) CH3CH2CH2CH   CH2

Chapter 26_Alcohols Phenols and Ethers.indd 683

(2) CH3CH2CH2C

O

OH

(2) O

H

(3) O

O

O

O

O

OH

(4)

1/4/2018 5:27:29 PM

684

OBJECTIVE CHEMISTRY FOR NEET

20. What is the product of the following reaction?

25. The product of the following reaction is (CH3)3C

1. CH3CH2MgBr, Et2O 2. H3O+

(1) Orange (2) Blue (3) Red (4) No color

OH

(3)

conc. HCl ZnCI2

26. What color is expected from the neopentyl alcohol in Victor Meyer’s test?

(2)

HO

CH3

(1) CH3C   CHCl   CH3 (2) (CH3)C   CH2CH2Cl (3) (CH3)CCl   CH(CH3)2 (4) (CH3)2 C   C(CH3)2

O

(1)

CHOH

(4)

Phenols 27. What is the IUPAC name for the following compound?

OH

OH

OH

21. Which of the following pair of alcohol can be distinguished by Victor Meyer test? (1) (2) (3) (4)

O

Methanol and ethanol Ethanol and 1-propanol 2-Pentanol and 3-pentanol 1-Propanol and 2-propanol

22. Which of the following reagents will give alkyl chloride from alcohol? (1) HCl and anhyd. ZnCl2 (2) SOCl2 (3) PCl3 (4) All of these 23. What would be the major product of the following reaction sequence? OH

Br

(1) p-Hydroxy-m-bromoacetophenone. (2) m-Bromo-p-hydroxypropiophenone. (3) 3-Bromo-4-hydroxypropiophenone. (4) 2-Bromo-4-propanoylphenol. 28. Carbolic acid is (1) phenol. (2) phenyl benzoate. (3) phenyl acetate. (4) salol. 29. Which of these species is the strongest base? (1)

H CH3SO2Cl base

mesylate

O-

O-

(2)

NaI Ethanol

H3C

CH3

H

I

H I

(1)

H

(3)

(2) CH3

H H

H

CH3

I

CH3

O2N

30. Which of the following substances would have the smallest pKa?

SO2l

(4)

(3)

NO2 O

(1) H

O-

(4)

Cl H

I

O-

CH3

(2)

OH O

24. Bouveault–Blanc reduction reaction involves (1) (2) (3) (4)

reduction of an acyl halide with H2/Pd. reduction of an anhydride with LiAlH4 reduction of an ester with Na/C2H5OH. reduction of a carbonyl compound with Na/Hg and HCl.

Chapter 26_Alcohols Phenols and Ethers.indd 684

HO

NH2 (3)

O OH

(4) HO

1/4/2018 5:27:31 PM

Alcohols, Phenols and Ethers 31. Which one of the following statements is not correct?

OH

OH

(1) Alcohols are weaker acids than water. (2) Acid strength of alcohols decreases in the following order RCH2OH > R2CHOH > R3COH. (3) Carbon-oxygen bond length in methanol, CH3OH is shorter than that of C   O bond length in phenol.

Br

Br

(1) OH

(2)

CHO

CHO OH

O (4) The bond angle C

H in methanol is 108.9°.

(3)

32. Which of the following isomer of nitrophenol is steam volatile?

(4)

CHO Br

OH

33. The major product formed in the following reaction is

NaOH + CHCI3

OH

(1)

Br



36. In the following reaction sequence,

(1) ortho (2) meta (3) para (4) None of these

NaOH

CH3I

NaOH + CCI4

(2)

OCH3

685



OH

Zn dust

C

B

the compounds C and B are, respectively, (1)

CH3

A

CHO

OH COOH

(3)

(4)

OH

OH

(2)

OH

OH CHO

COOH

CH3 CH3

34. What is the product of the reaction of phenol and chloroacetic acid in basic solution, followed by acidification? O

(1) O

(2)

Cl

(3)

OH

COOH CHO

OH

O Cl

(4) None of these

(3)

OH

(4)

O

O

OH O

O

35. What should be the product in the following reaction? OH Br2/Fe

CHO

Chapter 26_Alcohols Phenols and Ethers.indd 685

Product

37. What is the IUPAC name of the final product obtained via the following reaction sequence? 2-Bromophenol

1. 2-bromopropane, AlCl3 2. NaOH 3. (CH3CO)2O

(1) 1-Isopropyl-3-bromo-4-acetylbenzene (2) 2-Bromo-4-isopropyl -acetophenone (3) 3-Bromo-1-isopropyl-4-phenyl acetate (4) (2-Bromo-4-isopropyl) phenyl acetate 38. Which position is predicted to be the chief site of substitution when the following substance reacts with bromine in carbon disulfide at 10°C?

1/4/2018 5:27:33 PM

686

OBJECTIVE CHEMISTRY FOR NEET II

III

(2) Ethers generally have lower boiling points than alcohols of a corresponding molecular weight. (3) Ethers generally have much lower water solubilities than alcohols with a corresponding molecular weight. (4) Ethers can generally be cleaved by heating them with strong acids.

IV

I

V

HO

CH3

(1) I (2) II (3) III (4) IV 39. In the following reaction, P is

44. How many ethers will be formed when a mixture of ethyl and methyl alcohols is treated with conc. H2SO4?

OH HNO3

(1) 1 (2) 2 (3) 3 (4) 4

P

H2SO4

45. Conc. H2SO4 heated with excess of C2H5OH at 140°C to form

CH3

(1) H3C

OH

(1) CH3CH2OCH2CH3 (3) CH3OCH2CH2CH3

O2N

(2)

OH

H3C

Level II

NO2

(3) Both (1) and (2)

(4)

(2) CH3CH2OCH3 (4) CH2   CH2

Alcohols

No reaction

1. What is the correct IUPAC name for the following compound?

40. In the following reaction, identify (B):

CH3

Br

CH3CHOHCHCHCH(CH3)2 Mg Ether

(1)

OMgBr

(A)

H2O

(2)

CH3

(B)

(1) 4-Isopropyl-3,4-dimethyl-2-butanol (2) 2,3,4-Trimethyl-4-pentanol (3) 1,1,2,3-Tetramethyl-4-pentanol (4) 3,4,5-Trimethyl-2-hexanol

MgBr

2. Glycerine has (3)

OH

(1) (2) (3) (4)

(4)

one primary and two secondary    OH groups. one secondary and two primary    OH groups. three primary    OH groups. three secondary    OH groups.

3. Methanol and ethanol are miscible in water due to 41. The product of 2-phenylpropene is

acid-catalyzed

hydration

of

(1) 3-phenyl-propan-2-ol. (2) 1-phenyl-propan-2-ol. (3) 2-phenylpropan-2-ol. (4) 2-phenylpropan-1-ol.

(1) (2) (3) (4)

covalent character. hydrogen bonding character. oxygen bonding character. none of these.

4. Which of the following has maximum pKa value? (1)

OH

OH

(2)

Ethers 42. Oxygen atom in ether is (1) very active. (2) replaceable. (3) comparatively inert. (4) active. 43. Which of the following statements is not true of ethers? (1) Ethers are generally unreactive molecules toward reagents other than strong acids.

Chapter 26_Alcohols Phenols and Ethers.indd 686

(3)

(4) OH

OH

5. Which of the following is true for the following diol? OH OH 2 5

1/4/2018 5:27:34 PM

687

Alcohols, Phenols and Ethers (1) (2) (3) (4)

OH at C2 is more basic than that at C5. OH at C2 is more acidic than that at C5. Both OH groups act as equally strong bases. Both OH groups act as equally strong acids.

6. The correct order of boiling points of the following alcohols: (I) pentan-1-ol; (II) 2-methylbutan-2-ol; (III) 3-methylbutan-2-ol is

10. What is the major product for the following reaction? 1. Hg(OAc)2, H2O 2. NaBH4, NaOH

O

(1)

(1) I > II > III (2) I > III > II (3) III > II > I (4) II > III > I 7. What is the product formed in the following reaction?

H O

(2)

+ enantiomer

1. BH3:THF 2. H2O2, OH-

OH OH

OH

(3) (1)

OH

(2)

OH

+ enantiomer

(4) OH

(3)

OH

(4)

8. What is the product of the following reaction?

11. Conversion of isobutene into isobutyl alcohol can be made possible with (1) HOH/H+ (2) BH3 followed by H2O2/–OH (3) conc. H2SO4 followed by HOH (4) All of these

OH PBr3

(1)

(3)

Br

Br

Br

(2)

Br

(4)

Br

12. An organic compound X on treatment with acidified K2Cr2O7 gives a compound Y which reacts with I2 and sodium carbonate to form triiodomethane. The compound X is (1) CH3OH (2) CH3COCH3 (3) CH3CHO (4) CH3CHOHCH3 O

13. Ph 9. Identify B in the following reaction: 1. PBr3 2. Mg, ether

A

CH3CH2CHO

CH

C

OHH2O

Q

OH (P) B



P and Q are isomers. Identify Q

OH

O

O

(1)

OH

(1) Ph

(2)

OH

(3) Ph

(4)

OH

Chapter 26_Alcohols Phenols and Ethers.indd 687

CH2

C

OH

(2) Ph

O

OH

(3)

H

C

C

OCH3

O CH2OH

(4) H

C

CH2

O

Ph

14. The red-colored compound formed during Victor Meyer’s test for ethanol is

1/4/2018 5:27:37 PM

688

OBJECTIVE CHEMISTRY FOR NEET (1) CH3

C

NO2−Na+

(2) CH3

NOH

(3) CH3CH

C

NO2

N

O−Na+

(4) CH3CH2

NOH

NHOH

15. trans-3-Methylcyclopentanol is treated with CH3SO2Cl in the presence of base. The product of this reaction is then heated with KI in methanol. What is the final product? (1) trans-1-Iodo-3-methylcyclopentane (2) cis-1-Iodo-3-methylcyclopentane (3) 1-Methylcyclopentene (4) 2-Methylcyclopentene 16. Haloform reaction does not take place with (1) acetone. (2) 2-chloropropane. (3) ethanol. (4) methanol.

Phenols OH

20. 2-Phenylethanol may be prepared by the reaction of phenyl magnesium bromide with (1) HCHO

(2) CH3CHO

(3) CH3COCH3

(4)

O

21. Each of the following mixtures was added to a flask or a separation funnel that contained diethyl ether (as an organic solvent) and mixed well. In which case, would the organic compound be present in the aqueous phase (soluble in aqueous NaHCO3)? (1) (2) (3) (4)

p-Cresol + aqueous NaHCO3 p-Chlorophenol + aqueous NaHCO3 p-Nitrophenol + aqueous NaHCO3 Picric acid + aqueous NaHCO3

22. What product is likely to be obtained by the action of Ag+ or Fe3+ on the following substance? OH

17. What is the IUPAC name for

?

CH3 Ag+ or Fe3+

(1) p-Hydroxyphenol (2) p-Dihydroxybenzene (3) Resorcinol (4) 4-Methylphenol 18. The product formed in the following reaction is

OH

OH

OH

(2)

OCH3

OH

OH OH

CH3

CH3

(4)

(4) HO H

19. Which of the following phenols will give lower yield during Kolbe–Schmidt reaction? (1)

O-

O

(3)

O

(2)

O

CH3

OH OCH3

CH3

(1)

CH3

(3)

OH

O CH2N2

(1)

?

(2)

OH

O

23. What is the final product? Phenol

t-butyl chloride AlCl3

C10H14O (para-isomer)

1. NaOH 2. CH3CH2I

?

(1) 1-tert-Butyl-4-ethoxybenzene (2) 1-tert-Butyl-4-ethylbenzene (3) 1-tert-Butoxy-4-ethoxybenzene (4) tert-Butyl ethyl ether CH3

(3)

OH

OCH3

24. The major product of the reaction is

(4) All will give good yield

HNO3, H2SO4

OH NO2

Chapter 26_Alcohols Phenols and Ethers.indd 688

1/4/2018 5:27:38 PM

Alcohols, Phenols and Ethers (1)

(2)

28. Choose the reagent to carry out the following reaction:

NO2

OH

NO2

OH Br

OH

OH

(3) HO

(4) HO NO2

(1) (2) (3) (4)

Bromine water Bromine in CCl4 Either (1) or (2) Reaction is not possible

29. Which of these reactions does not produce phenol?

NO2

O

25. The stability towards dehydration of the following compounds decreases in the order (I)

689

1. NaOH, H2O 2. H3O+

O ONa

(II)

OH

(1)

H3O+

(2)

OH

O

(III)

(IV) OH

HI

(3)

OH

Cl

H2O

(4)

(1) I > II > III > IV (2) I > IV > III > II (3) IV > II > I > III (4) II > III > IV > I 26. The reaction of the following compound with HBr gives

100°C

Ethers 30. Select the structure of benzyl methyl ether.

CH3

CH

CH

(1) CH3CHBrCH2

OH

OH (2) CH3CH2CHBr

(3) CH3CHBrCH2

Br

(4) CH3CH2CHBr

OH

(1)

CH3

(2)

CH3O

(3)

CH3

C6H6

(1)

2H2SO4, heat

A

NaOH, 270˚C

(2)

SO3H

B

CH2

CH

O

CH3

CH CH3

31. Which is a correct IUPAC name for CH3CH2OCH2CH2CH2OCH2CH3?

OH

OH

O

CH3

Br

(4) 27. Identify the end product in the given reaction:

O

OH

(1) 1,4-Dioxane (2) Ethylene glycol diethyl ether (3) 1,3-Diethoxypropane (4) 1,2-Diethoxyethane 32. What is the product of the following reaction?

(3)

(4)

ONa

OH OH

OH 1. NaH

ONa

Chapter 26_Alcohols Phenols and Ethers.indd 689

2. Isopropyl iodide

1/4/2018 5:27:40 PM

690

OBJECTIVE CHEMISTRY FOR NEET CH3

(1)

(1) CH 3

O

OCH3

CH

CH2OH + CH3

CH2

I

CH3

(2)

(2) CH 3

CH

CH2

(3) CH3

CH

CH3 + CH3CH2OH

I + CH3 CH2OH

CH3 O

(4) CH3

O

CH2OH + CH3CH3

CH CH3

(AIPMT 2007) (3)

(4)

CH2OH

2. CH2OH on heating with periodic acid gives

33. The product(s) of the following reaction is(are):

O

(1)

(1) 2

1 equiv. HI Heat

C

(2) 2CO2

O

(4) CHO

O

CHO OH and

(AIPMT 2009)

I

3. Consider the following reaction,

(3) I

Ethanol

OH I

(4)



O

34. What would be the major product of the following reaction? O

1. NaBH4 2. Ethylene oxide, HA

A

O

X

alc. KOH

Y

H2SO4 H2O, heat

Z

the product Z is (1) CH3CH2OH (2) CH2   CH2 (3) CH3CH2   O   CH2   CH3 (4) CH3   CH2   O   SO3H

(AIPMT 2009)

(1) II > III > I > IV (2) II > III > IV > I (3) III > IV > II > I (4) III > II > IV > I

O

(2)

PBr3

4. Given are cyclohexanol (I), acetic acid (II), 2,4,6-trinitrophenol (III) and phenol (IV). In these, the order of decreasing acidic character will be

OH

(1)

O

(AIPMT PRE 2010) HO

O

OH

5. Which one of the following compounds has the most acidic nature?

O

(4)

(1)

OH

Previous Years’ NEET Questions (3)

CH3 CH3

CH

CH2

O

CH2

CH3 + HI

(2)

OH

CH

(4)

CH2OH

∆

Which of the following compounds will be formed?

Chapter 26_Alcohols Phenols and Ethers.indd 690

OH

OH

1. In the reaction



H

(3) 2HCOOH

(2)

(3)

H

(AIPMT PRE 2010)

1/4/2018 5:27:43 PM

691

Alcohols, Phenols and Ethers 6. Among the following four compounds: (II)   methyl phenol (IV)   para-nitrophenol

(I) phenol (III) meta-nitrophenol

10. Which of the following will not be soluble in sodium hydrogen carbonate? (1) 2,4,6-Trinitrophenol (2) Benzoic acid (3) o-Nitrophenol (4) Benzenesulphonic acid

The acidity order is (1) III > IV > I > II (2) III > IV > III > II (3) II > I > III> IV (4) IV > III > I > II (AIPMT MAINS 2010)

(AIPMT 2014) 11. The reaction is called CH3

7. When glycerol is treated with excess of HI, it produces (1) allyl iodide. (2) propene. (3) glycerol triiodide. (4) 2-iodopropane.

CH3

(1) (2) (3) (4)

8. Among the following ethers, which one will produce methyl alcohol on treatment with hot concentrated HI? CH3

CH2

CH2

(2)

CH3

CH2

CH

CH2

O

ONa + CH3CH2CI

C

−NaCI

CH3

CH3

(AIPMT MAINS 2010)

(1)

CH3

CH3

C

O

CH2

CH3

CH3

Williamson continuous etherification process. Étard reaction. Gattermann–Koch reaction. Williamson synthesis. (AIPMT 2015)

O

CH3

12. Reaction of phenol with chloroform in presence of dilute sodium hydroxide finally introduces which one of the following functional group?

CH3 CH3

(3)

CH3

C

O

(1)    CHCl2 (2)    CHO (3)    CH2Cl (4)    COOH

CH3

CH3

(4)

CH3

CH

(RE AIPMT 2015) CH2

O

CH3

13. The reaction

CH3

OH

(NEET 2013) 9. Among the following sets of reaction which one produces anisole? (1) CH3CHO; RMgX (2) C6H5OH; NaOH; CH3I (3) C6H5OH; neutral FeCl3 (4) C6H5-CH3; CH3COCl:AlCl3



NaH

+

O− Na

Me−I

O

Me

can be classified as (1) (2) (3) (4)

Williamson ether synthesis reaction. Alcohol formation reaction. Dehydration reaction. Williamson alcohol synthesis reaction. (NEET I 2016)

(AIPMT 2014)

Answer Key Level I  1. (2)

 2. (4)

 3. (2)

 4. (1)

 5. (3)

 6. (1)

 7. (1)

 8. (4)

 9. (1)

10. (4)

11. (4)

12. (3)

13. (2)

14. (1)

15. (4)

16. (2)

17. (4)

18. (2)

19. (3)

20. (1)

21. (4)

22. (4)

23. (1)

24. (3)

25. (3)

26. (3)

27. (3)

28. (1)

29. (2)

30. (2)

31. (3)

32. (1)

33. (1)

34. (4)

35. (2)

36. (1)

37. (4)

38. (1)

39. (1)

40. (4)

41. (3)

42. (3)

43. (3)

44. (3)

45. (1)

Chapter 26_Alcohols Phenols and Ethers.indd 691

1/4/2018 5:27:44 PM

692

OBJECTIVE CHEMISTRY FOR NEET

Level II  1. (4)

 2. (2)

 3. (2)

 4. (4)

 5. (1)

 6. (2)

 7. (2)

 8. (4)

 9. (2)

10. (3)

11. (2)

12. (4)

13. (3)

14. (2)

15. (2)

16. (4)

17. (4)

18. (4)

19. (3)

20. (4)

21. (4)

22. (1)

23. (1)

24. (1)

25. (4)

26. (2)

27. (3)

28. (2)

29. (4)

30. (3)

31. (3)

32. (2)

33. (3)

34. (1)

 5. (1)

 6. (4)

 7. (4)

 8. (3)

 9. (2)

10. (3)

Previous Years’ NEET Questions  1. (1)

 2. (1)

 3. (4)

11. (4)

12. (2)

13. (1)

 4. (4)

Hints and Explanations Level I 2. (4) Alkanols consist of an alkane that contains a hydroxyl (OH) group. The general formula of homologous series of alkanes is CnH2n + 2O.

17. (4) Reduction with sodium borohydride reduces only the ketone, not the acid. 18. (2) The reaction is

6. (1) This is because alcohols are weaker acids than water. While going from primary to secondary to tertiary, the acidity of alcohols decreases. This decrease in acidity is due to two factors that increase the activation energy for proton removal: Increase of electron den­sity on the oxygen atom of the more highly-substi­ tuted alcohol, and steric hindrance (because of the alkyl groups, which inhibit solvation of the resulting alkoxide ion). 9. (1) Alcohol (B) is a tert-alcohol as it reacts fastest with Lucas reagent. Its structure is CH3 CH3

OH

C CH3

11. (4) Alcohol and carboxylic acid react with phosphorous pentachloride in which the hydroxyl group is replaced by chlorine atom to give alkyl chloride and acyl chloride respectively in good yield. The reactions involved are R R

OH + PCl5 COOH + PCl5

Cl + POCl3 + HCl

R R

COCl + POCl3 + HCl

14. (1) Lithium aluminium hydride reduces carboxylic acids, esters, aldehydes and ketones to alcohols.

Chapter 26_Alcohols Phenols and Ethers.indd 692

CH3

CH

CH3

PCI5

CH3

Nucleophilic substitution

OH

CH

CH3

CI

(Z) alc. KOH Elimination

CH3

CH

1. conc. H2SO4

CH2

2. H2O, boil Electrophilic addition

CH3

(X)

CH

CH3

OH (Z)

19. (3) Oxidation with PCC oxidizes primary alcohols to aldehydes and secondary alcohols to ketones. 20. (1) Grignard reagents reduce ketones to alcohols while adding the alkyl group once. 21. (4) Victor Meyer test is used to distinguish between 1°, 2° and 3° alcohols.

1-Propanol and 2-propanol can be distinguished by Victor Meyer test since these alcohols have different degrees.

22. (4) All reagents will convert alcohol into alkyl chloride. R

OH + HCI

ROH + SOCl2 ROH + PCl3

ZnCl2

R

Cl + H2O

RCI + SO2 + HCI RCl + H3PO3

1/4/2018 5:27:45 PM

Alcohols, Phenols and Ethers 2 3. (1) OH

OSO2CH3 H

H

I NaI

Base

H

and is more volatile than para isomer where intermolecular hydrogen bonding takes place. The boiling point of meta isomer lies between that of ortho and para isomers.

H

CH3SO2Cl

CH3 H

CH3

O

Acetone (SN1)

O N

H

693

H

CH3

O

Mesylate

2 4. (3) Bouveault–Blanc reduction involves the reduction of esters to primary alcohols in the presence of sodium and alcohol. RCOOR’

Na/C2H5OH

o-Nitrophenol

3 3. (1)

RCH2OH + R’OH

O−

OH

O

CH3

2 5. (3) (CH3)3C

CH

CH3

conc. HCI ZnCl2

+

(CH3)3C

CH

OH

H3C

+ NaOH

CH3

1,2-Methyl shift

I

3 4. (4)

CI H3C

O

C

CH

H3C

CH3

CI−

CH3

CH3

+

C

CH

H3C

CH3

O

CH3

H +

Cl

CH2

C

OH−

(White turbidity)

2 6. (3) Neopentyl alcohol is a primary alcohol, and hence in Victor Meyer test gives red coloration.

O O

CH3 CH3

C

CH2

CH3

O

O−

CH2

COOH

3 5. (2) Electrophile will attack at ortho position with respect to activating group (   OH). The reaction is

30. (2) Both —NO2 and —CH3CO in the compound

OH

OH

NO2

Br

HO

Br2/Fe

CHO O



C

H+

CH2OH

2 9. (2) Basic character is inversely proportional to the stability of the anion. Compound (2) is least stable anion due to the presence of electron donating methyl group.

OH

are electron withdrawing, hence, it will be most acidic and will have smallest pKa value.

3 1. (3) In phenol, due to resonance bond strength between C   O bond become stronger than C   O bond in alcohol hence C   O bond length in phenol is shorter than C   O bond in alcohol. 3 2. (1) Intramolecular hydrogen bonding takes place in ortho isomer. Therefore, this has a lower boiling point

Chapter 26_Alcohols Phenols and Ethers.indd 693

CHO

3 6. (1) In the first reaction, the first step involves Reimer– Tiemann reaction on phenol in the presence of chloroform followed by hydrolysis to yield salicylaldehyde. In the next step, treatment with zinc dust removes the hydroxyl group to give benzaldehyde. OH

OH

CHO CHO

NaOH + CHCI3





Zn dust

(A)

(C)

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OBJECTIVE CHEMISTRY FOR NEET

The second reaction also involves Reimer–Tiemann reaction on phenol, but in the presence of carbon tetrachloride, to yield salicylic acid. OH

40. (4) The reaction is Br

MgBr

OH

H2O

Mg Ether

COOH NaOH + CCl4

(B)



Grignard reagent

41. (3)

Compound C is benzaldehyde and B is salicylic acid.

OH

37. (4) OH

OH Br

CH2

O− Br

2-Bromopropane AlCl3

C

CH3

CH3

+ C

CH3

CH3

C

CH3

Br H+ H2O

NaOH

H+ H2O

44. (3) (CH3CO)2O

O

C2H5OH + CH3OH

conc. H2SO4

CH3

O

+ CH3

O Br



CH3 + CH3 CH2

O

O

CH2CH3

CH2CH3

Therefore, total 3 (self + crossed condensation products).

45. (1) 38. (1)  OH is a better electron donating group than of  CH3. It is also ortho- and para-directing. Since, the para position is already occupied hence, substitution will take place at position I. HO

CH2

CH3 Br2/CS2

Br HO

CH2

CH3

Strong activator OH NO2 HNO3

CH3 Weak activator

Chapter 26_Alcohols Phenols and Ethers.indd 694

4. (4) OH has maximum number of alkyl groups attached to it (tertiary ROH). Therefore, +I effect increases and acidic strength decreases or pKa increases. 5. (1) OH group at C2 forms a tertiary alcohol while OH at C5 forms a secondary alcohol. Therefore, OH at C2 is more basic than OH at C5.

6. (2) Greater the extent of hydrogen bonding, greater is the boiling point. Therefore, ROH > R2COH > R3COH. 7. (2) Hydroboration-oxidation gives the syn, anti-­ Markovnikov product.

H2SO4

CH3

Level II

1 OH OH 2 34 5 6

39. (1) The reaction is OH

. H 2 SO4 C 2H5OH + C 2H5OH conc   → CH 3CH 2OCH 2CH 3 + H 2O 140° C

8. (4) Addition of PBr3 to the alcohol inverts the stereochemistry.

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695

Alcohols, Phenols and Ethers 9. (2) The reaction is 1. PBr3

16. (4) For haloform reaction, the group should have  CH3CO    or get oxidized to it, or an intermediate of haloform reaction. CH3OH does not have the acetyl group so it does not undergo haloform reaction.

2. Mg, ether

OH

Br

MgBr (A) CH3

CH3

CH2

CH

18. (4) CH2NH2 is a strong electrophile and readily releases N2. CH2N2

CH3CH2CHO

CH

CH2

CH2 + N2

OH

CH3

OCH3 CH2N2

OH (B)

10. (3)

19. (3) Electron withdrawing groups decrease the yield in Kolbe-Schmidt reaction. OH

Hg(OAc)2 H2O

OH

NaBH4

H

CH2 + (H

BH2)2

(CH3

CH

Diborane

CH3

CH

CH2

CH3

CH2)3B

C6H5MgBr C6H5MgBr

CH3CHO

1. HCHO

CH3

CH3

K2Cr2O7

CH3

C

CH3

OH

O

(X)

(Y)

C6H5MgBr

Na2CO3/I2

CH3COCH3

S

O O

O

Cl

S O

Base

H3C

C6H5MgBr

Iodoform (Yellow ppt.)

CH3

trans-3-Methyl cyclopentanol

CH3

I

C

C6H5

CH2

C6H5

CH2

CH2

OH

2-Phenylethanol

21. (4) In general, phenols are more acidic than alcohols but less acidic than carboxylic acid. Carboxylic acids can react with weak base NaHCO3, and are soluble in aqueous sodium bicarbonate. So, only those phenols can be dissolved in aqueous NaHCO3 which are sufficiently acidic to react with weak base, NaHCO3.

KI/Methanol SN2

H2C

CH3

CH3 2-Phenylpropan-2-ol

O

CHI3 + CH3COONa

O CH3

C6H5

OH

15. (2) OH

CH

OH 1-Phenylethanol

12. (4) The reaction is CH

C6H5CH2OH

2. H3O+

OH + B(OH)3

CH3 Isobutyl alcohol

CH3

OCH3

20. (4) The reactions involved are

CH3

3 CH3


methyl phenol > phenol. 7. (4) The reaction is

H+

CH2OH

O

2.

CHOH + 3HI

OH

Glycerol

Previous Years’ NEET Questions CH3  CH 3  CH   CH 2   O   CH 2   CH 3 + HI ∆ → CH 3  CH 3  CH  CH 2OH + CH 3  CH H2  I

OH

+ HIO4

2HCHO

3. (4) The reactions involved are: C2H5OH

PBr3

C2H5Br (X)

alc. KOH ∆ −HBr

CH2

CH2I

CH2I

(Y)

CH3CH2

OSO3H (Z)

4. (4) Acetic acid is stronger than phenol, as acetate ion is more stable than phenoxide ion. But in case of picric acid, (2,4,6-trinitrophenol), because of strong electron withdrawing nature (–I and –R) of nitro groups it becomes even more acidic than CH3COOH. Cyclohexanol is an even weaker acid than H2O. 5. (1) Phenol is more acidic than alcohols, because its conjugate base phenoxide ion is much more stable as com­ pared to the conjugate base of ROH, that is alkoxide ion.

CH3 −I2

CHI

CH

CH3

CH2

CH2I

Propane

(Unstable)

CHI

8. (3) The reaction is CH3 CH3

C

CH3 O

CH3 + HI

CH3OH + CH3 Methanol

CH3

C

I

CH3

9. (2) Phenols can be converted to ethers through Williamson synthesis. Because phenols are more acidic than alcohols, they can be converted to sodium phenoxides through the use of sodium hydroxide. C6H5OH + NaOH → C6H5ONa

CH2

H2SO4

Chapter 26_Alcohols Phenols and Ethers.indd 697

CH + I2

CH3

2-Iodopropane

2. (1) 1,2- or Vicinal diols are cleaved by periodic acid, HIO4, into two carbonyl compounds. This reaction is selective for 1,2-diols.

CH2

CHI

1,2,3-Triodopropane Allyl iodide (unstable) HI

CH3

1. (1) The reaction involved is

OH

CH2

CH2OH

O

CH2

Warm

CH2I

CH3I

C6H5OCH3 + NaI

10. (3) Most phenols are not soluble in aqueous sodium bicarbonate (NaHCO3), but acids are soluble. 2,4,6Trinitrophenol is a stronger acid due to the presence of three electron withdrawing nitro groups. 11. (4) The reaction is known as Williamson synthesis. CH3 CH3

C CH3



CH3 O−Na+ + CH3CH2

CI

CH3

C

OCH2CH3 + NaCI

CH3

It consists of an SN2 reaction of a sodium alkoxide with an alkyl halide. It is an important route to synthesize the unsymmetrical ethers.

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OBJECTIVE CHEMISTRY FOR NEET

12. (2) Treatment of phenol with chloroform in presence of aqueous sodium or potassium hydroxide at 340 K followed by hydrolysis of the resulting product gives 2-hydroxybenzaldehyde (salicylaldehyde). This reaction is called Reimer–Tiemann reaction. OH

13. (1) The given reaction depicts Williamson synthesis of ether. In this reaction, sodium alkoxide reacts with alkyl halide to form ether.

OH CHO + CHCI3

dil. NaOH



Chapter 26_Alcohols Phenols and Ethers.indd 698

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27

Aldehydes and Ketones

Chapter at a Glance 1. Introduction   Aldehydes (RCHO) and ketones (R2CO) are similar in structure in that both classes of compounds possess a C   O bond, called a carbonyl group. The carbonyl group of an aldehyde is flanked by a hydrogen atom, while the carbonyl group of a ketone is flanked by two carbon atoms. Carbonyl group O

O

R H An aldehyde

R

R A ketone

2. Structure of Carbonyl Group   The carbon atom of the carbonyl group is sp2 hybridized and carbon is joined to three other atoms by sigma bonds. The fourth bond is made by the overlap of p-electrons of carbon with oxygen to form a p-bond. 120°

sp2 C

120° sp2

O

120°

3. Nomenclature of Aldehydes and Ketones (a) A  ldehydes are named substitutively by replacing the final – e of the name of the corresponding alkane with - al. Example, methanal, ethanal and propanal. O

O

O

C H H Methanal (formaldehyde)

H Ethanal (acetaldehyde)

H Propanal (propionaldehyde)

(b) A  ldehydes in which the CHO group is attached to a ring system are named substitutively by adding the suffix carbaldehyde. O O C

O

C

H

C H

Benzencarbaldehyde (benzaldehyde)

H Cyclohexanecarbaldehyde

2-Napthalenecarbaldehyde

(c) A  liphatic ketones are named substitutively by replacing the final - e of the name of the corresponding alkane with - one. The chain is then numbered in the way that gives the carbonyl carbon atom the lowest possible number, and this number is used to designate its position.

Chapter 27_Aldehydes and Ketones.indd 699

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OBJECTIVE CHEMISTRY FOR NEET

O

O

O

Butanone (ethyl methyl ketone)

Pent-4-en-1-one (not 1-penten-4-one) (allyl methyl ketone)

2-Pentanone (methyl propyl ketone)

4. Methods of Preparation of Aldehydes R’−H R’−ΟH (where R’ =

)

1. CrO2Cl2/CS2

Reimer–Tiemann reaction

2. H3O+

R’−H

(where R’ =

)

CHCl3 /KOH

Etard reaction CO, HCl

(where R’ =

)

R’

R’

H

H Alkene

1. O2 2. Zn, HOAc

Gatterman–Koch reaction

O

Ozonolysis

PCC

R'

R'

Oxidation

OH

1°alcohol

H 1. ∆/DIBAL-H 2. H2O

1. LiAlH(O-t-Bu)3 2. H2O

R'

Stephen reaction

O

reduction

R'

Cl

Acyl chloride

R’

Esters

R′

SnCl2/HCl

Reduction

Rosenmund

O

H2SO4/HgSO4 H2 O

C

C C H Alkyne

N

Nitriles

OR”

5. Methods of Preparation of Ketones O

R’

R’ Cl Acyl chloride

R”

Friedel-Crafts acylation

R’’

R’

Ozonolysis

Alkene

1. O3 2. Me2S (Products depend on R group)

OH

1. ArH, AlCl3;

R’

2. HOH, (leads to R’ Ar)

1. R’MgBr R”Li 2. H3O+

O

KMnO4

R’

C N Nitrile

R’

Oxidation

R’

R’

2°alcohol 400°C

R’COO Ca

dil. H2SO4/ HgSO4

R’

C

R’

R’COO Calcium salt of fatty acid

Chapter 27_Aldehydes and Ketones.indd 700

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Aldehydes and Ketones

701

6. Physical Properties (a)  Aldehydes and ketones have relatively higher boiling points in comparison to the corresponding hydrocarbons, due to the presence of polar carbonyl group. (b)  The lower members of aldehydes and ketones (up to four carbon atoms) are soluble in water. It is due to their capability of forming hydrogen bonds with water molecules. Their solubility in water decreases with increasing length of the alkyl chain. 7. Chemical Reactions of Aldehydes and Ketones (a) T  he presence of carbonyl group in both aldehydes and ketones is responsible for similar chemical reactivity exhibited by the two groups of compounds. (b) The most characteristic reaction of aldehydes and ketones is nucleophilic addition to the carbon–oxygen double bond. General Reaction Nu

R O +H

C

Nu

R

OH

C

H

H

H OH 1. +

O H3

gX

2.

RM

Cl

Al ka

Cl

ne Zn-

Hg

Cle

mm

HO

/ co

nc.

Na OSO 2 OH

l2

OC

HC l en red uct ion

ens

l5

PC

O3

NaHS

O

[H+

S or

], H2O

OH

Hydrate formation

ine fo r Enam

N Raney nickel or Raney Ni Desulphurization

H H

] R2NH −H2O

Oxime formation

ati

for m

ine

lm

S

[H+

HCN, KCN

on

S

[H+] RNH2 −H2O

on

OH

n

−H2O

ati

cti du Re

HS

O

for m

atio

O

orm ne f

−H2O

in

razo

HO

H2C=PPh3

n

oh yd r

matio n

OR

Witt ig re actio

Cy an

Hyd

RO

n atio orm tal f Ace on ati [H+] orm n f l io ta OH at ace rm clic o y f C l ta [H+] ce oa i h ct SH cli Cy

on

[H+] ROH′ − H2O

[H+] NH2OH H2O

CN

1. LAH 2. H2O

[H+] NH2NH2 −H2O

OH

R R

N

Aldimine (Schiff’s base)

R N

OH

N

NH2

NaOH, H2O, heat Wolff–Kishner Reduction

Chapter 27_Aldehydes and Ketones.indd 701

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OBJECTIVE CHEMISTRY FOR NEET

8. Some Other Important Reactions (a) Oxidation (i) Aldehydes are oxidized to carboxylic acids by a variety of common oxidizing agents. RCHO + [O]

RCOOH O

O H2CrO4

H

Hexanal

Hexanoic acid

OH

(ii) Ketones are more resistant to oxidation than aldehydes. According to Popoff ’s rule, during oxidation of ketones, keto group always stays with the smaller alkyl group. R

1 CH2

2 C

3 CH2

[O]

R’

COOH + R’

R

O

(By cleavage of C2 CH3

CH2COOH

(By cleavage of C1 C2 bond) + R CH2COOH + R’ COOH

CH2COCH2

[O]

C2H5

C3 bond)

CH2COOH + C2H5CH2COOH +CH3CH2COOH + C2H5COOH

(b) Aldol condensation: Aldehydes and ketones containing a-hydrogen atoms react to form b-hydroxyl aldehydes (aldol) and b-hydroxyl ketones (ketol). OH

O 2CH3CH

10% NaOH (aq.)

O ∆

CH3CHCH2CH

5˚C

CH3CH

3-Hydroxybutanal (50%)

CH

CHO

2-Butenal

(c) Crossed aldol condensation O

O

CH3CH + CH3CH2CH Acetaldehyde Propanal

OH OH− H2O

O

OH

CH3CHCH2CH + CH3CH2CH OH 3-Hydroxybutanal

CH

C

CH3

O

O

H + CH3CHCHCH

3-Hydroxy-2methylpentanol

OH

O

+ CH3CH2CHCH2CH

CH3 3-Hydroxy-2methylbutanal

3-Hydroxy pentanal

(d) Cannizzaro’s reaction: Aldehydes lacking a-hydrogen undergo disproportionation in the presence of a strong base to give an alcohol and sodium salt of an acid. 2HCHO Formaldehyde

NaOH

CH3OH + HCOONa Methyl Sodium alcohol formate

(e) Haloform reaction: Methyl ketones react with sodium hypohalites to form sodium salts of carboxylic acids and haloforms. O R

C

O CH3

NaOX

R

C

ONa + CHX3 (X = Cl, Br, l)

 Iodoform reaction is used for the detection of CH3CO group or CH3CH(OH) group which produces CH3CO group on oxidation.

Chapter 27_Aldehydes and Ketones.indd 702

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Aldehydes and Ketones

NaOH

CH3CHO + l2 R

C

CHl3 lodoform (yellow ppt)

NaOH

CH3 + l2

703

O

9. Chemical Tests to Distinguish Between Aldehydes and Ketones (a) T  ollen’s test (Silver mirror test): Aldehydes reduce Tollen’s reagent, [Ag(NH3)2]+ to Ag that appears in the form of silver mirror. It is given by aldehydes and reducing sugars. R − CHO + 2[Ag(NH3)2]OH

∆

RCOO− + Ag (Silver)

+ 4NH3 + H2O

 Tollen’s reagent test is negative for all ketones, except a-hydroxy ketones. (b) Fehling’s test: This reagent contains copper (II) ions complexed with tartrate ions in sodium hydroxide solution. Aldehydes reduce Fehling’s solution and on heating give reddish-brown precipitate of Cu2O. R − CHO + 2Cu2+ + 5OH−

RCOO− + Cu2O + 3H2O Reddishbrown ppt.

 This test is not given by benzaldehyde and other aromatic aldehydes. (c) Benedict solution: This reagent contains copper (II) ions complexed with citrate ions in sodium carbonate solution. Aldehydes reduce complexed Cu2+ ions in Benedict’s solution to Cu+.

Solved Examples 1. What is the correct IUPAC name for the following compound?

Solution (4) The IUPAC name of the compound is 2,3-dimethyl-2hepten-4-one. 1

O

(1) 5,5-Dimethyl-2-heptanone (2) 5-Ethyl-5, 5-dimethyl-methyl-2-octanone (3) 5-Ethyl-5-methyl- 2-hexanone (4) 5,5-Dimethyl-2-octanone

7

3 6 5 4

2

4

5

7 6

3. Which is the correct structure for 2-hydroxybenzophenone? (1) CH3

OH

O

CH

CH

O

(2)

C OH

O

(3)

C

1

(4)

O CH

CH3 OH

OH

O

2. What is the correct IUPAC name for the following compound?

3

O

Solution (1) The IUPAC name of the compound is 5, 5-dimethyl2-heptanone.

2

Solution (2) The structure of 2-hydroxybenzophenone is O C

O OH

(1) 1,1,2-Trimethyl-1,3-hexenone (2) 1,2-Dimethyl-1,3-hexenone (3) 2,3-Dimethyl-1,3-heptenone (4) 2,3-Dimethyl-2-hepten-4-one

Chapter 27_Aldehydes and Ketones.indd 703

4. The compound having least solubility in water is (1) methanol. (2) acetaldehyde. (3) acetone. (4) acetophenone.

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OBJECTIVE CHEMISTRY FOR NEET

Solution

Solution

(4) In case of acetophenone, that is, C6H5COCH3, there are two hydrophobic alkyl groups present in the molecule, so its solubility will be less.

(2) The presence of electron donating methoxy group facilitates the release of hydride ion.

5. Which compound is a Schiff base? (1)

(2) NH2

(3)

9. Select the structure of the major product in the following reaction

+

-

N

1. KMnO4, H+, ∆

NH2

2. H3O+

(1)

(4)

CH3CH2NH3CH3

(3) OH

NH

(2)

Solution

(4)

OH OH

O

(4) A molecule containing a carbon–nitrogen double bond is called an imine or, alternatively, a Schiff base. 6. Among the following, the least reactive aldehyde is (1) HCHO (2) C6H5CHO (3) CH3CHO (4) C2H5CHO

Solution (4) 1. KMnO4, H+, ∆

Solution (2) C6H5CHO is least reactive towards nucleophilic addition due to steric hinderance. 7. What, in general, is the order of decreasing reactivity of these carbonyl compounds towards nucleophilic reagents? O CH3CCH3

(I)

H CH3C

O O

(II)

HC

(III)

Cyclopentanone

10. Select the structure of the major product in the following reaction. 1. BH3 2. H2O2, OH−, H2O

O

(1) Ethylbenzene (2) 1-Phenylethanol (3) Acetophenone (4) 2-Phenylethanal

(CH3)3CCC(CH3)3

(IV)

(V)

(1) I > III > IV > II > V (2) IV > II > I > III > V (3) V > III > I > II > IV (4) I > IV > II > III > V

Solution (4)

C

CH

CH

ΟΗ

2. H2O2, OH−, H2O

(2) As the number of alkyl group (a-methyl group) increases in a carbonyl group; the electron density of carbonyl carbon increases because alkyl group is electron releasing group which reduces its ability to react towards nucleophilic substitution.

Tautomerization

2

H

(3)

H O OH

−

O−

O−

H

(4) O

−

O− O2 N

It is an example of hydroboration oxidation reaction which gives anti Markovnikov product.

11. Which of the following procedures would not yield 3-pentanone as a major product?

MeO H

(2-Phenylethanal)

O−

O-

H3O+

CH3CH2MgBr

(1) CH3CH2CN

Ether 1. O3

(2)

(3) CH3CH2CN Chapter 27_Aldehydes and Ketones.indd 704

1

CH2CΗΟ

8. In a Cannizzaro reaction, the intermediate that will be the best hydride donor is

(2)

CH

1. BH3

Solution

(1)

?

O

H

(CH3)3CCCH3

Oxidative Ozonolysis

O

2. H3O+

2. Zn/H+ CH3CH2Li

(4) CH3CH2CO2H

H3O+

Ether CH3CH2MgBr Ether

H3O+

1/4/2018 5:28:47 PM

C

CH

CH2

C

HgSO4, H2SO4 H2O

Tautomerization H3O+

CH3CH2MgBr

(1) CH3CH2CN

O

Ether

C

CH3

705

Aldehydes and Ketones

1. O3

(2)

2. Zn/H+ CH3CH2Li

(3) CH3CH2CN

H3O+

CH2CN

Ether

(4) CH3CH2CO2H

CH2

DIBAL-H/−78°C

CH

NH

H3O+

CH3CH2MgBr Ether

H2O

Solution

CH2CHO

(4) 1. CH3CH2MgBr/ether

CH3CH2CN

CH3

2. H3O+

CH2

C

CH2CH3 C

O Pentan-3-one

2. Zn/H+

O Pentan-3-one

CH2

CH2

CH3CH2COOH

CH3CH2MgBr

H

H3O+

(1) Ammonia (2) Hydrazine (3) Nitrogen (4) Phenylhydrazine

CH3CH2COOH

Solution

H

(1)

(2) The reaction involved is

would be

2 Ph

H

CH

H2O, H2SO4

CN

1. DIBAL-H/−78°C

O

2. H2O

H

(3)

HCN

(1)

1. O3

LiAlH4

OH

CH

N

N

CH

Ph

H3O+

(2) CN

CN

2. Zn/H+

(4)

Ph

O + NH2 NH2 Hydrazine

14. What is the major product of the following reaction sequence?

HgSO4

(2)

H

13. The compound C6H5C N N CC6H5 is produced by the reaction of an excess of benzaldehyde with which compound?

CH3CH3 + CH3CH2COO

O

CH2COOH

2. H3O+

CH3

−

12. A good synthesis of

OH 1. KMnO4, NaOH, ∆

CH3CH2 C CH2 Pentan-3-one

H3O+

CHO

O

O CH3CH2Li/ether

CH3CH2CN

C

O3/Zn /H+

+ CH2 = Ο

1. O3

CH2

CH

OH

(3)

OH NH2

1. KMnO4, NaOH, heat 2. H3O+

(4) OH

Solution

Solution

(3)

(2)

OH C

CH HgSO4, H2SO4

O

CH2

C

H2O

CN

O

CH2CN DIBAL-H/−78°C

Chapter 27_Aldehydes and Ketones.indd 705

CH2

H2O

OH CH3

CH

LiAlH4 H3O+

OH CH2NH2

15. What would be the product of the following reaction sequence?

Tautomerization

C

OH

HCN

O

1. PCl3 2. C6H6, AlCl3 3. Zn(Hg), HCl, heat

NH

1/4/2018 5:28:49 PM

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OBJECTIVE CHEMISTRY FOR NEET O

O

N

(1)

OH

N

(2)

(2)

(1) OH

(3)

N

(3)

(4)

N

(4)

Solution Solution

(4) OH

Cl

PCl3

O

O

(3) The reaction involves conversion of a ketone to an enamine.

/AlCl3

O

N

OH (CH3CH2)NH

Friedel–Crafts acylation

H3O+

O Zn(Hg)/HCl; ∆

1 8. What is the product of the following reaction?

Clemmenson reduction

+ (C6H5)3P

O

1 6. Identify the final product Z in the following reaction sequence: Me2C

O + HCN

H3O+

X

(1) (CH3)2C(OH)COOH

(2) CH2

(3) HO

(4) CH3CH

CH2CH(CH3)COOH

H2SO4 ∆

Y

Z

(2)

(1)

C(CH3)COOH CHCOOH

− CH2

Ph

(3)

(4)

Solution (2) The reaction is H3C

C

H3C

O

HCN

H3C H3C

C

OH CN

Solution

Cyanohydrin (X)

(4) This is a Wittig reaction.

H3O+

CH2

C

H2SO4

COOH

∆

CH3 (Z)

H3C

C

H3C

+ − (C6H5)3P–CH2

OH O

COOH

Acid (Y)

1 9. Which is the product of the following sequence of reactions?

1 7. What is the product of the following reaction?

H2N

O

O

Ethanol

H

H3O+

(1)

H2/Pt

H2N

O

(CH3CH2)NH

N

(2)

HN

N

(3)

HN

(4) NH

Chapter 27_Aldehydes and Ketones.indd 706

NH HN HN 1/5/2018 12:49:36 PM

H2N

O

Ethanol

H

N

(1)

(2)

D NH

HN

(4)

(1) D

C

D

O

C

O , D −

(2) D

OH

C

C

O , D −

H

O

HN

OH

D

D

(3) H

HN

NH

If D is deuterium, the products formed are

HN

N

(3)

707

Aldehydes and Ketones

H2/Pt

H2N

O

C

C

O− , D

H

O

Solution (2)

(4) None of these.

OH

Solution (2)

O H2N

O Ethanol

OH−

O

C

D

N

H2N

H

O

OH

N D

D

C

−+

O

D

C

D

D

H2 / Pd

HN O− NH

20. Which is the major product of the reaction of ethyl Grignard and propiophenone (ethyl phenyl ketone)? CH2CH3

(1)

C

CH2CH3

CH2CH3

(2)

OH CH3

(3)

C

C

D

(4)

CH3

C

(1)

Proton exchange

D

C

+ D

O

C

D

D

O

CH

CH3

(2)

C

CH3

CH3

OH

(3) CH3

O

CH3

C

C

O CH3

(4)

C

CH2

CH3

(1) −+

O

O MgBr δ−

δ+

+ CH3CH2MgBr

C

C2H5

C CH2CH3

Solution (4) NaOI is the reagent formed during iodoform test and it will produce yellow coloration only with ketones which contain CH3CO    group.

OH C2H5

C

H+

CH2CH3 D

21. 2D

D

OH

Solution

C2H5

C

O−

OH

22. Which of the following will not give yellow coloration with NaOI?

CH3

OH

O +D

D

OH CH2CH3

C

OH

C

−

O + OH

Chapter 27_Aldehydes and Ketones.indd 707

Products

NaOI also works as a mild oxidizing agent that converts methylene alcohols into methylene carbonyl (acetyl) groups, and then NaOI converts them into iodoform. So compounds (1), (2) and (3) will produce yellow color. Compound (4) cannot produce yellow color because it lacks methylene carbonyl group.

23. The product of acid hydrolysis of P and Q can be distinguished by

1/4/2018 5:28:53 PM

708

OBJECTIVE CHEMISTRY FOR NEET

H2C

OCOCH3

H3C

CH3 (P)

(Q)

OCOCH3

OCOCH3 H2O

H2C

(1) Lucas reagent (2) 2,4-DNP (3) Fehling’s solution (4) NaHCO3

H3C

H2O

(3) Fehling’s and Tollens’ reagent can be used as aldehydes respond to them positively and ketones do not.

OH

O

CH3

CH

CH3

CH3

CH3

CH3

OCOCH3 (Q)

Solution

Practice Exercises

H2C

CH3 (P)

O

OH

CH

O

CH

C Ketone

CH3

OH

O

CHCH C 2

H Aldehyde CH3

OH

O

CH

C

OH

ortho-Hydroxybenzaldehyde 2-Methylbenzaldehyde 2-Hydroxypropanoic acid

(I)

Level I

(III)

(II)

H C

Nomenclature

O

1. What is the correct IUPAC name for the following compound?

CH3

C

CH2

O

O

CH

C

H

C C

CH3

O

CH3

1,3-Butanedione

Benzophenone

Cinnamaldehyde

(IV)

(V)

(VI)

O

(1) I, II, III (2) IV, V, VI (3) II, IV, V (4) III, IV, VI

(1) 2-Methyl-5-heptanone (2) 7-Methyl-4-octanone (3) 6-Isopropyl-4-octanone (4) Isobutyl propyl ketone

Structure and Isomerism

2. What is the correct IUPAC name for the following compound? O

4. Which of the following is the maximum contributing form of carbonyl group? (1)

– C

+ O

+ C

(2)

– O

(3)

C

(4)

O

C

O

5. Which of the following compounds is an acetal? O

O

(1) (1) 2,4-Dimethyl-2-pentenone (2) 2,5-Dimethylcyclopenten-3-one (3) 2,4-Dimethylcyclopent-2-enone (4) 3,5-Dimethylcyclopent-2-enone

O CH

CH

CH3

OH

O

CH

C

(III)

(II)

H C

O CH3

C

CH2

O

O

CH

C

(1) OH

O

O

O

H C

CH3

(2)

(3)

O

OH

O

(4)

O

OCH3 OH

O

C O

CH3

1,3-Butanedione

Benzophenone

Cinnamaldehyde

(IV)

(V)

(VI)

Chapter 27_Aldehydes and Ketones.indd 708

OH

6. Which of the following compounds is a hemiacetal?

ortho-Hydroxybenzaldehyde 2-Methylbenzaldehyde 2-Hydroxypropanoic acid

(I)

(4)

O

O

O

CH3

(2)

O

(3)

3. Which compounds are named correctly? OH

O OCH3

1/4/2018 5:28:56 PM

Aldehydes and Ketones 7. Which of the following compounds is a mixed ketone? (1) 3-Pentanone (2) Benzophenone (3) Acetophenone (4) All of these 8. The number of constitutional isomers of ketones with formula C6H12O is (1) 6 (2) 2 (3) 5 (4) 4 9. Which of the following is not a resonating form of benzaldehyde? H

C

O

H

C

(2)

(1)

H

C

(3)

−

O− +

O+

H

C

O−

(4) +

10. Which structure is the major tautomer of 2-pentanone in aqueous acid? (1)

(2)

+

OH

OH

(1) benzenone. (2) carboxybenzene. (3) hydroxybenzene. (4) benzene carbaldehyde. 14. The most suitable reagent for the conversion of R   CH2   OH ® R   CHO is (1) KMnO4 (2) K2Cr2O7 (3) CrO3 (4) PCC (Pyridinium Chlorochromate) 15. An alkene, C7H14, on reductive ozonolysis gave propanal and a ketone. The probable ketone is (1) acetone. (2) ethyl methyl ketone. (3) pentan-2-one. (4) pentan-3-one. 16. The best reagent to convert pent-3-en-2-ol into pent-3en-2-one is (1) (2) (3) (4)

acidic permanganate. acidic dichromate. chromic anhydride in glacial acetic acid. conc. HNO3

Physical Properties

OH

(3)

OH

(4)

11. Which of these gem-diols is expected to be the most stable? OH

(1) CF3CCF3

17. Which of the compounds listed below would you expect to have the highest boiling point? (They all have approximately the same molecular weight.) (1) Pentane (2) 1-Butanol (3) Butanal (4) 1-Fluorobutane 18. Arrange the compounds in order of increasing solubility in water (least first). O

(2) CH3CH(OH)2

CH

OH OH

O

(3) CH3CCH3 (4) C6H5CH(OH)2 O

Methods of Preparation

CH3CH2

12. Identify the reagent(s) that would bring about the following reaction: CH3CH2CH2COCl

CH3CH2CH2CHO

(1) H2/Ni (2) Li/liq. NH3 (3) LiAlH[OC(CH3)3]3, ether (4) NaBH4, CH3OH 13. CO + HCI anhyd. AICI3



The compound X is

Chapter 27_Aldehydes and Ketones.indd 709

(II)

(I)

OH



709

X + HCI

CH3 CHCH2CH2CH2CH2

CH

(III)

CH3

O C

CH3

(IV)

(1) I, II, III, IV (2) IV, I, II, III (3) IV, III, II, I (4) II, III, I, IV

Chemical Reactions 19. Formation of cyanohydrin from a ketone is an example of (1) (2) (3) (4)

electrophilic addition. nucleophilic addition. nucleophilic substitution electrophilic substitution.

1/4/2018 5:28:57 PM

710

OBJECTIVE CHEMISTRY FOR NEET

20. The least reactive compound towards nucleophilic addition reactions is

29. The increasing order of the rate of HCN addition to compounds I – IV is

(1) propanone. (2) pentan-3-one. (3) pentan-2-one. (4) 2,4-dimethylpentan-3-one.

(I) HCHO (II) CH3COCH3 (III) PhCOCH3 (IV) PhCOPh (1) I < II < III < IV (2) IV < II < III < I (3) IV < III < II < I (4) III < IV < II < I

21. Reaction of acetaldehyde with HCN followed by hydrolysis gives a compound which shows (1) optical isomerism. (2) geometrical isomerism. (3) metamerism. (4) tautomerism.

30. In the given transformation, which of the following is the most appropriate reagent? CH

(1) C6H5NHCH3 (2) (CH3)3N (3) C6H5NHC6H5 (4) C6H5NHNH2

(3)

CH3

(2)

(4)

CH3

(1) Iodoacetone (2) Acetic acid (3) Iodoform (4) Acetophenone 32. The reaction of phenyl magnesium bromide with tertiary butanol results in the formation of (1) (2) (3) (4)

CH2

(1)    CH2OH (2)    COCl (3)    CN (4)    NC

LAH is not sufficiently reactive. RCOOH is converted into RCOOLi. RCOOH is reduced to RCH2OH. RCOOH is reduced to RCH3.

26. Which of the following does not undergo aldol condensation? (1) ClCH2CHO (2) CCl3   CHO (3) C6H5CH2CHO (4) CH3CHO 27. Which of the following aldehydes will undergo Cannizzaro’s reaction? (1) Ethanal (2) Propanal (3) m-Chlorobenzaldehyde (4) Phenylacetaldehyde 28. In which of the following reactions, both oxidized and reduced forms of the same compound are obtained? Aldol condensation Cannizzaro reaction Reimer–Tiemann reaction Kolbe–Schmidt reaction

Chapter 27_Aldehydes and Ketones.indd 710

benzene. phenol. tert-butyl benzene. tert-butylmethyl ether.

Tests to Distinguish Between Aldehydes and Ketones 33. Which of the following does not react with Fehling’s solution? (1) CH3CHO (2) C6H5CHO (3) C6H12O6 (4) HCOOH

25. LiAlH4 (LAH) cannot be used to convert carboxylic acids to the corresponding aldehydes because

(1) (2) (3) (4)

HO

31. Which is the major product formed when acetone is heated with iodine and potassium hydroxide?

24. In Stephen’s reduction, which of the following group is converted to    CHO group?

(1) (2) (3) (4)

CHCH2CH3

(1) NH2NH2, OH- (2) Zn–Hg/HCl (3) Na (liq. NH3) (4) NaBH4

23. Benzaldehyde on reaction with CH2   CH   CH   PPh3 forms CH2

CH Reagent

HO

22. Which of the following will react with acetone to give a product containing C   N   ?

(1)

CHCOCH3

34. The reagent used for separation of acetaldehyde and acetophenone is (1) Fehling solution (2) C6H5NHNH2 (3) NH2OH (4) NaOH - I2 35. In the reaction

CH3

CH

CH

CHO

oxidizing agent

CH3

CH

CH

COOH,

the oxidizing agent can be (1) alkaline KMnO4. (2) acidified K2Cr2O7. (3) Benedict’s solution. (4) all of the above. 36. Which of the following ketones will not respond to iodoform test? (1) 3-Methylbutan-2-one (2) Ethyl tert-butyl ketone (3) Methyl phenyl ketone (4) Dimethyl ketone

1/4/2018 5:28:59 PM

711

Aldehydes and Ketones 37. Which of the following is present in Fehling’s solution? (1) (2) (3) (4)

1. H2, Lindlar catalyst

Potassium tartrate Sodium oxalate Sodium potassium oxalate Sodium potassium tartrate

2. O3 3. Zn, CH3CO2H

(1) 4-Methylhexanal (2) 4-Methyl-1-hexanol (3) 3-Methylhexanal (4) 4, 10-Dimethyldodecane-6,7-dione

38. Which reagent will not differentiate between 3-butenal and 2-butanone? (2) Ag2O,OH(4) KMnO4,OH-

(1) Br2/CCl4 (3) H2NNHC6H5

39. Which one is used in the manufacture of silver mirror? (1) (2) (3) (4)

5. An optically active organic compound has molecular formula C5H12O (X). X on oxidation with CrO3/H2SO4 gives an achiral C5H10O. Hence, X could be OH

Red lead Ammoniacal AgNO3 Ammoniacal AgNO3 + red lead Ammoniacal AgNO3 + HCHO

O

(2)

O

(2)

HO OH

(3)

40. To which of these can the haloform reaction be applied for the synthesis of a carboxylic acid? (1)

OH

(1)

(4) OH

Chemical Reactions 6. Which of the following will be oxidized by HIO4? (I) R

O

(3)

(4)

O

(III) R

C

C

O

O

CH

CH2

OH

CH

(IV) R

R

OH

C

CH

O

OH

CH

CH

OH

OH

R

R

(1) I, II and III (2) I, III and IV (3) I, II and IV (4) II, III and IV

Level II Methods of Preparation 1. When a mixture of calcium benzoate and calcium acetate is dry distilled, the resulting compound is

7. For obtaining ethyl methyl ketone from acetyl chloride, which of the following reagents, can efficiently be employed? (1) One mole of CH3MgI (2) Reaction with HI and P (3) H2, Pd-BaSO4 (4) One mole of (C2H5)2Cd

(1) acetophenone. (2) benzaldehyde. (3) phenol. (4) acetaldehyde. 2. Which of the following on heating with aqueous KOH, produces acetaldehyde? (1) CH3COCl (2) CH3CH2Cl (3) CH2ClCH2Cl (4) CH3CHCl2

8. Which of the following reagents reacts differently with HCHO, CH3CHO and CH3COCH3? (1) HCN (2) NH2NH2 (3) NH2OH (4) NH3

3. Which synthesis or syntheses would yield propanal? (1) CH3CH2CH2OH

PCC CH2Cl2

9. The structure of compound B in the following reaction is

O

(2) CH3CH2CCl (3) CH3C

(II) R

R

CH

LiAlH[OC(CH3)3]3 Ether, −78 °C 1. Sia2BH





CH3CHO + HCHO

(1) CH2

CH

2. H2O2, OH−

dil. NaOH Heat

CH

A

HCN H3O+

COOH

B

(2) CH3CH2

OH

CH

OH

CN

(4) All of these 4. Select the structure of the major product in the following reaction.

Chapter 27_Aldehydes and Ketones.indd 711

(3) CH2

CH

CH CN

OH

(4) CH3

CH

COOH

OH

1/4/2018 5:29:01 PM

712

OBJECTIVE CHEMISTRY FOR NEET

10. In the reaction sequence, the product B is:

OH−

2CH3CHO

(1) CH3

CH2

∆

A CH2

O

B CH2

(I) Ph OH

C

CH3

CH

(4) CH3

CH2

CH

CHO

CH2

(III) CH3

17. Predict the major organic product of the following reaction: O

12. A certain carbonyl compound X on reaction with Zn-Hg/ HCl gives methycyclopentane. The compound X is (1) cyclopentane carbaldehyde. (2) cyclohexanone. (3) 2-methyl cyclopentan-1-one. (4) both (1) and (3).



H

H3O+

H3C

(3) H3C

(4) H3C

(3) H3C

19. Which reagent(s) could be used to carry out the following transformation? O ?

(4) None is correct

OH

(1) Zn(Hg), HCl, reflux (2) LiAlH4, ether (3) HSCH2CH2SH, BF3; then Raney Ni (H2) (4) Two of the above

14. The crossed-aldol product formed when propanal acts as the electrophile and butanal as nucleophile is (1) 3-hydroxy-2-methylpentanal. (2) 3-hydroxy-2-methylhexanal. (3) 2-ethyl-3-hydroxypentanal. (4) 2-ethyl-3-hydroxyhexanal.

20. What new compound will eventually be formed when HCl is added to a solution of pentanal in methanol? (1) 1-Chloro-1-methoxypentane (2) 1,1-Dimethoxypentane (3) 1,1-Dihydroxypentane (4) 1,1-Dichloropentane

15. Which of the following pairs on aldol condensation followed by dehydration give methyl vinyl ketone? (1) HCHO + CH3COCH3 (2) HCHO + CH3CHO (3) CH3CHO + CH3CHO (4) CH3COCH3 + CH3COCH3 16. The correct order of reactivity of PhMgBr with the following compounds is

Chapter 27_Aldehydes and Ketones.indd 712

OH

O−

CH3 O

O

(2) H3C

O

product

H

O

(1) HO2C

O−

(1) aldol. (2) a, b-unsaturated ester. (3) b-ketoaldehyde. (4) acid.

(2) H3C

OH

H2O

18. If the enolate ion combines with the carbonyl group of an ester, we get a (an)

Identify the product formed. (1) H2C

Ag(NH3)2+

O−

O C

H

CH3

H

(1) The reaction involves hydride transfer from one aldehyde to another. (2) The ArCHO is reduced in the reaction, which results in the major products (3) Aldol addition product is not possible for the given reactants. (4) One of the major products of the reaction is ethanol.

CH

C

conc. NaOH ∆

ArCHO + HCHO

C

(1) I > II > III (2) III > II > I (3) II > III > I (4) I > III > I

CH3

11. Which of the following statements is incorrect for the reaction given below?

13. CH3MgBr + CH2

(II) CH3

Ph

O

(2) CH3

O

(3) CH3

C

O

21. What is the major product of the following reaction sequence? O





HCN

HCl H2O Heat

1/4/2018 5:29:04 PM

Aldehydes and Ketones (1)

OH

(2) CN

CN

(3)

OH

(3) a beta-hydroxy aldehyde or a beta-hydroxy ketone. (4) an alpha-hydroxy aldehyde or ketone. (AIPMT 2007)

(4)

OH

COOH

OH

3. Which one of the following on treatment with 50% aqueous sodium hydroxide yields the corresponding alcohol and acid? O

22. Which reaction does not lead to 3-methyl-3-hexanol? (1) (2) (3) (4)

2-Butanone + propylmagnesium bromide 3-Hexanone + methyl magnesium bromide 2-Pentanone + ethylmagnesium bromide 3-Pentanone + ethylmagnesium bromide

23. Which is the major product of the following reaction? O C

HCl / H2O

CH3 + CH3 MgBr

diethyl ether

OH

(1)

CCH3

OH

(2)

CHCH2OH

(1) CH3

C

OH

(4)

CHCH2CH3

(AIPMT 2007) 4. Acetophenone when reacted with a base, C2H5ONa and then heated yields a stable compound which has the structure (1)

(2)

CH

OH

OH

C

(3)

CH

CH

(1) cyclobutanol. (2) cyclobutyraldehyde. (3) n-butane. (4) cyclobutane. 25. The product obtained as a result of aldol condensation of acetaldehyde and acetone in the presence of dilute NaOH contains how many number of s bonds, p bonds, lone pairs? p bonds

C

13

2

2

(2)

16

0

4

(3)

16

1

4

(4)

13

1

4

Previous Years’ NEET Questions

CH2C

CH3

(4)

O

CH3

CH3

C

C

OH

OH

(AIPMT 2008) 5. Which of the following reaction will not result in the formation of carbon-carbon bonds? (1) (2) (3) (4)

Cannizaro reaction Wurtz reaction Friedel–Crafts acylation Reimer–Tieman reaction (AIPMT PRE 2010)

1. Reduction of aldehydes and ketones into hydrocarbons using zinc amalgam and conc. HCl is called Wolff–Kishner reduction. Clemmensen reduction. Cope reduction. Dow reduction.

6. Which one of the following compounds will be most readily dehydrated? (1)

O

(2)

O

(4)

O

OH

OH

(AIPMT 2007) 2. The product formed in aldol condensation is (1) an a, b-unsaturated ester. (2) a beta-hydroxy acid.

O

lone pairs

(1)

Chapter 27_Aldehydes and Ketones.indd 713

CH

CH3

24. Wolff–Kishner reduction converts cyclobutanone to

(1) (2) (3) (4)

(4) CH3CH2CH2CHO

(3) C6H5CHO

CH3

s bonds

(2) C6H5CH2CHO

CH3

CH2CHCH3

CH3

(3)

713

(3)

OH

O

OH

(AIPMT MAINS 2010)

1/4/2018 5:29:06 PM

714

OBJECTIVE CHEMISTRY FOR NEET

7. Following compounds are given: (I) CH3CH2OH CH CHOH

(II) CH3COCH3

11. CH3CHO and C6H5CH2CHO can be distinguished chemically by (1) Benedict test. (2) Iodoform test. (3) Tollen’s reagent test. (4) Fehling solution test.

3

(III)



CH3

(IV) CH3OH

(AIPMT PRE 2012)

Which of the above compound(s), on being warmed with iodine solution and NaOH, will give iodoform? (1) Only I (2) I, II and III (3) I and II (4) I, III and IV

12. Acetone is treated with excess of ethanol in the presence of hydrochloric acid. The product obtained is O

(1)

(AIPMT MAINS 2010)

(1) H2 and Pt as catalyst (2) Glycol with KOH (3) Zn-Hg with HCl (4) LiAlH4 (AIPMT PRE 2011)

(2)

CH3CH2CH2

(3)

(CH3)2C

(4)

9. The order of reactivity of phenyl magnesium bromide (PhMgBr) with the following compounds is

H

O

C

CH3 CH3

C

O

(II)

(I)

Ph Ph

C

O

OH OC2H5 OC2H5 OC2H5

(AIPMT MAINS 2012) 14. Consider the reaction: RCHO + NH2NH2



50% KOH

CH2COO−

+

RCH

Electrophilic addition – elimination reaction Free radical addition – elimination reaction Electrophilic substitution – elimination reaction Nucleophilic addition – elimination reaction (AIPMT MAINS 2012)

Cl

Cl CH2OH

OH

+

15. Consider the following reaction: COCl H2

(3)

CH2OH

COO−

+

CH2OH

COO−

+



A

The product A is (1) C6H5CHO (2) C6H5OH (3) C6H5COCH3 (4) C6H5Cl

Cl

Cl

(4)

Pd-BaSO4

OH

OH

NH2

N

What sort of reaction is it? (1) (2) (3) (4)

Cl

(2)

CH2CH2CH3

(1) Acetophenone (2) Methyl acetate (3) Acetamide (4) 2-Hydroxypropane

CHO

CH2OH

(CH3)2C

C

13. Which of the following compounds will give a yellow precipitate with iodine and alkali?

(1) I> II > III (2) III > II > I (3) II > I > III (4) I > III > II

(1)

CH3

(AIPMT PRE 2012)

(III)

10. Predict the product in the given reaction.

C O

8. Clemmensen reduction of a ketone is carried out in the presence of which of the following?

CH3

CH3CH2CH2

(AIPMT MAINS 2012) 16. Reaction by which benzaldehyde cannot be prepared? CH3

OH

OH

(1) (AIPMT PRE 2012)

+ CrO2Cl2 in CS2 followed by H3O+ COCl

(2)

Chapter 27_Aldehydes and Ketones.indd 714

(3)

+ H2 in presence of Pd-BaSO4

+ CO + HCl in presence of anhydrous AlCl3

1/4/2018 5:29:08 PM

CH3

(1)

Aldehydes and Ketones

+ CrO2Cl2 in CS2 followed by H3O+

21. Which of the following reagents would distinguish cis-cyclopenta-1,2-diol from the trans-isomer?

COCl

(2)

+ H2 in presence of Pd-BaSO4

(3)

(1) Acetone (2) Ozone (3) MnO2 (4) Aluminum isopropoxide

+ CO + HCl in presence of anhydrous AlCl3

(NEET I 2016)

COOH

(4)

+ Zn/Hg and conc. HCl

22. The correct statement regarding a carbonyl compound with a hydrogen atom on its alpha-carbon is (1) a carbonyl compound with a hydrogen atom on its alpha-carbon never equilibrates with its corres­ ponding enol. (2) a carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as aldehyde-ketone equilibration. (3) a carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as carbonylation. (4) a carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as keto-enol tautomerism.

(NEET 2013) 17. Which one is most reactive towards nucleophilic addition reaction? COCH3

CHO

(2)

(1) CHO

CHO

(4)

(3) CH3

NO2

(AIPMT 2014) 18. An organic compound X having molecular formula C5H10O yields phenyl hydrazone and gives a negative response to the iodoform test and Tollen’s test. It produces n-pentane on reduction. X could be

(NEET I 2016) 23. The product formed by the reaction of an aldehyde with a primary amine is (1) Schiff base. (2) ketone. (3) carboxylic acid. (4) aromatic acid.

(1) 2-pentanone. (2) 3-pentanone. (3) n-amyl alcohol. (4) pentanal.

(NEET I 2016)

(AIPMT 2015) 19. Treatment of cyclopentanone

O with methyl

24. The correct structure of the product A formed in the following reaction is O

lithium gives which of the following species? (1) (2) (3) (4)

Cyclopentanonyl cation Cyclopentanonyl radical Cyclopentanonyl biradical Cyclopentanonyl anion

H2 (gas, 1 atmosphere) Pd/carbon, ethanol

(AIPMT 2015)

hydrocyanic acid. sodium hydrogen sulphite. a Grignard reagent. hydrazine in presence of feebly acidic solution. (RE AIPMT 2015)

Chapter 27_Aldehydes and Ketones.indd 715

A

is (1)

20. Reaction of carbonyl compound with one of the following reagents involves nucleophilic addition followed by elimination of water. The reagent is (1) (2) (3) (4)

715

(3)

O

(2)

OH

(4)

OH

OH

+

(NEET II 2016)

1/4/2018 5:29:09 PM

716

(1)

OBJECTIVE CHEMISTRY FOR NEET

O

25. Of the following, which is the product formed when cyclohexanone undergoes aldol condensation followed by heating? (1)

(2)

O

(3)

(4)

(2) O

OH

O

O

OH

(NEET 2017)

OH O

Answer Key (3) Level I

O

(4)

O

OH

 1. (2)

 2. (3)

 3. (1)

 4. (3)

 5. (3)

 6. (1)

 7. (3)

 8. (1)

 9. (2)

10.  (3)

11.  (1)

12.  (3)

13.  (4)

14.  (4)

15.  (2)

16.  (3)

17.  (2)

18.  (2)

19.  (2)

20.  (4)

21.  (1)

22.  (4)

23. (4)

24.  (3)

25. (3)

26.  (2)

27.  (3)

28. (2)

29.  (3)

30.  (1)

31.  (2)

32.  (1)

33.  (2)

34.  (1)

35. (3)

36.  (2)

37.  (4)

38.  (3)

39.  (4)

40.  (4)

 1. (1)

 2. (4)

 3. (4)

 4. (1)

 5. (1)

 6. (3)

  7. (4)

 8. (4)

 9. (1)

10. (2)

11. (4)

12. (4)

13. (1)

14. (3)

15. (1)

16. (3)

17. (2)

18. (3)

19. (4)

20. (2)

21. (3)

22. (4)

23. (1)

24. (4)

25. (1)

Level II

Previous Years’ NEET Questions  1. (2)

 2. (3)

 3. (3)

 4. (2)

 5. (1)

 6. (2)

 7. (2)

 8. (3)

 9. (1)

10. (3)

11. (2)

12. (4)

13. (1),(4)

14. (4)

15. (1)

16. (4)

17. (4)

18. (2)

19. (4)

20. (4)

21. (1)

22. (4)

23. (1)

24. (1)

25. (1)

Hints and Explanations Level I

CH3

C

CH2

C

CH3

Benzophenone O

6. (1) When an alcohol (   OH) and ether (   OR) group is attached to one carbon atom, it is called hemiacetal.

Chapter 27_Aldehydes and Ketones.indd 716

CH2

3-Pentanone

5. (3) When two ether groups are attached to one carbon atom, it is called acetal.

7. (3) The general formula for the ketone is RCOR, in case of mixed ketone both the R groups are different. The structures of the given ketones are as follows:

O

O

4. (3) >C   O structure has no formal charge. Structures having least formal charge are the most stable.

C

CH3

Acetophenone



Therefore, acetophenone is the mixed ketone.

1/4/2018 5:29:10 PM

717

Aldehydes and Ketones 8. (1)

15. (2) The reaction is O

O

CH3

CH2

CH3

Zn dust O3

O + CH3

CH2

C

CH

C

CH3 (C7H 14 )

O

O

CH2

CH3

O

C

CH2

H

CH3 Ethylmethyl ketone (4 carbons)

O

C

O

H

C

O−

H

Propanal (3 carbons)

16. (3) H2CrO4 can be used for oxidation of secondary alcohol to ketone because it is a mild oxidant.

9. (2) The resonating structures are as follows: H

O

C

O−

H C

+ +

C

C

C

C

OH

H

Pent-3-ene-2-ol

H

Chromic anhydride CH3COOH (glacial)

O H

C

O

H

C

O−

C

+

C

C

C

C

Pent-3-ene-2-one

19. (2) The reaction is 11. (1) (CF3)2C(OH)2 is most stable due to intramolecular hydrogen bonding. F

H

C F F

O

F

F

C

C F

C

HCN

O–

C

O

Adduct nucleophile

H

2. H2O

20. (4) Compound 2,4-dimethylpentan-3-one is most sterically hindered, hence is least reactive. O

CH3CH2CH2CHO

CH3

C

CH

CH

CH 3

13. (4) This is Gattermann–Koch reaction for aldehyde synthesis.

14. (4) All others except PCC are stronger oxidizing agents and will convert R   CH2   OH R   COOH. PCC is a mild oxidizing agent and will convert R   CH2   OH R   CHO.

Chapter 27_Aldehydes and Ketones.indd 717

CH 3

OH

O + HCl

CH 3

21. (1)

CHO CO + HCl

OH

CN

Carbonyl group

1. LiAIH(O-t-Bu)3, Et2O, -78°C

anhyd. AICI3

C

CN

O

12. (3) Acyl chlorides can be reduced to aldehydes by treating them with LiAlH[OC(CH3)3]3 at -78°C. CH3CH2CH2COCl

Nucleophile CN−

CH3

C

H + HCN

CH3

C CN

OH H

H3O+

CH3

*C

H

COOH

22. (4) Acetone will react with phenylhydrazine to give a product that contains C   N   

1/4/2018 5:29:12 PM

718

H3C H3C

OBJECTIVE CHEMISTRY FOR NEET 31. (3)

H3C O + H2N

C

NH

C6H5

N

N H

H3C

23. (4) Aldehydes and ketones react with phosphorus ylides to yield alkenes and triphenylphosphine oxide. CHO + CH2

+

−

CH

CH

O

C6H5

CH3 C



Iodoform and potassium acetate are formed as products in given reaction.

32. (1) The reaction involved is

PPh3

C6H5MgBr + (CH3)3 CH

CH

24. (3) Stephen’s reduction involves reduction of nitriles to aldehydes. 25. (3) When carboxylic acid is treated with LiAIH4, it is reduced all the way to 1° alcohol. This happens because LAH is a very powerful reducing agent.

C

LiAlH4

OH

LiAlH4

[RCHO]

Acid

R

Aldehyde

35. (3) Benedict solution contains CuSO4, sodium carbonate and sodium citrate which is specific for oxidation of aldehydes. CH3CH

Benedict (Cu2+) solution

CHCHO

CH2OH

CH

O

OH−

CH2OH

COONa +

Cl

Cl

Cl

28. (2) Cannizzaro reaction is an example of auto oxidation reduction reaction. For example, 2HCHO

conc. NaOH

HCOONa + CH3OH

29. (3) Aldehydes are more reactive than ketones because they have only one electron donating group attached to the carbonyl carbon. Secondly a more sterically hindered ketone is less stable than an aldehyde or a less hindered ketone. 30. (1) The reaction involved is Wolff–Kishner reduction. CH

CHCOCH3 KOH

CH

Chapter 27_Aldehydes and Ketones.indd 718

But-2-one O

Decolorization of brown color, positive

Br2 /CCl4 (Red-brown) Ag2O, OH − (Mild oxidizing agent)

4

No decolorization

Oxidized to carboxylic acid

No oxidation

Forms phenyl hydrazone Forms phenyl hydrazone

(Phenyl hydrazine) KMnO /OH −

Forms vicinal diol

Does not form diol

Level II 1. (1) The reaction is C6H5

COO

C6H5

COO

Ca +

CH3

COO

CH3

COO

O C6H5

C

Ca

Dry distillation

O C6H5 +

CH3

C

CHCH2CH3 CH3CHCI2

HO

COOH

O CH3 + C6H5

C

CH3

2. (4) Acetaldehyde is formed when geminal halides react with water because these are unstable and readily release water molecule.

NH2⋅NH2

HO

CH

H

NH2NH C6H5

conc. NaOH

2

CH

But-3-en-1-al O Reagent

CHCl3 + HCOO−

27. (3) Since m-chlorobenzaldehyde does not have a-­ hydrogen, therefore, it undergoes Cannizzaro reaction. CHO

CH3

38. (3)

1° alcohol

26. (2) In case of chloral, haloform is formed as chloral does not have an a-hydrogen attached to it. CCl3

C6H6 + (CH3)3 COMgBr

OH

36. (2) (CH3)3COCH2CH3 (ethyl tert-butyl ketone) does not contain CH3CO– group required for iodoform reaction.

O R

C

33. (2) Aromatic aldehydes do not react with Fehling’s solution.

PPh3

CH2 + O

CH

CH3COOK + CHI3

CH3 + KOH + I2

aq. KOH

OH CH3CH

OH

−H2O

CH3CHO

1/4/2018 5:29:15 PM

719

Aldehydes and Ketones 3. (4)

10. (2)

(1) CH3 CH2 CH2 OH

O

PCC/CH2 Cl2

CH3 CH2 CHO

O

(2) CH3

CH2

C

Cl

CH3

LiAIH[OC(CH3)3]3

OH

CHO

− CH2

−

CH3

CHO

C

C

Ether, -78°C

1. Sia2BH

CH

CH3

2. H2O2/OH-

CH3

CH

CH3 CH2 CHO

C

CH

CH3

Tautomerization

4. (1)

H2/Pd/BaSO4 Lindlar catalyst

H2O

OH

OH

CH3CH2CHO

CHO

CH2

H

(3) CH3

O−

H

CH

∆

CHO

CH

CH3

−H2O

(B)

CH

CHO

CH2

(A)

11. (4) Since both aldehydes lack alpha hydrogen, aldol condensation is not possible. In this case, Cannizzaro’s reaction will take place. Two aldehydes without alpha hydrogens will give crossed Cannizzaro’s products. However, HCHO transfers hydride more easily as compared to ArCHO, so the major products of the reaction are ArCH2OH and HCOONa. conc. NaOH ∆

ArCHO + HCHO

ArCH2OH + HCOONa

12. (4) The reactions are as follows: Zn-Hg HCl, heat Clemmensen reduction

O3/Zn/CH3COOH

CHO CH ⇓ 5

6

4

2

5. (1) OH

O

CH

C

CH2

H

CH

CI + (C2H5)2Cd Diethyl cadmium

CH2

— O + H2NH (CH3 )2 C —

CH3

2CH3

C

CH2

CH3 + CdCI2

CH3CH2CHO + CH3

CH3

CH2

CH2

(CH3 — CHNH)3 + 3H2O Acetaldehyde ammonia trimer (CH3 )2 C — NH Ketimine

CH

CH3

CH2

CHO

CH3

− CH

CH2

Nucleophile CH

O But-2-one

CH

14. (3)

Electrophile

(CH2)6N4 Hexamethylene tetramine

3CH3 — CHO + 3H2 NH

CH

OH

7. (4) The reaction is

Chapter 27_Aldehydes and Ketones.indd 719

CH2

C5H10O Achiral

6. (3) 1,2-Diols, a-hydroxyl ketones and 1,2-diketones are oxidized by HIO4, whereas, other diols remain unaffected.

8. (4) 6HCHO + 4NH3

+

+

OMgBr

H 3 O+

H2SO4

C5H12O (X) Chiral

O Acetyl chloride

d −d + CH3MgBr

−

O

CrO3

*

C

CH3

13. (1) The polarity of C   O is very high as compared to C   C bond and therefore the Grignard’s reagent will preferably attack on carbonyl carbon.

1 H

4-Methylhexanal

2CH3

HCl, heat Clemmensen reduction

CH3

O

3

Zn-Hg

O

O

CH3

C

CHO

CH2

CHO

CH3CH2CHO −H2O

CH3

CH2

CH3

CH

CH

CHO

OH

CH2

CH3

15. (1) The reaction is O HCHO + CH3

C

O CH3

OH−

CH3

Attack will take place on this carbon

C

CH2

CH2OH

CH

CH2

H+ ∆

(will form enolate)

O CH3

C

Methyl vinyl ketone

1/4/2018 5:29:19 PM

720

OBJECTIVE CHEMISTRY FOR NEET

16. (3) Greater the deficiency of electron density on card+ dbonyl carbon, C O , greater is its reactivity and less

20. (2)

hindrance also favors it. Therefore, the order of reactivity on carbonyl carbon is CH3CHO > CH3COCH3 > C6H5COC6H5.

OH

CH3OH

CH3CH2CH2CH2CHO

CH3CH2CH2CH2CH

Dry HCl

CH3OH; dry HCl

17. (2) Tollen’s reagent oxidizes aldehyde to corresponding acid. CHO

H2Ο

O + CH3

C

OCH2

CH3

CH3

C

2

1

OCH3

CN

OH

HCl, H2O ∆

COOH

22. (4) OCH2

CH2

H Enolate ion

OH

HCN

O

O−

O C

3

OCH3 1,1-Dimethoxy pentane

18. (3) This is an example of Claisen condensation. − CH2

4

21. (3)

H 3C

H3C

5

CH3CH2CH2CH2CH

COO−

Ag (NH3)2+

OCH3

CH3

CHO

+ - OMgBr

O CH3

CH2

C

d- d+

CH2 CH3 + CH3 CH2 MgBr

CH3 CH2

CH2CH3

C

CH2 CH3

−OC2H5−

H+ O CH3

C

OH CH2

CHO

CH3 CH2

b-Ketoaldehyde

C

CH2 CH3

CH2 CH3

24. (4) The Wolff–Kishner reduction of cyclobutanone is as follows:

19. (4)

O

Zn(Hg) / HCl

N

NH2

Clemmenson reduction

+ NH2NH2 Hydrazone

OH

O

alc. KOH

LiAlH4 , ether

S

CH2–SH CH2–SH

BF3

S

25. (1) The reaction is CH3

C

CH3 + CH3

O H2/Ni

H

O

dil. NaOH Heat Aldol condensation

OH CH3

CH

CH2

C O

Chapter 27_Aldehydes and Ketones.indd 720

C

CH3

D

CH3

CH

CH

C

CH3

O Contains 13 sigma bonds, 2 lone pairs and 2 pi bonds

1/4/2018 5:29:22 PM

721

Aldehydes and Ketones

Previous Years’ NEET Questions

O

C

O

Zn−Hg HCl

O 2CH3CH

OH 5°C

NaOH

Sodium benzoate

O

O

OH

CH

CH2



All of the above compounds contain these groups, so they will give iodoform when warmed with I2 and NaOH.

8. (3) Clemmenson reduction involves the refluxing of ketone with hydrochloric acid containing amalgamated zinc.

CH

CH

(Aldol)

Zn−Hg conc. HCl

O

CH3

C

H

5. (1) No new bond formation takes place in Cannizaro reaction, instead it is a disproportion chain reaction.

Hydrocarbon

O >

H3C

C

H3C

(II)

O >

Ph Ph

C

O

(III)

10. (3) The reaction involved is Cannizzaro reaction CHO

6. (2) The rate of dehydration of given compounds can be explained on the basis of stability of the carbocation formed. In compound (1), (3) and (4), the positive charged generated on carbon is adjacent to the electron withdrawing    COCH3 group which destabilizes the carbocation.

C

(I)

O HCOONa + CH3OH

CH2

9. (1) Aldehydes are more reactive than ketones towards nucleophilic addition reactions due to both steric and electronic reasons. Therefore, the order is

CH3 a , b -unsaturated ketone

H + conc. NaOH

O

Carbonyl group

C

OH

CH3

O

CH3

Chapter 27_Aldehydes and Ketones.indd 721

O

CH3CH2OH, CH3 COCH3, CH3

C

∆ −H2O

C

(2)

7. (2) Compounds having either CH3 C group or CH3CHOH    group, give iodoform when warmed with I2 and NaOH.

C2H5ONa

2H

+

CH3

CH3

CH

(4)

Although, the carbocation cation formed from compound (2) is relatively stable as the electron withdrawing species is not next to the carbon bearing positive charge. Therefore, compound (2) will be most readily dehydrated.

CH3CHCH2CH 3-Hydroxybutanal (Aldol) (50%)

Benzyl alcohol

C



O

4. (2) This is an example of aldol condensation.

2

+

C6H5CH2OH + C6H5COONa

Benzaldehyde

O

(3)

O CH3

3. (3) The disproportionation of aldehydes lacking a-hydrogens (e.g., HCHO, C6H5CHO, R3C   CHO, etc.) in the presence of a strong base to form salt of an acid and a corresponding primary alcohol is known as the Cannizzaro reaction. 2C6H5CHO

CH3

(1)

CH2 + H2O

2. (3) Aldehydes and ketones containing a-hydrogen react to form b -hydroxyl aldehydes (aldol) and b -hydroxyl ketones (ketol). Because ketones are similar to aldehydes, so the term “aldol condensation” is applicable to ketones as well. 10% NaOH, H2O

+

H3C

1. (2) The reduction of aldehydes and ketones to the corresponding hydrocarbons with amalgamated zinc and concentrated hydrochloric acid is called Clemmensen reduction.

+

CI

conc. KOH

CH2OH +

Cannizzaro reaction

CI

COO-

CI

11. (2) CH3CHO will give yellow precipitate of iodoform on reaction with I2 and NaOH as it contains methyl ketone group, that is,    CH3CO; whereas C6H5CH2CHO will not give positive iodoform test as it does not contain    CH3CO group.

1/4/2018 5:29:27 PM

722

OBJECTIVE CHEMISTRY FOR NEET

12. (4)



H3C

O + CH3CH2OH(excess)

C

H3C

HCl gas

H3C

COCI

OCH2CH3

C

H3C

Rosenmund reaction CHO

OH Hemiacetal

CH3CH2OH H+

H3C



Gatterman–Koch reaction CHO

OCH2CH3

C

H3C

H2 Pd/BaSO4

OCH2CH3

CO + HCI

Acetal

AICI3 (Anhy.)

13. (1), (4) The reactions involved are COCH3

COONa + CHI3 + 2NaOH

I2

+ 3NaOH

Iodoform (Yellow ppt.)

Sodium benzoate

Acetophenone CH3CHOHCH3 + 4I2 + 6NaOH

Heat

CHI3 + CH3 COONa + 5NaI + 5H2O

2-Propanol

14. (4) Nucleophile    NNH2 attacks the carbonyl carbon and elimination of H2O also takes place. So, it is a nucleophilic addition–elimination reaction. R R

O + H2N

C

R

NH2

R

C

Zn-Hg in presence of HCl would not reduce the COOH group.

17. (4) The characteristic reaction of aldehydes and ketones is the nucleophilic addition to the carbon-oxygen bond. Aldehydes are more reactive in nucleophilic additions than are ketones. Both steric and electronic factors favor aldehydes. Alkyl groups are electron releasing group, thus, reduces the reactivity of carbonyl while electron withdrawing substituents (   NO2) cause the carbonyl carbon to be more positive and cause the addition reaction to be more favorable. 18. (2) The reactions are:

O− +

NH2

NH2



Option (1): N NHC6H5

Addition R R

NH2NHC6H5

H− A

+

C

NH

NH2

−H2O

−H+

N

CH

R

R

C

O

NH2

CH3

CH2CH2CH3

2-Pentanone Reduction

15. (1) The reaction involved is Rosenmund reaction.

CI + H2

Benzoyl chloride

Pd-BaSO4 Boiling xylene

C

N NHC6H5 NH2NHC6H5

H + HCl

Benzaldehyde (A)

CH3CH2

Etard reaction

CrO2CI2 CS2/H3O +

Chapter 27_Aldehydes and Ketones.indd 722

CH3CH2

C

CH2CH3

O C

CH2CH3

3-Pentanone (X) CHO

CH2CH3

Option (2):

16. (4) The reactions are as follows:

CH3

CH2

Gives positive iodoform (methyl ketone present) and negative Tollen’s test n-Pentane

O

C



C

NH2

O

C

O

H

NH R Elimination

CH3



Gives negative iodoform test and negative Tollen’s test Reduction

n-Pentane

Due to the absence of CH3CO    group, it does not give positive iodoform test. Ketones are not oxidized by Tollen’s reagent. Option (3): All the given tests are characteristic of aldehydes and ketones.

1/4/2018 5:29:31 PM

723

Aldehydes and Ketones

Option (4): NH2NHC6H5

NNHC6H5

CH3 CH2CH2 CH2 CH

23. (1) Schiff bases are formed when aldehydes react with primary amines. O C

R CH3CH2 CH2CH2CHO

Gives negative iodoform and negative Tollen’s test

Pentanal Reduction

n-Pentane

19. (4) The reaction involves the addition of organolithium reagent to the carbonyl carbon of ketone that leads to the formation of cyclopentanonyl anion. d− d+ O + CH3 Li

OH H + R’NH2

Aldehyde

C

R

Primary amine

N

R’ + H2O

Schiff base

24. (1) In a, b-unsaturated carbonyl compounds, when reduction is carried out using hydrogen gas with palladium as catalyst, the carbon-carbon double bond is selectively reduced. O

O

−

O Li+ CH3

H2 gas, (1 atmosphere) Pd/carbon, ethanol

20. (4) Carbonyl compounds undergo nucleophilic addition elimination reaction with ammonia and its derivative in the presence of feebly acidic solution. C

O + H 2N

Aldehyde or Ketone

NH2

25. (1) The reaction involved is as follows: H

H 3O +

C

Hydrazine

NH2 + H2O

N

H3C

O

+ H2O

O

Hydrazone Enolate anion

21. (1) Acetone usually condenses with the cis hydroxyl groups on the adjacent carbon atoms. H3C

− −

HO− + O

C

O + −O

O

−

O

O+ OH

O

OH

O

C

CH3

CH3

Ketal formation

22. (4) A carbonyl compound with a hydrogen atom on its alpha-carbon rapidly equilibrates with its corresponding enol and this process is known as keto-enol tautomerism.

O + OH− H

O

−

O

+ H2O HO ∆ −H2O

H R

C

C

H

O

R’

(Keto form)

Chapter 27_Aldehydes and Ketones.indd 723

R

CH

C

R’

O

OH (Enol form)

1/4/2018 5:29:34 PM

Chapter 27_Aldehydes and Ketones.indd 724

1/4/2018 5:29:34 PM

28

Carboxylic Acids and its Derivatives

Chapter at a Glance 1. Introduction (a) Carboxylic acids are characterized by the presence of carboxylic group. C

O (abbreviated

CO2H or

COOH)

HO

(b) Carboxylic acids can be aliphatic or aromatic. R C

Ar

O

HO Aliphatic

C

O

OH Aromatic

2. Structure of Carboxylic Acids (a) T  he carbon atom of a carboxylic acid moiety is sp2 hybridized and therefore exhibits trigonal planar geometry with bond angles that are nearly 120°. (b) The carboxylic carbon is stabilized by the following resonance structures, and hence is more stable and less electrophilic than carbonyl carbon. O−

O R

C

R O

O−

+

R

C

H

O

C

H

O +

H

3. Nomenclature of Carboxylic Acids (a) In IUPAC system, systematic or substitutive names for carboxylic acids are obtained by dropping the final -e of the name of the alkane corresponding to the longest chain in the acid and by adding -oic acid. The carboxyl carbon atom is assigned number 1. CH3

CH2CH2COOH

CH3

CH2

CH

COOH

Br 2-Bromo butanoic acid

Butanoic acid

(b) F  or compounds containing more than one carboxyl group, the alkyl chain is numbered and the number of carboxyl groups is indicated by adding the multiplicative prefix, dicarboxylic acid, tricarboxylic acid, etc. to the name of parent alkyl chain. The position of  COOH groups are indicated by the arabic numeral before the multiplicative prefix. HOOC

(CH2)3

COOH

Pentanedioic acid

HOOC

CH2

CH(COOH)

CH2

COOH

Propane-1,2,3-tricarboxylic acid

4. Carboxylic Acids Derivatives The carboxyl group is the parent group of a large family of related compounds called acyl compounds or carboxylic acid derivatives.

Chapter 28_Carboxylic Acids and its Derivatives.indd 725

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OBJECTIVE CHEMISTRY FOR NEET

5. Preparation of Carboxylic Acids (a) From primary alcohols: A variety of strong oxidizing agents can be used to oxidize primary alcohols and produce carboxylic acids. -

RCH2OH

1. KMnO4, OH , D 2. H3O

RCO2H

+

(b) From aldehydes and ketones R

C

O

H

[O]

RCOOH

KMnO4 +

R

1. Ag2O or [Ag(NH3)2] OH

CHO

R

−

RCO2H

2. H3O+

C

O

conc. HNO3

RCOOH

R

(c) From alkyl benzenes: Primary and secondary alkyl groups (but not 3° groups) directly attached to a benzene ring are oxidized by KMnO4 to a  CO2H group. CH3CH3 R CH

COOH

R

KM

nO 4

KMnO4

CH2CH2CH3

Acidified

KM

nO

4

CH3 KMnO4 Acidified

(d) From oxidative cleavage of alkynes An internal alkyne O R

C

C

R’

KMnO4 or O3

C

R

O OH

+

C

HO

R’

A terminal alkyne O R

C

C

H

KMnO4 or O3

C

R

OH

+ O

C

O

(e) From hydrolysis of nitriles O + HCN R’

R

HO

CN

R

R’

HA H2O

HO

CO2H

R

R’

(f ) From carboxylation of Grignard reagent R

Chapter 28_Carboxylic Acids and its Derivatives.indd 726

Cl

Mg Et2O

R

MgCl

O

CO2

R

O

H3O+

OMgCl

R

OH

1/4/2018 5:28:56 PM

Carboxylic Acids and its Derivatives

727

(g) From acyl halides, anhydrides and esters O

(i)

RCCl Acyl chloride

O

(ii) (iii)

OR’

RCO − + Cl− O

O

OH − /H O 2

+ H2O

O

RCOH + HOCR’

H 2O

RC O CR’ Anhydride O R

O

OH − /H O 2

O

C

RCOH + HCl

H 2O

O

O

RCO + OCR’ O −

H3O+

R

C

−

OH

+ R’

OH

6. Physical Properties of Carboxylic Acids (a)  Carboxylic acids are soluble in water due to their ability to form strong intermolecular hydrogen bonding. (b) Carboxylic acids can form two hydrogen-bonding interactions, allowing molecules to associate with each other in pairs. O

H

O

R

R O

H

O

 hese hydrogen-bonding interactions explain the relatively high boiling points of carboxylic acids than the aldeT hyde, ketones, alcohols, hydrocarbon, etc. (c)  Carboxylic acids have no regular system of melting points. The first ten members show oscillation effect, that is, compounds with even number of carbon atoms have higher melting points.

7. Chemical Properties of Carboxylic Acids (a) Acidic character of carboxylic acids (i) Carboxylic acids exhibit mildly acidic protons. Treatment of a carboxylic acid with a strong base, such as sodium hydroxide, yields a carboxylate salt. 2RCOOH + 2Na



2RCOO−Na+ + H2 Sodium carboxylate

(ii) Carboxylic acids are stronger acids than phenols    Carboxylate ion is more stabilized as compared to phenoxide ion, because in carboxylate ion, the negative charge is equally distributed over two electronegative atoms (oxygen atoms) while in phenoxide ion, it is present only on one oxygen. dO

d-

dO d-

R

C

dO Carboxylate ion (More stable)

dPhenoxide ion (Less stable)

(iii) M  ost unsubstituted carboxylic acids have Ka values in the range of 10−4−10−5 (pKa = 4−5). Carboxylic acids having electron-withdrawing groups are stronger than unsubstituted acids. The chloroacetic acids, for example, show the following order of acidic character: Cl Cl

pKa

Chapter 28_Carboxylic Acids and its Derivatives.indd 727

O OH Cl 0.70

O

Cl > Cl

OH H

O

Cl > H

OH H

1.48

> H

O

H

OH H

2.86

4.76

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OBJECTIVE CHEMISTRY FOR NEET

(iv)  Since inductive effect is not transmitted very effectively through covalent bonds, the acid-strengthening effect decreases as the distance between the electron-withdrawing group and the carboxyl group increases. Of the chlorobutanoic acids that follow, the strongest acid is 2-chlorobutanoic acid: O

Cl

O

O Cl OH

OH Cl 2-Chlorobutanoic acid (pKa = 2.85)

OH

3-Chlorobutanoic acid 4-Chlorobutanoic acid (pKa = 4.05) (pKa = 4.50)

(b) Reactions of aliphatic carboxylic acids   The reactions of carboxylic acids can be classified into any of the following types: (i) Cleavage of O   H bond. (ii) Cleavage of C   OH bond. (iii) Involving    COOH group. (iv) Substitution reaction on the hydrocarbon part ( Hell–Volhard–Zelinsky reaction). Reactions of aliphatic carboxylic acids are summarized below. Litmus paper Na NaOH

NaHCO3

RCOOH

NH3 ∆ Soda lime NaOH + CaO PCl5 PCl3 SOCl2 LiAlH4 N3H ∆ P2O5 ∆

R’OH CH2N2

X2/Red P (where X = Cl, Br)

Blue turns red RCOONa + 1/2 H2 RCOONa + H2O RCOONa + CO2 + H2O RCOONH4

∆

RCONH2

RH + CO2 RCOCl + POCl3 + HCl RCOCl + H3PO3 RCOCl + SO2 + HCl RCH2 OH N2 + CO2 + R

NH2

O R

C

R

C

O

O RCOOR’ + H2O RCOOCH3 + N2 CH2

COOH (HVZ reaction)

Cl

Chapter 28_Carboxylic Acids and its Derivatives.indd 728

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Carboxylic Acids and its Derivatives

(c) Reactions of aromatic carboxylic acids

729

COONa

Na

NaOH CaO

CONH2

COOH NH3

COOH

conc. HNO3 H2SO4

COOH

NO2

H2SO4 SO3

SO3H COOH

Cl2 FeCl3 (anhyd.)

Cl

8. Chemical Properties of Carboxylic Acid Derivatives (a) The overall order of reactivity of carboxylic acid derivatives is O R

O >

C

R

C

O >

R

>

C

O

Cl R′

Amide

O

O −

O Na+ (–Na+Cl−)

Carboxylic acid

or PCI5

O

R

O

R'

Anhydride

SOCI2 or PCI3

OH

NH2

Ester

R'

R

C

C

(b) Reactions of acyl chlorides

O

R

OR′

O Acid anhydride

Acyl chloride

O

O

O R

Cl

Acyl chloride

R’— OH, base (–HCI) H

R

OR’ Ester

R’

O

N R’’ (–R’ R’’NH2 Cl ) +

−

R

N

R’

R’’ Amide

Chapter 28_Carboxylic Acids and its Derivatives.indd 729

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OBJECTIVE CHEMISTRY FOR NEET

(c) Reactions of carboxylic acid anhydrides O

R’OH

O

R

R

R

Ester

O

O

+

OR’

O

OH Acid

R

Anhydride

O Excess R’R’’ NH

O

R’ R

+

N

−

R

ONH2R’R’’+

R’’ Amide

By-products

(R’, R’’ = H or C)

(d) Reaction of esters O

O

R

OR’

+ NaOH

H 2O

Ester

− ONa+ +

R

R’OH

Sodium carboxylate

Alcohol

Solved Examples 1. Which of the following is the best name for the following compound?

Solution (3) The structure of potassium oxalate is

O O

O

(1) (2) (3) (4)

+

Isobutyl ethanoate Ethyl isopropanoate 3-Methylbutyl ethanoate Ethyl 3-methylbutanoate

(1) (2) (3) (4)

O O

2. Which of the following is the correct structure for potassium hydrogen oxalate? +

O O

(1) Boiling point of amides is the maximum due to dipole– dipole interactions and intermolecular hydrogen bonding.

O

−

O

R

Dipole-dipole interactions OH

(4)

O

O −

O K

Chapter 28_Carboxylic Acids and its Derivatives.indd 730

R

O

OH

+ −

Solution

(2) K O

OH

(3) K O

Primary amides > carboxylic acids > nitriles > esters. Carboxylic acids > primary amides > nitriles > esters. Carboxylic acids > nitriles > primary amides > esters. Amides > carboxylic acids > esters > nitriles.

HO

O

Oxalate ion

3. The correct order of the boiling point of comparable molecular weight of acid and its derivatives is

(4) The name of the compound is ethyl 3-methylbutanoate.

+ −

O O

Solution

(1) K O

OH

−

K

+



+ N O− C R

R C

O− R N + R

−O

R C

R

N +

H

O− C R

H N +

R

Intermolecular H-bond

Carboxylic acids also have strong hydrogen bonding due to which their boiling point are more than that of nitriles and esters but lower than that of primary amides.

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731

Carboxylic Acids and its Derivatives O

Intermolecular H-bond Cl R

C OH



R

6. What would be the final organic product of the following reaction?

O

Nitriles contain strong dipole-dipole interactions, so, their boiling points are more than that of esters but lower than that of acids. Dipole-Dipole interactions R

Br

C N d+ d−

2. H2O

?

Solution (3) Br

CO2H

1. Mg, Et2O

2. CO2

MgBr

COO

+

(2)

1. LAH, Et2O

2. CO2

(1) (CH3)3CCO2H (2) (CH3)3CCOCH3 (3) (CH3)3CCH2OH (4) (CH3)3COCH3

4. Which of the following would be the strongest acid? (1)

1. Mg, Et2O 3. H3O+

d− d+ N C R

CO2H

OH

2. H3O+

Et2O

HO

C

1. CO2

−

O

MgCl

Mg

H3O

Cl Cl CO2H

(3) Cl

(4)

Cl

Cl

Solution

5. Consider the two attempts given below to prepare the carboxylic acid (CH3)3C−COOH. Cl

(I)

Mg

1. CO2

Et2O

2. H3O +

COOH

H2O

7. What would be the final product of the following reaction? CO2H

(3) Acidic character is directly proportional to the presence of electron withdrawing group, so, compound (3) is the strongest acid due to the presence three electron withdrawing groups.

LAH, Et2O

CH2OH

CO2H

1. PCl5 2. NH3 3. P4O10, heat

(1) CH3CH2CH2NH2 (3) CH3CH2CONHCOCH2CH3

(2) CH3CH2CONH2 (4) CH3CH2CN

Solution (4) CH3CH2COOH

1. PCl5

2. NH3

CH3CH2COCl

CH3CH2CONH2

(Dehydration)

P4O10, ∆

CH3CH2CN Cl

(II)

1. HO-, H2O

NaCN (80%)

2. H3O+

8. In a set of reactions, acetic acid yielded product D.

(75−80%)



Select the correct statement about the outcome of reactions: (1) (2) (3) (4)

Both reactions produce desired acid in good yield. Only I produces desired acid in good yield. Only II produces desired acid in good yield. None of them is a good method to prepare acid.

CH3COOH



Chapter 28_Carboxylic Acids and its Derivatives.indd 731

A

Benzene anhyd. AlCl3

B

HCN

C

HOH H+

D

The structure of D would be (1)

COOH

(2)

OH C

COOH

CH3

CH3

Solution (2) CN− being a strong base gives elimination reaction instead substitution, so, nitrite is not produced in the second reaction and carboxylic acid is not produced in good yield.

SOCl2

(3)

OH

CH3

OH

CN C

C

OH

(4) CH3

CH2

C

CH3

CN

1/4/2018 5:29:05 PM

732

OBJECTIVE CHEMISTRY FOR NEET

Solution

11. What is the ultimate product of this sequence of reactions?

(1) The reaction is

O O

SOCl2

CH3COOH

CH3

C

O CH3

Cl Cl

Benzene

CH3

(1)

O

Cl

HOH H+

CH3

CN

H N O O

(4)

N H

OH

C

H N

(2)

N H

O

(3)

OH

(1 eq.)

O

Friedel-Crafts acylation

HCN

C

CH3NH2

CH3CH2OH

Cl

O

O

Solution

C COOH

(1) O

9. Identify the product(s) of the following reaction.

Cl

O

O CH3CH2OH

C

(1 eq.)

Cl

Cl

CH3NH2

C

OCH2CH3

NH

CH3

COOH

OH

12. What would be the final organic product of the following reaction?

O COOH

1. DIBAL-H HO 1. DIBAL-H

C6H5COCl H3O+ C6H2. 5COCl

(2)

COOH

(3)

Et

O

∆

(1)

O

C

COOH

(4)

(1) C6H5

H+

OH HO

2. H3O+

OH H+

O O (2) C H CH3 CH 6 5 C6H5 O CH O

(3) C6H5COOCH2CH2OH (4) C6H5CHO Solution (2) b-Keto carboxylic acid undergoes decarboxylation (eliminate CO2) on heating.

Solution (1) The reaction is OH

O

O

C6H5COCl

COOH

1. DIBAL-H 2. H3O+

C6H5CHO

∆ − CO2

10. Which of the following statements concerning nitriles is incorrect? (1) Nitriles can be hydrolyzed to carboxylic acids. (2) Nitriles can be formed from (many) alkyl halides by nucleophilic substitution by cyanide ion. (3) Nitriles can be reduced with excess lithium aluminum hydride to primary amines, RNH2. (4) Nitriles react with Grignard reagents to form tertiary alcohols.

RCN R

C

CN

R

SN2 LiAlH4 (Excess)

N

R′MgX H3O

CH

O O

(1) (2) (3) (4)

Benzyl alcohol + acetic anhydride → Benzyl alcohol + acetic acid + H3O+ → Benzyl alcohol + acetyl chloride → All of the above

Solution (4) C6H5CH2OH + CH3

O

O C

CH3

OCOCH3

C

O CH2

C6H5

(Benzyl acetate)

R

C6H5CH2OH + CH3COOH

CN

R

O

COOH

RCH2 NH2 O +

X

H3O

N −

R

C

+

(4) R

C6H5

13. Which of the reactions listed below would serve as a synthesis of benzyl acetate, CH3CO2CH2C6H5?

C

Chapter 28_Carboxylic Acids and its Derivatives.indd 732

+

Solution

OH H+ Acetal formation

H3O

Esterification

O

R’

C6H5CH2OH + CH3

C

Cl

C6H5CH2 O C CH3 (Benzyl acetate) O C6H5

CH2

O

C

CH3

(Benzyl acetate)

1/4/2018 5:29:07 PM

Carboxylic Acids and its Derivatives 14. What is the product of this reaction?

O

O

(1)

CH3OH

COOH

O

(3)

HO

(2)

O

O

(3)

O

(4)

O

O O

O

(4)

O

(2)

O O

O O

O

(1)

O

733

O O

O

Solution (4)

Solution

O

(2) O

O CH3OH

O

CH3O

O O− Na+ + CH3

C

C

CH2

CH2

CH2

C

Cl

OH O C

15. Which of the following is the product from the reaction of sodium benzoate and acetyl chloride?

O O

C

CH3

Practice Exercises Level I

(3) 1-Chloropentanoyl chloride (4) 1,2-Dichloropentanal

Nomenclature

4. Which of the following structures is N-benzyl-N-propyl2,3-dimethylbutanamide?

1. Which is the IUPAC name for the following compound? O

H N

O

(1) O

OH

(1) (2) (3) (4)

2-Oxohexanoic acid 5-Oxohexanoic acid Methyl butyroxo ketone 4-Ketopentanoic acid

N

(2)

2. The IUPAC name for the following compound is CH3(CH2)7 H

(1) (2) (3) (4)

C

C

O

(CH2)7COOH H

trans-9-octadecenoic acid cis-9-octadecenoic acid cis-7-nonadecenoic acid trans-7-decadecenoic acid

(3)

N O

Cl Cl

3. What is the IUPAC name for

?

O

(1) a-Chlorovaleryl chloride (2) 2-Chloropentanoyl chloride

Chapter 28_Carboxylic Acids and its Derivatives.indd 733

(4)

N O

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OBJECTIVE CHEMISTRY FOR NEET

5. Which of the following structures is 3,4-dimethylpentyl chloroformate? (1)

Acidic Character of Carboxylic Acids 10. Which of the following is most acidic?

(2)

H

O

Cl

(1) o-Toluic acid (2) m-Toluic acid (3) p-Toluic acid (4) All of these

O

O

11. Which of the following statements is incorrect about HCOOH?

O

Cl

(3)

(4)

Cl

O

Cl

(1) It is a stronger acid than CH3COOH. (2) It forms formyl chloride with PCl5. (3) It gives CO2 and H2O on heating with concentrated H2SO4. (4) It does not reduce Tollens’ reagent.

O

O

O

Structure and Physical Properties of Carboxylic Acids

12. Arrange the compounds in order of increasing acidity (lowest first).

6. Picric acid is

Cl

COOH

COOH

CH3

O

C

C

(2) OH

OH

COOH

(3)

OH

C

Cl

(II)

O CH3CH2

C

OH

OH

(III)

(IV)

(1) IV, I, II, III (2) III, II, IV, I (3) IV, II, I, III (4) II, I, III, IV (1) CH3COOH (2) HCOOH

(4) NH2 NO2

7. Carboxyl functional group (

C

CH3

C

O

13. Among the following acids which has the lowest pKa value?

NO2

NO2

CH2CH2

(I)

NO2

Cl

O

OH

H

(1)

Cl

(3) (CH3)2CH   COOH (4) CH3CH2COOH 14. The correct acidity order of the following is OH

OH

(I)

(II)

COOH

COOH

(III)

(IV)

 COOH) is present in

(1) picric acid. (2) barbituric acid. (3) ascorbic acid. (4) aspirin. 8. Carboxylic acids and amides have in general higher boiling points than esters and anhydrides because of which property? (1) (2) (3) (4)

Dispersion forces Resonance stabilization Conjugated functional groups Hydrogen bonding

O O-K+

OH

(I)

(II) O

O

(1) III > IV > II > I (2) IV > III > I > II (3) III > II > I > IV (4) II > III > IV > I (1) p-nitrophenol. (2) p-hydroxybenzoic acid. (3) o-hydroxybenzoic acid. (4) p-toluic acid. 16. Which of the following would be the strongest acid?

(IV)

CO2H

(1)

CO2H

(2) O2N

OH

OH

(III)

CH3

15. Among the following compounds, the most acidic is

9. Arrange the compounds in order of increasing solubility in water (least soluble first). O

Cl

NO2

(3)

CO2H

O2N

NO2

(4)

CO2H

(1) II, III, I, IV (2) IV, I, III, II (3) I, IV, II, III (4) II, III, IV, I

Chapter 28_Carboxylic Acids and its Derivatives.indd 734

NO2

1/4/2018 5:29:10 PM

735

Carboxylic Acids and its Derivatives 17. Which compound would be the strongest acid?

23. What is correct reagent for the following reaction?

(1) CHCl2CH2CH2CO2H (2) ClCH2CHClCH2CO2H

OH

(3) CH3CCl2CH2CO2H (4) CH3CH2CCl2CO2H

OH O

18. Which of the following would be the weakest acid? CO2H

(1)

 (I) (i) KMnO4, OH-, heat (ii) H3O+ (II) PCC, CH2Cl2 (III) (i) O3, CH3CO2H (ii) H2O2 (IV) H2CrO4

CO2H

(2) H3C

CH3

(3)

CO2H

H3C

CH3

(1) I and IV (2) I and II (3) I and III (4) none of these.

CO2H

(4)

24. Which of the following reaction produces benzoic acid? CH3 (1)

Br

Methods of Preparation of Carboxylic Acids 19. Which of the following does not form benzoic acid on oxidation with KMnO4/H+? (1) Toluene (2) Cumene (3) tert-Butylbenzene (4) Acetophenone

(1)

1. KMnO4, HO−

(3)

Br NaCN

(3)

1. HO-, H2O 2. H3O+

1. Cl2/OH−(excess) 2. H3O+

(4)

(2)

1. CO2 2. H3O+

heat 2. H3O+

20. Which of the following would serve as syntheses of (CH3)3CCO2H? O

(2)

Mg Et2O

Br

Br

1. O3, CH3CO2H

1. CN−

2. H2O2

2. H3O+ (heat)

25. Predict the major organic product of the reaction sequence below:

1. Mg, Et2O 2. CO2 3. H3O+

+ KMnO4

(4) Both (1) and (3) 21. Which of these combinations will not produce benzoic acid? (1) C6H5CH2OH + KMnO4/OH /H2O, heat; then H3O (2) C6H5CH3 + KMnO4/OH−/H2O, heat; then H3O+ (3) C6H6 + CO2, high pressure −

+

(4) C6H5COCH3 + Cl2/OH−/H2O; then H3O+

OH CO2H

OH

(1)

(2) OH CHO

(3)

22. What is X in the following reaction?

H3O+

Heat H2O, OH−

(4)



COOH

26. What is the reactant of the following reaction sequence?

CHO +X

CH3COONa

MeO

(1) CH3COOH (2) BrCH2COOH (3) (CH3CO)2O (4) OHC–COOH

Chapter 28_Carboxylic Acids and its Derivatives.indd 735

OH

1. Mg/ether

MeO

2. CO2 3. H+

O

(1) HCO2CH2C6H5 (2) C6H5CH2COOH (3) C6H5CH2Cl (4) C6H5CHClCOOH

1/4/2018 5:29:12 PM

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OBJECTIVE CHEMISTRY FOR NEET

Chemical Reactions of Carboxylic Acids

Cl

27. The rate of soda-lime decarboxylation is in the order COOH

COOH

COOH

COOH

CO2H

COOH

1. OH- (2eq.), H2O heat

H2

1. SOCl2

2. H3O+

cat.

2. CH3OH

Cl

O

(1)

(2)

O

CO2H

O NO2

Cl

(I)

(II)

CH3 (III)

OCH3

(IV)

(1) I > II > III > IV > V (2) II > I > III > V > IV (3) III > IV > I > V > II (4) IV > V > II > I > III 28. Which of the following reactions proceed nearly to completion as written? (I)

O

O

+ NaOH OH

(II)

O

O

OH

(III)

O OH

(IV)

O

+ H2O + CO2

O

+ H2SO3 O Na −

+

+

O O−Na+

OH O + O−Na+

O

29. When CH2   CH   COOH is reduced with LiAlH4, the compound obtained will be (1) CH3   CH2   COOH (2) CH2 CH   CH2OH (3) CH3   CH2   CH2OH (4) CH3   CH2   CHO 30. Choose the reagent(s) that would bring about the following reaction: CH3CH2CH2COOH

CH3CH2CH2CH2OH

(1) H2/Ni (2) Li/liq NH3 (3) LiAlH[OC(CH3)3]3 (4) LiAlH4, ether 31. When propionic acid is treated with aqueous sodium bicarbonate, CO2 is liberated. The ‘C’ of CO2 comes from (1) methyl group. (2) carboxylic acid group. (3) methylene group. (4) bicarbonate. 32. Predict the major organic product, P, of the following sequence of reactions:

Chapter 28_Carboxylic Acids and its Derivatives.indd 736

O

33. Which carboxylic acid would decarboxylate when heated to 100−150°C? O CO2H

HO2C

(I)

CO2H

HO2C

(II)

CO2H

(III)

34. A carboxylic acid reacts with an isotopically labeled methanol to produce (1) (2) (3) (4)

methyl acetate having the labeled oxygen. water having all the labeled oxygen. both methyl acetate and water contain isotopic oxygen. no esterification.

35. A liquid was mixed with ethanol and a drop of concentrated H2SO4 was added. A compound with a fruity smell was formed. The liquid was (1) CH3OH (2) HCHO (3) CH3COCH3 (4) CH3COOH

OH

(1) I, II (2) III, IV (3) I, III (4) II, IV



O

(1) I (2) II (3) III (4) More than one of these

O−Na+ + NaHSO4

(4)

O O

+ H2O O−Na+

+ Na2CO3

(3)

(V)

P

Chemical Reactions of Carboxylic Acid Derivatives 36. Choose the reagent(s) that would bring about the following reaction: CH3CH2CH2COCl

CH3CH2CH2CHO

(1) H2/Ni (2) Li/liq NH3 (3) LiAl(OC(CH3)3)3H (4) NaBH4, CH3OH 37. Choose the reagent(s) that would bring about the following reaction: CH3C

CCH2CO2CH2CH3

CH3C

CCH2CH2OH

(1) H2/Ni (2) Li/liq NH3 (3) LiAlH[OC(CH3)3]3 (4) LiAlH4, ether 38. What would be the final product of the following reaction? C6H5CH2CONH2

(1) C6H5CH2CO2CH3 (3) C6H5CH2COCH3

P4O10

1. CH3MgI, Et2O

Heat

2. H3O+

(2) C6H5CH2CH2NHCH3 (4) C6H5CH2CH(CH3)CN

1/4/2018 5:29:14 PM

737

Carboxylic Acids and its Derivatives 39. Hydrolysis of acetamide produces (1) acetic acid. (2) acetaldehyde. (3) methylamine. (4) formic acid.

3. Which of the following would serve as a synthesis of 2,2-dimethylpropanoic acid? 1. Mg, Et2O

(1)

40. Which of the following products is not possible in the reaction given below? (C2H5COO)2Ca + (C6H5COO)2Ca

(2)

Dry

42. What is the product of the following reaction?

(1) hypochlorous acid. (2) chlorine. (3) hydrochloric acid. (4) phosphorous pentachloride.

O

(2)

OH

O

(4)

CH3COOH + PCl5

O

43. On mixing ethyl acetate with aqueous sodium chloride, the composition of the resultant solution is

1. Predict the major organic product in the following reaction,

(1) (3) *

* COOH

H3O+

CH3CH2CH2MgBr Et2O

(2)

COOH (4)

*

COOH

2. 70% H2SO4, reflux

A + NH4+

(2) C6H5CH2COOH

(3) C6H5CH2OSO3H (4) C6H5CHClCOOH

Chapter 28_Carboxylic Acids and its Derivatives.indd 737

(1) CH3CH(OH)C2H5

C

ether

(2) CH3COC6H5 CH2CH3

(4) CH3

C (OH)C6H5

7. Which reagent would best serve as the basis for a simple chemical test to distinguish between C6H5CH   CHCOOH and C6H5CH   CHCH3? (1) Conc. H2SO4

(2) Br2/CCl4

(3) CrO3/H2SO4

(4) NaHCO3/H2O

8. What is the product of the following reaction? O

O Heat

HO

*

1. NaCN

(1) HCO2CH2C6H5

C2H5MgBr

product C would be

COOH

2. What is the expected product, A, of the following reaction sequence? Cl

B

(1) tartaric acid. (2) pyruvic acid. (3) cinnamic acid. (4) propionic acid.

Methods of Preparation of Carboxylic Acids

X(gas)

C6H6 Anhy.AlCl3

6. Lactic acid on oxidation by alkaline potassium permanganate gives

Level II

HCl



A

(3) CH3CH(OH)C6H5

(1) CH3COCl + C2H2OH + NaOH (2) CH3COONa + C2H5OH (3) CH3COOC2H5 + NaCl (4) CH3Cl + C2H5COONa

* BaCO3

5. In a set of the given reactions, acetic acid yielded a product C.

O N

(3)

2. OH-, H2O, heat 3. H3O+

4. The    OH group of an alcohol or the carboxylic acid can be replaced by    Cl, using

O

O

Br

Chemical Reactions of Carboxylic Acids

Isopropyl alcohol

(1)

2. H3O+

(4) Answers (1) and (2) only

O

O

1. KMnO4, OH-, heat

1. CN-

(3)

41. The calcium salt of which of the following acid on dry distillation gives 2,4-dimethylpentan-3-one? (1) Propionic acid (2) Isobutyric acid (3) Butyric acid (4) Adipic acid

2. CO2 3. H3O+

OH

distillation

(1) C2H5COC2H5 (2) C2H5COC6H5 (3) C6H5COC6H5 (4) C6H5CH2COCH3

O

Br

OH O

(1)

(2) O

O

O

OH

(3) EtO

O

(4) O

O

O

O

OEt

9. In the presence of a small amount of phosphorous, aliphatic carboxylic acids react with chlorine or bromine

1/4/2018 5:29:16 PM

738

OBJECTIVE CHEMISTRY FOR NEET to yield a compound in which a-hydrogen has been replaced by halogen. This reaction is known as (1) (2) (3) (4)

O CH3

Wolff–Kishner reaction. Etard reaction. Hell–Volhard–Zelinsky reaction. Rosenmund reaction.

CH2

C

CH3

11. What would be the final product, F, of the following sequence of reactions? 1. Mg, Et2O

PBr3

(2)

Br CO2H

(3)

(4)

CH2

CH2

CH3

O

C

NH2

CH3

CH2

C

OH

(IV)

O

O C

− Z + OH

(1)

(3)

Physical Properties and Chemical Reactions of Carboxylic Acids Derivatives

O

(II)

O− + HZ

C

(2) CH3

O

O

CO2H

(4) CH3 CO

CH3O

16. Self-condensation of two moles of ethyl acetate in the presence of sodium ethoxide yields

12. Characterize the amides I and II in caffeine.

(1) ethyl butyrate. (2) acetoacetic ester. (3) methyl acetoacetate. (4) ethyl propionate.

N

O

C

15. The following reaction is fastest when Z is which group?

CO2H

Br

CH2

(1) I, III, II, IV (2) III, I, IV, II (4) I, IV, II, III (4) II, I, III, IV

3. H3O+

(1)

CH3

(III)

F

2. CO2

2. H2O

CH3

O

(1) benzoic acid. (2) benzenesulphonic acid. (3) salicylic acid. (4) carbolic acid.

1. LAH, Et2O

O

(I)

10. The compound that does not liberate CO2, on treatment with aqueous sodium bicarbonate solution, is

CO2H

O

17. In the following reaction

I

N

N

COOK

N

II O Caffeine

(1) (2) (3) (4)

Electrolysis

I is a primary amide, II is a tertiary amide I is a secondary amide, II is a secondary amide I is a tertiary amide, II is a secondary amide I is a tertiary amide, II is a tertiary amide

COOK



A is

13. Which functional groups are correctly named? O

O

O

O

O

Amide

Anhydride

Ester

(I)

(II)

(III)

O

O HC

Lactone

(IV)

(1)

(2)

(3)

(4)

O CH3CH2COCH3

O

CH3 N

CH3 Amide

(V)

n



18. The product, Z, of the following sequence of reactions is which compound?

(1) I, II, III, IV (2) I, III, IV, V (3) III, IV, V (4) I, II, IV, V

p - Chlorotoluene

14. Which is the order of increasing solubility in water of the following compounds (least first)?

1. KMnO4, OH−, heat 2. H3O+ 3. SOCl2 4. CH3CH2OH, base

Z

O

(1)

(2) Cl

Chapter 28_Carboxylic Acids and its Derivatives.indd 738

A

(3)

O Cl

O

(4)

O 1/4/2018 5:29:19 PM

1. KMnO4, OH−, heat 2. H3O+

p - Chlorotoluene

Z

3. SOCl2 4. CH3CH2OH, base

O

(1)

(2)

O

O

OCH2CH3

(1) HO

O

(2)

O Cl

Cl O

(3)

O

(4)

O

23. What is the product of this reaction?

1. Mg, Et2O

A

2. CO2 3. H3O+

PCl5

NH3

B

C

O

P4O10

H2O

HN

D H N

(1)

The compound D is (1) an amide. (2) a nitrile. (3) an amine. (4) a carboxylic acid.

OH-

O

(2) H N

CO2H

2

OO

O O

(3)

(4)

N H

20. Identify the product(s) of the following reaction.

O N H

O

24. The product of the following reaction is

O H3O

CO2CH3

O

+

Heat

O O

(2)

O

O

+ CO2 + CH3OH

(1 mol) O

(4)

CO2H

+ CO

O

Heat

(1 mol) O

O

(1)

O

(2) CO2H

CO2H

21. Compound (A), C8H9Br, gives a white precipitate when warmed with alcoholic AgNO3. Oxidation of (A) gives an acid (B), C8H6O4. (B) easily forms anhydride on heating. Identify the compound (A).

O O

(3)

O

O OH

CI

C2H5

O O

(4)

CH2Br

(1)

OH

+

O

OH

(3)

OCH2CH3

O

19. Consider the reaction sequence given below:

(1)

O

OCH2CH3O (4) H3CH2CO

(3) H3CH2CO Cl



OH

HO

O

O

Br

739

Carboxylic Acids and its Derivatives

H

O

Previous Years’ NEET Questions

(2) Br

1. Consider the following compounds: CH3

(I) C6H5COCl

(II) O2N

COCI

(IV) OHC

COCI

CH2Br

CH2Br

(4)

(3) CH3

(III) H3C CH3



22. Predict the major organic product of the reaction sequence,

COCI

The correct decreasing order of their reactivity towards hydrolysis is (1) II > IV > I > III (2) II > IV > III > I (3) I > II > III > IV (4) IV > II > I > III

O O

1. CH3CH2OH 2. dilute HCl, cold

O

Chapter 28_Carboxylic Acids and its Derivatives.indd 739

M

(AIPMT 2007) 2. Which of the following presents the correct order of the acidity in the given compounds?

1/4/2018 5:29:22 PM

740

OBJECTIVE CHEMISTRY FOR NEET (1) FCH2COOH > ClCH2COOH > BrCH2COOH > CH3COOH (2) CH3COOH > BrCH2COOH > ClCH2COOH > FCH2COOH (3) FCH2COOH > CH3COOH > BrCH2COOH > ClCH2COOH (4) BrCH2COOH > ClCH2COOH > FCH2COOH > CH3COOH (AIPMT 2007)

3. The relative reactivities of acyl compounds towards nucleophilic substitution are in the order of (1) (2) (3) (4)

acyl chloride > ester > acid anhydride > amide. acyl chloride > acid anhydride > ester > amide. ester > acyl chloride > amide > acid anhydride. acid anhydride > amide > ester > acyl chloride. (AIPMT 2008)

4. Propionic acid with Br2/P yields a dibromo product. Its structure would be (having restricted amount of Br2/P) Br

(1) CH3

(2) CH2Br

COOH

C

CHBr

COOH

Br

7. Match the compounds given in List-I and List-II and select the suitable option using the code given below List-I

List-II

(a) Benzaldehyde (b) Phthalic anhydride

(i) Phenolphthalein (ii) Benzoin condensation (c) Phenyl benzoate (iii) Oil of wintergreen (d) Methyl salicylate (iv) Fries rearrangement Code:

(a) (b) (c) (d)

(1) (ii) (i) (iv) (iii) (2) (iv) (i) (iii) (ii) (3) (iv) (ii) (iii) (i) (4) (ii) (iii) (iv) (i) (AIPMT MAINS 2011) 8. The correct order of decreasing acid strength of trichloroacetic acid I, trifluoroacetic acid II, acetic acid III and formic acid IV is (1) II > I > IV > III (2) II > IV > III > I (3) I > II > III > IV (4) I > III > II > IV (AIPMT 2012)

Br

(3) H

C

(4) CH2Br

CH2COOH

CH2

9. Which one of the following esters gets hydrolyzed most easily under alkaline conditions?

COBr

Br

(AIPMT 2009) 5. Among the given compounds, the most susceptible to nucleophilic attack at the carbonyl group is (1) CH3CONH2 (2) CH3COOCOCH3 (3) CH3COCl (4) CH3COOCH3 (AIPMT PRE 2010)

OCOCH3

OCOCH3

(1)

(2) Cl OCOCH3

OCOCH3

(3)

(4) H3CO

O2N

6. In a set of reaction, ethyl benzene yields a product D CH2CH3

KMnO4 KOH

B

Br2 FeCl3

C

C2H5OH H+

D

(RE AIPMT 2015) 10. The correct order of strengths of the carboxylic acids is COOH



COOH

COOH

D would be (1)

Br

(2)

COOH

(I) Br

COOC2H5

(4)

CH2

CH

(II)

O

(III) (NEET II 2016)

OCH2CH3

CH2COOC2H5

(3)

O

COOC2H5

(1) II > III > I (2) III > II > I (3) II > I > III (4) I > II > III

Br Br

(AIPMT PRE 2010)

Chapter 28_Carboxylic Acids and its Derivatives.indd 740

1/4/2018 5:29:24 PM

Carboxylic Acids and its Derivatives

741

Answer Key Level I  1. (2)

 2. (2)

 3. (2)

 4. (4)

 5. (4)

 6. (3)

 7. (4)

 8. (4)

 9. (2)

10. (1)

11. (4)

12. (3)

13. (2)

14. (3)

15. (3)

16. (3)

17. (4)

18. (2)

19. (3)

20. (4)

21. (3)

22. (3)

23. (1)

24. (1)

25. (2)

26. (3)

27. (1)

28. (1)

29. (2)

30. (4)

31. (4)

32. (4)

33. (4)

34. (1)

35. (4)

36. (3)

37. (4)

38. (3)

39. (1)

40. (4)

41. (2)

42. (1)

 1. (1)

 2. (2)

 3. (4)

 4. (4)

 5. (4)

 6. (2)

 7. (4)

 8. (4)

 9. (3)

10. (4)

11. (4)

12. (4)

13. (3)

14. (4)

15. (4)

16. (2)

17. (1)

18. (2)

19. (2)

20. (3)

21. (3)

22. (1)

23. (2)

24. (1)

4. (1)

5. (3)

6. (3)

7. (1)

43. (3)

Level II

Previous Years’ NEET Questions 1. (1)

2. (1)

 3. (2)

8. (1)

9. (3)

10. (1)

Hints and Explanations 5. (4)

Level I 3. (2)

5

4

3

2

Cl

1

2-Chloropentanoyl chloride

O





CH3 CH3

O

Cl

Cl

C

O

H2C 1’

H2C 2’

HC 3’

HC 4’

CH3 5’

3,4-Dimethylpentyl

Chloroformate

 COCl is the main functional group and numbering starts from carbon of COCl.

4. (4)

7. (4) Only aspirin contain –COOH group. The structures of the compounds are shown below:

Propyl N

1

2

3

6. (3) 2, 4, 6-Trinitrophenol is known as picric acid.

OH

4 Amide

NO2

NO2

Contains no

COOH group

O NO2 (Picric acid)

Benzyl N-Benzyl-N-propyl-2,3-dimethylbutanamide

HO HO

O

O Contains no

COOH group

OH

HO

(Ascorbic acid) O HN

NH Contains no

Chapter 28_Carboxylic Acids and its Derivatives.indd 741

O

O

COOH group 1/4/2018 5:29:27 PM

OH NO2

742

NO2

Contains no

COOH group

OBJECTIVE CHEMISTRY FOR NEET NO2 (Picric acid)

HO

O

HO

O Contains no

COOH group

18. (2) Compound (3) and (4) are more acidic than compound (1) and (2) due to ortho effect. Between (1) and (2), compound (2) is the least acidic due to +I effect of methyl group. COOH

OH

HO

COOH

(Ascorbic acid) O HN

>

NH Contains no

COOH group

O

O

CH3 (+H) (less acidic)

(more acidic)

(Barbituric acid) COOH

19. (3) As it does not contain any a hydrogen.

OCOCH3

Contains

20. (4)

COOH group

O C

(Aspirin)

CH3

Cl2 / OH

O

-

O- + CHCl3

C

(Haloform reaction)

+ H3O

10. (1) o-Toluic acid is most acidic due to ortho effect. 11. (4) HCOOH > CH3COOH (as CH3COOH has greater +I effect)

COOH

O HCOOH + PCl5

H

HCOOH HCOOH



conc. H2SO4 ∆ [Ag(NH3)2]+ Tollens’ reagent

C

Cl + POCl3 + HCl

Br

CO2 + H2O

Br

1. CN-

(Elimination)

1. Mg, Et2O

COO-

CO2

MgBr

H3O+

Silver mirror

It is the only acid which reduces Tollens’ reagent.

13. (2) As there is no alkyl group in HCOOH, whereas in other acids the presence of electron donating group (i.e. alkyl group) doesn’t assist the release of proton easily. 14. (1) Phenols are weaker acids than carboxylic acids. Electron withdrawing group like -Cl stabilizes negative charge developed after removal of proton and hence increases the acidity of the compound. While electron donating group like -CH3 decreases the acidity of the compound. 15. (3) o-Hydroxybenzoic acid is the strongest acid among the given compounds. It is due to the presence of ortho effect (in this both the groups are ortho to each other which causes the steric repulsions.) OH

COOH

COOH

COOH

21. (3) CH2OH

COO

Chapter 28_Carboxylic Acids and its Derivatives.indd 742

H3O+

KMnO4 / OH−, ∆ Oxidation

COO −

CH3 KMnO4 / OH−, ∆

COOH H3O+

Oxidation

CHO + CO

HCl / AlCl3 Gattermann-Koch reaction

o-Hydroxy benzoic acid

COO−

COCH3 + Cl2

OH p-Hydroxy benzoic acid

COOH

COOH OH

NO2 p-Nitro phenol

−

OH−

+ CHCl3 COOH

Haloform reaction

CH3 p-Toluic acid

H3 O +

1/4/2018 5:29:29 PM

743

Carboxylic Acids and its Derivatives 22. (3)

27. (1) Presence of electron withdrawing groups increases the rate of decarboxylation while electron donating groups decrease the rate of decarboxylation. Hence, the rate of decarboxylation is I > II > III> IV > V.

CHO + (CH3CO)2O Acetic MeO anhydride Aromatic aldehyde (X) CH

CH

CH3COONa

29. (2) LiAlH4 reduces carboxylic acid to a primary alcohol.

COOH

CH2

CH

COOH

a , β-unsaturated acid

H2/Ni

23. (1) PCC will oxidize alcohol into an aldehyde. Ozonolysis only works on alkenes. CH3CH2CH2COOH

CH2OH

Br

MgBr

OH

Mg

1. CO2

Et2O

2. H3O

No reaction

LiAlH [OC(CH3)3]3 NaBH4 , CH3OH

O



CH

No reaction

Li/liq NH3

24. (1) Only reaction in first option produces benzoic acid.



CH2

30. (4)

MeO



LiAlH4

LiAlH4 , ether

+

The alkyl group on benzene ring can be oxidized to carboxylic acid group if it contains benzylic hydrogen atom. In given reaction in option (2) benzene ring doesn’t bear benzylic hydrogen atom, so, no oxidation occurs in this case. In the reaction in option (3), aryl halide cannot give SN2 reactions, so, no nitrile is possible which on hydrolysis produces carboxylic acid otherwise. In option (4), ozonolysis oxidizes C   C double and triple bonds only, so, it produces acetic acid instead benzoic acid.

2. H2O2

H3C

No reaction CH3CH2CH2CH2OH

31. (4) The ‘C’ of CO2 comes from bicarbonate as can be explained from the reaction below. CH3CH2COOH + NaHCO3

CH3CH2COONa + CO2 + H2O

32. (4) Cl

OH COOH

1. OH -

H2O/∆

COO-

+

H2/Catalyst

H3O

O 1. O3, CH3CO2H

CH3CH2CH2CHO

-

COOH

COO

OH COOH

25. (2)

1. SOCl2

COCl 2. CH3OH

COO− + H2O + CO2

KMnO4 / OH −

C

(oxidative ozonolysis)

OCH3

O

+ H3O

33. (4) b-keto acid and gem dicarboxylic acid decarboxylate (eliminate CO2) on heating whereas 1,4-dicarboxylic acids eliminate water.

COOH

O

26. (2) CH2Cl 1. Mg/Ether

CH2COO−

CH2MgCl 2. CO2

H

COOH COOH (I) Succinic acid

+

O

D -H2O

O Succinic anhydride

O

O

CH2COOH

COOH

D -CO2

b-keto acid CH2

COOH COOH

D -CO2

CH3COOH

Gem dicarboxylic acid Chapter 28_Carboxylic Acids and its Derivatives.indd 743

1/4/2018 5:29:32 PM

COOH

O

D -H2O

COOH (I) Succinic acid

O Succinic anhydride

O

O D

COOH -CO2 OBJECTIVE CHEMISTRY FOR NEET

744

b-keto acid CH2

COOH

D -CO2

COOH

42. (1) Addition of an alcohol to an anhydride produces an ester.

CH3COOH

Gem dicarboxylic acid 18

18

34. (1) CH3COOH + CH3OH

43. (3) There is no reaction hence the resultant mixture contains CH3COOC2H5 + NaCl.

CH3COOCH3 + H2O

This is an example of Fischer esterification, where because of resonance in acid and reaction mechanism, the OH of the acid is lost in the form of H2O.

Level II 1. (1) -

35. (4) Alcohol reacts with acid in presence of H2SO4 to form ester. Fruity smell formed is of ester and the reaction is CH3COOH + C2H5OH + H+

CH3COOC2H5 + H2O

Acetic acid Ethanol

Ethyl acetate

C

C

CH2

COOCH2CH3

LiAlH4, ether

CH3

C

C

* CO 2

HCl

CH3CH2CH2MgBr Et2O

CH3CH2CH2

C

O

H3O+

* CH3CH2CH2COOH

37. (4) Esters will be reduced to primary alcohol by LiAlH4. CH3

* BaCO 3

+

OMgBr

2. (2) Cl

CH2CH2OH

CH2CN

1. NaCN

38. (3) C6H5CH2CONH2

P4 O10

C6H5CH2CN

Heat

1. CH3MgI, Et2O 2. H2SO4 (70%) / D

C6H5

CH2

C

CH2

C6H5

O 2. H3O

CH3

CH

NH

CH2COOH

CH3

+

+

+ NH4

39. (1) O CH3

Phenyl acetic acid

O

C

NH2

OH-

CH3

C

OH

40. (4) On dry distillation calcium salt produces ketones as shown below. Calcium carbonate is also produced in the reaction. (C2H5COO)2 Ca

Dry distillation

C2H5COC2H5 + CaCO3

(C6H5COO)2 Ca

Dry distillation

C6H5COC6H5 + CaCO3

(C6H5COO)2 Ca + (C2H5COO)2 Ca



Dry distillation

CH3

CH

COO

OH + PCl5

R

Cl + POCl3 + HCl

COOH + PCl5

R

COCl + POCl3 + HCl

C6H5COC2H5 + CaCO3

5. (4) The reaction involved is O CH3COOH + PCl5

OH

CH3 Ca

CH3CCl (A)

CH3 COO

The reactions involved are

R

41. (2) The reaction is

CH



R

Therefore, the compound C6H5CH2COCH3 cannot be formed in the reaction.

CH3

4. (4) Alcohol and carboxylic acid react with phosphorous pentachloride in which the hydroxyl group is replaced by chlorine atom to give alkyl chloride and acyl chloride respectively in good yield.

Heat

CH3 Calcium salt of isobutyric acid

Chapter 28_Carboxylic Acids and its Derivatives.indd 744

CH3

CH

C

CH

O

CH3

CH3

CH3

C

C6H6 Anhy. AlCl3

C2H5

O CCH3

C2H5MgBr Ether

2,4-Dimethylpentan-3-one (C)

(B)

1/4/2018 5:29:36 PM

745

Carboxylic Acids and its Derivatives 7. (4) C6H5

18. (2) CH

CH

COOH

NaHCO3

C6H5

acid base reaction

CH

Cl

-

CH

COO

+ H2O + CO2 C6H5



CH

CH

NaHCO3

CH3

Cl

Cl 2. H3O+

1. KMnO4, OH-, D

No reaction

Carboxylic acid gives effervescence of CO2 on reaction with NaHCO3.

-

CH3

COO

COOH 3. SOCl2 (Darzon process)

8. (4) Diacids will cyclize on heating to form a cyclic anhydride. 9. (3) Hell–Volhard–Zelinsky reaction a

CH3COOH

Cl2, P

ClH2C

-HCl

Acetic acid

Cl2, P

COOH

Cl2CH

-HCl

Monochloroacetic acid

Cl

Cl COOH

4. CH3CH2OH Base

Dichloroacetic acid

COCl

COOCH2CH3

Cl2, P -HCl

Cl3 - COOH Trichloroacetic acid

19. (2) The complete reaction is

Br

10. (4) Comparatively phenol is a weaker acidic than carbonic acid (H2CO3), and hence does not liberate CO2 on treatment with aqueous NaHCO3 solution. The rest are more acidic than H2CO3, so liberate CO2.

O

1. Mg, Et2O 2. CO2 3. H3O

O

PCl5

Cl

OH (A)

(B) NH3

11. (4) COOH

1. LiAlH4, Et2O

PBr3

CH2OH

2. H2O

CH2Br

2-Methyl propanoic acid

1. Mg, Et2O

O COOH

H3O+

CH2

C

CO2

-

O

CH2MgBr

16. (2) The  acetoacetate ester  is produced mainly by self-condensation of ethyl acetate in the presence of sodium ethoxide. CH3

C

CH2CH3 + CH3

O

C

O

O

CH2CH3

O -C2H5OH

CH3

(D)

O

20. (3) O CH3

CH2

C

O CH

C

Ester hydrolysis

O CH3

CH2

C

O D (−CO2)

CH2CH3

CH3

CH2

a CH

C b

COOH + CH3OH

CH3

(b-keto acid) C

O

CH2CH3

21. (3)

O

17. (1) It is a Kolbe’s electrolysis reaction.

CH2Br

CH2+

AgNO3(alc.)

+

CH3

(A)

CH3

AgBr (Pale yellow) O

COOK

COOH [O] Electrolysis

(B)

Chapter 28_Carboxylic Acids and its Derivatives.indd 745

H3O+

OCH3

CH3

Acetoacetic ester

COOK

NH2

(C)

C2H5ONa

CH2

C

O

P4O10

CN

COOH

D -H2O

O O Phthalic anhydride

1/4/2018 5:29:39 PM

746

OBJECTIVE CHEMISTRY FOR NEET

22. (1) O

decreases in the order F > Cl > Br. Because fluorine is the most electronegative, the O   H bond is most polarized, and the proton in O  H is most positive. Therefore, O  H loses a proton most readily in compound FCH2COOH and its the most acidic among the given compounds followed by ClCH2COOH > BrCH2COOH > CH3COOH.

O O

CH2CH3

O

1. CH3CH2OH

-

O O

O

3. (2) Acyl chlorides are the most reactive toward nucleophilic addition–elimination, and amides are the least reactive. In general, the overall order of reactivity is

dil HCl, cold

O CH2

C

OCH2CH3

CH2

COOH

O R

C

(P) O NH

OH

-

Acyl chloride

OH

Intramolecular, acid base reaction

NH -



O − O

CH2

C

CH2

CH2

NH2

24. (1) O O

+ CH3CH2CH2OH

COOCH2CH2CH3

D

COOH O

Previous Years’ NEET Questions 1. (1) The rate of hydrolysis enhances with the increase in magnitude of positive charge on carbonyl group. Electron with drawing group like  NO2 and  CHO increases the positive charge and electron releasing group like −CH3 decreases the negative charge on carbonyl carbon. Amongst  NO2 and  CHO groups nitro has more  I effect than  CHO. Therefore, the reactivity towards hydrolysis is O Cl

Cl

R’

C

O

>R O

NO2 > Cl

O

O > Cl

C

Ester

2. (1) Carboxylic acids having electron-withdrawing groups are stronger than unsubstituted acids. Electronegativity

NH2

Amide

Br CH3CH2COOH

Br2/P

COOH

CH3CH Br2/P

Br C

COOH

Br

CHO >

CH3

C

4. (1) In the presence of a small amount of phosphorus, aliphatic carboxylic acids react smoothly with chlorine or bromine to yield a compound in which a-hydrogen is replaced by halogen. This is called Hell–Volhard–Zelinsky reaction.

CH3

C

O

>R OR’

The general order of reactivity of acid derivatives can be explained by taking into account the basicity of the leaving groups. When acyl chlorides react, the leaving group is a chloride ion. When acid anhydrides react, the leaving group is a carboxylic acid or a carboxylate ion. When esters react, the leaving group is an alcohol, and when amides react, the leaving group is an amine (or ammonia). Of all of these bases, chloride ions are the weakest bases and acyl chlorides are the most reactive acyl compounds. Amines (or ammonia) are the strongest bases and so amides are the least reactive acyl compounds.

5. (3) Cl has a strong electron with drawing nature and is the best leaving group, so CH3COCl is the most reactive. -

O

Chapter 28_Carboxylic Acids and its Derivatives.indd 746

C

O Acid anhydride

O

C

C

C

Cl

23. (2) O

O >R

CH3

C -

Nu

O

O Z

CH3

C

Z

CH3

C

-

Nu + Z

Nu

1/4/2018 5:29:41 PM

747

Carboxylic Acids and its Derivatives 6. (3) The reactions involved are

O

COOH

C

COOH

OCH3

HO

CH2CH3 Oxidation

Halogenation

(B)

(C)

Br

COOC2H5

8. (1) CF3COOH > CCl3COOH > HCOOH > CH3COOH

C2H5OH/H

+

Esterification

(D)

Br

7. (1) Option (a)–(ii): Benzoin condensation: This reaction involves an aromatic aldehyde, Ph   CHO, and is similar to Cannizzaro reaction. O C

H KCN

C

CH

O

OH

Acidic character increases with the electronegativity of the group attached as F is more electronegative than Cl, so CF3COOH is more acidic than CCl3COOH; whereas in CH3COOH, the CH3 group being an electron donating group decreases the acidic character.

9. (3) Esters undergo base-promoted hydrolysis known as saponification. It involves nucleophilic addition-­ elimination at the acyl carbon, thus, the presence of electron withdrawing group increases the rate of hydrolysis. The reaction involved is O CH3



C

O

NO2

Option (b)-(i): The reaction is O

OH O

H

O H

O-

O

H+

CH3

-H2O

C

O

NO2

OH + O OH

OH

OH

OH

CH3

C

-

OH + O

NO2

Phenolpthalein



Option (c)-(iv): Fries rearrangement: It is one of the methods for the preparation of acyl phenols. O O

C

C6H5

OH

O C

AICl3 D

OH C6H5 +

10. (1) The presence of electron withdrawing groups increases the acidic strength of carboxylic acids due to a combination of inductive and mesomeric effects. Compounds II and III are more acidic than carboxylic acid I due to  I effect of oxygen present in the ring. However, II is more acidic than III as the inductive effect reduces with increase in distance. COOH

C O



Option (d)-(iii): Methyl salicylate is known as oil of wintergreen.

Chapter 28_Carboxylic Acids and its Derivatives.indd 747

>

C6H5 O (-I effect is more) (II)

COOH O (-I effect is less) (III)

COOH

>

(+I effect) (I)

1/4/2018 5:29:43 PM

Chapter 28_Carboxylic Acids and its Derivatives.indd 748

1/4/2018 5:29:43 PM

29

Organic Compounds Containing Nitrogen

Chapter at a Glance 1. The alkyl or aryl derivatives of ammonia are known as amines. They are obtained by replacing one or more hydrogen atoms of ammonia by alkyl or aryl groups. Amines are classified as being primary (1°), secondary (2°) or tertiary (3°) on the basis of the number of alkyl or aryl groups attached to the nitrogen. 2. Structure of Aliphatic Amines (a) The nitrogen atom of amines is sp3 hybridized and the geometry is pyramidal. (b) Each of the three sp3 hybridized orbitals of nitrogen overlap with orbitals of hydrogen or carbon depending upon the composition of the amines.

N H

H H

3. Nomenclature (a) In the IUPAC system, the amines are regarded as alkanamines, for example CH3

CH2

NH2

CH3CH2

NH

CH3

N-Methylethanamine

Ethanamine

N(CH3)2 C2H5 CH3

CH2

N C2H5

N, N-Diethylethanamine

N, N-Dimethyl benzenamine

The IUPAC name for simplest aryl amine C6H5NH2 is benzenamine but common name aniline is also accepted by IUPAC. NHCH3

N-Methylbenzenamine (N-Methylaniline)

Chapter 29_Organic Compounds Containing Nitrogen.indd 749

NH2

NH2

CH3

OCH3

4-Methylbenzenamine (p-Toluidine)

4-Methoxybenzenamine (p-Anisidine)

1/4/2018 5:29:19 PM

750

OBJECTIVE CHEMISTRY FOR NEET

4. Methods of Preparation of Primary Amines (a) Reduction of nitro compounds: Aromatic nitro compounds get reduced to aromatic amines in the presence of hydrogen gas (passed in the presence of finely divided Ni/Pt/Pd) or when reduced with metals in acidic medium. Ar

H2, Catalyst

NO2

or 1. Fe, HCl 2. OH−

Ar

NH2

In case of a dinitro compound, selective reduction of one nitro group is often achieved through the use of measured amount of hydrogen sulphide in aqueous (or alcoholic) ammonia. NO2

NO2 H2S NH3,C2H5OH

NO2 m-Dinitrobenzene

NH2 m-Nitroaniline (70-80%)

(b) Ammonolysis of alkyl halides NH3 + R

R

R N+H3X− Substituted ammonium salt R NH2 + H2O + Na+X− N+H3X− + NaOH

RNH2 (1°)

X

RX

RX

R2NH (2°)

R3N (3°)

RX

R4N+X−

Hofmann ammonolysis

Quaternary ammonium salt

Ammonolysis has the disadvantage of salt yielding a mixture of primary, secondary, tertiary amines and also a quaternary ammonium salt. However primary amine is obtained as a major product by taking large excess of NH3. (i) Order of reactivity of halides with amines is RI > RBr > RCl (ii) Aromatic amines cannot be prepared by this method since aryl halides are much less reactive towards nucleophilic substitution reactions. (c) Reduction of nitriles (Mendius reaction) R

C Nitrile

N + 4[H]

[H]

R CH2NH2 1° amine

(d) Reduction of amides O R

C

LiAlH4

NH2

H2O

R

CH2NH2

(e) Gabriel’s phthalimide synthesis: Primary amines can be prepared from potassium phthalimide using Gabriel synthesis. The method avoids the complication of multiple alkylations observed in ammonolysis of alkyl halides. O N O

O

O H

1. KOH 2. R X

N O

R

NH2NH2 Ethanol, reflux

R

N

NH2 +

N

H H

O

(f ) R  eduction of azides (Curtius rearrangement): In this reaction, the amine formed has one carbon less than the parent amide. To obtain primary amine with same number of carbon atoms for primary amide, reduction is done with LiAlH4/ether.

Chapter 29_Organic Compounds Containing Nitrogen.indd 750

1/4/2018 5:29:21 PM

Organic Compounds Containing Nitrogen

1. SOCI2

RCOOH

2. NaN3

Carboxylic acid

RCON3

Heat

Acyl azide

R

N

C

O

H2O/H+

Alkyl isocyanate

R

751

NH2

Primary amine

(g) Hoffmann bromamide degradation reaction: In this reaction amides with no substituent on the nitrogen react with solutions of bromine or chlorine in sodium hydroxide to yield primary amines. O R

C

NH2 + Br2 + 4NaOH

RNH2 + Na2CO3 + 2NaBr + 2H2O

5. Physical Properties of Aliphatic Amines (a)  Primary amines with three or more carbon atoms are liquids and higher members are all solids. (b)  Lower aliphatic amines are water soluble because they can form hydrogen bonds with water molecules. However, the solubility decreases with increase in size of hydrophobic alkyl group. (c)  The boiling points of amines are higher than that of the corresponding alkanes. The order of boiling points is primary > secondary > tertiary. (d)  Tertiary amines do not have intermolecular association due to the absence of hydrogen atom available for hydrogen bond formation. 6. Basic Strength of Amines (a)  Amines act as Lewis bases due to the presence of lone pair of electrons on the nitrogen atom. Due to their basic nature, they turn red litmus paper blue. (b) The basic character of amines can be determined by their Kb and pKb values. +

RNH2 + H2O

-

RNH3 + OH

 Greater is the value for dissociation constant of the base (Kb), higher is the basicity of amines. Lesser the pKb value, higher is the basicity of amines. (c) Aliphatic amines (CH3NH2) are stronger bases than NH3, due to electron releasing effect of the alkyl group.  In gas phase, the basicities of the methyl substituted amines increases with increase in methyl substitution. (C2H5)2NH > (CH3)2NH > CH3NH2 > NH3

(d) In aqueous solutions, the order of basic strength of ethyl and methyl substituted amines is as follows. (C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3 (CH3)2NH > CH3NH2 > (CH3)3N > NH3

 This order of basicity can be explained on the basis of combination of inductive effect, solvation effect and steric hindrance of the alkyl groups. (e)  Aromatic amines are weaker bases than aliphatic amines and NH3, due to the fact that the electron pair on the nitrogen atom is involved in resonance with the p- electron cloud of the ring. (f ) Ortho-substituted aromatic amines are usually weaker bases than aniline irrespective of the nature of substituent, that is, electron releasing or electron withdrawing. This is called ortho- effect and is probably due to steric and electronic factors. 7. Chemical Reactions of Amines (a) Alkylation C2H5NH2 + C2H5Br 1° amine

Chapter 29_Organic Compounds Containing Nitrogen.indd 751

−HBr

(C2H5)2NH 2° amine

C2H5Br −HBr

(C2H5)3N 3° amine

C2H5Br

(C2H5)4N+Br− Quartenary ammonium salt

1/4/2018 5:29:22 PM

752

OBJECTIVE CHEMISTRY FOR NEET

(b) Acylation C2H5

NH2 + CH3COCl

Base

C2H5

N

C

H

O

Acetyl chloride

CH3 + HCl

Amide NH2 + CH3COCl

C6H5

N

C6H5

Acetyl chloride

CH3 + HCl

C

H O Acetanilide

(c) C  arbylamine reaction: Primary amines when heated with chloroform and ethanolic potassium hydroxide form pungent isocyanides or carbylamines. This reaction is exclusive to primary amines and is used as a test to distinguish them from secondary and tertiary amines. Heat

NH2 + CHCl3 + 3KOH

R

NC + 3KCl + 3H2O

R

(d) Reaction with nitrous acid: The products of this reaction depend on whether the amine is primary, secondary or tertiary and whether the amine is aliphatic or aromatic. (i) With primary aliphatic amines: These undergo diazotization reaction with nitrous acid to yield highly unstable aliphatic diazonium salts. R

NH2 + NaNO2 + 2HX

(HONO)

1° aliphatic amine

+

R

H2O

N X− + NaX + 2H2O

N

Aliphatic diazonium salt (Highly unstable)

(ii) With primary aromatic amine: These react to form diazonium salts at low temperatures (273–278 K). C6H5

NaNO2 + 2HCl

NH2

C6H5

273−278 K

N2+ Cl− + NaCl + 2H2O

Benzenediazonium chloride

Aniline

(iii) With secondary amines: Both aryl and alkyl react with nitrous acid to yield N-nitrosoamines which usually separate from the reaction mixture as oily yellow liquids. (CH3)2NH + HCl + NaNO2

(HONO) H2O

(CH3)2N

N

O

N-Nitrosodimethylamine (A yellow oil)

Dimethylamine

(iv) With tertiary amines: For tertiary aliphatic amine, equilibrium is established among the tertiary amine, its salt and an N-nitrosoammonium compound. + -

2R3N: + HX + NaNO2

R3NHX

Tertiary aliphatic amine

Amine salt

+

+ R3N

N

OX−

N-Nitrosoammonium compound

(e) Reaction with arylsulfhonyl chloride: Benzenesulfonyl chloride (C6H5SO2Cl, Hinsberg’s reagent), reacts with primary and secondary amines to form sulphonamides. H R

R

N

O H + Cl

1º amine

O Sulphonyl chloride

R

O

N

H + Cl

2º amine

Chapter 29_Organic Compounds Containing Nitrogen.indd 752

S

S O

Ar

−HCl

R

H

O

N

S

Ar

O N-Substituted sulphonamide

Ar

−HCl

R

R

O

N

S

Ar

O N,N-Disubstituted sulphonamide 1/4/2018 5:29:25 PM

H R

O

N

H + Cl

S

−HCl

R

R

O

N

O

N

S

Ar

O O N-Substituted Containing Nitrogen Sulphonyl Organic Compounds sulphonamide chloride

1º amine

R

Ar

H

S

H + Cl

Ar

−HCl

R

O

2º amine

R

O

N

S

753

Ar

O N,N-Disubstituted sulphonamide

(f ) Electrophilic substitution in aromatic amines: The presence of the    NH2 group increases electron density at the positions ortho and para to the group. This is because the lone pair of electron present on nitrogen atom becomes delocalized as a result of resonance making NH2 group highly ring activating. Some examples of electrophilic substitution are: (i) Bromination NH2

NH2 Br + 3Br2

+ 3HBr

Br 2,4,6-Tribromoaniline

Aniline

  

Br

Br2/H2O

For preparing monosubstituted aniline derivative, the NH2 group is protected by acetic anhydride by acylation. O H

NH2

N

O CH3

C

H

CH3

C

NH2 1

OH- or H+

(CH2CO)2O

Br2

Pyridine

CH3COOH

2 3

Br (Major)

N-Phenylethanamide (Acetanilide)

Aniline

N

Br 4-Bromoaniline

(ii) Nitration NH2

NH2 HNO3, H2SO4, 288 K

NH2

NH2 +

NO2

+ NO2

NO2 (51%)

(47%)

(2%)

    The -NH2 group can be protected by acetylation to produce p-nitroaniline as the major product. NH2

NHCOCH3 (CH3CO)2O

NHCOCH3

NH2

OH− or H+

HNO3, H2SO4, 288 K

Pyridine

NO2 p-Nitroacetanilide

Acetanilide

NO2 p-Nitroaniline

(iii) Sulphonation +

NH2

−

NH3HSO4 H2SO4

NH3

SO3H

SO3

Sulphanilic acid

Zwitter ion

453−473 K

Anilinium hydrogen sulphate

Chapter 29_Organic Compounds Containing Nitrogen.indd 753

+

NH2

−

1/4/2018 5:29:27 PM

754

OBJECTIVE CHEMISTRY FOR NEET

(iv) Friedel–Crafts reaction: Aniline forms salt with aluminium chloride, a Lewis acid used as a catalyst and hence it does not undergo Friedel–Crafts reaction (alkylation and acylation). 8. Diazonium Salts (a) Primary aryl amines react with nitrous acid to give arenediazonium salts. +

NH2

NCl−

N NaNO2 + dil. HCl 273−278 K (0−5°C)

Benzenediazonium chloride

Aniline

(b) T  he naming of diazonium salts is done by suffixing diazonium to the name of the parent hydrocarbon followed by the name of anion. (c)  Alkyldiazonium salts formed from primary aliphatic amines are highly unstable while arenediazonium salts formed from primary aromatic amines are stable for a short time in solution at low temperatures (273–278 K). Their stability can be explained on the basis of following resonance structures: +

N

+

N

N

−

+

N

N

−

−

+

N

N

N

+

+ +

(d) Chemical reactions (i) The reactions of diazonium salts involving displacement of nitrogen can be summarized as follows: ArNH2 HONO 0−5°C

ArN+2 (Arenediazonium salt) Sandmeyer reactions CuCN

Ar-CN

CuCl

Ar-Cl

(Gattermann reaction) Cu/HCl

(BalzSchiemann reaction) HBF4, ∆

CuBr

Ar-Br

Ar-Cl

Ar-F

KI

H2O

Ar-I

HBF4

H3PO2, H2O

Cu2O, Cu2+,

Ar-OH

+

Ar-N2BF4−

Ar-H Ar-Ar

Ar-H GombergBachmann reaction

NaNO2, Cu ∆

Ar-NO2 + N2 + NaBF4

(ii) Reactions involving retention of diazo group: These reactions are known as diazo coupling reactions and lead to formation of intensely colored azo compounds. Arenediazonium ions are weak electrophiles; they react with highly reactive aromatic compounds – with phenols and tertiary arylamines – to yield azo compounds. +

OH

N2CI- + Benzenediazonium chloride

Chapter 29_Organic Compounds Containing Nitrogen.indd 754

Phenol

0°C NaOH H2O

N

N

OH

p-(Phenylazo) phenol (Orange soild)

1/4/2018 5:29:29 PM

Organic Compounds Containing Nitrogen

755

Solved Examples 1. What is the classification of the amines I and II in the following compound? O

I N

(1) 1° amine. (2) 2° amine. (3) 3° amine. (4) quaternary salt. Solution

O II NH2

(1) (2) (3) (4)

4. C 3H9N cannot represent

(4) C3H9N cannot represent quaternary structure whereas it can represent 1°, 2° and 3° amines. Their structures are as follows:

I: primary aliphatic; II: primary aromatic. I: tertiary aliphatic, II: primary aromatic. I: tertiary aliphatic, II: tertiary aromatic. I: tertiary aromatic, II: tertiary aliphatic.

Solution (2) In the given compound, I is tertiary aliphatic amine and II is primary aromatic amine. 2. Which of the following is a tertiary amine? (1) CH3CH2CH2CH2NH2 (2) CH3CH2NHCH2CH(CH3)2 (3) (CH3CH2)2NCH2CH(CH3)2 (4) (CH3CH2)4N+ OH-

CH3CH2CH2NH2 1°

H3C H2C

CH2

N

CH3

3°

5. Which of the following is properly termed a quaternary ammonium salt? (1) (CH3)3CCH2CH2NH3+Cl(2) (CH3CH2CH(CH3)CH2)2NH2+Cl(3) (CH3CH2CH2)3NH+Cl(4) (CH3CH2CH2)4N+ClSolution (4) In quaternary ammonium salt, nitrogen is attached with four carbon atoms. CH3

(3) If nitrogen is attached with three carbon atoms, then, it known as called tertiary amine. N

CH3

2°

Solution

CH3 CH2

CH3

H CH3CH2NCH3

CH

CH3 CH3

3. Which is a correct common name for the following substance?

CH2

CH2 CH2 CH3 − Cl CH2 CH2 CH3 CH2 CH2 CH3 + N

CH2

6. Which of these compounds is expected to possess the lowest boiling point? (1) CH3CH2CH2CH2CH2NH2 (2) CH3CH2CH2NHCH2CH3 (3) (CH3CH2)2NCH3 (4) (CH3CH2)2CHOH Solution

N

(3) Compound (CH3CH2)2NCH3 cannot form hydrogen bond, thus, will have the lowest boiling point. (1) Ethylethylisobutylamine (2) Diethylisobutylamine (3) sec-Butyldiethylamine (4) Ethylethyl-sec-butylamine Solution (3) The name of the compound is sec-butyldiethylamine.

7. Which is the correct sequence of reagents for the following conversion? R   CH2OH ® R[CH2]2NH2 (1) NH3, H3O-, KCN (2) PBr3, KCN, H3O+ (3) PBr3, KCN, H2/Pt (4) KCN, H3O+, H2/Pt Solution (3) The reaction involved is

N Ethyl group

Chapter 29_Organic Compounds Containing Nitrogen.indd 755

RCH2OH

PBr3 SN2

R

CH2

Br

KCN SN2

R

CH2

CN

CH2

NH2

sec-butyl grp. R

CH2

H2 /Pt reduction

1/4/2018 5:29:32 PM

756

OBJECTIVE CHEMISTRY FOR NEET

8. Which one of the following methods is neither meant for the synthesis nor for separation of amines?

Option (3): Benzaldehyde on reaction with methylamine in the presence of reducing agent forms N-­benzylmethanamine.

(1) Hinsberg method (2) Hofmann method (3) Wurtz reaction (4) Curtius reaction

Option (4): Nitrobenzene reacts with nitrous acid at low temperature to yield diazonium salt.

Solution (3) Wurtz reaction is used to prepare symmetrical alkanes.

11. What is the product of the following reaction?

9. An organic compound (A) upon reacting with NH3 gives (B). On heating (B) gives (C). (C) in presence of KOH react with Br2 to give CH3CH2NH2. (A) is (2) H3C

(1) CH3CH2CH2COOH

CH

CI

1. NH3 2. Br2, NaOH, H2O

O

COOH

CH3

(4) CH3COOH

(3) CH3CH2COOH

(1)

NH2

(2)

NH2

Solution (3) The reactions can be elucidated as

CH3CH2

C (A)

OH + NH3

CH3CH2

C (B)

- +

ONH4

Hofmann degradation

CH3CH2

C (C)

(1) Hofmann degradation turns amides into primary amines.

NH2 NH2

10. Which is the best method to prepare

NH2

Solution

O CH3CH2NH2

(4)

O

D (-H2O)

Br2/KOH

NH2

(3)

O

O

CI

NH2

1. NH3

O

O

?

O

2. Br2, NaOH, H2O

(1)

NH3 (excess) H2, Ni

Br

NH2

(2)

NH3

12. Identify the best method(s) to prepare (3)

O

H N

CH3NH2 H2, Ni

NO2

(4)

NH2

HONO 0-5°C

(1 mol)

(1)

CH3I (1 mol)

Solution (1) Reductive amination of a ketone is almost always a better method for the preparation of amines of the type  R’ CH(R) NH2 than treatment of an alkyl halide with ammonia. NH

O NH3

Chapter 29_Organic Compounds Containing Nitrogen.indd 756

NH2 H2, Ni

(2)

(3) (4)

O

(CH3)2NH H2, Ni

O

CH3NH2 H2, Ni

Both (1) and (2)

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757

Organic Compounds Containing Nitrogen Solution

14. Which of the following groups does not decrease the basic strength of aniline?

(3) CH2CHO

CH

CH2

N

CH3

+ CH3NH2

(1) OCH3 (2) NO2 (3) CN (4) halogen Solution (1) This is because electron releasing groups due to their +R effect increase the electron density and thus increases the basic character, whereas electron withdrawing groups such as CN, NO2, halogens decrease the basic character.

H2 /Ni

CH2 CH2 NHCH3

13. What would be the product of the following reaction sequence? Cl 1.

, AlCl3

2. (CH3)2NH 3. LiBH3CN

(3) The order of basicity of amines (in aqueous solution) is secondary > primary > tertiary and in case of aniline, lone pair is involved in resonance, so it is a weaker base than aliphatic amines.

O O

(1)

16. Arrange the following amines in the order of increasing basicity.

N

(2)

(1) Trimethylamine (2) Aniline (3) Dimethylamine (4) Methylamine Solution

O

N

15. Which one of the following is the strongest base in aqueous solution?

NH2

N

OH

(3)

(1)

NH2
II > III (2) II > III > I (3) II < III < I (4) III < I < II

Rubbers 23. Natural rubber is a(an) (1) condensation polymer. (2) addition polymer. (3) co-ordination polymer. (4) None of these. 24. Buna-S is a polymer of (1) butadiene. (2) butadiene and styrene. (3) styrene. (4) butadiene and nitryl.

Biodegradable Polymers 30. Which of the following is a biodegradable polymer? (1) Cellulose (2) Polythene (3) Polyvinyl chloride (4) Nylon-6

Uses of Polymers 31. The polymer used for making contact lenses for eyes is (1) polymethylmethacrylate. (2) polyethylene. (3) polyethylacrylate. (4) nylon-6. 32. Wash and wear clothes are manufactured using (1) (2) (3) (4)

nylon fibers. cotton mixed with nylon. terylene fibers. wool fibers.

Level II Classification of Polymers 1. Acrilan is a hard, horny and a high melting point material. Which of the following represent its structure? CH3

25. Heating of rubber with sulphur is known as (1) galvanization. (2) vulcanization. (3) bessemerization. (4) sulphonation. 26. The synthetic polymer which resembles natural rubber is (1) neoprene. (2) chloroprene. (3) glyptal. (4) nylon. 27. Trans-form of polyisoprene is (1) Gutta-percha. (2) hydrochloride rubber. (3) Buna-N. (4) synthetic rubber. 28. Which of the following statement is correct regarding the drawbacks of raw rubber? (1) (2) (3) (4)

It is plastic in nature. It has little durability. It has large water-absorption capacity. All of these.

Molecular Mass of Polymers 29. The mass average molecular mass and number average molecular mass of a polymer are respectively 40,000 and 30,000. The polydispersity index of polymer will be (1) 1 (3) 1 (4) 0

Chapter 30_Polymers.indd 789

789

(1)

CH2

CH CN

(3)

CH2

(2)

CH2

COOC2H5

n

C COOC2H5

C

(4)

CH2

n

CH

n

Cl

n

2. The molecular formula of hexamethylene diamine adipate (monomer of nylon-6,6) is (1) C12H22O2N2 (2) C10H26O4N2 (3) C12H26O4N2 (4) C12H24O3N2 3. Starch is the condensation polymer of (1) a-glucose. (2) b-glucose. (3) a-fructose. (4) b-fructose. 4. Which of the following is a copolymer? (1) Natural rubber (2) Synthetic rubber (3) Gutta-percha (4) Saran 5. Which of the following is a branched polymer? (1) Low-density polyethene (2) Polyester (3) Nylon (4) PVC 6. Which of the following is an elastomer? (1) Bakelite (2) Polyethene (3) Nylon-6 (4) Natural rubber

1/4/2018 5:30:06 PM

790

OBJECTIVE CHEMISTRY FOR NEET

7. Monomer of the following polymer is

(1) polysaccharide. (2) polyester. (3) polyamide. (4) polypeptide.

CH3 C

CH2

CH3

n

(1) 2-methylpropene. (2) styrene. (3) propylene. (4) ethane. 8. Which of the following is a copolymer? (1) Buna-S (2) PAN (3) Polyethene (4) PTFE 9. Thermoplastics (1) are linear polymers. (2) soften or melt on heating. (3) are molten polymers, can be moulded into desired shape. (4) All of these. 10. Natural silk is a (1) polypeptide. (2) polysaccharide. (3) polychloroprene. (4) polyacrylonitrile. 11. Which of the following sets contain only addition homopolymers? (1) (2) (3) (4)

Polyethene, natural rubber, cellulose Starch, nylon, polyester Teflon, bakelite, Orlon Neoprene, PVC, polyethene

12. Which of the following polymer is hard? (1) Linear (2) Cross-linked (3) Branched-chain (4) Thermoplastic 13. Among the following, the weakest inter-particle forces of attraction are present in (1) thermosetting polymers. (2) thermoplastic polymers. (3) fibers. (4) elastomers. 14. Homopolymers are made from (1) (2) (3) (4)

only one type of monomers. two different types of monomers. three different types of monomers. several different types of monomers.

15. Styrene at room temperature is (1) solid. (2) liquid. (3) gas. (4) colloidal solution. 16. Which of the following type of forces are present in nylon-6,6? (1) (2) (3) (4)

van der Waals forces of attraction Hydrogen bonding Three-dimensional network of bonds Metallic bonding

Chapter 30_Polymers.indd 790

17. Wool is a

18. Artificial silk is a (1) polypeptide. (2) polyvinyl chloride. (3) polysaccharide. (4) polyethene.

Types of Polymerization Reactions 19. Low-density polyethene is prepared by (1) (2) (3) (4)

free radical polymerization. cationic polymerization. anionic polymerization. Ziegler-Natta polymerization.

20. Glyptal polymer is obtained from ethylene glycol by reacting it with (1) malonic acid. (2) phthalic acid. (3) maleic acid. (4) acetic acid. 21. Which of the following polymer has ester linkages? (1) Nylon (2) Bakelite (3) Terylene (4) PVC 22. Which of the following is a step growth polymer? (1) Polyacrylonitrile (2) Polyisoprene (3) Nylon (4) Polyethene 23. Benzoyl peroxide has a role in the following type of addition polymerization? (1) Cationic (2) Anionic (3) Free-radical (4) None of these.

Rubbers 24. A copolymer of acrylonitrile and 1,3-butadiene is called (1) Buna-N. (2) polystrene. (3) neoprene. (4) Buna-S. 25. The monomer of synthetic rubber is (1) butadiene. (2) 2-chloro-1,3-butadiene. (3) 2-methyl-1,2-butadiene. (4) 2-methyl-1,3-butadiene. 26. Which of the following is used in vulcanization of rubber? (1) SF6 (2) CF4 (3) Cl2F2 (4) C2F2 27. Vulcanized rubber resists (1) (2) (3) (4)

wear and tear due to friction. cryogenic temperature. high temperature. action of acids.

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791

Polymers 28. In vulcanization of rubber (1) (2) (3) (4)

sulphur reacts to form a new compound. sulphur cross-links are introduced. sulphur forms a very thin protective layer over rubber. All of these.

29. The S in Buna-S stands for (1) sodium. (2) sulphur. (3) styrene. (4) Just a trade name. 30. Gutta–percha is a polymer related to which of the following? (1) Proteins (2) Cellulose (3) Rubber (4) PVC 31. Extent of stiffness of vulcanized rubber depends upon (1) (2) (3) (4)

temperature of vulcanization. time of vulcanization. amount of sulphur. All of these.

Molecular Mass of Polymers 32. Number average molecular mass (M n ) and weight average molecular mass (M w ) of synthetic polymers are related as: (1) M n < M w (2) M n > M w (3) M n = M w (4) None of these. 33. Polydispersity index for natural polymers is generally close to (1) 0 (2) 100 (3) 1 (4) 10

(2) Polystyrene (3) Poly(ethylene terephthalate) (4) Polytetrafluoroethylene 38. Which of the following polymers can be used for lubrication and as an insulator? (1) SBR (2) PVC (3) PTFE (4) PAN 39. To make PVC a flexible plastic, the additive used is called (1) filler. (2) antioxidant. (3) stabilizer. (4) plasticizer. 40. Polymer used in bullet proof glass is (1) PMMA. (2) Lexan. (3) Nomex. (4) Kevlar. 41. Which of the following is used in paints? (1) Terylene (2) Nylon (3) Glyptal (4) Chloroprene 42. Which of the following polymers is used for making magnetic recording tapes? (1) Dacron (2) Bakelite (3) Glyptal (4) Acrilan

Previous Years’ NEET Questions 1. Which one of the following polymers is prepared by condensation polymerization? (1) Styrene (2) Nylon-6,6 (3) Teflon (4) Rubber (AIPMT 2007)

34. Select the incorrect statement among the following. (1) For natural polymers, PDI is generally 1. (2) For natural polymers are more homogeneous than synthetic polymers. (3) For synthetic polymers, PDI is generally 1. (4) The polymers whose molecules have nearly same molecular masses, PDI is 1.

Biodegradable Polymers

2. Which one of the following statements is not true? (1) Natural rubber is a 1, 4-polymer of isoprene. (2) In vulcanization, the formation of sulphur bridges between different chains makes rubber harder and stronger. (3) Natural rubber has the trans-configuration at every double bond. (4) Buna-S is a copolymer of butadiene and styrene.

35. Which of the following is not a biopolymer? (1) Proteins (2) Nucleic acids (3) Cellulose (4) Neoprene

Uses of Polymers 36. Soft drinks and baby feeding bottles are generally made up of

(AIPMT 2008) 3. Structures of some common polymers are given. Which one is not correctly presented? NH(CH2)6NHCO(CH2)4 (1) Nylon-6,6 CF2 n Teflon CF (2) 2

(3) Neoprene

CH2

(1) polyester. (2) polyurethane. (3) polystyrene. (4) polyamide. 37. Which one of the following is used to make non-stick cookware? (1) PVC

Chapter 30_Polymers.indd 791

C

CH

CO

CH2

2

CH2

n

Cl

(4) Terylene

OC

COOCH2

CH2

O n

(AIPMT 2009)

1/4/2018 5:30:07 PM

792

OBJECTIVE CHEMISTRY FOR NEET

4. Which of the following structure represents neoprene polymer? CN

(1)

(

CH2

(

CH2

(4) ( CH2

(n

(1) CH2

CH C

(3) CH2

CH

CH2

(n

C

(1) Nylon-6,6 (2) Terylene (3) Bakelite (4) Melamine

6. Which one of the following is not a condensation polymer? (1) Melamine (2) Glyptal (3) Dacron (4) Neoprene (AIPMT PRE 2012)

CH

C

CH2

(4) CH2

CH

(AIPMT PRE 2012) 8. Which one of the following sets forms the biodegradable polymer?

(2) HN2

CH2

COOH and HN2

(3) HO

CH2

CH2 CH

CH

CH

(CH2)5

CH2

(1)

CH2 and CH2

COOH

CH

CH

( CH2

C

CH

CH2 (n

(2)

CH (n

( CH2

Cl H

(3)

(4)

(N

(CH2)8

H

O

O

N

C (CH2)4

C (n

OH

OH CH2

n

(AIPMT 2014) 12. Which of the following organic compounds polymerizes to form the polyester Dacron? (1) (2) (3) (4)

Propylene and para-HO   (C6H4)   OH. Benzoic acid and an ethanol. Terephthalic acid and ethylene glycol. Benzoic acid and para-HO   (C6H4)   OH. (AIPMT 2014)

COOH

OH and HOOC

CH2

11. Which one of the following is an example of a thermosetting polymer?

CH2

Artificial silk is derived from cellulose. Nylon-6,6 is an example of elastomer. The repeat unit in natural rubber is isoprene. Both starch and cellulose are polymers of glucose.

CN and CH2

CH

Cl

7. Which of the following statements is false?

CH

CH2

Cl

(AIPMT PRE 2011)

(1) CH2

CH

(NEET 2013)

5. Of the following which one is classified as polyester polymer?

CH2

(AIPMT MAINS 2012)

13. Biodegradable polymer which can be produced from glycine and aminocaproic acid is (1) PHBV. (2) Buna-N. (3) Nylon-6,6. (4) Nylon-2-nylon-6. (AIPMT 2015) 14. Caprolactam is used for the manufacture of

9. Nylon is an example of (1) polyester. (2) polysaccharide. (3) polyamide. (4) polythene. (NEET 2013)

Chapter 30_Polymers.indd 792

(2) CH2

CH

CH3

(n

(AIPMT PRE 2010)

(4)

C

Cl

C6H5

(1) (2) (3) (4)

CH

Cl

CH (n

CH2

(3) ( CH

(2)

10. Which is the monomer of neoprene in the following?

(1) terylene. (2) nylon-6,6. (3) nylon-6. (4) teflon. (RE AIPMT 2015)

1/4/2018 5:30:08 PM

(1)

H2 C

H2 C H C

H C NH2

15. Natural rubber has

(2)

(1) all cis-configuration. (2) all trans-configuration. (3) alternate cis– and trans-configuration. (4) random cis– and trans-configuration.

H2 C

(3)

O C

16. Which one of the following structures represents nylon-6,6 polymer? (1)

H2 C

H2 C H C

H C NH2

(2)

H2 C

2

 1. (3) 11. (3) 21. (4) (4) 31. (1)

Level II

O C

NH2

H2 C C 2C  2. H2 (4) 12. (4) O

Cl

H2 C

C H2

H2 C

H2 C

H2 C H C

COOH

CH3

NH n

H2 C H C CH3

66

H2 H C C

H C

COOH

CH3

6

6

H N ( CH2)6

2C

H2 H C C

H C

O

NH2

66

H2 H C C

H C

Answer Key NH (3) I Level

(4)

Cl

(NEET II 2016) 6

6

H N ( CH2)6

NH

 3. (1)

 4. (4)

 5. (2)

 6. (4)

 7. (1)

 8. (4)

 9. (2)

10. (1)

13. (4)

14.n(4)

15. (1)

16. (1)

17. (4)

18. (3)

19. (1)

20. (1)

23. (2) H222. (1)H 2 C H C H 32.C(3) C

24. (2)

25. (2)

26. (1)

27. (1)

28. (4)

29. (2)

30. (1)

NH2

793

66

H2 H C C

H C NH2

(NEET I 2016)

Polymers

NH2

CH3

66

 1. (1)

 2. (4)

 3. (1)

 4. (4)

 5. (1)

 6. (4)

 7. (1)

 8. (1)

 9. (4)

10. (1)

11. (4)

12. (2)

13. (4)

14. (1)

15. (2)

16. (2)

17. (3)

18. (3)

19. (1)

20. (2)

21. (3)

22. (3)

23. (1)

24. (1)

25. (2)

26. (4)

27. (1)

28. (2)

29. (3)

30. (3)

31. (3)

32. (1)

33. (3)

34. (3)

35. (4)

36. (3)

37. (4)

38. (3)

39. (4)

40. (2)

41. (3)

42. (2)

 7. (2)

 8. (2)

 9. (3)

10. (3)

Previous Years’ NEET Questions  1. (2)

 2. (3)

 3. (3)

 4. (2)

 5. (2)

 6. (4)

11. (4)

12. (3)

13. (4)

14. (3)

15. (1)

16. (3)

Hints and Explanations Level I 1. (3) Bakelite is thermosetting polymer. It becomes infusible on heating and cannot be remolded. 3. (1) It is present in the cell wall of plant. 4. (4) Silk is protein fiber. Dacron is polyester fiber and Nylon-6,6 is polyamide fiber.

Chapter 30_Polymers.indd 793

5. (2) Polyethylene is a homopolymer.



nCH2   CH2® (   CH2   CH2)n

6. (4) Terylene is fiber not a thermosetting plastic because on heating it melts and does not show plastic property while rest of the options is true regarding to terylene.

1/4/2018 5:30:09 PM

794

OBJECTIVE CHEMISTRY FOR NEET

7. (1) Sucrose is a disaccharide which upon acid or enzymatic hydrolysis gives only two molecules of monosaccharides. 11. (3) Nylon is a polyamide and hence contains nitrogen.

1 N

6 N 5

Level II 1. (1) The structure of acrilan is

12. (4) Melamine is 1,3,5-triazine-2,4,6-triamine. H2N

wool (Terywool) to increase their resistance to wear and tear.

CH2

NH2 2

4 NH2

CN

N 3



14. (4) Nylon is the copolymer of hexamethylenediamine and adipic acid. It is not a homopolymer because homopolymer formed by the same monomer units. 18. (3) Dacron (Terylene) is a condensation polymer. It is formed by the condensation polymerization of terephthalic acid and ethylene glycol.

4. (4) Vinylidene dichloride is polymerized with vinyl chloride to prepare copolymer Saran. 5. (1) Low-density polyethene is a branched polymer whereas nylon, PVC and polyesters are linear polymers. 6. (4) Natural rubber is an elastomer. 7. (1) CH3 nCH3

CH

CH2

Polymerization

CH3

(CH2

C

CH

CH2)n

C

CH3 Polymerization

CH2

C

2-Methylpropene

23. (2) Natural rubber is addition polymer of isoprene (2-methyl-1,3-butadiene). C

n

The other name for the polymer is polyacrylonitrile (PAN).

21. (4) Tetrafluoroethene (CF2   CF2).

nCH2

CH

CH3

CH nCH2

CH

CH

CH2

CH2 + n

CH

1,3-Butadiene

27. (1) Gutta-percha rubber is very hard, horny material consisting of trans 1,4-polyisoprene polymer. 28. (4) The raw rubber is plastic in nature. It becomes soft at high temperature, has little durability and large water absorption capacity.

n

8. (1)

CH3

24. (2) Buna-S is an addition co-polymer of butadiene and styrene. It is an aritificial rubber.

CH2

Styrene Na, D Copolymerization

CH2

CH

CH

CH2

CH2

CH

29. (2) Average number molecular weight M n = 30, 000

Average mass molecular weight M w = 40, 000



Polydispersity index (PDI) =

M w 40, 000 = = 1.33 M n 30, 000

30. (1) Cellulose is the natural fiber which are biodegradable polymer rest are synthetic polymer which are not biodegradable. 31. (1) It is also known as PMMA. It is a transparent, excellent light transmitter and its optical clarity better than glass so it is used in the preparation of lenses for eyes. 32. (3) The fiber of terylene is highly crease-resistant, durable and has low moisture content. It is also not damaged by pests like moths and mildew. It is therefore used for the manufacture of wash and wear fabrics. It is also blended with cotton (Terycot) and

Chapter 30_Polymers.indd 794

(Buna-S)

n

10. (1) Natural silk is made of protein fiber consisting of fibroin, secreted insect larvae, held together by various peptide linkages. 11. (4) Homopolymers are formed from only one repeating unit. Neoprene, PVC and polyethene contain only one type of monomer and undergo addition polymerization. 12. (2) Cross-linked polymer is hard because the monomer units are linked together to form three-dimensional network like polymers. 13. (4) Elastomers have very weak intermolecular forces which can be overcome easily and the polymer chains can be stretched by applying small stress and they can regain shape on removing stress.

1/4/2018 5:30:10 PM

Polymers

795

14. (1) Homopolymers are formed from only one repeating unit. For example, polyethene, styrene, etc.

27. (1) Because of presence of sulphur and additives like ZnO, the vulcanized rubber resists wear and tear due to friction.

16. (2) In nylon-6,6, there is orientation of linear polyamide molecules so that they are parallel to the fiber axis. This allows hydrogen bonds to form between    NH    and    C   O groups on adjacent chains.

29. (3) Buna-S is copolymer of 1,3-butadiene and styrene. In Buna-S, Bu represents butadiene, Na (Natrium) stands for sodium which is a polymerizing agent and S stands for styrene.

17. (3) Wool and silk are two naturally occurring polyamides.

30. (3) Gutta-percha is trans-polyisoprene and hence the geometrical isomer of natural rubber.

18. (3) Artificial silk is a cellulose derivative, hence a polysaccharide. It gives thread of ash on burning.

31. (3) Rubber containing about 5% sulphur is used in making rubber tyres. Similarly, vulcanized rubber containing about 30% sulphur is employed for making cases of batteries, etc.

20. (2) COOH nHOH2C

CH2OH + n

Ethylene glycol

32. (1) From the expressions of number average and weight average mass

COOH

Phthalic acid

Mn =

åM N åN i

i

i

Zn(OCOCH3)2 + Sb2O3 420−460 K



O OCH2

CH2

O

C

+ (2n–1)H2O O C n

Glyptal

22. (3) Nylon is a step growth polymer because polymerization proceeds in steps and is called step growth polymerization. 23. (3) It acts as an initiator and is used to generate a pair of radicals. 24. (1) nCH2

CH

CH

CH2 + nCH2

CN Acrylonitrile

1,3-Butadiene

Heat

CH2

CH

CH

CH

CH2

CH2

Peroxide

CH

n

CN

Mw =

åw M åw i

i

i

=

åN M åN M i

2 i

i

i

We have Mn < Mw .

33. (3) The ratio of weight average and number average molecular masses is called polydispersity index (PDI). In natural polymers, which are generally monodispersed, the PDI is unity. 34. (3) Polydispersity index is a measure of distribution of mass in a given sample of polymer. It has value greater than or equal to one. It is given by Mw/M n . For synthetic polymers, PDI is generally greater than one because Mw is always higher than Mn . 35. (4) Neoprene is a synthetic manmade polymer whereas proteins, nucleic acids and cellulose are natural biopolymers. 36. (3) Soft drinks and baby feeding bottles are generally made up of polystyrene because it is a transparent thermoplastic material that floats over water. 37. (4) Polytetrafluoroethylene is used to make “non-stick” cookware because it is very hard and tough. It is resistant to heat, acids and bases. 38. (3) PTFE can be used for lubrication and as an insulator because it is very tough and hard and a bad conductor of electricity.

Nitrile rubber (Buna-N)

25. (2) The monomer of neoprene rubber (or synthetic rubber) is chloroprene or 2-chloro-1,3-butadiene. Cl CH2 C CH CH2 Chloroprene(2-chloro-1,3-butadiene)

Chapter 30_Polymers.indd 795

39. (4) Plasticizers are substances which are added in the formation of polymers in order to alter their physical properties. These are normally added to increase the softness of hard polymers. 40. (2) Lexan is used in bullet proof glass because it is very hard and tough.

1/4/2018 5:30:13 PM

796

OBJECTIVE CHEMISTRY FOR NEET

41. (3) Glyptal is used in paints. It is a network cross-linked polymer. When dissolved in suitable organic solvent, the solution formed on evaporation leaves a tough non-­ flexible film.

8. (2) Nylon-2-nylon-6 is an alternating copolymer of glycine (NH2   CH2   COOH) and amino capric acid (NH2   (CH2)5   COOH) and is biodegradable. nH2N

COOH+ nNH2 ( CH2 )5 COOH

CH2

42. (2) Bakelite is used for making combs, phonograph records, electrical switches and handles of various utensils.

Glycine

Previous Years’ NEET Questions

C

N

N

(CH2)6

C

Polymerization

n

C

CH

CH2

CH2

CH

CH2

CH2

Cl

n

Neoprene

11. (4) Phenol-formaldehyde (bakelite) is an example of thermosetting polymer.

nHOH2C

CH2OH + nHOOC

Ethylene glycol CH2

CH

n

4. (2) The structure of neoprene is CH

C

12. (3) The reaction involved in the preparation of Dacron is

Cl

C

9. (3) Polyamides are polymers possessing an amide linkage and are prepared by condensation polymerization of dicarboxylic acids with diamines or amino acids with their lactams as in nylon-6,6 and nylon-6 respectively.

Cl

3. (3) The correct structure of neoprene is

CH2

O

2-Chloro-1,3-butadiene

2. (3) Natural rubber is a polyisoprene, but not all bonds have trans configuration.

C

C ( CH2 )5 NH

Nylon-2-nylon-6

nCH2

Nylon-6,6

CH2

NH

Amino caproic acid

10. (3) Chloroprene is the monomer of neoprene.

C

(CH2)4

CH2

O

1. (2) The polymers formed by intermolecular condensation reaction by the functional groups of monomers with continuous elimination of byproducts are called condensation polymers. Nylon-6,6 is prepared by condensation polymerization of adipic acid and hexamethylenediamine. nNH2(CH2)6NH2 + nHOOC(CH2)4COOH Adipic acid Hexamethylene diamine H O O H

−H2O

COOH

Tetraphthalic acid

n

Cl

5. (2) Polyesters are polymers formed from a dicarboxylic acid and a diol. Among the given polymers, terylene is a polyester polymer. O

O

C

C

O

CH2

CH2

6. (4) Neoprene is an addition polymer of isoprene. C

CH

Cl Chloroprene

CH2

Polymerization

CH2

C

CH

Cl

CH2

C

C n

Dacron

NH n

Neoprene

7. (2) Nylon-6,6 is an example of fiber. Fibers are the polymers which have strong intermolecular forces between the forces.

Chapter 30_Polymers.indd 796

CH2

O

13. (4) Nylon-2-nylon-6 is an alternating polyamide copolymer of glycine (H2N   CH2   COOH) and aminocaproic acid [NH2(CH2)5COOH]. It is biodegradable polymer and its structure is

O n

nCH2

OCH2

O

CH2

C O

NH ( CH2 )5 C O n

14. (3) Nylon-6 can be prepared by a ring-opening polymerization of e-caprolactum in the presence of water at 553–543 K.

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Polymers O

16. (3)

O NH H2O

OC ( CH2 )5

NH

(−H2O) 250°C

C

O

C ( CH2 )5 NH C ( CH2 )5 NH Nylon-6

CH

CH2

Polymerization

O n

C

H N

(CH2)6

H

O

N

C

O (CH2)4

C n

Nylon-6,6

15. (1) Natural rubber is cis polyisoprene. H2C

nNH2(CH2)6NH2 + nHOOC(CH2)4COOH Hexamethylene Adipic acid diamine

NH

+ NH3 +

O

e-Caprolactam (a cyclic amide)

797

Polymerization

CH3 Isoprene CH3 CH2

Chapter 30_Polymers.indd 797

C

C

H

CH2

CH3 CH2 cis-Polyisoprene

C

C

CH2 H

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31

Biomolecules

Chapter at a Glance Carbohydrates 1. Carbohydrates are defined as polyhydroxy aldehydes or ketones or those compounds which yield these compounds on hydrolysis. They are thus polyhydroxy compounds that have an aldehyde or a ketonic functional group either in free form or as hemiacetal or acetal. Their empirical formula is Cx(H2O)y or (CH2O)n, where n and y are integers. 2. Classification (a) Carbohydrates can be classified on the basis of their behavior on hydrolysis as: (i) Monosaccharides: These cannot be hydrolyzed further to give simpler unit of polyhydroxy aldehyde or ketone. Monosaccharide can be further classified on the basis of number of carbon atoms they contain and the functional group present in them as shown below Number of carbon atoms Term 3 4 5 6 7

Triose Tetrose Pentose Hexoses Heptoses

Aldehyde

Ketone

Aldotriose Aldostetrose Aldopentose Aldohexose Aldoheptose

Ketotriose Ketotetrose Ketopentose Ketohexose Ketoheptose

(ii) Oligosaccharides: These are carbohydrates that yield up to ten monosaccharide units on hydrolysis. They can be further classified as diasaccharides, trisaccharides, tetrasaccharides, etc. depending upon the number of monosaccharide units they yield on hydrolysis. For example, sucrose, raffinose, etc. (iii) Polysaccharides: These on hydrolysis yield large number of monosaccharides, for example, starch, cellulose, glycogen, etc. They are not sweet in taste, therefore they are also termed as non-sugars. (b) Carbohydrates are also classified as reducing or non-reducing sugars based on their behavior towards Fehling’s solution and Tollens’ reagent. (i) Carbohydrates that reduce these reagents are called reducing sugars and these contain free aldehydic or ketonic groups as in all monosaccharaides. (ii) Carbohydrates that do not reduce these reagents are called non-reducing sugars. In these compounds, the aldehydic or ketonic groups are bonded, for example, sucrose. 3. D and L Designation: A monosaccharide whose highest numbered stereogenic center (the penultimate carbon) has the same configuration as D-(+)-glyceraldehyde is designated as D sugar; one whose highest numbered stereogenic center has the same configuration as L-glyceraldehyde is designated as L-sugar. 4. Monosaccharides (a)  If the monosaccharide ring is six membered, the compound is called pyranose (e.g., glucose) and if it is fivemembered it is designated as furanose (e.g., fructose). (b) Epimers are isomers that differ in configuration at one stereogenic center (e.g., glucose and galactose).

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OBJECTIVE CHEMISTRY FOR NEET

(c)  Change in value of specific rotation of carbohydrates is called mutarotation. (d) Monosaccharides are converted to their ester derivatives when treated with acid chloride or anhydride (excess). (e)  On oxidation, aldoses yield an aldonic acid. On treatment with HNO3, an aldose is oxidized to give a dicarboxylic acid called aldaric acid. 5. Glucose (C6H12O6) (a) Glucose is obtained from ripe grapes, sweet fruits and honey. (b) Methods of preparation (i) From sucrose +

C12 H22 O11 H→ C 6 H12 O6 + C 6 H12 O6 Fructose

Glucose

(ii) From starch

+

(C 6 H10 O5 )n + nH2 O H → C 6 H12 O6 ∆ Glucose

(c) Chemical properties: Glucose exhibits the following properties Hl NH3

CH3(CH2)4CH3 Hexane CH

N

OH

(CHOH)4 CH2OH HCN

CN CH OH (CHOH)4

CHO

CH2OH Bromine water

(CHOH)4

COOH (CHOH)4

CH2OH

CH2OH

Glucose

Gluconic acid Acetic anhydride

CHO

O

(CH

O

C

CH3)4

CH2

O

C

CH3

(O)

O Oxidation

COOH (CHOH)4 COOH Saccharic acid

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Biomolecules

(d) Structure of glucose (i) Fischer (open-chain) structure

801

CHO CHO

H

H

C

OH

HO

C

H

OH H

HO H

OH

H

C

OH

H

OH

H

C

OH

CH2OH CH2OH

Fischer projection formula

Wedge-linedashed wedge formula

(e) Cyclic structure: The following properties of glucose can be explained only on the basis of cyclic structure of glucose: (i) The aldehyde group in glucose does not give 2,4-DNP test, Schiff ’s base test and reaction with NaHSO3. (ii) The pentacetate formed does not react with hydroxylamine. (iii) It exists in two different forms a and b. (f ) Cyclic forms: The cyclic forms of D-(+)-glucose are hemiacetals formed by intramolecular reaction of the    OH group at C5 with the aldehyde group. This creates a stereogenic center at C1 and two cyclic forms are thus possible. The cyclic isomers differing in configuration only at C1 are diasteroemers and are called anomers. The hemiacetal carbon atom is called anomeric carbon atom and the two anomers are designated as a- and b- anomers. HOCH2

CH2OH O

O H HO

H OH

H

H

OH

H

H

+

HO

HO

OH

H OH

H

H

OH

H

Haworth formulas

HO HO

HOCH2

O

HOCH2

HO

O OH

HO

HO

OH

OH a-D-(+)-Glucopyranose

b-D-(+)-Glucopyranose

6. Fructose (C6H12O6) (a)  Fructose, also known as levulose, is a ketohexose and occurs in fruit juices, honey, and (along with glucose) as a constituent of the disaccharide sucrose. Fructose is the major constituent of the polysaccharide inulin which is a starch-like substance present in many plants. (b) Structure: Fructose has an open-chain structure, belongs to the D-series and is a laevorotatory compound. CH2OH C

O

HO

H

H

OH

H

OH CH2OH D-(−)-Fructose

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OBJECTIVE CHEMISTRY FOR NEET

It also exists as a five-membered cyclic furanose ring. 1 HOH2C

2 C

OH

3

HO

O

H

4

H

O

H

4

H

OH

1 CH2OH

3

HO

5

H

2 C

OH

OH

5

H

6 CH2OH

6 CH2OH a -D-(−)-Fructofuranose

b -D-(−)-Fructofuranose

7. Disaccharides In a disaccharide, two monosaccharide units are joined by a glycosidic bond between the anomeric carbon of one unit and an –OH of the other by the loss of water molecule. Three important disaccharides are sucrose, lactose and maltose. (a) Sucrose: It is the ordinary table sugar and is found in all photosynthetic plants and is obtained commercially from sugarcane or sugar beets. (i)  On hydrolysis, sucrose gives an equimolar mixture of D-(+)-glucose and D-(–) fructose. C12 H22 O11 + H2 O ® C 6 H12 O6 + C 6 H12 O6 D-(+)-Glucose

D-( - )-Fructose

The product is known as invert sugar. (ii) The structure of sucrose is as follows: 6 HOCH2 5 From D-glucose

1 CH2OH

O 1

4

H 2

HO

HO

HO

5

O

2

3

O

4

OH

6 CH2OH

From D-fructose

OH

a -Glucosidic linkage

b -Fructosidic linkage

(iii)  Since the anomeric carbons of both the monosaccharide units are involved in formation of the glycosidic bond, neither of the unit is in equilibrium with its open-chain form. Thus, sucrose is a non-reducing sugar. (b) Maltose: It is made up of two a-D-glucose units in which C1 of one glucose unit is linked with C4 of another glucose unit. It is a reducing sugar as the free aldehyde group can be produced at C1 of the second glucose unit in solution. 6 HOCH2 H 4 HO

5 H HO 3 H

6 HOCH2 O 1 H 2 OH

H H 4 O

5 H HO 3 H

O

OH

H 1 2 OH

H

a -Glucosidic linkage

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Biomolecules

803

(c) Lactose: It is a disaccharide present in the milk and is known as milk sugar. It is a reducing sugar that hydrolyses to yield D-glucose and D-galactose. The glycosidic linkage is b with C1 of galactose is linked with C4 of glucose. 6 HOCH2 5 From D-glucose

1 CH2OH

O

4 HO

OH 3

O

H 1

5

2

HO

O

2 OH

3 OH

a -Glucosidic linkage

6 CH2OH

From D-fructose

4

b -Fructosidic linkage

8. Polysaccharides These consist of monosaccharides joined together by glycosidic linkages. Polymers of a single monosaccharide are called homopolysaccharides; those made up of more than one type of monosaccharide are called heteropolysaccharides. (a) Three important polysaccharides, all of which are glucans, are starch, glycogen and cellulose. (i) Starch: It is a polymer of a-glucose that occurs as microscopic granules in the roots, tubers and seeds of plants. • I t is a polymer of a-glucose. Heating starch with water produces a colloidal suspension with two major components amylose and amylopectin. • T  he water soluble amylose has glucopyranoside units connected in a-linkages between C1 of one unit and C4 of the next. • W  ater-insoluble amylopectin has a structure similar to that of amylose links, but the chains are branched. Branching takes place between C6 of one glucose unit and C1 of another and occurs at intervals of 20–25 glucose units. (ii) Glycogen It occurs in liver, muscles and brain and is broken down into glucose for further use in the body. Its structure is similar to that of amylopectin but is only more branched. (iii) Cellulose: It serves as structural material in plants. It contains D-glucopyranoside units linked in (1 → 4) fashion in very long unbranched chains. Unlike starch and glycogen, however, the linkages in cellulose are b-glycosidic linkages. Proteins 1. Amino Acids (a)  Amino acids are carboxylic acids that contain an amino group. When the amino group is attached to C-2 (a-carbon atom), these are called a-amino acids. a-carbon

Variable group Amino group

R

CH

COOH

Carboxyl group

NH2

Some amino acids may have two amino groups and some may have two acid groups. (b)  Depending on the number of amino or carboxylic groups, amino acids are classified as acidic (more carboxyl groups than amino); basic (more amino groups than carboxyl) or neutral (same number of amino and carboxyl groups). Some common amino acids with their structure; full and abbreviated names are listed as follows.

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OBJECTIVE CHEMISTRY FOR NEET

Structure Neutral amino acids

Name

Abbreviation

Glycine

G or Gly

Alanine

A or Ala

Valine

V or Val

OH

Leucine

L or Leu

OH

Isoleucine

I or Ile

Phenylalanine

F or Phe

Tyrosine

Y or Tyr

Tryptophan

W or Trp

O H2N

OH O OH

NH2 O OH NH2 O

NH2 O

NH2 O OH NH2 O OH NH2

HO

O OH NH2

N H

Side-chains containing an acidic (carboxyl) group O HO OH O O

Aspartic acid

D or Asp

Glutamic acid

E or Glu

NH2 O

HO

OH NH2

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Biomolecules

Structure Name Side-chains containing a basic (amino) group

805

Abbreviation

O H2N

OH

Lysine

K or Lys

Arginine

R or Arg

Histidine

H or His

NH2 NH H2N

O

N H

OH NH2 O

H N N

OH NH2

(c)  There are about 200 amino acids known but only 20 of these are found in almost all proteins. Ten amino acids (valine, leucine, isoleucine, arginine, lysine, threonine, methionine, phenylaniline, tryptophan and histidine) are considered essential because these cannot be synthesized by human body. (d)  Amino acids, when dissolved in water, form a dipolar ion called Zwitterion. They exhibit amphoteric behavior in zwitterion form and react with both acids and bases. O

O R

CH

C

O

H

CH

R +

NH2

C

O−

NH3

Zwitterion

(e) Isoelectric point is the pH at which the particular amino acid molecule carries no net electrical charge. It can affect the solubility of the molecule at a given pH. 2. Structure of Proteins – Polypeptide Bond The bond connecting the amino acids in a protein is called a peptide linkage or peptide bond, formed between    COOH group and    NH2 group. O

O CH2C NH2

+

OH H

Glycine

CH2COOH N

H

Glycine

−H2O

CH2C NH2

N

Peptide linkage CH2COOH + H2O

H

Glycylglycine (Gly - Gly) (a dipeptide)

3. Classification of Proteins Proteins can be classified into two types on the basis of molecular shape. (a) Fibrous proteins: In these proteins, the polypeptide chains run parallel and are held together by hydrogen and disulphide bonds. They are water insoluble, for example, keratin, myosin, etc. (b) Globular proteins: In these proteins, the chains of polypeptide coil around to give a spherical shape. They are water soluble, for example, insulin and albumins. 4. Structure of Proteins The structure and shape of proteins can be described in terms of following four levels of organization.

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OBJECTIVE CHEMISTRY FOR NEET

(a) Primary structure: The sequence of amino acid residues in a polypepetide is called its primary structure. The protein may have one or more polypeptide chains, in which basic unit amino acids are linked with each other in a particular sequence. If we bring any change in the structure or the sequence of amino acids, it creates a different protein. (b) Secondary structure: It is defined by the local conformation of its polypeptide backbone. It refers to the shape in which a long polypeptide chain can exist which can be a-helix and b-pleated sheet. a-Helix secondary structure is stabilized by intramolecular hydrogen bonding. (c) Tertiary structure: It represents the three-dimensional shape that arises due to further folding of polypeptide chains, that is, folding superimposed on the coils of a-helices or the secondary structure of the protein. (d) Quaternary structure: Some of the proteins are composed of two or more polypeptide chains. The spatial arrangement of these sub-units with respect to each other is known as quaternary structure. 5. Denaturation of Proteins: Whenever a protein is subjected to physical change like change in temperature or pH, hydrogen bonding gets disturbed and protein loses it biological activity. This is called denaturation. It causes disruption of the non-covalent forces and disulphide bonds that impact the secondary and tertiary structures but primary structure is maintained. 6. Enzymes (a)  Enzymes are catalysts of biochemical reactions. They are named after the reactions they catalyze, for example, oxidoreductase that catalyzes oxidation–reduction reactions. (b)  Many enzymes show stereospecific activity, that is, they react with or produce a particular stereoisomer. 7. Hormones (a) These function as chemical messengers in the body. (b)  Chemically, hormones may be steroids (e.g., estrogen, androgens), polypeptides (e.g., insulin, endorphins), amines (thyroxine) amino acids (e.g., epinephrine). Nucleic Acids 1. The hereditary characters are transmitted through chromosomes present in the nucleus, which are made up of proteins and nucleic acids. There are two types of nucleic acids, deoxy ribonucleic acid (DNA) and ribonucleic acid (RNA). 2. Nucleic acids are polynucleotides formed by simple units called nucleotides. (a) (b)

Chemical composition Nucleotide = Nucleosides + Phosphoric acid (at 5′carbon of pentose) Nucleoside = Nitrogenous base + Sugar (D-ribose or 2-deoxy-D-ribose) There are two types of nitrogenous bases: (i) Purine: Adenine, Guanine (DNA, RNA) (ii) Pyrimidine: Thymine (DNA), uracil (RNA), cytosine (c)  DNA is the genetic material of the cell and forms chemical basis of heredity. Proteins are synthesized in through RNA molecules (translation) and RNA is synthesized by DNA (transcription). (d)  Two important differences between DNA and RNA are (i) The sugar in DNA is D-2-deoxyribose while that in RNA is D-ribose. (ii) DNA contains base thymine while RNA contains a base called uracil. Vitamins 1. Vitamins are compounds not synthesized by human body but required in small amounts for normal biochemical functions to take place. These are classified as water soluble (Vitamin C, B) and fat soluble (Vitamin D, E, A, K). 2. Chemical name of some vitamins, their sources and diseases arising from their deficiency are given as follows.

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Biomolecules

Name of Vitamin Vitamin A (Retinol) Vitamin B1 (Thiamine) Vitamin B2 (Riboflavin) Vitamin B6 (Pyridoxine) Vitamin B12 (Cyanocobalamine)

Source Green leafy vegetables, fish, liver oil Green leafy vegetables, cereals, milk, yeast Egg white, milk, liver, kidney Yeast, milk, egg yolk, cereals Meat, fish, egg, curd

Deficiency Disease Xerophthalmia (Night blindness) Beri beri Cheilosis Convulsions Pernicious anaemia.

Vitamin C (Ascorbic acid) Vitamin D (Calciferol)

Citrus fruits Exposure to sunlight and fish

Vitamin E (Tocopherol)

Oils – vegetable sunflower

Vitamin K (Phylloquinone)

Green leafy vegetables

Scurvy Rickets (children), Osteomalacia (Soft bones in adults). Increases fragility of RBC and muscular weakness. Blood clotting time is increased.

807

Solved Examples 1. Which of the following will not show mutarotation?

4. What is the product of the following reaction? CHO

(1) Maltose (2) Lactose (3) Glucose (4) Sucrose

H H HO H

Solution (4) Mutarotation is the interconversion of a and b anomers. Since the anomeric carbons of both the glucopyranose and fructofuranose units are involved in formation of the glycosidic bond, neither monosaccharide unit is in equilibrium with its open-chain form. Thus, sucrose is a non-reducing sugar.

(1)

(2)

CO2H OH OH H OH

H H HO H

(3)

H H HO H

(3) The body generates L-lactic acid as a result of vigorous exercise.

OH OH H OH CH2OH

(4)

CHO H H HO H

Solution

OH OH H OH

H H HO H

CO2H OH OH H OH CO2H

CO2H

Solution (4) CHO

CO2H

Solution

H

OH

(4) In order for a sugar to be reduced by NaBH4, which is capable of reducing a carbonyl group to an alcohol, the sugar must contain at least one anomeric carbon that is in equilibrium with the open-chain form. For this to be the case, the sugar must have a hemiacetal. Maltose and lactose both contain a monosaccharide that is a hemiacetal, but sucrose does not. Both of the anomeric carbons in sucrose are glycosides (acetals).

H

OH

Chapter 31_Biomolecules.indd 807

CH2OH

CH2OH

(1) Glucose (2) Glycogen (3) L-lactic acid (4) Pyruvic acid

(1) Sucrose (2) Lactose (3) Maltose (4) Both (2) and (3)

HNO3

CH2OH

2. Accumulation of which of the following molecules in the muscles occurs as a result of vigorous exercise?

3. Which disaccharides are reduced by NaBH4?

OH OH H OH

HO H

H OH CH2OH

HNO3

H

OH

H

OH

HO H

H OH CO2H

5. The disaccharide _______ is found in table sugar. _______ is the sugar found in milk. _______ is the polysaccharide

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808

OBJECTIVE CHEMISTRY FOR NEET consisting of highly branched polymers of glucose that human and other animals use to store sugars in the body. (1) fructose, maltose, glycosaminoglycans (2) sucrose, lactose, glycogen (3) sucrose, maltose, starch (4) fructose, lactose, cellulose

(2) Enzymes are specific biological catalysts that possess the well-defined active sites. (3) Enzymes are specific biological catalysts that cannot be poisoned. (4) Enzymes are normally heterogeneous catalysts that are very specific in their action.

Solution

Solution

(2) Sucrose, the most widely occurring disaccharide, is found in all photosynthetic plants and is obtained commercially from sugarcane or sugar beets. Lactose is a disaccharide present in the milk of humans, cows, and almost all other mammals. The chains of glycogen are much more highly branched and animals store energy as fats (triacylglycerols) as well as glycogen.

(2) Enzymes are specific biological catalysts that possess well-defined active sites.

6. Which one of the following statements is correct?

(1) Vitamin B6 is known as pyridoxine.

(1) All amino acids except lysine are optically active. (2) All amino acids are optically active. (3) All amino acids except glycine are optically active. (4) All amino acids except glutamic acid are optically active.

10. Vitamin B6 is known as (1) pyridoxine. (2) thiamine. (3) tocopherol. (4) riboflavin. Solution

11. The following structure is a _______. It contains the following amino acids, _______, _______ and _______. OH O

H

Solution (3) In amino acids if all the groups attached to a carbon are different then it is optically active. In glycine, two identical groups are attached (H’s) to the carbon atom, so there is no chiral carbon and hence the compound is optically inactive. H2N

CH2

H N

H2N

N H

H

O

O

H

H N

OH

H

O

O

COOH

Glycine

7. The secondary structure of protein refers to (1) (2) (3) (4)

fixed configuration of the polypeptide backbone. a-helical backbone. hydrophobic interactions. sequence of a-amino acids.

Solution (2) Secondary structure of proteins involves a-helical backbonding and b-sheet structures. These are formed as a result of hydrogen bonding between different peptide groups. 8. An amino acid that does not form an a-helix is (1) asparagine. (2) tyrosine. (3) tryptophan. (4) proline. Solution (4) If proline is used in a-helix, then the helix formed will be slightly bent as there will be no hydrogen bonds, and overall the helix will become unstable. 9. Identify the correct statements regarding enzymes. (1) Enzymes are specific biological catalysts that can normally function at very high temperature (T ~ 1000 K).

Chapter 31_Biomolecules.indd 808

NH2

(1) (2) (3) (4)

tripeptide, alanine, phenylalanine, arginine tripeptide, alanine, phenylalanine, glutamine tetrapeptide, alanine, threonine, glutamic acid tetrapeptide, alanine, serine, glutamine

Solution (4) The following structure is tetrapeptide. It contains the alanine, serine and glutamine amino acids. 12. Which of the following amino acids contains two carboxylic acid groups? (1) Cystine (2) Cysteine (3) Glutamic acid (4) Both (1) and (3) Solution (4) O

O HO

S

S

NH2 Cystine

NH2

OH NH2

HO

OH O

O

Glutamic acid

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809

Biomolecules 13. _______ contain heterocyclic bases attached to a  _______  that has a  _______  attached to one of the OH’s. (1) nucleosides, 5 carbon monosaccharide, phosphate ion (2) nucleotides, 5 carbon monosaccharide, phosphate ion (3) nucleosides, 6 carbon monosaccharide, phosphate ion (4) nucleotides, 6 carbon monosaccharide, phosphate ion Solution (2) O P OH

O

5’

N O

b -N-Glycosidic linkage

1’

4’

Solution (4) Removal of the phosphate group of a nucleotide converts it to a compound known as a nucleoside. 15. The presence or absence of hydroxyl group on which carbon atom of sugar differentiates RNA and DNA? (1) 2nd (2) 3rd (3) 4th (4) 1st Solution

Heterocyclic base HO

(2) nucleic acids, ribose ring (3) DNA, nucleic acid (4) nucleosides, phosphate

3’

2’ OH OH A nucleotide

(1) In RNA, the sugar is b-D-(−) ribose and in DNA the sugar is b-2-deoxy-D-(−) ribose. So presence or absence of hydroxyl group on C2 of sugar moiety differentiates RNA and DNA. 5 HOH2C

OH

4 H

5 HOH2C

1 H 3

14. _______  are formed by removing a  _______  from a nucleotide. (1) purines, deoxyribose ring

O

OH

H

H

O

1

4 H

H

2

3 H

b -2-Deoxy-D-(−)-ribose

OH

H

H

2

OH

OH

b -D-(−)-Ribose

Practice Exercises Level I Carbohydrates 1. General formula for carbohydrates is (1) CnH2nO2n + 2 (2) Cx(H2O)2x (3) Cx(H2O)y (4) None of these 2. The commonest disaccharide has the molecular formula (1) C10H18O9 (2) C10H20O10 (3) C18H22O11 (4) C12H22O11 3. The term anomers of glucose refers to (1) isomers of glucose that differ in configurations at carbons one and four (C-1 and C-4). (2) a mixture of D-glucose and (L)-glucose. (3) enantiomers of glucose. (4) isomers of glucose that differ in configuration at carbon one (C-1). 4. Glucose contains in addition to aldehyde group (1) (2) (3) (4)

one secondary OH and four primary OH groups. one primary OH and four secondary OH groups. two primary OH and three secondary OH groups. three primary OH and two secondary OH groups.

Chapter 31_Biomolecules.indd 809

5. Which of the following statements about ribose is incorrect? (1) (2) (3) (4)

It is a polyhydroxy compound. It is an aldehyde sugar. It has six carbon atoms. It exhibits optical activity.

6. Carbohydrates are stored in human body as (1) glucose. (2) glycogen. (3) starch. (4) fructose. 7. An example of a disaccharide made up of two units of the same monosaccharides is (1) sucrose. (2) maltose. (3) lactose. (4) none of these. 8. a-D-glucose and b-D-glucose differ from each other due to difference in one of the carbons with respect to its (1) size of hemiacetal ring. (2) number of OH groups. (3) configuration. (4) conformation. 9. Which one of the following sets of monosaccharides forms sucrose?

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810

OBJECTIVE CHEMISTRY FOR NEET (1) (2) (3) (4)

a-D-Galactopyranose and a-D-Glucopyranose. a-D-Glucopyranose and b-D-Fructofuranose. b-D-Glucopyranose and a-D-Fructofuranose. a-D-Glucopyranose and b-D-fructopyranose.

10. Sucrose is (1) laevorotatory. (2) dextrorotatory. (3) racemic mixture. (4) optically inactive. 11. The hydrolysis of sucrose produces a mixture which is (1) laevorotatory. (2) dextrorotatory. (3) equally both (+) and (–) rotatory. (4) optically inactive. 12. Pick out the incorrect statement(s) from the following. (I) Glucose exists in two different crystalline forms, a-D-glucose and b-D-glucose. (II) a-D-glucose and b-D-glucose are anomers. (III) a-D-glucose and b-D-glucose are enantiomers. (IV) Cellulose is a straight chain polysaccharide made of only b-D-glucose units. (V) Starch is a mixture of amylose and amylopectin, both contain unbranched chain of a-D-glucose units. (1) I and II (2) II and III (3) III and IV (4) III and V 13. Glucose and fructose form (1) (2) (3) (4)

the same osazone. the same acid on oxidation. the same alcohol when reduced. different osazone.

14. On heating with conc. H2SO4, sucrose gives (1) CO and CO2 (2) CO and SO2 (3) CO, CO2 and SO2 (4) None of these 15. Glucose gives silver mirror with tollen’s reagent. It shows the presence of (1) an acidic group. (2) an alcoholic group. (3) a ketonic group. (4) an aldehydic group. 16. Hydrolysis of sucrose is called (1) esterification. (2) saponification. (3) inversion. (4) hydration. 17. Glucose when heated with CH3OH in presence of dry HCl gas gives a and b-methyl glucosides because it contains (1) an aldehyde group. (2) a  CH2OH group. (3) a ring structure. (4) five hydroxyl groups. 18. Which one of the following compounds is found abundantly in nature? (1) Fructose (2) Starch (3) Glucose (4) Cellulose

Chapter 31_Biomolecules.indd 810

19. Which of the following carbohydrates is an essential constituent of plant cells? (1) Cellulose (2) Sucrose (3) Vitamins (4) Starch 20. The linkage between the two monosaccharide units in lactose in (1) C1 of b-D-glucose and C4 of b-D-galactose. (2) C1 of b-D-galactose and C4 of b-D-glucose. (3) C1 of a-D-galactose and C4 of b-D-glucose. (4) C1 of b-D-galactose and C4 of b-D-glucose.

Proteins 21. The peptide linkage in amino acids is (1)    CO   NH    (2)    C   NH2 (3) SO   NH    (4)    CO   N    22. The functional group which is found in amino acid is (1)    COOH group. (2)  NH2 group. (3)    CH3 group. (4) Both (1) and (2). 23. Irreversible precipitation of proteins is called (1) denaturation. (2) hydrolysis. (3) rearrangement. (4) electrophoresis. 24. Amino acids are (1) (2) (3) (4)

liquids. volatile solids. non-volatile crystalline compounds. mixture of amines and acids.

25. Amino acids usually exist in the form of zwitter ions. This means that it consists of (1) The basic group  NH2 and the acidic group  COOH. (2) The basic group - NH +2 and the acidic group  CO2. (3) The basic group - CO -2 and the acidic group - NH +2. (4) No acidic or basic group 26. Which of the following is not a function of proteins? (1) (2) (3) (4)

Nails formation Skin formation Muscle formation Providing energy for metabolism

27. Among following, the achiral amino acid is (1) 2-ethylalanine. (2) 2-methylglycine. (3) 2-hydroxymethyl serine. (4) tryptophan. 28. Which of the following sets consists only of essential amino acids? (1) Alanine, tyrosine, cysteine (2) Leucine, lysine, tryptophan (3) Alanine, glutamine, lysine (4) Leucine, proline, glycine

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Biomolecules 29. Enzymes in the living systems (1) (2) (3) (4)

provide energy. provide immunity. transport oxygen. catalyze biological processes.

30. Enzymes belong to which of the following classes of compounds? (1) Polysaccharides. (2) Polypeptides. (3) Polynitrogen heterocyclic compounds. (4) Hydrocarbons. 31. Albumin proteins are most abundant in (1) meat. (2) milk. (3) egg. (4) soyabean. 32. The protein which maintains blood sugar level in the human body is (1) haemoglobin. (2) oxytocin. (3) insulin. (4) ptyalin. 33. Which of the following proteins acts as a messenger in the living system? (1) Hormone (2) Enzyme (3) Protective protein (4) Transport protein 34. Insulin regulates the metabolism of (1) minerals. (2) amino acids. (3) glucose. (4) vitamins.

39. Nucleic acid is a polymer of (1) nucleosides. (2) a-amino acids. (3) nucleotides. (4) glucose.

Vitamins 40. Ascorbic acid is a (1) vitamin. (2) enzyme. (3) protein (4) carbohydrate. 41. Deficiency of which vitamin causes rickets (1) vitamin D. (2) vitamin B. (3) vitamin A. (4) vitamin K. 42. Which of the following vitamins has isoprene units in its structure? (1) vitamin A. (2) vitamin C. (3) vitamin B2. (4) vitamin D. 43. Deficiency of vitamin H causes (1) skin diseases. (2) scurvy. (3) burning of eyes. (4) anaemia. 44. Which one of the following vitamins is water soluble? (1) Vitamin B (2) Vitamin E (3) Vitamin K (4) Vitamin A 45. The human body does not produce (1) hormones. (2) enzymes. (3) DNA. (4) vitamins.

Nucleic Acids

Level II

35. Which of the following is responsible for heredity character?

Carbohydrates

(1) DNA (2) RNA (3) Proteins (4) Hormones 36. The base adenine occurs in (1) DNA only. (2) RNA only. (3) DNA and RNA both. (4) protein. 37. Which of the following statements about the assembly of nucleotides in a molecule of deoxyribose nucleic acid (DNA) is correct? (1) A pentose of one unit connects to a pentose of another. (2) A pentose of one unit connects to the base of another. (3) A phosphate of one unit connects to a pentose of another. (4) A phosphate of one unit connects to the base of another. 38. The function of DNA in an organism is (1) to assist in the synthesis of RNA molecule. (2) to store information of heredity characteristics. (3) to assist in the synthesis of proteins and polypeptides. (4) all of these.

Chapter 31_Biomolecules.indd 811

811

1. The letter D in D-glucose signifies (1) configuration at all chiral carbons. (2) dextrorotatory. (3) that is a monosaccharide. (4) configuration at a particular chiral carbon 2. Which of the following sugars is least sweet? (1) Glucose (2) Fructose (3) Maltose (4) Sucrose 3. Which among the following has the simplest structure? (1) Glucose (2) Cellulose (3) Starch (4) None of these 4. Indigestible carbohydrate, which is also a constituent of our diet, is (1) cellulose. (2) galactose. (3) maltose. (4) starch. 5. The disaccharide present in milk is (1) maltose. (2) lactose. (3) sucrose. (4) cellobiose.

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OBJECTIVE CHEMISTRY FOR NEET

6. Which of the following does not show any reducing test of aldehyde? (1) Sucrose (2) Fructose (3) Maltose (4) Lactose

(1) Glucose (2) Fructose (3) Maltose (4) Sucrose

(1) mouth. (2) stomach. (3) liver. (4) intestine.

17. The number of chiral carbons in b-D(+)-glucose is

8. Sucrose is a

(1) five. (2) six. (3) three. (4) four.

reducing sugar. non-reducing sugar. mixture of glucose. monosaccharide and fructose.

Proteins 18. Which element is absent in protein?

9. In Haworth projection, the a-anomer of glucose contains the OH group (1) (2) (3) (4)

above the plane of ring. below the plane of ring. in the plane of ring pointing outwardly. inside the plane of ring pointing inwardly.

(1) (2) (3) (4)

20% of amylose and 80% of amylopectin. 30% of amylose and 80% of amylopectin. 80% of amylose and 20% of amylopectin. 70% of amylose and 30% of amylopectin.

(1) dipeptide bonds. (2) hydrogen bonds. (3) ether bonds. (4) peptide bonds. 21. Which of the following hormones contains iodine? (1) Insulin (2) Thyroxine (3) Adrenaline (4) Testosterone

(1) Amylopectin (2) Glycogen (3) Starch (4) Amylose 12. Methyl a-D-glucoside and methyl b-D-glucoside are

23. Antibodies are (1) carbohydrates. (2) proteins. (3) lipids. (4) enzymes.

13. Milk changes after digestion into (1) cellulose. (2) fructose. (3) glucose. (4) lactose.

24. Starting with three different amino acid molecules, how many different tripeptide molecules are formed?

14. The correct statement about the following disaccharide is CH2OH

OH

OH

(a)

H

(1) (2) (3) (4)

H H OH

OCH2

O

CH2OH CH2O

H OH

(b)

H OH H

Ring (a) is pyranose with a-glycosidic link. Ring (a) is furanose with a-glycosidic link. Ring (b) is furanose with a-glycosidic link. Ring (b) is pyranose with b-glycosidic link.

Chapter 31_Biomolecules.indd 812

22. An amino acid contains an amino group attached to (1) a-carbon atom. (2) b-carbon atom. (3) g-carbon atom. (4) d-carbon atom.

(1) epimers. (2) anomers. (3) enantiomers. (4) conformational diastereomers.

H

exclusively D-form of amino acids. exclusively L-form of amino acids. 50% of each of D-form and L-form of amino acids. 33% of D-form and 67% of L-form of amino acids.

20. The helical structure of protein is stabilized by

11. Which of the following is a linear polymer?

H

(1) C (2) N (3) S (4) P 19. Proteins contain

10. Starch contains (1) (2) (3) (4)

(1) a reducing sugar. (2) a non-reducing sugar. (3) an aldose. (4) a furanose. 16. Which of the following sugars is most sweet?

7. Galactose is converted into glucose in the

(1) (2) (3) (4)

15. In alkaline medium, fructose is

CH2OH

(1) 12 (2) 9 (3) 8 (4) 6 25. The sequence in which the a-amino acids are linked to one another in a protein molecule is called its (1) primary structure. (2) secondary structure. (3) tertiary structure. (4) quaternary structure. 26. The hormone which controls the processes of burning of fats, proteins and carbohydrates and liberates energy in the body is (1) thyroxine. (2) adrenaline. (3) insulin. (4) cortisone.

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Biomolecules 27. Which of the following structures represents the peptide chain? H (1)

N

O C

N

O

H

C

NH

H (2)

N

C

C

C

C

C

N

(1) (2) (3) (4)

O C

C

(3)

H

O

N

N

C

C

C

C

(1) (2) (3) (4)

O H (4)

N

O C

C

C

H N

C

C

N

H

C

C

O

(1) H3N+CHRCOOH (2) H3N+CHRCOO-  (3) H2NCHRCOO-  (4) H2NCHRCOOH 29. Which statement is incorrect about peptide bond? (1) C   N bond length in proteins is longer than usual bond length of C  N bond. (2) Spectroscopic analysis shows planar structure of CO  NH  group. (3) C  N bond length in proteins is smaller than usual bond length of C  N bond. (4) None of these. 30. Lysine is least soluble in water in the pH range (1) 3 to 4 (2) 5 to 6 (3) 6 to 7 (4) 8 to 9 31. Rice is deficient in (1) lysine. (2) alanine. (3) glycine. (4) isoleucine. 32. All of the following are sulphur-containing amino acids are found in proteins except (1) cysteine. (2) threonine. (3) cystine. (4) methionine.

Nucleic Acids 33. The sugar present in DNA is (1) glucose. (2) deoxyribose. (3) ribose. (4) fructose. 34. The pyrimidine bases present in DNA are

Chapter 31_Biomolecules.indd 813

It has a single strand. It does not undergo replication. It does not contain any pyrimidine base. It controls the synthesis of proteins.

37. The nucleic acid base having two possible binding sites is C

28. At low pH, an amino acid exists as

(1) cytosine and adenine. (2) cytosine and guanine.

It has a double helix structure. It undergoes replication. The two stands in DNA molecule are exactly similar. It contains the pentose sugar, 2-dexoyribose.

36. Which of the following statements about RNA is not correct?

O H

(3) cytosine and thymine. (4) cytosine and uracil. 35. Which of the following statement about DNA is not correct?

NH

H

813

(1) thymine. (2) cytosine. (3) guanine. (4) adenine. 38. Consider the double helix structure for DNA. The base pairs are (1) (2) (3) (4)

part of the backbone structure. inside the helix. outside the helix. None of these.

39. Which of the following sets of bases is present both in DNA and in RNA? (1) (2) (3) (4)

Adenine, uracil, thymine Adenine, guanine, cytosine Adenine, guanine, uracil Adenine, guanine, thymine

40. In both DNA and RNA, heterocyclic base and phosphate ester linkages are at (1) C5′ and C2′, respectively, of the sugar molecule. (2) C2′ and C5′, respectively, of the sugar molecule. (3) C1′ and C5′, respectively, of the sugar molecule. (4) C5′ and C1′ respectively of the sugar molecule. 41. Which of the following is not a pyrimidine? (1) Uracil (2) Guanine (3) Cytosine (4) Cytosine

Vitamins 42. The vitamins absorbed from intestine along with fats are (1) A, D. (2) A, B. (3) A, C. (4) D, B. 43. The deficiency of vitamin C causes (1) scurvy. (2) rickets. (3) pyorrhea. (4) pernicious anaemia.

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OBJECTIVE CHEMISTRY FOR NEET

44. The best source of vitamin A is

6. Fructose reduces Tollen’s reagent due to

(1) oranges. (2) beans. (3) carrots. (4) wheat. 45. The disease beri beri is caused by deficiency of vitamin (1) A (2) B1 (3) C (4) D

7. Which one of the following statements is incorrect about enzyme catalysis?

Previous Years’ NEET Questions 1. RNA and DNA are chiral molecules, their chirality is due to (1) (2) (3) (4)

D-sugar component. L-sugar component. chiral bases. chiral phosphate ester units.

(1) Enzymes are mostly proteinous in nature. (2) Enzyme action is specific. (3) Enzymes are denatured by ultraviolet rays at high temperature. (4) Enzymes are least reactive at optimum temperature. (AIPMT PRE 2012)

(AIPMT 2007) 2. In DNA, the complementary bases are (1) (2) (3) (4)

(1) primary alcoholic group. (2) secondary alcoholic group. (3) enolization of fructose followed by conversion to aldehyde by base. (4) asymmetric carbons. (AIPMT MAINS 2010)

uracil and adenine; cytosine and guanine. adenine and thymine; guanine and cytosine. adenine and thymine; guanine and uracil. adenine and guanine; thymine and cytosine.

8. D-(+)-glucose reacts with hydroxyl amine and yield an oxime. The structure of the oxime would be CH

(1)

(AIPMT 2008) 3. Which one of the following is an amine hormone?

OH

HO

C

H

HO

C

H

HO

C

H

HO

C

H

H

C

OH

H

C

OH

H

C

OH

CH2OH

(AIPMT 2009) 5. Which of the following statements about denaturation are correct? Statements (I) Denaturation of proteins causes loss of secondary and tertiary structures of the protein. (II) Denaturation leads to the conversion of double strand of DNA into single strand. (III) Denaturation affects primary structure which gets distorted. Options: (1) I, II and III (2) II and III (3) I and III (4) I and II (AIPMT MAINS 2011)

Chapter 31_Biomolecules.indd 814

CH

(3)

NOH

C

(AIPMT 2008)

(1) nucleoside. (2) nucleotide. (3) ribose. (4) gene.

CH

(2)

H

(1) Progesterone (2) Thyroxine (3) Oxypurin (4) Insulin

4. The segment of DNA which acts as the instrumental manual for the synthesis of the protein

NOH

CH2OH NOH

HO

C

H

H

C

OH

HO

C

H

C

CH

(4)

NOH

H

C

OH

HO

C

H

H

H

C

OH

OH

H

C

OH

CH2OH

CH2OH

(AIPMT 2014) 9. Which of the following hormones is produced under the condition of stress which stimulates glycogenolysis in the liver of human being? (1) Thyroxine (2) Insulin (3) Adrenaline (4) Estradiol (AIPMT 2014) 10. In a protein molecule various amino acids are linked together by (1) a-glycosidic bond. (2) b-glycosidic bond. (3) peptide bond. (4) dative bond. (NEET I 2016)

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Biomolecules 11. The correct statement regarding RNA and DNA, respectively is (1) the sugar component in RNA is arabinose and the sugar component in DNA is 2′-deoxyribose. (2) the sugar component in RNA is ribose and the sugar component in DNA is 2′-deoxyribose. (3) the sugar component in RNA is arabinose and the sugar component in DNA is ribose. (4) the sugar component in RNA is 2′-deoxyribose and the sugar component in DNA is arabinose. (NEET I 2016) 12. The central dogma of molecular genetics states that the genetic information flows from

815

(1) DNA → Carbohydrates → Proteins (2) DNA → RNA → Proteins (3) DNA → RNA → Carbohydrates (4) Amino acids → Proteins → DNA (NEET II 2016) 13. Which of the following statements is not correct? (1) Ovalbumin is a simple food reserve in egg-white. (2) Blood proteins thrombin and fibrinogen are involved in blood clotting. (3) Denaturation makes the proteins more active. (4) Insulin maintains sugar level in the blood of a human body. (NEET 2017)

Answer Key Level I  1. (3)

 2. (4)

 3. (4)

 4. (2)

 5. (3)

 6. (2)

 7. (2)

 8. (3)

 9. (2)

10. (2)

11. (1)

12. (4)

13. (1)

14. (4)

15. (4)

16. (3)

17. (3)

18. (4)

19. (1)

20. (2)

21. (1)

22. (4)

23. (1)

24. (3)

25. (3)

26. (4)

27. (3)

28. (2)

29. (4)

30. (2)

31. (3)

32. (3)

33. (1)

34. (3)

35. (1)

36. (3)

37. (3)

38. (4)

39. (3)

40. (1)

41. (1)

42. (1)

43. (1)

44. (1)

45. (4)

 1. (4)

 2. (3)

 3. (1)

 4. (1)

 5. (2)

 6. (1)

 7. (3)

 8. (2)

 9. (2)

10. (1)

11. (4)

12. (2)

13. (3)

14. (1)

15. (1)

16. (2)

17. (1)

18. (4)

19. (2)

20. (2)

Level II

21. (2)

22. (1)

23. (2)

24. (4)

25. (1)

26. (3)

27. (3)

28. (1)

29. (1)

30. (4)

31. (1)

32. (2)

33. (2)

34. (3)

35. (3)

36. (3)

37. (3)

38. (2)

39. (2)

40. (3)

41. (2)

42. (1)

43. (1)

44. (3)

45. (2)

 6. (3)

 7. (4)

 8. (2)

 9. (3)

10. (3)

Previous Years’ NEET Questions  1. (1)

 2. (2)

 3. (2)

11. (2)

12. (2)

13. (3)

 4. (4)

 5. (4)

Hints and Explanations Level I 1. (3) Carbohydrates are hydrates of carbon. Their general formula is Cx(H2O)y.

Chapter 31_Biomolecules.indd 815

3. (4) Anomers are such distereomeric molecules which differ in the configuration of C-1 carbon only.

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OBJECTIVE CHEMISTRY FOR NEET

4. (2)

39. (3) Nucleic acid is a polymer of nucleotides. CHO

44. (1) Vitamin B is water soluble vitamin.

CHOH CHOH

Level II

CHOH CHOH CH2OH Primary OH group

1. (4) Glyceraldehyde has been chosen as the standard configuration in sugar chemistry, D- or L signifies configuration at a particular chiral carbon.

¾¾¾ ® glu cos e + glu cos e. 7. (2) Maltose ¾Hydrolysis Maltase

2. (3) Maltose is the least sweet sugar.

9. (2) Sucrose is a disaccharide of a-D-glucopyranose and b-D-fructopyranose.

3. (1) Glucose is the simplest carbohydrate, that is, monosaccharide rest is polysaccharide.

12. (4) Starch obtained from the same source consists of two fractions, that is, amylose (20%) and amylopection (80%). Amylose is a linear polymer while amylopectin is a highly branched polymer. Both are composed of a-Dglucose units linked by glycosidic linkages. a-D-glucose and b-D-glucose are anomers to each other not the enantiomers.

5. (2) Lactose is present in milk (Glucose + Galactose). 6. (1) In sucrose the two monosaccharide units joined by a-1,2-glycoside bond. Since sucrose does not have hemiacetal carbon, it is non-reducing sugar. 8. (2) Sucrose is a non-reducing sugar as it does not reduce Tollens’ reagent and Fehling’s solution.

15. (1) Glucose + Tollen’s reagent → Gluconic acid + Ag mirror

9. (2) For the D-sugars, the anomeric hydroxyl group is below the ring in the a-anomer and above the ring in the 1. (1) Two or more amino acids unite through a bond   2 b-anomer. CO  NH  . HOCH2

23. (1) Protein ¾Heated ¾¾® Denatured protein

H

or change in pH

24. (3) Amino acids are non-volatile crystalline compounds. HO

25. (3) Zwitter ion is a dipolar ion containing both a positive and negative charge in the following form NH +3  R  CH  COO -



HOCH2

O H OH

H

H

OH

H

O

H HO

OH

Haworth formulas

H OH

H

H

OH

a-anomer

OH H

b-anomer

10. (1) Starch is a mixture of two components: water soluble amylose (20%) and water insoluble amylopectin (80%).

Zwitter ion

26. (4) Protein is a body building substance not energy giving substance.

11. (4) Amylose is a linear polymer made of D-glucose units. 12. (2) Anomers differ at C1 carbon atom.

28. (2) Essential amino acids are those amino acids which are supplied to our bodies by food because they cannot be synthesized in the body. Essential amino acids are listed as valine, leucine, isoleucine, phenyl-alanine, tryptophan, threonine, methionine, lysine, arginine and histidine. 34. (3) Insulin is essential for metabolism of carbohydrates in the body. It facilitates the entry of glucose into the cells by increasing the penetration of cell membranes and phosphorylation of glucose. This decreases glucose concentration in blood. 36. (3) Adenine is a purine base common in both RNA and DNA.

Chapter 31_Biomolecules.indd 816

H H

1

C

2

C

3

HO

C

H

C

H

4

OH H

O

OH

5

C

6



OCH3

CH2OH Methyl-a-D-glucoside

H3CO

1

C

2

H

C

HO

C

H

C

H

3 4

H OH H

O

OH

5

C

6

CH2OH Methyl-b-D-glucoside

13. (3) Milk contains lactose which is a disaccharide. Disaccharide on hydrolysis yields two monosaccharides. 2O Lactose HH + → Glucos e + Galactose

( Milk )

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Biomolecules 14. (1) Ring (a) is pyranose with a-glycosidic linkage and ring (b) is furanose with b-glycosidic linkage. 15. (1) Fructose is converted into glucose and mannose under alkaline conditions and gives positive test with Fehling’s solution. 16. (2) Fructose is the sweetest naturally occurring sugar.

1

H

C

HO

C

H

C

H

C

∗

∗

R

−H2O

O R CH2 C NH CH COOH

R'

NH2 R' Peptide linkage

28. (1) At low pH, the amino and carboxyl groups will be protonated and the molecules will be in the acid form. As the pH increases, the molecule will become neutral with both basic and acidic groups, hence is called as zwitterion. However, when the pH increases further, molecule becomes basic.

OH H O OH

∗

+

H3N

C H2OH b -D-Glucose



O CH2 C OH + H HN CH COOH



b C∗ H ∗

27. (3) Peptide linkage is formed as follows:

NH2

17. (1) There are 5 chiral carbons as shown: HO

26. (3) Insulin secreted by pancreas, controls the process of burning of fats, proteins and carbohydrates and liberates energy in the body.



18. (4) Phosphorus is not present in any amino acid. C and N are found in every amino acid, and there are certain sulphur-containing amino acids as well, such as cysteine, cystine, etc. 19. (2) Proteins are condensation polymers which contains L-amino acids as monomeric units.

CH

COOH

OH− H+

+

CH

H3N

COO−

OH− H+

H 2N

R Zwitterion

R (I)

CH

COO−

R (II)

29. (1) In peptide linkage, the C   N bond is shorter than usual C   N bond due to resonance. O C



OH NH

C

N

20. (2) Helical structure of protein is stabilized by hydrogen bonds between amino group and carboxylic group.

30. (4) Any amino acid has its lowest solubility on its isoelectronic point and the isoelectronic point is the pH at which the amino acid carries no charge.

21. (2) Thyroxine contains four iodine atoms. It is produced by attaching iodine atoms to the ring structure of tyrosine molecules.

31. (1) Rice is deficient in lysine and threonine which are essential amino acids required for the growth and maintenance of health.

a- carbon

22. (1)

32. (2) The structures of sulphur-containing cysteine, cystine and methionine are as follows:

O

H2N CH C OH carboxyl group a- amino group R Side chain

NH2

O OH

HS

23. (2) Antibodies are proteins found in animal cells.

NH2

24. (4) Six types of tripeptide molecules are formed.

S

OH

O NH2 Cystine O

MeS O

OH

C

O

S

Cysteine

25. (1) The primary structure of protein is the linear sequence of amino acid. Carboxyl end

O

HO

NH2 Methionine

Amino end

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Specific sequence of amino acids



Threonine does not contain sulphur

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818

OBJECTIVE CHEMISTRY FOR NEET 43. (1) The deficiency of vitamin C causes scurvy. It causes general weakness, anaemia, gum and skin problems.

O OH

HO

44. (3) b-carotene present in carrots is metabolized into vitamin A.

NH2 Threonine



33. (2) Deoxyribose is the monosaccharide used to make DNA. The deoxyribose sugar molecule is shown as follows: 5 HOH2C

O

OH

4 H

1 H 3

H 2

H

OH H b −2−Deoxy-D(−)ribose



34. (3) Pyrimidine bases present in DNA are cytosine and thymine, whereas purine bases are adenine and guanine.

45. (2) Beri beri is caused due to deficiency of thiamine, and it causes general weakness and paralysis of legs.

Previous Years’ NEET Questions 1. (1) Nucleic acids (DNA and RNA) are long-chain polymers of nucleotides and are also known as polynucleotides. Mild degradations of nucleic acids yield monomeric units called nucleotides. It consists of a heterocyclic base, a five-carbon monosaccharide and a phosphate ion. The sugar moiety of RNA is b-Dribose and that of DNA is b-D-2-deoxyribose that arise the chirality. 5 HOH2C

35. (3) The two strands of DNA are never similar instead they are complementary to each other. This complimentary structure of both the strands keeps them affixed in a double helical shape.

O 4

H

H

3 OH

5 HOH2C

OH H 2

O 4

1 H

H

H

H

3 OH

OH

b −D−Ribose

OH 1

2

H

H

b −D−2−Deoxyribose

36. (3) The pyrimidine bases in RNA are cytosine and uracil.



38. (2) In DNA double helix structure, sugar–phosphate backbone is on the outside of the helix whereas the base pairs (adenine, thymine, cytosine, guanine) are present inside the helix, stacking perpendicular to the helical axis.

2. (2) Complementary bases in DNA are adenine and thymine and guanine and cytosine.

39. (2) Uracil is present in RNA in place of thymine.

3. (2) Thyroxine is 3,5,3′,5′–tetraidothyronine, which is basically an amine hormone. I

40. (3) In both DNA and RNA heterocyclic base and phosphate ester linkage are at C1¢ and C¢5, respectively, of the sugar molecule. O O−

O

5 H2C

−

O 4

H

Base

‚

1 H

H ‚

3



O I

‚

P O

HO

OH

‚

H

‚

2

OH



I

I

H

O OH NH2 Amine group

4. (4) DNA carries all of the information for physical characteristics and is the genetic material in most of the organisms. The information is determined by proteins. Each protein is encoded by a gene. The order of nucleotides within a gene specifies the order and types of amino acids that make a protein.

41. (2) In DNA and RNA nucleobases are subdivided into purines and pyrimidines. Adenine (1) and guanine (G) are purines, while cytosine (3), thymine (T) and uracil (U) are pyrimidines.

5. (4) During denaturation, primary structure remains intact while secondary and tertiary structures of proteins are destroyed.

42. (1) Intestine absorbs vitamins A and D along with fats and these vitamins as well as vitamins E and K are called fat soluble vitamins. Vitamins B and C are water soluble vitamins.

6. (3) In alkaline solution ketoses (fructose) are converted into aldoses, which are then oxidized by the Tollen’s reagent.

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Biomolecules

response to these hormones, liver cells perform glycogenolysis (breakdown of glycogen to glucose), and adipose tissue cells perform lipolysis (breakdown of triglycerides to fatty acids and glycerol).

O H

CH2OH O

C

Fructose

C

OH

C

OH

H

C

OH

C

OH

Aldose

Enol tautomer

Ag+ + OH(Tollen´s reagent)

O H

C

O-

H

C

OH



H2NCH − COOH + H2N − CH − COOH −H2O

R

H

C

OH

HO

C

H

H

C

OH

H

C

OH

CH2OH D-glucose

+ NH2OH

-H2O

CH

NOH

H

C

OH

HO

C

H

H

C

OH

H

C

OH

CH2OH Oxime

9. (3) In stressful situations and during exercise, impulses from the hypothalamus stimulate sympathetic preganglionic neurons, which in turn stimulate the chromaffin cells to secrete epinephrine and norepinephrine, also called adrenaline and noradrenaline, respectively. In

Chapter 31_Biomolecules.indd 819

R

H2N − CH − C − NH − CH − COOH R

8. (2) Glucose reacts with hydroxylamine to form an oxime.



10. (3) In biopolymer proteins, monomers amino acids are joined to each other by peptide bonds. This bond is formed when the carboxyl group of one amino acid reacts with the amino group of the other, and releases a molecule of water (H2O).

+ Ag↓ (Silver mirror)

7. (4) Enzymes are most active at optimum temperature. At this temperature, the rate of an enzyme reaction becomes maximum.

CHO

819

O

R

Peptide bond



11. (2) The sugar component of RNA is b-D-ribose and that of DNA is b-D-2-deoxyribose. HOCH2 H



OH

O

H

H

OH

OH

H

b -D-ribose Sugar component of RNA

HOCH2 H

OH

O

H

H

OH

H

H

b -D-2-deoxyribose Sugar component of DNA

12. (2) DNA ¾Transcription ¾¾¾¾ ® RNA ¾Translation ¾¾¾® Proteins 13. (3) The three-dimensional shapes of proteins are dependent on the temperature. If the environment changes, a protein may lose its characteristic shape (secondary, tertiary and quaternary structure). This process is called denaturation. Denatured proteins are no longer functional or become inactive.

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32

Chemistry in Everyday Life

Chapter at a Glance 1. Drugs: Chemicals that have a physiological effect on a living system are called drugs. When these are used for therapeutic purposes, they are called medicines. 2. Classification of Drugs Drugs are classified on the basis of the following: Drugs Classification

Pharmacological effect: Based on use of specific drug for a particular type of problem. For example, analgesics have pain killing effect while antiseptics kill or inhibit the growth of microorganisms.

Drug action: Based on the action of drug on a particular biochemical process. For example, antihistamines are used to inhibit the action of histamine.

Chemical structure: Based on the chemical structure of the drug. For example, sulpha drugs

Molecular targets: This classification is used for medicinal chemists. For example, in this drugs interact with biomolecules such as carbohydrates, proteins, etc.

3. Drug–Target Interaction Some important biomolecules which fulfill vital functions of the body are called target molecules. For example, lipids and carbohydrates form structures of the cell membrane, nucleic acids code genetic information, etc. Proteins that function as biological catalysts are called enzymes; those which are involved in receiving communication signals are called receptors and those which transport polar molecules across cell membranes are called carriers. The target molecules and receptors are attacked by the drugs to either block the unrequired activity or perform certain changes to improve their biological functions and this is known as drug–target interaction. (a) Enzymes as drug targets: Enzymes catalyze the biochemical processes either by holding the substrate or providing certain chemicals or signal to improve its biological activity. (b) Receptors as drug targets: Receptors are the proteins that are very crucial to body communication system. Drug can interact with the receptors and either block or modify its activity to reduce disease. 4. Therapeutic Action of Drugs (a) Analgesics: These are the chemical substances used for relieving pain in the body. They can be of two types: (i) Non-narcotic: These relieve skeletal pain and bring down body temperature. For example, aspirin, paracetamol, acetaminophen and naproxen. (ii) Narcotic: These act on the central nervous system as a depressant and cause an overall state of euphoria. For example, morphine, heroin and codeine.

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OBJECTIVE CHEMISTRY FOR NEET

(b) A  ntipyretics: These are the chemicals used to bring down the body temperature during high fever. For example, paracetamol, aspirin, etc. They can also relieve pain. (c) Tranquilizers: These are the drugs given to patients suffering from anxiety and mental tension. For example, barbituric acid and its derivatives that are known as barbiturates. (d) Antidepressants: These inhibit the enzymes which catalyze the degradation of noradrenaline (neurotransmitter). For example, iproniazid and phenelzine. (e) Antifertility drugs: These drugs are used to control pregnancy. They are also called oral contraceptives. For example, norethindrone. (f ) Antimicrobials: These are the chemical substances which inhibit, destroy and prevent the development of microbes. These can be antibacterial, antiviral, antifungal or antiparasitic depending on the type of microorganism they target. For example, salvarsan. (g) Antibiotics: These are the chemical substances produced by chemical synthesis, in low concentration, which can inhibit the growth or even destroy other microorganisms. These are of two types: (i) Bactericidal: These are chemical substances which kill the disease-causing microorganisms in the body. For example, penicillin. (ii) Bacteriostatic: These are chemical substances which inhibit or arrest the growth of disease-causing microorganisms. For example, tetracycline. (iii)  They can also be classified on the basis of their spectrum of action as: · B  road spectrum antibiotics which can be used to treat a wide range of infections caused by both Gram positive and Gram negative bacteria. For example, ampicillin, amoxycillin, vancomycin, ofloxacin, chloramphenicol. · N  arrow spectrum antibiotics which are only effective against Gram positive or Gram negative bacteria. For example, Penicillin G. · L  imited spectrum antibiotics which are effective only against a single organism or disease. For example, Dysidazirine is toxic against cancer cells. (h) Antiseptics: They are the chemical substances which prevent the growth of microorganisms and may even kill them. They are safe to be applied on living tissue. For example, furacin, soframycin. (i) Disinfectants: They are chemical substances which kill microorganisms but are not safe to be applied on living tissue. For example, 1% solution of phenol. (j) Antacids: These are the drugs which can reduce or neutralize the acidity in stomach and raise the pH to some appropriate level. They are mildly basic in nature and can neutralize HCl in a proton-transfer reaction. For example, sodium bicarbonate deprotonates HCl to form carbonic acid that degrades into carbon dioxide and water very quickly. (k) Antihistamines: These are drugs which diminish the action of histamine released in the body. Histamine is involved in the regulation of gastric acid secretion in the stomach. For example, cimetidine (Tegamet) and ranitidine (Zantac) are used to control over production of gastric acid associated with acid reflux diseases and ulcers. 5. Chemicals in Food There are few chemicals used to improve the table value (food colors), nutritive value (minerals, vitamins, amino acids) and increase the shelf-life (antioxidants or preservatives) of food. (a) Preservatives   These are used to prevent the spoilage of food due to microbial growth. These are classified as: (i) Class I preservatives: These include simple preservatives such as vinegar, salt, sugar, honey, vegetable oil. (ii) Class II preservatives: These include chemicals such as sodium benzoate, EDTA and salts of sorbic acid and propanoic acid, sulphur dioxide (used in squashes and juices). Butylated hydroxyl anisole (BHA) and butylated hydroxyl toluene (BHT) are antioxidants used as preservatives to prevent the oxidation of food containing oils and fats. (b) Food colors are used to improve the table value. (c) Artificial sweetening agents are used to provide sweet taste to the food, for example, saccharin, sucralose, aspartame and alitame, etc. These are preferably used at normal room temperature conditions as at higher temperatures, they lose their activity.

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Chemistry in Everyday Life

823

6. Cleansing Agents Soaps

Detergents

Preparation

Soaps are prepared by alkaline hydrolysis (i.e., saponification) of triacylglycerols which yield glycerol and a mixture of salts of long-chain carboxylic acids (soaps).

Rather than being sodium salts of carboxylic acids, synthetic detergents are sodium salts of alkylbenzenesulphonates.

Cleansing action

When soap comes in contact with grease on a soiled surface, the hydrocarbon end of the soap dissolves in the grease, leaving the negatively charged carboxylate end exposed on the grease surface. Since the negatively charged carboxylate groups are strongly attracted by water, small droplets are formed, and the grease is literally lifted or floated away from the soiled object.

A detergent consists of two parts: a long hydrocarbon part which is hydrophobic and a short ionic part (containing    COO-Na+) group which is hydrophilic. When the detergent is added to dirty clothes, the greasy and oily parts stick to the hydrophobic part while the ionic part remains attached to the water. When the dirty clothes are agitated in detergent solution, the hydrophobic part gets washed away and the clothes get cleaned.

7. Types of Detergents (a) Cationic detergents They are quaternary ammonium salts with long hydrocarbon chain, a positive charge on the N atom and acetates, chlorides or bromides as anions. CH3(CH2)14CH2

+

N(CH3)3

Grease soluble, Water soluble, hydrophilic hydrophobic

Example, cetyltrimethylammonium bromide CH3 CH3(CH2)15

N

+

Anionic detergents

Non-ionic detergents

These are sodium salts of sulphonated long-chain alcohols or hydrocarbons, for example, sodium salts of alkylbenzenesulphonates. These detergents are used in toothpastes and shampoo formulations.

These do not contain an ionic group but have a grease-soluble component and a water-soluble component. They have a polar part to provide required water solubility:

Example, sodium lauryl sulphate

Example, cetyl palmitate.

CH3(CH2)10CH2SO−3Na+

Sodium lauryl hydrogen sulphate

CH3 Br −

CH3(CH2)10CH2

(CH2CH2O)7

O

CH2CH2OH

Water soluble, hydrophilic

Grease soluble, hydrophobic

O C15H31C OC16H33 Cetyl palmitate (hexadecyl hexadecanoate) (Non-ionic detergent)

CH3 Cetyltrimethyl ammonium bromide (Cationic detergent)

(b) Biodegradable detergents These are detergents that can be decomposed by microorganisms. They are usually straight-chain hydrocarbons. These detergents do not get accumulated in water bodies. Example, soaps and sodium lauryl sulphate

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Non-biodegradable detergents These detergents cannot be decomposed by microorganisms. They contain branched hydrocarbon chains. These detergents get accumulated in water supplies and cause severe pollution problems due to excessive foaming and other undesirable effects. Example, sodium 4-(1,3,5,7-tetramethyloctyl) benzene sulphonate.

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OBJECTIVE CHEMISTRY FOR NEET

Solved Examples 1. Which classification of drugs is useful for general physicians? (1) (2) (3) (4)

Classification based on chemical structure. Classification based on drug use. Classification based on pharmacological effect. All of these.

Solution (3) Classification of drugs based on their pharmacological effect is the most useful for physicians. 2. Phenol can be used as a(an) (1) antiseptic. (2) disinfectant. (3) analgesic. (4) both (1) and (2) Solution (4) In low concentrations phenol is an antiseptic (up to 0.2%) whereas in higher concentration (upto 1%) it is acts as a disinfectant. 3. Aspirin is known as (1) acetyl salicylic acid. (2) phenyl salicylate. (3) acetyl salicylate. (4) methyl salicylic acid. Solution (1) Aspirin is common name of acetyl salicylic acid. The reaction involved in the synthesis of aspirin is O OH

O C 2

COOH + (CH3CO)2O Salicylic acid

Acetic anhydride

CH3 COOH

1

6. Use of medicines for curing diseases is known as (1) pharma therapy. (2) chemical therapy. (3) disease management. (4) any of these. Solution (2) Use of medicines for curing diseases is known as chemical therapy. 7. Which of the following compounds is not an antacid? (1) Cimetidine (2) Phenelzine (3) Ranitidine (4) Aluminium hydroxide Solution (2) This is because phenelzine is a tranquilizer, whereas rest of all is antacids. 8. Which one of the following types of drugs reduces fever? (1) Analgesic (2) Antipyretic (3) Antibiotic (4) Tranquilizer Solution (2) Antipyretic are the chemicals used to bring down the body temperature during fever. 9. Chemically heroin is (1) morphinediacetate. (2) morphinemonoacetate. (3) morphinedibenzoate. (4) morphinemonobenzoate. Solution (1) The chemical name of heroin is morphinediacetate.

2-Acetoxy benzoic acid or Acetyl salicylic acid or aspirin

4. The drug which suppresses ovulation is an (1) antiseptic. (2) antimicrobial. (3) antibiotic. (4) antifertility. Solution (4) Antifertility drugs are those which suppress ovulation or kill the fertilized egg. 5. Further growth of cancerous cells in the body is arrested by

10. Pick the odd one amongst the following on the basis of their medicinal properties. (1) Chloroxylenol (2) Phenol (3) Chloramphenicol (4) Bithional. Solution (3) Chloramphenicol. It is an antibiotic while the other three are used as antiseptic. 11. Match the drug with its use (a) Bithional

(p) Used in the prevention of tuberculosis.

(b) Chloramphenicol

(q) Used to relieve pain due to arthritis

Solution

(c) Streptomycin

(2) Chemotherapy is the use of medication to destroy cancer cells. It also inhibits the further growth of cancerous cells in the body.

(r) Used in the treatment of typhoid, dysentery, etc.

(d) Paracetamol

(s) Used to impart antiseptic properties to soap.

(1) physiotherapy. (2) chemotherapy. (3) electrotherapy. (4) psychotherapy.

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Chemistry in Everyday Life (1) (a) → r; (b) →p; (c) → q; (d) → s (2) (a) → q; (b) → s; (c) → p; (d) → r (3) (a) → s; (b) →r; (c) → p; (d) → q (4) (a) → q, s; (b) → r; (c) → q; (d) → p Solution (3) Bithional is added to soap to impart antiseptic properties.

Chloramphenicol is used in the treatment of typhoid, dysentery, etc.



Streptomycin is used in the prevention of tuberculosis.



Paracetamol belongs to non-narcotic analgesics and is used to relieve pain due to arthritis.

16. Which of the following food additives is an antioxidant? (1) Butylated hydroxyanisole (2) Cyclamate (3) Sodium metabisulphite (4) Amaranth Solution (1) Butylated hydroxyanisole acts as an antioxidant. It retards the degradation of food in contact with O2 and heat. OH C(CH3)3

12. Antiseptics are different from disinfectants as (1) antiseptics merely inhibit the growth and disinfectant kill the microorganisms. (2) antiseptics are used against microorganisms while disinfectants are used against insects. (3) antiseptics are used only over skin while disinfectants can be taken orally also. (4) antiseptics are used over living tissues while disinfectants cannot be used over living tissues. Solution (4) Disinfectants are harmful to the human tissue. 13. Among the following organic acids, the acid present in rancid butter is (1) pyruvic acid. (2) lactic acid. (3) butyric acid. (4) acetic acid. Solution (3) Rancidity is development of any disagreeable flavor in fat. When butter becomes rancid it breaks down into glycerol and fatty acid. Rancid smell is due to formation of butyric acid (CH3   CH2   CH2   COOH). 14. Which artificial sweetener contains chlorine?

OCH3 BHA

17. Food colors are added to improve (1) vitamin content. (2) nutritional value. (3) table value. (4) Both (1) and (2). Solution (3) Food colors are added to improve the table value of food. 18. Sodium lauryl sulphate is a (1) soap. (2) detergent. (3) food color. (4) none of these. Solution (2) Sodium lauryl sulphate is a detergent. 19. Which of the following is an example of non-­biodegrad­ able detergent? (1) CH3

(CH2)11

(2) CH3

(CH2)9CH

(1) Aspartame (2) Saccharin (3) Sucralose (4) Alitame Solution (3) Sucralose is an artificial sweetener that contains chlorine. 15. Carotene is (1) (2) (3) (4)

a food preservative. an artificial sweetener. synthetic dye. none of these.

Solution (4) Carotene is a natural dye. Carotene is also used as edible color substance in food industry.

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825

SO3Na

SO3Na

CH3

(3) CH3

CH3

CH3

CHCH2

CH

SO3Na

(4) CH3(CH2)10CH2OSO3Na Solution (3) It is an alkylbenzene sulphonates detergent and is nonbiodegradable as it contains branched chains. 20. Cetyl ammonium bromide is a (1) soap. (2) anionic detergent. (3) cationic detergent. (4) none of these.

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OBJECTIVE CHEMISTRY FOR NEET

Solution

(1) CH3CH(C4H9)

CH2CH2CH(C4H9)(CH2)3CH(C4H9)CH2CH2SO3Na

(3) The structure is +

CH3 CH3(CH2)15

N

CH3 Br −

CH3 Cetyltrimethyl ammonium bromide (Cationic detergent)

21. Which one of the following substances is a good detergent in water? (1) C16H33N(CH3)+3Cl− Hexadecyltrimethyl ammonium chloride

(2) C16H34 Hexadecane

(3) C15H31COOH Palmitic acid (hexadecanoic acid) O

(4) C15H31 C

(2) CH3(CH2)11

SO3Na

(3)

SO3Na

(4) Detergents are always pollution-free Solution (1) Since it has a long carbon chain, it undergoes slow biodegradation. 24. Which one of the following substances is a good detergent in water? (1) Hexadecyltrimethyl ammonium chloride: C16H33N(CH3)3+Cl(2) Hexadecane:C16H34 (3) Hexadecanoic acid: C15H31COOH (4) Hexadecyl hexadecanoate: C15H31COOC16H33 Solution

OC16H33

(4) Cetylpalmitate is a good detergent and is non-ionic.

(1) Only compound given in option (1), hexadecyltrimethyl ammonium chloride would be a good detergent in water. It has both hydrophilic and hydrophobic ends. It is a cationic detergent because it contains long carbon chain of positively charged ions.

22. Which of the following statement is true?

25. Synthetic detergents contain

Cetylpalmitate (hexadecyl hexadecanoate)

Solution

(1) Synthetic detergents contain only hydrophilic part. (2) Synthetic detergents contain only hydrophobic part. (3) Synthetic detergents contain large amount of chlorine. (4) Synthetic detergent contains both hydrophobic and hydrophilic part. Solution

(1) 70% or more soap. (2) 50–70% of soap. (3) 10–50% soap. (4) no trace of soap. Solution (4) Synthetic detergents do not contain any trace of soap. 26. Which of the following compounds is soap? (1) CH3CH2CH2COONa (2) CH3(CH2)14CH2ONa (3) CH3(CH2)14COONa (4) [CH3(CH2)14COO]2Ca

(4) Synthetic detergents contain both hydrophilic and hydrophobic part.

Solution

23. Which detergent can cause maximum pollution?

(3) Soaps are the sodium salts of fatty acids containing long carbon chain. Thus, compound in option (3) satisfies the conditions for soap.

Practice Exercises Level I Medicines 1. Which of the following term means pain killing? (1) Antibiotic (2) Analgesic (3) Antipyretic (4) Penicillin 2. Bithional is an example of (1) disinfectant. (2) antiseptic. (3) antibiotic. (4) analgesic.

Chapter 32_Chemistry in Everyday Life.indd 826

3. The antibiotic used for the treatment of typhoid is (1) penicillin. (2) chloramphenicol. (3) terramycin. (4) sulphadiazine. 4. Which of the following is used as a “morning after pill”? (1) Norethindrone (2) Ethynylestradiol (3) Mifepristone (4) Bithional 5. In the following sets of compounds, the one which contains only medicinal compounds is

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Chemistry in Everyday Life (1) (2) (3) (4)

Alizarin, phenacetin, morphine. Aspirin, genatin, violet, phenolphthalein. Boric acid, chloramphenicol, aspirin. 9-Oxdecanoic acid, boric acid, morphine.

6. Which of the following statements is not true? (1) Some disinfectants can be used as antiseptic at low concentration. (2) Sulphadiazine is a synthetic antibacterial. (3) Aspirin is analgesic and antipyretic. (4) Norethindrone is a pheromone. 7. Select the incorrect statement. (1) Equanil is used to control depression and hypertension. (2) Mifepristone is a synthetic steroid used as “morning after pill”. (3) A 0.2% solution of phenol is an antiseptic while its 1.0% solution is a disinfectant. (4) A drug which kills the organism in the body is called bacteriostatic. 8. Which of the following is not a tranquilizer? (1) Equanil (2) Veronal (3) Salvarsan (4) Serotonin 9. Barbiturates are used as (1) analgesics. (2) food preservatives. (3) antipyretics. (4) tranquilizers. 10. Arsenic containing medicine used for the treatment of syphilis, is (1) Erythromycin. (2) Ofloxacin. (3) Tetracycline. (4) Salvarsan. 11. Which set has different class of compounds? (1) (2) (3) (4)

Tranquilizers: Equanil, heroin, valium Antiseptics: Bithional, Dettol, boric acid Analgesics: Naproxen, morphine, aspirin Bactericidal: Penicillin, aminoglycosides, ofloxacin

12. Terfenadine is commonly used as a/an (1) tranquilizer. (2) antihistamine. (3) antimicrobial. (4) antibiotic. 13. Which one of the following is an antacid? (1) Iproniazid (2) Salvarsan (3) Zantac (4) Chloramphenicol 14. An ester used as medicine is (1) ethyl acetate. (2) methyl acetate. (3) methyl salicylate. (4) ethyl benzoate. 15. The bacteriostatic antibiotic among the following is (1) Erythromycin. (2) Penicillin. (3) Aminoglycoside. (4) Ofloxacin.

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827

16. Which of the following is not an antipyretic? (1) Aspirin (2) Paracetamol (3) Barbituric acid (4) Phenacetin 17. Which of the following drugs is an analgesic? (1) Sulphaguanidine (2) Paludrine (3) Analgin (4) Iodex 18. An antibiotic with a broad spectrum (1) (2) (3) (4)

kills the antibodies. acts on a specific antigen. acts on different antigens. acts on both the antigens and antibodies.

19. Antiallergy drugs are (1) antimicrobials. (2) antihistamines. (3) antivirals. (4) antifungals. 20. Antiseptic chloroxylenol is (1) (2) (3) (4)

4-chloro-3, 5-dimethyphenol. 3-chloro-4, 5-dimethylphenol. 4-chloro-2, 5-dimethylphenol. 5-chloro-3, 4-dimethylphenol.

21. Which of the following is used as antipyretic? (1) Paracetamol (2) Chloroquine (3) Chloramphenicol (4) LSD 22. A drug that is antipyretic as well as analgesic is (1) Chloropromazine hydrochloride. (2) Paracetomol. (3) Chloroquin. (4) Penicillin. 23. Which of the following antibiotic contains NO2 group attached to aromatic nucleus in its structure? (1) Penicillin (2) Streptomycin (3) Chloramphenicol (4) All of these 24. Drugs that bind to the receptor site and inhibit its natural functions are called (1) anatagonists. (2) agonists. (3) enzymes. (4) molecular targets. 25. Which of the following is not an antacid? (1) Histamine (2) Ranitidine (3) Omeprazole (4) All of these 26. Which among the following is not an antibiotic? (1) Penicillin (2) Oxytocin (3) Erythromycin (4) Tetracyclin

Chemicals in Food 27. Substance used for the preservation of colored fruit juices is

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OBJECTIVE CHEMISTRY FOR NEET

Previous Years’ NEET Questions

(1) benzene. (2) benzoic acid. (3) phenol. (4) sodium metabisulphite.

1. Which one of the following is employed as a tranquilizer drug?

28. Which of the following artificial sweetening agents is the least sweet? (1) Dulcin (2) Sucralose (3) Aspartame (4) Alitame 29. Aspartame is one of the good artificial sweeteners whose use is limited to cold foods and soft drinks because (1) (2) (3) (4)

it has very low boiling point. it gets dissociated at cooking temperature. it is sweetener at low temperature only. it is insoluble at higher temperatures.

Cleansing Agents 30. Which one of the following is not used as a filler in laundry soaps? (1) Sodium silicate (2) Glycerol (3) Sodium rosinate (4) Borax 31. Which of the following is not a surfactant? (1) CH3  (2) CH3  (3) CH3  (4) OHC 

 (CH2)15   (CH2)14   (CH2)16   (CH2)14 

 N (CH3)3Br   CH2NH2  CH2OSO2Na+  CH2   COO−Na+ +

−

32. The detergent which is used as a germicide is (1) sodium laurylsulphate. (2) cetyltrimethylammonium chloride. (3) lauryl alcohol ethoxylate. (4) sodium-2-dodecylbenzenesulphonate. 33. The cationic detergent that is used in hair conditioners is (1) (2) (3) (4)

sodium dodecylbenzene sulphonate. sodium lauryl sulphate. tetramethyl ammonium chloride. sodium stearyl sulphate.

34. What chemical is added to washing powders to keep them dry? (1) Sodium perborate (2) Sodium carbonate (3) Sodium sulphate (4) None of these 35. What should be the feature of detergent molecule structure so as to be biodegradable? (1) (2) (3) (4)

It should be saturated. It should be unsaturated. Branching should be maximum. Branching should be minimum.

Chapter 32_Chemistry in Everyday Life.indd 828

(1) Chlorpheninamine (2) Equanil (3) Naproxen (4) Tetracycline (AIPMT 2009) 2. Which one of the following is employed as a tranquilizer drug? (1) Valium (2) Naproxen (3) Mifepristone (4) Promethazine (AIPMT PRE 2010) 3. Chloroamphenicol is an (1) antifertility drug. (2) antihistaminic. (3) antiseptic and disinfectant. (4) antibiotic-broad spectrum. (AIPMT MAINS 2012) 4. Antiseptics and disinfectants either kill or prevent growth of microorganisms. Identify which of the following statements is not true? (1) A 0.2% solution of phenol is an antiseptic while 1% solution acts as a disinfectant. (2) Chlorine and Iodine are used as strong dis­infectants. (3) Dilute solutions of boric acid and hydrogen peroxide are strong antiseptics. (4) Disinfectants harm the living tissues. (NEET 2013) 5. Artificial sweetener which is stable under cold conditions only is (1) saccharine. (2) sucralose. (3) aspartame. (4) alitame. (AIPMT 2014) 6. Bithionol is generally added to the soaps as an additive to function as a/an (1) dryer. (2) buffering agent. (3) antiseptic. (4) softener. (AIPMT 2015) 7. Which of the following is an analgesic? (1) Novalgin (2) Penicillin (3) Streptomycin (4) Chloromycetin (NEET I 2016) 8. Mixture of chloroxylenol and terpineol acts as (1) antiseptic. (2) antipyretic. (3) antibiotic. (4) analgesic. (NEET 2017)

1/4/2018 5:32:45 PM

Chemistry in Everyday Life

829

Answer Key Level I  1. (2)

 2. (2)

 3. (2)

 4. (3)

 5. (3)

 6. (4)

 7. (4)

 8.  (3)

 9. (4)

10.  (4)

11.  (1)

12.  (2)

13.  (3)

14.  (3)

15.  (1)

16.  (3)

17.  (3)

18.  (3)

19.  (2)

20.  (1)

21.  (1)

22.  (2)

23.  (3)

24.  (1)

25.  (1)

26.  (2)

27.  (2)

28.  (3)

29.  (2)

30.  (2)

31.  (2)

32.  (2)

33.  (2)

34.  (3)

35.  (4)

 4. (3)

 5. (3)

 6. (3)

 7. (1)

 8. (1)

Previous Years’ NEET Questions  1. (2)

 2. (1)

 3. (4)

Hints and Explanations Level I 3. (2) It is a specific antibiotic used for the treatment of typhoid. 7. (4) A drug which kills the organism in the body is called bactericidal and not bacteriostatic. 9. (4) Barbituric acid along with its 5,5-disubsituted derivatives are used as transqulizers, that is, sleep inducing agents.

29. (2) Aspartame has the strong tendency to react with other food ingredients forming unique chemical compounds; also the free amino acids present in it undergo racemization at high temperatures producing significant amounts of unnatural D-type amino acid. 31. (2) For a molecule to behave as an surfactant, it should have both hydrophobic as well as hydrophilic part.    NH2 is the hydrophilic part and the hydrocarbon part is the hydrophobic part. 32. (2) Cetyltrimethylammonium bromide is a popular cationic detergent possessing germicidal properties.

11. (1) Heroin is a narcotic drug. 14. (3) Methyl salicylate (oil of winter green) is used to treat joint and muscular pains. 16. (3) Aspirin and paracetemol are antipyretics. Phenacetin is a very moderate and safe antipyretic, whereas barbituric acid acts as an antipyretic tranquilizer.

CH3 CH3(CH2)15

N

+

CH3 Br −

CH3

17. (3) Analgin is used for relieving pain and fever (analgesic).

Cetyltrimethyl ammonium bromide (Cationic detergent)

19. (2) Antihistamines are medicines used to reduce allergy caused by proteins.

34. (3) Na2SO4 acts like silica gel and hence keeps washing powders free from moisture.

21. (1) Paracetamol is used as antipyretic drug to reduce fever. 23. (3) Chloramphenicol contains    NO2 and    NH   groups. OH O2N

CH

CH

CH2

OH

NH

CO

CHCl2

Chloramphenicol

25. (1) Histamine is a vasodilator which causes allergy, nasal congestion associated with common cold. 28. (3) Aspartame is the least sweet, and alitame is the sweetest.

Chapter 32_Chemistry in Everyday Life.indd 829

Previous Years’ NEET Questions 1. (2) Equanil is tranquilizer used in the depression and hypertension. 2. (1) Valium is commonly used tranquilizer which is a sedative and muscle relaxant. 3. (4) Chloramphenicol is a broad spectrum antibiotic, it is rapidly absorbed from the gastrointestinal tract and is given orally in case of typhoid, dysentery, acute fever and certain form of urinary infections.

1/4/2018 5:32:45 PM

830

OBJECTIVE CHEMISTRY FOR NEET

4. (3) Dilute solutions of boric acid and H2O2 are mild antiseptics. 5. (3) Aspartame cannot be used for cooking and baking because it decomposes with heat and, therefore, stable under only cold conditions. 6. (3) Bithionol is added to the soaps as an additive to function as an antiseptic.

Chapter 32_Chemistry in Everyday Life.indd 830

7. (1) Novalgin is an analgesic. It is used for the treatment of pain. Penicillin and streptomycin are antibiotics. Chloromycetin is used for the treat of various infections. 8. (1) Antiseptics are chemical substances that prevent the growth of microorganisms or kill them but are not harmful when applied to human tissues. Mixture of chloroxylenol and terpineol is known as dettol which is an example of antiseptic.

1/4/2018 5:32:45 PM

Mock Test I

1. Bond angle decreases except when + 4  2 

(1) NH is converted into NH3. (2) NH is converted to N3 -. (3) SO3 is converted to SO2 . (4) CO2 is converted to CO3 2-. 2. Markonikov’s rule is empirical but may be explained theoretically on the basis that the addition occurs by a (1) (2) (3) (4)

carbanion intermediate. carbocation intermediate. carbene intermediate. carbon free radical intermediate.

3. When the acids HClO3, H3BO3, H3PO3 are arranged in order of increasing acidic strength, which is the correct order? (1) H3BO3 < H3PO4 < HClO3 (2) HClO3 < H3BO3 < H3PO4 (3) H3PO4 < HClO3 < H3BO3 (4) H3BO3 < HClO3 < H3PO4 4. For the first order reaction A(g) ® 2B(g) + C(g), the initial pressure is pA = 90 mm of Hg. The pressure after 10 min is found to be 180 mm of Hg. The half-life period of the reaction is (1) 1.15 × 10–3 s (2) 600 s (3) 3.45 × 10–3 s (4) 200 s 5. Which of the following is not applicable to chemisorption? (1) (2) (3) (4)

Heat of adsorption is negative. Takes place at high temperature. Is reversible. Forms monomolecular layers.

6. The radius of isoelectronic species (1) (2) (3) (4)

increases with increase in nuclear charge. decreases with increase in nuclear charge. is same for all. first increases and then decreases.

7. MgO and CaO are used for lining furnaces because (1) (2) (3) (4)

they have high melting point. they are good conductors of heat. they are electrical insulators. all are correct.

8. The dipole moment of H2O2 is 2.1 D while that of water is 1.84 D. But water H2O is a better solvent than that of H2O2 because

Mock Test 1.indd 831

(1) its dipole moment is less than of H2O. (2) it is less corrosive. (3) H2O2 gets ionized during chemical reactions. (4) H2O2 gets decomposed during chemical reactions. 9. Which of the following statements is correct? (1) Metamerism belongs to the category of structure isomerism. (2) Tautomeric structure is the resonating structure of a molecule. (3) Keto form is always more stable than the enol form. (4) Geometrical isomerism is shown only by alkenes. 10. At 298 K, the solubility of PbCl2 is 2 ´ 10−2 mol L−1, then its Ksp will be (1) 1 ´ 10−7 (3) 1 ´ 10−5



(2) 3.2 ´ 10−7 (4) 3.2 ´ 10−5

11. The glycosidic linkage involved in the linking of the glucose units in amylase part of starch is (1) C1-C4 b-linkage. (2) C1-C6 a-linkage. (3) C1-C4 a-linkage. (4) C1-C6 b-linkage. 12. Which of the following methods of separation is based upon difference in distribution between a stationary and a moving phase? (1) Chromatography (2) Electrolysis (3) Ultracentrifugation (4) Counter-current distribution 13. The radioactive lanthanide is (1) ytterbium (Yb). (2) iron (Fe). (3) promethium (Pm). (4) copper (Cu). 14. Which of the isoelectronic ions K+, Ca2+ , Cl- and S2- has the lowest ionization energy? (1) K+ (2) Ca2+ (3) Cl- (4) S2 15. Magnetic moments of  V(Z = 23), Cr(Z = 24), Mn(Z = 25) are x, y, z. Hence (1) x = y = z (3) x < z < y

(2) x < y < z (4) z < y < x

16. Which of the following complexes is responsible for the brown color of the ring formed in the ring test for nitrates? (1) [Fe(CN)5NO]3+ (2) [Fe(H2O)4NO]2+NO+ (3) [Fe(CN)5NO]2+ (4) [Fe(H2O)5NO]2+

1/4/2018 5:33:10 PM

832

OBJECTIVE CHEMISTRY FOR NEET

17. Which of the following reagents would distinguish ciscyclopenta-1,2-diol from the trans-isomer? (1) HIO4 (2) MnO2 (3) Aluminium isopropoxide (4) Ozone 18. Given that



A(s) ® A(l);   ∆H = x





A(l) ® A(g);   ∆H = -y



The enthalpy of sublimation of A will be (1) x + y (2) x – y (3) x or y (4) –(x + y)

19. An element (Molar mass = 100 g) having bcc structure has unit cell edge length = 400 pm. Then density of the unit cell is (1) 10.376 g cm-3 (2) 5.188 g cm-3 (3) 7.289 g cm-3 (4) 2.144 g cm-3 20. In the following statements, which combination of true (T) and false (F) options is correct? (I) Ionic mobility is the highest for I– in water as compared to other halides. (II) Stability order is: Cl 3- > Br3- > I 3(III) Reactivity order is: F < Cl < Br < I (IV) Oxidizing power order is: F2 < Cl2 < Br2 < I2 (1) TFTF (2) TFFF (3) TFFT (4) FTFT 21. Which of the following is an antiseptic? (1) Furacine (2) Ethynyl estradiol (3) Novalgin (4) Chloromycetin 22. A compressed air tank carried by scuba divers has a volume of 8.0 L and a pressure of 140 atm at 20°C. What is the volume of air in the tank (in liters) at STP? (1) 1.1 ´ 103 L (2) 1.0 ´ 10-3 L (3) 1.0 ´ 102 L (4) 1 ´ 10-2 L 23. In the following reaction, X is

Boiling 2 + HCl X Bromination  → Y NaNO  → Z C  → 2 H5 OH

(1) Aniline (2) Salicylic acid (3) Phenol (4) Benzoic acid 24. In molecules of the type AX2Ln, where L represents lone pair n its number, there exists a bond between elements A and X. Then, XAX angle (1) (2) (3) (4)

Mock Test 1.indd 832

always decreases if n increases. always increases if n increases. will be maximum for n = 3. will always be less than 180o if n = 0.

25. In N2O, the N   N distance corresponds to (1) N N bond. (2) N N bond. (3) N   N bond. (4) Intermediate of N

N and N

N.

26. If x1 and x2 represent the mole fraction of a component A in the vapor phase and liquid mixture, respectively, po A and po B represent vapors pressures of pure A and pure B, then total vapor pressure of the liquid mixture is (1)

p Ao x1 po x (2) A 2 x2 x1

(3)

pBo x1 po x (4) B 2 x2 x1

27. Which of the following contains minimum number of lone pairs around Xe atom? (1) XeF4 (2) XeF6 (3) XeOF2 (4) XeF2 28. How many oxides are formed by nitrogen? (1) 1 (2) 3 (3) 5 (4) 4 29. Calculate EMF for the cell at 298 K Zn | Zn 2+ || Zn 2+ | Zn ( 0.01M )

( 0.1M )

(1) 0.0395 V (2) –0.0395 V (3) 0.0295 V (4) 1.0345 V 30. Which of the following will undergo homolytic fission? (1) Br2 (2) HBr (3) H2O (4) HCl 31. If x is the fraction of molecules having energy greater than Ea, it will be given by (1) x = -

E Ea (2) ln x = - a RT RT

(3) x = e Ea /RT (4) None of these 32. A metal forms two complexes in the same oxidation state. In one complex, the magnetic moment is 4.9 BM in another it is 0 BM. Which of the following metals fits this description? (1) Cr3+ (2) Mn2+ (3) Fe2+ (4) Fe3+ 33. PH3 can be obtained by heating (1) (2) (3) (4)

white phosphorus with hot concentrated alkali. phosphinic acid (H3PO2). metaphosphoric acid (HPO3). phosphoric acid (H3PO4).

1/4/2018 5:33:11 PM

833

Mock Test I 34. Sorption is the term used when

39. Which of the following biphenyls is optically active?

(1) adsorption takes place. (2) absorption takes place. (3) desorption takes place. (4) it is not possible to differentiate between adsorption and absorption. 35. The van’t Hoff factor of 0.1 M Ca(NO3)3 is 2.74. Thus, percentage dissociation of salt is (1) 85% (2) 58% (3) 65.8% (4) 56.8% 36. The number of optical isomers of the compound CH3CH(Br)CH(Br)COOH is (1) 0 (2) 1 (3) 3 (4) 4

H2SO4

NO2

+ NaO NO2

NO2

(4)

NO2

O CHO + OH-

CH

COO-

(1)

C

C OH

(3)

Mock Test 1.indd 833

Both (1) and (2)

H

CH3

C

O

KCN

OK

CH3 C H

CN

PBr3

Br

(2)

C

C

O

O

(4) None of these

(1) (2) (3) (4)

messenger RNA, translational RNA, structural RNA. messenger RNA, ribosomal RNA, transfer RNA. cytosine RNA, nucleoside RNA, nucleotide RNA. primary RNA, secondary RNA, tertiary RNA.

43. Which one of the following is less basic?

OH O-

D

42. RNA molecules are of three types which are based on their different function. These are

38. The correct intermediate for the following intramolecular Cannizzaro reaction is

O

NaNH2

(I)

(1) 64.225°C (2) 6.4225°C (3) 62.2°C (4) 161.2°C

+ HO

C

40. For the following reactions

41. Determine the boiling point of a solution in which 25 g urea (Mol. wt. = 60) is dissolved in 500 g chloroform. The boiling point of pure chloroform is 61.2°C, Kb of chloroform = 3.63 K molal–1

NO2

F

NO2 NO2

(1) I is elimination, II is addition, III is substitution reaction. (2) II is elimination, III is addition, I is substitution reaction (3) III is addition, II is elimination, III is substitution reaction (4) I is elimination, II is substitution, III is addition

2HNO3

ONa +

NO2 CH3

OH

NO2

(3)

(4)

(III)

NO2

O

NO2

COOH NO2

(3)

H

O

F

CH3 CH3 NO2

(II)

NO2

(2)

(2)

(1)

Br

37. Which of the following would work best for the synthesis of the ether shown below?

(1)

CH3 CH3

O-

(1) NO2

(3) C6H5

NH2

NH2

(2) CH3O

(4) H2N

NH2

NH2

1/4/2018 5:33:12 PM

834

OBJECTIVE CHEMISTRY FOR NEET

44. An analysis of a food sample involves smearing a small amount of the food sample onto newsprint. This is done to test for the presence of (1) starch. (2) carbohydrates. (3) protein. (4) fat.

Mock Test 1.indd 834

−

45. The pair of electron in the given carbanion, present in which of the following orbitals? (1) sp2 (3) sp

is

(2) sp3 (4) 2p

1/4/2018 5:33:12 PM

Mock Test II

1. Liquid A and B form an ideal solution. Which of the following is the correct statement? (1) The enthalpy of mixing is zero. (2) The entropy of mixing is zero. (3) The free energy of mixing is zero. (4) The free energy as well as entropy of mixing are each zero. 2. A solution is a mixture of 0.05 M NaCl and 0.05 M NaI. The concentration of iodide ion in the solution when AgCl just starts precipitating is equal to (Ksp(AgCl) = 10-10, Ksp(AgI) = 4 × 10-16) (1) 4 × 10-6 M (2) 2 × 10-8 M (3) 2 × 10-7 M (4) 8 × 10-15 M N N 3. 40 mL KOH are mixed together. HCl and 60 mL 20 10 Calculate the normality of the salt formed.

(3) A white precipitate of Ca(HCO3)2 is formed. (4) A white precipitate of CaCO3 is formed. 9. The state of hybridization of boron and oxygen atom in boric acid H3BO3 is respectively (1) sp3 , sp3 (2) sp2 , sp3 (3) sp3, sp2 (4) sp2 , sp2 10. Pure phosphine is not combustible while impure phosphine is combustible; this combustibility is due to the presence of (1) P2H4 (2) N2 (3) PH5 (4) P2O5 11. Which of the following statement regarding sulphur is incorrect? (1) At 600°C, the gas mainly consists of S2 molecules. (2) The oxidation state of sulphur is never less than +4 in its compounds. (3) S2 molecule is paramagnetic. (4) The vapor at 200°C consists mostly of S8 rings.

(1) 0.06 (2) 0.09 (3) 1 (4) 0.03 4. Calculate the volume of Cl2 at NTP produced during electrolysis of molten MgCl2 which produces 6.5 g Mg. (1) 3.03 L (2) 2.02 L (3) 6.06 L (4) 5.05 L

12. Which of the following elements does not form hydride by direct heating with dihydrogen? (1) Be (2) Mg (3) Sr (4) Ba

5. H3BO3 is (1) (2) (3) (4)

13. For the reaction

monobasic and weak Lewis acid. monobasic and weak Bronsted acid. monobasic and strong Lewis acid. tribasic acid and weak Bronsted acid.



6. Aluminium (III) chloride forms a dimer because aluminium (1) (2) (3) (4)

cannot form a trimer. has high ionization energy. belongs to third group. can have higher coordination number.

(1) it is cheaper than iron. (2) E (3) E

o Zn 2 + /Zn

E

M M (4) 11 20

14. Number of spherical nodes in 3p orbitals is o Fe2 + /Fe o Fe2 + /Fe

8. Which of the following changes occurs when excess of CO2 gas is passed into a clear solution of lime water? (1) A white precipitate containing both CaCO3 and Ca(HCO3)2 is formed. (2) Initially a white precipitate of CaCO3 is formed which changes into soluble Ca(HCO3)2 on passing excess CO2 gas.

Mock Test II.indd 835

(1) M (2) M 8 (3)

7. Zinc is coated over iron to prevent rusting of iron because o Zn 2 + /Zn o Zn 2 + /Zn

11 O2 → Fe2O3 + 4SO2 2 What is the equivalent weight of FeS2, if its molecular weight is M. 2FeS 2 +

(1) 1 (2) 3 (3) 2 (4) 0 15. A solution of ammonia in water contains (1) H+ (2) OH(3) Only NH4+  (4) OH- , NH4+ and NH4OH 16. Which of the following statements is incorrect? (1) Single N   N bond is stronger than the single P   P bond.

1/4/2018 5:33:22 PM

836

OBJECTIVE CHEMISTRY FOR NEET (2) PH3 can act as a ligand in the formation of coordination compound with transition elements. (3) NO2 is paramagnetic in nature (4) Covalency of nitrogen in N2O5 is four.

23. The major product of the following reaction is Cr2O3–Al2O3 600°C

17. Formation of complex compound can be detected by (1) (2) (3) (4)

change in color. change in solubility. change in pH. change in electrical conductivity.

18. The gradual decrease in radius of M3+ ion for lanthanoids is not obeyed by (1) Eu only. (2) Yb only. (3) Both Eu and Yb. (4) None of these.

Cl

+ CH3

Cl

(1) (2) (3) (4)

AlCl3

(4)

N

a -D-Galactopyranose and a -D-galactopyranose. a -D-Galactopyranose and b -D-fructofuranose. b -D-Galactopyranose and a -D- fructofuranose. a -D-Galactopyranose and b -D-fructofuranose.

27. What is the amount of catalyst required in the Friedel– Crafts acylation? + Ph

Cl

+ CH3

Cl

(1) (2) (3) (4)

AlCl3

AlCl3

(1) [SiH6]2- has sp3d2 hybridization. (2) PF5 has sp3d hybridization. (3) SF6 has sp3d2 hybridization. (4) XeO4 has sp2 hybridization. 22. Acetate rayon is prepared from (1) acetic acid. (2) glycerol. (3) starch. (4) cellulose.

One equivalent Two equivalent Three equivalent More than one equivalent

28. Among the following compounds nitrobenzene, benzene, aniline and phenol, the strongest basic behavior in acidic medium exhibited by (1) phenol. (2) aniline. (3) nitrobenzene. (4) benzene.

21. Select the incorrect statement:

Mock Test II.indd 836

NH

26. Which one of the following set of monosaccharides forms sucrose?

Me

(4)

N

(2)

(1) translation. (2) transduction. (3) replication. (4) transcription.

AlCl3

Me

(3)

NH2

25. DNA multiplication is called

COOH

(2)

(4)

(3)

NH2 + CH3

(3)

(1)

(1) 22s and 15p (2) 23s and 15p (3) 22s and 16p (4) 15s and 8p

(1)

(2)

24. Which of the following amines give isocyainde on reaction with CHCl3/KOH?

19. The number of s and p-bonds in Fe2(CO)9, respectively are

20. Which one of the following can show Friedel–Crafts reaction?

(1)

29. Consider the reaction CH 3CH 2CH 2CH 2Cl + NaOH → CH 3CH 2CH 2CH 2OH + NaCl

The reaction is fastest in (1) ethanol. (2) N,N-dimethyl acetamide. (3) water. (4) methanol.

1/4/2018 5:33:23 PM

Mock Test II 30. Which is the most suitable reagent for the following conversion? O CH3

CH

CH

CH2

C

CH

CH

CH2

C

CH3

(3) Sn 4+ > Al 3+ > Ba 2+ > Na + (4) PO34- > SO24- > Cl 38. In the reaction Cr2O72- + 14H + + 6e - → 2Cr 3+ + 7H 2O, calculate the quantity of electricity (coloumbs) needed to reduce 1 mol of Cr2O72-.

OH

(1) Tollen’s reagent (2) Benzoyl peroxide (3) I2 and NaOH solution (4) Sn and NaOH solution

(1) 6 F (2) 5 F (3) 4 F (4) 3 F 39. For a d electron, the orbital angular momentum is

31. Which among the following compound will show tautomerism? O

O

(1)

O C

(3)

6 (2)

2

(3)  (4)

2

p  p  (1) ∆S = nR ln  f  (2) ∆S = nRT ln  f   pi   pi  p  (3) ∆S = nR ln  i  (4) 0  pf 

O H

(1)

40. For a sample of an ideal gas when its pressure is changed from pi to pf in reversible adiabatic process, the entropy change is given by

O

(2)

O

37. Gold sol is negatively charged. The flocculation value of effective ions varies in the order. (1) Sn 4+ < Al 3+ < Ba 2+ < Na + (2) PO34- < SO24- < Cl -

O CH3

837

C

(4)

CH3

41. The ratio of any colligative property of KCl to that of sugar is nearly (assuming m and solvent as same) (1) 1 (2) 0.5 (3) 2 (4) 2.5

32. Which of the following acids cannot be prepared by Grignard reagent? (1) Acetic acid (2) Succinic acid (3) Formic acid (4) All of these 33. Cracking of propane is expected to yield (1) (2) (3) (4)

propene, ethene, methane, and hydrogen. propene and hydrogen. ethene and methane. ethyne, methane and hydrogen.

-5

42. The pH of 0.2 M NH4Cl at 25°C is ( K b(NH4 OH) = 1.8 × 10 ) (1) 3.977 (2) 4.977 (3) 2.977 (4) 5.977 43. Coordination number of Ni in [Ni(C 2O 4 )3 ]4(1) 3 (2) 6 (3) 4 (4) 5 44. What will be the EMF of the given cell? Pt | H 2( p1 )| H + (aq )| H 2( p2 )| Pt

34. Which of the following molecule has intramolecular hydrogen bonding? (1) ortho-Nitrophenol (2) ortho-Boric acid (3) Both (1) and (2) (4) None of these

(1)

RT  p1  RT  p1  ln ln   (2) 2 F  p2  F  p2 

35. The molar conductivity of 0.025 mol L–1 methanoic acid is 46.1 S cm2 mol–1. Its degree of dissociation (a) is [Given: ∞ 2 -1 l H∞+ = 349.6 S cm 2mol -1 and l HCOO ] - = 54.6 S cm mol

(3)

RT  p2  RT  p2  ln ln   (4) F 2 F  p1   p1 

(1) a = 0.214 (2) a = 0.114 (3) a = 0.150 (4) a = 0.314 36. Thermal decomposition of a compound is of first order. If 50% of a sample of compound is decomposed in 120 min, How long will it take for 90% of compound to decompose? (1) 299 min (2) 399 min (3) 99 min (4) 9.9 min

Mock Test II.indd 837

45. In the compound given below, the correct order of the acidity of the positions I, II and III is +

+

H3N II

NH3 III COOH I

(1) I > II > III (2) III > I > II (3) II > I > III (4) III > II > I

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Mock Test II.indd 838

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Mock Test III

1. The pollutants which come directly in the air from source are called primary pollutants. Primary pollutants are sometimes converted into secondary pollutants. Which of the following belongs to secondary pollutants? (1) CO (2) Hydrocarbon (3) Peroxyacetyl nitrate (4) NO 2. An element with atomic number 106 has been discovered. Which of following electronic configuration will it possess? (1) [Rn] 5f14 6d4 7s2 (2) [Rn] 5f14 6d5 7s1 (3) [Rn] 5f14 6d6 7s0 (4) [Rn] 5f14 6d1 7s1 7p3 3. What is the major product when m-bromotoluene reacts with KNH2/NH3? (1) o-Toludine (2) m-Toludine (3) p-Toludine (4) Benzylamine 4. 25 mL of HCl solution on reaction with excess AgNO3 solution gave 2.125 g of AgCl. The molarity of HCl solution is (1) 0.25 (2) 0.6 (3) 1.0 (4) 0.75 5. The maximum % of s-character in N–H bond is observed in (1) NH3 (2) NH +4 (3) N2H4 (4) N2H2 6. Which is not an ore of copper? (1) Atacamite (2) Copper glance (3) Chalcopyrite (4) Cerrusite 7. In SiF62- and SiCl 62- which one is known and why? (1) (2) (3) (4)

SiF62- because of small size of F. SiF62- because of large size of F. SiCl 62- because of small size of Cl. SiCl 62- because of large size of Cl.

→ 2NO 2 NO + NO 3  K2



(fast)

Order of the reaction is (1) 0 (2) –1 3 (3) 1 (4) 2

Mock Test III.indd 839

(I) 2 NO(g )  N 2 (g) + O2(g ) K1 = 2.4 × 1030 1 (II) NO(g )+ Br2(g )  NOBr(g ) K1 = 1.4 2 (1) 3.15 × 10 -9 (2) 6.35 × 10 -18 (3) 9.03 × 10 -16 (4) 17 × 10 -17 10. Which of the following statements is correct? (1) The difference in potential energies of any two energy levels is always more than difference in kinetic energies of these two levels. (2) According to Bohr theory, energy decreases as n increases. (3) In Rutherford’s a-scattering experiment, the distance of closest approach is of the order 10–16 m. (4) The uncertainty in position and momentum in Heisenberg’s principle is not due to electron wave. 11. Which of the following statements is true? (1) Catalyst does not alter DH of the reaction. (2) DG° of the catalyzed and uncatalyzed reaction changes. (3) Catalyzed reaction alters equilibrium constant even at constant temperature. (4) Catalyst increases e Ea/RT value. 12. When KMnO4 reacts with H2O2 in a slightly alkaline and acidic medium, the respective products obtained are (1) K2MnO4 and Mn2+ (3) MnO2 and Mn2+

(2) MnO2 and MnO2 (4) Mn2+ and MnO2

(1) zeise’s salt. (2) ferrocene. (3) dibenzene chromium. (4) tetraethyl tin.

(fast)

1 → NO2 + NO + O2 NO2 + NO3 K

from the following information

14. The p-bonded organometallic compound which has ethene as one of its component is

The following mechanism is suggested K



(1) OCH3 (2) NO2 (3) CN (4) halogen

1 N 2O5 → 2NO2 + O2 2 eq   N 2O 5    NO2 + NO3

1 1 1 N 2(g ) + O2(g ) + Br2(g )  NOBr(g ) 2 2 2

13. Which of the following groups does not decrease the basic strength of aniline?

8. The decomposition of N2O5 is



9. Calculate KC for the reaction,

(slow)

15. The root mean square velocity of a gas is urms. If average kinetic energy = E of the gas, then (Molar mass of gas = M) (1) urms =

3E 3E (2) urms = 2M 3M

(3) urms =

2E E (4) urms = M 3M

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840

OBJECTIVE CHEMISTRY FOR NEET 25. A first order reaction has a rate constant 6.93 ´ 10−2 s−1. How much time will it take for 20 g of the reactant to reduce to 2.5 g?

16. Glycerol is not used in (1) explosive. (2) cosmetics. (3) soaps. (4) matches.

(1) 10 s (2) 20 s (3) 30 s (4) 40 s

17. The following drug is used as an H N

26. Which of the following statement is not correct from the view point of molecular orbital theory? N

CH2 CH2

NH2

(1) antacid. (2) analgesic. (3) antimicrobial. (4) antiseptic. 18. For the preparation of KMnO4 from K2MnO4 which of the following reagents is the best? (1) Dil. H2SO4 (2) SnCl2 (acidified) (3) CO2 is passed (4) Cl2 is passed 19. The correct order of pseudo halide, polyhalide and interhalogen are (1) BrI 2- , OCN - , IF5 (2) IF5 , BrI 2- , OCN (3) OCN - , IF5 , BrI 2- (4) OCN - , BrI 2- , IF5

O

21. CH3

C

CH

1% HgSO4

A

CH3

C

CH3

Structure of A and types of isomerism in the above reaction are respectively (1) (2) (3) (4)

Prop-1-en-2-ol, metamerism. Prop-1-en-1-ol, tautomerism. Prop-2-en-1-ol, geometrical isomerism. Prop-1-en-2-ol, tautomerism.

22. Henry’s law constant for molality of methane in benzene at 298 K is 4.27 × 105 mm Hg. Determine the solubility of methane is benzene at 298 K under 760 mm of Hg. (1) 178 × 10-5 (2) 178 × 10-3 (3) 356 × 10-5 (4) 356 × 10-3 23. Decreasing order of reactivity of following alkyl halides in the Williamson’s ether synthesis is (I) (CH3)3CCH2Br (III) ClCH2CH2CH3

 (II) ClCH2CH CH2 (IV)  BrCH2CH2CH3

(1) II > IV > III > I (2) I > II > III > IV (3) IV > III > II > I (4) III > IV > II > I 24. What is the standard reduction potential of hydrogen electrode in contact with a solution whose pH = 10 (at 25°C)? (1) 0.581 V (2) −0.591 V (3) 0.591 V (4) 5.91 V

Mock Test III.indd 840

27. The weaker p-bond formation is responsible for the non-existence of which of the following compounds as discrete molecule? (1) CO2 (2) CS2 (3) CSe2 (4) None of these

(1) S2 (3) 6 S2

(1) 99° (2) 120° (3) 93° (4) 180° Isomerization

s 2s < s * 2s < s 2pz < (p 2px ≈ p 2p y ) < (p * 2px ≈ p * 2p y ) < p * 2pz

28. For a sparingly soluble salt A2B3, the relationship of its solubility products (Ksp) with its solubility (S) is given by

20. In ethene, the bond angle(s) is/are

40% H2SO4

(1) Be2 is not a stable molecule. (2) He2 is not stable but He+2 is expected to exist. (3) Bond strength of N2 is maximum amongst the homonuclear diatomic molecules belonging to the second period. (4) The order of energies of molecular orbitals in N2 is

(2) 108 S2 (4) S5

29. The reducing power of a metal depends on various factors. Suggest the factor which makes Li, the strongest reducing agent in aqueous solution. (1) (2) (3) (4)

Sublimation enthalpy. Ionization enthalpy. Hydration enthalpy. Electron gain enthalpy.

30. Calculate q for the isothermal reversible expansion of 1 mol of an ideal gas from initial pressure 1 bar to final pressure 0.1 bar at constant temperature of 273 K. (1) −5227 J (2) +5227 J (3) 3120 J (4) None 31. The magnetic moment of Am5+ (At. no. 95) is (1)

24 BM (2)

35 BM

(3)

15 BM (4)

3 BM

32. Which of the following amounts to repeated distillations? (1) Vacuum distillation (2) Steam distillation (3) Fractional distillation (4) Simple distillation 33. Which of the following coordination compounds would exhibit optical isomerism? (1) trans-Dicyanobis(ethylenediamine) chromium (III) chloride (2) tris-Ethylenediamine cobalt (III) bromide

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841

Mock Test III (3) Pentaamminenitrocobalt (III) iodide (4) Diamminedicholoplatinum (II)

(1) I > II > III > IV (2) II > III > IV > I (3) II > I > IV > III (4) III > II > I > IV

34. The complex ion having minimum magnitude of D0 (CFSE) is

41. Which of the following statements is correct regarding self-reduction?

(1) [CoCl 6 ]3- (2) [Co(CN)6 ]3(3) [Cr(H 2O)6 ]3+ (4) [Co(NH 3 )6 ]3+

(1) Partial roasting and self-reduction occur together. (2) First self-reduction occurs followed by partial roasting. (3) First partial roasting occurs followed by self-­ reduction. (4) Partial roasting is done in the reverberatory furnace first, and then self-reduction is done in the blast furnace.

35. Which one of the following is most acidic compound? COOH

COOH

(1)

(2) Cl

42. Which of the following is the correct statement regarding nucleophile?

NO2 COOH

(1) It is positively charged species and can form a bond by donating a pair of electron to another nucleophile. (2) It is generally neutral species only and can form a bond by donating a pair of electron to electrophile. (3) It can be either negatively or neutral charged species and can form a bond by donating a pair of electron to electrophile. (4) It is positively charged species and can form a bond by donating a pair of electron to electrophile.

SO3H

(3)

(4) Cl CN

36. Perxenate ion is (1) XeO64- (2) HXeO6(3) XeO24 (4) XeO 4

37. Which of the following reaction is appropriate for converting ethylamine to ethyl isocyanide? (1) (2) (3) (4)

Stephens reaction Carbylamine reaction Hofmann bromamide reaction Gabriel pthalimide synthesis

38. Identify the compound that cannot give Cannizzaro reaction. (1) CCl3CHO (2) (CH3)2CHCHO (3) (CH3)3CCHCl2 (4) C6H5CHO 39. Which one of the following metal ions is essential inside the cell for the metabolism of glucose/synthesis of proteins? (1) Na+ (2) K+ (3) Mg2+ (4) Ca2+ 40. The correct order of C compounds H2N

(I)

O

H2N

MeO

(II)

(1) CH3CH2CCl2CH2CH3 (2) CH3CH2CH2CCl2CH3 (3) CH3CH2CH2CH2CHCl2 (4) CH3CH2CH2CHClCH2Cl 44. The IUPAC name of the compound H2N

CH



COOH COOH

is (1) (2) (3) (4)

CH

CHO

3-amino-2-formyl butane. 3-amino-2, 3-dicarboxy propanal. 2-amino-3-formyl butane-1, 4-dioic acid. 1-amino-2-formyl propanoic acid.

45. If atomic radius of barium is 217.6 pm and it crystallizes in bcc unit cell, the edge length of unit cell is

N bond length in the following H2N

43. An organic compound A of the molecular formula C5H10Cl2 is hydrolyzed to compound B, C5H10O, which gives an oxime with hydroxylamine and yellow precipitate with a mixture of iodine and sodium hydroxide. The compound A should

(1) 351.2 pm (2) 248.3 pm (3) 503.1 pm (4) None of these.

H2N (III)

(IV)

O NO2

Mock Test III.indd 841

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Mock Test III.indd 842

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Answer Key Mock Test I  1. (2)

 2. (2)

 3. (1)

 4. (2)

 5. (3)

 6. (2)

 7. (4)

 8. (4)

 9. (1)

10. (4)

11. (3)

12. (1)

13. (3)

14. (4)

15. (3)

16. (4)

17. (1)

18. (2)

19. (2)

20. (2)

21. (1)

22. (1)

23. (1)

24. (3)

25. (4)

26. (2)

27. (2)

28. (3)

29. (3)

30. (1)

31. (2)

32. (3)

33. (2)

34. (4)

35. (2)

36. (4)

37. (3)

38. (3)

39. (3)

40. (1)

41. (1)

42. (2)

43. (1)

44. (4)

45. (1)

Mock Test II  1. (1)

 2. (3)

 3. (4)

 4. (3)

 5. (1)

 6. (4)

 7. (3)

 8. (2)

 9. (2)

10. (1)

11. (2)

12. (1)

13. (3)

14. (1)

15. (4)

16. (1)

17. (4)

18. (3)

19. (1)

20. (4)

21. (4)

22. (4)

23. (4)

24. (1)

25. (3)

26. (2)

27. (4)

28. (2)

29. (2)

30. (3)

31. (4)

32. (3)

33. (1)

34. (3)

35. (2)

36. (2)

37. (1)

38. (1)

39. (1)

40. (4)

41. (3)

42. (2)

43. (2)

44. (2)

45. (1)

Mock Test III  1. (3)

 2. (1)

 3. (2)

 4. (2)

 5. (4)

 6. (4)

 7. (1)

 8. (3)

 9. (3)

10. (1)

11. (1)

12. (3)

13. (1)

14. (1)

15. (3)

16. (4)

17. (1)

18. (4)

19. (4)

20. (2)

21. (4)

22. (1)

23. (1)

24. (2)

25. (3)

26. (4)

27. (3)

28. (2)

29. (3)

30. (2)

36. (1)

37. (2)

38. (2)

39. (3)

40. (2)

31. (1)

32. (3)

33. (2)

34. (1)

35. (4)

41. (4)

42. (3)

43. (2)

44. (3)

45. (3)

Answere Keys_Mock Test.indd 843

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NEET–2018 CHEMISTRY PAPER 1. The correct order of N-compounds in its decreasing order of oxidation states is (1) HNO3, NH4Cl, NO, N2 (2) HNO3, NO, NH4Cl, N2 (3) HNO3, NO, N2, NH4Cl (4) NH4Cl, N2, NO, HNO3 Solution (3) The decreasing order of oxidation states of N-compounds is as follows: ( +5 )

( +2 )

0

( −3 )

H N O3 > N O > N 2 > N H 4 Cl 2. Which one of the following elements is unable to form MF63− ion? (1) B (2) Al (3) Ga

(4) In

Solution (1) Boron does not contain vacant d orbitals hence it cannot extend its covalency beyond 4. 3. Considering Ellingham diagram, which of the following metals can be used to reduce alumina? (1) Mg (2) Zn (3) Fe (4) Cu Solution (1) In a number of processes, one metal is used to reduce the oxide of another metal. Any metal will reduce the oxide of other metals which lie above it in the Ellingham diagram because the Gibbs energy will become more negative by an amount equal to the difference between the two graphs at that particular temperature. Mg is below Al2O3 in the Ellingham diagram hence it can reduce alumina. 4. The correct order of atomic radii in group 13 elements is (1) B < Ga < Al < Tl < In (2) B < Al < Ga < In < Tl (3) B < Al < In < Ga < Tl (4) B < Ga < Al < In < Tl Solution (4) The metallic radii of the atoms do not increase regularly on descending the group. Ga, In and Tl follow immediately after a row of ten transition elements. They therefore have ten d electrons, which are less efficient at shielding the nuclear charge than the s and p electrons. (Shielding is in the order s > p > d > f.) Poor shielding of the nuclear charge results in the outer electrons being more firmly held by the nucleus. Thus atoms with a d10 inner shell are smaller than would otherwise be expected. Hence, the correct order is B < Ga < Al < In < Tl. 5. Which of the following statements is not true for halogens? (1) All but fluorine show positive oxidation states. (2) All are oxidizing agents. (3) All form monobasic oxyacids. (4) Chlorine has the highest electron-gain enthalpy. Solution (1) The ionization energy for fluorine is appreciably higher than for the others, because of its small size. F always has an oxidation state of (−I) except in F2. It forms compounds either by gaining an electron to form F−, or by sharing an electron to form a covalent bond. 6. In the structure of ClF3, the number of lone pairs of electrons on central atom ‘Cl’ is (1) four. (2) two. (3) one. (4) three.

NEET_2018.indd 1

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P-2     NEET–2018 Chemistry Paper

Solution (2) The structure of ClF3 is as follows:

F F

Cl F

There are two lone pairs of electrons on central atom Cl. 7. Identify the major products P, Q and R in the following sequence of reactions: + CH3CH2CH2Cl

Q

P

OH

,

,

CH2CH2CH3 (3)

R

CH(CH3)2 (1)

,

Anhydrous AlCl3

P

1. O2

Q+R

2. H3O+/D

P CH2CH2CH3

CH3CH(OH)CH3

(2)

Q CHO

R COOH

,

,

CHO

OH ,

CH3CH2

OH

CH(CH3)2

(4)

, CH3

,

CO

CH3

Solution (4)

CH3

CH3 CH

CH3

CH3

O

H

O

OH O

1. O2

CH3CH2CH2Cl AlCl3 (Friedel crafts alkylation)

C

2. H3O+/D

+ CH3 (Q)

(P) Cumene

8. Which of the following compounds can form a zwitterion? (1) Benzoic acid (2) Acetanilide (3) Aniline

C

CH3

(R)

(4) Glycine

Solution (4) Amino acids exist as dipolar ions, a form in which the carboxyl group is present as a carboxylate ion, −CO2−, and the amino group is present as an aminium ion, − NH 3+. Dipolar ions are also called zwitterions. Among the given compounds, glycine can exist as zwitterion. +

-

H2NCH2CO2H

H3NCH2CO2

Glycine

Zwitterion

9. Regarding cross-linked or network polymers, which of the following statements is incorrect? (1) Examples are bakelite and melamine. (2) They are formed from bi-and tri-functional monomers. (3) They contain covalent bonds between various linear polymer chains. (4) They contain strong covalent bonds in their polymer chains.

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NEET–2018 Chemistry Paper     P-3

Solution (4) When the functionality of monomer is 3 and above, cross-linked three-dimensional network of polymer is formed. They are linked by strong covalent bonds. In fact, whenever a multifunctional monomer is polymerized, the polymer ultimately forms a network polymer. For example, bakelite, urea–formaldehyde, etc. 10. Nitration of aniline in strong acidic medium also gives meta-nitroaniline because (1) In absence of substituents nitro group always goes to meta-position. (2) In electrophilic substitution reactions amino group is meta-directive. (3) In spite of substituents nitro group always goes to only meta-position. (4) In acidic (strong) medium aniline is present as anilinium ion. Solution (4)

+ NH3

NH2 H+

Aniline

Anilinium ion

- NH 3+ is a meta directing group, hence besides ortho and para, meta product also formed. 11. The difference between amylose and amylopectin is (1) amylopectin have 1 → 4 a-linkage and 1 → 6 b-linkage. (2) amylose have 1 → 4 a-linkage and 1 → 6 b-linkage. (3) amylopectin have 1 → 4 a-linkage and 1 → 6 a-linkage. (4) amylose is made up of glucose and galactose. Solution (3) Amylose typically consists of more than 1000 D-glucopyranoside units connected in a linkages between C1 of one unit and C4 of the next. Amylopectin has a structure similar to that of amylose [i.e., a (1 → 4) links], except that in amylopectin the chains are branched. Branching takes place between C6 of one glucose unit and C1 of another and occurs at intervals of 20–25 glucose units. 12. A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H2SO4. The evolved gaseous mixture is passed through KOH pellets. Weight (in g) of the remaining product at STP will be (1) 2.8 (2) 3.0 (3) 1.4 (4) 4.4 Solution (1)

2.3 = 0.05 mol 46 4.5 Number of moles of oxalic acid = = 0.05 mol 90 Number of moles of formic acid =

The reactions involved are as follows: HCOOH Initial moles Finall moles

0.05 mol 0 H 2 C2 O4

Initial moles Final moles

NEET_2018.indd 3

0.05 mol 0

H 2 SO4 ¾conc. ¾¾¾ ® CO(g) + H 2 O(l)

0 0.05 mol

0 0.05 mol

H 2 SO4 ¾conc. ¾¾¾ ® CO 2 (g) + CO(g) + H 2 O(l)

0 0 0 0.05 mol 0.05 mol 0.05 mol

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P-4     NEET–2018 Chemistry Paper

KOH will absorb CO2 and H2SO4 will absorb water. Hence, the remaining product will be CO. Total number of moles of CO = (0.05 + 0.05) mol = 0.1 mol Therefore, weight of CO = 0.1 mol × 28 g mol−1 = 2.8 g 13. Which of the following oxides is most acidic in nature? (1) BaO (2) BeO (3) MgO

(4) CaO

Solution (2) As the atoms get larger, the ionization energy decreases and the elements also become more basic. Therefore, the basic strength increases on moving down the group while acidic strength decreases. Hence, BeO will be most acidic. 14. Which oxide of nitrogen is not a common pollutant introduced into the atmosphere both due to natural and human activity? (1) N2O (2) NO2 (3) N2O5 (4) NO Solution (3) The major nitrogen oxide produced from soil micro-organisms includes nitrous oxide (N2O). Major anthropogenic releases of nitrogen oxides come primarily from fuel combustion, biomass burning and certain production processes. These processes emit a mixture of nitric oxide (NO) and nitrogen dioxide (NO2). Thus, N2O5 is the least common pollutant.  15. The compound A on treatment with Na gives B, and with PCl5 gives C. B and C react together to give diethyl ether. A, B and C are in the order (1) C2H5Cl, C2H6, C2H5OH (2) C2H5OH, C2H5Cl, C2H5ONa (3) C2H5OH, C2H6, C6H5Cl (4) C2H5OH, C2H5ONa, C2H5Cl Solution (4) C2H5OH

Na

C2H5O-Na+

(A)

(B)

PCl5

C2H5Cl (C) C2H5O-Na+

+

(B)

C2H5Cl

C2H5OC2H5

(C)

16. The compound C7H8 undergoes the following reactions: Cl2 / D / HCl 2 / Fe C7 H8 3 → A Br  → B Zn  →C

The product ‘C’ is (1) 3-bromo-2,4,6-trichlorotoluene. (3) m-bromotoluene.

(2) o-bromotoluene. (4) p-bromotoluene.

Solution (3)

CH3

CCl3 3Cl2 /D

CH3

CCl3 Br2

Zn

Fe

Toluene

(A)

HCl

Br

Br (B)

m-Bromotoluene (C)

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NEET–2018 Chemistry Paper     P-5

17. Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is converted to gaseous hydrocarbon containing less than four carbon atoms. (A) is (1) CH3—CH3 (2) CH2   CH2 (3) CH   CH (4) CH4 Solution (4) Formation of higher alkanes from alkyl halides in the presence of sodium and dry ether is named Wurtz reaction. CH4 (A) Methane

Br2

CH3Br

hn

Na/dry ether

CH3

Wurtz reaction

CH3

18. Which of the following molecules represents the order of hybridization sp2, sp2, sp, sp from left to right atoms? (1) CH2   CH—CH   CH2 (2) CH2   CH—C   CH (3) HC   C—C   CH (4) CH3—CH   CH—CH3 Solution (2) (a) Carbon–carbon double bonds are comprised of sp2-hybridized carbon atoms. (b) Carbon–carbon triple bonds are comprised of sp-hybridized carbon atoms. sp2

CH2

sp2

CH

sp

C

sp

CH

19. Which of the following carbocations is expected to be most stable? (1)

NO2

NO2

(2)

(3)

NO2

(4)

NO2 H Y

H Y

Y

H

Y

H

Solution (1) Structure (1) is most stable because —I effect of NO2 is exerted to least extent in (1), which make the carbocation most stable. 20. Which of the following is correct with respect to —I effect of the substituents? (R = alkyl) (1) —NH2 > —OR > —F (2) —NR2 < —OR < —F (3) —NH2 < —OR < —F (4) —NR2 > —OR > —F Solution (2), (3) —I effect increases with increase in the electronegativity of the atom. Fluorine is most electronegative followed by alkoxy group. The electronegativity of amine and their derivative would be less than that of alkoxy group as oxygen is more electronegative than that of nitrogen. 21. In the reaction OH

O-Na+ CHO

+ CHCl3 + NaOH

The electrophile involved is +

(1) dichloromethyl anion ( C HCl 2 ) +

(3) dichloromethyl cation ( C HCl 2 )

NEET_2018.indd 5

+

(2) formyl cation ( C HO) (4) dichlorocarbene ( :CCl 2)

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P-6     NEET–2018 Chemistry Paper

Solution (4) The reaction is Reimer–Tiemann reaction. The electrophile involved in the reaction is dichlorocarbene ( :CCl 2). + NaCCl3 + H2O

Fast

CHCl3 + NaOH

CCl3

Slow

CCl2 + Cl-

22. Carboxylic acids have higher boiling points than aldehydes, ketones and even alcohols of comparable molecular mass. It is due to their (1) more extensive association of carboxylic acid via van der Waals force of attraction. (2) formation of carboxylate ion. (3) formation of intramolecular H-bonding. (4) formation of intermolecular H-bonding. Solution (4) Carboxylic acids generally have boiling points higher than other types of organic compounds of comparable molecular weights, such as alcohols, aldehydes and ketones. For example, butanoic acid has a higher boiling point than either 1-pentanol or pentanal. The higher boiling points of carboxylic acids result from their polarity and from the fact that they form very strong intermolecular hydrogen bonds which are not broken completely even in the vapor phase. In the liquid and solid states, carboxylic acids are associated by intermolecular hydrogen bonding into dimers, as shown for acetic acid. Hydrogen bonding in the dimer d-

O

H3C

d+

O

H

C

C O

CH3

O

H

d+

d-

23. Compound A, C8H10O, is found to react with NaOI (produced by reacting Y with NaOH) and yields a yellow precipitate with characteristic smell. A and Y are respectively (1)

CH

CH3 and I2

OH

(3)

H3C

CH2

CH2

(2)

OH and I2

CH3 CH2

OH and I2

(4)

OH and I2

CH3

Solution (1) All compounds containing the grouping CH3CHOH when treated with a halogen and excess of alkali (i.e., sodium hypohalite, NaOX) form haloforms. If the halogen used is iodine yellow precipitates of iodoform are formed and the reaction is called iodoform reaction. I2 + NaOH

NaOI + NaI + H2O

(Y)

CH (A)

CH3

NaOI

C

CH3

O

OH

I2 /NaOH

O C

ONa + CHI3 ¯

Sodium benzoate

NEET_2018.indd 6

(Yellow ppt.)

23-05-2018 19:35:44

NEET–2018 Chemistry Paper     P-7

24. Match the metal ions given in Column I with the spin magnetic moments of the ions given in Column II and assign the correct code: Column I

(p)

8 BM

3+

(q)

35 BM

3+

(r)

3 BM

2+

(s)

24 BM

(t)

15 BM

(a) Co (b) Cr (c) Fe

(d) Ni

(a)

(b)

(c)

(d)

(1)

s

p

q

r

(2)

p

q

r

s

(3)

s

t

q

p

(4)

r

t

p

q

Column II

3+

Solution (3) Spin magnetic moment can be calculated as m = n(n + 2) where n is the number of unpaired electrons. Co3+ = [Ar]3d6; Unpaired electrons (n) = 4; m = 4(4 + 2) = 24 BM Cr3+ = [Ar]3d3; Unpaired electrons (n) = 3; m = 3(3 + 2) = 15 BM Fe3+ = [Ar]3d5; Unpaired electrons (n) = 5; m = 5(5 + 2) = 35 BM Ni2+ = [Ar]3d8; Unpaired electrons (n) = 2; m = 2(2 + 2) = 8 BM 25. Which one of the following ions exhibits d-d transition and paramagnetism as well? 2− 2− (1) MnO 4− (2) CrO2− 7 (3) CrO 4 (4) MnO 4

Solution (4)

MnO 4− : Mn 7 + = [Ar ]d 0 ; Unpaired electrons = 0; Diamagnetic



Cr2 O27 − : Cr 6 + = [Ar ]d 0 ; Unpaired electrons = 0; Diamagnetic



CrO24 − : Cr 6 + = [Ar ]d 0 ; Unpaired electrons = 0; Diamagnetic



MnO24 − : Mn 6 + = [Ar ]d 1 ; Unpaired electrons = 1; Paramagnetic

Ions containing d 0 and d 10 configuration do not exhibits d-d transition. 26. Iron carbonyl, Fe(CO)5 is (1) trinuclear. (2) mononuclear.

(3) tetranuclear.

(4) dinuclear.

Solution (2) Based on the number of metal atoms present in a complex, they can be categorized into mononuclear, dinuclear, trinuclear and tetranuclear. Hence, Fe(CO)5 is a mononuclear complex. 27. The type of isomerism shown by the complex [CoCl2(en)2] is (1) ionization isomerism. (2) coordination isomerism. (3) geometrical isomerism. (4) linkage isomerism.

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P-8     NEET–2018 Chemistry Paper

Solution (3) Complex [CoCl2(en)2] shows geometrical isomerism, that is, cis and trans. Cl en

Cl

Co

en

en

Cl

Co en

Cl Cis

Trans

28. The geometry and magnetic behavior of the complex [Ni(CO)4] are (1) square planar geometry and paramagnetic. (2) tetrahedral geometry and diamagnetic. (3) square planar geometry and diamagnetic. (4) tetrahedral geometry and paramagnetic. Solution (2) [Ni(CO)4] : The arrangement of electrons in Ni0 (3d8 4s2) will be 3d

4s

4p

Ni [Ni(CO)4] sp3 hybridization (Tetrahedral shape)

29. Following solutions were prepared by mixing different volumes of NaOH and HCl of different concentrations: M M M M HCl + 40 mL NaOH ( II) 55 mL HCl + 45 mL NaOH 10 10 10 10 M M M M  (III) 75 mL HCl + 25 mL NaOH (IV) 100 mL HCl + 100 mL NaOH 5 5 10 10    (I) 60 mL

pH of which one of them will be equal to 1? (1) (IV) (2) (I)

(3) (II)

(4) (III)

Solution M M HCl + 25 mL NaOH is (75 + 25) mL = 100 mL 5 5 M M 25 mL of NaOH will neutralize 25 mL of HCl 5 5 (4) Total volume of 75 mL

Remaining

M HCl = 75 − 25 = 50 mL 5

[H+] = [HCl] =

M 50 M × = 5 100 10

pH can be calculated as   pH = −log[H+] = −log

1 =1 10

30. On which of the following properties does the coagulating power of an ion depend? (1) Both magnitude and sign of the charge on the ion. (2) Size of the ion alone. (3) The magnitude of the charge on the ion alone. (4) The sign of charge on the ion alone.

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NEET–2018 Chemistry Paper     P-9

Solution (1) According to Hardy-Schulze rule, greater the valency of the active ion or flocculating ion, greater will be its coagulating power. 31. Given van der Waals constant for NH3, H2, O2 and CO2 are respectively 4.17, 0.244, 1.36 and 3.59, which one of the following gases is most easily liquefied? (1) O2 (2) H2 (3) NH3 (4) CO2 Solution (3) a is a van der Waal’s constant. The value of a depends upon intermolecular force of attraction. Higher the value of a, easier will be the liquefaction of gas. Therefore, NH3 will be liquefied most easily. 32. The solubility of BaSO4 in water is 2.42 × 10−3 gL−1 at 298 K. The value of its solubility product (Ksp) will be (Given molar mass of BaSO4 = 233 g mol−1) (1) 1.08 × 10−14 mol2 L−2 (2) 1.08 × 10−12 mol2 L−2 −10 2 −2 (3) 1.08 × 10 mol L (4) 1.08 × 10−8 mol2 L−2 Solution (3) Solubility of BaSO4 (in mol L−1) =

2.42 ´ 10 -3 g L-1 = 1.04 ´ 10 -5 ( mol L-1 ) 233 g mol -1

     BaSO 4 (s)  Ba 2 + (aq ) + SO24 - (aq ) -5 -5 2+ mol L-1 and [SO2− mol L-1. SubstitutThe equilibrium concentrations of Ba2+ and SO2− 4 are [Ba ] = 1.04 ´ 10 4 ] = 1.04 ´ 10 ing the values in the expression for solubility product, we get

K sp = [ Ba 2 + ][SO24 − ] = (1.04 × 10 −5 )2 = 1.08 × 10 −10 mol 2 L−2 33. In which case is the number of molecules of water maximum? (1) 0.00224 L of water vapors at 1 atm and 273 K (2) 0.18 g of water (3) 18 mL of water (4) 10−3 mol of water Solution 0.00224 = 10 −4 mol 22.4       Number of molecules of water = NA × Number of moles (3) Option (1): Number of moles of water =

         = 10−4 NA Option (2): Number of molecules of water =

0.18 g N A = 10 -2 N A 18 g mol -1

Option (3): Number of molecules of water =

18 g NA = NA 18 g mol -1

Option (4): Number of molecules of water = 10 −3 N A 34. The correct difference between first and second order reactions is that (1) a first order reaction can be catalyzed; a second order reaction cannot be catalyzed. (2) the half-life of a first-order reaction does not depend on [A]0; the half-life of a second order reaction does depend on [A]0.

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P-10     NEET–2018 Chemistry Paper

(3) the rate of a first order reaction does not depend on reactant concentrations; the rate of a second order reaction does depend on reactant concentrations. (4) the rate of a first order reaction does depend on reactant concentrations; the rate of a second order reaction does not depend on reactant concentrations. Solution 0.693 k Thus, the half-life time in a first order reaction is independent of concentration of the reactant. (2) For a first order reaction, half-life is given by t1/ 2 =

For a second order reaction, half-life is given by t1/ 2 =

1 k[ A]0

Hence, t1/2 of second order reaction is inversely proportional to the initial concentration of reactant. 35. Among CaH2, BeH2, BaH2, the order of ionic character is (1) BeH2 < BaH2 < CaH2 (2) CaH2 < BeH2 < BaH2 (3) BeH2 < CaH2 < BaH2 (4) BaH2 < BeH2 < CaH2 Solution (3) On moving down the group, the metallic character of the metals increases, hence ionic character of metal hydrides increases. 36. Consider the change in oxidation state of bromine corresponding to different emf values as shown in the diagram below: BrO-4 Br-

1.82 V

BrO3-

1.0652 V

Br2

1.5 V

HBrO

1.595 V

Then the species undergoing disproportionation is (2) BrO 4− (3) BrO3−

(1) Br2

(4) HBrO

Solution +1 0 (4) H Br O → Br2 ; E o = 1.595 V +1

+5

H Br O ® BrO3- ; E o = -1.5 V o Ecell for the disproportionation of HBrO can be calculated as o o o = EHBrO/Br − EBrO Ecell − /HBrO 2 3

= 1.595 − 1.5 = 0.095 V o o Since, the value of Ecell is positive, DG = –nFEcell therefore, HBrO will undergo disproportionation.

37. For the redox reaction MnO 4- + C2 O24 - + H + ® Mn 2 + + CO2 + H 2 O The correct coefficients of the reactants for the balanced equation are MnO 4−

C2 O2− 4

H+

(1)

2

16

5

(2)

2

5

16

(3)

16

5

2

(4)

5

16

2

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NEET–2018 Chemistry Paper     P-11

Solution (2) The balanced redox reaction is 2MnO 4− + 5C2 O24 − + 16 H + → 2 Mn 2 + + 10CO2 + 8H 2 O 38. Which one of the following conditions will favor maximum formation of the product in the reaction

A 2 (g) + B2 (g)  X 2 (g) ; D r H = - X kJ?

(1) High temperature and high pressure. (3) Low temperature and high pressure.

(2) Low temperature and low pressure. (4) High temperature and low pressure.

Solution (3) We have A 2 (g) + B2 (g)  X 2 (g) ; D r H = - X kJ If we remove heat or decrease the temperature, the reaction shifts in a direction to increase that heat. Therefore, for an exothermic reaction, cooling the reaction mixture increases the concentration of products when equilibrium is re-established. An increase in pressure shifts the equilibrium in the direction where the number of moles decreases. As there are two moles of gaseous reactant and one mole of gaseous product. Therefore, the increase in pressure shifts the equilibrium in the forward reaction. 39. When initial concentration of the reactant is doubled, the half-life period of a zero order reaction (1) is tripled. (2) is doubled. (3) is halved. (4) remains unchanged. Solution (2) Half-life for zero order reaction is given by t1/ 2 =

[ A]0 2k

The half-life time in a zero order reaction is directly proportional to the initial concentration of the reactants. Hence, half-life will be doubled on doubling the concentration of reactant. 40. The bond dissociation energies of X2, Y2 and XY are in the ratio of 1: 0.5: 1. ∆H for the formation of XY is −200 kJ mol −1. The bond dissociation energy of X2 will be (1) 800 kJ mol −1 (2) 100 kJ mol −1 (3) 200 kJ mol −1 (4) 400 kJ mol −1 Solution (1)

1 1 X 2 (g) + Y2 (g) → XY(g) 2 2

The ratio of bond dissociation energy of X2, Y2 and XY is z: 0.5 z: z We know

D r H o = ∑ BE reactant − ∑ BE product

 z z = +  −z  2 4     z −200 = − ⇒ z = 800 4 Bond dissociation energy of X2 = z = 800 kJ mol −1 41. The correction factor ‘a’ to the ideal gas equation corresponds to (1) electric field present between the gas molecules. (2) volume of the gas molecules. (3) density of the gas molecules. (4) forces of attraction between the gas molecules. Solution (4) The correction factor ‘a’ to the ideal gas equation corresponds to forces of attraction between the gas molecules. Higher the value of a greater is the intermolecular force of attraction.

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P-12     NEET–2018 Chemistry Paper

42. Consider the following species: CN+, CN−, NO and CN Which one of these will have the highest bond order? (1) CN+ (2) CN− (3) NO

(4) CN

Solution NO = s 1s 2 s * 1s 2 s 2 s 2 s * 2 s 2 s 2 pz2 p 2 px2 = p 2 py2p * 2 p1x = p * 2 py0

(2)

  Bond order = (Nb − Na)/2 = (10 − 5)/2 = 5/2 = 2.5 CN = s 1s 2 s * 1s 2 s 2 s 2 s * 2 s 2 p 2 px2 = p 2 py2s 2 p1z



  Bond order = (Nb − Na)/2 = (9 − 4)/2 = 5/2 = 2.5 CN − = s 1s 2 s * 1s 2 s 2 s 2 s * 2 s 2 p 2 px2 = p 2 py2s 2 pz2



  Bond order = (Nb − Na)/2 = (10 − 4)/2 = 6/2 = 3 CN + = s 1s 2 s * 1s 2 s 2 s 2 s * 2 s 2 p 2 px2 = p 2 py2



  Bond order = (Nb − Na)/2 = (8 − 4)/2 = 4/2 = 2 43. Magnesium reacts with an element (X) to form an ionic compound. If the ground state electronic configuration of (X) is 1s2 2s2 2p3, the simplest formula for this compound is (1) Mg2X (2) MgX2 (3) Mg2X3 (4) Mg3X2 Solution (4) Electronic configuration of X is 1s2 2s2 2p3. Therefore, the valency of X is 3 while the valency of Mg is 2. Hence, Mg3X2 compound will be formed. 44. Iron exhibits bcc structure at room temperature. Above 900°C, it transforms to fcc structure. The ratio of density of iron at room temperature to that at 900°C (assuming molar mass and atomic radii of iron remains constant with temperature) is (1)

3 3 4 2

(2)

4 3 3 2

(3)

3 2

(4)

1 2

Solution (1) Density is given by r =

z×M a3 × N A 4r

For bcc,

z = 2, a =

For fcc,

z = 4, a = 2 2r

Therefore,

zM / a 3 N A r25° C = r900° C zM / a 3 N A

3

( (

) )

2  2 2r  =  4  4r / 3      3 3 = 4 2

bcc fcc

3

45. Which one is a wrong statement? (1) The electronic configuration of N atom is

↓

↓

↓

↓

2p1x 2p1y 2p1z

↓

NEET_2018.indd 12

2s2

↓



1s2

↓

23-05-2018 19:35:55

NEET–2018 Chemistry Paper     P-13

(2) An orbital is designated by three quantum numbers while an electron in an atom is designated by four quantum numbers. (3) Total orbital angular momentum of electron in ‘s’ orbital is equal to zero. (4) The value of m for d x2 is zero. Solution (1) According to Hund’s rule, the degenerate orbitals (i.e. orbitals having same energy) will be filled by one electron each having same spin and then only pairing of electrons will take place. Hence the correct electronic configuration of N will be:

↓

↓

2s2

2p3

↓

1s2

↓ ↓

N :

↓

↓

7

or

↓ NEET_2018.indd 13

↓

↓

↓

↓

↓ ↓

23-05-2018 19:35:55

NEET_2018.indd 14

23-05-2018 19:35:56

Chapter at a Glance with flowcharts and tables for quick revision of all important concepts.

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Solved Examples placed in the same order as the topics in the feature Chapter at a Glance.

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Practice Exercises based on latest NEET pattern in big number, arranged topic-wise.

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Previous Years' NEET Questions (2007-2017) with solutions covered chapter-wise.

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Hints and Explanations for tricky and difficult questions.

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Three Mock Tests to develop examination temperament.

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Solved NEET 2018 Chemistry paper

Chapter at a Glance, an attractive feature with concepts summarized in a systematic flow, has been incorporated to enhance the quick-learning. The enormous number of practice exercises based on latest NEET pattern has been extensively developed based on NCERT Chemistry books of Class 11 and 12. These are arranged topic-wise under two levels of difficulty and include Previous Years’ NEET Questions. The designing of questions is strictly in accordance with the topic that aids students in approaching the corresponding problems of the topic under study. Apart from the Answer Key, the distinctive feature, Hints and Explanations of tricky and difficult questions have also been included to simplify learning. At the end of the book, three Mock tests have been provided to give the students an experience of attempting the actual examination. About Maestro Series l Idea: World-class content developed by “Master teachers” adapted to the needs of medical aspirants. l

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Objective

CHEMISTRY

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SINGH MEHTA ASOKAN

Features of the book include: Focused on NEET examination Chapters summarized in a systematic flow for quick revision Selected topic-wise Practice Questions to cover all important concepts Previous Years' NEET Questions (2007-2017) with solutions covered chapter-wise Three NEET Mock Tests for self-evaluation Includes solved NEET 2018 Chemistry paper

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Objective CHEMISTRY

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ABOUT THE BOOK The book Objective Chemistry for NEET is an endeavour to help students to prepare for NEET (National Eligibility cum Entrance Test) and other medical entrance examinations. NEET-UG has replaced All India Pre-Medical Test (AIPMT), and all individual MBBS and BDS exams conducted by states or colleges themselves for admissions into medical and dental courses.

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