Oswaal NTA CUET (UG)| Question Bank Chapterwise & Topicwise Chemistry For 2024 Exam 9789359584645, 9359584649

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For 2024 Exam

Highly Recommended

CHAPTER-WISE

QUESTION BANK Includes SOLVED PAPERS (2021 - 2023)

CHEMISTRY Section-II (Domain Specific Subject) Strictly as per the Latest Exam Pattern issued by NTA

The ONLY book you need to #AceCUET(UG)

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100% Updated

Previous Years’ Questions

Revision Notes

Concept Videos

800+ Questions

With 2023 CUET Exam Paper

(2021-2023) for Better Exam Insights

for Crisp Revision with Smart Mind Maps

for Complex Concepts Clarity

for Extensive Practice

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1st EDITION

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YEAR 2024 "9789359584645"

CUET (UG)

SYLLABUS COVERED

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This book is published by Oswaal Books and Learning Pvt Ltd (“Publisher”) and is intended solely for educational use, to enable students to practice for examinations/tests and reference. The contents of this book primarily comprise a collection of questions that have been sourced from previous examination papers. Any practice questions and/or notes included by the Publisher are formulated by placing reliance on previous question papers and are in keeping with the format/pattern/ guidelines applicable to such papers. The Publisher expressly disclaims any liability for the use of, or references to, any terms or terminology in the book, which may not be considered appropriate or may be considered offensive, in light of societal changes. Further, the contents of this book, including references to any persons, corporations, brands, political parties, incidents, historical events and/or terminology within the book, if any, are not intended to be offensive, and/or to hurt, insult or defame any person (whether living or dead), entity, gender, caste, religion, race, etc. and any interpretation to this effect is unintended and purely incidental. While we try to keep our publications as updated and accurate as possible, human error may creep in. We expressly disclaim liability for errors and/or omissions in the content, if any, and further disclaim any liability for any loss or damages in connection with the use of the book and reference to its contents”.

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Preface Welcome to the ultimate resource for your Common University Entrance Test (CUET) preparation! The Common University Entrance Test (CUET) marks a significant shift in the admission process for UG programs in Central Universities across India. The introduction of CUET aims to create a level playing field for students nationwide, regardless of their geographical location, and revolutionize the way students connect with these prestigious institutions. CUET (UG), administered by the esteemed National Testing Agency (NTA), is a prestigious all-India test that serves as a single-window opportunity for admissions. The NTA consistently provides timely notifications regarding the exam schedule and any subsequent updates. The curriculum for CUET is based on the National Council of Educational Research and Training (NCERT) syllabus for class 12 only. CUET (UG) scores are mandatory required while admitting students to undergraduate courses in 44 central universities. A merit list will be prepared by participating Universities/organizations. Universities may conduct their individual counselling on the basis of the scorecard of CUET (UG) provided by NTA. Oswaal CUET (UG) Question Bank is your strategic companion designed to elevate your performance and simplify your CUET journey for success in this computer-based test.

Here’s how this book benefits you: • 100% Updated with 2023 CUET Exam Paper • Previous years Questions (2021-2023)for Better Exam insights • Revision Notes for Crisp Revision with Smart Mind Maps • Concept Videos for complex concepts clarity • 800+ questions for Extensive Practice Almost 1.92 million candidates registered for CUET (UG) in 2023. Candidates have been quite anxious about appearing for CUET (UG), however, with the right preparation strategy and resources, you can secure a good rank in CUET (UG). We believe that with dedication, hard work, and the right resources, you can conquer CUET and secure your place in the Central Universities of your choice. Good luck with your preparations, with this trusted companion on your journey to academic success! All the best! Team Oswaal Books

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Contents l Know your CUET(UG) Exam

6 - 6

l Latest CUET (UG) Syllabus

7 - 8

l Examination Paper CUET 2023



Chapter-1 – The Solid State

11 - 15

1 - 11

Chapter-2 – Solutions

12 - 22

Chapter-3 – Electrochemistry

23 - 33

Chapter-4 – Chemical Kinetics

34 - 45

Chapter-5 – Surface Chemistry

46 - 55

Chapter-6 – General Principles and Processes of Isolation of Elements

56 - 66

Chapter-7 – p-Black Elements

67 - 76

Chapter-8 – d and f Block Elements 

77 - 86

Chapter-9 – Coordination Compounds 

87 - 95

Chapter-10 – Haloalkanes and Haloarenes 

96 - 105

Chapter-11 – Alcohols, Phenols and Ethers

106 - 117

Chapter-12 – Aldehydes, Ketones and Carboxylic Acid

118 - 130

Chapter-13 – Organic Compounds Containing Nitrogen

131 - 144

Chapter-14 – Biomolecules

145 - 155

Chapter-15 – Polymers

156 - 166

Chapter-16 – Chemistry in Everyday Life

167 - 176

qqq

LOOK OUT FOR 'HIGHLY LIKELY QUESTIONS'

These questions are selected by Oswaal Editorial Board. These are highly likely to be asked in the upcoming examinations.

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2 Languages + 6 Domain Specific Subject + General Test

OR

3 Languages + 5 Domain Specific Subjects + General Test

Subject/Language Choice

Objective Type with MCQs

CBT

Mode of Test

Test Pattern

Know Your CUET (UG) Exam SECTIONS

SECTION I (A) 13 Languages

Tested through reading Comprehension (i) Factual (ii) Literary (iii) Narrative

SECTION III SECTION I (B)

SECTION II

20 Languages

Domain Specific Subjects ( 27 Subjects)

General Test (Compulsory)

INCLUDES : NCERT Model syllabus (only of 12th Standard) is available on all the Subjects

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• • • • • •

General Knowledge Current Affairs General Mental Ability Numerical Ability Quantitative Reasoning Logical & Analytical Reasoning

Latest CUET (UG) Syllabus CHEMISTRY - 306 Note: There will be one Question Paper which will have 50 questions out of which 40 questions need to be attempted. Unit I: Solid State

Unit VI: General Principles and Processes of Isolation of Elements

Classification of solids based on different binding forces: molecular, ionic covalent, and metallic solids, amorphous and crystalline solids(elementary idea), unit cell in two dimensional and three-dimensional lattices, calculation of density of unit cell, packing in solids, packing efficiency, voids, number of atoms per unit cell in a cubic unit cell, point defects, electrical and magnetic properties, Band theory of metals, conductors, semiconductors and insulators and n and p-type semiconductors.

Principles and methods of extraction – concentration, oxidation, reduction electrolytic method, and refining; occurrence and principles of extraction of aluminum, copper, zinc, and iron.

Unit VII: p-Block Elements Group 15 elements: General introduction, electronic configuration, occurrence, oxidation states, trends in physical and chemical properties; nitrogen – preparation, properties, and uses; compounds of nitrogen: preparation and properties of ammonia and nitric acid, oxides of nitrogen ( structure only); Phosphorous-allotropic forms; compounds of phosphorous: preparation and properties of phosphine, halides (PCl3, PCl5) and oxoacids (elementary idea only). Group 16 elements: General introduction, electronic configuration, oxidation states, occurrence, trends in physical and chemical properties; dioxygen: preparation, properties, and uses; classification of oxides; ozone. Sulphur – allotropic forms; compounds of sulphur: preparation, properties, and uses of sulphur dioxide; sulphuric acid: industrial process of manufacture, properties and uses, oxoacids of sulphur (structures only). Group 17 elements: General introduction, electronic configuration, oxidation states, occurrence, trends in physical and chemical properties; compounds of halogens: preparation, properties and uses of chlorine and hydrochloric acid, interhalogen compounds, oxoacids of halogens (structures only). Group 18 elements: General introduction, electronic configuration, occurrence, trends in physical and chemical properties, uses.

Unit II: Solutions Types of solutions, expression of concentration of solutions of solids in liquids, the solubility of gases in liquids, solid solutions, colligative properties – the relative lowering of vapour pressure, Raoult’s law, elevation of B.P., depression of freezing point, osmotic pressure, determination of molecular masses using colligative properties, abnormal molecular mass, Vant Hoff factor.

Unit III: Electrochemistry Redox reactions; conductance in electrolytic solutions, specific and molar conductivity variations of conductivity with concentration, Kohlrausch’s Law, electrolysis and laws of electrolysis (elementary idea), dry cell – electrolytic cells and Galvanic cells; lead accumulator, EMF of a cell, standard electrode potential, Nernst equation and its application to chemical cells. Relation between Gibbs energy change and EMF of a cell, fuel cells; corrosion.

Unit IV: Chemical Kinetics

Rate of a reaction (average and instantaneous), factors affecting rates of reaction: concentration, temperature, catalyst; order and molecularity of a reaction; rate law and specific rate constant, integrated rate equations, and half-life (only for zero and first-order reactions); concept of collision theory (elementary idea, no mathematical treatment).Activation energy, Arrhenius equation.

Unit VIII: d and f Block Elements General introduction, electronic configuration, occurrence and characteristics of transition metals, general trends in properties of the first-row transition metals – metallic character, ionization enthalpy, oxidation states, ionic radii, colour, catalytic property, magnetic properties, interstitial compounds, alloy formation. Preparation and properties of K2Cr2O7 and KMnO4. Lanthanoids – electronic configuration, oxidation states, chemical reactivity, and lanthanoid contraction and its consequences. Actinoids – Electronic configuration, oxidation states, and comparison with lanthanoids.

Unit V: Surface Chemistry Adsorption – physisorption and chemisorption; factors affecting adsorption of gases on solids; catalysis: homogenous and heterogeneous, activity and selectivity: enzyme catalysis; colloidal state: the distinction between true solutions, colloids, and suspensions; lyophilic, lyophobic multimolecular and macromolecular colloids; properties of colloids; Tyndall effect, Brownian movement, electrophoresis, coagulation; emulsions – types of emulsions.

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Contd... Unit IX Coordination Compounds

Unit XIII: Organic Compounds Containing Nitrogen

Coordination compounds: Introduction, ligands, coordination number, colour, magnetic properties and shapes, IUPAC nomenclature of mononuclear coordination compounds, bonding, Werner’s theory VBT, CFT; isomerism (structural and stereo)importance of coordination compounds (in qualitative analysis, extraction of metals and biological systems).

Amines: Nomenclature, classification, structure, methods of preparation, physical and chemical properties, uses, identification of primary secondary, and tertiary amines. Cyanides and Isocyanides – will be mentioned at relevant places in context. Diazonium salts: Preparation, chemical reactions, and importance in synthetic organic chemistry.

Unit X: Haloalkanes and Haloarenes Haloalkanes: Nomenclature, nature of C-X bond, physical and chemical properties, mechanism of substitution reactions. Optical rotation. Haloarenes: Nature of C-X bond, substitution reactions (directive influence of halogen for monosubstituted compounds only). Uses and environmental effects of – dichloromethane, trichloromethane, tetrachloromethane, iodoform, freons, DDT.

Unit XIV: Biomolecules Carbohydrates – Classification (aldoses and ketoses), monosaccharide (glucose and fructose), D-L configuration, oligosaccharides (sucrose, lactose, maltose), polysaccharides (starch, cellulose, glycogen): importance. Proteins - Elementary idea of a-amino acids, peptide bond, polypeptides, proteins, primary structure, secondary structure, tertiary structure and quaternary structure (qualitative idea only), denaturation of proteins; enzymes. Hormones –Elementary idea (excluding structure). Vitamins – Classification and functions. Nucleic Acids: DNA and RNA

Unit XI: Alcohols, Phenols, and Ethers Alcohols: Nomenclature, methods of preparation, physical and chemical properties (of primary alcohols only); identification of primary, secondary, and tertiary alcohols; mechanism of dehydration, uses, with special reference to methanol and ethanol. Phenols: Nomenclature, methods of preparation, physical and chemical properties, acidic nature of phenol, electrophilic substitution reactions, uses of phenols. Ethers: Nomenclature, methods of preparation, physical and chemical properties, uses.

Unit XV: Polymers Classification – Natural and synthetic, methods of polymerization (addition and condensation), copolymerization. Some important polymers: natural and synthetic like polythene, nylon, polyesters, bakelite, rubber. Biodegradable and nonbiodegradable polymers.

Unit XII: Aldehydes, Ketones, and Carboxylic Acids

Unit XVI: Chemistry in Everyday Life 1. Chemicals in medicines – analgesics, tranquilizers, antiseptics, disinfectants, antimicrobials, antifertility drugs, antibiotics, antacids, antihistamines. 2. Chemicals in food – preservatives, artificial sweetening agents, elementary idea of antioxidants. 3. Cleansing agents – soaps and detergents, cleansing action.

Aldehydes and Ketones: Nomenclature, nature of carbonyl group, methods of preparation, physical and chemical properties, mechanism of nucleophilic addition, the reactivity of alpha hydrogen in aldehydes; uses. Carboxylic Acids: Nomenclature, acidic nature, methods of preparation, physical and chemical properties; uses.

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TELANGANA

Sri Balaji Book Depot, (040) 27613300, 9866355473, Shah Book House, 9849564564 Vishal Book Distributors, 9246333166, Himalaya Book World, 7032578527

(9)

0808

VIJAYAWADA

Contd... UTTARAKHAND

GORAKHPUR

Central Book House, 9935454590, Friends & Co., 9450277154, Dinesh book depot, 9125818274, Friends & Co., 9450277154

DEHRADUN

Inder Book Agencies, 9634045280, Amar Book Depot , 8130491477, Goyal Book Store, 9897318047, New National Book House, 9897830283/9720590054

JHANSI

Bhanu Book Depot, 9415031340

MUSSORIE

Ram Saran Dass Chanda kiran, 0135-2632785, 9761344588

KANPUR

Radha News Agency, 8957247427, Raj Book Dist., 9235616506, H K Book Distributors, 9935146730, H K Book Distributors, 9506033137/9935146730

UTTAR PRADESH

LUCKNOW

Vyapar Sadan, 7607102462, Om Book Depot, 7705871398, Azad Book Depot Pvt. Ltd., 7317000250, Book Sadan, 9839487327, Rama Book Depot(Retail), 7355078254,

Ashirwad Book Depot, 9235501197, Book.com, 7458922755, Universal Books,

9450302161, Sheetla Book Agency, 9235832418, Vidyarthi Kendra Publisher & Distributor Pvt Ltd, (Gold), 9554967415, Tripathi Book House, 9415425943

AGRA

Sparsh Book Agency, 9412257817, Om Pustak Mandir, (0562) 2464014, 9319117771,

MEERUT

ALLAHABAD

Mehrotra Book Agency, (0532) 2266865, 9415636890

NOIDA

Prozo (Global Edu4 Share Pvt. Ltd), 9318395520, Goyal Books Overseas Pvt.Ltd., 1204655555 9873387003

AZAMGARH

Sasta Sahitya Bhandar, 9450029674

PRAYAGRAJ

Kanhaiya Pustak Bhawan, 9415317109

ALIGARH

K.B.C.L. Agarwal, 9897124960, Shaligram Agencies, 9412317800, New Vimal Books, 9997398868, T.I.C Book centre, 9808039570

MAWANA

Subhash Book Depot, 9760262264

BULANDSHAHAR

Rastogi Book Depot, 9837053462/9368978202

BALRAMPUR

Universal Book Center, 8933826726

KOLKATA

BAREILLY

Siksha Prakashan, 9837829284

RENUKOOT

HARDOI

Mittal Pustak Kendra, 9838201466

DEORIA

Kanodia Book Depot, 9415277835

COOCH BEHAR

S.B. Book Distributor, Cooch behar, 9002670771

VARANASI

Gupta Books, 8707225564, Bookman & Company, 9935194495/7668899901

KHARAGPUR

Subhani Book Store, 9046891334

MATHURA

Sapra Traders, 9410076716, Vijay Book House , 9897254292

SILIGURI

Agarwal Book House, 9832038727, Modern Book Agency, 8145578772

FARRUKHABAD

Anurag Book Agencies, 8844007575

DINAJPUR

Krishna Book House, 7031748945

NAJIBABAD

Gupta News Agency, 8868932500, Gupta News Agency, ( E & C ), 8868932500

MURSHIDABAD

New Book House, 8944876176

DHAMPUR

Ramkumar Mahaveer Prasad, 9411942550

Sanjay Publication, 8126699922 Arti book centre, 8630128856, Panchsheel Books, 9412257962, Bhagwati Book Store, (E & C), 9149081912

Ideal Book Depot, (0121) 4059252, 9837066307

WEST BENGAL Oriental Publishers & Distributor (033) 40628367, Katha 'O' Kahini, (033) 22196313, 22419071, Saha Book House, (033), 22193671, 9333416484, United Book House, 9831344622, Bijay Pustak Bhandar, 8961260603, Shawan Books Distributors, 8336820363, Krishna Book House, 9123083874

Om Stationers, 7007326732

Entrance & Competition Distributors PATNA

BIHAR

CUTTAK

A.K.Mishra Agencies, 9437025991

Metro Books Corner, 9431647013, Alka Book Agency, 9835655005, Vikas Book Depot, 9504780402

BHUBANESHWAR

M/s Pragnya, 9437943777

CHATTISGARH KORBA

Kitab Ghar, 9425226528, Shri Ramdev Traders, 9981761797

PUNJAB JALANDHAR

Cheap Book Store, 9872223458, 9878258592

DELHI

RAJASTHAN

DELHI

Singhania Book & Stationer, 9212028238, Radhey Book depot, 9818314141, The KOTA Book Shop, 9310262701, Mittal Books, 9899037390, Lov Dev & Sons, 9999353491

Vardhman Book Depot, 9571365020, Raj Traders, 9309232829

NEW DELHI

Anupam Sales, 9560504617, A ONE BOOKS, 8800497047

Goyal Book Distributors, 9414782130

JAIPUR

HARYANA AMBALA

BOKARO

UTTAR PRADESH

Bharat Book Depot, 7988455354

AGRA

BHAGWATI BOOK STORE, 9149081912, Sparsh Book Agency, 9412257817, Sanjay Publication, 8126699922

JHARKHAND

ALIGARH

New Vimal Books, 9997398868

Bokaro Student Friends Pvt. Ltd, 7360021503

ALLAHABAD

Mehrotra Book Agency, (532) 2266865, 9415636890

MADHYA PRADESH

GORAKHPUR

Central Book House, 9935454590

INDORE

Bhaiya Industries, 9109120101

KANPUR

Raj Book Dist, 9235616506

CHHINDWARA

Pustak Bhawan, 9827255997

LUCKNOW

Azad Book Depot PVT LTD, 7317000250, Rama Book Depot(Retail), 7355078254 Ashirwad Book Depot , 9235501197, Book Sadan, 8318643277, Book.com , 7458922755, Sheetla Book Agency, 9235832418

MAHARASHTRA

PRAYAGRAJ

Format Center, 9335115561, Garg Brothers Trading & Services Pvt. Ltd., 7388100499

NAGPUR

Laxmi Pustakalay and Stationers, (0712) 2727354

PUNE

Pragati Book Centre, 9850039311

MUMBAI

New Student Agencies LLP, 7045065799

ODISHA

Inder Book Agancies, 9634045280

WEST BENGAL KOLKATA

Bijay Pustak Bhandar Pvt. Ltd., 8961260603, Saha Book House, 9674827254 United Book House, 9831344622, Techno World, 9830168159

Trimurti Book World, 9437034735

0808

BARIPADA

UTTAR PRADESH DEHRADUN

( 10 )

CUET (UG) Exam Paper 2023 National Testing Agency Held on 26th May 2023

CHEMISTRY

[This includes Questions pertaining to Domain Specific Subject only] Max. Marks: 200

Time allowed: 45 Minutes

General Instructions: (i) (ii) (iii) (iv) (v)

This paper consists of 50 MCQs, attempt any 40 out of 50 . Correct answer or the most appropriate answer: Five marks (+5) . Any incorrect option marked will be given minus One mark (–1) . Unanswered/Marked for Review will be given No mark (0) . If more than one option is found to be correct then Five marks (+5) will be awarded to only those who have marked any of the correct options . (vi) If all options are found to be correct then Five marks (+5) will be awarded to all those who have attempted the question . (vii) If none of the options is found correct or a Question is found to be wrong or a Question is dropped then all candidates who have appeared will be given five marks (+5). (viii) Calculator / any electronic gadgets are not permitted . OCOCH3 1. The compound

COOH is used as:

(1) Antiseptic (2) Antibiotic (3) Analgesic (4) Pesticide 2. A small amount of CaF2 is added in electrolytic reduction of Al2O3 dissolved in fused cryolite (A) To decrease the rate of oxidation of carbon at anode (B) To act as catalyst (C) To make fused mixture conducting (D) To lower the fusion temperature of melting. Choose the correct answer from the options given below: (1) (A), (B), (D) only (2)  (B), (C) only (3) (A), (B) only (4)  (C), (D) only 3. Match List -I with List - II. List -1 Complex (A) (B) (C) (D)

[CoF6]3– [Co(NH3)6]3+ [NiCl4]2– [Ni(CN)4]2–

List - II hybridisation (I) (II) (III) (IV)

sp3 dsp2 d2sp3 sp3d2



Choose the correct answer from the options given below (1) (A)-(I), (B)-(II), (C)-(III), (D)-(IV) (2) (A)-(II), (B)-(III), (C)-(lV), (D)-(I) (3) (A)-(III), (B)-(IV), (C)-(I), (D)-(II) (4) (A)-(IV), (B)-(III), (C)-(I), (D)-(II) 4. Match List -I with List - II. List -I (A) (B) (C) (D)



Chloroxylenol + terpineol Penicillin Iproniazid Aspirin

List - II (I) (II) (III) (IV)

Tranquillizer Analgesic Antiseptic Antibiotic

Choose the correct answer from die options given below

(1) (A)-(III), (B)-(II), (C)-(I), (D)-(IV) (2) (A)-(III), (B)-(IV), (C)-(I), (D)-(II) (3) (A)-(II), (B)-(IV), (C)-(III), (D)-(I) (4) (A)-(l), (B)-(III), (C)-(II), (D)-(IV) 5. Zone refining is mainly used for: (1) Ni (2) Ga (3) Zr (4) Ta 6. An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molecular mass of the solute? (1) 41.35 g/mol (2) 10 g/mol (3) 23.4 g/mol (4) 20.8 g/mol 7. Solution of KMnO4 is reduced to various products P , Q , R depending upon the pH of the solution. At pH 7 it forms a green coloured solution R . P , Q , R , will be: (1) (2) (3) (4)

P

Q

R

Mn2+ MnO2 MnO42– Mn2+

MnO2 MnO42– Mn2+ Mn4+

MnO42– Mn2+ MnO2 MnO2

8. Molecules are the constituent particles of molecular solids. Identify the molecular solid amongst the following; (1) SO2 (2) C (3) Fe (4) NaCl 9. Which of the following reaction will give t-Butyl ethyl ether by Williamson synthesis? CH3 |

(1) CH3–C–ONa + CH3 CH2Cl |

CH3

12

OSWAAL CUET (UG) Chapterwise Question Bank CHEMISTRY

17. Match List-I with List-II.

CH3

List – I

|

(2) CH3–CH2ONa + CH3–C–Cl

(A) (B) (C) (D)

|

CH3 CH3

CH3

|

|

(3) CH3–C–Br + CH3–C–ONa |

|

H

H

[Ni(CN)4 bidentate chelate ligand [NiCl4]2-

(I) (II) (III) (IV)

EDTA paramagnetic diamagnetic oxalate

CH3 |

(4) CH3–CH2CH2ONa + CH3–CH–Br 10. Electrolysis of molten NaCl gives: (1) H2 at cathode, CI2 at anode and NaOH solution (2) Na at cathode, Cl2 at anode and NaOH solution (3) H2 at anode, Cl2 at cathode (4) Na at cathode and Cl2 at anode 11. Which two transition metal ions has same 3d electronic configuration? (A) Mn3+ (B) Fe2+ 2+ (C) Mn (D) Fe3+ 3+ (E) Cr Choose the correct answer from the options given below: (1) (A) and (B) only (2) (B) and (C) only (3) (C) and (D) only (4) (D) and (E) only 12. Limiting molar conductivity of H2O is equal to: (1) Λ°m NaCl + Λ°m HCl – Λ°m NaOH (2) Λ°m HC1 + Λ°m NaOH – Λ°m NaCl (3) Λ°m HCl + Λ°m NH4OH – Λ°m NaCl (4) Λ°m NaCl + Λ°m HCl – Λ°m NH4OH 13. A cell is prepared by dipping a Cu rod in 1 M CuSO4 solution and Sn rod in 1 M SnCl2 solution. The standard electrode potential of Cu is +0.34 V and Sn is -0.14 V. Emf of cell will be: (1) 0.48 V (2) 0.20 V (3) 0.34 V (4) 0.14 V 14. On passing ammonia gas through a solution of copper sulphate, a deep blue solution is obtained. The deep blue colour of the solution is due to the formation of: (1) [Cu(NH3)2]2+ (2) [Cu(NH3)4]2+ + (3) [Cu(NH3)6] (4) [Cu(NH3)6]2+ 15. The correct increasing basicity in aqueous medium for (I) NH3 (II) CH3CH2CH2  (III) (CH3CH2)2NH (IV) (CH3CH2)3N is: (A) (II) < (I) < (IV) < (III) (B) (II) < (I) < (III) < (IV) (C) (I) < (II) < (IV) < (III) (D) (I) > (II) > (IV) > (III) Choose the correct answer from the options given below: (1) (C) (2) (B) (3) (D) (4) (A) OH 16. (1) NaOH (2) CO2 (3) HCl

List – II 2–

A

Product 'A' is: (1) Salicylaldehyde (2) Benzoic acid (3) p-Chloro benzoic acid (4) Salicylic acid

Choose the correct answer from the options given below (1) (A)-(III), (B)-(IV), (C)-(l), (D)-(II) (2) (A)-(II), (B)-(IV), (C)-(I), (D)-(III) (3) (A)-(III), (B)-(I), (C)-(IV), (D)-(II) (4) (A)-(II), (B)-(I), (C)-(IV), (D)-(III) 18. Identify the biodegradable polymer in the following: (1) Nylon -6, 6 (2) PVC (3) Polythene (4) PHBV 19. Which of the following statements are correct about alkyl halides: (A) Alkylhalides are polar so they are soluble in water (B) RX on reaction with alc.KOH give alkenes (C) RX on reaction with AgNO2 give RONO (D) SNl proceeds with inversion in configuration (E) R—X have higher boiling point than parent alkanes Choose the correct answer from the options given below: (1) (B), (E) only (2) (A), (C) only (3) (D), (E) only (4) (B), (C) only 20. When 1.5g of a non-volatile solute is dissolved in 30g of a solvent, the boiling point of the solvent is raised by 2 K. Calculate the molar mass of the solute, given that Kb for the solvent is 1.85 K kg mol-1. (1) 46.25 g mol–1 (2) 103 g mol–1 –1 (3) 94 g mol (4) 23 g mol–1 21. Which of the following statements are true? (A) Noble gases have low M.P. or B.P. due to weak dispersion forces. (B) All noble gases have completely filled ns2np6 electronic configuration. (C) XeF2 and XeF6 are colourless crystalline solid. (D) XeO3 has pyramidal geometry. Choose the correct answer from the options given below: (1) (A), (C) only (2) (B), (C), (D) only (3) (A), (B), (C), (D) only (4) (A), (C), (D) only 22. The entities which do not dissociate into simple ions are: (A) [Fe(CN)6]4

(B) KCl∙MgCl2∙6H2O



(C) K4[Fe(CN)6]



(D) KAl(SO4)2∙12H2O



(E) FeSO4∙(NH4)2SO4∙6H2O



Choose the correct answer from the options given below: (1) (A) and (C) only (2) (B) and (D) only (3) (D) and (E) only (4) (A) and (B) only

13 13

CUET (UG) Exam Paper 2023

23. Ethanol cannot be prepared from: COOC2H5 (1)



(2) CH3CH2CH2COOC2H5 (3) C2H5COOCH3 (4) CH3COOC2H5 24. Arrange the following in the increasing order of their van't Hoff factors (A) Very dilute MgCl2 solution (B) Very dilute AlCl3 solution (C) Very dilute NaCl solution (D) Very dilute Al2O3 solution (E) Very dilute urea solution

Choose the correct answer from the options given below:

Choose the correct answer from the options given below: (1) (A)-(II), (B)-(IV), (C)-(III), (D)-(I) (2) (A)-(III), (B)-(II), (Q-(IV), (D)-(I) (3) (A)-(II), (B)-(III), (C)-(IV), (D)-(I) (4) (A), (I), (B)-(II), (C)-(III), (D)-(IV) 27. Manganese ions on reaction with peroxydisulphate ions form: (1) MnO42– (2) MnO2 (3) Mn2O7 (4) MnO4– 28. Glylosidic linkage is formed by: (1) Elimination of water molecule (2) Addition of water molecule (3) Elimination of ammonia molecule (4) Addition of ammonia molecule 29. t99%with respect to t90% for a first order reaction is (1) four times of t50% (2) one and half time (3) It is same (4) Double 30. Match List -1 with List - II.

(1) (E) < (C) < (A) < (B) < (D)

List -I

(2) (E) < (C) < (A) < (D) < (B)

Number of moles of solute/kilogram of solvent (B) Molality (II) Osmotic pressure (C) Coiligative property (III) Number of moles of solute/litre of solution (D) Non-ideal solution (IV) Deviation from Raoult's Law

(3) (E) < (C) < (B) < (A) < (D) (4) (E) < (B) < (A) < (D) < (C) 25. Match List - I with List - II. List - I

List - II

(A) Denaturation (B) Primary structure of protein (C) Quaternary structure of protein (D) Secondary structure of protein

List - II

(A) Molarity

(I)

Sequence of amino acids in a protein (II) a-Helix and b-pleated sheets (III) 2o and 3o structures gets destroyed (IV) The spatial arrangement of subunits (containing 2 or more polypeptide chain) with respect to each other.

(I)



Choose the correct answer from the options given below: (1) (A)-(III), (B)-(I), (C)-(IV), (D)-(II) (2) (A)-(III), (B)-(I), (C)-(II), (D)-(IV) (3) (A)-(I), (B)-(III), (C)-(II), (D)-(IV) (4) (A)-(I), (B)-(III), (C)-(IV), (D)-(II) 31. The decomposition of ammonia on platinum surface is a zero order reaction. How much time it will take for 1 × 10–4 mol L–1 of ammonia to reduce into half of its concentration? Choose the correct answer from the options given (K=0.5×l0–4 mol L–1 s–1) below: (1) 1 s (2)  10 s (1) (A)-(I), (B)-(II), (C)-(III), (D)-(IV) (3) 100 s (4)  5 s 32. Identify the correct statement: (2) (A)-(III), (B)-(I), (C)-(IV), (D)-(II) (1) CrO is acidic, Cr2C3 is basic, CrO3 is amphoteric. (3) (A)-(III), (B)-(II), (C)-(IV), (D)-(I) (2) CrO is basic, Cr2C3 is acidic, CrO3 is amphoteric. (4) (A)-(I), (B)-(IV), (C)-(II), (D)-(III) (3) CrO is basic, Cr2C3 is amphoteric, CrO3 is acidic. (4) CrO is amphoteric, Cr2C3 is basic, CrO3 is acidic. 26. Match List -1 with List - II. 33. Which of the following is not a property of α-sulphur? List -I List -II (1) It is readily soluble in CS2 (2) It has monoclinic structure. + Br /H O – (A) NH2 2 2 (I) NH3 SO3 (3) This allotrope is yellow in colour. (4) It's melting point is 385.8 K. 34. The following pentose sugar is obtained by complete + – (1) (CH CO) O, Pyridine Br (B) N2Cl (2) HNO3 , H2 SO hydrolysis of DNA, along with phosphoric acid and 3 2 4 (II) Br (3) OH– NH2 nitrogenous base: Br +

– (1) HBF 4

(C)

N2Cl

(D)

NH2

(2) NaNO2, ∆

H2SO4 ∆

NH2

(II) NO2 (IV)

NO2

O

CH2OH H

(1)

OH H H

H OH

OH

14

OSWAAL CUET (UG) Chapterwise Question Bank CHEMISTRY

O

CH2OH H

(2)

OH H H

H OH

H O

CH2OH OH

(3)

OH OH

41. H

H H

H

(4)

(A)

(C)

H

(4) time vs Ro 37. The correct order of reactivity of following halides towards SN1 reaction is:



(I)

(II)

CH2Br

(III)

CH3 H

H

CH2Cl

(B)

H

35. Lustre, electrical and thermal conductivity of metals can be explained by the theory of: (1) dipole-dipole interactions (2) Coulombic (electrostatic) forces (3) Sea of free electrons (4) London forces 36. The integrated rate equation for a first order reaction kt is log Ro— log Rt. The straight line graph is 2.303 obtained by plotting: (1) time vs Rt (2) time vs log Rt 1 (3) time vs Rt

Cl

CH3 H

OH

H H

List - I

H O

CH2OH

Read the passage below and answer the question: Aldehydes and Ketones, having atleast one methyl group linked to the carbonyl carbon atom (methyl Ketones), are oxidised by sodium hypohalite to sodium salts of corresponding carboxylic acids having one carbon atom less them that of carbonyl compound is converted to haloform. This oxidation does not affect carbon – carbon double bond, if present in a molecule.

CH2I

(IV)

(1) (IV) > (III) > (II) > (I) (2) (I) > (II) > (III) > (IV) (3) (IV) > (II) > (III) > (I) (4) (I) > (III) > (II) > (IV) 38. The reaction of 3-methylbutan-2-ol with HBr, will give: (1) 2-Bromo-2-methylbutane (2) 3-Bromo-2-methylbutane (3) 3-Bromo-3-methylbutane (4) 2-Bromo-3-methylbutane 39. The shape of XeF6 is: (1) Regular octachedral (2) Square planar (3) Distorted octahedral (4) Square pyramidal 40. The relative order of reactivity of 1°, 2°, 3° alcohol towards dehydration is: (1) 1° > 2° > 3° (2) 2° > 1° > 3° (3) 2° > 3° > 1° (4) 3° > 2° > 1°

CH3 H

(D)

CH3 H

List - II (I)

Hydrazones

(II)

oximes

(III)

Ethane

(IV)

phenyl hydrazones

C=O+H2NOH

C=O+H2NNH2

C=O+H2NNHC6H5

C=O+Zn – Hg/HCl

Choose the correct answer from the options given belowr: (1) (A)-(II), (B)-(IV), (C)-(III), (D)-(I) (2) (A)-(II), (B)-(I), (C)-(IV), (D)-(III) (3) (A)-(III), (B)-(IV), (C)-(I), (D)-(II) (4) (A)-(III), (B)-(I), (C)-(II), (D)-(IV) 42. Which of the following aldehyde will give Aldol condensation: O || CH3 CH=C–C–H | O | || CH3–C–CHO CH3 | —CHO CH3 (A)

(B)

(C)

(D)

(1) (A) (2) (B) (3) (C) (4) (D) 43. Arrange in decreasing order of nuelophilic addition: (A) CH3CHO (B) C2H5COCH3 (C) CH3COCH3 (D) HCHO Choose the correct answer (1) (D) > (A) > (C) > (B) (2) (D) > (A) > (B) > (C) (3) (C) > (B) > (A) > (D) (4) (B) > (C) > (A) > (D) 44. Which of the following statements are true? (A) CH3CHO, and CH3CH2CHO can be distinguished by Tollen's test. (B) CH3CH2CHO and CH3COCH3 can be distinguished by Fehling's test. (C) CH3CHO,

—CHO can be distinguished by

Fehling's test. (D) HCHO, CH3CHO can be distinguished by NaHCO3 test.

15 15

CUET (UG) Exam Paper 2023

(E) HCOOH, CH3COOH can be distinguished by NaHCO3 test. Choose the correct answer from the options given below: (1) (A), (B) only (2) (B), (C) only (3) (C), (D) only (4) (E), (A) only 45. Predict the correct product (A): O O || || H+ —CH=CH–C–H+NH2–C–NHNH2 A

(1)

O || —CH=CH–CH=N–C–NHNH2

(2)

O || —CH=CH–CH=NNH–C–NH2

(3) (4)

OH O | + || —CH=CH–C–H+NH3–C–NHNH2 O || —CH–CH2–C–H | NHNH–C–NH 2 || O

Read the passage below and answer the question (Q. No. 46 to 50): The accumulation of molecular species at the surface rather than in the bulk of a solid or liquid is termed as adsorption. There are two types of adsorption. In physiorsorption, attractive forces are mainly Van der Waals while in chemical adsorption adsorbate

is held with chemical bonds adsorbent. Adsorption increases with increase in pressure and decreases as temperature is increased. 46. Adsorption is spontaneous because: (1) ∆S is – ve and is more than ∆H (2) ∆S is + ve and is more than ∆H (3) ∆H is – ve and is more than T∆S (4) ∆H – ∆S 47. Which of the following is not characteristic of chemical adsorption? (1) It is irreversible (2) It is highly specific (3) High temperature is favourable (4) It results into multi molecular layers 48. Which of the following is not an application of adsorption? (1) Control of humidity using silica gel (2) Separation of noble gases using charcoal (3) Removal of colouring matter from solutions (4) Inversion of cane sugar using mineral acids as catalysts 49. For Freundlich isotherm, graph of log(x/m) is plotted against log U. The slope of line and its Y axis intercepts respectively are: 1 1 (1) , K (2) , log K n n 1 (3) n, 1/K (4) n, log K 50. Critical temperature of few gases are given SO2 (630K), CH4, (190K), H2 (33K). What is the correct order of case of physisorption of these gases? (1) CH4 > H2 > SO2 (2) SO2 > CH4 > H2 (3) H2 > CH4 > SO2 (4) H2 > SO2 > CH4 

WRITING YOUR NOTES Just in case you have forgotten today, takedown your notes! But why is it so important? Tools for the hands are tools for the brain writes Hetty Roessingh. Handwritten notes are a powerful tool for encrypting embodied cognition and in turn supporting the brain’s capacity for recuperation of information. If that sounds so scientific then in simple words: Writing notes by hand help you in: 

Increasing your comprehension  Strengthening your memory  Igniting your creativity  Engaging your mind  Increasing your attention span Are these reasons enough to get you started?

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Study Time

CHAPTER

1

Max. Time: 1:50 Hours Max. Questions: 50

THE SOLID STATE points in three-dimensional space of crystal.

 Revision Notes

 Solids:  Properties: { Have definite mass, shape, volume, rigidity, high density and low compressibility. { Have high melting and boiling points.  Types: (i) Amorphous solids  Properties: { Constituent particles do not have an ordered arrangement (short range order) e.g., plastic, glass, etc. { No regular arrangement of particles, no sharp melting or boiling points and no definite shape. { Behave as super cooled liquids and have isotropic properties. (ii) Crystalline solids:  Properties: { Constituent particles have an ordered arrangement (long range order) e.g., NaCl, diamond { Have sharp melting or boiling points and definite shape.  Properties: { Molecular solids: The constituent particles are molecules which are joined with: • London forces • Dipole-dipole interactions • Hydrogen bonding { Ionic solids: The constituent particles are ions (positive and negative) which are joined with coulombic or electrostatic forces. { Metallic solids: The constituent particles are positive ions (kernels) in a sea of delocalised electrons joined by the metallic bond. { Covalent or Network solids: The constituent particles are atoms joined by the covalent bonds. { Unit cell: Smallest, fundamental repeating threedimensional unit of a crystal lattices. { Types

of unit cell:

{ Primitive cell: Constituent particles present only at the

Scan to know more about this topic

Types of closed packings:

(i)  P  articles touch each other in a row, packing is in one dimension. (ii)  Particles are placed in rows of close-packed spheres, packing is two dimen­sional.

Arrangement in fcc

Fig. 1.1 Possible unit cells in two dimensions { These are of two types:

(a)  Square close packed layers: ƒ ƒ

 oth the layers are perfectly aligned horizontally B as well as vertically. (‘AAA’ type pattern, C.N. = 6)

(b)  Hexagonal close packed layers:  he central sphere is in contact with six other T spheres in two dimensions. ƒ More efficient packing. (iii) Packing is by placing two dimensional Scan to know layers one above other. more about this ƒ

topic

Hcp packing

corner positions of a unit cell.

{ Centred unit cell or non-primitive unit cell: Particles

(or points) present at corners and at other positions also. different types of lattices are possible known as Bravais lattice based on the arrangement of lattice

{ Fourteen

Fig 1.2. A portion of a three dimentional cubic space of a crystal lattice and its unit cell.

2 Oswaal CUET (UG) Chapterwise Question Bank

CHEMISTRY

3

THE SOLID STATE

  Some important terms: (i) Co-ordination number: Number of closest neighbours of any constituent particle. • Simple cubic = 6

(v) Density of unit cell (d): Mass of unit cell Density of unit cell = Volume of unit cell Z×M d= 3 a ´ NA where, d = density of unit cell (vi) Defects:

•

Face centred cubic lattice (fcc) = 12

•

Body centred cubic lattice (bcc) = 8.

 Point defects: When there is irregularities in ideal arrangement of a constituent particle in a crystal lattice is called point defect.



Three types: a. Stoichiometric defects: Do not disturb the stoichiometric of the solid. ƒ

Fig 1.3 (ii) Packing Fraction: The volume of the unit cell that is occupied by the spheres the volume of the unit cell. Scan to know { Primitive

cubic unit cell: 52.4% { Face centred cubic unit cell:74% { Body centred cubic unit cell: 68%

Scan to know more about this topic

ƒ ƒ

more about this topic

ƒ

(iii) Voids/Interstitial sites: Holes left in the close packing of spheres. Voids a. Trigonal voids: � Enclosed by three spheres in contact. � Radius ratio: 0.155. b. Tetrahedral voids:

Vacancy defect-When a constituent particle missing from crystal lattice. Interstitial defect- When a constituent particle occupies an interstitial site in a crystal lattice. Frenkel defect- When a cation leaves its site and occupy a interstitial position in a crystal lattice. Schottky defect- When a cation and anion leave their position, create vacancy in the crystal lattice.

� S urrounded �

c.

by four points lying at the vertices of a regular tetrahedron. formula = 2 × Number of particles in close packing.

Defects in solids

Vacancy defects

Octahedral voids: � � �

 urrounded by six spheres lying at the S vertices of a regular octahedron. qual to number of particles in close E packing. Radus ratio: 0.414.

Frenkel defect Schottky defect Fig.1.5: Stoichiometric defects b.

c.

Impurity defects: Impurity defects arise when foreign atoms are present in the lattice site or in the interstitial site. Non–stoichiometric defects: Do not obey the law of constant proportions. ƒ

Fig. 1.4 Radius of cation (iv) Radius ratio = Radius of anion

ƒ ƒ

Metal deficiency defect. Metal excess defect due to anionic vacancies. Metal excess defect due to presence of extra cations.

4 Oswaal CUET (UG) Chapterwise Question Bank

CHEMISTRY

ƒ n-type semiconductors: Silicon and germanium (Group 14) doped with donor impurity (like P or As of Group 15). ƒ p-type semiconductors: Silicon and germanium (Group 14) doped with acceptor impurity (like B, Al or Ga of Group 13).

Introduction of cation vacancy in NaCl by substitution of Na+ by Sr2+ Fig.1.6: Impurity defect

Fig 1.8 (viii) Magnetic properties of solids: { Paramagnetic

An F-centre in a crystal Fig.1.7: Non stoichiometric defects d.

Line defects: Irregularities in the entire rows of lattice points.

substances are attracted in external magnetic field. { Diamagnetic substances are repelled in external magnetic field. { Ferromagnetic substances are most easily attracted in external magnetic field. { Anti-Ferromagnetic substances have zero magnetic moment. { Ferrimagnetic substances are paramagnetic with magnetic moment less than that of ferromagnetic.

(vii) Electrical properties: Band theory explains conductivities of solids. a. Conductors: Negligible energy gap such that electrons can jump from valence band to conduction band easily. b. Insulators: Large energy gap such that electrons cannot jump from valence band to conduction band. c. S  emi-conductors: Small energy gap such that electrons can jump from valence band to conduction band on providing little energy. Doping is mixing of impurity in semi-conductors.

Schematic alignment of magnetic moments in (a) ferromagnetic (b) antiferromagnetic and (c) ferrimagnetic. Fig 1.9

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS 1. Among the following statements, choose The correct statements. (A) In ionic solid, ions are the constituent particles. (B) Ionic solids are soft. (C) Ionic solids are electrical insulators in the solid state. (D) Ionic solids conduct electricity in molten state. (E) Ionic solids have low melting and boiling points.

 hoose the correct answer from the options given C below: (1) A, C & D only (2) A, D & E only

(3) A, B & C only

(4) A, C & E only [CUET 2022, 10th Aug]

2. A  toms of element B form hcp lattice and those of element A occupy 2/3rd of tetrahedral voids. What is the formula of the compound formed by the elements A and B? (1) A3B4 (2) A4B3 (3) A2B3 (4) A3B2 [CUET 2022, 10th Aug] 3. W  hich of the following arrangements represents align­ ment of magnetic moments of ferrimagnetic substance?

THE SOLID STATE

(1) (2) (3) (4) [CUET 2022, 17th Aug] 4. Atoms of B form an fcc lattice and those of element A occupy alternate tetrahedral voids. The formula of the compound formed by elements A and B: (1) AB2 (2) A2B (3) AB (4) A2B3 [CUET 2022, 18th Aug] 5. T  he total number of atoms in body centered cubic unit cell (bcc) is: (1) 2 (2) 4 (3) 9 (4) 1 [CUET 2022, 20th Aug] 6. AIN is an example of: 1. Molecular solid 2. Covalent solid 3. Metallic solid 4. Ionic solid (1) 1 (2) 2 (3) 3 (4) 4  [CUET 2022, 21st Aug] 7. Which of the following is correct for a hexagonal crystal system? (1) a = b ≠ c, α = ß = γ = 90° (2) a ≠ b ≠ c, α = ß = γ = 90° (3) a = b ≠ c, α = ß = 90°, γ =120° (4) a ≠ b ≠ c, α = ɣ = 90°, ß ≠ 90° [CUET 2022, 23rd Aug] 8. Efficiency of packing body-centered cubic structures is found to be: (1) 33%   (2)  74%   (3)  52.4%   (4)  68% [CUET 2022, 23rd Aug] 9. An element has a cubic structure with a cell edge of 288 pm. The density of the element is 7.2 g cm–3. 208 g of the element has 24.16×1022 numbers of atoms. The unit cell of this cubic structure is: (1) primitive (2) body-centered (3) face-centered (4) hexagonal [CUET 2021, 23rd Sept] 10. Which stoichiometric defect does not change the den­ sity of the crystal? (1) Frenkel defect (2) Schottky defect (3) Interstitial defect (4) F-centres 11. In which of the following structures coordination num­ ber for cations and anions in the packed structure will be same? (1) Cl– ion form fcc lattice and Na+ ions occupy all octahedral voids of the unit cell. (2) Ca2+ ions form fcc lattice and F– ions occupy all the eight tetrahedral voids of the unit cell. (3) O2– ions form fcc lattice and Na+ ions occupy all the eight tetrahedral voids of the unit cell.

5 (4) S2– ions form fcc lattice and Zn2+ ions go into alternate tetrahedral voids of the unit cell. 12. Elemental silicon to be used as a semiconductor is purified by: (1) heating under vacuum (2) floatation (3) zone refining (4) electrolysis 13. Examine the given defective crystal: A+ B– A+ B– A+ – – + B 0 B A B– + – + A B A 0 A+ – + – + B A B A B– How is the density of the crystal affected by this defect? (1) Density increases (2) Density decreases (3) No effect on density (4) Density first increases then decreases 14. The sharp melting point of crystalline solids is due to: (1) a regular arrangement of constituent particles observed over a short distance in the crystal lattice. (2) a regular arrangement of constituent particles observed over a long distance in the crystal lattice. (3) same arrangement of constituent particles in different directions. (4) different arrangements of constituent particles in different directions. 15. A compound is formed by two elements M and N. The element N forms ccp lattice and atoms of M occupy two atoms and Mercury 1/3rd of tetrahedral voids. What is the formula of the compound. (1) MN2  (2) M2N3    (3) M3N2  (4) M2N2 16. Which of the following defects is also known as dis­ location defect? (1) Frenkel defect (2) Schottky defect (3) Non-stoichiometric defect (4) Simple interstitial defect 17. Interstitial compounds are formed when small atoms are dropped under the curved lattice of metals. Whether the following is not the characteristics property of interstitial compounds? (1) They have high melting points in to pure metals (2) They are very hard (3) They retain metallic conductivity (4) They are chemically very reactive 18. Structure of a mixed oxide is cubic close packed (ccp). The cubic unit cell of 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  (2)  A2BO2  (3)  A2B3O4  (4)  AB2O2 19. A compound formed by elements X and crystallizes in a cubic structure in which the X atoms are at the corners of a cube and the Y atoms are at the face- centres. The formula of the compound is: (1) XY3  (2)  X3Y   (3)  XY   (4)  XY2

6 Oswaal CUET (UG) Chapterwise Question Bank 20. A ferromagnetic substance becomes a permanent mag­ net when it is placed in a magnetic field because: (1) All the domains get oriented in the direction of magnetic field (2) All the domains get oriented in the direction oppo­site to the direction of magnetic field (3) Domains get oriented randomly (4) Domains are not affected by magnetic field 21.  Which of the following is not a characteristic of a crystalline solid? (1) Definite and characteristic heat of fusion. (2) Isotropic nature. (3) A regular periodically repeated pattern of arrangement of constituent particles in the entire crystal. (4) A true solid. [NCERT Exemp. Q 2, Page 1] 22.  Which of the following is true about the value of refractive index of quartz glass? (1) Same in all directions (2) Different in different directions (3) Cannot be measured (4) Always zero [NCERT Exemp. Q 5, Page 2] 23.  Iodine molecules are held in the crystals lattice by ____________. (1) London forces (2) Dipole-dipole interactions (3) Covalent bonds (4) Coulombic forces [NCERT Exemp. Q 8, Page 3] 24. Which of the following is a network solid? (1) SO2 (solid) (2) I2 (3) Diamond (4) H2O (ice) [NCERT Exemp. Q 9, Page 2] 25. Graphite is a good conductor of electricity due to the presence of __________. (1) Lone pair of electrons (2) Free valence electrons (3) Cations (4) Anions  [NCERT Exemp. Q 12, Page 3] 26.  Which of the following solids is not an electrical conductor? (i) Mg (s) (ii) TiO (s) (iii) I2 (s) (iv) H2O (s) Choose the correct answer from the options given  below: (1) (i) only (2) (ii) Only (3) (iii) and (iv) (4) (ii), (iii) and (iv) [NCERT Exemp. Q 10, Page 2] 27. Which of the following is true about the charge acquired by p-type semiconductors? (1) positive (2) neutral (3) negative (4) depends on concentration of p impurity [NCER Exemp. Q 19, Page 4]

CHEMISTRY

28. Which of the following statement is not true about the hexagonal close packing? (1) The coordination number is 12. (2) It has 74% packing efficiency. (3) Tetrahedral voids of the second layer are covered by the spheres of the third layer. (4) In this arrangement spheres of the fourth layer are exactly aligned with those of the first layer [NCERT Exemp. Q 25, Page 5] 29. Which kind of defects are introduced by doping? (1) Dislocation defect (2) Schottky defect (3) Frenkel defects (4) Electronic defects [NCERT Exemp. Q 28, Page 5] 30. Which of the following statements is not true? (1) Paramagnetic substances are weakly attracted by magnetic field. (2) Ferromagnetic substances cannot be magnetised permanently. (3) The domains in antiferromagnetic substances are oppositely oriented with respect to each other. (4) Pairing of electrons cancels their magnetic moment in the diamagnetic substances. [NCERT Exemp. Q 30, Page 6] [B] ASSERTION REASON QUESTIONS Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (1) Both A and R are correct and R is The correct explanation of A. (2) Both A and R are correct but R is NOT The correct explanation of A. (3) A is correct but R is not correct. (4) A is not correct but R is correct. 1. Assertion (A): The total number of atoms present in a simple cubic unit cell is one. Reason (R): Simple cubic unit cell has atoms at its corners, each of which is shared between eight adjacent unit cells.  [NCERT Exemp. Q 69, Page 13] 2. Assertion (A): Graphite is a good conductor of electricity. However, diamond belongs to the category of insulators. Reason (R): Graphite is soft in nature on the other hand diamond is very hard and brittle.  [NCERT Exemp. Q 70, Page 13] 3. Assertion (A): Total number of octahedral voids present in unit cell of cubic close packing including the one that is present at the body centre, is four. Reason (R): Besides the body centre there is one octahedral void present at the centre of each of the six faces of the unit cell and each of which is shared between two adjacent unit cells.  [NCERT Exemp. Q 71, Page 13]

7

THE SOLID STATE

4. A  ssertion (A): The packing efficiency is maximum for the fcc structure. Reason (R): The coordination number is 12 in fcc structures.  [NCERT Exemp. Q 72, Page 131] 5. Assertion (A): Semiconductors are solids with conductivities in the intermediate range from 10–6 – 104 Ohm–1 m–1 Reason (R): Intermediate, conductivity in semicon­ ductor is due to partially filled valence band.  [NCERT Exemp. Q 73, Page 131] 6. Assertion (A): In crystalline solids the value of resistance is different in different directions. Reason (R): Crystalline solids are isotropic in nature. 7. Assertion (A): Glass panes fixed to windows or panes of old building are found to be slightly thicker at the bottom. Reason (R): Amorphous solids have a tendency to flow. 8. Assertion (A): CsCl has body-centred cubic arrange­ ment. Reason (R): CsCl has one Cs ion and 8 Cl– in its unit cell. 9. Assertion (A): On heating ferromagnetic or ferr­ imagnetic substances, they become paramagnetic. Reason (R): The electrons change their spin on heating. 10. Assertion (A): The total number of atoms present in a simple cubic unit cell is one. Reason (R): Simple cubic unit cell has atoms at its corner, each of which is shared between 8 adjacent unit cell. [C] COMPETENCY BASED QUESTIONS I. Based on following passage answer questions from 1-5. Solids have definite mass, volume and shape. Depending upon order, they are classified as crystalline solids and amorphous solids. Crystalline solids are further classified into molecular, ionic, metallic and covalent solids. Crystalline solids are made up of constituent particles which are arranged in a regular pattern, which is depicted in the form of three dimensional array of points in spaces called lattice points. Small characteristic repeating portion of lattice is called unit cell. Unit cells can be primitive and also centred unit cells like face centred, body centred, end centred. Solids are not perfect. There are various types of defects (i) line defect (ii) point defect. Point defects are of three types: stoichiometric, impurity and non-stoichiometric defects. Solids show magnetic properties depending upon the orientation of electron. [CUET 30th Aug 2022] 1. F  or a primitive unit cell, the edge lengths are a ≠ b ≠ c and axial angles  ≠  ≠  ≠ 90°, then find the crystal system: (1) Hexagonal (2) Rhombohedral (3) Tetragonal (4) Triclinic 2. CaF2 is an example of: (1) Metallic solid (2) Covalent solid (3) Ionic solid (4) Molecular solid

3. I dentify the type of magnetic property from the schematic alignment of magnetic moment given below:

(1) Diamagnetic (3) Antiferromagnetic 4. Match List I with List II. List-I

(2) Ferromagnetic (4) Ferrimagnetic List-II

A. Schottky defect

I.

B. Dislocation defect

II. Decreases the density of the substance

C. Point defect

III. Frenkel defect

D. Vacancy defect

IV. Number of missing cations and anions are equal

Irregularities from an ideal arrangement around an atom

Choose the correct answer from the options given below: (1) (A)-(I), (B)-(II), (C)-(III), (D)-(IV) (2) (A)-(IV), (B)-(III), (C)-(I), (D)-(II) (3) (A)-(II), (B)-(III), (C)-(I), (D)-(IV) (4) (A)-(III), (B)-(II), (C)-(IV), (D)-(I) 5. Certain properties of crystalline solids are given: (A) They have sharp melting points. (B) They have long range order. (C) They do not have definite enthalpy of fusion. (D) They show anisotropy. (E) Rubber, plastic are examples of crystalline solids.

Incorrect statements are:



 hoose the correct answer from the options given C below: (1) A and C only (2) B and C only (3) C and E only (4) A, B and D only II. Based on following passage answer questions from 6-10 . Point defects play an important role in determining the physical properties of most crystalline substances, most notably those controlling the transport of matter and the properties that stem from it. Even a crystal of high purity under conditions of no irradiation contains point defects in thermal equilibrium some. Some lattice sites are vacant and some atoms are displaced from their normal lattice sights into interstitial positions or onto ‘wrong’ lattice sites. For stoichiometric compounds of high purity the concentrations of these point defects are very low, even at temperatures up to melting point. A meaningful model, then, is to consider the crystal as a solvent containing a very dilute solution of simple, individual vacancies and interstitials. Long range in fractions among the defect and with impurity atoms and short range interactions that produce pair or other clusters can be produced in a firstorder approximation. 6. W  hich one of the given below statement is wrong about Frenkel defect?

8 Oswaal CUET (UG) Chapterwise Question Bank (1) It is a combination of vacancy and interstitial defect. (2) Cations leave their actual lattice site and occupy the interstitial space in the solid. (3) Density remains the same. (4) Density of the crystal increases. 7. Which one of the following is an interstitial void? (1) Octahedral void (2) Tetrahedral void (3) None of the above (4) Both (1) and (2) 8. This type of defect arises due to absence of equal number of cations and anions from lattice sites in the crystalline solid of the type A B and it lowers the den­ sity of the crystal. (1) Vacancy defect (2) Schottky defect (3) Interstitial defect (4) Frenkel defect

CHEMISTRY

9. W  hich one of the following cannot be called as a nonstoichiometric defect. (1) Metal excess defect due to anion vacancies. (2) Metal excess defect due to presence of extrication (3) Metal deficiency due to absence of cations. (4) Combination of vacancies and interstitial defects. 10. The correct statement regarding stoichiometric defects is: (1) It does not disturb the stoichiometry of the compound. (2) It makes the compound not obey Law of conservation of mass. (3) It makes the compound not obey Law of definite proportion. (4) It makes the compound not obey Law of constant composition.

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (1)

2. (2)

3. (1)

4. (3)

5. (1)

6. (2)

7. (3)

8. (4)

9. (2)

10. (1)

11. (1)

12. (3)

13. (2)

14. (2)

15. (2)

16. (1)

17. (4)

18. (4)

19. (1)

20. (1)

21. (2)

22. (1)

23. (1)

24. (3)

25. (2)

26. (3)

27. (2)

28. (4)

29. (4)

30. (2)

9. (1)

10. (1)

9. (4)

10. (1)

[B] ASSERTION REASON QUESTIONS 1. (1)

2. (2)

3. (3)

4. (2)

5. (3)

6. (3)

7. (1)

8. (3)

[C] COMPETENCY BASED QUESTIONS 1. (4)

2. (3)

3. (4)

4. (2)

5. (3)

6. (4)

7. (4)

8. (2)

ANSWERS WITH EXPLANATION [A] MULTIPLE CHOICE QUESTIONS: 1. Option (1) is correct. Explanation: Ionic compound forms by the transfer of electrons from one atom to another, which create ions (cation and anion) and they become the main constituents of an ionic compound. These ions are strongly packed together in a three-dimensional arrangement by a strong electrostatic force of attraction which makes ionic crystals hard and brittle in nature. These compounds have high melting points and boiling points because of the large amounts of energy is needed to break many strong bonds. The ionic solids are insulators in the solid state because the ions are entrapped in fixed places within the crystal lattice and cannot move on applying an electric field. However, in a molten state, the well-ordered arrangement of ions is destroyed and the ions are in a position to move about in the liquid medium when an electric field is applied. So, ionic solids conduct electricity in the molten state. 2. Option (2) is correct. Explanation: In hcp lattice, if atoms of element B = n Number of tetrahedral voids = 2n Then, atoms of element A occupy two third of the tetrahedral voids

2 4n ´ 2n = 3 3 4n 3 Atomic ratio = =4 3 n Thus, the formula of the lattice is A4B3.

3. Option (1) is correct.

Explanation: For ferrimagnetic substances, below Tc (Curie temperature), spins are aligned antiparallel but do not cancel. 4. Option (3) is correct. Explanation: No. of atoms per unit cell in fcc = 4 The total no. of tetrahedral voids is double the number of atoms. 8 voids for 4 atoms As, B occupies alternate tetrahedral voids half of the voids will be filled. For 4 atoms of A, 4 voids are filled by B. So, A: B = 4:4 = 1:1 so, the formula = AB 5. Option (1) is correct. Explanation: The total number of atoms per unit cell of bcc is 2. In the bcc unit cell, one atom is present at body centre.. Its contribution to the unit cell is 1. All the eight atoms are present at eight corners of the unit cell where each corner atom contributes one eight to the unit cell.

9

THE SOLID STATE

Hence, the net contribution of eight corner atoms and body center atom is: (8 × 1/8) + 1 = 1 + 1 = 2 6. Option (2) is correct. Explanation: Aluminium nitride (AlN) is covalently bonded network crystalline solid. 7. Option (3) is correct. Explanation: Tetragonal a = b ≠ c,  =  =  = 90° Orthorhombic a ≠ b ≠ c,  =  =  = 90° Hexagonal a = b ≠ c,  =  = 90°,  = 120° Monoclinic a ≠ b ≠ c,  =  =90° ,  ≠ 90° 8. Option (4) is correct. Explanation: for body centered cubic lattice:

11. Option (1) is correct. Explanation: NaCl unit cell has a fcc structure of Cl– ions, and Na+ ions occupy octahedral voids. The radius ratio of 0.524 for NaCl suggests an octahedral void. 12. Option (3) is correct. Explanation: Zone refining is used for metals which are required in very high purity. Semiconductor grade silicon is purified by this method. 13. Option (2) is correct. Explanation: The given defective crystal shows that there is missing of one cation and one anion from their lattice positions which is Schottky defect. Due to missing of ions, density of the crystal decreases. 14. Option (2) is correct. Explanation: The sharp melting point of crystalline solids is due to a regular arrangement of constituent particles observed over a long distance in the crystal lattice. 15. Option (2) is correct.

a2 + b2 = c2 a2 + 2a2 = (4r)2 √3a= 4r Packing efficiency = Volume occupied by two spheres in unit cell Volume of unit cell 2 ´ 4 3p r 3 = ´ 100 a3 8 3p r 3 = ´ 100 = 68% 4r 3

9. Option (2) is correct. Explanation: Edge length ‘a’ = 288 pm = 288 × 10–10 cm –3 Density (d) = 7.2 g cm No. of atoms (N) = 24.16 × 1023 Mass of an element (m) = 208 g m ´ NA N 208 ´ 6.022 ´ 1023 = 24.16 ´ 1023 = 51.8 g

M=

d ´ a ´ NA M

Explanation: Suppose the atoms N in the ccp = a No. of tetrahedral voids = 3 → 2a 2a No. of atoms M = : a = 2: 3 3 Hence compound as M2 : N3 16. Option (1) is correct. Explanation: Frenkel defect is also known as dislocation defect because in this defect one of the ion is missing from its lattice site and occupies an interstitial site. 17. Option (4) is correct. Explanation: Interstitial compounds are usually nonstoichiometric and are neither typically ionic nor covalent. Hence, interstitial compounds are chemically inert. 18. Option (4) is correct. Explanation: Number of atoms in cubic close packing = 4 = O2– Number of tetrahedral voids = 2N = 2 × 4 = 8 Number of A2+ ions = 8 × 1 / 4 = 2 Number of octahedral voids = Number of B+ ions = N = 4 Ratio of ions will be O2– : A2+: B+ = 4 : 2 : 4 = 2 : 1 : 2 Formula of oxide = AB2O2 19. Option (1) is correct. Explanation: X atoms at the corners =

3

Z=

7.2 ´ ( 288 ´ 10 ) ´ 6.022 ´ 1023 = 2 51.8 Hence, body-centered arrangement. =

-10 3

10. Option (1) is correct. Explanation: In Frenkel defect, one of the ions is missing from its lattice site and occupies an interstitial site. So, density of the crystal does not change.

Y atoms at the face centres =

1´ 8 =1 8

1 ´6 = 3 2

Ratio of atoms, X: Y = 1: 3 Hence, formula is XY3 20. Option (1) is correct. Explanation: A ferromagnetic substance can be magnetised permanently. So, the domains get oriented in the direction of applied magnetic field.

10 Oswaal CUET (UG) Chapterwise Question Bank 21. Option (2) is correct. Explanation: Crystalline solids are anisotropic in nature, that is, some of their physical properties like electrical resistance or refractive index show different values when measured along different directions in the same types of crystals. 22. Option (1) is correct. Explanation: Since quartz glass is an amorphous solid having short range order of constituents. Hence, value of refractive index is: • same in all directions, • can be measured and • not be equal to zero always. 23. Option (1) is correct. Explanation: Iodine is a type of molecular solids which are held by weak dispersion forces or London forces. 24. Option (3) is correct. Explanation: Diamond is a three-dimensional network solid in which each carbon atom is bonded tetrahedrally with four carbon atoms. 25. Option (2) is correct. Explanation: Graphite is a good conductor of electricity due to the presence of free valence electrons. 26. Option (3) is correct. Explanation: I2 (s) is a non-polar molecular solid and hence non-conductor of electricity. In H2O (s), molecules are bonded together because of hydrogen bonds and show nonionic nature. Thus, water in solid state is also a non-conductor of electricity. 27. Option (2) is correct. Explanation: p-type semiconductors are neutral but they conduct electricity through positive holes. 28. Option (4) is correct. Explanation: Hexagonal close packing can be arranged by two layers A and B one over another which can be diagrammatically represented as;

CHEMISTRY

2. Option (2) is correct. Explanation: Diamond is bad conductor of electricity because all valence e of carbon is involved in bonding. In graphite however 3 out of 4 valence electrons are involved in bonding, fourth electron remains free between adjacent layers which makes it a good conductor. Graphite is soft because parallel layers are held together by week van der Waals force. However, diamond is hard due to compact three-dimensional network of bonding. 3. Option (3) is correct. Explanation: (i) All edge centres and body centres represent octahedral voids. 1 (ii) Total number of octahedral voids = 12 × + 1 = 4 4 4. Option (2) is correct. Explanation: In fcc unit cell, there is ccp arrangement with packing efficiency of 74.01% which is maximum. In ccp arrangement, coordination number is 12. 5. Option (3) is correct. Explanation: Conductivity of semiconductors lies bet­ween metals and insulators, i.e., in the range of 10–6 – 104 Ohm–1 m–1. 6. Option (3) is correct. Explanation: Crystalline solids are anisotropic in nature. In crystalline solids the value of electrical resistance or refractive index is different in different directions. 7. Option (1) is correct. Explanation: Solids have a tendency to flow, though very slowly. Glass is sometimes called a supercooled liquid because it does not form a crystalline structure, but instead forms an amorphous solid that allows molecules in the material to continue to move. 8. Option (3) is correct. Explanation: CsCl has one Cs+ and one Cl– in its unit cell. 9. Option (1) is correct. Explanation: All magnetically ordered solids transform to the paramagnetic state at high temperature due to the randomisation of the spins. 10. Option (1) is correct. Explanation: In the simple cubic unit cell, the total number of atoms is 8  (1/8) = 1

29. Option (4) is correct. Explanation: When electron rich or electron deficient impurity is added to a perfect crystal, it introduces electronic defect in them. 30. Option (2) is correct. Explanation: Ferromagnetic species are strongly attra­cted in the magnetic field and can be permanently mag­netised.

[C] COMPETENCY BASED QUESTIONS 1. Option (4) is correct. Explanation: Crystal system

Edge lengths

Axial angles

Hexagonal

a=b≠c

α = β = 90° γ = 120°

Rhombohedral

a=b=c

α = β = γ ≠ 90°

[B] ASSERTION REASON QUESTIONS

Tetragonal

a=b≠c

α = β = γ = 90°

1. Option (1) is correct.

Triclinic

a≠b≠c

α ≠ β ≠ γ = 90°

Explanation: The total number of atoms present in a simple cubic unit cell is one. Simple cubic unit cell has atoms at its corners, each of which is shared between eight adjacent unit cells.

2. Option (3) is correct. Explanation: CaF2 is ionic solid having ions, Ca2+ and F1– in crystal lattice.

11

THE SOLID STATE

3. Option (4) is correct. Explanation: Ferrimagnetism: This effect is observed when the magnetic moments of the domains in the substance are aligned in parallel and anti-parallel directions in unequal numbers. 4. Option (2) is correct. Explanation: Schottky defect: A vacancy defect in ionic solids maintaining electrical neutrality by the equal number of missing cations and anions. Dislocation defect: Frenkel defect is also called dislocation defect. Point defect: Irregularities from ideal arrangement around a point or an atom in a crystalline substance. Vacancy defect: Some of the lattice sites are vacant and it decreases the density of substance. 5. Option (3) is correct. Explanation: Rubber and plastic: Amorphous solids Crystalline solids: Definite enthalpy of fusion. 6. Option (4) is correct.

Explanation: This is the only incorrect statement. All other statements explain the features of Frenkel defect. 7. Option (4) is correct. Explanation: Voids can be of two types, octahedral and tetrahedral. 8. Option (2) is correct. Explanation: In Schottky defect, due to the absence of equal number of cations and anions from lattice sites, there is lowering of the density of the crystal. 9. Option (4) is correct. Explanation: The other statements are stoichiometric defects. The compounds which obey the law of definite proportions, the law of constant composition and the law of conservation of mass are called stoichiometric compounds. The defects in crystals which do not disturb the stoichiometry of the compound or crystal are called stoichiometric defects. 10. Option (1) is correct. Explanation: The stoichiometric defects do not disturb the laws of chemical reactions of a compound.

Study Time

CHAPTER

2

Max. Time: 1:50 Hours Max. Questions: 50

SOLUTIONS Molality of CH3COOH =

  Revision Notes

 Solutions: {  Solutions are homogeneous mixtures of two or more than two components. In a binary solution, each component may be solid, liquid or in gaseous state. { Concentration of solutions can be expressed in terms of: ƒ Molarity (M): The number of moles of solute dissolved in one litre (or one cubic decimetre) of solution. Moles of solute M= Volume of solution in litre

gram equivalents of solute Volume of solution in litre

Molality (m): The number of moles of the solute per kilogram (kg) of the solvent. Moles of solute m= Mass of solvent in kg

Mass percentage (w/W): = Mass of the component in the solution ´ 100 Total mass of the solution ƒ Volume percentage (v/V) = Volume of the component ´ 100 Total volume of solution ƒ

ƒ ƒ

 Henry’s Law: {  Statement: The partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution. p = KH x, KH is the Henry’s constant.

Normality (N): Number of equivalents per litre of solution. N=

ƒ

= 0.556 mol kg–1

Y-axis

ƒ

Moles of CH 3COOH 0.0417 mol ´ 1000 g kg -1 = kg of benzene 75 g

X-axis Fig. 2.1  Experimental results for the solubility of HCl gas in cyclohexane at 293 K. The slope of the lime is the Henry's Law constant, KH Example 2: If N2 gas is bubbled through water at 293 K, how many millimoles of N2 gas would dissolve in 1 litre of water? Assume that N2 exerts a partial pressure of 0.987 bar. Given that Henry’s Law constant for N2 at 293 K is 76.48 kbar. Solution: The solubility of gas is related to the mole fraction in aqueous solution. The mole fraction of the gas in the solution is calculated by applying Henry’s law.

 ass by volume percentage (w/V): It is the mass M of solute dissolved in 100 mL of the solution. Parts per million: When solute is present in trace quantities

X(Nitrogen) =

X(Nitrogen) =

Mole fraction(x): = Number of moles of the component Total number of moles of alll the components

ƒ

Mass of benzene in kg = 75 g/1000 g kg–1 = 75 × 10–3 kg

KH

=

0.987 bar =1.29 × 10–5 76,480 bar

As 1 litre of water contains 55.5 mol of it, therefore if n represents number of moles of N2 in solution,

Number of parts of the component ´ 106 Total number of parts of alll components of the solution

Example 1: Calculate molality of 2.5 g of ethanoic acid (CH3COOH) in 75 g of benzene. Solution: Molar mass of CH3COOH: 12 × 2 + 1 × 4 + 16 × 2 = 60 g mol–1 2.5 g Moles of CH3COOH = = 0.0417 mol 60 g mol-1

p (nitrogen)

n n = = 1.29 × 10–5 n + 55.5 55.5

(n in denominator is neglected as it is < < 55.5) Thus, n = 1.29 × 10–5 × 55.5 mol = 7.16 × 10–4 mol = 7.16 ´ 10-4 mol ´ 1000 mmol 1 mol 

= 0.716 mmol

Vapour pressure:

Vapour pressure of a pure solvent is always more than the vapour pressure of a solution. Presence of a non-volatile solute lowers the vapour pressure of a solution.

SOLUTIONS

13

14 Oswaal CUET (UG) Chapterwise Question Bank 

Raoult’s Law:

Scan to know more about this topic

{ Vapour

pressure of liquid-liquid solutions: Statement: For any solution the partial vapour pressure of each volatile component in the solution is directly proportional to its mole fraction present Vapour pressure in solution For component 1: p1 µ χ1, p1 = χ1 p10 For component 2 : p2 µ χ2 , p2 = χ2 p20 ptotal = p1 + p2 = χ1 p10 + χ2 p20 = (1 – χ2 ) p10 + χ2 p20 = p10 + (p20 – p10) χ2

Vapour Pressure

P0A

0

PB

=PB

  Azeotropes: Binary mixtures having the same composition in liquid and vapour phase and boil at a constant temperature. It is not possible to separate the components by fractional distillation. (i) Maximum boiling azeotropes: Solutions that show a large positive deviation from Raoult’s law form minimum boiling azeotrope at a specific composition. For example, ethanol-water mixture (ii) Minimum boiling azeotropes: Solutions that show large negative deviation from Raoult’s law form maximum boiling azeotrope at a specific composition. For example, Nitric acid - water mixture. 

P02

P P1 =PA + B

.χ B



\

χA =0 χB=1

Mole Fraction

Vapour pressure of solution

p

2

χ1 = 0 Mole fraction χ1 = 1 χ1 χ2 = 1 χ2 = 0 χ2

Po - P W2 ´ M 1 = Po W1 ´ M 2

OR



DP W2 ´ M 1 = Pn W1 ´ M 2

Example 3: The vapour pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, nonelectrolyte solid weighing 0.5 g when added to 39.0 g of benzene (molar mass 78 g mol–1). Vapour pressure of the solution, then, is 0.845 bar. What is the molar mass of the solid substance? Solution: p10 = 0.850 bar; p = 0.845 bar; M1 = 78 g mol–1; w2 = 0.5 g; w1 = 39 g Substituting these values in the above equation: 0.850 bar – 0.845 bar 0.5 g × 78 g mol–1 = 0.850 bar M2 × 39 Therefore, M2 = 170 g mol–1 (ii) Elevation of boiling point:

Vapour pressure of solution

Vapour pressure

Vapour pressure

on the number of solute particles. of the nature of total number of particles present in the solution. They are: (i) Relative lowering of vapour pre­ssure of the solvent: DP Po - P = = x2 Pn Po

Fig. 2.2  Vapour pressure for ideal solution The plot of vapour pressure and mole fraction of an ideal solution at constant temperature. ƒ  Liquid-liquid solutions ideal solutions obey Raoult’s law over the entire range of concentration. The enthalpy of mixing of the pure components to form the solution is zero and the volume of mixing is also zero, i.e., Δmix H = 0, Δmix V = 0 {  Non ideal solutions do not obey and deviate from Raoult’s law over the entire range of concentration. (i) Positive deviation: The intermolecular attractive forces between the solute-solvent molecules are weaker than those between the solute-solute and solvent-solvent molecules leading to increase in vapour pressure. An example of this type of mixture is ethanol and acetone. (ii) Negative deviation: The intermolecular attractive forces between solute-solute and solvent-solvent are weaker than those between solute-solvent and leads to decrease in vapour pressure. An example of this type is a mixture of phenol and aniline

1

{ Depend

A

Vapour Pressure Diagram for ideal Solution

p

Colligative properties: { Independent

PA = PA 0. χ

χA =1 χB=0

CHEMISTRY

p

2

p

1

χ1 = 0 χ2 = 1

Mole fraction

χ1 χ2

χ1 = 1 χ2 = 0

   (i) Positive deviation     (ii) Negative deviation Fig 2.3

Fig. 2.4  The vapour pressure curve for solution lies below the curve for pure solvent The diagram shows that ΔTb denotes the elevation of boiling point of a solvent in solution. Molality = m ΔTb µ m Therefore, ΔTb = Kbm m=

w2 ´ 1000 w1M 2

15

SOLUTIONS

(where Kb – boiling point elevation constant) So,



ΔTb = kb ´

w2 ´ 1000 w1 ´ M 2

where, w1 = Mass of a solvent in gram, w2 = Mass of soule in gram, M2 = Molecular mass of the solute. Kb is Boiling Point Elevation Constant or Molal Elevation Constant (Ebullioscopic constant). The unit of Kb is K kg mol–1. Example 4: 18 g of glucose, C6H12O6, is dissolved in 1 kg of water. At what temperature will water boil at 1.013 bar? Kb for water is 0.52 K kg mol–1. Moles of glucose = 18 g/ 180 g mol–1 = 0.1 mol Solution: Number of kilograms of solvent = 1 kg, molality of glucose solution = 0.1 mol kg–1 For water, change in boiling point. ΔTb = Kb × m = 0.52 K kg mol–1 × 0.1 mol kg–1 = 0.052 K Water boils at 373.15 K at 1.013 bar pressure, Therefore, the boiling point of solution will be 373.15 + 0.052 = 373.202 K. (iii) Depression of freezing point of the solvent

Thus, molar mass of the solute = 256 g mol–1 (iv) Osmotic pressure of the solution: The process of flow of the solvent through semipermeable membrane is called osmosis. The flow will continue till the equilibrium is attained. The pressure that just stops the flow of solvent is called osmotic pressure of the solution. Scan to know more about this topic

Osmotic pressure

Fig.2.6 : Osmotic pressure The excess pressure equal to the osmotic pressure must be applied on the solution side to prevent osmosis. \pV = RT for 1 mole \pV = nRT ... (a) for n moles n \p = RT V



\p = CRT .. (b) Example 6: 200 cm3 of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure of such a solution at 300 K is found to be 2.57 × 10–3 bar. Calculate the molar mass of the protein? Solution: Given: P = 2.57 × 10–3 bar, V = 200 cm3 = 0.200 litre T = 300 K, R = 0.083 L bar mol–1 K–1 1.26 g × 0.083 L bar K–1 mol–1 × 300 K Substituting, M2 = 2.57×10 bar × 0.200 L = 61,03.8 g mol–1 { Two solutions having same osmotic pressure at a given temperature are called isotonic solutions. {  A hypertonic solution has a higher concentration of solute than the solution it’s being compared to. This means that water will move into the hypertonic solution. { A hypotonic solution has a lower concentration of solute than the solution it’s being compared to. This means that water will move out of the hypotonic solution.



Van’t Hoff factor:

Fig.2.5 : Diagram showing Δ Tf, depression of the freezing point of a solvent in a solution ΔTf µ m Þ ΔTf = Kf ´ m Þ ΔTf = kb ´

WB 1000 ´ M B WA

where, ΔTf = depression in freezing point, Kf = freezing point depression constant. m = molality of the solution. WB = mass of solute MB = molar mass of solute and WA = mass of solvent. Kf is Freezing point depression Constant or Molal Depression Constant (Cryoscopic Constant). The unit of Kf is K kg mol–1. Example 5: 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol–1. Find the molar mass of the solute. Solution: Substituting the values, M2 =

5.12 K kg mol–1 × 1.00 g × 1000 g kg–1 0.40 K × 50 g

In 1880 van’t Hoff introduced a factor i, known as the van’t Hoff factor, to account for the extent of dissociation or association. { Some compounds when dissolved in water dissociate into ions, resulting in more number of moles of particles in the solution. Thus, when there is dissociation of solute into ions, the experimentally determined molar mass is always lower than the true value. { Some compounds when dissolved in water associate to form bigger particles, resulting in lesser number of

16 Oswaal CUET (UG) Chapterwise Question Bank moles of particles in the solution. Thus, when there is association of solute, the experimentally deter­ mined molar mass is always higher than the true value.

Scan to know more about this topic

i= i=

van’t Hoff factor

{ Such

a molar mass that is either lower or higher than the expected or normal value is called as abnormal molar mass. This factor i is defined as: normal molar mass i= Abnormal molar mass

CHEMISTRY

Observed colligative property Calculated colligative property

Total number of moles of particles after association/ dissociation Number of moles of particles before association/ dissociation

{ Inclusion

of van’t Hoff factor modifies the equations for colligative properties as follows: (i) Elevation of Boiling point, ΔTb = i Kb m (ii) Depression of Freezing point, ΔTf = i Kf m (iii) Osmotic pressure of solution, P = i n2 R T / V Colligative properties have been used to determine the molar mass of solutes.

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS 1. Consider the 1 M aqueous solution of the following com-­ pounds and arrange them in the increasing order of elevation in the boiling points. (A) C6H12O6 (B) NaCl (C) MgCl2 (D) AlCl3 (E) Al2(SO4)3 Choose the correct answer from the options given below: (1) B < C < D < E < A (2) A < E < D < C < B (3) A < B < C < D < E (4) E < D < C < B < A [CUET 2022, 10th Aug] 2. Adding a salt to water leads to increase in the boiling point of solution with respect to water because: (1) Vapour pressure of solution is increased. (2) Solubility of salt in water is increased. (3) Solubility of salt in water is decreased. (4) Vapour pressure of solution is decreased. [CUET 2022, 17th Aug] 3. Which of the following is a colligative property? (1) Vapour pressure (2) Osmotic pressure (3) Freezing point (4) Boiling point [CUET 2022, 17th Aug] 4. The incorrect statement about azeotropes is: (1) They are binary mixtures of liquids. (2) The composition remains the same in liquid and vapour phase, at constant temperature. (3)  The components can be separated by fractional distillation. (4) They show deviations from Raoult’s law. [CUET 2022, 21st Aug] 5. T  he values of Henry’s law constant for the given gases in water, at 293 K, are: (A) He - 144.97/K bar (B) H2 - 69.16/K bar (C) N2 - 76.48/K bar (D) O2 - 34.86/K bar

Which gas has the lowest solubility in water?

(1) He (3) N2

(2) (4)

H2 O2 [CUET 2022, 21st Aug]

6. W  hich one of the following will have the highest value of van’t Hoff factor, i? 1. K2SO4 2. MgSO4 3. KCl 4. NaCl [CUET 2022, 21st Aug] 7. Observe the given graph and identify the correct statement for the solution.

(1) Component 2 is more volatile than component 1. (2) Component 1 is more volatile than component 2. (3) Boiling point of component 1 is lower than that of component 2. (4) Volatility of a component depends upon its mole fraction. [CUET 2022, 23rd Aug] 8. A18 g of a non-volatile solution A is dissolved in 1 kg of water, the boiling point of water is raised to 373.51 K. Given Kb for water is 0.52 K kg mol–1, boiling point for water is 373.15 K at 1.013 bar pressure. The molecular weight of the solid A is: (1) 58.0 g mol–1 (2) 26.0 g mol–1 –1 (3) 55.0 g mol (4) 110.0 g mol–1 [CUET 2022, 23rd Aug] 9. At a certain place, pure water boils at 99.725°C. Calculate the boiling point of 0.69 m solution of urea in water at the same place.

17

SOLUTIONS

(Given Kb of water = 0.513°C kg mol –1) (1) 0.3539°C (2) 100.079°C (3) 99.37°C (4) 99.725°C [CUET 2022, 30th Aug] 10. Which one of the following one molal aqueous solution will show maximum freezing point depression? (1) [Co (H2O)3Cl3] × 3H2O (2) [Co (H2O)6]Cl3 (3) [Co (H2O)5Cl]Cl2 × H2O (4) [Co (H2O)4Cl]Cl2 × 2H2O [CUET 2022, 30th Aug] 11. The vapour pressure of a two components system as a function of composition shows positive deviation from Raoult’s Law. Identify the correct curve to show the positive deviation. (1)

(2)

(3 )

(4)

[CUET 2022, 20th Aug] 12. A beaker contains a solution of substance ‘A’. Precipitation of substance ‘A’ takes place when small amount of ‘A’ is added to the solution. The solution is ______. (1) saturated (2) supersaturated (3) unsaturated (4) concentrated [NCERT Exemp. Q 4, Page 17] 13. Low concentration of oxygen in the blood and tissues of people living at high altitude is due to ____________. (1) low temperature (2) low atmospheric pressure (3) high atmospheric pressure (4) both low temperature and high atmospheric pressure [NCERT Exemp. Q 6, Page 18]

14.  Considering the formation, breaking and strength of hydrogen bond, predict which of the following mixtures will show a positive deviation from Raoult’s law? (1) Methanol and acetone (2) Chloroform and acetone (3) Nitric acid and water (4) Phenol and aniline [NCERT Exemp. Q 7, Page 18] 15. Colligative properties depend on ________ (1) the nature of the solute particles dissolved in solution. (2) the number of solute particles in solution. (3) the physical properties of the solute particles dissolved in solution. (4) the nature of solvent particles. [NCERT Exemp. Q 8, Page 18] 16. Which of the following aqueous solutions should have the highest boiling point? (1) 1.0 M NaOH (2) 1.0 M Na2SO4 (3) 1.0 M NH4NO3 (4) 1.0 M KNO3 [NCERT Exemp. Q 9, Page 18] 17. The unit of ebullioscopic constant is _______________. (1) K kg mol–1 or K (molality)–1 (2) mol kg K–1 or K–1(molality) (3) kg mol–1 K–1 or K–1(molality)–1 (4) K mol kg–1 or K (molality) [NCERT Exemp. Q 10, Page 18] 18.  In comparison to a 0.01 M solution of glucose, the depression in freezing point of a 0.01 M MgCl2 solution is _____________. (1) the same (3) about three times

(2) about twice (4) about six times [NCERT Exemp. Q 11, Page 19] 19. An unripe mango placed in a concentrated salt solution to prepare pickle, shrivels because _____________. (1) it gains water due to osmosis. (2) it loses water due to reverse osmosis. (3) it gains water due to reverse osmosis. (4) it loses water due to osmosis [NCERT Exemp. Q 12, Page 19] 20. The values of Van’t Hoff factors for KCl, NaCl and K2SO4, respectively, are _____________. (1) 2, 2 and 2 (2) 2, 2 and 3 (3) 1, 1 and 2 (4) 1, 1 and 1 [NCERT Exemp. Q 15, Page 19] 21. The value of Henry’s constant KH is _____________. (1) greater for gases with higher solubility. (2) greater for gases with lower solubility. (3) constant for all gases. (4) not related to the solubility of gases. [NCERT Exemp. Q 117, Page 20] 22. Raoult’s law states that: (1) The lowering of vapour pressure is equal to the mole fraction of solute. (2) The relative lowering of vapour pressure is equal to the mole fraction of solute.

18 Oswaal CUET (UG) Chapterwise Question Bank

CHEMISTRY

(3) The relative lowering of vapour pressure is proportional to the amount of solute in solution. (4) The vapour pressure of the solution is equal to the mole fraction of the solute 23. When 1 mole of benzene is mixed with 1 mole of toluene the vapour will contain: (Given: vapour of benzene = 12.8kPa and vapour pressure of toluene = 3.85 kPa). (1) equal amount of benzene and toluene as it forms an ideal solution (2) unequal amount of benzene and toluene as it forms a non ideal solution. (3) higher percentage of benzene (4) higher percentage of toluene 24. Which of the following units is useful in relating con­ centration of solution with its vapour pressure? (1) Mole fraction (2) Parts per million (3) Mass percentage (4) Molality 25. 4L of 0.02 M aqueous solution of NaCl was diluted by adding one litre of water. The molality of the resultant solution is __________ (1) 0.004 (2) 0.008 (3) 0.012 (4) 0.016 26. The values of colligative properties of colloidal solution are of small order in comparison to those shown by true solutions of same concentration because of colloidal particles. (1) exhibit enormous surface area. (2) remain suspended in the dispersion medium. (3) form lyophilic colloids. (4) are comparatively less in number. 27.  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) 2.63 (2) 3.11 (3) 0.381 (4) 2.05 28. pA and pB are the vapour pressure of pure liquid components A and B, respectively of an ideal binary solution. If ×A represents the mole fraction of component A, the total pressure of the solution will be: (1) pA + χA (pB – pA) (2) pA + χA (pA – pB) (3) pB + χA (pB – pA) (4) pB + χA (pA – pB) 29. Molarity of liquid HCl, if density of solution is 1.17 g/cc is: (1) 36.5 (2) 18.25 (3) 32.05 (4) 42.10 30. All form ideal solution except: (1) C6H6 and C6H5CH3 (2) C2H5Cl and C2H5I (3) C6H5Cl and C6H5Br (4) C2H5I and C2H5OH

(3) A is correct but R is not correct. (4) A is not correct but R is correct. 1. Assertion (A): Molarity of a solution in liquid state changes with temperature. Reason (R): The volume of a solution changes with change in temperature. 2. Assertion (A): When methyl alcohol is added to water, boiling point of water increases. Reason (R): When a volatile solute is added to a volatile solvent elevation in boiling point is observed.  [NCERT Exemp. Q 52, Page 26] 3. Assertion (A): When NaCl is added to water a dep­ression in freezing point is observed. Reason (R): The lowering of vapour pressure of a solution causes depression in the freezing point.  [NCERT Exemp. Q 53, Page 27] 4. Assertion (A): When a solution is separated from the pure solvent by a semipermeable membrane, the solvent molecules pass through it from pure solvent side to the solution side. Reason(R): Diffusion of solvent occurs from a region of high concentration solution to a region of low concentration solution.  [NCERT Exemp. Q 54, Page 27] 5. Assertion (A): 0.1 M solution of KCl has greater osmotic pressure than 0.1 M solution of glucose at same temperature. Reason (R): In solution KCl dissociates to produce more number of particles. 6. Assertion (A): Pressure have any effect on solubility of solids in liquid. Reason (R): Solids and liquids are not incompressible. 7. Assertion (A): Elevation in boiling point is a colligative property. Reason (R): Elevation in boiling point is directly proportional to molarity. 8. Assertion (A) Azeotropic mixtures are not formed only by non-ideal solutions and they may have boiling points either greater than both the components or less than both the components. Reason (R): The composition of the vapour phase is not same as that of liquid phase of azeotropic mixture. 9. Assertion (A): The concentration of pollutants in water or atmosphere is often expressed in terms of PPM. Reason (R): Concentration in parts per million can be expressed as mass to mass volume to volume and mass to volume.

[B] ASSERTION REASON QUESTIONS Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (1) Both A and R are correct and R is The correct explanation of A. (2) Both A and R are correct but R is NOT The correct explanation of A.

[C] COMPETENCY BASED QUESTIONS I. Based on following passage answer questions from 1-5.  [CUET 2022,18 Aug] Solubility of gases in liquids is greatly affected by pressure and temperature. Henry gave the quantitative relationship between pressure and solubility of gas in a solvent. According to Henry’s law, partial pressure of a gas above a liquid is directly proportional to its mole fraction in solution and is expressed as

10. Assertion (A): An ideal solution obeys Henry’s law. Reason (R): In an ideal solution, solute-solute as well as solvents-solvent interaction are similar to solute-solvent interaction.

19

SOLUTIONS

P = KH.x, where KH is Henry’s constant and x is mole fraction of gas. KH is a function of nature of gas. F M Raoult gave a quantitative relationship between partial pressures and mole fractions in the binary solution of volatile liquids. Raoult’s law states that for a binary solution of volatile liquids, the partial pressure of each component in the solution is directly proportional to its mole fraction. Thus, for a solution of component 1 and 2, partial pressure of each component P1 = p1°x1, where p1° is the vapour pressure of pure component 1 at the same temperature. Similarly, P2 = p2° x2 1. Solubility of carbon dioxide in soda water increases with: (1) Increase in pressure and temperature both. (2) Increase in pressure and decrease in temperature. (3) Decrease in pressure and increase in temperature. (4) Decrease in pressure and decrease in temperature. 2. Which of the following show negative deviation from Raoult’s law? (1) Water and Ethanol (2) Ethanol and Acetone (3) Acetone and Chloroform (4) Carbon disulphide and Acetone 3. Identify the correct order of their increasing solubility in water: (1) HCHO < CH4 < CO2 < Ar (2) HCHO < CO2 < CH4 < Ar (3) Ar < CO2 < CH4 < HCHO (4) Ar < CH4 < CO2 < HCHO 4. If a solution exhibits positive deviation from Raoult’s Law then sign of ∆mix H and ∆mix V will be (1) Dmix H = positive; Dmix V = positive (2) Dmix H = positive; Dmix V = negative (3) Dmix H = negative; Dmix V = positive (4) Dmix H = negative; Dmix V = negative 5.  Henry’s law constant for O2 in water at 293 K is 46.82 bar. The mole fraction of oxygen in air is 0.21. What mole fraction of oxygen is dissolved in 1 litre of water at 2.026 bar pressure and 293 K? (1) 0.4254 (2) 0.850 (3) 0.018 (4) 0.009 II.  Based on following passage answer questions from 6-10. Read the passage given below and answer the following questions: Raoult’s law states that for a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution. Dalton’s law of partial pressure states that the total

pressure (ptotal) over the solution phase in the container will be the sum of the partial pressures of the components of the solution and is given as: ptotal = p1 + p2.

6. W  hat type of deviation from Raoult’s law does the above graph represent? (1) First positive then negative (2) Negative deviation (3) Positive deviation (4) First negative then positive 7. In comparison to a 0.01 M solution of glucose, the depression in freezing point of a 0.01 M MgCl2 solution is _____________. (1) the same (2) about twice (3) about three times (4) about six times 8. A  solution of two liquids boils at a temperature more than the boiling point of either of them. What type of deviation will be shown by the solution formed in terms of Raoult’s law? (1) Negative deviation (2) Positive deviation (3) First positive then negative (4) First negative then positive 9. W  hich of the following aqueous solutions should have the highest boiling point? (1) 1.0 M NaOH (2) 1.0 M Na2SO4 (3) 1.0 M NH4NO3 (4) 1.0 M KNO3 10. When the intermolecular attractive forces between solutesolute and solvent-solvent are weaker than those between solute-solvent then: (1) there is decrease in vapour pressure of solution. (2) there is increase in vapour pressure of solution. (3) there is no change in vapour pressure of solution. (4) it is independent of vapour pressure of solution.

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (3)

2. (4)

3. (2)

4. (3)

5. (1)

6. (1)

7. (1)

8. (2)

9. (2)

10. (2)

11. (1)

12. (2)

13. (2)

14. (1)

15. (2)

16. (2)

17. (1)

18. (3)

19. (4)

20. (2)

21. (2)

22. (2)

23. (3)

24. (1)

25. (4)

26. (4)

27. (1)

28. (4)

29. (3)

30. (4)

8. (2)

9. (4)

10. (4)

[B] ASSERTION REASON QUESTIONS 1. (1)

2. (4)

3. (1)

4. (2)

5. (1)

6. (1)

7. (3)

20 Oswaal CUET (UG) Chapterwise Question Bank

CHEMISTRY

[C] COMPETENCY BASED QUESTIONS 1. (2)

2. (3)

3. (3)

4. (1)

5. (4)

6. (2)

7. (3)

8. (1)

9. (2)

10. (1)

ANSWERS WITH EXPLANATION [A] MULTIPLE CHOICE QUESTIONS

7.

1.

Explanation: The plot of vapour pressure and mole fraction of an ideal solution is at a constant temperature. As, P20 > p10, that means component 2 is more volatile than component 1. As high vapour pressures indicate high volatility, while high boiling points indicate low volatility. So, boiling point of component 1 is greater than component 2.

Option (3) is correct.

Explanation: Elevation in boiling point depends upon van’t Hoff factor and molarity at a specific temperature.ΔTb = i Kbm For the same molarity 1M aqueous solution, the extent to which a substance associates or dissociates in a solution is described by the van’t Hoff factor. ΔTb is proportional to i (A) C6H12O6   non electrolyte (B) NaCl i=2 (C) MgCl2 i=3 (D) AlCl3 i=4 (E) Al2(SO4)3 i=5 The increasing order of elevation in boiling point is: A 0, DSmix > 0 5.

Option (4) is correct.

Explanation: Partial pressure of O2 = 2.026 × 0.21 = 0.425 Now, applying Henry’s law pO = KH × XO 2 2 pO2 = 0.425 XO = = 0.009 46.82 2 KH II.  Based on following passage answer questions from 6-10 6.

Option (2) is correct.

Explanation: Negative deviation 7.

Option (3) is correct.

Explanation: ΔTf = iKf m, where i = 1 for glucose. ΔTf glucose = 1 × Kf × 0.01 In case of MgCl2 → Mg2+ + 2Cl−, where i = 3, ∆Tf MgCl2 = 3 × 0.01 × Kf ⇒ ∆T f MgCl2 = 3 × ΔTf glucose Hence, the depression in freezing point of MgCl2 is three times that of glucose. 8.

Option (1) is correct.

Explanation: Since the boiling point of the solution is more than the boiling point of the individual components in the solution, it indicates that the vapour pressure exerted by the solution is less than the expected, as boiling starts when vapour pressure equals the atmospheric pressure. Hence, the solution shows a negative deviation from the Raoult’s law. 9.

Option (2) is correct.

Explanation: Na2SO4 will release 3 moles of ions/ moles of Na2SO4 in the aqueous solution, and Boiling point being a colligative property, the boiling point of this solution will be the highest as other solutions release only 2 ions each. 10. Option (1) is correct. Explanation: Due to stronger interaction between solutesolvent, the solute is unable to escape and there is decrease in vapor pressure of solution.

Study Time

CHAPTER

3

Max. Time: 1:50 Hours Max. Questions: 50

ELECTROCHEMISTRY z∝E where z = equivalent weight E = electrochemical equivalent Also, E = Atomic mass/No. of e– required to reduce cations of the metal { The charge on one mole of electrons is approximately equal to 96500 coulombs. This quantity of electricity is called Faraday constant (F).

 Revision Notes 

Electrochemical cell: {  Electrochemical

cell is the ionic conductor or electrolyte. {  It consists of two metallic electrodes dipped in electrolytic solutions. The cells are of two types. 

Electrolytic cells: { Electrical energy is used to carry out a non-spontaneous

redox reaction. { In an electrolytic cell external source of voltage is used to bring about a chemical reaction. {  Sodium and magnesium metals are produced by the electrolysis of their fused chlorides and aluminium. 





Galvanic cells: of two half cells. half cell contains an electrolytic solution and a metallic electrode. { The electrode at which oxidation takes place is called an anode and the electrode at which reduction takes place is called the cathode. { The half-cells are separated from each other by means of a porous pot or a salt bridge. { Difference of potential which causes current to flow from the electrode of higher negative potential is called the electromotive force (emf). {  Each



          = 0.2938 g. (b) Second law: When the same quantity of electricity is passed through solutions of different electrolytes, the weight of different substances deposited or liberated at the respective electrodes are proportional to their chemical equivalent weights.

Galvanic cell

{ Consists

Faraday’s laws of electrolysis:

(a) First law: The amount of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed through the electrolyte. Q=Ixt m ∝ Q m = Z.Q m = Z.I.t I = current in amperes T = time in seconds Z = electrochemical constant where Q = charge in coulombs where, m = mass of substance deposited or liberated.  Example 1: A solution of CuSO4 is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode? Solution: t = 600 s charge = current × time = 1.5 A × 600 s = 900 C According to the reaction: Cu2+(aq) + 2e– = Cu(s) We require 2F or 2 × 96500 C to deposit 1 mol or 63 g of Cu. (63 g mol–1 × 900 C) For 900 C, the mass of Cu deposited = (2 × 96500 C mol–1)

Scan to know more about this topic

Scan to know more about this topic

Fig: 3.1: Daniell cell having electrodes of zinc and copper dipping in the solutions of their respective salts 

Electrode Potential

Electrode potential: {  The

standard electrode potential for any electrode dipping in an appropriate solution is defined with respect to hydrogen. { The potential of SHE is assigned a value of zero. E° = 0 V. { The standard potential of the cell is the difference of the standard potentials of cathode and anode

24 Oswaal CUET (UG) Chapterwise Question Bank

CHEMISTRY

25

ELECTROCHEMISTRY

and each having an area = 1 cm2 κ is reciprocal of specific resistance or resistivity. The SI units of conductivity are S m–1 but quite often, κ is expressed in S cm–1 { Molar conductance (Λm): The conductance of the volume of solution which contains one mole of the solute and is placed between two parallel electrodes which are one centimetre apart and having sufficient area to hold the whole of the solution. Unit of molar conductance is Ω–1 cm2 mol–1 or S cm2 mol–1 Λm = 1000 ´ κ C C = concentration of solution in moles per litre (or Molarity). {  The electrical conductance through metals decreases with increase in temperature. { The electrolytic conductance increases with increase of temperature. { Equivalent conductance: The value of equivalent conductance increases with dilution and attains a maximum value at infinite dilution. { Specific conductance: The value of specific conductance decreases with dilution as the number of current carrying particles i.e., ions present per cm3 of solution decreases on dilution. { Molar conductance: The value of molar conductance increases with dilution and finally attains a maximum value at infinite dilution.

E° cell= E° cathode – E° anode { The standard potential of the cells are related to: Standard Gibbs energy: ΔrG° = –nF E° cell Equilibrium constant: ΔrG° = – RT ln K 0 { A negative E means that the redox couple is a stronger reducing agent than the H+/H2 couple. 0 { A positive E means that the redox couple is a weaker reducing agent than the H+/H2 couple. 

Nernst equation: {  Concentration

dependence of the potentials of the electrodes and the cells are given by Nernst equation. { Potential of half electrode: Mn+ + ne → M(s) E cell = E°Mn+/ M + ln [Mn+] {  For an electrochemical cell for which the overall reaction is: aA + bB ⇋ cC + dD E cell = E°cell –2.303RT log [C]c[D]d nF

[A]a[B]b

At 298K, E cell = E°cell –0.059 log [C]c[D]d

n [A]a[B]b {  The equilibrium constant, of cell can be related to standard emf of cell: E°cell = 2.303RT log Kc nF = 0.0591 log Kc at 298 K n Example 2: The standard electrode potential for Daniell cell is 1.1V. Calculate the standard Gibbs energy for the reaction: Solution: Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) ∆rG° = – nF E° (cell) n in the above equation is 2, F = 96500 C mol –1 and E° cell = 1.1 V Therefore, ∆rG° = – 2 × 1.1V × 96500 C mol –1 = – 21230 J mol–1 = – 212.30 kJ mol–1 

Conductance (G)

{ Conductance

is reciprocal of resistance.

Scan to know G = 1/R more about this R = ρl/A topic G = A/ρl where A = area of cross section ρ = Resistivity Variation of l = length of the conductor conductivity -1 { Unit of conductance is ohm or mho with dilution or Siemens. { Specific conductance (κ) or conductivity: the conductance of a solution taken in a cell whose electrodes are at unit distance apart from each other

Fig: 3.2: Molar conductivity versus c½ for acetic acid (weak electrolyte) and potassium chloride (strong electrolyte) in aqueous solutions. 

Kohlraush’s law:

{ expressed

as the sum of the contributions of the cation and the anion of the electrolyte. λ0 = υ + λ0+ + υ – λ0 Λ0 = υ+ λ0+ + υ – λ0– where λ+, λ– are the number of cat ions and anions per unit formula of the electrolyte respectively; λ0+, and λ0– are the limiting molar conductivities.

 Batteries: {  Consists of two or more galvanic cells connected in series. { Primary batteries: When the reactants have been converted into products,

26 Oswaal CUET (UG) Chapterwise Question Bank no more electricity is produced. The cell reaction cannot be reversed and the battery becomes dead. { Secondary batteries: The cell reaction can be reversed by passing electricity through the battery (charging). The battery can be used through a large number of discharging and charging cycles. {  Electrical cells that are discharged to convert the energy from the combustion of fuels (hydrogen, carbon monoxide, methane, etc.) directly into the electrical energy are called fuel cells.

CHEMISTRY

Fig: 3.3: Commonly used mercury cell. The reducing agent is zinc and the oxidising agent is mercury (II) oxide.

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS: 1. Among the following statements related to ionic con­ ductance, choose the correct statements. (A) Ionic conductance depends on the nature of the electrolyte. (B) Ionic conductance is due to the movements of electrons. (C) Ionic conductance is also called electronic conductance. (D) Ionic conductance depends on temperature. (E) Ionic conductance also depends on the nature of the solvent. Choose the correct answer from the options given below: (1) A, B and C only (2) B, C and D only (3) B, C and E only (4) A, D and E only [CUET 2022, 10th Aug] 2. The products formed at the cathode and anode by electrolysis of aqueous NaCl solution respectively are:  (1) Na, Cl2 (2) Na, O2 (3) H2, Cl2 (4) H2, O2 [CUET 2022, 10th Aug] 3. How many electrons flow when a current of 5 amperes is passed through a metal for 193 s? (Given: F = 96500 C mol–1, NA = 6.022 × 1023 mol–1 ) (1) 6.022 × 1023 electrons (2) 6.022 × 1021 electrons (3) 3.011 × 1021 electrons (4) 3.011 × 1023 electrons [CUET 2022, 17th Aug] 4. Kohlrausch law of independent migration of ions is applicable: (1) Only to weak electrolytes at a certain concentration. (2) Only to strong electrolytes at all concentrations. (3) To both - strong and weak electrolytes. (4) To non-electrolytes [CUET 2022, 17th Aug] 5. When a lead storage battery is recharged: (A) Pb acts as cathode. (B) PbO2 acts as anode. (C) H2SO4 is produced. (D) 38% NH4Cl is used as electrolyte. (E) Lead storage cell has a longer life than nickel-cadmium cell.

Choose the correct answer from the options given below: (1) A only (2) B only (3) C only (4) D, E only [CUET 2022, 18th Aug] 6. Molar conductivity of ____________ increases drastically on dilution. (1) NH4Cl (2) NaCl (3) CH3COOH (4) CH3COONa [CUET 2022, 18th Aug] 3+ 7. A cell represented as : Al(s) | Al (aq) || Ni2+ (aq) | Ni(s) has a standard electrode potential 1.41 V. Calculate the standard Gibb’s energy for the cell reaction occurring at this cell: (1) –136065 J mol–1 (2) –272130 J mol–1 –1 (3) –408195 J mol (4) –816390 J mol–1 [CUET 2022, 20th Aug] 8. Λm for NaCl, HCl and CH3COONa are 126.4, 425.9 and 91.0 S cm2 mol–1, respectively. Calculate Λ°m for CH3COOH. (1) 516.9 S cm2 mol–1 (2) 653.3 S cm2 mol–1 2 –1 (3) 390.5 S cm mol (4) 460.9 S cm2 mol–1 [CUET 2022, 20th Aug] 9. The variation of Λm with √C for a strong electrolyte is shown in the graph. Identify the correct representation for the same. (1)   (2)

(3)

  (4)

[CUET 2022, 21st Aug] 10. Identify the correct equations for the emf of a cell at equilibrium. (A) E 0 cell =

-2303 log K C nF

ELECTROCHEMISTRY

(B) E 0 cell =

DG nF

(C) E cell = E 0 cell -

RT ln [Pr oducts] nF [Reactants]

(D) E°cell = E° right – E° left Choose the correct answer from the options given below: (1) (A) and (C) only (2) (A) and (B) only (3) (B) and (C) only (4) (C) and (D) only [CUET 2022, 21st Aug] 11. Which of the following method cannot prevent corrosion? (1) Covering the surface with paint. (2) Covering the surface with other more reactive metal. (3) Covering the surface with sacrificial electrode of another metal. (4) Covering the surface with non-metal. [CUET 2022, 30th Aug] 12. Dissociation constant and molar conductance of an acetic acid solution are 1.78 × 10–5 mol L–1 and 48.15 S cm–2 mol–1 respectively. The conductivity of the solution is (considering molar conductance at infinite dilution is 390.5 S cm–2 mol–1) (1) 4.9 × 10–2 S cm–1 (2) 4.9 × 102 S cm–1 –5 –1 (3) 4.9 × 10 S cm (4) 4.9 × 105 S cm–1 [CUET 2021, 23rd Sept] 13. Which cell will measure standard electrode potential of copper electrode? (1) Pt (s)|H2 (g,0.1 bar) | H+ (aq.,1 M) || Cu2+(aq.,1M) | Cu (2) Pt(s) | H2 (g, 1 bar) | H+ (aq.,1 M) || Cu2+ (aq.,2 M) | Cu (3) Pt(s) | H2 (g, 1 bar) | H+ (aq.,1 M) || Cu2+ (aq.,1 M) | Cu (4) Pt(s) | H2 (g, 1 bar) | H+ (aq.,0.1 M) || Cu2+ (aq.,1 M) | Cu [NCERT Exemp. Q 1, Page 33] 14. Which of the following statement is correct? (1) E Cell and ∆r G of cell reaction both are extensive properties. (2) E Cell and ∆r G of cell reaction both are intensive properties. (3) E Cell is an intensive property while ∆r G of cell reaction is an extensive property. (4) E Cell is an extensive property while ∆r G of cell reaction is an intensive property. [NCERT Exemp. Q 3, Page 34] 15. The difference between the electrode potentials of two electrodes when no current is drawn through the cell is called ___________. (1) Cell potential (2) Cell emf (3) Potential difference (4) Cell voltage [NCERT Exemp. Q 4, Page 34] 16. Which of the following statement is not correct about an inert electrode in a cell? (1) It does not participate in the cell reaction. (2) It provides surface either for oxidation or for reduction reaction. (3) It provides surface for conduction of electrons. (4) It provides surface for redox reaction. [NCERT Exemp. Q 5, Page 34]

27 17. An electrochemical cell can behave like an electrolytic cell when ____________. (1) E cell = 0 (2) E cell > E ext (3) E ext > E cell (4) E cell = E ext [NCERT Exemp. Q 6, Page 34] 18. Which of the statements about solutions of electrolytes is not correct? (1) Conductivity of solution depends upon size of ions. (2) Conductivity depends upon viscosity of solution. (3) Conductivity does not depend upon solvation of ions present in solution. (4) Conductivity of solution increases with temperature. [NCERT Exemp. Q 7, Page 34] 19. The quantity of charge required to obtain one mole of aluminium from Al2O3 is ___________. (1) 1F (2) 6F (3) 3F (4) 2F [NCERT Exemp. Q 13, Page 35] 20. The cell constant of a conductivity cell _____________. (1) changes with change of electrolyte. (2) changes with change of concentration of electrolyte. (3) changes with temperature of electrolyte. (4) remains constant for a cell. [NCERT Exemp. Q 14, Page 35] 21. While charging the lead storage battery ______________. (1) PbSO4 anode is reduced to Pb. (2) PbSO4 cathode is reduced to Pb. (3) PbSO4 cathode is oxidised to Pb. (4) PbSO4 anode is oxidised to PbO2. [NCERT Exemp. Q 15, Page 35] 22. Λ0m (NH OH) is equal to ______________. 4 (1) Λ0m (NH OH) + Λ0m(NH Cl) – Λ0m(HCl) 4 4 (2) Λ0m (NH Cl) + Λ0m(NaOH) – Λ0m(NaCl) 4 (3) Λ0m (NH Cl) + Λ0m(NaCl) – Λ0m(NaOH) 4 (4) Λ0m (NaOH) + Λ0m(NaCl) – Λ0m(NH OH) 4 [NCERT Exemp. Q 15, Page 32] 23. 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) – 44.5 kJ (2) 89.0 kJ (3) – 89.0 kJ (4) –98.0 kJ 24. In an electrochemical process, a salt bridge is used: (1) as a reducing agent (2) as an oxidizing agent (3) to complete the circuit so that current can flow (4) None of these 25. An electrochemical cell behaves like an electrolytic cell when: (1) Ecell = Eexternal (2) Ecell = 0 (3) Eexternal > Ecell (4) Eexternal < Ecell 26. What will happen during the electrolysis of aqueous solution of CuSO4 by using platinum electrodes? (1) Copper will deposit at cathode. (2) Copper will deposit at anode.

28 Oswaal CUET (UG) Chapterwise Question Bank (3) Oxygen will be released at anode. (4) Copper will dissolve at anode. 27. Which of the statements about solutions of electrolytes is not correct? (1) Conductivity of solution depends upon size of ions. (2) Conductivity does not depend upon solvation of ions present in solution (3) Conductivity depends upon viscosity of solution. (4) Conductivity of solution increases with temperature. 28. When 0.1 mol 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 29. The number of electrons delivered at the cathode during electrolysis by a current of 1 ampere in 60 second is : (charge on electron = 1.60 × 10–19 C) (1) 6 × 1023 (2) 6 × 1020 (3) 3.75 × 1020 (4) 7.48 × 1023 30. 5A current is passed through a solution of zinc sulphate for 40 min. The amount of zinc deposited at the cathode is: (1) 40.65 g (2) 0.4065 g (3) 4.065 g (4) 65.04 g [B]ASSERTION REASON QUESTIONS Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (1) Both A and R are correct and R is The correct explanation of A. (2) Both A and R are correct but R is NOT The correct explanation of A. (3) A is correct but R is not correct. (4) A is not correct but R is correct. 1. Assertion (A): Cu is less reactive than hydrogen. Reason (R): E°Cu2+/Cu is negative.  [NCERT Exemp. Q 56, Page 42] 2. Assertion (A): Ecell should have a positive value for the cell to function. Reason (R). E cathode < E anode  [NCERT Exemp. Q 57, Page 42] 3. Assertion (A): Conductivity of all electrolytes decreases on dilution. Reason (R): On dilution number of ions per unit volume decreases.  [NCERT Exemp. Q 58, Page 42] 4. Assertion (A): Mercury cell does not give steady potential. Reason (R): In the cell reaction, ions are not involved in solution.  [NCERT Exemp. Q 60, Page 42] 5. Assertion (A): Electrolysis of NaCl solution gives chlorine at anode instead of O2. Reason (R): Formation of oxygen at anode requires overvoltage.  [NCERT Exemp. Q 61, Page 42]

CHEMISTRY

6. Assertion (A): For measuring resistance of an ionic solution an AC source is used. Reason (R): Concentration of ionic solution will change if DC source is used.  [NCERT Exemp. Q 62, Page 42] 7. Assertion (A): Current stops flowing when E cell = 0. Reason (R): Equilibrium of the cell reaction is attained.  [NCERT Exemp. Q 63, Page 42] 8. Assertion (A): E Ag+ /Ag increases with increase in concentration of Ag+ ions. Reason (R): E Ag+ /Ag has a positive value  [NCERT Exemp. Q 64, Page 42] 9. Assertion (A): Copper sulphate can be stored in zinc vessel. Reason (R): Zinc is less reactive than copper.  [NCERT Exemp. Q 65, Page 42] 10. Assertion (A): Λm for weak electrolytes shows a sharp increase when the electrolytic solution is diluted. Reason (R): For weak electrolytes degree of dissociation increases with dilution of solution. [C] COMPETENCY BASED QUESTIONS I. Based on following passage answer questions from 1-5.

A potential difference developed between the electrode and electrolyte is called electrode potential. When the concentrations of all the species involved in a half cell is unity, then the electrode potential is known as standard electrode potential. In a galvanic cell, the half-cell in which oxidation takes place is called anode and it has a negative potential with respect to solution. The other half cell in which reduction takes place, is called cathode and it has a positive potential with respect to solution. Thus, there exists a potential difference between the two electrodes, cathode and anode. This difference is called cell potential and is measured in volts. It is called the cell electromotive force when no current is drawn through the cell. A galvanic cell is represented by putting a vertical line between metal and electrolyte solution and putting a double vertical line between the two electrolytes solution by salt bridge. Under this convention, emf of cell is positive and is given as E cell = Eright – Eleft



The standard electrode potential is very important. The value at standard electrode potential of an electrode is greater than zero, then its reduced form is more stable compared to hydrogen gas. The value at some standard electrode potentials at 298 K are given below (ions are present as aqueous species and H2O as liquid) E° V + Ag / Ag(s) 0.80 Cu 2+ / Cu(s) 0.34 Pb 2+ / Pb(s) –0.13 Fe 2+ / Fe(s) –0.44 Mg 2+ / Mg(s) –2.36 [CUET 2022, 23rd Aug]

29

ELECTROCHEMISTRY

1. The strongest oxidising agent amongst the following Ag+, Cu2+, Fe2+, Pb2+ is: (1) Ag+ (2) Cu2+ (3) Fe2+ (4) Pb2+ 2. Amongst the following the weakest reducing agent is: (1) Mg (2) Pb (3) Fe (4) Cu 3. The emf of the cell: Ag(s) |Ag+(1m) || Pb2+(1m)| Pb(s), is: (1) 0.67 V (2) 1.06 V (3) –0.93 V (4) 0.93V 4. When Pb is added to an aqueous solution of a mixture of Cu 2+ and Mg 2+ ions, it is: (1) Cu2+ is reduced (2) Mg2+ is reduced (3) Cu is reduced (4) Pb2+ is reduced 5. The combination of electrodes which will give maximum value of E° cell at 298 K is: (1) Anode Cathode Ag Mg (2) Anode Cathode Cu Fe (3) Anode Cathode Mg Ag (4) Anode Cathode Pb Mg II. Based on following passage answer questions from 6-10: The cell constant is usually determined by measuring the resistance of the cell containing a solution whose

6. (1) (3) 7. (1) (3) 8. (1) (2) (3) (4) 9. (1) (3) 10. (1) (2) (3) (4)

conductivity is already known. For this purpose, we generally use KCl solutions whose conductivity is known accurately at various concentrations and at different temperatures. Consider the resistance of a conductivity cell filled with 0.1 M KCl solution is 200 Ohm, if the resistance of the same cell when filled with 0.02 M KCl solution is 420 Ohm. (Conductivity of 0.1M KCl solution is 1.29 Sm–1) What is the conductivity of 0.02 M KCl solution? 0.452 S m–1 (2) 0.215 S m–1 0.614 S m–1 (4) 0.433 S m–1 What will happen to the conductivity of the cell with the dilution? First decreases then increases (2) Increases First increases then decreases (4) Decreases The cell constant of a conductivity cell: changes with change of electrolyte. changes with change of concentration of ele­ctrolyte. changes with temperature of electrolyte. remains constant for a cell. SI unit for conductivity of a solution is: S m–1 (2) S m2 mol–1 mol cm–3 (4) S cm2 mol–1 Normally KCl solution is used to determine the resistance of a cell because: it is easily available it is stable at all temperatures it is easy to handle its conductivity is known accurately

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (4)

2. (3)

3. (2)

4. (3)

5. (3)

6. (3)

7. (4)

8. (3)

9. (4)

10. (1)

11. (4)

12. (3)

13. (3)

14. (3)

15. (2)

16. (4)

17. (3)

18. (3)

19. (3)

20. (4)

21. (1)

22. (2)

23. (3)

24. (3)

25. (3)

26. (3)

27. (2)

28. (2)

29. (3)

30. (3)

8. (2)

9. (4)

10. (1)

8. (4)

9. (1)

10. (4)

[B] ASSERTION REASON QUESTIONS 1. (3)

2. (3)

3. (1)

4. (4)

5. (1)

6. (1)

7. (1)

[C] COMPETENCY BASED QUESTIONS 1. (1)

2. (4)

3. (3)

4. (1)

5. (3)

6. (3)

7. (4)

ANSWERS WITH EXPLANATION [A] MULTIPLE CHOICE QUESTIONS: 1. Option (4) is correct. Explanation: Ionic conductance is electrical conductance due to the motion of ionic charge. Elementary science introduces this phenomenon as a property of liquid electrolyte solution. The conductance of a solution depends on:

•  the nature of the electrolyte added, •  size of the ions produced •  their solvation •  the nature of the solvent •  its viscosity •  concentration of the electrolyte • temperature.

30 Oswaal CUET (UG) Chapterwise Question Bank For example, with increases in dilution, the number of ions present in the solution increases and the conductance of the solution increases. 2. Option (3) is correct. Explanation: When an electric current is passed through a concentrated sodium chloride solution, hydrogen gas forms at the negative electrode, chlorine gas forms at the positive electrode, and a solution of sodium hydroxide is also formed. NaCl(aq) → Na+(aq) + Cl –(aq) At cathode: H2O (l) + e- → 1/2 H2 (g) + OH-(aq) At anode: Cl–(aq) → 1/2 Cl2 (g) +e– Net reaction: NaCl(aq) + H2O (l) → Na+(aq) + OH–(aq)+ 1/2 Cl2 (g) + 1/2 H2 (g) 3. Option (2) is correct.

Explanation: When current flows through a conductor, the relation between charge and the current flowing through conductor per unit time is given by: Q = I  t , Where, I = current flowing through the conductor, A Q = amount of charge flowing, C = charge t = time Q = 5  193 = 965 C Now, 1 F = charge of 1 mole electrons Hence, 0.01 F = charge of 0.01 mole electrons = charge of 6.02  1023  0.01 electrons (as 1 mole of electrons) = (6.02  1023 atoms) = 6.02  1021 electrons 4. Option (3) is correct. Explanation: Kohlrausch's law of independent migration of ions states that the limiting molar conductivity of an electrolyte can be represented as the sum of individual contributions of its cations and anions. This law is applicable to both strong and weak electrolytes. 5. Option (3) is correct. Explanation: When a lead storage battery is recharged, At anode, oxidation takes place. PbSO4 + 2H2O → PbO2 + SO4 2− + 4H+ + 2e− Net reaction: 2PbSO4 + 2H2O → Pb + PbO2 + 2H2SO4 Thus, sulphuric acid is formed. 6. Option (3) is correct. Explanation: Since only acetic acid is a weak electrolyte among them. When more amount of solvent is added to the solution of a weak electrolyte, it helps in dissociating the molecules into ions of a weak electrolyte. Thus, the degree of dissociation of the weak electrolyte increases upon dilution. The rest of the salts are strong electrolytes 7. Option (4) is correct. Explanation: 2Al + 3Ni2+ → 2Al3+ + 3Ni (n = 6) ∆G° = –nFE° = –6 × 96500 × 1.41 = –816390 J mol–1

CHEMISTRY

Λ°m (CH3COOH) = Λ°m (CH3COONa) + Λ°m (HCl) – Λ°m NaCl = 91 + 425.9 – 126.4 = 390.5 cm2 mol–1 9. Option (4) is correct. Explanation: Molar conductivity of solution of strong electrolyte increases rapidly with dilution & hence, the graph is linear with negative slope. 10. Option (1) is correct. Explanation: RTQ Nernst eqn. is given as E cell = °E cell nF [Pr oducts] Where, Q = [Reactants]

RT ln K c nF Also at equilibrium, Gibb’s energy is given by ∆G° = –nFE°cell Where, E°cell = E°cathode – E°anode E°cell = E°right – E°left 11. Option (4) is correct. But at equilibrium, Q = K c , E 0cell =

Explanation: Corrosion is the deterioration of a material that affects its durability. To prevent corrosion use of barrier coatings like paint, plastic, or powder including epoxy, nylon, and urethane, adhere to the metal surface to create a thin film, with some other reactive metal (Sn, Zn etc.) or with the sacrificial electrode of another metal (Mg, Zn, etc.) which corrodes itself but saves the object. Galvanisation: Galvanised metal is coated with a thin layer of zinc to protect it against corrosion. Alloying: A method of improving the properties of a metal by mixing the metal with another metal or nonmetal. Painting: Rusting of iron can be easily prevented by coating the surface with paint which protects iron from air and moisture. 12. Option (3) is correct. Explanation: Ka = 1.78 × 10–5 mol L–1 Λm = 48.15 S cm2 mol–1 Λm° = 390.5 S cm2 mol–1 α = Λm /Λm° = 4.185/ 390.5 =0.1233 . . . c 2a 2 ca 2 Since, K a = = c(1 - a) c(1 - a) = -1.78 ´ 10-5 =

c(0.1233) 2 (1 - 0.1233)

C = 0.001027 molL–1 1000 ´ k Now, Lm = c L mc k= 1000

8. Option (3) is correct.

κ = 48.15 × 0.001027/ 1000 κ = 4.9 × 10–5 S cm–1

Explanation:

13. Option (3) is correct.

31

ELECTROCHEMISTRY

Explanation: When copper electrode is connected to standard hydrogen electrode, it acts as cathode and its standard electrode potential can be measured. E° = E°R – E°L = E°R – 0 = E°R Pt(s) | H2 (g., 1bar)| H+ (aq.,1M | | Cu2+ (aq.,1M)|Cu will measure standard electrode potential of copper electrode. So, to calculate the standard electrode potential of the given cell it is coupled with the standard hydrogen electrode in which pressure of hydrogen gas is one bar and H+ ion in the solution is one molar and also the concentrations of the oxidized and the reduced forms of the species in the right hand half-cell is unity.

PbSO4(s) + 2H2O → PbO2(s) + SO42- + 4H+ + 2e- (Oxidation) Overall reaction: 2PbSO4(s) + 2H2O → Pb(s) + PbO2(s) + SO42- (aq) + 4H+(aq) + 2e-

14. Option (3) is correct.

23. Option (3) is correct.

Explanation: Ecell is an intensive property as it does not depend upon mass of species (number of particles) but ∆rG of the cell reaction is an extensive property because this depends upon mass of species (number of particles). 15. Option (2) is correct. Explanation: EMF is the difference between the electrode potentials of two electrodes, when no current is drawn through the cell. 16. Option (4) is correct. Explanation: Inert electrode does not participate in redox reaction and acts only as source or sink for electrons. It provides surface either for oxidation or for reduction reaction. 17. Option (3) is correct. Explanation: If an external opposite potential is applied on the galvanic cell and increased reaction continues to take place till the opposing voltage reaches the value 1.1 V. At this stage no current flow through the cell and if there is any further increase in the external potential then reaction starts functioning in opposite direction. Hence, this works as an electrolytic cell. 18. Option (3) is correct. Explanation: Conductivity depends upon salvation of ions present in the solution. The greater the salvation of ions, the lesser is the conductivity. 19. Option (3) is correct. Explanation: Al2O3 → 2Al+3 + 3O-2 3+ Al + 3e → Al (for 1 mole) 3F charge is required to obtain 1 mole of al from Al2O3. 20. Option (4) is correct.

Explanation: The cell constant of a conductivity cell remains constant for a cell. 21. Option (1) is correct. Explanation: While charging the lead storage battery the reaction occurring on cell is reversed and PbSO4(s) on anode and cathode is converted into Pb and PbO2 respectively. Hence, option (A) is the correct choice The electrode reactions are as follows: At cathode: PbSO4 (s) + 2e- → Pb(s) + SO42- (aq) (Reduction) At anode:

22. Option (2) is correct. Explanation: NH4Cl ⇋ NH4+ + Cl– NaCl ⇋ Na+ + Cl– NaOH ⇋ Na+ + OH– NH4OH ⇋ NH4+ + OH– Hence, Λ0m (NH Cl) + Λ0m(NaOH) – Λ0m(NaCl) = Λ0m (NH4OH) 4

Explanation: The relation between standard Gibbs free energy and E°cell is ∆G° = – nFE0cell For the cell reaction 2Ag+ + Cu Cu+2 + 2Ag ∆E°cell = 0.46 V   ∆G° = –nFE°cell   ∆G° = –2 × 96500 × 0.46    = –88780 J    ∆G° = –89.0 kJ 24. Option (3) is correct. Explanation: In an electrochemical cell, a salt bridge is used to complete the circuit so that current can flow. 25. Option (3) is correct. Explanation: If an external opposite potential is applied on the electrochemical cell, the reaction continues to take place till the opposite voltage reaches the value 1.1V. At this stage, no current flow through the cell and if there is any further increase in the external potential (Eexternal), then reaction starts functioning in opposite direction i.e., an electrochemical cell behaves like an electrolytic cell. Eexternal > Ecell 26. Option (3) is correct. Explanation: CuSO4 ⇋ Cu2+ + SO42H2O ⇋ H+ + OH− At cathode, Cu2+ + 2e− → Cu; E0cell = 0.34 V H+ + e− → 1/2 H ; E0cell = 0.00 V This reaction will take place due to higher reduction potential. At anode, 2SO42- + 2e- → S2O82-; E0cell =1.96 V + 0 2H2O → O2 + 4H + 4e ; E cell = 1.23 V The reaction with lower value of E° will be preferred at anode, hence O2 is released at anode. 27. Option (2) is correct.

Explanation: Conductivity depends upon solvation of ions present in solution. Greater the solvation of ions of an electrolyte, lesser will be the electrical conductivity of the solution.

32 Oswaal CUET (UG) Chapterwise Question Bank 28. Option (2) is correct. Explanation: When 0.1 mole of CoCl3(NH3)5 was reacted with excess of AgNO3, we get 0.2 moles of AgCl. So, there are two chloride ions that are free and not part of the complex. The formula for complex has to be [Co(NH3)5Cl]Cl2. [Co(NH3)Cl]Cl2 → [Co(NH3)5Cl]2+ + 2Cl− Therefore, the conductivity of the solution will be 1: 2 electrolyte. 29. Option (3) is correct. Explanation: We know that, Q = It , Q = 1 × 60 = 60 C Now, 1.60 × 10–19 C = 1 electron

CHEMISTRY

Explanation: Ag+ + e → Ag E Ag+ /Ag = E0 Ag+ /Ag – 0.059 log 1 1  [Ag]+ 0 E Ag+ /Ag = E Ag+ /Ag + 0.059 log[Ag]+ On increasing [Ag]+, E Ag+ /Ag will also increase and will have a positive value. 9. Option (4) is correct. Explanation: Zinc will get dissolved in CuSO4 solution since zinc is more reactive than copper. 10. Option (1) is correct.

= 3.75 × 1020 electrons

Explanation: Weak electrolytes dissociate partially in concentrated solution. On dilution, their degree of dissociation increases hence, their Λm increases sharply

30. Option (3) is correct.

[C] COMPETENCY BASED QUESTIONS

60 = 3.75 ´ 1020 60C = 1.6 ´ 10-19

Explanation: Given that I = 5 ampere t = 40 min = 40 × 60 = 2400 sec Amount of electricity passed Q = It = 5 × 2400 = 12000 C Zn2+ + 2e– → Zn (n = 2e–) From Faraday's first law, W = Z It (Z = equivalent mass) Mass 65.39 = g of Zn nF 2 ´ 96500 Therefore, 12000 C charge will deposit 65.38 ´ 12000 = = 4.065 g of Zn 2 ´ 96500 [B]ASSERTION REASON QUESTIONS 1. Option (3) is correct. Explanation: Cu is less reactive than hydrogen because E°Cu2+ /Cu is positive. 2. Option (3) is correct. Explanation: E cell = E cathode – E anode. To have positive value of E cell, E cathode > E anode 3. Option (1) is correct. Explanation: Conductivity depends on number of ions per unit volume which decreases on dilution of electrolytes. 4. Option (4) is correct. Explanation: Mercury cell gives a steady potential because in the cell reaction ions are not involved in the solution. 5. Option (1) is correct. Explanation: Formation of oxygen has lower value of E° than formation of chlorine even then it is not formed because it requires overvoltage. 6. Option (1) is correct. Explanation: Alternating current is used in the measurement of resistance of electrolyte solution because concentration changes with DC current due to electrolysis. 7. Option (1) is correct. Explanation: At equilibrium, Ecell = 0 and no current flows. 8. Option (2) is correct.

1. Option (1) is correct. Explanation: The one with the more positive reduction potential is the stronger oxidising agent. E°RP: Ag+ >Cu2+ >Pb2+ >Fe2+ >Mg2+ 2. Option (4) is correct. Explanation: Less the oxidation potential the element will have less tendency to get oxidised and hence lesser reducing power. E0OP: Mg>Fe>Pb>Cu>Ag 3. Option (3) is correct. Explanation: E cell = E 0 cell -

0.059 [Products] log n [Reactants]

E cell = E 0 ( Pb + 2 / Pb ) - E 0 ( Ag + / Ag ) cell = = (-0.13) - (0.8) -

0.059 [Ag + ]2 log 2+ 1 [Pb ] 3

0.059 log1 2

= –0.93 V 4. Option (1) is correct. Explanation: When Pb is added to an aqueous solution of a mixture of Cu2+ and Cu2+ ions, Cu2+ is reduced. As reduction potential of Cu2+ is higher than Pb2+ and Mg2+, so Pb will reduce Cu2+ into Cu. 5. Option (3) is correct. Explanation: E0cell = E0(Mg+2/Mg) – E0(Ag+/Ag) = –3.26 V E0cell = E0(Fe+2/Fe) – E0(Cu+2/Cu) = –0.78 V E0cell = E0(Ag+1/Ag) – E0(Mg+2/Mg) = +3.16 V E0cell = E0(Mg+2/Mg) – E0(Pb+2/Pb) = –2.23 V 6. Option (3) is correct.

Explanation: Conductivity of 0.02 mol L–1 KCl Solution = Cell constant resistance = 258 /420 = 0. 614 S m–1 7. Option (4) is correct. Explanation: The conductivity decreases with dilution.

33

ELECTROCHEMISTRY

8. Option (4) is correct.

Explanation: SI unit for conductivity of a solution is S m–1.

Explanation: The cell constant of a conductivity cell remains constant for a cell

10. Option (4) is correct.

9. Option (1) is correct.

Explanation: The conductivity values of KCl is known accurately at different temperatures and concentrations.

Study Time

CHAPTER

Max. Time: 1:50 Hours Max. Questions: 50

4

CHEMICAL KINETICS

 Revision Notes

  Chemical kinetics is the study of chemical reactions with respect to reaction rates, effects of various variables, rearrangement of atoms and formation of intermediates. Scan to know more about this topic

  Rate of reaction:

For the reaction: A + B → C of diappearence of A = – ΔA /Δt { Rate of diappearence of B= – ΔB /Δt { Rate of diappearence of C= + ΔC /Δt Factors: Rate of reaction { Increases with increase in temperature. { Decreases with the passage of time as the concentration of reactants decrease. { Increase with increase in the concentration of products per unit time. Scan to know more about this { Presence of catalyst. topic { Nature of the reactants. { Surface area of the reactants. { Effect of radiations. { Rate



  Instantaneous rate: Factors { The rate of change in concentration affecting rate of any of the reactants or products at of reaction a particular time. { For the reaction: A + B → C Instantaneous rate: dx/dt= –d[A]/dt = –d[B]/dt = +d[C]/dt   Average rate of reaction: { The rate of reaction measured over a longtime interval. { Δx /Δt , Δx = change in concentration in given time , Δt = time taken

Second order reaction

2

mol L1 1 ´ = mol-1 Ls -1 2 s mol L1

 Molecularity: { The number of reacting species taking part in an elementary reaction, which must collide simultaneously in order to bring about a chemical reaction. { Cannot be zero or a non-integer. { It is applicable only for elementary reactions. { For complex reaction it has no meaning   Order of the reaction: { The sum of powers of the concentration of the reactants in the rate law expression. { An experimental quantity, can be zero and even a fraction. { It is applicable to elementary as well as complex reactions. x y { Rate = k [A] [B] , order of the reaction = x + y Example 1: Calculate the overall order of a reaction which has the rate expression: (a) Rate = k [A]1/2 [B]3/2 (b) Rate = k [A]3/2 [B]–1 Solution: (a) Rate = k [A]x [B]y order = x + y, Order = 1/2 + 3/2 = 2, i.e., second order (b) order = 3/2 + (–1) = 1/2, i.e., half order. a. Zero order reaction: { The rate of the reaction is proportional to zero power of the concentration of reactants. 0 { Rate = k [NH3] = k { Graphical representation:

  Rate Law: x y { Rate = k [A] [B] { k is a proportionality constant called rate constant. { Units of rate constant k is depends on the overall reaction order. Reaction

Order

Zero order reaction

0

First order reaction

1

Units of rate constant mol L1 1 = mol L-1 s -1 ´ s mol L10

mol L1 1 = s -1 ´ s mol L1

Fig 4.1: Variation in the concentration vs time plot for a zero-order reaction

CHEMICAL KINETICS

35

36 Oswaal CUET (UG) Chapterwise Question Bank

b. First order reaction: { The rate of the reaction is proportional to the first power of the concentration of the reactant R. dR { Rate = = - k[R ] dt 2.303 [R ]0 log { k = t [R ]

[R]0 = initial concentration of reactant [R] = concentration of reactant at time t { Graphical representation:

Fig: 4.2: Plot of log [R]0/[R] vs time for a first order reaction c. Pseudo first order reactions: ƒ Are not truly first order but become first order under certain conditions. ƒ Unit: s ƒ Example: inversion of sugar and acidic hydrolysis of an ester. Example 1: Hydrolysis of methyl acetate in aqueous solution has been studied by titrating the liberated acetic acid against sodium hydroxide. The concentration of the ester at different times is given below. t/min

0

30

60

CHEMISTRY

We can now determine k from k´ [H2O] = 2.004 × 10–3 min–1 k´ [55 mol L–1] = 2.004 × 10–3 min–1 k´ = 3.64 × 10–5 mol–1 L min–1   Half life: {  The time in which the concentration of a reactant is reduced to one half of its initial concentration. { Represented as t1/2. { For a zero order reaction, t1/2 = R°/ 2k, dependent on the concentration of reactant. { For a first order reaction, t1/2= 0.0693/k, independent of the concentration of reactant. Example 2: A first order reaction is found to have a rate constant, k = 5.5 × 10–14 s–1. Find the half-life of the reaction. Solution: Half-life for a first order reaction is t1/2= 0.693/k 0.693 = ×10–14 s = 1.26 × 1013 s 5.5   Arrhenius equation: –Ea /RT { k = A e { A is the Arrhenius factor or the frequency factor. { Ea is activation energy measured in joules/mole. k2 Ea [T2 - T1 ] = { log k 2 2.303 T1T2

Scan to know more about this topic

90

C/mol 0.8500 0.8004 0.7538 0.7096 L–1 Show that it follows a pseudo first order reaction, as the concentration of water remains nearly constant (55 mol L–1), during the course of the reaction. What is the value of k′ in this equation? Rate = k′ [CH3COOCH3][H2O] Solution: For pseudo first order reaction, the reaction should be first order with respect to ester when [H2O] is constant. The rate constant k for pseudo first order reaction is k = 2.303 log [C]0 / [C] t k = k′ [H2O] t/min

C/ mol L–1

k′ / min–1

0 30 60 90

0.8500 0.8004 0.7538 0.7096

– 2.004  10–3 2.002  10–3 2.005  10–3

It can be seen that k´ [H2O] is constant and equal to 2.004 × 10–3 min–1 and hence, it is pseudo first order reaction.

Fig: 4.3: Diagram showing plot of potential energy vs reaction coordinate.

Activation energy

Factors:

a. Activation energy: ƒ Lowering Ea will increase in the rate of reaction. b. Temperature: ƒ For a chemical reaction with rise in temperature by 10°, the rate constant is nearly doubled.

Fig: 4.4: Distribution curve showing temperature dependence of rate of a reaction

37

CHEMICAL KINETICS





Fig: 4.5: Effect of positive catalyst on activation energy c. Catalyst ƒ Presence of a positive catalyst lowers the activation energy by providing an alternate path for the reaction. 2 2KClO3 ¾MnO ¾¾ ® 2 KCl + 3O 2

d. Collision theory: Scan to know more about this ƒ  Refers to the orientation of topic molecules leading to effective collisions. ƒ The number of collisions per second per unit volume of the reaction mixture is known as Collision collision frequency (Z). theory ƒ Activation energy and proper orientation of the molecules together determine the criteria for an effective collision and hence the rate of a chemical reaction. –Ea /RT ƒ Modified Arrhenius equation; k = PZ AB e

Intermediate Fig: 4.6: Reaction happens only when molecules collide forming an unstable intermediate

ƒ Presence

of a negative catalyst elevates the activation energy thus hindering the reaction.

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS 1. The decomposition of NH3 on the platinum surface is zero order reaction. If k = 2.5 × 10–4 mol L–1 s–1 the rate of production of H2 is: (1) 2.5 × 10–4 mol L–1 s–1 (2) 7.5 × 10–4 mol L –1 s–1 (3) 3 5.0 × 10–4 mol L–1 s–1 (4) 10.0 × 10–4 mol L–1 s–1  2. (1) (3) 

th

[CUET 2022, 10 Aug] The molecularity of the following elementary reaction is: NH4NO2 → N2 + 2H2O Zero (2) One Two (4) Three

The rate law is : r = k[A]1 [B]1, Molecularity of the above reaction will be: (1) 21/2 (2) 1 (3) 2 (4) 0



[CUET 2022, 17th Aug]

4. When temperature changes from 293 K to 313 K, the rate of reaction: (1) Remains same (2) Gets halved (3) Gets doubles (4) Gets quadruples

(2)

(3)

[CUET 2022, 18th Aug] 5. The decomposition of dimethyl ether leads to formation of CH4, H2 and CO. The reaction rate is given by: Rate = k[PCH3 OCH3]3/2 If pressure is measured in bar and time in minutes, what is the unit of rate constant? (1) bar-1/2 min-1 (2) bar 3/2 min-1 (3) bar-1 min-1 (4) bar-3/2 min-1 

[CUET 2022, 18th Aug]

[CUET 2022, 20th Aug]

7. Which of the following graphs are correct for the first order reaction? (1)

[CUET 2022, 10th Aug]

3. In the first-order reaction the concentration of the reactant is reduced 1/4 th in 60 minutes, what will be its half-life?  (1) 120 minutes (2) 40 minutes (3) 30 minutes (4) 25 minutes 

6. For an elementary reaction given below: A + B → Product

(4)

38 Oswaal CUET (UG) Chapterwise Question Bank Choose the correct answer from the options given below. (1) B and C only (2) A and B only (3) A and C only (4) C and D only  [CUET 2022, 20th Aug] 8. When the temperature of a reaction is increased by 20°C, the rate of reaction increases by: (1) 3 times (2) 4 times (3) 2 times (4) 1.5 times [ CUET 2022, 23rd Aug]

 9. For a chemical reaction:

A + B → Products Experiment

[ A] mol L-1

[ A] mol L-1

Initial rate mol L-1 s -1

1

0.1

0.1

2.0  10–3

2

0.2

0.2

4.0  10–3

3

0.1

0.2

2.0  10–3

Which is the overall order of chemical reaction? (1) 3 (2) 1 (3) 2 (4) 0 

10. The rate of a gaseous reaction becomes half when the volume of the vessel is doubled. The order of the reaction is: (1) Zero (2) First (3) Second (4) Third th

[CUET 2022, 30 Aug]

11. In the presence of a catalyst, heat evolved or absorbed during reaction : (1) increases. (2) decreases. (3) remains unchanged. (4) may increase or decrease 



[NCERT Exemp. Q 2, Page 37]

12. A graph of volume of hydrogen released vs. time for the reaction between zinc and dilute HCl is given in figure. On the basis of this mark the correct option:

13. Which of the following statement is correct? (1) The rate of a reaction decreases with passage of time as the concentration of reactants decreases. (2) The rate of a reaction is same at any time during the reaction. (3) The rate of a reaction is independent of temperature change. (4) The rate of a reaction decreases with increase in conc­ entration of reactant(s). 

[NCERT Exemp. Q 11, Page 50]

14. Rate law for the reaction A+2B → C, is found to be Rate = k[A][B]. If the concentration of reactant B is doubled keeping the concentration of A constant, the value of rate constant will be : (1) the same. (2) doubled. (3) quadrupled. (4) halved. [NCERT Exemp. Q 14, Page 51]

15. A first-order reaction is 50% completed in 1.26 × 1014 s. How much time would it take for 100% completion?  (1) 1.26 × 1015 s (3) 2.52 × 1028 s

(2) 2.52 × 1014 s (4) Infinite



[NCERT Exemp. Q 16, Page 51]

16. Compounds ‘A’ and ‘B’ react according to the following chemical equation :

A(g) + 2B(g) → 2C(g)



Concentration of either ‘A’ or ‘B’ were changed keeping the concentration of one of the reactants constant and rates were measured as function of initial concentration. Following results were obtained.



Choose the correct option for this reaction: Experimnent Initial Initial Initial concentration concentration concentration of of [A]/mol L-1 of [B]/mol L-1 [C]/mol L-1 s-1

1. 2. 3.

0.30 0.30 0.60

(1) Rate = k [A]2[B] (3) Rate = k [A][B] 

(1) Average rate up to 40 seconds is

V3 - V2 40

(2) Average rate up to 40 seconds is

V3 - V2 40 - 30

(3) Average rate up to 40 seconds is

V3 40

V3 - V2 40 - 20

[NCERT Exemp. Q 8, Page 49]



[CUET 2022, 23rd Aug]



(4) Average rate up to 40 seconds is

CHEMISTRY

0.30 0.60 0.30

0.10 0.40 0.20

(2) Rate = k [A][B]2 (4) Rate = k [A]2[B]0 [NCERT Exemp. Q 17, Page 51]

17. Consider the reaction A → B. The concentration of both the reactants and the products varies exponentially with time. Which of the following figures correctly describes the change in concentration of reactants and products with time? (1)

39

CHEMICAL KINETICS

(2)

(4)

(3)

 [NCERT Exemp. Q 5, Page 48] 21. The half-life period for a zero-order reaction is equal to 2k 0.693 (1) (2) [ R ]0 k (3)

k=

P 2.303 log i Pi + x t

2.303 (4) k

[ R ]0

2k (where [R]0 is initial concentration of reactant and k is rate constant). 22. The slope in the plot of ln[R] vs. time gives (1) +k (2) +k/2.303 (3) –k (4) −k/2.303

(4)

23. In a chemical reaction X → Y, it is found that the rate of reaction doubles when the concentration of X is increased four times. The order of the reaction with respect to X is (1) 1 (2) 0 (3) 2 (4) ½ 

[NCERT Exemp. Q 20, Page 52]

18. Which of the following statements is not correct about order of a reaction? (1) The order of a reaction can be a fractional number. (2) Order of a reaction is experimentally determined quantity. (3) The order of a reaction is always equal to the sum of the stoichiometric coefficients of reactants in the balanced chemical equation for a reaction. (4) The order of a reaction is the sum of the powers of molar concentration of the reactants in the rate law expression.  [NCERT Exemp. Q 9, Page 49] 19. The value of rate constant of a pseudo-first-order reaction: (1) depends on the concentration of reactants present in small amount. (2) depends on the concentration of reactants present in excess. (3) is independent of the concentration of reactants. (4) depends only on temperature. 

[NCERT Exemp. Q 19, Page 52]

20. Consider a first-order gas-phase decomposition reaction given below : A(g) → B(g) + C(g)

The initial pressure of the system before decomposition of A was ‘Pi.’ After lapse of time ‘t’, total pressure of the system increased by x units and became ‘Pt’. The rate constant k for the reaction is given as ___________.

(1)

k=

P 2.303 log i Pi - x t

(2)

k=

2.303 Pi log t 2Pi - Pt

(3)

k=

2.303 Pi log t 2Pi + Pt

24. How long will 5g of a reactant take to reduce to 3g? (Given: Rate constant, k = 1.15 × 10–3 s–1) (1) 222.189 s (2) 444.379 s (3) 111.095 s (4) 888.789 s 25. Under which condition a bimolecular reaction is kine­ tically first order reaction: (1) When two reactants are involved. (2) When one of the reactants is in excess. (3) When one of the reactants does not involve in reaction. (4) None of these 26. For the reaction, 2N2O5 → 4NO2 + O2, rate and rate constant are 1.02 × 10–4 and 3.4 × 10–5 s–1 respectively, then concentration of N2O5 at that time will be : (1) 1.732 (2) 3 (3) 1.02 × 10–4 (4) 3.4 × 105 27. The rate of first order reaction is 1.5 × 10–2 mol L–1 min–1 at 0.5 M concentration of the reactant. The half-life of the reaction is: (1) 0.383 min (2) 23.1 min (3) 8.73 min (4) 7.53 min 28. A substance A decomposes by a first order reaction starting initially with [A] = 2.00 m and after 200 min, [A] becomes 0.15 m. For this reaction t1/2 is: (1) 53.49 min (2) 50.49 min (3) 48.45 min (4) 46.45 min 29. What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20°C to 35°C? (H = 8.314 J mol–1 K–1) (1) 342 kJ mol–1 (2) 269 kJ mol–1 –1 (3) 34.7 kJ mol (4) 15.1 kJ mol–1 30. (1) (2)

A chemical reaction is catalysed by a catalyst X. Hence, X: reduces enthalpy of the reaction. decreases rate constant of the reaction.

40 Oswaal CUET (UG) Chapterwise Question Bank Reason (R): The unit of k is mole L–1s–1.

(3) increases activation energy of the reaction. (4) does not affect equilibrium constant of the reaction.



[B] ASSERTION REASON QUESTION: Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark The correct choice as: (i) Both A and R are correct and R is The correct explanation of A. (ii) Both A and R are correct but R is NOT The correct explanation of A. (iii) A is correct but R is not correct (iv) A is not correct but R is correct



1. Assertion (A): Order of the reaction can be zero or fractional.

Reason (R): We cannot determine order from balanced chemical equation.



[NCERT Exemp. Q 57, Page 59]

2. Assertion (A): Order and molecularity are same.

Reason (R): Order is determined experimentally and molecularity is the sum of the stoichiometric coefficient of rate determining elementary step.



[NCERT Exemp. Q 58, Page 59]

CHEMISTRY

9. Assertion (A): Reactions can occur at different speeds. Reason (R): Rate of reaction is also called speed of reaction.

10. Assertion (A): Rate of reaction doubles when concen­ tration of reactant is doubled if it is a first order reaction.

Reason (R): Rate constant also doubles.

[C] COMPETENCY BASED QUESTIONS I. Based on following passage answer questions from 1-5 The rate of a reaction is concerned with decrease in concentration of reactants or increase in concentration of products per unit time. It can be expressed as instantaneous rate at a particular instant of time and average rate over a large interval of time. A number of factors such as temperature, concentration of reactants, catalyst, affect the rate of a reaction. Mathematical representation of rate of a reaction is given by rate law. It has to be determined experimentally and cannot be predicted. Order of a reaction with respect to a reactant is the power of its concentration which appears in the rate law equation. Molecularity is defined only for an elementary reaction. Molecularity and order of an elementary reaction are same. [CUET 2022, 21st Aug]



3. Assertion (A): The enthapy of reaction remains constant in the presence of a catalyst.

1. Which of the following expressions is correct for the rate of the reaction given below?





Reason (R): A catalyst participating in the reaction, forms different activated complex and lowers down the activation energy but the difference in energy of reactant and product remains the same.



[NCERT Exemp. Q 59, Page 59]

4. Assertion (A): All collision of reactant molecules lead to product formation.

Reason (R): Only those collisions in which molecules have correct orientation and sufficient kinetic energy lead to compound formation.



[NCERT Exemp. Q 60, Page 59]

5. Assertion (A): Rate constants determined from Arrhenius equation are fairly accurate for simple as well as complex molecules.

Reason (R): Reactant molecules undergo chemical change irrespective of their orientation during collision.



[NCERT Exemp. Q 61, Page 59]

6. Assertion (A): Rate of reaction is a measure of change in concentration of reactant with respect to time.

Reason (R): Rate of reaction is a measure of change in concentration of product with respect to time.

7. Assertion (A): For a reaction: P + 2Q → Products, 1/2

1



Rate = k [P] [Q] , so the order of reaction is 1.5



Reason (R): Order of reaction is the sum of stoichiometric coefficients of the reactants.

8. Assertion (A): The unit of k is independent of order of reaction.

H2O2(aq) + 3I– (aq) + 2H+ → 2H2O(l) + I3–

(A)

D éë I - ùû

=3

(C)

D éë I - ùû

=

Dt

Dt

(1) 1 (3) 3

D éë H + ùû Dt

  (B) 

D éë I - ùû Dt

=

+ 2 D éë H ùû 3 Dt

+ D éH+ ù D éI- ù 3 D éë H ùû   (D)  ë û = 2 ë û Dt Dt 2 Dt



(2) 2 (4) 4

2. Rate constant for a first order reaction is 0.3465 × 10– 3 s –1. The t1/2 for the same reaction is: 1. 2000 s 2. 33.33 hr 3. 0.555 × 103 hr 4. 2 × 10-3 s (1) 1 (2) 2 (3) 3 (4) 4 3. 75% of a first order reaction was completed in 32 minutes. 50% of the reaction was completed in: 1. 24 minutes 2. 16 minutes 3. 8 minutes 4. 4 minutes (1) 1 (2) 2 (3) 3 (4) 4 4. Given below is a biomolecular elementary reaction.

2HI → H2 + I2

The order of the above reaction is: (1) 1 (2) 2 (3) 3 (4) 4 5. Which of the following statement is incorrect? (A) Order and molecularity for an elementary reaction are always same.

41

CHEMICAL KINETICS

(B) Rate of reaction can be expressed as increase in concentration of products formed in unit time. (C) Rate law can be written with the help of balanced chemical equation. (D) Rate of reaction depends upon temperature of the reaction. (1) 1 (2) 2 (3) 3 (4) 4 II. Based on following passage answer questions from 6-10:

The rate of the reaction is proportional to the concentration of the reactant. Hydrogenation of ethene results in the formation of ethane. The rate constant, k for the reaction was found to be 2.5 × 10–15s–1. The concentration of the reactant reduces to one-third of the initial concentration in 5 minutes.

6. Find the order of reaction: (1) Zero order (3) Second order

(2) First order (4) Fractional order

7. The rate law equation is: (1) Rate = k [C2H6] (3) Rate = k [C2H4]

(2) Rate = k [C2H4]2 (4) Rate = k [C2H4]2

8. The half-life for the reaction is: (1) 2.772 × 10–24 s (2) 2.772 × 10–12 s –24 (3) 1.386 × 10 s (4) 1.386 × 10–12 s 9. The rate constant of the reaction after 5 minutes is: (1) 0.4290 min–1 (2) 0.1297 min–1 –1 (3) 0.2197 min (4) 0.6591 min–1 10. What happens if concentration of C2H4 is doubled: (1) Remains same (2) Doubles (3) Becomes half (4) Is independent

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (2)

2. (2)

3. (3)

4. (4)

5. (1)

6. (3)

7. (3)

8. (2)

9. (2)

10. (2)

11. (3)

12. (3)

13. (1)

14. (1)

15. (4)

16. (2)

17. (2)

18. (3)

19. (2)

20. (2)

21. (4)

22. (3)

23. (4)

24. (2)

25. (2)

26. (2)

27. (2)

28. (1)

29. (3)

30. (4)

8. (4)

9. (2)

10. (3)

8. (1)

9. (3)

10. (2)

[B] ASSERTION REASON QUESTIONS 1. (2)

2. (4)

3. (1)

4. (4)

5. (3)

6. (2)

7. (3)

[C] COMPETENCY BASED QUESTIONS 1. (3)

2. (1)

3. (2)

4. (2)

5. (3)

6. (2)

7. (3)

ANSWERS WITH EXPLANATION [A] MULTIPLE CHOICE QUESTIONS 1. Option (2) is correct. Explanation: For chemical reaction, 2NH3 (g) → N2 (g) + 3H2 (g) As it is a zero-order reaction, the rate is independent of the rate of change in the reactant’s concentration. Rate = k[NH3]2 - DNH 3 1 DH Rate = k = 1 = 2 Dt 3 Dt 1 DH = 3k = 3 ´ 2.5 ´ 10-4 3 Dt = 7.5 × 10 –4 mol L –1 s –1 2. Option (2) is correct. Explanation: Molecularity is the number of molecules that participates in an elementary (single-step) reaction. Molecularity is equal to the sum of stoichiometric coefficients of reactants in the elementary reaction with effective collision (sufficient energy) and correct orientation. NH4NO2 → N2 + 2H2O Here, only one reactant is involved in the reaction, therefore, it is a unimolecular reaction. Hence, its molecularity is 1.

3. Option (3) is correct. Explanation: Half of half-life will be t1/4 = 2t1/2 t1/4 = time for the concentration of reactant to be reduced to 1/4th of its original concentration = 60 min (given) So, 2 t1/4 = 60 Hence, t1/4 = 30 min

4. Option (4) is correct. Explanation: In general, increase in temperature increases rate. Rate is approximately doubled when temperature increases by 10°C. So, from 293K to 313 K, for a 20 °C changes, the rate will be increased four times. So, the rate gets quadrupled. 5. Option (1) is correct. Explanation: For an n th order reaction Units of k = ( mol L–1) 1–n s –1 where n = order of the reaction. Here, n = 3/2 Units of k = (bar)1–3/2 min–1 = bar–1/2 min–1 6. Option (3) is correct. Explanation: The molecularity of a reaction is defined as the number of reacting molecules which collide simultaneously to bring about a chemical reaction. In other words, the molecularity

42 Oswaal CUET (UG) Chapterwise Question Bank of an elementary reaction is defined as the number of reactant molecules taking part in the reaction. Therefore, Molecularity of the reaction is 1+1=2 7. Option (3) is correct. Explanation: First order reaction, k t = or, kt = ln[R]0 – ln[R] ln[R ]0 From (i), we can write = R

ln[R ]0  R

...(i) …(ii)

log [R ]0 kt = ...(iii) 2.303 [R ] •   In[R] versus t graph will be obtained from equation (ii)

R→P Rate, r1 = k[A]n r2 = r1/2, = k [ A/2]n Comparing (i) and (ii), n = 1 Therefore, it is a first order reaction.

CHEMISTRY

…(i) …(ii)

11. Option (3) is correct. Explanation: There is no effect on heat evolved or absorbed during the reaction in the presence of a catalyst. It is because catalyst influence the rate of reaction and does not participate in the reaction. 12. Option (3) is correct. Explanation : Average rate of reaction up to 40 seconds on the basis of the graph is : V3 - 0 V3 = 40 - 0 40 13. Option (1) is correct. Explanation: The rate of a reaction depends upon the concen­ tration of reactants.

log [R ]0 versus t graph obtained from equation (iii), will be [R ]

14. Option (1) is correct. Explanation: Rate constant of a reaction does not depend upon concentrations of the reactants. 15. Option (4) is correct. Explanation: The reaction will be 100% complete only after infinite time. 16. Option (2) is correct.

8. Option (2) is correct. Explanation: For a chemical reaction with the rise in 10°C the fraction of molecules having energy equal or greater than activation energy is doubled and hence the rate of reaction is doubled or we may say rate constant is doubled. Thus, for 20°C rise in temperature, the rate constant increases by four times. 9. Option (2) is correct. Explanation: If, r = k[A]x[B]y From the data given in the table we get, 2  10–3 = k[0.1]x[0.1]y...(i) 4  10–3 = k[0.2]x[0.2]y...(ii) 2  10–3 = k[0.1]x[0.2]y...(ii) From (i) and (iii), 2 ´ 10-3 k [0.1] [0.1] æ1ö =ç ÷ = 2 ´ 10-3 k [0.1]x [0.2]y è 2 ø x

y

y

y=0 From (i) and (ii), 2 ´ 10-3 k [0.1] [0.1] æ1ö =ç ÷ = 4 ´ 10-3 k [0.2]x [0.2]0 è 2 ø x

0

x

Explanation: When concentration of B is doubled keeping the concentration of A constant, the rate of formation of C will increase by a factor of four. This shows that the rate of reaction depends upon the square of concentration of B. When concentration of A is doubled then the rate of formation of C also doubles from the initial value. Hence, this represents that the rate depends on first power of concentration of A. 17. Option (2) is correct. Explanation: In this reaction, A → B. The concentration of both reactants and the products varies exponentially with time. 18. Option (3) is correct. Explanation : Order of reaction is equal to the sum of powers of concentration of the reactants in rate law expression. For any chemical reaction, xA + yB → Product Rate = k [A]x[B]y, Order = x + y Order of reaction can be a fraction also. Order of reaction is not always equal to sum of the stoichiometric coefficients of reactants in the balanced chemical equation. For a reaction it may or may not be equal to sum of stoichiometric coefficient of reactants. 19. Option (2) is correct.

x =1

Explanation: Rate constant of a pseudo-first-order reaction depends on the concentration of reactants present in excess.

Overall order of reaction, x + y = 1

20. Option (2) is correct.

10. Option (2) is correct.

Explanation: Let us consider a first-order gas-phase decomposition reaction : A(g) → B(g) + C(g)

Explanation: When volume of vessel is doubled shows the concentration being halved. As a result, rate of reaction is also halved.

43

CHEMICAL KINETICS

The initial pressure of the system before decomposition of A is ‘Pi.’ After lapse of time ‘t’, total pressure of the system increased by x units and became ‘Pt’. Hence, the pressure of A decreased by x atom. A(g) → B(g) + C(g) Initial pressure: Pi atm 0 0 Pressure after time t: (Pi-x) +x + x Pt = Pi + x atm X = P t – Pi Pressure of A after time t = Pi – x = Pi– (Pt – Pi) = 2Pi – Pt 2.303 Pi          k = log t 2Pi - Pt 21. Option (4) is correct. Explanation: Half life period of a zero order reaction = Where [R]0 = initial concentration of reactant  k = Rate constant 22. Option (3) is correct. Explanation:

[ R ]0 2k

Explanation: When one of the reactants is in excess, a bimolecular reaction is kinetically first order reaction. 26. Option (2) is correct. Explanation: For the reaction, 2N2O5 → 4NO2 + O2 -d[ N 2O5 ] = k[ N 2O5 ] dt 1.02 × 10–4 = 3.4 × 10–5 s–1 × [N2O5] ∴ [N2O5] = 3 27. Option (2) is correct.

Explanation: Rate for the first order reaction will be [dx ] Rate = k [A] dt [A] = 0.5 M, k = ? ∴ 1. 5 × 10–2 = k × 0.5 ⇒ k = 3 × 10–2 min–1 t 1/2 of reaction =      

0.693 0.693 = k 3 ´ 10-2 min -1

= 23.1 min

28. Option (1) is correct. Explanation: Given that, [A]0 = 2 m, t = 200 min, [A] = 0.15 m For first order reaction, Rate constant k = k=

23. Option (4) is correct. Explanation: X → Y Rate(r) ∞ [X]n [Where, n = Order of reaction] If the concentration X is increased by 4 times X’ = 4X Then, Rate(r’) α [X’]n r¢ [4X]n = =2 r [X]n r’ is new rate, X’ is a new concentration [4]n = 2, n = 1/2 Order of reaction = 1/2 24. Option (2) is correct. Explanation: Initial amount= 5 g Final concentration = 3 g Rate constant= 1.15 × 10–3 s–1 We know that for a first order reaction 2.303 [R ] k= log initial t [R final ] =

2.303 5 log = 444.397 secc -3 1.1510 3

25. Option (2) is correct.

0 2.303 [A]0 log t [A]

0 2.303 log 2 / 0.15 t

k = 0.01295 min–1 t1/2 = 0.693/ k = 0.693/ 0.01295 = 53.50 min 29. Option (3) is correct. Explanation: Initial temperature, T1 = 20 + 273 = 293 K Final temperature, T2 = 35 + 273 = 308 K R = 8.314 JK–1 mol–1 As rate becomes double on raising temperature ∴ r2 = 2r1 or r2/r1 = 2 As rate constant, k α r ∴ k2/k1 = 2 According to Arrhenius equation we know that: log

k2 - Ea [T1 - T2 ] = k1 2.303R [T1T2 ]

Ea = 34.7 kJ mol–1

30. Option (4) is correct. Explanation: Catalyst speed up the reaction but it does not shift the position of equilibrium because catalyst reduces the height of barrier by providing an alternative path for the reaction and lowers the activation energy. However, the lowering in activation energy is to the some extent for the forward as well as the backward reaction.

44 Oswaal CUET (UG) Chapterwise Question Bank

CHEMISTRY

[B] ASSERTION REASON QUESTION:

10. Option (3) is correct.

1. Option (2) is correct.

Explanation: Rate of first order reaction depends on the concentration of reactant. Rate constant is a fixed value for a particular reaction and does not change with change in concentration of reactant.

Explanation: Order of a reaction may be zero or fractional. Order can be determined by rate law expression. 2. Option (4) is correct. Explanation: Order and molecularity may not be necessarily same. Order is determined experimentally but molecularity is calculated using balanced stoichiometric equation. 3. Option (1) is correct. Explanation: ΔH = Activation Energy of forward reaction – Activation Energy of reverse reaction. Catalyst does not alter heat of reaction because it affects activation energy of forward and reverse reactions equally. 4. Option (4) is correct. Explanation: Every collision among reactant molecules does not lead to the formation of product. Only effective collision brings out the formation of product. 5. Option (3) is correct. Explanation: According to Arrhenius equation (k = Ae -Ea/RT ); it is found almost accurate for single as well as complex reaction. However, orientation is essential for the reactant molecules participating in the reaction. 6. Option (2) is correct. Explanation: Rate of a reaction is the rate of change of concentration of reactant or product with time Mathematical expression of rate of reaction, R→P DC R DC P R= = DT DT 7. Option (3) is correct. Explanation: For the given chemical reaction, order is the sum of powers of the reactants which is calculated experimentally i.e., 1/2 + 1 = 1.5 In an elementary reaction, order can be the sum of stoichiometric coefficients of the reactants but the given reaction is a complex reaction.

[C] COMPETENCY BASED QUESTIONS 1. Option (3) is correct. Explanation: Rate depends on the disappearance of reactants or appearance of products. 2. Option (1) is correct. Explanation: Half-life period for first order reaction is given as: t1/2 = 0.693/k k=

0.693 = 2000 seconds 0.3465 10-3

3. Option (2) is correct. Explanation: For first order reaction, 2.303 k= log a / (a - x ) t k1 =

2.303 100 log (100 - 75) 32

k2 =

2.303 100 log (100 - 50) 32

k1 = k 2 =

2.303 100 2.303 100 log = log (100 - 50) t 32 (100 - 75)

t = 16 mins 4. Option (2) is correct. Explanation: For bimolecular reaction 2HI → H2 + I2 Rate = K [HI]2 Order of reaction is 2. 5. Option (3) is correct. Explanation: Rate law cannot be determined by a balanced chemical equation.

8. Option (4) is correct.

6. Option (2) is correct.

Explanation: For a general reaction, aA + bB → cC + dD Rate = k[A]a × [B]b Where a+b = n = order of the reaction unit of k = [mol L–1]1–n s–1 Thus, k is dependent on the concentration of the reactants so, unit of k for different orders have different unit. For zero order, n = 0, thus units of k is mol L–1s–1.

Explanation: Since the rate of the reaction is proportional to the concentration for the reactant i.e., ethene so, it is first order reaction.

9. Option (2) is correct. Explanation: Rate of reaction is the speed with which a reaction takes place which can occur at different speeds. The frequency of collisions such that, more often molecules collide with each other, the faster the reaction proceeds and the energy of collisions where more forcefully molecules collide with each other, the more likely they are to react and the faster the reaction proceeds

7. Option (3) is correct. Explanation: C 2 H 4 + H 2 ¾catalyst ¾¾® C 2 H 6 Rate law equation, Rate = K [C2H4] 8. Option (1) is correct.

Explanation: For first order reaction, t1/2 = 0.693/k 9. Option (3) is correct. Explanation: t = 5 min For first order reaction,

45

CHEMICAL KINETICS

2.303 [R ]0 log t R 2.303 log 3 = 5

k=

= 0.2197 min–1

10. Option (2) is correct. Explanation: For first order reaction: Rate = k [A]

Study Time

CHAPTER

5

Max. Time: 1:50 Hours Max. Questions: 50

SURFACE CHEMISTRY

 Revision notes

Surface chemistry is the branch of chemistry which deals with the phenomenon that occurs at the surfaces.   Adsorption: {  The phenomenon of attracting and retaining the molecules of a substance on the surface of a solid. { A higher concentration on the surface than in the bulk a surface. { The extent of adsorption of a gas on a solid depends upon: i. nature of gas ii. nature of solid iii. surface area of the solid iv. pressure of gas v. temperature of gas   Important terms: a. Adsorbent: Is the surface that adsorbs b. Adsorbate: The one that gets adsorbed is called an adsorbate. c.  Adsorption isotherm: The relationship between the extent of adsorption (x/m) and pressure of the gas at constant temperature. d.  Desorption: The process of removing an adsorbed substance from a surface on which it is adsorbed. e. Sorption: When adsorption and absorption take place simultaneously. f. Adsorption isobar: A plot of extent of adsorption (x/m) vs. temperature at constant pressure.   Types of adsorptions: a. Physical adsorption: ƒ Adsorbate is held to the adsorbent by van der Waals forces. ƒ It lacks specificity. ƒ It depends on the nature of adsorbate. ƒ It is reversible in nature. ƒ  It increases with increase of surface area of the adsorbent. ƒ It decreases with increase of temperature. ƒ Enthalpy of adsorption is quite low. ƒ  Results into multimolecular layers on adsorbent surface under high pressure.



For example, dihydrogen is first adsorbed on nickel. b. Chemical adsorption: ƒ  Adsorbate is held to the adsorbent by strong chemical bond. ƒ Chemical bonds are involved. ƒ High specificity. ƒ Usually irreversible in nature. ƒ  Increases with increase of surface area of the adsorbent. ƒ Increases with the increase of temperature. ƒ Enthalpy of chemisorption is high. ƒ Results into unimolecular layer. For example, oxygen is adsorbed on metals by virtue of oxide formation.   Adsorption isotherm: {  The variation in the amount of gas adsorbed by the adsorbent with pressure at constant temperature

Fig: 5.1: Adsorption Isotherm

Freundlich Adsorption isotherm: { Relationship

between the quantity of gas adsorbed by unit mass of solid adsorbent and pressure at a particular temperature Scan to know more about this

topic   Catalysis: { Process of using catalyst a. Types: i. Homogenous catalysis: ƒ Catalyst is in the same phase as Catalysts and are the reactants enzymes Ex: 2SO2(g) + O2(g) → 2 SO3(g) ii. Heterogeneous catalysis: ƒ The catalyst is in a different phase from that of the reactants.

SURFACE CHEMISTRY

47

48 Oswaal CUET (UG) Chapterwise Question Bank Ex: N2(g) + 3H2(g)

Fe(s)/ Mo(s)

2 NH3(g)

Fig: 5.2: Freundlich adsorption Isotherm

b. Mechanism of catalysis: - Adsorption theory of catalysis: ƒ Diffusion of reactants to the surface of the catalyst. ƒ Adsorption of reactant molecules on the surface of the catalyst. ƒ  Occurrence of chemical reaction on the catalyst’s surface through formation of an intermediate. ƒ  Desorption of reaction products from the catalyst surface. ƒ Diffusion of reaction products away from the catalyst’s surface.

  Colloids: - Solutions are intermediate between true solutions and suspensions. a. Properties: ƒ The size of the colloidal particles range from 1 to 1000 nm. ƒ Consists of two phases - the dispersed phase and the dispersion medium. ƒ Exhibit optical, mechanical and electrical properties. ƒ Colloids find several applications in industry as well as in daily life. Dispersed Dispersion Type of phase medium colloid Solid

Solid

Solid Solid Liquid Liquid Liquid

Liquid Gas Solid Liquid Gas

Gas Gas

Solid Liquid



Examples

Scan to know c. Types: more about this i. Lyophilic colloids: liquid-loving topic ƒ Reversible sols ƒ Are quite stable ƒ Cannot be easily coagulated ii. Lyophobic colloids: liquid-hating Properties of ƒ Readily coagulated colloids ƒ Are not stable ƒ Irreversible sols d. Properties: i. Colligative properties: ƒ  The values of colligative properties are of small order as compared to true solutions ii. Tyndall effect: ƒ Scattering light in all directions in space iii. Colour: ƒ Depends on the wavelength of light scattered by the dispersed particles iv. Brownian motion: ƒ A state of continuous zig-zag motion all over the field of view v. Charge on particles: ƒ Particles carry an electric charge. The nature of this charge is either positive or negative

Positively charged sols

Negatively charged sols

1. Hydrated metallic oxides, e.g., Al2O3. xH2O, CrO3.xH2O and Fe2O3.xH2O, etc 2. Basic dye stuffs, e.g., methylene blue sol. 3. Haemoglobin (blood) Oxides, e.g., TiO2 sol.

1. Metals, e.g., copper, silver, gold sols. Metallic sulphides, e.g., As2S3, Sb2S3, CdS sols. 2. Acid dye stuffs, e.g., eosin, congo red sols. 3. Sols of starch, gum, gelatin, clay, charcoal, etc.

ƒ  The

existence of charge on colloidal particles is confirmed by electrophoresis experiment.

Solid sol

Some coloured glasses and gem stones Paints, cell fluids Sol Smoke, dust Aerosol Cheese, butter, jellies Gel Emulsion Milk, hair cream Fog, mist, cloud, Aerosol insecticide sprays Solid sol Pumice stone, foam rubber Froth, whipped cream, Foam soap lather

b. Preparation of colloids: ƒ Mechanical dispersion ƒ Electro dispersion ƒ Peptization

CHEMISTRY

Fig:5.3: Electrophoresis

49

SURFACE CHEMISTRY

vi. Coagulation: When the charge on colloidal particles is removed, they come nearer to form aggregates ƒ By electrophoresis ƒ By mixing two oppositely charged sols ƒ By boiling ƒ By persistent dialysis ƒ By addition of electrolytes e. Purification: ƒ Dialysis ƒ Electrodialysis ƒ Ultrafiltration Gold number: It is the minimum weight (in milligrams) of a

protective colloid required to prevent the coagulation of 10 ml of a standard hydro gold sol when 1 ml of a 10% sodium chloride solution is added to it. The higher the gold number, the lower the protective power of the colloid, because a greater amount of colloid is required to prevent coagulation.   Emulsions: { Both dispersed phase and dispersion medium are liquids. {  To stabilise emulsifying agents are added. {  Soaps and detergents are most frequently used as emulsifiers.

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Emulsions

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS 1. Which of the following is not the characteristic of physisorption? (1) It arises because of van Der Waals forces. (2) It is not specific in nature. (3) Enthalpy of adsorption is high. (4) It results into multimolecular layers on the adsorbent surface under high pressure.  [CUET 2022, 10th Aug] 2. Which one of the following is an emulsion? (1) Smoke (2) Hair cream (3) Paint (4) Cheese [CUET 2022, 10th Aug] 3. Which of the following is a positively charged sol? (1) Starch (2) Gum (3) Gold sol (4) Blood  [CUET 2022, 10th Aug] 4. Consider the case when a highly diluted solution of KI is added to AgNO3 solution. Arrange the following in the increasing order of ease of coagulation of the resulting sol. (A) BaSO4 (B) NaCl (C) Na3PO4 (D) K4[Fe(CN)6] Choose the correct answers from the options given below: (1) A < C < B < D (2) D < C < A < B (3) A < B < C < D (4) B < A < C < D  [CUET 2022, 17th Aug] 5. Match List-I with List-II: List I State of dispersed phase and dispersion medium A.

B.

List II Example of colloidal system

Dispersed phase: Solid Dispersion medium: Liquid

(I)

Dispersed phase: Liquid Dispersion medium: Solid

(II)

Soap lather

Paints

(1) (2) (3) (4)  6. (1) (2) (3) (4)  7. A. B. C. D.  (1) (3) 8. (1) (2) (3) (4) 

C.

Dispersed phase: Gas Dispersion medium: Liquid

(III)

Pumice stones

D.

Dispersed phase: Gas Dispersion medium: Solid

(IV)

Jellies

Choose the correct answer from the options given below: A-III,  B-IV, C-II, D-I A-IV, B-I,   C-III, D-II A-II, B-IV,  C-I, D-III A-I,  B-II,  C-IV,  D-III [CUET 2022, 18th Aug] A gas that is readily adsorbed on 1 g of activated charcoal by physical adsorption Dihydrogen (critical temperature 33 K) Helium (critical temperature 2.15 K) Methane (critical temperature 190 K) Sulphur Dioxide (critical temperature 630 K) [CUET 2022, 18th Aug] When KI solution is added to AgNO3 solution, then: Ag+ ions form dispersion medium get adsorbed on AgI Positively charged sol is formed. Negatively charged sol is formed. K+ ions of dispersed phase get adsorbed on AgI. [CUET 2022, 20th Aug] Choose the correct answer from the options given below: A and C only (2) D and B only C only (4) A and B only Which one of the following is NOT applicable to the phenomenon of adsorption? It is an endothermic process. Entropy change is negative. Gibb’s free energy change is less than zero. Enthalpy of chemisorption is higher than that of physisorption [CUET 2022, 20th Aug]

50 Oswaal CUET (UG) Chapterwise Question Bank 9. The movement of colloidal particles from a colloidal solution, under the influence of applied electric potential towards one or the other electrode is called: (1) Brownian movement (2) Electro-osmosis (3) Electro-dialysis (4) Electrophoresis  [CUET 2022, 21st Aug] 10. The colloids that cannot be easily coagulated are: (1) Lyophobic colloids (2) Lyophilic colloids (3) Irreversible sols (4) Associated colloids  [CUET 2022, 23rd Aug] 11. Match List-I with List-II. List I (Example of colloidal system)

List II (Types of colloid)

A.

Smoke

(I)

Foam

B.

Cheese

(II)

Aerosol

C.

Soap lather

(III)

Emulsion

D.

Milk

(IV)

Gel

(1) (2) (3) (4)  12. (1) (2) (3) (4)  13. A. C. (1) (3)  14. A. C. (1) (3)  15. (1) (2) (3) (4)  16. (1) (3) 

Choose the correct answer from the options given below: A-II, B-IV, C-I, D-III A-I, B-II, C-III, D-IV A-I, B-III, C-II, D-IV A-IV, B-III, C-II, D-I [CUET 2022, 23rd Aug] Which among the following is a false statement? Adsorption is a thermodynamically favourable process. Adsorption is an entropically favourable process. Adsorption is an enthalpically favourable process. Adsorption process is always favourable in dark condition. [CUET 2021, 23rd Sept] Which of the following is an example of negatively charged sol? Al2O3. xH2O B. As2S3 TiO2 D. Blood Choose the correct answer from the options given below: 1 (2) 2 3 (4) 4 [CUET 2023, 7th June] Which has the highest flocculating power? [Fe(CN)6]4- B. PO432- SO4 D. ClChoose the correct answer from the options given below: 1 (2) 2 3 (4) 4 [CUET 2023, 7th June] Which of the following process does not occur at the interface of phases? Crystallisation Heterogeneous catalysis Homogeneous catalysis Corrosion [NCERT Exemp. Q 1, Page 63] Which of the following interface cannot be obtained? Liquid-liquid (2) Solid-liquid Liquid-gas (4) Gas-gas [NCERT Exemp. Q 3, Page 63]

17. (1) (2) (3) (4)  18. (1) (2) (3) (4)  19.

CHEMISTRY

Extent of physisorption of a gas increase with increase in temperature. decrease in temperature. decrease in surface area of adsorbent. decrease in strength of van der Waals forces. [NCERT Exemp. Q 5, Page 64] Which of the following is an example of absorption? Water on silica gel. Water on calcium chloride. Hydrogen on finely divided nickel. Oxygen on metal surface. [NCERT Exemp. Q 11, Page 64] In which of the following reactions heterogeneous catalysis involved? VO (i) 2SO2(g) + O2(g) 2 5 2SO3(g) NO(g) (ii) 2SO2(g) + O2 2SO3(g) Fe (iii) N2(g) + 3H2(g) 2NH3(g) HCl(aq) (iv) CH3COOCH3(l) + H2O(l) CH3COOH(aq) + CH3OH(aq) (1) (i), (iii) (2) (ii), (iii), (iv) (3) (i), (ii), (iii) (4) (iv)  [NCERT Exemp. Q 13, Page 65] 20. Which of the following will show Tyndall effect? (1) Aqueous solution of soap below critical micelle concen­ tration. (2) Aqueous solution of soap above critical micelle concen­ tration. (3) Aqueous solution of sodium chloride. (4) Aqueous solution of sugar.  [NCERT Exemp. Q 15, Page 65] 21. Method by which lyophobic sol can be protected? (1) By addition of oppositely charged sol. (2) By addition of an electrolyte. (3) By addition of lyophilic sol. (4) By boiling.  [NCERT Exemp. Q 16, Page 65] 22. Freshly prepared precipitate sometimes gets converted to colloidal solution by: (1) coagulation (2) electrolysis (3) diffusion (4) peptisation  [NCERT Exemp. Q 17, Page 65] 23. Which of the following process is responsible for the formation of delta at a place where rivers meet the sea? (1) Emulsification (2) Colloid formation (3) Coagulation (4) Peptisatiotin  [NCERT Exemp. Q 22, Page 67] 24. Which of the following process is not responsible for the presence of electric charge on the sol particles? (1) Electron capture by sol particles (2) Adsorption of ionic species from solution (3) Formation of Helmholtz electrical double layer (4) Absorption of ionic species from solution  [NCERT Exemp. Q 24, Page 67] 25. In the presence of a catalyst, heat evolved or absorbed during reaction

SURFACE CHEMISTRY

(1) (2) (3) (4) 26.

increases. decreases. remains unchanged. may increase or decrease Which one of the following is not applicable to the phenomenon of adsorption? (1) ΔH > 0 (2) ΔG < 0 (3) ΔS < 0 (4) ΔH < 0 27. Extent of adsorption of adsorbate from solution phase increases with: (1) increase in amount of adsorbate in solution. (2) decrease in surface area of adsorbent. (3) increase in temperature of solution. (4) decrease in amount of adsorbate in solution. 28. Which of the following is most effective in coagulating negatively charged hydrated ferric oxide sol? (1) NaNO3 (2) MgSO4 (3) AlCl3 (4) KCl 29. The coagulation value in millimoles per litre of the electrolytes used for the coagulation of As2S3 are given below: I. (NaCl) = 52 II. (BaCl2) = 0.69 III. (MgSO4) = 0.22 The correct order of their coagulating power is: (1) I > II > III (2) II > I > III (3) III > II > I (4) III > I > II 30. Which of the following forms cationic micelles above certain concentration? (1) Sodium ethyl sulphate (2) Sodium acetate (3) Urea (4) Cetyl trimethyl ammonium bromide [B] ASSERTION REASON QUESTIONS Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (1) Both A and R are correct and R is The correct explanation of A. (2) Both A and R are correct but R is NOT The correct explanation of A. (3) A is correct but R is not correct. (4) A is not correct but R is correct. 1. Assertion (A): An ordinary filter paper impregnated with colloidal solution stops the flow of colloidal particles. Reason (R): Pore size of the filter paper becomes more than size of colloidal particle.  [NCERT Exemp. Q74, Page 73] 2. Assertion (A): Colloidal solutions show colligative properties. Reason (R): Colloidal particles are large in size.  [NCERT Exemp. Q 74, Page 73] 3. Assertion (A): Colloidal solutions do not show Brownian motion. Reason (R): Brownian motion is responsible for stability of sols.  [NCERT Exemp. Q 73, Page 75]

51 4. Assertion (A): Coagulation power of Al3+ is more than Na+ Reason (R): Greater the Valency of the flocculating ion added, greater is its power to cause precipitation (HardySchulze rule).  [NCERT Exemp. Q 73, Page 76] 5. Assertion (A): Detergents with low CMC are more economical to use. Reason (R): Cleansing action of detergents involves the formation of micelles. These are formed when the concentration of detergents becomes equal to CMC.  [NCERT Exemp. Q 32, Page 77] 6. Assertion (A) : Colloids are homogeneous mixtures. Reason (R): The particle size dispersed is too large for it to be homogeneous mixtures. 7. Assertion (A): Positive catalysts increase the rate of a chemical reaction. Reason (R): An alternate pathway is taken that has lower energy of activation. 8. Assertion (A): Chemisorption is not specific and is reversible. Reason (R): Chemisorption will only occur if there is some possibility of chemical bonding between adsorbent and adsorbate. 9. Assertion (A): Colligative properties of colloids are generally higher than normal Reason (R): Colloidal particles forms aggregates. 10. Assertion (A): Colloidal particles carry charge. Reason (R): The sol particles have Brownian motion. [C] COMPETENCY BASED QUESTIONS I. Based on following passage answer questions from 1-5.  There are mainly two types of adsorption of gases on solids. If accumulation of gas on the surface of a solid occurs on account of weak van der Waals’ forces, the adsorption is termed as physical adsorption or physisorption. When the gas molecules or atoms are held to the solid surface by chemical bonds, the adsorption is termed chemical adsorption or chemisorption. The chemical bonds may be covalent or ionic in nature. Chemisorption involves a high energy of activation and is, therefore, often referred to as activated adsorption. Sometimes these two processes occur simultaneously and it is not easy to ascertain the type of adsorption. When gaseous molecules come in contact with the surface of a solid catalyst, a weak chemical combination takes place between the surface of the catalyst and the gaseous molecules, which increases the concentration of reactants on the surface. Different molecules adsorbed side by side have better chance to react and form new molecules. This enhances the rate of reaction. Also, adsorption is an exothermic process. The heat released in the process of adsorption is utilised in enhancing the reaction rate. A physical adsorption at low temperature may pass into chemisorption as the temperature is increased. 1. The correct statement is:

52 Oswaal CUET (UG) Chapterwise Question Bank (1) Gaseous molecules are adsorbed by the catalysts on their surface. (2) Gaseous molecules are absorbed by the catalysts on their surface. (3) Gaseous molecules are absorbed by the catalysts inside their bulk. (4) Gaseous molecules are adsorbed by the catalysts inside their bulk. 2. Adsorption is: (1) Exothermic only during physisorption. (2) Endothermic only during chemisorption. (3) Always endothermic during physisorption and chemi­ sorption. (4) Always exothermic during physisorption and chemi­ sorption. 3. Physisorption involves: A. van der Waal’s forces B. low energy of activation C. is not specific Choose the correct answer from the options given below: (1) only (A) (2) only(B) (3) only (C) (4) all (A), (B) and (C) 4. Dihydrogen is first adsorbed on nickel by van der Waals’ forces. Molecules of hydrogen then dissociate to form hydrogen atoms. Hence, the phenomenon is : (1) Only physisorption (2) Ony chemisorption (3) Physisorption gets converted to chemisorption (4) Chemisorption gets converted to physisorption. 5. Rate of reaction enhances when: (1) concentration of reactants increases on surface of catalyst. (2) concentration of reactants decreases on surface of catalyst. (3) concentration of products increases on surface of catalyst. (4) concentration of products decreases on surface of catalyst. II. Based on following passage answer questions from 6-10 If FeCl3 is added to excess of hot water, a positively charged sol of hydrated ferric oxide is formed due to adsorption of Fe3+ ions. However, when ferric chloride is

6. (1) (2) (3) (4) 7. (1) (3) 8. (1) (2) (3) (4) 9. A. B. C. (1) (3) 10. (1) (2) (3) (4)

CHEMISTRY

added to NaOH a negatively charged sol is obtained with adsorption of OH- ions. A positive or a negative charge is acquired by selective adsorption on the surface of a colloidal particle. Fe2O3.xH2O/Fe3+ Fe2O3.xH2O/OH Positively charged Negatively charged The correct statement is: the first layer of ions is firmly held and the second layer is mobile the first layer of ions is mobile while the second layer is firmly held both the first layer and second layer of ions are firmly held both the first layer and second layer of ions are mobile. The potential difference between the fixed layer and the diffused layer of opposite charges is called: Electrophoresis (2) Emulsification Zeta potential (4) Dialysis The presence of equal and similar charges on colloidal particles is responsible for: destabilising the colloidal solution by forming aggregates. destabilising the colloidal solution by not forming aggregates. stabilising the colloidal solution by forming aggregates. stabilising the colloidal solution by not forming aggregates. The charge on the sol particles is due to; due to electron capture by sol particles during electro dispersion of metals. due to preferential adsorption of ions from solution. due to formulation of electrical double layer. Choose the correct answer from the options given below: Only (A) (2) Only (B) Only (C) (4) All (A), (B), (C) The existence of charge on colloidal particles is confirmed by : Electrolysis experiment Electrophoresis experiment Tyndall effect Emulsification

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (3)

2. (2)

3. (4)

4. (4)

5. (3)

6. (4)

7. (4)

8. (1)

9. (4)

10. (2)

11. (1)

12. (4)

13. (2)

14. (1)

15. (3)

16. (4)

17. (2)

18. (2)

19. (1)

20. (2)

21. (3)

22. (4)

23. (3)

24. (4)

25. (3)

26. (1)

27. (1)

28. (3)

29. (3)

30. (4)

8. (4)

9. (1)

10. (2)

8. (4)

9. (4)

10. (2)

[B] ASSERTION REASON QUESTIONS 1. (3)

2. (2)

3. (4)

4. (1)

5. (1)

6. (4)

7. (1)

[C] COMPETENCY BASED QUESTIONS 1. (1)

2. (3)

3. (4)

4. (3)

5. (1)

6. (1)

7. (3)

53

SURFACE CHEMISTRY

ANSWERS WITH EXPLANATION [A] MULTIPLE CHOICE QUESTIONS 1. Option (3) is correct Explanation: Specificity: The force of van Der Waals is global. As a result, an adsorbent’s surface does not display any preference for a particular gas. Hence, specificity is lacking. Reversible nature: The physisorption of a gas by a material is usually reversible. As an outcome, it can be denoted as: Solid + Gas ↔ Gas/Solid + Heat Surface Area: As the surface area of the adsorbent grows, the fraction of physisorption increases. As a result, finely split metals and porous materials with large surface areas are good adsorbents. Enthalpy: The process of physical adsorption is already exothermic. However, its enthalpy of adsorption is low, ranging between 20 and 40 kJ mol–1 2. Option (2) is correct Explanation: Emulsion is a colloidal mixture made up of two immiscible phases in which the dispersed phase and dispersion medium both are liquids. Type of colloid

Example

Dispersed phase

Dispersed medium

Emulsion

Hair cream

Liquid

Liquid

Aerosol

Smoke

Solid

Gas

Sol

Paint

Solid

Liquid

Gel

Cheese

Liquid

Solid

3. Option (4) is correct Explanation: Positively charged sol: Blood Negatively charged sol: Starch, Gum, Gold 4. Option (4) is correct Explanation: The colloidal solution will be positively charged due to an excess of Ag+ ions. These positively charged metal ions will attract the anion of the electrolyte. The higher the negative charge on the electrolyte, the higher will be the coagulation. BaSO4 contains –2 charge on SO4 ion. NaCl contains –1 charge on Cl ion. Na3PO4 contains –3 charge on PO4 ion and K4 [Fe (CN)6] contains –4 charge on [Fe(CN)6] So, the order is: NaCl (B) < BaSO4 (A) < Na3PO4 (C) < K4[Fe (CN)6] (D) 5. Option (3) is correct

Explanation: Paints are a dispersion of solid in a liquid solvent. Jellies are formed by dispersion of a liquid in a solid medium. Soap lather is formed by mixing of air (gas) in liquid soap solution. Pumice stone is basically formed when gas molecules are trapped in a solid. 6. Option (4) is correct Explanation: Higher is the critical temperature of a gas, the

greater is the ease of liquefication, i.e., the greater are the van der Waals' forces of attraction and hence greater is the adsorption. 7. Option (4) is correct Explanation: When a dilute solution of potassium iodide (KI) is added to a dilute solution of silver nitrate (AgNO3), a positive charge sol to silver iodide (AgI) is generated, i.e., AgI/Ag+. This is due to adsorption of Ag+ ions from the dispersion medium. 8. Option (1) is correct Explanation: Physical adsorption decreases with an increase in temperature because an increase in temperature leads to an increase in kinetic energy of molecules and hence, their desorption occurs. Since physical adsorption is an exothermic process, it occurs more readily at lower temperatures and decreases with an increase in temperature (Le-Chatelier’s Principle). If the adsorption is a spontaneous phenomenon, it should be exothermic, at least for adsorption from the gaseous phase. A molecule in the gas phase will have an entropy higher than in the adsorbed state, then the ΔS should be negative, the only way to have a ΔG negative (spontaneous) is to have a ΔH negative (exothermic process). Therefore, the process will be spontaneous (ΔG= −ve) if the temperature is low. 9. Option (4) is correct Explanation: Electrophoresis is the migration of electrically charged colloidal particles in one direction under the influence of electric field 10. Option (2) is correct Explanation: Coagulation of lyophilic sols cannot be possible because there are two factors responsible for the stability of lyophilic sols. These factors are the charge and solvation of the colloidal particles. 11. Option (1) is correct Explanation: Type of colloid

Example

Dispersed phase

Dispersed medium

Aerosol

Smoke

Solid

Gas

Gel

Cheese

Liquid

Solid

Foam

Soap lather

Gas

Liquid

Emulsion

Milk

Liquid

Liquid

12. Option (4) is correct Explanation: All other statements are true, adsorption is favoured at low temperatures. 13. Option (2) is correct Explanation: All others are positively charged sol. 14. Option (1) is correct. Explanation: The order of flocculation is [Fe(CN)6]4– > PO43– > SO42– > Cl – 15. Option (3) is correct.

Explanation: There is no homogeneous catalysis. 16. Option (4) is correct.

54 Oswaal CUET (UG) Chapterwise Question Bank

CHEMISTRY

Explanation: Gas-gas interface cannot be obtained as they are completely miscible in nature.

So, the enthalpy ΔH as well as entropy ΔS of the system is negative.

17. Option (2) is correct.

27. Option (1) is correct.

Explanation: Physical adsorption of a gas increases with decrease in temperature because, at higher temperature weak van der Waals forces between gas and the surface become difficult to exist.

Explanation: Extent of adsorption of adsorbate from solution phase increases with increase in the amount of adsorbate in the solution. Therefore, as amount of adsorbate in the solution increases interaction of adsorbent increases which leads to increase in extent of adsorption.

18. Option (2) is correct. Explanation: Calcium chloride absorbs water. Other examples show adsorption. 19. Option (1) is correct. Explanation: In reaction (i) and reaction (iii), catalysts are in solid state, and reactants and products are gases. 20. Option (2) is correct. Explanation: Aqueous solution of soap above critical micelle concentration leads to the formation of colloidal solution. Tyndall effect is a characteristic of colloidal solution in which colloidal particles show a coloured appearance when sunlight is passes through it and seen perpendicularly. 21. Option (3) is correct. Explanation: Lyophobic sol can be protected by adding lyophilic sol which is known as protective colloid. 22. Option (4) is correct. Explanation: Peptisation is the process of converting freshly prepared precipitate into colloid. 23. Option (3) is correct. Explanation: A delta is formed at a place where rivers meet the sea due to the process of setting down of colloidal particles. The ions which are present in sea water are responsible for coagulation. 24. Option (4) is correct. Explanation: Absorption of ionic species from solution is not responsible for the presence of electric charge on the sol particles. Charge on the sol particles is due to: (i) electrons capture by sol particles during electro dispersion of metal. (ii) preferential adsorption of ionic species from solution. (iii) formation of Helmholtz electrical double layer.

28. Option (3) is correct. Explanation: The coagulating power of Al3+ is highest due to charge and small size. Therefore, aluminium chloride will be required in minimum amount to coagulate negatively charged sol of As2S3. 29. Option (3) is correct.

Explanation: Coagulating power ∞ 1/ Coagulating value Lower the coagulating value, higher is the coagulating power so, the correct order is: MgSO4 > BaCl2 > NaCl III II I 30. Option (4) is correct. Explanation: Cetyl trimethyl ammonium bromide forms cationic micelles above certain concentration. In the molecules of detergents, the negative ions aggregate to form a micelle of colloidal size. In polar medium like water the negative ion has a long hydrocarbon chain and a polar group –Br– at one end and on the other hand it has N+ ion thus cationic micelle is formed. [B] ASSERTION REASON QUESTIONS 1. Option (3) is correct. Explanation: Pore size of the filter paper becomes less than the size of colloidal particles hence colloidal particles do not flow through it. 2. Option (2) is correct Explanation: Colloidal particles are large in size and hence the number of particle is lesser than the true solution. The lesser number of particles causes lower colligative properties.

25. Option (3) is correct.

3. Option (4) is correct.

Explanation: There is no effect on heat evolved or absorbed during the reaction in the presence of a catalyst. It is because catalyst influence the rate of reaction and does not participate in the reaction.

Explanation: Colloidal particles show Brownian movement and it is responsible for the stability of colloidal solution.

26. Option (1) is correct. Explanation: Adsorption is an exothermic process, so the ΔH of adsorption is always negative. ΔH < 0 ΔG = ΔH – TΔS ΔG = Change in Gibbs free energy ΔH = Change in enthalpy T = Temperature in Kelvin ΔS = Change in entropy Since, adsorption is a spontaneous process, the thermodynamic requirement is at constant temperature and pressure, ΔG must be negative.

4. Option (1) is correct Explanation: According to Hardy-Schulze law, the greater is the valency of the coagulating ion, the more is the power to coagulate the colloidal solution. Thus, coagulation power of Al3+ is greater than that of Na+. 5. Option (1) is correct. Explanation: Cleansing of clothes takes place by micellisation, it starts at CMC. The lesser is the CMC, the better and more economical is the detergent. 6. Option (4) is correct Explanation: Colloids are heterogenous because their particle size is bigger than those of true solutions. 7. Option (1) is correct.

55

SURFACE CHEMISTRY

Explanation: By taking an alternate pathway with lower energy of activation the rate of reaction increases.

Explanation: The reactant molecules come closer and their concentration increases, enhancing rate of reaction.

8. Option (4) is correct. Explanation: chemisorption involves formation of chemical bonds and hence is irreversible and specific.

6. Option (1) is correct

9. Option (1) is correct. Explanation: Colligative properties of colloids are generally lower than normal because colloidal particles forms aggregates. 10. Option (2) is correct. Explanation: The sol particles acquire positive or negative charge by preferential adsorption of positive or negative ions. [C] COMPETENCY BASED QUESTIONS 1. Option (1) is correct. Explanation: Adsorption of gaseous molecules on catalyst is a surface phenomenon. 2. Option (3) is correct. Explanation: Adsorption is an exothermic process. 3. Option (4) is correct. Explanation: All the points are characteristics of physisorption. 4. Option (3) is correct. Explanation: Molecules come closer and tend to form new molecules. 5. Option (1) is correct.

Explanation: According to Helmholtz electrical double layer, the first layer of ions is firmly held and the second layer is mobile. 7. Option (3) is correct. Explanation: Separation of charge is a seat of potential, the charges of opposite signs on the fixed and diffused parts of the double layer results in a difference in potential between these layers known as Zeta potential. 8. Option (4) is correct. Explanation: Similar charges prevent association forming bigger molecules, thus stabilises the colloidal solution. 9. Option (4) is correct. Explanation: The charge on the sol arises because of all the given reasons. 10. Option (2) is correct. Explanation: When electric potential is applied across two platinum electrodes dipping in a colloidal solution, the colloidal particles move towards one or the other electrode. The movement of colloidal particles under an applied electric potential is called electrophoresis.

Study Time Max. Time: 1:50 Hours Max. Questions: 50

CHAPTER

6

GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS

 Revision Notes

Ores:

  Minerals and ores: { Minerals: ƒ The chemical substances present in the earth’s crust obtained by mining.

ƒ Minerals

from which metals can be extracted quantitatively and economically by the process of metallurgy.

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Extraction process of aluminium

Chief Ores and Methods of Extraction of Some Common Metals: Metals

Common ores

Extraction method

Must know

Copper

Copper pyrites, CuFeS2 Cuprite, Cu₂O Malachite, CuCO3.Cu(OH)2 Copper glance, Cu2S Azurite, 2CuCO3.Cu(OH)3

Roasting of sulphide partially and reduction 2Cu2O+Cu2S→ 6Cu+ SO2

It is self reduction in a specially designed converter. Sulphuric acid leaching is also employed.

Aluminium

Bauxite, Al2O3.xH2O Cryolite, Na3AIF Kaolinite, [Al2(OH)4Si5O] Aluminosilicates

Electrolysis of Al2O3 dissolved in molten cryolite or in Na3AIF6.

A good source of electricity is needed in the extraction of Al.

Zinc

Zinc blende or Sphalerite, ZnS Roasting and then reduction with C. Zincite, ZnO Calamine, ZnCO3

The metal may be purified by fractional distillation.

Iron

Haematite , Fe2O3 Magnetite, Fe3O4 Siderite, FeCO3 Iron pyrites, FeS2 Limonite, Fe2O3.3H2O

Limestone is added as flux which removes SiO2 as calcium silicate (slag) floats over molten iron and prevents its oxidation. Temperature approaching 2170 K is required.

Reduction with the help of CO and coke in blast furnace. Chemical reduction with CO. Calcination followed by reduction with CO. Roasting followed by reduction. Chemical reduction with CO.

  Concentration: Removal of gangue from ore is called concentration. { Methods: a. Hydraulic washing: ƒ Based on differences in gravities (densities) of ore and gangue particles. b. Magnetic separation ƒ Based on differences in magnetic and non-magnetic properties of ore components. c. Froth floatation process:

ƒ  Based

on the facts that sulphide ore is wetted by pine oil and gangue particles are wetted by water. d. Leaching: ƒ  Based on the fact that the ore is soluble in some suitable reagent and gangue is insoluble in the same reagent. ƒ During extraction of aluminium from bauxite: Al2O3(s) + 2NaOH(aq) + 3H2O(l) → 2Na[Al(OH)4](aq)   Conversion to metal oxides: a. Calcination:

GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS

57

58 Oswaal CUET (UG) Chapterwise Question Bank

CHEMISTRY

ƒ Heating of ore in absence of air below melting point

of metal.

ƒ Moisture

and volatile impurities escape. metal oxide is left behind. ZnCO3 (s) → ZnO(s) + CO2(g) b. Roasting: ƒ Heating of ore in regular supply of air at a temperature below the melting point of the metal. ƒ Used for sulphide ores and lower oxides. 2ZnS + 3O2 → 2ZnO + 2SO2 ƒ The



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Fig 6.1: Magnetic separation

Magnetic separation

Fig 6.3 c. Aluminothermic reduction: Cr2 O3 + 2 Al → Al2 O3 + 2 Cr   Electrometallurgy: { Done in isolation of metals where the sum of two redox couples is positive. { Electrolysis of the molten electrolyte is carried out in an electrolytic cell. { Gibb’s energy change is negative.   Refining: { To obtain metals of high purity.

Methods: a. Distillation ƒ For low boiling metals like zinc and mercury. ƒ The impure metal is evaporated. ƒ The pure metal is collected as distillate. b. Liquation ƒ Low melting metal like tin can be made to flow on a sloping surface. ƒ Can be separated from higher melting impurities. c. Electrolytic refining ƒ The impure metal is made to act as anode. ƒ The same metal in pure form is used as cathode. ƒ  Electrolyte is soluble salt solution of the same metal.

Fig 6.2: Froth floatation process   Reduction of metal oxides to the metals: { Methods: a. Oxide heated with reducing agent: Ex 4.1: reduction of Zinc oxide: using C as reducing agent. ZnO + C → Zn + CO Ex 4.2: reduction of iron oxide in blast furnace °C 3Fe 2O3 + CO ¾300 ¾¾ ® 2Fe3O 4 + CO 2 °C Fe3O 4 + 4CO ¾500 ¾¾ ® 3Fe + 4CO 2 C FeO + CO ¾700° ¾¾ ® Fe + CO 2 °C CaCO3 ¾800 ¾¾ ® CaO + CO 2

® CaO + 2CO CaCO3 + C ¾¾ 00° C CaO + SiO 2 ¾10¾¾ ® CaSiO3 (Fusible slag) Fe 2CO ¾¾® C + CO 2

® Carbon saturated iron Fe + C ¾¾ b. Auto-reduction: 2Cu2O+Cu2S → 6Cu+ SO2



Fig.6.4 : Zone refining d. Zone refining: ƒ Based on the principle that the impurities are more

59

GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS





soluble in the melt than in the solid state of the metal. ƒ A circular mobile heater is fixed at one end of a rod of the impure metal. ƒ The molten zone moves along with the heater which is moved forward. ƒ The pure metal crystallises out of the melt. e. Vapour phase refining: ƒ Metal is converted into volatile compound. ƒ On decomposition gives pure metal. e.g., Ni. f. Chromatographic method: ƒ Different components of a mixture are differentially adsorbed on the adsorbent. ƒ The mixture is put in a liquid or gaseous medium which is moved through the adsorbent. ƒ The adsorbed components are removed (eluted) by using suitable solvents (eluant).

Types: { Paper

chromatography chromatography { Gas chromatography A column of Al2O3 is prepared in a glass tube and the moving medium containing a solution of the components is in liquid form. { Column

Scan to know more about this topic

Column Chromatography

Fig 6.5: Column chromatography

  Uses of metals: Metal

Use

1. Aluminium

• Foils as wrappers • Paints and lacquers • Extraction of Cr and Mn • Wires for electricity conduction • Alloys

2. Copper

• Making wires used in electrical industry • Water and steam pipes. • Alloys that are tougher than the metal e.g., brass (with zinc), bronze (with tin)

3. Zinc

• For galvanising iron. • in batteries • Alloys, e.g., brass, (Cu 60%, Zn 40%) • Dust as a reducing agent in the manufacture of dye-stuffs and paints

4. Iron

• Cast iron used for casting stoves, railway sleepers, gutter pipes, toys, etc. • In the manufacture of wrought iron and steel. • Wrought iron for making anchors, wires, bolts, chains and agricultural implements. • Nickel steel is used for making cables, automobiles and aeroplane parts, pendulum, measuring tapes. • Chrome steel for cutting tools crushing machines. • Stainless steel for cycles, automobiles, utensils.

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS 1. Match list I with list II. List I

List II

A.

Siderite

(I)

Aluminium

B.

Malachite

(II)

Iron

C.

Calamine

(III)

Copper

D.

Bauxite

(IV)

Zinc

Choose the correct answer from the options given below. (1) A(I), B-(III), C(III), D(IV) (2) A(II), B-(III), C(IV), D(I)

(3) A(IV), B-(III), C(II), D(I) (4) A(III), B-(II), C(IV), D(I) [CUET 2022, 10th Aug] 2. The metal refined by the Van Arkel method is: (1) Ni (2) Zr (3) Cu (4) Sn [CUET 2022, 10th Aug] 3. Consider the statements for the metallurgical processes and select the correct statements: (A) Malachite is an ore of copper. (B) Bauxite is an ore of aluminium.

60 Oswaal CUET (UG) Chapterwise Question Bank (C) Calamine is an ore of zinc. (D) Haematite is an ore of iron. (E) Siderite is an ore of zinc. Choose the correct statement from the options given below: (1) A, B, E and D only (2) A and B only (3) A, B, C and D only (4) A only [CUET 2022, 17th Aug] 4. (1) (2) (3) (4) 5. (1) (2) (3) (4)

Nickel is refined by: Zone Refining Electrolytic Refining Van Arkel Method for Refining Mond Process for Refining [CUET 2022, 18th Aug] Which of the following reaction is an example of auto reduction?  Fe3O4 + 4CO → 3Fe + 4CO2 Cu2O + C → 2Cu + CO Cu2+ (aq) + Fe(s) → Cu + Fe3+ (aq) Cu2O + 1 /2 Cu2S → 3Cu + 1 /2 SO2 th

[CUET 2022, 18 Aug]

6. Match List-I with List-II: List I (Alloy) A.

Cast iron

(I)

Cutting tools and crushing machines

B.

Nickel steel

(II)

Railway sleepers

C.

Chrome steel

(III)

Cables, automobiles, aeroplane parts

D. (1) (2) (3) (4)

List II (Uses)

Stainless steel

(IV)

[CUET 2022, 21st Aug] 7. Match List-I with List-II: List I (Alloy)

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

List II (Constituents) (I)

A.

Brass

B.

German Silver (II)

Copper + Zinc

C.

Coinage alloy

(III)

Copper + Zinc + Nickel

D.

Bronze

(IV)

Copper + Nickel

Copper + Tin

Choose the correct answer from the options given below: A.-(IV), B.-(I), C.-(II), D.-(III) A.-(III), B.-(IV), C.-(I), D.-(II) A.-(II), B.-(III), C.-(IV), D.-(I) A.-(IV), B.-(I), C.-(III), D.-(II) [CUET 2022, 21st Aug]

8. The metals present in brass are: (1) Cu & Zn (3) Fe & Zn

9. Which one of the following is the ore of copper? (1) Calamine (2) Siderite (3) Malachite (4) Kaolinite [CUET 2022, 30th Aug] 10. Which one of the following reactions does not take place in blast furnace? (1) 3Fe2O3 + CO → 2Fe3O4 + CO2 (2) Fe3O4 + 4CO → 3Fe + 4 CO2 (3) Fe2O3 + CO → 3FeO + CO2 (4) FeO + SiO2 → FeSiO3 [CUET 2022, 30th Aug] 11. Match List-I with List -II: List I

(2) (4)

Cu & Fe Cu & Ni

[CUET 2022, 30th Aug]

List II

A.

PbS → PbO

(I)

Roasting

B.

CaCO3 → CaO

(II)

Calcination

C.

ZnS → Zn

(III)

Carbon reduction

D.

Cu2S → Cu

(IV)

Self reduction

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

Choose the correct answer from the options given below: (A)-(IV), (B)-(III), (C)-(I), (D)-(II) (A)-(1), (B)-(II), (C)-(III), (D)-(IV) (A)-(III), (B)-(IV), (C)-(II), (D)-(1) (A)-(II), (B)-(III), (C)-(IV), (D)-(1) [CUET 2023, 7th Jun] 12. Match List-I with List -II. List I

Cycles, autombile parts

Choose the correct answer from the options given below: A.-(IV), B.-(III), C.-(II), D.-(I) A.-(II), B.-(I), C.-(III), D.-(IV) A.-(III), B.-(II), C.-(IV), D.-(I) A.-(II), B.-(III), C.-(I), D.-(IV)

CHEMISTRY

(1) (2) (3) (4) 13. (1) (2) (3) (4)  14. (1) (3)  15. (1) (2)

List II

A.

Copper pyrites

(I)

Cu2O

B.

Malachite

(II)

Cu2S

C.

Cuprite

(III)

CuCO3.Cu(OH)2

D.

Copper glance

(IV)

CuFeS2

Choose the correct answer from the options given below: (A)-(1), (B)-(III), (C)-(IV), (D)-(II) (A)-(1), (B)-(III), (C)-(II), (D)-(IV) (A)-(IV), (B)-(III), (C)-(I), (D)-(II) (A)-(III), (B)-(II), (C)-(IV), (D)-(I) [CUET 2023, 7th Jun] When copper ore is mixed with silica in a reverberatory furnace, copper matte is produced. The copper matte contains: sulphides of copper (II) and iron (II) sulphides of copper (II) and iron (III) sulphides of copper (I) and iron (II) sulphides of copper (I) and iron (III) [NCERT Exemp. Q 2, Page 77] A number of elements are available in the earth’s crust but most abundant elements are: Al and Fe (2) Al and Cu Fe and Cu (4) Cu and Ag [NCERT Exemp. Q 4, Page 78] Zone refining is based on the principle that ___________. impurities of low boiling metals can be separated by distillation. impurities are more soluble in molten metal than in solid metal.

61

GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS

(3) different components of a mixture are differently adsorbed on an adsorbent. (4) vapours of volatile compound can be decomposed in pure metal. 

[NCERT Exemp. Q 5, Page 78]

16. 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 

[NCERT Exemp. Q 10, Page 79]

17. In the extraction of aluminium by Hall-Heroult process, purified Al2O3 is mixed with CaF2 to: (1) lower the melting point of Al2O3 (2) to convert bauxite to alumina (3) reduce Al3+ into Al(s). (4) acts as catalyst. 

[NCERT Exemp. Q 16, Page 80]

18. In the froth floatation process, zinc sulphide and lead sulphide can be separated by (1) using collectors (2) adjusting the proportion of oil to water (3) using emulsifiers (4) using froth stabilisers 

[NCERT Exemp. Q 18, Page 81]

19. Common impurities present in bauxite are: (1) CuO (2) ZnO (3) Fe2O3 (4) PbO

20. 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 

[NCERT Exemp. Q 9, Page 78]

21. 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 

[NCERT Exemp. Q 6, Page 78]

22. 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. 

[NCERT Exemp. Q 8, Page 78]

23. Which one of the following elements constitutes a major impurity in pig iron? (1) Silicon (2) Oxygen (3) Sulphur (4) Graphite 24. Roasting of sulphides gives the gas X as a by-product. This is a colourless gas with chocking 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 insolated. The gas X is: (1) H2S (2) SO2 (3) CO2 (4) SO3 25. Purification of aluminium by electrolytic refining is known as: (1) Hall’s process (2) Baeyer’s process (3) Hoope’s process (4) Serpeck’s process 26. Aluminium is extracted from alumina (Al2O3) by elec­ trolysis of a molten mixture of: (1) Al2O3 + HF + NaAlF4 (2) Al2O3 + CaF2 + NaAlF4 (3) Al2O3 + Na3AlF6+ CaF2 (4) Al2O3 + KF + Na3AlF6 27. Sulphide ores of metals are usually concentrated by froth floatation process. Which one of the following sulphide ores offers an exception and is concentrated by chemical leaching? (1) Argentite (2) Galena (3) Copper pyrite (4) Sphalerite 28. When copper ore is mixed with silica, in a reverberatory furnace copper matte is produced. The copper matte contains _____________. (1) sulphides of copper (II) and iron (II). (2) sulphides of copper (II) and iron (III). (3) sulphides of copper (I) and iron (II). (4) sulphides of copper (I) and iron (III). 29. Bell-metal is an alloy of: (1) Cu + Pb (3) Cu + Zn

(2) (4)

Cu + Sn Cu + Ni

30. Elemental silicon to be used as a semiconductor is purified by ________ . (1) heating under vacuum (2) floatation (3) zone refining (4) electrolysis [B] ASSERTION REASON QUESTIONS Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (1) Both A and R are correct and R is The correct explanation of A. (2) Both A and R are correct but R is NOT The correct explanation of A. (3) A is correct but R is not correct. (4) A is not correct but R is correct. 1. Assertion (A): Nickel can be purified by Mond’s process.

Reason (R): Ni(CO)4 is a volatile compound which decomposes at 460 K to give pure Ni. [NCERT Exemp. Q 50, Page 85]

62 Oswaal CUET (UG) Chapterwise Question Bank 2. Assertion (A): Zirconium can be purified by van Arkel method.

Reason (R): ZrI4 is volatile and decomposes at 1800 K.

[NCERT Exemp. Q 51, Page 85]

3. Assertion (A): Sulphide ores are concentrated by froth floatation method.

Reason (R): Cresols stabilize the froth in froth floatation method.



[NCERT Exemp. Q 53, Page 86]

4. Assertion (A): Zone refining method is very useful for producing semiconductors. 

Reason (R): Semiconductors are of high purity. [NCERT Exemp. Q 54, Page 86]

5. Assertion (A): Hydrometallurgy involves dissolving the ore in a suitable reagent followed by precipitation by a more electropositive metal. 

Reason (R): Copper is extracted by hydrometallurgy. [NCERT Exemp. Q 55, Page 86]

6. Assertion (A): During concentration of iron ore, it is crushed into small pieces and is concentrated by gravity separation.

Reason (R): Heavier impurities like sand and clay are removed.

7. Assertion (A): The concentrate ore is strongly heated in a reverberatory furnace in excess of air

Reason (R): The impurities are removed as their volatile

8. Assertion (A): Chromium can be extracted by reduction of Cr2O3 with Al

Reason (R): Cr is a better reducing agent than Al

CHEMISTRY

Si, Mn). This is known as pig iron and cast into variety of shapes. 1. The lower portion of the blast furnace has a very high temperature because: (A) Hot air is blown (B) Coke is burnt to produce CO2 (C) Electrolyte melts Choose the correct answer from the options given below. (1) Only (A) (2) Only (B) (3) Only (C) (4) Both (A) and (B) 2. The oxide of iron that is reduced is: (A) Fe2O3 (B) Fe3O4 (C) Fe2O3 and FeO (D) FeO Choose the correct answer from the options given below. (1) Only (A) (2) Only (B) (3) Both (A) and (B) (4) Only (D) 3. (1) (2) (3) (4)

Slag is: Carbonate impurity Silicate impurity Oxide impurity Metallic impurity

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

The function of limestone is: To decompose to form CaO and combine with silicates. To increase temperature of blast furnace. To help in the reduction of iron ore. To lower the purify the molten metal.

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

The statement which is NOT correct is: The iron obtained from blast furnace contains impurities. The iron collected is called pig iron. The iron collected is called cast iron. The iron can be moulded in different shapes.

9. Assertion (A): Iron is extracted from haematite ore.

II. Based on following passage answer questions from 6-10





This method is based on the principle that different components of a mixture are differently adsorbed on an adsorbent. The mixture is put in a liquid or gaseous medium which is moved through the adsorbent.



Different components are adsorbed at different levels on the column. Later the adsorbed components are removed (eluted) by using suitable solvents (eluant). Depending upon the physical state of the moving medium and the adsorbent material and also on the process of passage of the moving medium, the chromatographic method is given the name. In one such method the column of Al2O3 is prepared in a glass tube and the moving medium containing a solution of the components is in liquid form. This is an example of column chromatography. This is very useful for purification of the elements which are available in minute quantities and the impurities are not very different in chemical properties from the element to be purified.

Reason (R): The iron is extracted as cast iron.

10. Assertion (A): Aluminium is extracted by electro­ metallurgy of molten alumina. Reason (R): Its oxide is very stable and cannot be reduced by any reducing agents like H2, CO, or C. [C] COMPETENCY BASED QUESTIONS I.

Based on following passage answer questions from 1-5



In the Blast furnace, reduction of iron oxides takes place in different temperature ranges. Hot air is blown from the bottom of the furnace and coke is burnt to give temperature up to about 2200K in the lower portion itself. The burning of coke therefore supplies most of the heat required in the process. The CO and heat move to upper part of the furnace. In upper part, the temperature is lower and the iron oxides (Fe2O3 and Fe3O4) coming from the top are reduced in steps to FeO.



Limestone is also decomposed to CaO which removes silicate impurity of the ore as slag. The slag is in molten state and separates out from iron.



The iron obtained from Blast furnace contains about 4% carbon and many impurities in smaller amount (e.g., S, P,

6. The principle of chromatographic separation is: (1) different components of a mixture are differently adsorbed on an adsorbent. (2) different components of a mixture are differently absorbed on an adsorbent.

63

GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS

(3) different components of a mixture are equally adsorbed on an adsorbent. (4) different components of a mixture are equally absorbed on an adsorbent. 7. (1) (2) (3) (4)

The components are removed by: Adding water By heating By eluting using suitable solvents (eluant). By making a paste

8. The chromatographic methods depend upon: (A) the physical state of the moving medium (B) the adsorbent material (C) the process of passage of the moving medium. Choose the correct answer from the options given below. (1) Only (A) (2) Only (B) (3) Only (C) (4) All (A), (B) and (C)

9. For purification of the elements which are available in minute quantities, the method used is: (1) Gas chromatography (2) Liquid chromatography (3) Solid chromatography (4) Column chromatography 10. The mobile phase and the stationary phase are chosen based on: (1) different solubilities of components of the samples in the two phases. (2) same solubilities of components of the samples in the two phases. (3) different solubilities of components of the samples in the one phase. (4) same solubilities of components of the samples in the one phase.

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (2)

2. (2)

3. (3)

4. (4)

5. (4)

6. (4)

7. (3)

8. (1)

9. (3)

10. (4)

11. (2)

12. (3)

13. (3)

14. (1)

15. (2)

16. (1)

17. (1)

18. (2)

19. (3)

20. (1)

21. (3)

22. (2)

23. (4)

24. (2)

25. (3)

26. (3)

27. (4)

28. (3)

29. (2)

30. (3)

8. (3)

9. (2)

10. (1)

8. (4)

9. (4)

10. (1)

[B] ASSERTION REASON QUESTIONS 1. (1)

2. (1)

3. (2)

4. (2)

5. (2)

6. (4)

7. (1)

[C] COMPETENCY BASED QUESTIONS 1. (4)

2. (3)

3. (2)

4. (1)

5. (3)

6. (1)

7. (3)

ANSWERS WITH EXPLANATION [A] MULTIPLE CHOICE QUESTIONS

Iron

Haematite Magnetite Siderite Iron pyrites

Fe2O3 Fe3O4 FeCO3 Fes2

Copper

Copper pyrites Malachite Cuprite Copper glance

CuFeS2 CuCO3Cu(OH)2 Cu2O Cu2S

Zinc

Zinc blende or Sphalerite Calamine Zincite

ZnS ZnCO3 ZnO

1. Option (2) is correct. Explanation: Siderite: FeCO3 Malachite: Cu₂CO₃(OH)₂ Calamine: Zinc oxide and 0.5% ferric oxide (Fe2O3). Bauxite: Al2O3 2. Option (2) is correct.

Explanation: Ti or Zr metal is refined by the Van-Arkel method. Metal is converted to volatile tetraiodide which is then decomposed by electrically heating above 1800 K with a tungsten filament to get pure metal. 3. Option (3) is correct. Explanation: Metal

Ores

Composition

Aluminium

Bauxite Kaolinite (a form of clay)

AlOx(OH)3.2x [Where 0 AsH3 > SbH3 > BiH3 (HPO)3 i. Acidic nature of oxides: ƒ N2O3 > P2O3 > As2O3 > Sb2O3 > Bi2O3 O OH O j. Nitrogen shows anomalous behaviour because of its: ƒ small size P P P ƒ high electronegativity O O O O ƒ high ionization enthalpy OH O OH ƒ absence of d-orbitals. Polymetaphosphoric aicd ƒ Nitrogen







  Group 16 elements (Oxygen family) :

O(8), S(16), Se(34), Te(52), Po(84)





(HPO3)n

Fig:7.1: structure of some important oxy acids of phosphorous

a. General electronic configuration: ns2np4 b. Oxidation state: show + 2, + 4 and + 6. ƒ Oxygen shows –2 state, with +2 only with Fluorine c. Physical state: ƒ O2 is a gas, all others are solids d. Atomic radii: ƒ Increases down the group. ƒ Oxygen being the smallest e. Ionization enthalpy: ƒ Decreases down the group.

Fig: 7.2: shape of S8 and S6 molecule

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Monoclinic crystals of sulphur

68 Oswaal CUET (UG) Chapterwise Question Bank

CHEMISTRY

69

p-BLOCK ELEMENTS







f. Electron gain enthalpy: ƒ Decreases down the group ƒ Exception being O < S due its small size g. Electronegativity: ƒ Decreases down the group. ƒ Oxygen being the most electronegative h. Metallic character: ƒ Increases down the group ƒ O and S are non-metals

ƒ Se

and Te are metalloids is metal i. Properties of hydrides: ƒ Thermal stability: Decreases down the group. ƒ Acidic character: Increases down the group. ƒ Reducing nature: All are reducing agents except H2O. ƒ Boiling point: From H2O to H2S there is a sudden drop which increases from H2S to H2Te. ƒ Po



Fig: 7.3 oxy acids of sulphur   Group 17 elements (halogens): F(9), Cl(17),

Br(35), I(53), At(85)









a. General electronic Configuration: ns2 np5 b. Oxidation state: show +1, +3, +5 and +7. ƒ F shows only –1 Scan to know more about this c. Physical state: topic ƒ F2 and Cl2 are gases ƒ Br2 is liquid ƒ I2 is solid d. Atomic radii: Properties of ƒ Increases down the group. Halogens ƒ Fluorine being the smallest e. Ionization enthalpy: decreases down the group. f. Electron gain enthalpy: ƒ decreases down the group ƒ Exception, with F having less electron gain enthalpy because of its small size. g. Oxidizing property: F2 > Cl2 > Br2 > I2





h. Acidic strength: HF < HCl < HBr < HI i. Stability and bond dissociation enthalpy: HF > HCl > HBr > HI j. Stability of oxides of halogens: I > Cl > Br k. Acidic strength of oxoacids containing the same halogen: ƒ HOCl > HClO2 < HClO3 < HClO4

Fig 7.4: oxy acids of Chlorine l. Interhalogen compounds: ƒ General formula- XX’, XX’3, XX’5 and XX’7 ƒ covalent in nature

70 Oswaal CUET (UG) Chapterwise Question Bank ƒ diamagnetic ƒ more

in nature reactive than halogens

  Group 18 elements (Noble gases) : He(2), Ne(10),

Ar(18), Kr(36), Xe(54), Ra(86).



a. General electronic configuration: ns2 np6 2 ƒ Except He, ns Scan to know b. Atomic radii: increase down the more about this topic group c. Ionization enthalpy: Highest in respective periods, decreases down the group d. Electron gain enthalpy: large positive Noble gases values e. Liquefaction: difficult to liquefy ƒ down the group, the ease of liquefaction increases f. Reactivity: ƒ Due to complete octet of outermost shell, they have less tendency to form compounds. ƒ xenon combines with fluorine and oxygen only under certain conditions.



CHEMISTRY

g. Uses: ƒ Helium (He) is used in filling balloons for meteorological observations in gas-cooled nuclear reactors. Liquid He is used as cryogenic agent. ƒ Neon (Ne) is used in discharge tubes and fluorescent bulbs. ƒ Argon (Ar) is used to provide an inert atmosphere in high temperature metallurgical processes, for filling electric bulbs, in the laboratory for handling substances that are air-sensitive.

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Fig: 7.5: compounds of Xe, O and F

Compounds of Xe and F

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS 1. Arrange the following hydrides in increasing order of thermal stability. (A) H2O (B) H2Se (C) H2Po (D) H2Te (E) H2S Choose the correct answer from the options given below: (1) A < B < C < D < E (2) C < D < B < E < A (3) C < D < E < B < A (4) A < E < B < D < C [CUET 2022, 10th Aug] 2. The formula of a noble gas species which is isostructural with BrO3– is: (1) XeOF4 (2) XeF2 (3) XeO3 (4) XeF4 [CUET 2022, 10th Aug] 3. Which of the following oxoacids of chlorine is most acidic? (1) HClO (2) HClO2 (3) HClO3 (4) HClO4 [CUET 2022, 18th Aug] 4. Identify the one that does not exist: (1) XeOF4 (2) NeF2 (3) XeF2 (4) XeF6 [CUET 2022, 18th Aug] 5. Which of the following has a square pyramidal structure? (1) IBr (2) IF7

(3) CIF3

(4) BrF5 [CUET 2022, 20th Aug] 6. Oxygen shows anomalous behaviour due to its: (A) Small size (B) Covalency of four (C) High electronegativity (D) Tendency to form weak H-bonds Choose the correct answer from the options given below: (1) (A) and (B) only (2) (B) and (C) only (3) (C) and (D) only (4) (A) and (C) only [CUET 2022, 21st Aug] 7. The presence of SO2 is not detected by: (1) Its characteristic pungent smell. (2) Turning of red litmus blue. (3) Turning of acidified potassium dichromate solution green. (4) Decolourisation of acidified potassium permanganate solution. [CUET 2022, 21st Aug] 8. The correct order of boiling points for hydrogen halides is: (1) HCl (2) HBr (3) HF (4) HI Choose the correct answer from the options given below : (1) HF < HI < HBr < HCl (2) HF < HCl < HBr < HI (3) HF > HCl > HBr > HI (4) HF > HI > HBr > HCl [CUET 2022, 23rd Aug] 9. In the reaction: N2(g) + 3H2(g) → 2NH3(g), ΔH° = –46.1 k J mol –1 The yield of ammonia is expected to be maximum at: (1) High temperature and low pressure

71

p-BLOCK ELEMENTS

(2) High temperature and high pressure (3) Low temperature and high pressure (4) Low temperature and low pressure [CUET 2022, 23rd Aug] 10. The correct structure of perchloric acid is  (1)

(3)

11. (1) (2) (3) (4) 12. (1) (3) 13. (1) (2) (3) (4) 14. (1) (3) 15. (1) (3) 16. (1) (2) (3) (4) 17.





(2)

(4)

[CUET 2022, 30th Aug] The correct order of oxidation states of nitrogen: in NO, N2O, NO2 and N2O3 respectively is: N2O < NO2 < NO < N2O3 N2O < NO < N2O3 < NO2 NO2 < N2O3 < N2O < NO NO2 < NO < N2O < N2O3 [CUET 2022, 30th Aug] Correct number of lone pairs in SF4 and H2O molecules, respectively, should be: 1 and 2 (2) 2 and 1 1 and 1 (4) 2 and 2 [CUET 2021, 23rd Sept] On addition of conc. H2SO4 to a chloride salt, colourless fumes are evolved but in case of iodide salt, violet fumes come out. This is because: H2SO4 reduces HI to I2 HI is of violet colour HI gets oxidised to I2 HI changes to HIO3 [NCERT Exemp. Q 1, Page 90] Which of the following elements can be involved in pπ–dπ bonding? Carbon (2) Nitrogen Phosphorus (4) Boron [NCERT Exemp. Q 4, Page 90] Which of the following pairs of ions are isoelectronic and isostructural? CO32–, NO3– (2) ClO3–, CO32– 2– – SO3 , NO3 (4) ClO3–, SO32– [NCERT Exemp. Q 5, Page 91] 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? It is highly poisonous and has smell like rotten fish. It’s solution in water decomposes in the presence of light. It is more basic than NH3. It is less basic than NH3. [NCERT Exemp. Q 8, Page 91] Which of the following acids forms three series of salts?

(1) H3PO2 (3) H3BO3 18. (1) (3) 19. (1) (3) 20. (1) (2) (3) (4) 21.

(1) (3) 22. (1) (2) (3) (4) 23. (1) (2) (3) (4) 24. (1) (2) (3) (4) 25. (1) (2) (3) (4) 26. (1) (2)

(2) H3PO4 (4) H3PO3 [NCERT Exemp. Q 9, Page 91] On heating lead nitrate forms oxides of nitrogen and lead. The oxides formed are ______. N2O, PbO (2) NO2, PbO NO, PbO (4) NO, PbO2 [NCERT Exemp. Q 11, Page 92] Which of the following elements does not show allotropy? Nitrogen (2) Bismuth Antimony (4) Arsenic [NCERT Exemp. Q 12, Page 92] Which of the following statements is wrong? Single N–N bond is stronger than the single P–P bond. PH3 can act as a ligand in the formation of coordination compound with transition elements. NO2 is paramagnetic in nature. Covalency of nitrogen in N2O5 is four. [NCERT Exemp. Q 14, Page 92] Hot conc. H2SO4 acts as moderately strong oxidising agent. It oxidises both metals and nonmetals. Which of the following element is oxidised by conc. H2SO4 into two gaseous products? Cu (2) S C (4) Zn [NCERT Exemp. Q 22, Page 93] In the preparation of compounds of Xe, Bartlett had taken O2+ Pt F6 – as a base compound. This is because: both O2 and Xe have same size. both O2 and Xe have same electron gain enthalpy. both O2 and Xe have almost same ionisation enthalpy. both Xe and O2 are gases. [NCERT Exemp. Q 24, Page 94] When KMnO4 solution is added to oxalic acid solution, the decolourisation is slow in the beginning but becomes instantaneous after some time because: CO2 is formed as the product. Reaction is exothermic. MnO4– catalyses the reaction. Mn2+ acts as auto-catalyst. Which of the following statements are correct for SO2 gas? It acts as bleaching agent in moist conditions. Its molecule has linear geometry. It can be prepared by the reaction of dilute H2SO4 with metal sulphide. All of the above In solid state, PCl5 is a: covalent solid. octahedral structure. ionic solid with [PCl6]+ octahedral and [PCl4]– tetrahedral. ionic solid with [PCl4]+ tetrahedral and [PCl6]– octahedral. Phosgene is a common name for: phosphoryl chloride thionyl chloride

72 Oswaal CUET (UG) Chapterwise Question Bank (3) (4) 27. (1) (3) 28.

(1) (3) 29. (1) (3) 30. (1) (2) (3) (4)

carbon dioxide and phosphine carbonyl chloride When XeF6 is partially hydrolysed, it yields: XeSO3 (2) XeOF2 XeOF4 (4) XeF2 A black compound of manganese reacts with a halogen acid to give greenish yellow gas. When excess of this gas reacts with NH3 an unstable trihalide is formed. In this process the oxidation state of nitrogen changes from: – 3 to +3 (2) – 3 to 0 – 3 to +5 (4) 0 to – 3 When (NH4)2Cr2O7 is heated, the gas evolved is: N2 (2) NO2 O2 (4) N2O Identify the incorrect statement, regarding the molecule XeO4: XeO4 molecule is square planar. There are four pπ - dπ bonds. There are four sp3 p, σ bonds. XeO4 molecule is tetrahedral

[B] ASSERTION REASON QUESTIONS Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (1) Both A and R are correct and R is The correct explanation of A. (2) Both A and R are correct but R is NOT The correct explanation of A. (3) A is correct but R is not correct. (4) A is not correct but R is correct. 1. Assertion (A): N2 is less reactive than P4. Reason (R): Nitrogen has more electron gain enthalpy than phosphorus.  [NCERT Exemp. Q 64, Page 99] 2. Assertion (A): HNO3 makes iron passive. Reason (R): HNO3 forms a protective layer of ferric nitrate on the surface of iron.  [NCERT Exemp. Q 65, Page 99] 3. Assertion (A): HI cannot be prepared by the reaction of KI with concentrated H2SO4. Reason (R): HI has lowest H–X bond strength among halogen acids.  [NCERT Exemp. Q 66, Page 100] 4. Assertion (A): Both rhombic and monoclinic sulphur exist as S8 but oxygen exists as O2. Reason (R): Oxygen forms pπ – pπ multiple bond due to small size and small bond length but pπ – pπ bonding is not possible in sulphur.  [NCERT Exemp. Q 67, Page 100] 5. Assertion (A): NaCl reacts with concentrated H2SO4 to give colourless fumes with pungent smell. But on adding MnO2 the fumes become greenish yellow. Reason (R): MnO2 oxidises HCl to chlorine gas which is greenish yellow.  [NCERT Exemp. Q 68, Page 100]

CHEMISTRY

6. Assertion (A): SF6 cannot be hydrolysed but SF4 can be. Reason (R): Six F atoms in SF6 prevent the attack of H2O on sulphur atom of SF6.  [NCERT Exemp. Q 69, Page 100] 7. Assertion (A): Valency of noble gas is 0. Reason (R): Noble gases possess complete octet. 8. Assertion (A): Interhalogen compounds are more reactive than halogens (except chlorine). Reason (R): They all are diamagnetic in nature. 9. Assertion (A): Ozone layer in the upper region of atmosphere protect earth from UV radiation. Reason (R): Ozone is a powerful oxidizing agent as compared to oxygen. 10. Assertion (A): Oxygen shows –2 oxidation state in all its compounds. Reason (R): Oxygen is less electronegative than fluorine. [C] COMPETENCY BASED QUESTIONS I. Based on following passage answer questions from 1-5 On heating compound (A) gives a gas (B) which is a constituent of air. This gas (B) when treated with 3 mol of hydrogen (H2) in the presence of a catalyst gives another gas (C) which is basic in nature. Gas C on further oxidation in moist condition gives a compound (D) which is a part of acid rain.  [NCERT Exemp. Q 72, Page 100] 1. Compound (A) is identified as: (1) Ammonium nitrate (2) Ammonium nitrate (3) Sodium nitrite (4) Sodium nitrate 2. Gas (B) is identified as: (1) Nitric oxide (2) Nitrogen dioxide (3) Nitrogen (4) Ammonia 3. The catalyst used in conversion of gas (B) to gas (C) is: (1) Mo (2) Fe (3) Pt (4) Cu 4. The correct products formed when gas (C) is oxidised are: (1) N2 and H2O (2) N2O and H2O (3) NO and H2 (4) NO and H2O 5. The correct statement is: (1) Acid rain is formed when nitrous oxide reacts with water. (2) Acid rain is formed when nitrous oxide gets oxidised to nitric oxide, which reacts with water. (3) Acid rain is formed when nitric oxide falls as rain. (4) Acid rain is formed when nitric oxide gets oxidised to nitrogen dioxide, which reacts with water. II. Based on following passage answer questions from 6–10 All noble gases have general electronic configuration ns2np6 except helium which has 1s2. Many of the properties of noble gases including their inactive nature are ascribed to their closed shell structures. Due to stable electronic configuration these gases exhibit very high ionisation enthalpy. However, it decreases down the group with increase in atomic size. Atomic radii increase down the group with increase in atomic number. Since noble gases have stable electronic configurations, they have no

73

p-BLOCK ELEMENTS



6. (1) (3) 7. (1) (3)

tendency to accept the electron and therefore, have large positive values of electron gain enthalpy. All the noble gases are monoatomic. A series of compounds like XeF2, XeF4, XeF6, XeO2F2, XeOF4 are formed. After this discovery, a number of xenon compounds mainly with most electronegative elements like fluorine and oxygen, have been synthesised. The compounds of krypton are fewer. Only the difluoride (KrF2) has been studied in detail. The general electronic configuration of Group 18 [except He] is: ns1np6 (2) ns2np5 2 6 ns np (4) ns2np5 The shape of XeF4 is: Linear (2) Triagonal Square planar (4) Square pyramidal

8. (1) (3) 9. (1) (3) 10. (1)

The ionisation enthalpy of Group 18 is: Highest in a period (2) Lowest in a period Highest in a group (4) Lowest in a group Three lone pairs are present in: XeO2F2 (2) XeOF4 XeF2 (4) XeF4 The correct statement is: Noble gases are diatomic and can accept and give electrons (2) Noble gases are diatomic and cannot accept and give electrons (3) Noble gases are monoatomic and can accept and give electrons (4) Noble gases are monoatomic and cannot accept and give electrons

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (2)

2. (3)

3. (4)

4. (2)

5. (4)

6. (4)

7. (2)

8. (4)

9. (3)

10. (1)

11. (2)

12. (1)

13. (3)

14. (3)

15. (1)

16. (3)

17. (2)

18. (2)

19. (1)

20. (1)

21. (3)

22. (3)

23. (4)

24. (1)

25. (4)

26. (4)

27. (2)

28. (1)

29. (1)

30. (1)

8. (2)

9. (2)

10. (4)

8. (1)

9. (3)

10. (4)

[B] ASSERTION REASON QUESTIONS 1. (3)

2. (3)

3. (2)

4. (1)

5. (1)

6. (1)

7. (1)

[C] COMPETENCY BASED QUESTIONS 1. (1)

2. (3)

3. (2)

4. (3)

5. (4)

6. (3)

7. (3)

ANSWERS WITH EXPLANATION [A] MULTIPLE CHOICE QUESTIONS 1.

Option (2) is correct.

Explanation: Thermal stability of group 16 elements decreases down the group. H2O > H2S > H2Se> H2Te > H2Po This is because the H-E bond length increases down the group, hence the bond dissociation enthalpy decreases down the group. 2. Option (3) is correct. Explanation: In BrO3–, the hybridisation of Br is sp3 and has a pyramidal structure while in XeO3, the hybridisation of Xe is also sp3 with pyramidal structure. BrO3– is isostructural with XeO3.

HClO < HClO2 < HClO3 < HClO4 4.

Option (2) is correct.

Explanation: Compound of neon does not exist. Neon does not form compounds like xenon because neon holds its electrons much more tightly than xenon. 5.

Option (4) is correct.

Explanation:



I — Br Linear 3.

Option (4) is correct.

Explanation: As the oxidation number increases, the acidic character increases. Hence, the order of acidity is acidic strength:

6.



Option (4) is correct.

Explanation: Oxygen shows anomalous behaviour due to (1) small atomic size, (2) high electronegativity & (3) absence of inner ‘d’ orbital.

74 Oswaal CUET (UG) Chapterwise Question Bank 7.

Option (2) is correct..

Explanation: According to the question, SO2 gas can be detected by A, (pungent smell), C (Turning of acidified potassium dichromate solution green) and D (decolourisation of acidified potassium permanganate solution). But being an acidic gas cannot be detected by B-turning of red litmus blue. When SO2 gas is passed through acidified K2Cr2O7, orange colour of K2Cr2O7 is turned into green due to formation of chromium sulphate. K2Cr2O7 (Orange) + H2SO4 + 3SO2 → K2SO4+ Cr2 (SO4)3(green) + H2O 8.

Option (4) is correct.

Explanation: HF has an exceptionally high boiling point due to intermolecular Hydrogen-bonding. The boiling points of the rest of the hydrogen halides increase as the molecular size increases. The extra electrons allow bigger temporary dipoles and so increase the amount of van der Waals dispersion forces between the molecules. Thus, the boiling point of HF, HCl, HBr and HI are 293 K, 189 K, 206 K and 238 K respectively. 9.

Option (3) is correct.

Explanation: The production of ammonia is exothermic reaction. According to Le Chatelier’s Principle, this will be favoured if lower the temperature. The system will respond by moving the position of equilibrium to counteract this in other words by producing more heat. 10. Option (1) is correct. Explanation: The correct structure of perchloric acid is: Rest all options are not valid as they do not follow the octet rule for stability. 11. Option (2) is correct. Explanation: N2O: 2x + (–2) = 0, x = +1 NO: x + (–2) = 0, x = +2 N2O3: 2x + 3(–2) = 0, x = +3 NO2 : x + 2(–2) = 0, x = +4 Correct order of O.S is: N2O < NO < N2O3 < NO2 12. Option (1) is correct. Explanation: SF4 13. Option (3) is correct. Explanation: HI formed during reaction is oxidized to I2 which is violet in colour. 2 NaCl + H2SO4 → Na2SO4 + 2HCl In case of Iodine, the halogen acid obtained (HI) is oxidised to free Iodine. 14. Option (3) is correct. Explanation: Phosphorus can be involved in pπ-dπ bonding due to presence of vacant d orbitals. C, N, and B do not have d orbitals.

CHEMISTRY

15. Option (1) is correct. Explanation: CO32– and NO3– are isoelectronic with 32 electrons and sp2 hybridisation hence, trigonal planar structure. 16. Option (3) is correct. Explanation: P4 + 3 NaOH + 3H2O → PH3 + 3NaH2PO2 PH3 is less basic than NH3. 17. Option (2) is correct.

Explanation: H3PO4 has 3 –OH groups i.e., has three ionisable H-atoms and hence forms three series of salts. These three possible series of salts of H3PO4 are as follows: NaH2PO4, NaHPO4 and Na3PO4. 18. Option (2) is correct.

Explanation: 2Pb(NO3)2 → 2PbO + 4NO2 + O2 19. Option (1) is correct.

Explanation: Nitrogen does not show allotropy due to its weak N – N single bond. Therefore, ability of nitrogen to form polymeric structure or more than one structure become less. Hence, nitrogen does not show allotropy. 20. Option (1) is correct. Explanation: Nitrogen forms pπ-pπ multiple bond and the bond strength is very high. Single N – N bond is weaker than single P – P bond. 21. Option (3) is correct. Explanation: [H2SO4 → H2O + SO2 + O]x 2 C + 2O → CO2 Net reaction: C + 2 H2SO4 → 2H2O + 2SO2 + C 22. Option (3) is correct.

Explanation; Bartlett had taken O2+ Pt F6– as a base compound because O2 and Xe both have almost same ionization enthalpy. The ionization enthalpies of noble gases are the highest in their respective periods due to their stable electronic configurations. 23. Option (4) is correct. Explanation: When KMnO4 solution is added to oxalic acid solution, the decolourisation is slow in the beginning but becomes instantaneous after sometime because Mn2+ acts as an auto-catalyst. Reduction half-reaction: [MnO4– + 8H+ +5e → Mn2+ + 4H2O ] × 2 Oxidation half-reaction: [C2O4 2– → 2CO2 + 2e–] × 5 Overall equation: 2MnO4– + 16H+ + 5C2O42– → 10CO2 + 2Mn2+ + 8H2O End point of this reaction: Colourless to light. 24. Option (1) is correct. Explanation: SO2 acts as a bleaching agent under moist conditions. SO2 (g) + 2H2O → H2SO4+ 2[H] SO2 is oxidized to sulphuric acid and releases nascent hydrogen which bleaches the material. But this is a temporary

75

p-BLOCK ELEMENTS

as atmospheric oxygen re-oxides the bleached matter after sometime.

Explanation: SF6 is sterically protected due to presence of six F atoms around S which prevents the attack of H2O on SF6.

25. Option (4) is correct.

7.

Explanation: the structure of PCl5 in solid state is:

Explanation: Noble gases possess the electronic configuration ns1np6 and has 8 electrons in their outer shell, hence their valency is 0. 8.

Option (1) is correct.

Option (2) is correct.

Explanation: Interhalogen compounds are more reactive than halogens because X-X’ bond in interhalogens in weaker than X-X bond in halogen (except F-F bond). 26. Option (4) is correct. Explanation: Carbonyl chloride is COCl2 (phosgene). 27. Option (2) is correct.

Explanation: Partial hydrolysis of XeF6 gives oxy difluorides, XeO2F2. XeF4 + H2O → XeOF2 + 2HF Xenon oxy difluoride 28. Option (1) is correct. Explanation: MnO2 + 4HCl → MnCl2 + 2H2O + Cl2 (Greenish yellow gas) NH3 + 3Cl2 → NCl3 + 3HCl When excess of chlorine reacts with ammonia then NCl3 and HCl will form. In this reaction on left-hand side Nitrogen has (−3) oxidation state and on the right-hand chlorine has (+3) oxidation state. 29. Option (1) is correct. Explanation: When (NH4)2Cr2O7 is heated then N2 gas is evolved. (NH4)2Cr2O7 + heat → Cr2O3 + 4H2O + N2 ↑ 30. Option (1) is correct. Explanation:

9.

Option (2) is correct.

Explanation: Ozone layer filters the radiation coming from sun, hence serves as the protective layers. 10. Option (4) is correct. Explanation: Oxygen is less electronegative than Fluorine and hence shows +2 oxidation state in OF2. [C] COMPETENCY BASED QUESTIONS 1.

Option (1) is correct.

Explanation: NH 4 NO 2 ¾Heat ¾¾ ® N 2 + 2H 2O 2.

Option (3) is correct.

Explanation: The gas is nitrogen. N 2 +3H 2 ¾Catalyst ¾¾® 2 NH 3 3.

Option (2) is correct.

Explanation: Fe acts a s a catalyst in Haber’s process in the conversion of N2 to NH3. 4.

Option (3) is correct.

Explanation: 4NH 3 +5O 2 ¾¾ ® 4 NO+6H 2O ® 2NO 2 4NO+O 2 ¾¾ 5.

Option (4) is correct.

Explanation: [B] ASSERTION REASON QUESTIONS 1.

Option (3) is correct.

Explanation: N2 is less reactive due to high bond dissociation energy. Its electron gain enthalpy is less than phosphorus. 2.

Option (3) is correct.

Explanation: Passivity is attained by formation of a thin film of oxide on iron. 3.

2NO+O 2 ¾¾ ® 2 NO 2 3NO 2 +H 2O ¾¾ ® 2HNO3 + NO 6.

Option (3) is correct.

Explanation: Group 18 have complete octet and hence their general electronic configuration except He is ns2np6 7.

Option (3) is correct.

Explanation:

Option (2) is correct.

Explanation: Both statements are correct but are independent of each other. 4.

Option (1) is correct.

Explanation: S exists as S8 but oxygen exists as O2 because oxygen forms pπ-pπ multiple bonds which is not possible in S. 5.

Option (1) is correct.

Explanation: Colourless fumes of HCl become greenish yellow because MnO2 oxidises HCl to chlorine gas. 6.

Option (1) is correct.

8.

Option (1) is correct.

Explanation: Ionization enthalpy is highest in respective periods and decreases down the group. This is because of their complete duplet/ octet.

76 Oswaal CUET (UG) Chapterwise Question Bank 9.

Option (3) is correct.

Explanation: the shape of XeF2 is linear with three lone pair of electrons.

CHEMISTRY

10. Option (4) is correct. Explanation: Since noble gases have stable electronic configurations, they have no tendency to accept the electron and therefore, have large positive values of electron gain enthalpy. All the noble gases are monoatomic.

Study Time

CHAPTER

8

 Revision Notes

Max. Time: 1:50 Hours Max. Questions: 50

d AND f BLOCK ELEMENTS

 d-Block elements: Group 3 – 12 (Transition elements) .  General electronic configuration of transition elements: (n–1)d1-10 ns1-2  d-block elements: The elements in which last electron enters in the d - sub-shell i.e. penultimate shell and lies in the middle of the periodic table belonging to groups 3-12.  Transition elements: The elements of d-block are known as transition elements as they possess properties that are transitional between the s-block and p-block elements. Transition elements are defined as the elements which have incompletely filled d-orbitals in their ground states or in its most common oxidation state.  Transition elements have four series: (i)  First transition series: These elements have incomplete 3d-orbitals and they are from Sc (21) to Zn (30). (ii) Second transition series: These elements have incomplete 4d-orbitals and they are from Y (39) to Cd (48). (iii)  Third transition series: These elements have incomplete 5d- orbitals and they are La (57) and then from Hf (72) to Hg (80). (iv)  Fourth transition series: This series is yet incomplete and these elements have incomplete 6d- orbitals. Known elements of this series are– actinium (89) and then from Rf (104) and other elements.  Physical properties: 1. All d-block elements are metals. 2. They are malleable, ductile, lustrous and sonorous except mercury which is a liquid. 3.  Atomic radii: In period, first decreases till the middle, becomes constant and then increases. 4.  Ionic radii: Decreases with increase in oxidation state. 5. Ionization enthalpy: From left to right, it increases. 6. They show variable oxidation state. 7. Complex and interstitial compounds are formed. 8. The magnetic moment increases with the increasing number of unpaired electrons.

9. Formation of colored compounds and alloys.  Potassium dichromate (K2Cr2O7) Preparation: Preparation: (i) Chromate ore is fused with sodium carbonate in the presence of air to give sodium chromate. 2 FeCr2O4 + 4Na2CO3 + 7/2O2 → Fe2O3 + 4Na2CrO4 + 4CO2 Sodium chromate (ii) Na2CrO4 is filtered and acidified with conc. H2SO4 to give Na2Cr2O7. 2Na2CrO4 + 2H+ → Na2Cr2O7 + 2Na+ + H2O. (iii) Sodium dichromate solution is treated with KCl to give K2Cr2O7. Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl Properties: Orange, crystalline solid. Cr2O72– + 2OH– ® 2CrO42– + H2O Dichromate ion Chromate ion (orange red) (Yellow) 1. Cr2O72– + 14H+ + 6e– →2Cr3+ + 7H2O [Fe2+ → Fe3+ + e] ´ 6 Cr2O72– + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O 2. Cr2O72– + 14H+ + 6e2– → 2Cr3+ + 7H2O [Sn2+ → Sn4+ + 2e]  3 2–

Cr2O7

+ 3Sn2+ +14H+ → 2Cr3+ + 7H2O + 3Sn4+

3. Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O [SO2 + 2H2O → SO42– + 4H+ 2e–]  3 2–

Cr2O7

+ 3SO2 + 2H+ → 2Cr3+ + 3SO42– + H2O

4. Cr2O72– + 14H + 6e– → 2Cr3+ + 7H2O [H2S → 2H+ + S + 2e–]  3 2–

Cr2O7

+ 3H2S + 8H+ → 2Cr3+ +3S + 7H2O

5. Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O [2I– → I2 + 2e–]  3 2–

Cr2O7

+ 6I– + 14H+ → 2Cr3+ 3I2 + 7H2O

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784 Oswaal CUETQuestion (UG) Chapterwise Question Bank CHEMISTRY OSWAAL CBSE Sample Papers, CHEMISTRY, Class-XII

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79

d AND f BLOCK ELEMENTS

Structure: 2–

O

O Cr O O

O

O

Chromate ion

9 17

 General electronic configuration of f - block ­elements is (n–2)f1-14 (n–1)d0-1 ns2  Lanthanoids: Ce (Z = 58) to Lu (Z = 71) Electronic Configuration: [Xe] 4f1–14 5d0–1 6s2  Properties: Scan to know more about this ƒ Highly dense metals, soft, mal­ topic leable and ductile. ƒ High melting point. ƒ Atomic and ionic size decreases with increasing atomic number. Lanthanoids This steady decrease is known as (f-block elements) lanthanide contraction. ƒ Oxidation state: Mainly +3 but some show +2 and +4 also.

2–

pm O

O

126° Cr 16Cr 3p m

O Dichromate ion

O O

 Potassium permanganate (KMnO4): In laboratory, (i) 2MnO2 +4KOH + O2 ® 2K2MnO4 + 2H2O 3MnO42– + 4H+ ® 2MnO4– + MnO2 + 2H2O 2– Fused with KOH (ii) MnO2 + 2e– ¾Oxidised ¾¾¾¾¾¾ ® MnO4 with air/KNO



3

Manganate ion – – oxidation MnO42– ¾Electrolytic ¾¾¾¾¾ ® MnO4 +1e in alkaline solution

+C, 2773 K

Permanganate ion

+O2

In laboratory, 2Mn2+ + 5S2O82– + 8H2O ® 2MnO4– + 10SO42– + 16H+

Ln2S3

Properties: Dark purple crystalline solid, sparingly soluble in water and powerful oxidising agent.

2.

MnO4– + 8H+ + 5e– ® Mn2+ + 4H2O [Fe2+ ® Fe3+ + e–]  5 MnO4– + 5Fe2+ + 8H+ ® Mn2+ + 5Fe3+ +4H2O

3.

[MnO4– + 8H + 5e– ® Mn2+ + 4H2O]  2 [C2O42– ® 2CO2 + 2e–]  5 2MnO4– + 5C2O42– + 16H+ ® 2Mn2+ + 10CO2 + 8H2O

4.

[MnO4– + 8H+ + 5e– ® Mn2+ + 4H2O]  2 [S2– ® S + 2e–]  5 2MnO4– + 5S2– + 16H+ ® 2Mn2+ + 5S + 8H2O

5.

[MnO4– + 8H+ + 5e– ® Mn2+ + 4H2O]  2 [SO32– + H2O ® SO42– + 2H+ + 2e–]  5 5SO32– + 2MnO4– + 6H+ ® 2Mn2+ + 5SO42– + 3H2O

6.

[MnO4– + 8H+ + 5e– ® Mn2+ + 4H2O]  2 [NO2– + H2O ® NO3– + 2H+ + 2e–]  5 2MnO4– + 5NO2– + 6H+ ® 2Mn2+ + 5NO3– + 3H2O

7.

Ln

[MnO4– + 2H2O + 3e– ® MnO2 + 4OH–]  2 I– + 6OH– ® IO3– + 3H2O + 6e– 2MnO4– + I– + H2O ® IO3– + 2MnO2 + 2OH– Structure: O–

O–

Mn

Mn

O OH O O OH O Tetrahedral manganate Tetrahedral manganate ion(Purple) ion(Green)

+N2 +H2O

2KMnO4 ® K2MnO4 + MnO2 + O2 [MnO4– + 8H+ + 5e– ® Mn2+ + 4H2O]  2 [2I– ® I2 + 2e–]  5 2MnO4– + 10I– + 16H+ ® 2Mn2+ + 5I2 + 8H2O

Ln2O3

+S

Peroxodisulphate

1.

Ln3C, LnC2 LN2 C3

+X2 +acid

LnN Ln(OH)3+HX LnX2 H2

 Actinoids: Ac (Z = 89) to Lr (Z = 103) Electronic configuration: [Rn] 5f1–14 6d0–1 7s2  Properties: ƒ  Highly dense metals and forms alloys. ƒ High electropositivity and mel­ting point. ƒ Atomic and ionic radii decreases with an increase in atomic size due to poor shielding effect of 5f electrons (actinoid contraction). ƒ Paramagnetic, ions are coloured and radioactive. ƒ Oxidation state: Commonly +3 but exhibit +4, +5, +6 and +7 also. ƒ Less reactive towards acids.  Difference between Lanthanoids and Actinoids: S. No.

Lanthanoids

Actinoids

(i)

4f orbital is progressively filled.

5f orbital is progressively filled.

(ii)

+ 3 oxidation state is most common along with + 2 and + 4.

+ 3 oxidation state is most common, but exhibit higher oxidation state of + 4, + 5, + 6, + 7.

(iii)

Except promethium, All are radioactive. all are non-radioactive.

(iv)

Less tendency of complex formation.

(v)

Chemically less More reactive than reactive than actinoids. lanthanoids.

Strong tendency of complex formation.

Uses: (i) Misch metal is used in lighter flints. (ii) Lanthanoid oxides are used for polishing glass. (iii) Cerium salts are used in dyeing cotton and as catalysts.

80 Oswaal CUET (UG) Chapterwise Question Bank

CHEMISTRY

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS 1. Which pair of element does not have almost identical size? (1) Sc/Y, (2) Zr/Hf (3) Nb/Ta, (4) Mo/W 2. The lanthanides do not form complexes with π bonding ligands due to: (1) Lack of pi bonding (2) Lack of sigma bonding (3) Lack of sigma and pi bonding both (4) None of the above 3. Lanthanides are: (1) 14 elements in the sixth period (Atomic number = 90 to 103) that are filling 4f sub-level. (2) 14 elements in the seventh period (Atomic number = 90 to 103) that are filling 5f sub-level. (3) 14 elements in the sixth period (Atomic number = 58 to 71) that are filling 4f sub-level. (4) 14 elements in the seventh period (Atomic number = 58 to 71) that are filling 4f sub-level. 4. The correct statement about 4f series is: (1) Paramagnetism falls to minimum in Nd. (2) Lu3+ ions are bright green in aqueous solution. (3) Promethium is artificially synthesised 4f series. (4) Absorption band of Lanthanides are broad. 5. The correct bond order for carbon oxygen bond in following metal carbonyl is [Cr(CO)6], [V(CO)6]–, [Mn(CO)6]+ is: (1) [V(CO)6]< [Cr(CO)6]+ < [Mn(CO)6]+ (2) V(CO)6]- = [Cr(CO)6]+ = [Mn(CO)6]+ (3) Mn(CO)6]+ > [Cr(CO)6]+ > [ V(CO)6](4) [Cr(CO)6]+ > [ V(CO)6]- > [Mn(CO)6]+ 6. The magnetic pair of element depends on the presence of unpaired electron. Identify the configuration of transition element which shows highest magnetic moment. (1) 3d7 (2) 3d5 8 (3) 3d (4) 3d2 7. Match List I with List II. List I



List II

(A) Elements with highest second ionisation enthalpy

(I) Ni

(B) Elements with highest third ionisation enthalpy

(II) Cr

(C) M in M(CO)6 is

(III) Cu

(D) Elements with highest heat (IV) Zn of atomisation Choose the correct options from the options given below: (1) (A)-(I), (B)-(II), (C)-(III), (D)-(IV) (2) (A)-(IV), (B)-(III), (C)-(II), (D)-(I) (3) (A)-(II), (B)-(III), (C)-IV, (D)-(I) (4) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)

8. The metal refined by the Van Arkel method is: (1) Ni (2) Zr (3) Cu (4) Sn 9. The formula of a noble gas species which is isostructural with BrO3– is: (1) XeOF4 (2) XeF2 (3) XeO3 (4) XeF4 10. Match list I with list II. List I (Transition Metals) (A) Ti

List II (Maximum Oxidation State) (I) 7

(B) V

(II) 4

(C) Mn

(III) 5

(D) Cu (IV) 2 Choose the correct answer from the options given below: (1) (A)-(II), (B)-(III), (C)-(I), (D)-(IV) (2) (A)-(I), (B)-(II), (C)-(III), (D)-(IV) (3) (A)-(III), (B)-(I), (C)-(II), (D)-(IV) (4) (A)-(II), (B)-(I), (C)-(III), (D)-(IV) 11. The metal from first transition series having positive E° M2+/ M value: (1) Cr (2) V (3) Cu (4) Ni 12. Magnetic moment of a divalent ion in aqueous solution of an element with atomic number 25 is: (1) 2.84 BM (2) 3.87 BM (3) 4.90 BM (4) 5.92 BM [NCERT Exemp. Page 74] 13. Match the items of Column I with the items of Column II and assign the correct code:

Column I

Column II

(A) Coloured bands

(I) Zone refining

(B) Impure metal to volatile complex

(II) Fractional distillation

(C) Purification of Ge and Si

(III) Mond Process

(D) Purification of mercury (IV) Chromatography (V) Liquation (1) (2) (3) (4)

(A) (I), (B) (II), (C) (IV), (D) (V) (A) (IV), (B) (III), (C) (I), (D) (II) (A) (III), (B) (IV), (C) (II), (D) (I) (A) (V), (B) (IV), (C) (III), (D) (II) [NCERT Exemp. Q 46, Page 84] 14. Match items of Column I with the items of Column II and assign the correct code: Column I

Column II

(A) Cyanide process

(I) Ultrapure Ge

(B) Forth floatation process

(II) Dressing of ZnS

81

d AND f BLOCK ELEMENTS

(C) Electrolytic reduction

(III) Extraction of Al

(D) Zone refining

(IV) Extraction of Au (V) Purification of Ni

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

(A) (IV), (B) (II), (C) (III), (D) (I) (A) (II), (B) (III), (C) (I), (D) (V) (A) (I), (B) (II), (C) (III), (D) (IV) (A) (III), (B) (IV), (C) (V), (D) (I) [NCERT Exemp. Q 47, Page 84] 15. Match the items of Column I with the items of Column II and assign the correct code : Column I

Column II

(A) Sapphire

(I) Al2O3

(B) Sphalerite

(II) NaCN

(C) Depressant

(III) Co

(D) Corundum

(IV) ZnS

21.

22.

23.

24.

(V) Fe2O3 (1) (2) (3) (4)

(A) (III), (B) (IV), (C) (II), (D) (I) (A) (V), (B) (IV), (C) (III), (D) (II) (A) (II), (B) (III), (C) (IV), (D) (V) (A) (I), (B) (II), (C) (III), (D) (IV) [NCERT Exemp. Q 48, Page 84] 16. Match the items of Column I with items of Column II and assign the correct code: Column I

26.

Column II

(A) Blistered Cu

(I) Aluminium

(B) Blast furnace

(II) 2Cu2O + Cu2S → 6Cu + SO2

(C) Reverberatory furnace (III) Iron (D) Hall-Heroult process

25.

(IV) 23 FeO + SiO FeSiO

27.

(V) 2Cu2S + 3O2 → 2Cu2O + 2SO2 (1) (2) (3) (4)

17.

18.

19.

20.

(A) (II), (B) (III), (C) (IV), (D) (I) (A) (I), (B) (II), (C) (III), (D) (V) (A) (V), (B) (IV), (C) (III), (D) (II) (A) (IV), (B) (V), (C) (III), (D) (II) [NCERT Exemp. Q 49, Page 85] Which of the following compounds should not be stored in glass container? (1) XeF2 (2) KrF2 (3) XeF4 (4) XeF6 There are 14 elements in actinoid series. Which of the following element does not belong to this series? (1) U (2) Tm (3) Np (4) Fm Which of the following will give a white precipitate upon reacting with AgNO3? (1) K2 [Pt(en)2Cl2] (2) [Co(NH3)3Cl3] (3) [Cr(H2O)6]Cl3 (4) [Fe(H2O)3Cl3] General electronic configuration of actinoids is (n – 2)f1–14 (n – 1)d0–2 ns2. Which of the following actinoids have one electron in 6d orbital?

28.



29.

30.

(1) U (Atomic no. 92) (2) Np (Atomic no. 93) (3) Pu (Atomic no. 94) (4) Bk (Atomic no. 97) The formula of the complex triamminetri (nitrito-O) Cobalt (III) is: (1) [Co(ONO)3(NH3)3] (2) [Co(NO2)3(NH3)3] (3) [Co(ONO2)3(NH3)3] (4) [Co(NO2)(NH3)3] In which of the following elements, 5f orbitals are progressively filled? (1) Alkaline earth metals (2) Actinoids (3) Lanthanoids (4) Transition elements The element of second row of transition element beneath Titanium in periodic table is: (1) Scandium (2) Copper (3) Zinc (4) Zirconium Ions formed from ligands that surrounds the central atom is: (1) cation (2) anion (3) interstitial ion (4) complex ion Which of the following is NOT a characteristic of transition element? (1) Lustrous (2) Low boiling point (3) Malleability (4) Ductility Transition metals are generally coloured because: (1) they absorb electro-magnetic radiations. (2) they undergo d – d – transition. (3) their penultimate d– subshells are fully filled. (4) completely filled d– orbitals. The basic character of the transition metal monoxides follow the order: (Atomic number Ti = 22, V = 23, Cr = 24, Fe = 26) (1) TiO > FeO > VO > CrO (2) TiO > VO > CrO > FeO (3) VO > CrO > TiO > FeO (4) CrO > VO > FeO > TiO Among the following series of transition metal ions, the one in which all metal ions have 3d2 electronic configuration is: (Atomic number Ti = 22, V = 23, Cr = 24, Mn = 25) (1) Ti3+, V2+, Cr3+, Mn4+ (2) Ti+, V4+, Cr6+, Mn7+ (3) Ti4+, V3+, Cr2+, Mn3+ (4) Ti2+, V3+, Cr4+, Mn5+ Which set of ions exhibit specific colours? (Atomic number of Sc = 21, Ti = 22, V=23, Mn = 25, Fe = 26, Ni = 28 Cu = 29 and Zn =30) (1) Sc3+, Ti4+, Mn3+ (2) Sc3+, Zn2+, Ni2+ 3+ 2+ 3+ (3) V , V , Fe (4) Ti3+, Ti4+, Ni2+ Which of the following statements concerning transuranium elements is incorrect? (1) Atomic number > 92 (2) Example is Thorium (3) Decay radioactively as they are unstable (4) Elements after Uranium

82 Oswaal CUET (UG) Chapterwise Question Bank [B] ASSERTION REASON QUESTIONS Direction: In the following questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (1) Both A and R are correct and R is The correct explanation of A. (2) Both A and R are correct but R is NOT The correct explanation of A. (3) A is correct but R is not correct. (4) A is not correct but R is correct 1. Assertion (A): Nickel can be purified by Mond’s process. Reason (R): Ni(CO)4 is a volatile compound which decomposes at 460 K to give pure Ni.  [NCERT Exemp. Q 50, Page 85] 2. Assertion (A): Zirconium can be purified by Van Arkel method. Reason (R): ZrI4 is volatile and decomposes at 1,800 K.  [NCERT Exemp. Q 51, Page 85] 3. Assertion (A): Sulphide ores are concentrated by froth flotation method. Reason (R): Cresols stabilize the froth in froth flotation method.  [NCERT Exemp. Q 52, Page 86] 4. Assertion (A): Zone refining method is very useful for producing semi-conductors. Reason (R): Semi-conductors are of high purity.  [NCERT Exemp. Q 53, Page 86] 5. Assertion (A): Hydrometallurgy involves dissolving the ore in a suitable reagent followed by precipitation by a more electropositive metal. Reason (R): Copper is extracted by hydrometallurgy.  [NCERT Exemp. Q 54, Page 86] 6. Assertion (A): The highest oxidation state of osmium is +8. Reason (R): Osmium is a 5d-block element. 7. Assertion (A): Transition metals have low melting points. Reason (R): The involvement of greater number of (n – 1) d and ns electrons in the interatomic metallic bonding. 8. Assertion (A): Transition metals have high melting point. Reason (R): Transition metals have completely filled d-orbitals 9. Assertion (A): Chromium is an actinoid. Reason (R): In chromium, 3d orbitals are filled. 10. Assertion (A): Magnetic moment values of actinides are lesser than the theoretically predicted values. Reason (R): Actinide elements are strongly paramagnetic. [C] COMPETENCY BASED QUESTIONS I. Based on following passage answer questions from 1-5. Within the 3d series, manganese exhibits oxidation states in aqueous solution from +2 to +7, ranging from Mn2+(aq) to MnO−4 (aq). Likewise, iron forms both Fe2+(aq) and Fe3+(aq) as well as the FeO−4 ion. Cr and Mn form oxyions CrO−4, MnO−4, owing to their willingness to form multiple bonds. The pattern with the early transition metals—in the 3d series up to Mn, and for the 4d, 5d metals upto Ru and Os—is that the maximum oxidation

1.

2.

3.

4.

5.

II.

CHEMISTRY

state corresponds to the number of ‘‘outer shell’’ electrons. The highest oxidation states of the 3d metals may depend upon complex formation (e.g., the stabilization of Co3+ by ammonia) or upon the pH (thus MnO42− (aq) is prone to disproportionation in acidic solution). Within the 3d series, there is considerable variation in relative stability of oxidation states, sometimes on moving from one metal to a neighbour; thus, for iron, Fe3+ is more stable than Fe2+, especially in alkaline conditions, while the reverse is true for cobalt. The ability of transition metals to exhibit a wide range of oxidation states is marked with metals such as vanadium, where the standard potentials can be rather small, making a switch between states relatively easy. Transition metals show variable oxidation state because of (1) small energy difference in 3d and 4s orbital. (2) higher energy difference in 3d and 4s orbital. (3) no energy difference is available. (4) All of the above. Fe3+ is more stable as compared to Fe2+ in alkaline medium. (1) patially filled d orbital (2) Half filled d-orbital are present (3) Fully filled d-orbital are present (4) No d-orbital are present Cr and Mn form oxyions due to(1) their ability to form single bonds (2) their ability to form multiple bonds (3) They do not form bond (4) None of the above 3d elements show colour in aqueous solution due to(1) d-d transition (2) Vacant d orbital (3) High wavelength (4) Fully filled d orbitals MnO42– show disproportionation in acidic medium beacuse(1) it can be oxidized as well reduced simultaneously (2) It can be oxidised (3) It can be reduced (4) No change take place Based on following passage answer questions from 6-10. The transition metals when exposed to oxygen at low and intermediate temperatures form thin, protective oxide films of up to some thousands of Angstroms in thickness. Transition metal oxides lie between the extremes of ionic and covalent binary compounds formed by elements from the left or right side of the periodic table. They range from metallic to semiconducting and deviate by both large and small degrees from stoichiometry. Since d-electron bonding levels are involved, the cations-exist in various valence states and hence give rise to a large number of oxides. The crystal structures are often classified by considering a cubic or hexagonal close-packed lattice of one set of ions with the other set of ions filling the octahedral or tetrahedral interstices. The actual oxide structures, however, generally show departures from such regular arrays due in part to distortions caused by packing of ions of different size and to ligand field effects. These

83

d AND f BLOCK ELEMENTS

distortions depend not only on the number of d-electrons but also on the valence and the position of the transition metal in a period or group. 6. Copper, which is in first series of transition metal exhibits +1 oxidation state most frequently? (1) It loose two electron from 4s orbital. (2) It lose one electron fron d orbital. (3) Copper readily loses one electron from its 4s orbital to form stable configuration. (4) it do not lose any electron. 7. The lowest oxide of transition metal is basic because (1) valence electrons of the metal atoms participate in bonding. (2) valence electrons of the metal atoms do not participate in bonding. (3) valence electrons of the metal atoms absent. (4) valence electrons of the metal atoms remain unreactive. 8. The variability in oxidation states of d-block different

from that of the p-block elements due to-  (1) to inert pair effect (2) stable structure (3) low ionization energy (4) small size 9. Crystal structure of oxides of transition metals often shows defects due to- (1) Loose pack structure (2) in part to distortions caused by packing of ions of different size and to ligand field effects. (3) Different shape (4) Geometry is unequal 10. When transition metal exposed to atmospheric oxygen? (1) transition metal forms metal oxides at low and moderate temprature. (2) do not react with oxygen. (3) make hydroxide. (4) make oxy-anions

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (1)

2. (1)

3. (3)

4. (3)

5. (4)

6. (2)

7. (3)

8. (2)

9. (3)

10. (1)

11. (3)

12. (4)

13. (2)

14. (1)

15. (1)

16. (1)

17. (4)

18. (2)

19. (3)

20. (2)

21. (1)

22. (2)

23. (4)

24. (4)

25. (2)

26. (2)

27. (2)

28. (4)

29. (3)

30. (1)

8. (3)

9. (4)

10. (2)

8. (1)

9. (2)

10. (2)

[B] ASSERTION REASON QUESTIONS 1. (1)

2. (1)

3. (2)

4. (2)

5. (2)

6. (2)

7. (4)

[C] COMPETENCY BASED QUESTIONS 1. (1)

2. (2)

3. (2)

4. (1)

5. (1)

6. (1)

7. (2)

ANSWERS WITH EXPLANATION [A] MULTIPLE CHOICE QUESTIONS 1. Option (1) is correct. Explanation: Zr/Hf, Nb/Ta, and Mo/W have identical size due to Lanthanide contraction. However, Sc /Y does not have identical sizes as we move down from Sc to Y then La, the size goes on increasing. 2. Option (1) is correct. Explanation: The lack of π bonding is attributed to unavailability of f orbital for bonding. 3. Option (3) is correct. Explanation: Lanthanides are the 14 elements of 3rd group and 6th period that are filling 4f subshell of antipenultimate shell from 1 to 14. 4. Option (3) is correct. Explanation: Promethium is found in Earth’s crust in very small quantity thats why promethium can be produced by irradiating neodymium and pasaeodymium with neutrons, deuterons and alpha particles.

5. Option (4) is correct. Explanation: The C-O bond order will be lowest in [Mn(CO)6]+. The metal bears a positive charge which makes the back donation process favourable. The electrons from the metal fill the pi antibonding orbital of CO, they weaken the C-O bond. The strongest C-O bond in [Cr(CO)6]+ is due to increased oxidation state of metal and low π back donation bonding tendency. 6. Option (2) is correct. Explanation: 3d5 configuration has maximum number of unpaired electron that’s why it shows the highest magnetic moment. 7. Option (3) is correct. 8. Option (2) is correct. Explanation: Ti or Zr metal is refined by the Van-Arkel method. Metal is converted to volatile tetra iodide which is then decomposed by electrically heating above 1800 K with a tungsten filament to get pure metal. 9. Option (3) is correct.

84 Oswaal CUET (UG) Chapterwise Question Bank Explanation: In BrO3–, the hybridisation of Br is sp3 and has a pyramidal structure while in XeO3, the hybridisation of Xe is also sp3 with pyramidal structure. Therefore, BrO3– is isostructural with XeO3. 10. Option (1) is correct. Explanation: Titatinium lies in group 4 therfore its electronic configuration is [Ar] 3d24s2, it shows maximum oxidation state of Ti is +4. Vanadium lies in group five and its electronic configuration is [Ar] 3d34s2. So, maximum oxidation state of V is +5. Mn is group 7 element it exhibit Electronic configuration of [Ar] 3d54s2. So, maximum oxidation state for Mn is +7. Cu is placed in group 11 of periodic table. Its electronic configuration is [Ar] 3d104s1 So, maximum oxidation state of Cu is +2 11. Option (3) is correct. Explanation: The reduction potential of Cu2+ is positive in value due to high enthalapy of atomization and low enthalpy of hydration. Since, high energy to transform Cu to Cu+ ions is not balanced by hydration enthalapy in 3d series and the rest Ni2+, V2+, Cr2+ have negative values. 12. Option (4) is correct. Explanation: Atomic number of element X=25; Electronic configuration of X: [Ar]3d54s2. µ = 5(5+2) = 5.92 BM 13. Option (2) is correct. Explanation: Column I

Column II

(A)

(4) C oloured bands are observed in chromatography.

(B)

(3) Impure metal is converted to volatile complex by using Mond’s process.

(C)

(1) Ge and Si are purified by zone refining method.

(D)

(2) Purification of mercury is done by fractional distillation.

14. Option (1) is correct. Explanation: Column I

Column II

(A)

(4) Cyanide process is used for extraction of Au through formation of anionic complex.

(B)

(2) Froth floatation process is used for dressing of ZnS.

(C)

(3) E lectrolytic reduction method is used for extraction of aluminium. Graphite electrode is used for this purpose.

(D)

(1) Zone refining is used for purification of Ge.

CHEMISTRY

(B)

(4) Molecular formula of sphalerite is ZnS.

(C)

(2) NaCN is used as a depressant in froth floatation method.

(D)

(1) Molecular formula of corundum is Al2O3.

16. Option (1) is correct. 17. Option (4) is correct. Explanation: XeF6 can not be stored in glass container as it react with SiO2 to give XeO3 which is highly explosive in nature. 18. Option (2) is correct. Explanation: There are 15 actinoides. Uranium (U), Neptunium (Np) and Fermium (Fm). Thulium (Tm) is a rare earth element of lanthanide series. 19. Option (3) is correct. Explanation: A complex which can disassociate Cl– in the solution will be able to form white ppt. with Silver nitrate which is [Cr(H2O)6]Cl3 20. Option (1) and (2) are correct.

Explanation: Actinoids are elements from 89 to 103 and fill their 5f shell. The electronic configuration of elements are – U (92) = 5f 3 6d17s2 Np (93) = 5f 4 6d17s2 Pu (94) = 5f 6 6d07s2 Bk (97) = 5f 9 6s26p67s2 U & and Np have one electron in 6d orbital. 21. Option (1) is correct. Explanation: The alphabetical order of the groups is followed which means ammine group followed by the nitro group with O linkage. Co is in +3 oxidation state. 22. Option (2) is correct. Explanation: Actinoids are 5f block elements so in actinoids, 5f orbitals are progressively filled. 23. Option (4) is correct. Explanation: Scandium, copper and zinc are from I series of transition elements whereas zirconium belong to second series beneath the titanium. 24. Option (4) is correct. Explanation: A complex ion has a central metal ion surrounded by a number of other molecules or ions. These are considered to be attached to the central ion by coordinate bonds. The molecules or ions surrounding the central metal ion are called ligands. 25. Option (2) is correct.

15. Option (1) is correct.

Explanation: They have all characteristics of metals. Thus, transition metals have high boiling point because their atoms are closely packed by the strong metallic bonds.

Explanation:

26. Option (2) is correct.

Column I (A)

Column II (3) Sapphire is a gemstone which contains Co.

Explanation: Due to the presence of unpaired d electrons and d–d– transition, electrons get excited to higher energy d–orbitals which gives colour to compounds. When visible (white) light falls on a compound, it absorbs certain radiations

85

d AND f BLOCK ELEMENTS

of white light and transmit the rests as complementary colour to that of the absorbed light. 27. Option (2) is correct. Explanation: The order of basic character of the transition metal monoxide is TiO > VO > CrO > FeO because basic character of oxides decreases with increase in atomic number. Oxides of transitional metals in low oxidation state, i.e., +2 and +3 are generally basic except Cr2O3. 28. Option (4) is correct.

Explanation: Ti2+ = [Ar]3d2 4s0 V3+ = [Ar]3d2 4s0 Cr4+ = [Ar]3d2 4s0 Mn5+ = [Ar]3d2 4s0 Ti2+, V3+, Cr4+, Mn5+ all have 3d2 electronic configuration. 29. Option (3) is correct. Explanation: V3+, V2+, Fe3+ ions exhibit specific colours. Atomic number of V = 23, Electronic configuration of V - [Ar]3d3 4s2 Electronic configuration of V2+ - [Ar]3d3 Electronic configuration of V3+ - [Ar] 3d2 Atomic number of Fe = 26 Electronic configuration of Fe - [Ar]3d6 4s2 Electronic configuration of Fe3+ - [Ar]3d5 Since these ions have partially filled d- subshells, they exhibit colour. Most transition-metal ions have a partially filled d subshell. As for other ions, Atomic number of Sc = 21 Electronic configuration of Sc - [Ar]3d1 4s2 Electronic configuration of Sc3+ - [Ar]3d0 Since d subshell is empty, it shows no colour. Atomic number of Ti = 22 Electronic configuration of Ti- [Ar]3d2 4s2 Electronic configuration of Ti4+ - [Ar]3d0 Since d subshell is empty, it shows no colour. Atomic number of Mn =25 Electronic configuration of Mn- [Ar]3d5 4s2 Electronic configuration of Mn2+ [Ar]3d5 Since d subshell is partially filled, it shows colour. Atomic number of Ni = 28 Electronic configuration of Ni- [Ar]3d8 4s2 Electronic configuration of Ni2+ [Ar]3d8 Since d subshell is partially filled, it shows colour. Atomic number of Zn =30 Electronic configuration of Zn- [Ar]3d10 4s2 Electronic configuration of Zn2+- [Ar]3d10 Since d subshell is full, it shows no colour 30. Option (1) is correct. Explanation: Transuranium elements are those that come after Uranium (Z = 92), are unstable, and decay radioactively into other elements. However, because Thorium has an atomic number of 90, it is not a transuranium element

[B] ASSERTION REASON QUESTIONS 1. Option (1) is correct. Explanation: When nickel (Ni) is treated with carbon monoxide it forms nickel tetra-carbonyl Ni(CO)4, while impurities are left behind. When the vapour of Ni(CO)4 is heated at 460 K, it is decomposed to give pure nickel while carbon monoxide is removed as gas. 2. Option (1) is correct. Explanation: Zirconium is also purified by vapour phase refining method in which it is treated with iodine to form ZrI4 which on heating decomposes to give pure zirconium. 3. Option (2) is correct. Explanation: Due to oil the sulphide ores gets wetted and becomes lighter and rises to the surface along with froths but impurities wetted by water become heavier and settle down. 4. Option (2) is correct. Explanation: The impurities of semi-conductors are more soluble in molten zone the ultrapure semiconductor crystallizes in zone refining method. 5. Option (2) is correct. Explanation: Copper (Cu) is extracted through the process of hydro-metallurgical process and in this process, salts of metal are dissolved in suitable solvent like water and then reduced by more electro-positive element. 6. Option (2) is correct. Explanation: The highest oxidation state of osmium (Os) is +8. It is due to its ability to expand octet by using its all 8 electrons. 7. Option (4) is correct. Explanation: Transition metals have high melting points because of the involvement of greater number of (n – 1)d and ns electrons in the interatomic metallic bonding. 8. Option (3) is correct. Explanation: Transition metals have high mealting point because of the involvement of greater number of (n-1)d and ns electrons in the interatomic metallic bonding. Hence, assertion is true but reason is false. 9. Option (4) is correct. Explanation: Chromium is transition metal and it have five electrons in 3d orbital rather than 4 because the have half filled orbital with extra stability. Assertion is false but reason is true. 10. Option (2) is correct. Explanation: The magnetic moment is less as the 5f electrons of actinides are less effectively shielded which results in quenching of orbital contributions, they are strongly paramagnetic due to presence of unpaired electrons. [C] COMPETENCY BASED QUESTIONS 1. Option (1) is correct. Explanation: Transition metal shows variable oxidation state due to availability of 3d and 4s electrons for bonding. The small difference between ionization energies makes easier for transition metal such as Mn having variable oxidation state.

86 Oswaal CUET (UG) Chapterwise Question Bank 2. Option (2) is correct. 3+

2+

3+

Explanation: Among Fe and Fe , Fe is more stable due to half –filled d- orbital. This can be explained by Aufbau principle. Half filled and completely filled d-orbitals are more stable than partially fileed d –orbitals. 3. Option (2) is correct. Explanation: Cr and Mn form oxyions CrO–4, MnO–4 due to their ability to form multiple bonds. The pattern with the early transition metals – in the 3d series upto Mn, and for the 4d, 5d metals upto Ru and Os – is that the maximum oxidation state corresponds to the number of outer shell electrons. 4. Option (1) is correct. Explanation: Transition metal ions generally possess one or more pair electrons. When visible light falls on a transition metal ions, the unpaired electrons present in the lower energy d-orbital get promoted to high energy d-orbital called d-d transition. Since the energy invoved d-d transition is quantized, only a definite wavelength gets absorbed, remaining wavelengths present in visible range get transmitted therefore transmitted light shows some colour complementary to the absorbed colour. 5. Option (1) is correct. Explanation: In MnO42-, Mn is in +6 oxidation state, which can be oxidized as well as can be reduced hence it shows

CHEMISTRY

disproportionation in acidic medium. II. Based on following passage answer questions from 6-10. 6. Option (1) is correct. Explanation: Copper readily loses one electron from its 4s orbital, to form stable 3d electronic configuration. 7. Option (2) is correct. Explanation: The lower oxide of transition metals has low oxidation states. This means some of the valence electrons of the metal atoms do not participate in bonding. Thus, they can donate electrons and can behave as bases. 8. Option (1) is correct. Explanation: In p-block, lower oxidation state is more stable due to inert pair effect, whereas in d-block elements higher oxidation states are more stable. In d-block, oxidation states differ by one, whereas in p-block, it differs by two. 9. Option (2) is correct. Explanation: The actual oxide structures generally show departures from regular arrays due in part to distortions caused by packing of ions of different size and to ligand field effects. 10. Option (2) is correct. Explanation: When exposed to atmospheric oxygen transition metal forms metal oxides at low and moderate temprature.

Study Time

CHAPTER

Max. Time: 1:50 Hours Max. Questions: 50

9

COORDINATION COMPOUNDS { Coordination

 Revision Notes {  The

compounds which contain a Scan to know central metal atom or ion surrounded more about this topic by number of oppositely charged ions or neutral molecules are called coordination compounds. e.g., [Cu(NH3)4]2+. ƒ According to Werner’s theory Transition of coordination compounds: elements and (i)  Metal possesses two types of coordination valencies – ionisable (Primary) compounds and non-ionisable (Secondary) valencies. (ii) Every metal ion has a fixed number of secondary valency and this is known as coordination num­ ber. (iii)  Primary valencies are ionisable which are satisfied by anions while secondary valencies are satisfied by ligands i.e., negative group or neutral molecules with lone pair of electrons. (iv)  Secondary valencies are directed in space towards internal positions. { Double Salt: When two or more salts are added to form a stable solid together and break into constituent ions when dissolved in water or any solvent. e.g., for FeSO4(NH4)2SO4.6H2O (Mohr’s salt). Co-ordinate spehre [Co(NH3)6] Central metal ion { The

Ligand

metal ion surrounded by fixed number of ions or molecules is called central metal atom or ion, e.g., in K4[Fe(CN6], Fe2+ is central metal ion. { A ligand is the ion or molecule bound to the central atom/ion in the coordination entity. For example, Cl–, OH–, CN–, CO, NH3, H2O, etc. A ligand may be neutral or charged species. { Monodentate ligands ligate through one donor atom, bidentate ligands ligate through two donor atoms and polydentate ligands ligate through more than two donor atoms. { When a ligand uses its two or more donor atoms to bind a single metal ion, it is said to be a chelate ligand. { Ambidentate ligand is a ligand which contains two donor atoms but only one of them forms a coordinate bond at a time with central metal/ion.

number: Number of ligands through which the metal is directly bonded. The number of such ligating groups is called the denticity of the ligand. { Charge on the complex ion: The charge on the com­ plex ion is equal to the algebraic sum of the charges on all the ligands coordinated to the central metal ion. { Oxidation number of central atom is defined as the charge it would carry if all the ligands are removed along with the electron pairs that are shared with the central atom. It is represented by Roman numerical. { Counter ions: The ions which are not included in the primary coordination sphere are known as counter ions, e.g., in K4[Fe(CN)6], K+ ions are counter ions. { Homoleptic complexes are the complexes in which metal atom or ion is bound to only one type of ligands e.g., [Co(NH3)6]3+, whereas complexes bound to more than one kind of ligands are known as hetero- leptic complexes. e.g., [Co(NH3)4Cl2]+. { The complexes in which only one metal atom is present are known as homonuclear complexes. e.g., [Co(NH3)6]Cl3 and [Cu(NH3)4]SO4. The complexes in which more than one metal atom is present are known as polynuclear complexes.  Nomenclature of coordination compounds: (i)  The cation whether simple Scan to know or complex is named first more about this topic followed by anion. (ii)  Ligands are named in alphabetical order. (iii) For indicating the number of individual ligand within the Trick to write coordination entity, numerical names of pre- fixes di, tri, tetra, etc. are coordination used. For ligands containing compounds any of these prefixes in their names, their numbers are indicated by prefixes bis, tris, tetrakis etc. Anionic ligands end in –o. Neutral ligands retain their names while cationic end in -ium. (iv) The coordination sphere is written in a square bracket. (v) In naming a coordination compound, ligands are named first in alphabetical order followed by metal atom and then the oxidation state of metal by a Roman numeral in parenthesis. (vi) Name of coordination compounds starts with a small letter and the complex part is written as one word.

[Cr(H2O)6]Cl3 [Cr(H2O)5Cl]Cl2.H2O

[Co(NH3)5SO4]Br [Co(NH3)5Br]SO4

[Co(NH 3)6] [Cr(CN)6] [Cr(NH3)6] [Co(CN) 6] and a lig rm

Pt NH3

NH3

Cl

cis NH3 Pt Cl NH3 trans

Cl

Cl en

dextro

en

Co

en

O

en laevo

Co

be

e ag

sw

Stereoisomerism : Different spatial arrangements

po

en

3+

rim pe su

: Im

al ptic

Structural: Different bonds

Mirror

3+

ds

an

lig

en

e th

es ul ec ol

ate n. fw /io ro e b tom um la n a t e r in om iffe dt a te- D ch e Solv a t t a

e isplac and can d

[Co(NO2)(NH3)5]Cl2 [Co(ONO)(NH 3)5]Cl2

Caused by d -d transition; The colour is complementary to wavelength absorbed.

Ligands are point charges and there is electrostatic force of attraction between ligands and metal atom/ion. Degeneracy of d orbitals is lifted causing splitting of d orbitals. o depends upon the field produced by the ligand and charge on metal ion.

Metal–carbon bond possess both  and  character

Metal atom/ion under the influence of ligands can use its (n–1)d, ns, np or ns, np, nd orbitals for hybridisation; sp3(Tetrahedral), dsp2(Square 2 planar) sp3d (Trigonal pyramidal), sp3d and d2sp3 (Octahedral); Magnetic moment = n(n+2) BM

Formulas of mononuclear

hi

 Cation is named first.  Naming of ligands in alphabetical order.  Anionic ligands end in -o, neutral and cationic are same.  Prefixes mono, di, tri etc. are used.  Followed by Roman numeral in parenthesis.

m

Te r

s

rdinatio Coompoundn o s

Dissociate completely into simple ions when dissolved in water. (Mohr’s salt FeSO4. (NH4)2 SO4.6H2O)

c

d

ch

an no t )3 ] 3+

se

en

o( [C

First Level

Second Level

Trace the Mind Map Third Level

In qualitative and quantitative chemical analysis. Estimation of hardness of water. In extraction of metals. In purification of metals. In biological systems. As catalysts for industrial processes. In black and white photography. In medicinal chemistry.

In coordination compounds metals show primary and secondary linkages valencies. Primary valencies are ionisable and are satisfied by negative ions. Secondary valencies are non-ionisable and are satisfied by neutral molecules or negative ions. Ions/groups bound by secondary linkages to metal have characteristic spatial arrangements corresponding to different coordination numbers.

 Coordination entity : A central metal atom/ion bonded to fixed number of ions or molecules. [Ni(CO)4]  Central atom/ion : Atom/ion to which a fixed number of ions/groups are bound in a definite geometrical arrangement.  Ligands : Ions or molecules bound to central atom/ion types : Unidentate – single donor, Didentate –two donors, Polydentate – several donors, Chelating – Di-or polydentate which forms more than one coordinate bonds. Ambidentate : Can ligate through two different atoms.  Coordination number : No. of ligand donor atoms to which metal is directly bonded.  Coordination sphere : Central atom/ion and the ligands attached to it and enclosed in square bracket.  Coordination polyhedron : Spatial arrangement of ligand atoms with central atom/ion.  Oxidation number : Charge of central atom if all ligands are removed along with e– pairs shared with central atom.  Homoleptic complex : Metal is bound to one type of donor 3+ groups. [Co(NH3)6]  Heteroleptic complex : Metal is bound to more than one type + of donor groups [Co(NH3)4Cl2]



Do not dissociate into simple ions when dissolved in water. K4[Fe(CN)6]



C

Central atom is listed first. Ligands in alphabetical order. Formula is enclosed in square bracket. Polyatomic ligands in parenthesis. No space between ligand and metal. Charge is indicated outside brackets. Charge of cation(s) balanced by charge of anion(s).



      

88 Oswaal CUET (UG) Chapterwise Question Bank CHEMISTRY

89

COORDINATION COMPOUNDS { Two

or more coordination compounds which have the same molecular formula but have their ligands attached to the isomers metal atom or ion in different ways are known as isomers. The phenomenon of different isomers is known as isomerism.

 Isomerism in co-ordination compounds: Scan to know (a) Structural isomerism more about this (i) Linkage isomerism: [Co(NH3)5 topic (NO2)]Cl2 (ii) Co-ordination isomerism: [Co(NH3)6][Cr(CN)6)] and [Cr(NH3)6] [Co(CN)6)], Stability of (iii)  Ionization isomerism: [Co (NH3)5SO4] Br and [Co(NH3)5 coordination compounds Br]SO4, (iv)  Solvate isomerism: [Cr(H2O)6]Cl3 [Cr(H3O)5Cl]2H2O (b) Geometrical Isomerism:

and

(i)

(ii)

(iii)

(c) Optical Isomerism : 3+

 Crystal field theory (CFT): (i) The ligands are considered as point charge or point dipole. (ii) Interaction between metal ion and ligand is con­ sidered as electrostatic in nature. (iii) Metal ion is supposed to be present at the origin of the axis. Ligands approach to metal ion along the axis of octahedral complex between the axis of tetrahedral complex and in the case of square planar complex four ligand approach to metal ion along x, y plane. (iv)  Due to the electrostatic interaction between ligands electrons and metal d-orbital electron degeneracy of d-orbital is lost and splitting of d-orbitals occurred. (v) Some ligands are able to produce strong fields in which case, the splitting will be large whereas others produce weak fields and consequently result in small splitting of d-orbitals.

3+

en

en

 Valence Bond theory: (i) A suitable number of vacant orbitals must be present in the central metal atom or ion for the formation of coordinate bond with the ligands. (ii) Central metal ion can use appropriate number of s, p and d-orbitals for hybridisation depending upon total number of ligands. (iii) The hybridised orbitals are allowed to overlap with those ligand orbitals that can donate an electron pair for bonding. (iv) The outer orbitals or inner orbitals complexes are formed depending upon whether outer d-orbitals or inner d-orbitals are used.

en en

Co

Co

dextro { d-orbital

Fig. 9.2: d-orbital splitting in a tetrahedral crystal field:

en

en mirror

{ Metal

carbonyls: Homoleptic carbonyls are formed by d-block elements and contain carbonyl ligands only. e.g., V(CO)6, Cr(CO)6, etc.

laevo

splitting in an octahedral crystal field:

 Number of Orbitals and Types of Hybridisation: Coordination number

Type of hybridising

Distribution of hybrid orbitals in space

4

sp3

Tetrahedral

4

dsp

5

3

sp d

Trigonal bipyramidal

6

sp3d2

Octahedral

2

6

d sp {  Bonding

Fig. 9.1: d-orbital splitting in an octahedral crystal field

2

Square planar

3

Octahedral

in metal carbonyls: It also involves both s-and p-bond. s-bond is formed by overlapping of lone pair on CO to the vacant d-orbitals of metal whereas p-bond is formed by the back donation of pair of d-electrons to vacant anti bonding orbital of carbonyl.

90 Oswaal CUET (UG) Chapterwise Question Bank

CHEMISTRY

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS 1. In the complex ion [AuXe4]2+, Xe acts as: (1) central atom (2) ligand (3) chelating agent (4) electrophile 2. Ions formed from ligands that surrounds the central atom is: (1) cation (2) anion (3) interstitial ion (4) complex ion 3. How many ions are produced from the complex [Co(NH3)5Cl]Cl2 in solution? (1) 2 (2) 3 (3) 6 (4) 8 4. The formula of the complex triamminetri(nitrito-O) Cobalt (III) is: (1) [Co(ONO)3(NH3)3] (2) [Co(NO2)3(NH3)3] (3) [Co(ONO2)3(NH3)3] (4) [Co(NO2)(NH3)3] [CUET 2023, Sep] 5. Which one of the following pairs has only paramagnetic species? (1) [Cu(NH3)4]Cl2 and O2 (2) [Cu(NH3)4]Cl2 and N2 (3) [Zn(H2O)6]Cl2 and O2 (4) [Cu(NH3)4]Cl2 and K4[Fe(CN)6] 6. Which one of the following will yield the highest splitting of d orbitals? (1) S2- (2) OH(3) CN (4) EDTA4– 7. Identify the correct IUPAC name for the complex. [(Co(H2NCH2CH2NH2)3]2 (SO4)3 (1) triethylene diamine cobalt(II) sulphate (2) triethaneammine cobalt(II) trisulphate (3) tris (ethane-1,2-diamine) cobalt(III) sulphate (4) tri (ethane-1,2-diamine) cobalt(II) sulphate 8. Which one of the following transition metal ion is colourless? (1) Sc3+ (2) V2+ 2+ (3) Mn (4) Co3+ 9. Which of the following elements can be involved in pπ– dπ bonding? (1) Carbon (2) Nitrogen (3) Phosphorus (4) Boron 10. 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. 11. Atomic number of Cr and Fe are respectively 24 and 26, which of the following is paramagnetic with the spin of electron? (1) [Cr(CO)6] (2) [Fe(CO)5] (3) [Fe(CN)6]4– (4) [Cr(NH3)6]3+

12. Crystal field stabilisation energy for high spin d4 octa­ hedral complex is: (1) – 1.8∆o (2) – 1.6∆o + P (3) – 1.2∆o (4) – 0.6∆o 13. [Cr(H2O)6]Cl3 (Atomic number of Cr = 24) has a mag­ netic moment of 3.83 B.M., the correct distribution of 3d electrons in the chromium of the complex is: (1) 3dxy, 3dyz,3dz2 (2) 3dxy, 3dyz,3dz2 2 (3) 3dxy, 3dyz,3dz (4) 3dxy, 3dyz,3dz2 14. Which one of the following is an outer orbital complex and exhibits paramagnetic behaviour? (1) [Ni(NH3)6]2+ (2) [Zn(NH3)6]2+ 3+ (3) [Cr(NH3)6] (4) [Co(NH3)6]3+ 15. The colour of the coordination compounds depends on the crystal field splitting. What will be the correct order 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]. 16. Match the compounds given in column I with the hybridisation and shape given in column II and mark the correct option. Column I



Column II

(A) XeF6

(i) Distorted octahedral

(B) XeO3

(ii) Square planar

(C) XeOF4

(iii) Pyramidal

(D) XeF4

(iv) Square pyramidal

Code: A B C D (1) (iv) (iii) (i) (ii) (2) (iv) (i) (ii) (iii) (3) (i) (iii) (iv) (ii) (4) (i) (ii) (iv) (iii) 17. Among the following complexes, the one which shows zero crystal field stabilization energy (CFSE) is: (1) [Mn(H2O)6]3+ (2) [Fe(H2O)6]3+ (3) [Co(H2O)6]2+ (4) [Co(H2O)6]3+ 18. Interstitial compounds are formed when small atoms are trapped inside the crystal lattice of metals. Which of the following is not the characteristic property of interstitial compounds? (1) They have high melting points in comparison to pure metals. (2) They are very hard. (3) They retain metallic conductivity. (4) They are chemically very reactive. 19. In the preparation of compounds of Xe, Bartlett had taken O2 +PtF6– as a base compound. This is because (1) both O2 and Xe have same size. (2) both O2 and Xe have same electron gain enthalpy. (3) both O2 and Xe have same ionisation enthalpy. (4) both Xe and O2 are gases.

COORDINATION COMPOUNDS

20. Which of the following will exhibit maximum ionic con­ ductivity? (1) K4[Fe(CN)6] (2) [Co(NH3)6]Cl3 (3) [Cu(NH3)4]Cl2 (4) [Ni(CO)4] 21. Which of the following complexes show linkage iso­ merism? (1) [Co(H2O)5CO]3+ (2) [Cr(NH3)5SCN]2+ + (3) [Fe(en)2Cl2] (4) All of the above 22. 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 oxidising agents in aqueous solution. (4) Ti2+ and Cr2+ are reducing agents in aqueous solution. 23. In which of the following coordination entities, the magnitude of ∆0 (CFSE in octahedral field) will be maximum? (Atomic number of Co = 27) (1) [Co(H2O)6]3+ (2) [Co(NH3)6]3+ 3– (3) [Co(CN6)] (4) [Co(C2O4)3]3– 24. The stabilisation of coordination compounds due to che­ lation is called the chelate effect. Which of the follow­ing is the most stable complex species? (1) [Fe(CO)5] (2) [Fe(CN)6]3– 3– (3) [Fe(C2O4)3] (4) [Fe(H2O)6]3+ [NCERT Exemp. Q 6, Page 121] 25. When 0.1 mol 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. [NCERT Exemp. Q 3, Page 120] 26. The correct IUPAC name of [Pt(NH3)2Cl2] is: (1) Diamminedichloridoplatinum (II). (2) Diamminedichloridoplatinum (IV). (3) Diamminedichloridoplatinum (I). (4) Dichloridodiammineplatinum (IV). [NCERT Exemp. Q 5, Page 121] 27. The stabilisation of coordination compounds due to che­ lation is called the chelate effect. Which of the follow­ing is the most stable complex species? (1) [Fe(CO)5] (b) [Fe(CN)6]3– 3– (3) [Fe(C2O4)3] (d) [Fe(H2O)6]3+ [NCERT Exemp. Q 6, Page 121] 28. The correct IUPAC name of [K2PdCl4] is: (1) Potassium tetrachlorinepalladium(II) (2) Potassium tetrachloridopalladate(II)Potassium (3) Potassium tetrachloridopalladium(II) (4) tetrachlorinepalladate(II) [NCERT Exemp. Q 5, Page 121] 29. Due to the presence of ambidentate ligands coordination compounds show isomerism. Palladium complexes of the type [Pd(C6H5)2(SCN)2] and [Pd(C6H5)2(NCS)2] are: (1) linkage isomers. (2) coordination isomers. (3) ionisation isomers. (4) geometrical isomers. [NCERT Exemp. Q 9, Page 121]

91 30. The compounds [Co(SO4)(NH3)5]Br and [Co(SO4) (NH3)5] Cl represent: (1) linkage isomerism. (2) ionisation isomerism. (3) coordination isomerism. (4) no isomerism. [NCERT Exemp. Q 10, Page 121] [B] ASSERTION REASON QUESTIONS Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark The correct choice as: (1) Both A and R are correct and R is The correct explanation of A. (2) Both A and R are correct but R is NOT The correct explanation of A. (3) A is correct but R is not correct. (4) A is not correct but R is correct. 1. Assertion (A): Toxic metal ions are removed by the chelating ligands. Reason (R): Chelate complexes tend to be more stable. 2. Assertion (A): [Cr(H2O)6]Cl2 and [Fe(H2O)6]Cl2 are reducing in nature. Reason (R): Unpaired electrons are present in their d-orbitals. 3. Assertion (A): Linkage isomerism arises in coordination compounds containing ambidentate ligand. Reason (R): Ambidentate ligand has two different donor atoms. [NCERT Exemp. Q 43, Page 127] 4. Assertion (A): Complexes of MX6 and MX5 L type (X and L are unidentate) do not show geometrical isomerism. Reason (R): Geometrical isomerism is not shown by complexes of coordination number 6 [NCERT Exemp. Q 44, Page 127] 5. Assertion (A): [Fe(CN)6]3– ion shows magnetic moment corresponding to two unpaired electrons. Reason (R): Because it has d2 sp3 type hybridisation. [NCERT Exemp. Q 45, Page 127] 6. Assertion (A): Among [Co(NH3)6]3+ and [Co(en)3]3+, co­ ordination compound [Co(en)3]3+ is a more stable com­ plex. Reason (R): Because (en) is a chelating ligand/bidentate ligand. 7. Assertion (A): Highest oxidation state is exhibited by transition metal lying in the middle of the series. Reason (R): The highest oxidation state exhibited corresponds to number of (n−1)d electrons. 8. Assertion (A): Fe3+ is more stable than Fe2+. Reason (R): Fe3+ has 3d5 configuration while Fe2+ has 3d6 configuration. 9. Assertion (A): Vanadium had the ability to exhibit a wide range of oxidation states. Reason (R): The standard potentials Vanadium are rather small, making a switch between oxidation states relatively easy. 10. Assertion (A): The primary valences are normally ionisable and are satisfied by negative ions.

92 Oswaal CUET (UG) Chapterwise Question Bank

Reason (R): The ions or molecules bound to the central atom/ion in the coordination entity are called ligands.

[C] COMPETENCY BASED QUESTIONS I. Based on following passage answer questions from 1-5. According to the valence bond theory, the metal atom or ion under the influence of ligands can use its (n-1)d, ns, np, nd orbitals for hybridisation to yield a set of equivalent orbitals of definite geometry. These hybridised orbitals are allowed to overlap with ligand orbitals that can donate electron pairs for bonding. It is usually possible to predict the geometry of a complex from the knowledge of its magnetic behaviour on the basis of the valence bond theory. Consider the formation of [Co(NH3)5Cl]Cl2. 1. The IUPAC name of the above coordination entity is: (1) Chloridopentaamminecobaltate (II) chloride (2) Chloridopentaamminecobaltate (II) dichloride (3) Pentaamminechloridocobalt (III) chloride (4) Pentaamminechloridocobalt (III) dichloride 2. The spin only magnetic moment of the complex [Co(NH3)5Cl]Cl2 in BM is: (1) 1.7 (2) 0.0 (3) 3.8 (4) 4.9 3. The hybridisation of cobalt in the above coordination entity is: (1) sp3d2 (2) d2sp3 3 (3) sp d (4) dsp3 4. The coordination number of cobalt in the above co­ ordination entity is (1) 2 (2) 4 (3) 5 (4) 6 5. The primary valence of Co in above coordination entity is (1) 1 (2) 2 (3) 3 (4) 4 II. Based on following passage answer questions from 6-10. The crystal field theory (CFT) is an electrostatic model which considers the metal-ligand bond to be ionic arising purely from electrostatic interactions between the metal ion and the ligand. Ligands are treated as point charges in case of anions or dipoles in case of neutral molecules. The five d orbitals in an isolated gaseous metal atom/ion have

6.

7.

8.

9.

10.

CHEMISTRY

same energy, i.e., they are degenerate. This degeneracy is maintained if a spherically symmetrical field of negative charges surrounds the metal atom/ion. However, when this negative field is due to ligands (either anions or the negative ends of dipolar molecules like NH3 and H2O) in a complex, it becomes asymmetrical and the degeneracy of the d orbitals is lifted. It results in splitting of the d orbitals. The colour of the coordination compounds depends on the crystal field splitting. What will be the correct order 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]3+ 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 −1 (3) 8,000 cm (4) 20,000 cm−1 An aqueous pink solution of cobalt(II) chloride changes to deep blue on addition of excess of HCl. This is because. (1) [Co(H2O)6]2+ is transformed into [CoCl6]4− (2) [Co(H2O)6]2+ is transformed into [CoCl4]2− (3) Tetrahedral complexes have larger crystal field splitting than octahedral complex. (4) None of the above 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? (1) Thiosulphato (2) Oxalato (3) Glycinato (4) Ethane-1,2-diamine Significance of Crystal Field Stablisation Energy (CFSE) for coordination compounds is in? (1) distribution of electrons in the d orbitals which lead to net stabilisation (decrease in energy) of some complexes depending on the specific ligand field geometry and metal d-electron configurations. (2) attraction in octahedral coordination compound yield two energy levels. (3) Optional filling up of orbitals by electrons. (4) None of the above.

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (2)

2. (4)

3. (2)

4. (1)

5. (1)

6. (3)

7. (3)

8. (1)

9. (3)

10. (1)

11. (4)

12. (4)

13. (4)

14. (1)

15. (3)

16. (3)

17. (2)

18. (4)

19. (3)

20. (1)

21. (2)

22. (1)

23. (3)

24. (3)

25. (2)

26. (1)

27. (3)

28. (3)

29. (1)

30. (4)

8. (1)

9. (1)

10. (2)

8. (2)

9. (1)

10. (1)

[B] ASSERTION REASON QUESTIONS 1. (1)

2. (2)

3. (1)

4. (3)

5. (4)

6. (1)

7. (3)

[C] COMPETENCY BASED QUESTIONS 1. (3)

2. (2)

3. (2)

4. (4)

5. (3)

6. (3)

7. (3)

93

COORDINATION COMPOUNDS

ANSWERS WITH EXPLANATION [A] MULTIPLE CHOICE QUESTIONS 1. Option (2) is correct. Explanation: Xe act as ligand as it surrounds the central metal atom. 2. Option (4) is correct. Explanation: A complex ion has a central metal ion surrounded by a number of other molecules or ions. These are considered to be attached to the central ion by coordinate bonds. The molecules or ions surrounding the central metal ion are called ligands.

↑

↑

↑

××

××

××

××

××

××

2 3 Number of unpaired electrons = 3 d sp Therefore, [Cr(NH3)6]3+ is paramagnetic while rest all others are diamagnetic.

12. Option (4) is correct. Explanation: For high spin d4 octahedral complex

eg 0.6 ∆0

3. Option (2) is correct. Explanation: Three ions are produced from the complex e.g., [Co(NH3)5Cl]Cl2 → [Co(NH3)5Cl]2++(aq) + 2Cl− 4. Option (1) is correct.

Explanation: According to Molecular Orbital Theory (MOT), in oxygen there is 1 unpaired electron in the pzpx anti-bonding orbital and another unpaired electron in pzpy anti-bonding orbital. As molecules containing unpaired electrons are strongly attracted by magnetic field, hence oxygen has paramagnetic nature. In [Cu(NH3)4]Cl2, Cu is in +2 oxidation state as Cu2+ : [Ar]3d9 4s0 Transfer of one electron from 3d to 4p takes place and dsp2 hybrid orbitals are formed. Since, it contains 1 unpaired electron , so the complex is paramagnetic in nature. 6. Option (3) is correct. Explanation: The separation of energies of two sets of d orbitals is expressed in ‘crystal field splitting’ parameter. (Denoted by ∆0 for octahedral complexes). The ligands are arranged according to their increasing fields strength as S2– < OH– < EDTA4– < CN–. 7. Option (3) is correct. Explanation: The correct IUPAC name of compound is Tris (ethane-1, 2-diamine) cobalt (III) sulphate.

t2g

CFSE = (-3×0.4) + (1×0.6)]∆0=-0.6∆0

Explanation: [Co(ONO)3 (NH3)3] 5. Option (1) is correct.

0.4 ∆0

Degenerate d orbitals

CFSE = [(–3  0.4) + (1  0.6)] ∆0 = –0.6∆0 13. Option (4) is correct.

Explanation µ for Cr in [Cr(H2O)6]Cl3 = 3.83 B.M

m = n(n + 2) 3.83 = n(n + 2) Þ n = 3 Thus, number of unpaired electrons in d-orbitals sub-shells of chromium (Cr = 24) = 3 ∴ Configuration of Cr = 1s2 2s2 2p6 3s2 3p6 3d3 In 3d3 the distrubution of electrons 3d1xy, 3d1yz, 3d1zx, 3d0x2–y2, 3d0z2, 14. Option (1) is correct.

Explanation: Those orbitals which utilises 3d-orbitals for bonding and exhibit paramagnetic behaviour forms outer orbital complex. In [Ni(NH3)6]2+ : Ni+2 = [Ar]3d3 4s0 3d

8. Option (1) is correct. Explanation: The ion will show colour if it has unpaired electrons in its electronic configuration. In Sc3+, electronic configuration [Ar]3d04s0 , there are no unpaired electrons due to completely vacant d orbital that cannot show d-d transition for colour spectra. Hence, it will be colourless.

In [Ni(NH3)6]2+

4s

3d

9. Option (3) is correct. Explanation: pπ-dπ bonding is present in phosphorus due to the presence of vacant d-orbitals and in carbon (C), nitrogen (N) and boron (B) do not have d orbitals. 10. Option (1) is correct. Explanation: Cu (I) should be stable as fully filled d-orbital (d10) than incompletely filled (d9) d-orbital of Cu(II). But Cu(II) has a greater charge density than Cu(I) ion and therefore forms much stronger bonds releasing more energy. 11. Option (4) is correct. Explanation: Atoms, ions or molecules containing unpaired electrons are paramagnetic. In [Cr(NH3)6]3+, Cr+3 = [Ar] 3d3 4s0 In excited state

unpaired electrons 4s ××

4p ××

××

4d ××

××

××

××

sp3d2 hybridisation Thus, it forms outer orbital complex and is paramagnetic 15. Option (3) is correct. Explanation: ∆0 values follow the order: [Co(H2O)6]3+ < [Co(NH3)6]3+ < [Co(CN)6]3– and therefore, absorption wave­ length follows the order: [Co(H2O)6]3+ > [Co(NH3)6]3+ > [Co(CN)6]3– 16. Option (3) is correct.

94 Oswaal CUET (UG) Chapterwise Question Bank

ligand. Here, only [Fe(C2O4)3]3– is a coordination compound which contains oxalate ion as a chelating ligand. Hence, it stabilises coordination compound by chelating Fe3+ ion. 25. Option (2) is correct.

Explanation: F F

F Xe

Xe

F

o o o

F

X eF3 Pyramidal sp 2

X eF6 Distorted Octahedral sp 3d 3

O F

F

F

F

F Xe

Xe F

XeOF4 Square Pyramidal sp2d 2

F

CHEMISTRY

F

XeF4 Square Planar sp 2d 2

17. Option (2) is correct. Explanation: CFSE for octahedral complex is given by general formula as follows: CFSE = [– 0.4 (t2g electrons) + 0.6 (eg electrons)] For Mn+3 ⇒ 3d4 → CFSE = [(– 0.4 × 3) + (0.6 × 1)] ∆o = – 0.6.∆o For Fe3+, 3d5 → CFSE = *(– 0.4 × 3) + (0.6 × 2)] = 0 For Co+2, 3d7 → CFSE = *(– 0.4 × 5) + (0.6 × 2)] = – 0.8∆o For Cd3+, *3d6 + → CFSE = *(– 0.4 × 4) + (0.6 × 2)] = – 0.4∆o 18. Option (4) is correct. Explanation: Interstitial compounds are chemically inert. 19. Option (3) is correct. Explanation: In the preparation of compounds of Xe, Bartlett had taken O2+ PtF6– as a base compound. This is because both O2 and Xe have almost same ionisation enthalpy. 20. Option (1) is correct.

Explanation: The coordination compound K4[Fe(CN)6], produces maximum 5 ions 4K+ + [Fe(CN)6]4– thus show more conductivity.

Explanation: One mole of AgNO3 precipitates one mole of chloride ion. In the above reaction, when 0.1 mole CoCl3(NH3)5 is treated with excess of AgNO3. 0.2 mole of AgCl are obtained and thus, there must be two free chloride ions in the solution of electrolyte. So, molecular formula of complex will be [Co(NH3)5Cl]Cl2 and electrolyte solution must contain [Co(NH3)5Cl2+] and two Cl– as constituent ions. Thus, it is 1 : 2 electrolyte. [Co(NH3Cl)]Cl2 → [Co(NH3)5Cl]2+ (aq) + 2Cl– (aq) 26. Option (1) is correct. Explanation: IUPAC name Diamminedichloridoplatinum (II). 27. Option (3) is correct.

of

[Pt(NH3)2Cl2]

is

Explanation : Chelation stabilises the coordination compound. The ligand, which chelates the metal ion, is known as chelating ligand. Here, only [Fe(C2O4)3]3– is a coordination compound which contains oxalate ion as a chelating ligand. Hence, it stabilises coordination compound by chelating Fe3+ ion. 28. Option (3) is correct. Explanation: IUPAC name of K2[PdCl4] is tetrachloridopalladium(II) 29. Option (1) is correct.

Potassium

Explanation: The ligands having two different bonding states are known as ambident ligands. For example : NCS, NO2, etc. Here, NCS has two binding states at N and S. Hence, NCS can bind to the metal ion in two ways : M → NCS or M → SNC. Thus, coordination compounds containing NCS as a ligand can show linkage isomerism, that is, [Pd(C6H5)2(SCN)2] and [Pd(C6H5)2(NCS)2] are linkage isomers. 30. Option (4) is correct. Explanation: [Co(SO4)(NH3)5]Br and [Co(SO4) (NH3)5]Cl represent no isomerism because they are different compounds.

21. Option (2) is correct.

[B] ASSERTION REASON QUESTIONS

Explanation: Linkage isomerism is the existence of co­ ordination compounds that have the same composition diffe­ ring with the connectivity of the metal to a ligand. Typical ligands that give rise to linkage isomers are: thiocyanate, SCN−; isothiocyanate, NCS−.

1. Option (1) is correct. Explanation: When a solution of chelating ligand is added to solution containing toxic metals ligands chelates the metal ions by formation of stable complex.

22. Option (1) is correct.

Explanation: In the complexes, Co-exists as Co2+ and Fe as Fe2+. Both of the complexes become stable by oxidation of metal ion to Co3+ and Fe3+.

Explanation: Copper does not liberate hydrogen from acids because copper lies below hydrogen in electrochemical series. So, copper does not have sufficient electrode potential to liberate elemental hydrogen form compounds in which oxidation state of hydrogen is +1. 23. Option (3) is correct. Explanation: In all the given options, the central metal atom is same and contain same number of d electrons. Thus, CFSE is decided by ligands. In case of strong field ligand, CFSE is maximum, as CN– is a strong field ligand, hence in [Co(CN)6]3– CFSE is maximum. 24. Option (3) is correct. Explanation: Chelation stabilises the coordination compound. The ligand, which chelates the metal ion, is known as chelating

2. Option (2) is correct.

3. Option (1) is correct. Explanation: Linkage isomerism arises due to two different donor atoms in ambidentate ligand. 4. Option (3) is correct. Explanation: For complexes of MX6 and MX5L type, different geometric arrangements of the ligands are not possible. MA4B2, M(AA)2B2 and MA3B3 type of complexes are the complexes with coordination number 6 which show geometrical isomerism. 5. Option (4) is correct. Explanation: [Fe(CN)6]3– ion shows magnetic moment corresponding to one unpaired electron.

95

COORDINATION COMPOUNDS

6. Option (1) is correct. Explanation: Since (ethylene diamine-en) is a chelating ligand/ bidentate ligand, [Co(en3)]3+ is a more stable complex as compared to the other one.

CH2

H2C

spin only magnetic moment. Hence, spin only magnetic moment is 0 BM. 3. Option (2) is correct. Explanation: In [Co(NH3)5Cl]Cl2, Co3+ Electronic configura­ tion of Co3+: [Ar]3d64s04p0

NH2

N2H M

NH 3 NH 3

Diagram should be created here, NH2-CH2—CH2-NH2 7. Option (3) is correct.

Explanation: Higher oxidation states are usually exhibited by the transition metals lying in the middle of the series because of higher number of unpaired electrons in (n – 1)d and ns orbitals at the middle of the series. 8. Option (1) is correct. Explanation: Fe3+ is more stable than Fe2+ because of the presence of half-filled d-orbital (3d5) in Fe3+ ion. 9. Option (1) is correct. Explanation: Both the assertion and reason are correct. Vanadium is known for its ability to exhibit multiple oxidation states due to its small standard potentials, allowing it to readily switch between these states during chemical reactions. 10. Option (2) is correct. Explanation: Primary valences in coordination chemistry correspond to the charge of the central metal ion and are typically satisfied by negative ions (ligands). Ligands are ions or molecules that form coordinate covalent bonds with the central metal ion. [C] COMPETENCY BASED QUESTIONS 1. Option (3) is correct. Explanation: There are 5 amine group, 1 chloride group atta­ ched to cobalt(III) in the coordination sphere. [Co(NH3)5Cl] Cl2 : Pentaamminechloridocobalt (III) chloride. 2. Option (2) is correct.

Explanation: Due to strong field ligand, NH3 electrons of d-orbital pair up so there is zero unpaired electrons to show

NH 3

NH 3 NH 3 Cl

(d 2 sp3 hybridised)

4. Option (4) is correct. Explanation: Coordination number is equal to the total number of donating sites within the coordination sphere 5 + 1 = 6. 5. Option (3) is correct. Explanation: x+5(0) +3(–1)=0 x = +3 Primary valency = +3 6. Option (3) is correct. Explanation: [Co(H2O)6]3+ > [Co(NH3)6]3+ > [Co(CN)6]3− 7. Option (3) is correct.

Explanation: Given that the CFSE for [CoCl6]4− is 18,000 cm−1, the CFSE for [CoCl4]2− will be: (18,000 cm−1) * (4/9) ≈ 8,000 cm−1 8. Option (2) is correct. Explanation: [Co(H2O)6]2+ is transformed into [CoCl4]2−. 9. Option (1) is correct.

Explanation: Thiosulphato is a monodentate ligand whereas oxalato, glycinato and ethylene diamine are bidentate ligands and can form rings with the central metal ion. So, they are also chelating ligands. Thiosulphato is a monodentate ligand and hence, cannot form chelate rings. Hence, it is not a chelating ligand. 10. Option (1) is correct. Explanation: CFSE used to determine the complex crystal structures and its stability. It also predict the stability of oxidation state and heat of hydration of transition metal elements.

Study Time

CHAPTER

Max. Time: 1:50 Hours Max. Questions: 50

10

HALOALKANES AND HALOARENES

 Revision Notes  Types: a.  Allylic halides: Halogen is bonded to allylic carbon.

 Sandmeyer's reaction:

b.  Benzylic halides: Halogen atom is bonded to a sp3 hybridised carbon atom next to an aromatic ring.

 Finkelstein Reaction: c.  Vinylic halides: Halogen is bonded to one of the carbon atoms of a vinylic carbon.

d.  Aryl halides: Halogen atom is directly bonded to sp2 hybridised carbon atom of an aromatic ring.

 Preparation:

ZnCl

2 + HX R–X + H2O (X = Cl, Br, I) ƒ R–OH + NaBr + H2SO4 ® RBr + NaHSO4 + H2O ƒ 3R–OH + PX3 ® 3R–X + H3PO3 (X = Cl, Br) ƒ R–OH + PCl5 ® R–Cl + POCl3 + HCl

ƒ R–OH

ƒ R–OH ƒ R–OH

Red P/X2

X2 = Br2, I2

R–X Cl2/UV light Or heat

CH3CH2CH2CH2Cl+ n-Butyl chloride

CH3CH2CHClCH3

sec-Butyl chloride (Major product)

3



Dry acetone

R–I + NaX

AgF, HgF, CoF or SbF

R–X R–F  Physical properties: ƒ The boiling points are higher than the corresponding hydrocarbons. ƒ There exists strong dipole-dipole and van der Waals forces of attraction. ƒ Are slightly soluble in water but  completely soluble in organic solvents.  Chemical properties: ƒ Undergoes nucleophilic substitution ƒ Elimination reaction:

Scan to know more about this topic

Introduction to the E2 mechanism

ƒ Forms

+ SOCl2 ® R–Cl + SO2 + HCl

ƒ CH3CH2CH2CH3

R–X + Nal  Swarts Reaction:

organometallic compounds with metals dry ether CH3CH2Br + Mg CH3CH2MgBr Grignard reagent

RMgX + H2O ® RH + Mg(OH)X ƒ Undergoes SN1 and SN2 on the basis of their kinetic properties. ƒ Reduction: Zn/HCl (conc) R – X + 2(H) R–H+H–X { Wurtz reaction: 2RX + 2Na ® R–R + 2NaX { Summary:

HALOALKANES AND HALOARENES

97

98 Oswaal CUET (UG) Chapterwise Question Bank { Friedel-Crafts

reaction:

{ Wurtz-Fitting

reaction:

{ Fittig

 Aryl halides:

CHEMISTRY

reaction:

 Nucleophilic substituiton reaction: (i) SN1 STEP I:

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SN1 reactions

{ Halogenation:

STEP II:

{ Nitration:

(ii) SN2

{ Sulphonation:

 Important compounds: ƒ Dichloromethane

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ƒ Chloroform ƒ Iodoform ƒ Carbon ƒ Freon

tetrachloride

and DDT

SN2 reactions

99

HALOALKANES AND HALOARENES

OBJECTIVE TYPE QUESTIONS [A] MULTPLE CHOICE QUESTION: 1. The structure of major mono halo product in the following reaction is:

7. Identify correct resonance structure of chlorobenzene: (1) (2)

(3) (1)

(2)

(3)

(4)

[CUET 2022, 10th Aug] 2. Aryl halides cannot be prepared by the reaction of aryl alcohols with PCl3, PCl5 or SOCl2 because: (1) Phenols are highly stable compounds. (2) Carbon-oxygen bond in phenols has a partial double bond character. (3) Carbon-oxygen bond is highly polar. (4) Phenyl cation is stabilised by resonance  [CUET 2022, 17th Aug] 3. Which one of the following species attacks the benzene ring in the above reaction?

(1) Cl- (3) AlCl3

(2) Cl+ (4) [AlCl4][CUET 2022, 18th Aug] 4. Identify the major product (P) formed in the chemical reaction is: Ethanol (CH3)3CBr + KOH P heat Major Product

(1) (CH3)3COH (3) (CH3)2C=CH2

(2) (CH3)2C=O (4) CH3CH2OCH2CH3 [CUET 2022, 18th Aug] 5. Monochlorination of toluene in sunlight followed by hydrolysis with aq. NaOH gives: (1) Benzyl alcohol (2) 2,4-Dihydroxytoluene (3) m-cresol (4) o-cresol [CUET 2022, 18th Aug] 6. Arrange the following compounds in order of decreasing boiling points: (A) CH3CH2CH2CH2Br (B) CH3–CH2–CH2Br (C) CH3 CH(Br) CH3 (D) (CH3)3CBr Choose the correct answer from the options given below: (1) (C), (B), (D), (A) (2) (A), (D), (B), (C) (3) (B), (D), (C), (A) (4) (D), (C), (B), (A) [CUET 2022, 20th Aug]

(4)



[CUET 2022, 20th Aug] 8. The organometallic compound from the following is (1) CH3COOAg (2) H3CCOONa (3) H3CMgBr

(4)

9. Predict the correct order of reactivity of the following compounds by SN2 reaction. Iso butyl iodide, n-Propyl chloride Ethyl chloride and Methyl iodide. (1) Ethyl chloride > Methyl iodide >n-Propyl chloride > Isobutyl iodide (2) Methyl iodide > Isobutyl iodide >Ethyl chloride > n-Propyl chloride (3) n-Propyl chloride > Isobutyl iodide >Ethyl chloride > Methyl iodide (4) Methyl iodide > Ethyl chloride >n-Propyl chloride > Isobutyl iodide [CUET 2022, 30th Aug] 10. Match the type of alkyl halide from column I with examples from column II and mark the appropriate answer. Choose the correct answer from the options given below: List-I

List-II

(A) Primary alkyl halide

(I) C  hlorophenyl methane

(B) S  econdary alkyl halide

(II) methylene chloride

(C) Geminal dihalide

(III) Isopropyl chloride

(D) Benzylic halide

(IV) Isobutyl chloride

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

(A)-(IV), (B)-(III), (C)-(II), (D)-(I) (A)-(II), (B)-(III), (C)-(I), (D)-(IV) (A)-(II), (B)-(IV), (C)-(I), (D)-(III) (A)-(IV), (B)-(II), (C)-(III), (D)-(I) [CUET 2022, 30th Aug] 11. Identify the following named reaction: 2Na/Dry ether

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

2C2H5Br Wurtz reaction Sandmeyer's reaction Williamson's synthesis Ullmann reaction

C2H5 – C2H5

[CUET 2022, 23rd Sept]

100 Oswaal CUET (UG) Chapterwise Question Bank 12. The products formed in the following reaction is:

16. Gammexane is: (1) bromobenzene (2) benzyl chloride (3) chlorobenzene (4) benzene hexachloride 17. In which of the following molecules, carbon atom marked with asterisk (*) is asymmetric?

(1)

(2) H3CCH2OH, HCOOH (3) H3CCH3, MgCl(OH) (4) H3CCH2CH3, MgCl(OH)  [CUET 2022, 23rd Aug] 13. The species which does not show high reactivity towards SN1 reaction is: (A) Primary halide (B) Allylic halide (C) Benzylic halide (D) Tertiary halide Choose the correct answer from the options given below. (1) (A) (2) (B) (3) (C) (4) (D)  [CUET 2022, 21st Aug] 14. Which of the following structures is enantiomeric with the molecule (A) given below?

(1)

(2)

(3)

(4)

15. The IUPAC name of the compound shown below is: Cl

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

Br 2-bromo-6-chlorocyclohex-1-ene 6-bromo-2-chlorocyclohexene 3-bromo-1-chlorocyclohexene 1-bromo-3-chlorocyclohexene

(2)



(3)

(1)

CHEMISTRY



(4)

18. Which reagent will you use for the following reaction? CH3CH2CH2CH3 → CH3CH2CH2CH2Cl + CH3CH2 CHClCH3  (1) Cl2/UV light (2) NaCl + H2SO4 (3) Cl2 gas in dark (4) Cl2 gas in the presence of iron in dark 19. Arrange the following compounds in increasing order of their boiling points. (i)

(ii) CH3CH2CH2CH2Br

(iii) Choose the correct answer from the options given below. (1) (ii) < (i) < (iii) (2) (i) < (ii) < (iii) (3) (iii) < (i) < (ii) (4) (iii) < (ii) < (i) 20. Complete the reaction: H3C-Br + AgF → (1) H3C-Br + AgF → H3C-F + AgBr (2) H3C-Br + AgF → Br-CH2-F + AgH (3) H3C-Br + AgF → [Ag(CH3)]F + Br (4) None of the above 21 The order of reactivity of following alcohols with halogen acids is: (i) CH3CH2–CH2–OH (ii)

(iii) (1) (i) > (ii) > (iii) (3) (ii) > (i) > (iii)

(2) (iii) > (ii) > (i) (4) (i) > (iii) > (ii)  [NCERT Exemp. Q 1, Page 133] 22. Toluene reacts with a halogen in the presence of iron (III) chloride giving ortho and para halo compounds. The reaction is: (1) Electrophilic elimination reaction (2) Electrophilic substitution reaction

101

HALOALKANES AND HALOARENES

(3) Free radical addition reaction (4) Nucleophilic substitution reaction  [NCERT Exemp. Q 4, Page 134] 23. Which of the following is halogen exchange reaction? (1) RX + NaI → RI + NaX (2) (3) R – OH + HX ® R – X + H2O (4)  [NCERT Exemp. Q 5, Page 134] 24. The position of –Br in the compound in CH3CH=CHC(Br) (CH3)2 can be classified as:  [NCERT Exemp. Q 12, Page 136] (1) Allyl (2) Aryl (3) Vinyl (4) Secondary 25. The reaction of toluene with chlorine in the presence of iron and in the absence of light yields:

(1)

(2)

(3) (4) mixture of (2) and (3)  [NCERT Exemp. Q 20, Page 137] 26. A primary alkyl halide would prefer to undergo _____________. (1) SN1 reaction (2) SN2 reaction (3) α–Elimination (4) Racemisation [NCERT Exemp. Q 16, Page 137] 27. Ethylidene chloride is a/an: (1) vic-dihalide (2) gem-dihalide (3) allylic halide (4) vinylic halide [NCERT Exemp. Q 14, Page 136] 28. Which is the correct IUPAC name for: CH3CH(C2H5)CH2Br (1) 1-Bromo-2-ethylpropane (2) 1-Bromo-2-ethyl-2-methylethane (3) 1-Bromo-2-methylbutane (4) 2-Methyl-1-bromobutane [NCERT Exemp. Q 18, Page 137] 29. Chloromethane on treatment with excess of ammonia yields mainly:

(1) N, N-Dimethyl methanamine (2) N–methyl methanamine (CH3—NH—CH3) (3) Methan amine (CH3NH2) (4) Mixture containing all these in equal proportion  [NCERT Exemp. Q 21, Page 138] 30. Molecules whose mirror image is non superimposable over them are known as chiral. Which of the following molecules is chiral in nature? (1) 2-Bromobutane (2) 1-Bromobutane (3) 2-Bromopropane (4) 2-Bromopropan-2-ol  [NCERT Exemp. Q 22, Page 133] [B] ASSERTION REASON QUESTIONS Direction: In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question: (1) Both A and R are correct and R is the correct explanation of A. (2) Both A and R are correct but R is NOT the correct explanation of A. (3) A is correct but R is not correct. (4) A is not correct but R is correct. 1. Assertion (A): SOCl2 is preferred over phosphorous tri chloride for the preparation of alkyl chlorides from alcohols. Reason (R): SOCl2 gives pure alkyl halides  [NCERT Exemp. Q 85, Page 148] 2. Assertion (A): Phosphorus chlorides (tri and penta) are preferred over thionyl chloride for the preparation of alkyl chlorides from alcohols. Reason (R): Phosphorus chlorides give pure alkyl halides. 3. Assertion (A): KCN reacts with methyl chloride to give methyl isocyanide. Reason (R): CN– is an ambident nucleophile. 4. Assertion (A): Tert-Butyl bromide undergoes Wurtz reaction to give 2, 2, 3, 3-tetramethylbutane. Reason (R): In Wurtz reaction, alkyl halides react with sodium in dry ether to give hydrocarbon containing double the number of carbon atoms present in the halide. 5. Assertion (A): 1,4 Dichloro benzene is polar. Reason (R): The dipole moments cancel each other because of the symmetrical structure. 6. Assertion (A): In monohaloarenes, further electrophilic substitution occurs at ortho and para positions. Reason (R): Halogen atom is a ring deactivator. 7. Assertion (A): Aryl iodides can be prepared by reaction of arenas with iodine in the presence of an oxidizing agent. Reason (R): Oxidising agent oxidises I2 into HI. 8. Assertion (A): It is difficult to replace chlorine by -OH in chlorobenzene in comparison to that in chloroethane. Reason (R): Chlorine-carbon (C – Cl) bond in chlorobenzene has a partial double bond character due to resonance. 9. Assertion (A): Hydrolysis of (-)-2-bromooctane proceeds with inversion of configuration.

102 Oswaal CUET (UG) Chapterwise Question Bank

Reason (R): This reaction proceeds through the formation of a carbocation. 10. Assertion (A): Nitration of chlorobenzene leads to the formation of m-nitro chlorobenzene. Reason (R): NO2 group is a m-directing group. [C] COMPETENCY BASED QUESTIONS I. Based on following passage answer questions from 1-5:  Compound ‘A’ with molecular formula C4H9Br is treated with aq. KOH solution. The rate of this reaction depends upon the concentration of the compound ‘A’ only. When another optically active isomer ‘B’ of this compound was treated with aq. KOH solution, the rate of reaction was found to be dependent on concentration of compound and KOH both. 1. The type of reaction that A undergo with aq. KOH is: (1) E1 (2) E2 (3) SN1 (4) SN2 2. In transition state when A reacts with aq. KOH involves the formation of: (1) A carbonium ion (2) A carbanion (3) A free radical (4) None of the above 3. Compound A is identified as: (1) 1 bromo butane (2) 2 bromo butane (3) 2 bromo, 2 methyl propane (4) 1 bromo 2 methyl propane 4. The product formed when A reacts with alcoholic KOH is: (1) But-1-ene (2) But-2-ene (3) Propene (4) 2 methyl propene 5. The product formed when A and B reacts with aq. KOH is: (1) An aldehyde (2) An acid (3) An ether (4) An alcohol II. Based on following passage answer questions from 6-10. Methyl chloride, methyl bromide, ethyl chloride and some chlorofluoromethanes are gases at room temperature. Higher members are liquids or solids. As we have already



6.

7.

8.

9.

10.

CHEMISTRY

learnt, molecules of organic halogen compounds are generally polar. Due to greater polarity as well as higher molecular mass as compared to the parent hydrocarbon, the intermolecular forces of attraction (dipole-dipole and van der Waals) are stronger in the halogen derivatives. That is why the boiling points of chlorides, bromides and iodides are considerably higher than those of the hydrocarbons of comparable molecular mass. The attractions get stronger as the molecules get bigger in size and have more electrons. For the same alkyl group, the boiling points of alkyl halides decrease in the order: RI> RBr> RCl> RF The lower members of alkyl halides are: (1) Gaseous (2) Liquids (3) Solids (4) Colloids The type of forces existing between haloalkanes are: (A) Ionic forces (B) Vander Waals forces (C) Hydrogen bonds Choose the correct answer from the options given below. (1) Only (A) (2) Only (B) (3) Only (C) (4) All (A), (B), (C) Boiling point of alkyl halides: (1) Increase with decrease in the number of carbon atoms (2) Increase with increase in the number of carbon atoms (3) Increase with branching of carbon atoms (4) Is independent of number of carbon atoms. The correct statement is: (1) Alkyl iodides have lowest boiling points (2) Alkyl chlorides have highest boiling points (3) Alkyl bromides have highest boiling points (4) Alkyl iodides have highest boiling points Alkyl halides are (1) Non polar (2) Less polar than alkanes (3) More polar than alkanes (4) Both alkanes and alkyl halides have same polarity

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (1)

2. (2)

3. (2)

4. (3)

5. (1)

6. (2)

7. (2)

8. (3)

9. (4)

10. (1)

11. (1)

12. (1)

13. (1)

14. (1)

15. (3)

16. (4)

17. (2)

18. (1)

19. (3)

20. (1)

21. (2)

22. (2)

23. (1)

24. (1)

25. (2)

26. (2)

27. (2)

28. (3)

29. (3)

30. (1)

8. (1)

9. (3)

10. (4)

8. (2)

9. (4)

10. (3)

[B] ASSERTION REASON QUESTIONS 1. (1)

2. (2)

3. (4)

4. (1)

5. (4)

6. (2)

7. (3)

[C] COMPETENCY BASED QUESTIONS 1. (3)

2. (2)

3. (3)

4. (4)

5. (4)

6. (1)

7. (2)

103

HALOALKANES AND HALOARENES

ANSWERS WITH EXPLANATION [A] MULTPLE CHOICE QUESTION: 1. Option (1) is correct. Explanation: Nucleophilic substitution chloride on sp3 carbon is faster than sp2 carbon. 8. Option (3) is correct.

2. Option (2) is correct. Explanation: Aryl halides cannot be prepared by replacing the hydroxyl group of phenols because the carbon-oxygen bond in phenols has a partial double bond character and is stronger than a single bond. Hence, difficult to break.

Explanation: Organometallic compounds are defined as compounds containing a covalent bond between a carbon atom and a metal e.g., Grignard reagent: CH3–Mg–Br. 9. Option (4) is correct.

3. Option (2) is correct.

Explanation: SN2 mechanism favours less hindered alkyl halide for reaction. Hence, correct order of reactivity is SN2 reaction is: CH3I > CH3CH2Cl > CH3CH2CH2Cl > (CH3)2CHCH2I

Explanation: Cl+ is an electrophile in halogenation of benzene and thus acts as an attacking species.

Explanation:

10. Option (1) is correct.



4. Option (3) is correct. Explanation:

5. Option (1) is correct. Explanation: Monochlorination of toluene in sunlight yields chloromethyl benzene which followed by hydrolysis with aq. NaOH gives benzyl alcohol.



Common name: Benzylic halide IUPAC name: Chlorophenyl methane

Methylene chloride

Common name: 2° alkyl halide IUPAC name: Isopropyl chloride CH3-CH2-CH2-Cl Common name: 1° alkyl halide IUPAC name: n-propyl chloride 11. Option (1) is correct. Explanation: When alkyl halides react with sodium in presence of dry ether, a hydrocarbon containing double the number of carbon atoms present in the halide is formed. This is known as Wurtz reaction. 12. Option (1) is correct.

6. Option (2) is correct. Explanation: More is the molar mass more is the boiling point. For isomeric alkyl halides more is the branching, lesser is the surface area and smaller will be van der Waals’ force of attraction and lower will be the boiling point. Correct order of decreasing boiling point: A, D, B, C. 7. Option (2) is correct.

Explanation: Reduction of aldehyde to alcohol by Grignard reagent.

13. Option (1) is correct. Explanation: The primary alkyl halide during SN1 reaction, forms least stable primary carbocation as: (1° carbocation) As stability of carbocation increases, SN1 mechanism is favourable. Order of stability of carbocation is as: 3°>2°>1°

104 Oswaal CUET (UG) Chapterwise Question Bank In allylic halides and benzylic halides carbocation formed is stablised by resonance effect whereas in tetiary halides carbocation is stabilised by +I effect of three alkyl groups attached to the carbocation.

23. Option (1) is correct.

14. Option (1) is correct.

24. Option (1) is correct.

Explanation: Compound (A) is enantiomer of compound (1) because the configuration of two groups, that is, CH3 and C2H5 in them is reversed at chiral carbon.

CHEMISTRY

Explanation: Halogen exchange reactions are those in which one halide replaces another. This reaction is known as Finkelstein reaction. Explanation: It is allylic compound in which Br is attached next to double bond carbon.

15. Option (3) is correct. Explanation: 25. Option (2) is correct.

IUPAC name: 3-bromo-1-chlorocyclohexene 16. Option (4) is correct.

Explanation: Cl– is an electrophile formed by the following reaction: AlCl3 + Cl2 → [AlCl4]- + Cl+ + Cl attacks the benzene ring to give chlorobenzene.

Explanation: Gammexane is an isomeric form of benzene hexachloride (BHC).

26. Option (2) is correct. BHC 17. Option (2) is correct. Explanation: Asymmetric/Chiral carbon atom is that in which all of its four valencies lie with four different groups or atoms. In molecules (i), (ii) and (iii), all have asymmetric carbon as each carbon has satisfied all four valencies with four different groups or atoms. In molecule (iv), carbon satisfies two of its valencies with two hydrogen atoms i.e., similar atom. So, it is not an asymmetric carbon atom. 18. Option (1) is correct. Explanation: The given reaction is a free radical substitution reaction. It occurs in presence of ultraviolet light or at high temperature or peroxides which are free radical generators. Free radical substitution cannot take place in dark. 19. Option (3) is correct. Explanation: Boiling points of isomeric haloalkane decrease with increase in branching as with increase in branching surface area decreases which leads to decrease in intermolecular forces. 20. Option (1) is correct. Explanation: H3C-Br + AgF → H3C-F + AgBr 21. Option (2) is correct.

Explanation: The reactivity order of alcohols towards halogen acids is 3°>2°>1°. Since the stability of carbocations is of the order of 3°>2°>1°. 22. Option (2) is correct. Explanation:

Explanation: SN2 reaction proceeds, via formation of transition state which is formed easily in primary alkyl halide due to less steric hindrance. 27. Option (2) is correct. Explanation: Ethylidene chloride is a gem-dihalide, CH3— CHCl2 in which both halogen atoms are attached to the same carbon atom. 28. Option (3) is correct. 29. Option (3) is correct. Explanation: Chloromethane on treatment with excess of ammonia yields mainly methanamine. CH3Cl + NH3 → CH3NH2 + HCl Excess

methanamine

However, if the two reactants are present in the same amount, the mixture of 1°, 2° and 3° amine is obtained. CH3CI + NH3 → CH3NH2 + HCI (Primary amine)

CH3NH2 + CH3C1 → CH3CI → (CH3)2NH + HCl (Secondary amine) (CH3)2NH + CH3C1→ (CH3)2N + HCl (Tertiary amine) (CH3)3N + CH3Cl → (CH3)4 NCl (Quaternary ammonium salt) 30. Option (1) is correct. Explanation: 2 bromo butane has a chiral carbon where the carbon atom is attached to four different groups. [B] ASSERTION REASON QUESTIONS 1. Option (1) is correct. Explanation: Thionyl chloride is best halogen carrier to convert alcohol to alkyl halide because it gives byproducts in gaseous state. Thus, we get pure alkyl halide in this reaction.

105

HALOALKANES AND HALOARENES

2. Option (2) is correct.

[C] COMPETENCY BASED QUESTIONS

Explanation: Thionyl chloride is best halogen carrier to convert alcohol to alkyl halide because it gives by-products in gaseous state. Thus, we get pure alkyl halide in this reaction.

1. Option (3) is correct.

3. Option (4) is correct.

2. Option (2) is correct.

Explanation: R—Cl + KCN –> R-CN + KCl KCN is ionic and hence isocyanide is not obtained in this reaction. CN– is an ambident nucleophile.

Explanation: Since A is tertiary halide it undergoes SN1 mechanism with aq. KOH. Explanation: In the first step of tertiary halide, if SN1 reaction undergoes heterolytic cleavage to form a carbonium ion in the Transition state.

4. Option (1) is correct.

3. Option (3) is correct.

Explanation: Wurtz reaction gives higher alkanes.

Explanation: Since A undergoes SN1 mechanism with aq. KOH, it has to be a tertiary halide.

5. Option (4) is correct. Explanation: because of symmetrical structure of 1,4 dichloro benzene it is non polar.

4. Option (4) is correct.

6. Option (2) is correct.

Explanation: Alcoholic KOH is the reagent for dehydrohalogenation. Hence product formed from A is 2 methyl, propene.

Explanation: Halogens are ortho-para directing due to (+M) or (+R) effect. Moreover, they are deactivating due to high electronegativity.

5. Option (4) is correct. Explanation: Alcohol is formed when alkyl halides undergo substitution reaction with aq. KOH

7. Option (3) is correct.

6. Option (1) is correct.

Explanation: Oxidising agent like (HIO3) converts HI to I2, otherwise HI may reduce aryl halide to arenes.

Explanation: First members of alkyl halides having lower number of carbon atoms are gaseous.

8. Option (1) is correct.

7. Option (2) is correct.

Explanation: Due to resonance, there is double bond character between benzene ring and halogen hence, nucleophilic substitution of aryl halide is difficult as compared to alkyl halide.

Explanation: Van der Waal’s forces -dipole-dipole interactions are present between alkyl halides.

9. Option (3) is correct.

Explanation: With increase in number of carbon atoms the intermolecular forces increase and boiling point increases.

Explanation: Hydrolysis of 2-bromooctane follows SN2 mechanism hence inversion in configuration takes place. This reaction does not involve the formation of carbocation intermediate. 10. Option (4) is correct. Explanation: Nitration of chlorobenzene gives ortho and para nitro chlorobenzene. However, nitro group is meta directing in nature.

8. Option (2) is correct.

9. Option (4) is correct. Explanation: Alkyl iodides have highest boiling points as compared to other halides. 10. Option (3) is correct. Explanation: Halides are electronegative and exert -I effect, thus making the C-X bond polar, inducing polarity in the molecule.

Study Time

CHAPTER

Max. Time: 1:50 Hours Max. Questions: 50

11

ALCOHOLS, PHENOLS AND ETHERS (vii)

 Revision Notes ALCOHOLS:  Types: { Alcohols

and phenols may be classified as mono–, di–, tri– or polyhydric compounds depending on whether they contain one, two, three or many hydroxyl groups. { Monohydric alcohols are classified as: a. Primary alcohols: RCH2OH b. Secondary alcohols: R2CHOH c. Tertiary alcohols: R3COH

{ Aromatic

(viii) RCOOH

LiAlH4, H2O

RCH2OH

 Preparation of phenols:

(i)

alcohols:

(ii)

 Preparation of Alcohols: (i) CH3 – Cl + aq.KOH → CH3 – OH + KCl

(iii)  Physical properties: { Boiling point increases with increase in number of carbon atoms. { Boiling point decreases with increase in branching due to decrease in Van der Waals forces. { Solubility of alcohols and phenols in water is due to their ability to form hydrogen bonds with water molecules. { The solubility decreases with increase in size of alkyl/ aryl groups. Several of the lower molecular mass alcohols are miscible with water in all proportion.

(ii) (iii) (iv)

HO

(v) CH2 = CH2 + H2SO4 → CH3CH2OSO3H 2 CH3CH2OH + H2SO4 (vi) 3R–CH=CH2+(BH3) → (R–CH2–CH2)3B 

H2O2

OH–, H2O

R – CH2 – CH2 – OH + H3BO3

 Chemical properties: { The carbon– oxygen bond length (136 pm) in phenol is slightly less than that in methanol. This is due to partial double bond character on account of the conjugation of unshared electron pair of oxygen with the aromatic ring.

ALCOHOLS, PHENOLS AND ETHERS

107

2,4,6- Tribromophenol

OH

OH Br

Br2

+

CS 2 108 Oswaal CUET (UG) Chapterwise Question Bank

CHEMISTRY

Br o- and p-Bromophenol Na

OH

C 6 H 5O – Na+ +H2

N aO H

Con c. H N O 3

C 6 H 5O – Na+ +H2

O2 N

NO2

Con c. H 2SO 4

Zn

+ZnO

Distillation

NO2 2,4,6-Trinitrophenol

NH 3

C 6H 5N H 2 + H 2O

ZnCl 2

OH

OH SO3 H

Conc. H 2SO 4

PCl 5

+

C6 H5 Cl

CH3 CO Cl

C 6H 5COOCH3

Pyridine

SO3 H OH

C6 H 5OH

C6 H5 C OCl

C 6H 5COOC6H5

NaOH

Na OH +CO2, 130 –140 °C

COOH

H + Kolbe - Schmidtreaction

o-Hydroxy benzoic acid

Na

C 6H 5OCH3

CH3 I

OH

OH

(i) CHCl3 + NaOH (ii) H +

Br

Br2 water

Br

CHO

Reimer- Tiemann reaction Salicylaldehyde

OH

Br 2,4,6- Tribromophenol

OH

HNO3

OH

+

At room temp.

Br

Br2

OH NO2

+

o-Nitrophenol

CS 2

O

NO 2 p-Nitrophenol

Na 2Cr2 O 7

Br o- and p-Bromophenol

H 2S O4

OH Con c. H N O 3

O2 N

NO2

O Benzoquinone

Con of c. HEthanol: 2SO 4  Reactions

NO2 2,4,6-Trinitrophenol

OH

OH SO3 H

Conc. H 2SO 4

+ SO3 H OH

Na OH +CO2, 130 –140 °C

COOH

H + Kolbe - Schmidtreaction

o-Hydroxy benzoic acid

OH (i) CHCl3 + NaOH (ii) H +

CHO

Reimer- Tiemann reaction Salicylaldehyde

OH

OH NO2

HNO3

+

At room temp. o-Nitrophenol

O

NO 2 p-Nitrophenol

Na 2Cr2 O 7 H 2S O4

O Benzoquinone

109

ALCOHOLS, PHENOLS AND ETHERS

Scan to know more about this topic

  Test to identify primary, secon­

dary and tertiary alcohols:

{ Lucas test: In tertiary alcohol turbid­

Scan to know more about this topic

{ Kolbe's

reaction:

Salicylic acid from aspirin

ity appears immediately, in second­ ary alcohol turbidity appears within five minutes and in primary alcohol Primary, turbidity appears on heating. secondary { Iodoform Test: Given by com­ and tertiary alcohol pounds with a methyl group next to a carbonyl group. Secondary alcohols with a CH3 on the carbon give a positive iodoform test.

ETHERS: 1. Preparation:

 Fries Rearrangement:

{ Williamson

synthesis: R - X + Na – O – Rʹ → R – O – Rʹ + NaX

H 2 SO 4 (i) C H OH + HOC H ¾conc. ¾¾¾¾ ® C 2 H 5OC 2 H 5 + H 2O 2 5 2 5 413 K

(ii) 2C 2 H 5Cl + Ag 2O ¾heat ¾¾ ® C 2 H 5 - O - C 2 H 5 + 2AgCl CH 3Cl + Ag 2O + C 2 H 5Cl ® CH 3 - O - C 2 H 5 + 2AgCl Scan to know more about this topic

(iii)

 Physical properties: { The C-O bonds in ethers are polar. { Absence of intermolecular hydrogen bonding results in lower boiling points.  Chemical properties:

(iv) C 2 H 5OH + CH 2 N 2 ¾BF ¾3 ® C 2 H 5OCH 3 + N 2 Williamson ether synthesis

K R - O - R + HX ¾373 ¾¾ ® ROH + RX (X = Br, I)

OCH3 Br2 inaceticacid (Bromination)

OCH3 Br + o-Bromoanisole (Minor)

Br p-Bromoanisole (Major)

OCH3 CH3Cl/anhyd.AlCl 3

O–CH3

OCH3 CH3 +

(Friedel-Craftsalkylation)

CH3

p-Methoxytoluene (Major)

Anisole

OCH3

OCH3 COCH3

=

O

CH3–C–Cl/anhyd.AlCl3

o-Methoxytoluene (Minor)

+

(Friedel-Craftsacylation)

COCH

o-Methoxy acetophenone

CH3

CH3Cl/anhyd.AlCl 3

O–CH3

+

(Friedel-Craftsalkylation)

o-Methoxytoluene

CH3 CUET (UG)(Minor) 110 Oswaal Chapterwise Question Bank p-Methoxytoluene

CHEMISTRY

(Major)

Anisole

OCH3

OCH3 COCH3

=

O

CH3–C–Cl/anhyd.AlCl3

+

(Friedel-Craftsacylation)

o-Methoxy acetophenone (Minor) p-Meth oxyacetophenone (Major)

COCH3 OCH3

HNO3 (conc.)/HSO 2 4 (conc.)

OCH3 NO2 +

Nitration

NO2

p-Nitroanisole (Major)

o-Nitroanisole (Minor)

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS 1. Among the following statements, choose the correct statements. (A) Boiling point of alcohols increases with increase in the number of carbon atoms. (B) In alcohols, boiling points increases with increase of branching in carbon chain. (C) Boiling points of alcohols are lesser in comparison to haloalkanes of comparable molecular mass. (D) Boiling points of alcohols are higher in comparison to hydrocarbons of comparable (E) The high boiling points of alcohols are mainly due to the presence of intramolecular hydrogen bonding. Choose the correct answer from the options given below: (1) (A), (D) and (E) only (2) (A), (B) and (C) only (3) (B), (C) and (D) only (4) (C), (D) and (E) only [CUET 2022, 10th Aug] 2. The structure of the product of the following reaction is: O

CH2—C—O—CH3 O

(1) OH

CH2—C—O—CH3

(2)

(3)

O

NaBH4

(4)  [CUET 2022, 10th Aug] 3. The compound formed on reaction of propanone with methyl magnesium bromide followed by hydrolysis is: (1)

(2) (3) CH3CH2CH2OH (4) CH3CH2CH2CH2OH

 [CUET 2022, 18th Aug] 4. Given below are three steps which are involved for conversion of alkene into alcohol. Arrange them in correct sequence. (A) Nucleophilic attack of water on carbocation resulting in deprotonation. (B) Deprotonation to form an alcohol. (C) Protonation of alkene to form carbocation by electrophilic attack of H3O+ Choose the correct answer from the options given below: (1) (A), (C), (B) (2) (A), (B), (C) (3) (C), (A), (B) (4) (C), (B), (A)  [CUET 2022, 20th Aug] 5. Arrange the following compounds in increasing order of their acid strengths:  (A) o-Cresol, pka = 10.2 (B) m-Nitrophenol, pka = 8.3 (C) Phenol, pka = 10 Choose the correct answer from the options given below: (1) (B), (C), (A) (2) (A), (B), (C) (3) (C), (B), (A) (4) (A), (C), (B) [CUET 2022, 20th Aug]

111

ALCOHOLS, PHENOLS AND ETHERS

(3) Propan-2-ol  13. The product formed upon potassium tert-butoxide: (1) Di-tert-butyl ether (3) isobutylene

6. The IUPAC name of the given compound.

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

2-Methylbutoxybenzene 3-Methylbutoxybenzene 3-Methylphenoxybenzene 2-Methyl-4-phenoxybutane

CH3

 [CUET 2022, 21st Aug] 7. Which set of reagents will be most suitable to bring about the following change? 2, 4, 6-Trinitochlorobenzene to picric acid (1) NaOH, 623 K, 300 atm (2) Hot conc. sulphuric acid (3) Warm water (4) Acidified water [CUET 2022, 23rd Aug] 8. In the following sequence of reactions, ‘A’ is:

14.

15.

H 2 /Pd ( i) NaNO 2 HCl Nal 2 A ¾95% ¾¾¾ ® B ¾AgNO ¾¾ ® C ¾ethanol ¾¾ ® D ¾(¾¾¾¾¾ ® H3 PO 4 ii) H 95% H O + 3

3

Methanol should come instead of ethanal

(1) Acetaldehyde (3) Dimethyl ether

(2) Diethyl ether (4) Nitroethane  [CUET 2022, 30th Aug] 9. The correct decreasing order of solubility of the following alcohols in water is:

16.

17.

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

(III) > (IV) > (II) > (I) (IV) > (II) > (III) > (I) (IV) > (III) > (II) > (I) (IV) > (II) > (I) > (III)

18.

(4) Butan-1-ol [CUET 2021, 23rd Sept] heating methyl bromide with

(2) Dimethyl ether (4) tert-butyl-methyl ether [CUET 2021, 23rd Sept] Monochlorination of toluene in sunlight followed by ­hydrolysis with aq. NaOH yields: (1) o-Cresol (2) m-Cresol (3) 2, 4-Dihydroxytoluene (4) Benzyl alcohol [NCERT Exemp. Q 1, Page 154] How many alcohols with molecular formula C4H10O are chiral in nature? (1) 1 (2) 2 (3) 3 (4) 4 [NCERT Exemp. Q 2, Page 154] CH3CH2OH can be converted into CH3CHO by: (1) catalytic hydrogenation (2) treatment with LiAlH4 (3) treatment with pyridinium chlorochromate (4) treatment with KMnO4 [NCERT Exemp. Q 4, Page 154] The process of converting alkyl halides into alcohols involves: (1) addition reaction (2) substitution reaction (3) dehydrohalogenation reaction (4) rearrangement reaction [NCERT Exemp. Q 5, Page 155] Which of the following compounds is aromatic alcohol?

[CUET 2022, 30th Aug] 10. Consider the following sequence of reactions: 



The final Product ‘C’ is: (1) Salicylic acid (3) Aspirin

(2) Salicylaldehyde (4) Ethyl salicylate [CUET 2022, 30th Aug] 11. Which of the following pairs can be distinguished by Lucas test? (1) Ethanol and ethylene glycol (2) o-and p-cresol (3) propan-1-ol and ethanol (4) Butan-1-ol and 2-methylpropan-2-ol [CUET 2021, 23rd Sept] 12. Which of the following does not give iodofom test? (1) Ethanol (2) Ethanal

Choose the correct answer from the options given below: (1) (A), (B), (C), (D) (2) (A), (D) (3) (B), (C) (4) (A) [NCERT Exemp. Q 6, Page 155 19. Which of the following compounds will react with sodium hydroxide solution in water? (1) C6H5OH (2) C6H5CH2OH (3) (CH3)3COH (4) C2H5OH [NCERT Exemp. Q 11, Page 156] 20. Phenol is less acidic than: (1) ethanol (2) o-nitrophenol (3) o-methyl phenol (4) o-methoxyphenol [NCERT Exemp. Q 12, Page 156] 21. Mark the correct order of decreasing acid strength of the following compounds:

112 Oswaal CUET (UG) Chapterwise Question Bank

CHEMISTRY

29. The reaction



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

(V) > (IV) > (II) > (I) > (III) (II) > (IV) > (I) > (III) > (V) (IV) > (V) > (III) > (II) > (I) (V) > (IV) > (III) > (II) > (I)  [NCERT Exemp. Q 14, Page 156] 22. Mark the correct increasing order of reactivity of the following compounds with HBr/HCl.

(1) a < b < c (3) b < c < a

(2) b < a < c (4) c < b < a

[NCERT Exemp. Q 15, Page 157] 23. Arrange the following compounds in increasing order of boiling point: Propan-1-ol, butan-1-ol, butan-2-ol, pentan-1-ol (1) Propan-1-ol, butan-2-ol, butan-1-ol, pentan-1-ol (2) Propan-1-ol, butan-1-ol, butan-2-ol, pentan-1-ol (3) Pentan-1-ol, butan-2-ol, butan-1-ol, propan-1-ol (4) Pentan-1-ol, butan-1-ol, butan-2-ol, propan-1-ol [NCERT Exemp. Q 16, Page 157] 24. Which of the following species can act as the strongest base? (1) ɵOH (2) ɵOR (3)

ɵ

OC6H5

(4)

25. Bond angle in ethers is slightly more than: (1) Square planar angle (2) Trigonal bipyramidal angle (3) Tetrahedral angle (4) None of the above 26. Williamson synthesis is used to obtain: (1) Primary alcohol (2) Ether (3) Aldehyde (4) Ketone 27. Formaldehyde reacts with methyl magnesium bromide followed by hydrolysis to form: (1) Methanol (2) Ethanol (3) Propanol (4) Butanol 28. Phenol can be distinguished from ethanol by the reaction with _____ (1) Br2/water (2) Na (3) Glycerol (4) All of these

can be classified as: (1) dehydration reaction (2) Williamson alcohol synthesis reaction (3) Williamson ether synthesis reaction (4) alcohol formation reaction. 30. In the reaction:



the electrophile involved is: (1) dichloromethyl cation (CHCl2) (2) formyl cation (CHO) (3) dichloromethyl anion (CHCl2) (4) dichlorocarbene (: CCl2)

[B] ASSERTION REASON QUESTIONS Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark The correct choice as: (1) Both A and R are correct and R is the correct explanation of A. (2) Both A and R are correct but R is NOT the correct explanation of A. (3) A is correct but R is not correct. (4) A is not correct but R is correct. 1. Assertion (A): Addition reaction of water to but-1-ene in acidic medium yields butan-1-ol. Reason (R): Addition of water in acidic medium proceeds through the formation of stable carbocation. 2. Assertion (A): p-nitrophenol is more acidic than phenol. Reason (R): Nitro group helps in the stabilisation of the phenoxide ion by dispersal of negative charge due to resonance.  [NCERT Exemp. Q 62, Page 163] 3. Assertion (A): IUPAC name of the compound.



is 2-ethoxy-2-methylethane. Reason (R): In IUPAC nomenclature, ether is regarded as hydrocarbon derivative in which a hydrogen atom is replaced by –OR or –OAr group [where R = alkyl group and Ar = aryl group] [NCERT Exemp. Q 63, Page 163] 4. Assertion (A): Bond angle in ethers is slightly less than the tetrahedral angle. Reason (R): There is a repulsion between the two bulky (–R) groups. [NCERT Exemp. Q 64, Page 163] 5. Assertion (A): Boiling points of alcohols is more than that of ethers with comparable molecular weight.  Reason (R): Alcohols can form intermolecular hydrogen bonding but ethers cannot.

113

ALCOHOLS, PHENOLS AND ETHERS

6. Assertion (A): Like bromination of benzene, bromination of phenol is also carried out in the presence of Lewis acid. Reason (R): Lewis acid polarises the bromine molecule. [NCERT Exemp. Q 66, Page 164] 7. Assertion (A): o-Nitrophenol is less soluble in water than the m- and p-isomers. Reason (R): m- and p-Nitrophenols exists as associated molecules. [NCERT Exemp. Q 67, Page 164] 8. Assertion (A): Ethanol is a weaker acid than phenol. Reason (R): Sodium ethoxide may be prepared by the reaction of ethanol with aqueous NaOH [NCERT Exemp. Q 68, Page 164] 9. Assertion (A): Phenol forms 2,4,6-tribromophenol on treatment with Br2 in water. Reason (R): Bromine polarises in carbon disulphide and so water is used. [NCERT Exemp. Q 69, Page 164] 10. Assertion (A): Phenols give o– and p-nitrophenol on nitration with conc. HNO3 and H2SO4 mixture. Reason (R): –OH group in phenol is o, p directing.  [NCERT Exemp. Q 70, Page 164] [C] COMPETENCY BASED QUESTIONS I. Based on following passage answer questions from 1-5: Ethers are class of organic compounds that contain ether group – an oxygen atom connected to two alkyl groups or aryl groups. They have the general formula R – O – R′, where R and R′ represents the alkyl or aryl groups. Ether, like water has a tetrahedral geometry, i.e., oxygen is sp3 hybridised. The C – O – C bond angle in ethers is slightly greater than the tetrahedral angle due to repulsive interactions between the two bulky groups when they are attached to oxygen. [CUET 2022, 17th Aug] 1. Which of the following cannot be made by using Williamson Synthesis? (1) Methoxybenzene (2) Benzyl p-nitro phenyl ether (3) Tert-Butyl methyl ether (4) Di-tert-Butyl ether 2. The IUPAC name of the ether CH2 = CH – CH2 – O – CH3 is: (1) Alkyl methyl ether (2) 1-Methoxyprop-2-ene (3) 3-Methoxyprop-1-ene (4) Vinyl dimethyl ether 3. Dehydration of alcohol to ethers is catalysed by: (1) Conc. H2SO4 at 413 K (2) Hot & NaOH (3) Hot & HBr (4) Hot & HNO3

(2)

(3) (4) CH3OH + CH3I

5.

(1)

(2)

(3)

(4)

II. Based on following passage answer questions from 6-10: Alcohols are versatile compounds. They act both as nucleophiles and electrophiles. The bond between O-H is broken when alcohols act as nucleophiles. (i) Alcohols as nucleophiles



are

4.

(1)

:

(ii) The bond between C-O is broken when they act as, electrophiles. Protonated alcohols react in this manner. Protonated alcohols as electrophiles. Based on the cleavage of O-H and C-O bonds, the reaction of alcohols and phenols may be divided into two groups: (a) Reactions involving cleavage of O-H bond (b) Reactions involving cleavage of C-O bond Acidity of alcohols and phenols (i) Reaction with metals: Alcohols and phenols react with active metals such as sodium, potassium and aluminium to yield corresponding alkoxide/phenoxides and hydrogen. 2 R-O-H + 2Na → 2R-O-Na + H2

114 Oswaal CUET (UG) Chapterwise Question Bank 6. Write down the decreasing order of reactivity of sodium metal towards primary, secondary and tertiary alcohols. (1) 1°alc < 2°alc < 3°alc (2) 1°alc > 2°alc > 3°alc (3) 3°alc < 1°alc < 2°alc (4) 3°alc > 1°alc < 2°alc 7. Name the following reaction: OH

ONa CHCl3 + aq NaOH

1

CH3

2

C—OH

3

CH3

CHO H

+

340 K Sodium salt of Salicylaidehyde

Phenol

(2) RCH2OH > RR’R”COH > RR’CHOH (3) RCH2OH < RR’CHOH < RR’R”COH (4) RCH2OH < RR’R”COH < RR’CHOH 9. Write the IUPAC name of the following compound:

OH CHO

CHEMISTRY

Salicyladehyde

(1) Williamson’s synthesis (2) Kolbe’s reaction (3) Reimer-Tiemann reaction (4) Sandmeyer’s reaction 8. Given the descending order of acid strength of alcohols. (1) RCH2OH > RR’CHOH > RR’R”COH

(1) 2-methyl-2-phenyl ethanol (2) 2-phenylbutanol (3) 2-Phenylopropan-2-ol (4) 1-methyl-1-phenyl ethanol 10. The correct order for acidic nature is: (1) H2O < CH3OH < C2H5OH < C6H5OH (2) H2O > CH3OH > C2H5OH > C6H5OH (3) C6H5OH > H2O > CH3OH > C2H5OH (4) C6H5OH < H2O > CH3OH < C2H5OH < C6H5OH

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (1)

2. (2)

3. (2)

4. (3)

5. (4)

6. (2)

7. (3)

8. (2)

9. (2)

10. (3)

11. (4)

12. (4)

13. (4)

14. (4)

15. (1)

16. (3)

17. (2)

18. (3)

19. (1)

20. (2)

21. (2)

22. (3)

23. (1)

24. (2)

25. (3)

26. (2)

27. (2)

28. (1)

29. (3)

30. (4)

8. (3)

9. (3)

10. (4)

8. (1)

9. (3)

10. (3)

[B] ASSERTION REASON QUESTIONS 1. (4)

2. (1)

3. (4)

4. (4)

5. (1)

6. (4)

7. (2)

[C] COMPETENCY BASED QUESTIONS 1. (4)

2. (3)

3. (1)

4. (2)

5. (1)

6. (2)

7. (3)

ANSWERS WITH EXPLANATION [A] MULTIPLE CHOICE QUESTIONS

O

1. Option (1) is correct. Explanation: Boiling points of alcohols increase with increase in the number of carbon atoms due to increase in van Der Waals forces. Also, it decreases with increase in branching due to decreased surface area. Alcohols have higher boiling point than hydrocarbons of their comparable mass due to intermolecular hydrogen bonding. (Note: Option E is correct if intra molecular is replaced by intermolecular) 2. Option (2) is correct Explanation: Sodium borohydride is a good reducing agent. Although not as powerful as lithium aluminium hydride (LiAlH4), it is very effective for the reduction of aldehydes and ketones to alcohols. By itself, it will generally not reduce esters, carboxylic acids, or amides (although it will reduce acyl chlorides to alcohols).

CH2—C—O—CH3

NaBH4

O OH CH2—C—O—CH3 O

3. Option (2) is correct. Explanation:

–

O

+

OMgBr

δ–δ CH3—C—CH3 + CH3MgBr +

CH3—C—CH3

OH CH3 | CH3 — C — CH3 +H2O | CH3

115

ALCOHOLS, PHENOLS AND ETHERS

4. Option (3) is correct. Explanation: The sequence of steps involved in the conversion of alkene to alcohol is: (i) Protonation of alkene to form carbocation by electrophilic attack of H3O+ (ii) Nucleophilic attack of water on carbocation. (iii) Deprotonation to form an alcohol. 5. Option (4) is correct. Explanation: In p-nitrophenol, nitro group shows −M effect which stabilises the conjugate anion hence, it will be most acidic. In p-cresol, methyl group shows +H effect which destabilises the conjugate anion hence, it will be least acidic. Therefore, acidity order will be: p-cresol < phenol < p-nitrophenol 6. Option (2) is correct. Explanation: The correct name of the given compound is 3-Methylbutoxybenzene. 7. Option (3) is correct. Explanation: 2, 4,6-trinitrochlorobenzene under mild hydroly­ sis conditions (H2O/323 K) gives 2,4,6-trinitrophenol or picric acid.

11. Option (4) is correct. Explanation: Lucas test is used to differentiate between primary, secondary and tertiary alcohols on reaction with anhydrous ZnCl2 and HCl. 12. Option (4) is correct.

Explanation: Iodoform test is given by aldehydes or ketones having a methyl group (−COCH3). Even alcohols having a methyl group can also show the iodoform test. Since, in butan-1-ol, no –COCH3 group is present, so it will not give iodoform test. 13. Option (4) is correct.

Explanation: The reaction of heating methyl bromide with potassium tert-butoxide is as follows:

14. Option (4) is correct.

8. Option (2) is correct.

Explanation: Monochlorination of toluene in sunlight gives benzyl chloride. On hydrolysis with aq. NaOH benzyl chloride shows nucleophilic substitution reaction to give benzyl alcohol.

Explanation: KI CH 3CH 2 - O - CH 2CH 3 ¾95% ¾¾¾ ® CH 3CH 2 I H3 PO 4

Diethyl ether (A)          Ethyl iodide (B) 2 CH 3CH 2 I ¾AgNO ¾¾ ® CH 3CH 2 NO 2

                  Nitroethane (C) H 2 /Pd CH 3CH 2 NO 2 ¾Ethanol ¾¾ ® CH 3CH 2 NH 2

                   Ethyl amine (D) 2 + HCl CH 3CH 2 NO 2 ¾NaNO ¾¾¾¾ ® CH 3CH 2OH H3 O+

                     Ethyl alcohol

9. Option (2) is correct. Explanation: With increase in the extent of hydrogen bonding, solubility of alcohols increases in water, but if hydrophobic part is increasing, then solubility decreases. Hence, more number of hydroxy groups increases the extent of H-bonding. Hence, solubility order is:

15. Option (1) is correct. Explanation: Only one alcohol contains chiral carbon atom (i) CH3CH2CH2CH2OH (ii)  (iii) Only 3 alcohols are chiral in nature. 16. Option (3) is correct. Explanation: Alcohols are oxidized to aldehydes and finally to acids. o] o] CH 3CH 2OH ¾[¾ ® CH 3CHO ¾[¾ ® CH 3COOH

10. Option (3) is correct. Explanation:

Oxidation can be stopped at aldehyde stage by using pyridinium chlorochromate (CrO3C5H5N– HCl). CH 3CH 2OH ¾PCC ¾¾ ® CH 3CHO 17. Option (2) is correct. Explanation: Conversion of alkyl halides into alcohols involves substitution reaction. RX ¾NaOH ¾¾® ROH+NaX

116 Oswaal CUET (UG) Chapterwise Question Bank

CHEMISTRY

18. Option (3) is correct. Explanation: Compound (A) i.e., phenol and compound (D), that is, a derivative of phenol cannot be considered as aromatic alcohol. As phenol is also known as carbolic acid and cannot be considered as aromatic alcohol. Compound (B) and (C) –OH group is bonded to sp3 hybridized carbon which in turn is bonded to benzene ring. 19. Option (1) is correct. Explanation: Phenol being more acidic reacts with sodium hydroxide solution in water to give sodium phenoxide which is resonance stabilised.Alcohols are very weak acids. C6H5OH + NaOH → C6H5ONa + H2O 20. Option (2) is correct.

Explanation: In o-nitrophenol nitro group is present at ortho position. Presence of electron withdrawing group at ortho position increases the acidic strength. On the other hand, in o-methylphenol and in o-methoxyphenol electron releasing group (–CH3–O–CH3). Presence of these groups at ortho or para positions of phenol decreases the acidic strength of phenols. So, phenol is less acidic than o-nitrophenol. 21. Option (2) is correct. Explanation: –NO2 is an electron withdrawing group which increases the acidity of phenol and the effect is more pronounced at ortho and para positions. Similarly, methoxy group is an electron releasing group which decreases the acidity of phenol and the effect is more pronounced at ortho and para positions.

28. Option (1) is correct. Explanation: Phenol decolourises bromine water to form white precipitate of 2,4,6-tribromophenol whereas ethanol does not precipitate. 29. Option (3) is correct. Explanation: The treatment of sodium alkoxide with a suitable alkyl halide to form an ether is called as Williamson ether synthesis reaction. 30. Option (4) is correct. Explanation: It is Reimer-Tiemann reaction. The electrophile formed is: CCl2 (Dichlorocarbene) according to the following reaction:

22. Option (3) is correct. Explanation: All three benzyl alcohols react with HBr/HCl through the formation of intermediate carbocation. The more stable the carbocation, the more reactive is the alcohol. The electron withdrawing groups, that is, NO2/Cl decreases the stability of the carbocation. Since, -NO2 group is stronger electron withdrawing than –Cl group, therefore, stability of carbocation increases in the order. 23. Option (1) is correct. Explanation: Boiling point increases with increase in molecular mass of the alcohols. Among isomeric alcohols 1o alcohols have higher boiling point than 2o alcohols. Thus, correct order is: Propan-1-ol, butan-2-ol, butan-1-ol, pentan-1-ol 24. Option (2) is correct. Explanation: Weakest acid has the strongest conjugate base. Since R-OH is the weakest acid, therefore, -OR is the strongest base. 25. Option (3) is correct. Explanation: Bond angle in ether is slightly more than the tetrahedral angle due to repulsion between the two bulky alkyl groups. 26. Option (2) is correct. Explanation: Williamson synthesis is used to obtain ether. For example, R – X + Na – O – R → R – O – R + NaX Alkyl Sodium alkoxide Ether halide

27. Option (2) is correct. Explanation:

[B] ASSERTION REASON QUESTIONS 1. Option (4) is correct. Explanation: Addition of water to but-1-ene in acidic medium yields butan-2-ol. Addition of water proceeds through the formation of stable secondary carbocation. 2. Option (1) is correct Explanation: p-nitrophenol is more acidic than phenol because Nitro group stabilizes phenoxide ion by dispersal of negative charge. 3. Option (4) is correct. Explanation: IUPAC name of the compound is 2-propoxypro­ pane. 4. Option (4) is correct. Explanation: Bond angle in ether is slightly more than the tetrahedral angle due to repulsion between the two bulky alkyl groups. 5. Option (1) is correct. Explanation: Boiling points of alcohols are higher than ethers. Alcohols can form intermolecular hydrogen bonding whereas ethers cannot. 6. Option (4) is correct. Explanation: Bromination of phenol carried out in the absence of Lewis acid. 7. Option (2) is correct. Explanation: In o-nitrophenol there is intramolecular

117

ALCOHOLS, PHENOLS AND ETHERS

H-bonding Thus, o- nitrophenol does not form H-bond with H2O but m-nitrophenol and p-nitrophenol form H-bonds with H2O. 8. Option (3) is correct.

Explanation: Phenol is stronger acid than ethanol as phenoxide ion is stabilized by resonance whereas no such stabilization occurs in ethoxide ion. Sodium ethoxide can be prepared by reaction of ethanol with sodium.

5. Option (1) is correct. Explanation:

9. Option (3) is correct. Explanation : Phenol forms 2,4,6-tribromophenol on treatment with bromine in water. In phenols, the polarization of bromine takes place even in the absence of Lewis acid. 10. Option (4) is correct. Explanation: Phenols give o- and p-nitrophenol on nitration with dil. HNO3 and with conc. HNO3, 2, 4, 6-trinitrophenol is formed. [C] COMPETENCY BASED QUESTIONS 1. Option (4) is correct. Explanation: In Williamson's synthesis, primary alkyl halide is always taken to produce ether. Hence, di-tert butyl ether cannot be made. 2. Option (3) is correct. Explanation: IUPAC name is 3-methoxy-1-propene. 3. Option (1) is correct. Explanation: Dehydration of alcohol takes place in the presence of conc. sulphuric acid at 413 K. Conc. sulphuric acid acts as a dehydrating agent.

6. Option (2) is correct. Explanation: Na metal is basic and alcohols are acidic in nature. Hence, reactivity of Na metal towards alcohols decreases as the acidic strength of alcohols decreases due to steric hinderance of alkyl groups in tertiary alcohol and increase in electron density on an oxygen atom in the hydroxyl bond. 7. Option (3) is correct. 8. Option (1) is correct. Explanation: The more stable the alkoxide ion, the more acidic is the alcohol. Electron releasing effect (+I effect) of alkyl group in secondary and tertiary alcohols makes the alkoxide ion less stable. 9. Option (3) is correct. Explanation:

10. Option (3) is correct. 4. Option (2) is correct. Explanation: Ether on reaction with HI gives alcohol and alkyl halides.

Explanation: Aromatic alcohols are more acidic because of stability of conjugate based due to resonance. While in aliphatic alcohols bigger alkyl group cause destabilisation of conjugate base because of increased +I effect. C6H5OH >H2O >CH3OH> C2H5OH

Study Time

CHAPTER

Max. Time: 1:50 Hours Max. Questions: 50

12

ALDEHYDES, KETONES AND CARBOXYLIC ACID containing group with hydrogen atom. Carbonyl group is attached with two alkyl or two aryl or one alkyl and one aryl group.  Preparation:

 Revision Notes

{ Ketones:

ALDEHYDES AND KETONES: { Carbonyl

group: Consist of a functional group C=O, where C atom is sp2 hybridized. { Aldehydes: Carbonyl group is attached to either two hydrogen atoms or one hydrogen atom and one carbon

(iii)

R—CH = CH—R' + O3

(iv) CH  CH+ H2O Ethyne (Acetylene)

O H 2 SO4 /HgSO4 333K

R — C — Cl + H2

Acyl chloride { Stephen

CH—R'

H2O, Zn – ZnO

O

OH | CH 2 = CH

R—CHO + R'—CHO Aldehyde

CH 3 —CHO Ethanal (Acetaldehyde)

Unstable

O (v)

Cu ¾¾ ® RCHO + H 2 (ii) R - CH 2 - OH ¾573 K

O

R—CH

Alkene

2 Cr2 O7 /H 2 SO 4 (i) RCH 2OH + [O] ¾K¾¾¾¾¾ ® R - CHO + H 2O or KMnO 4

O

Pd-BaSO4

R — C — H + HCl

Rosenmund

Aldehyde

Reduction

reaction:

2 2 + Cl RCN + H 2 ¾SnCl ¾¾ ® RCH = NH ¾H¾¾ ® RCHO + NH 4Cl HCl

{ Etard

reaction:

{ Gattermann

Koch reaction:

CHO

 Methods of preparation: (i) 2R - Mg - X + CdCl2 ® R 2Cd + 2Mg (X)Cl (ii)

2R'—C—Cl + R2Cd O

2R'—C—R + CdCl2 O

(iii)

(1-Phenylpropanone)

ALDEHYDES, KETONES AND CARBOXYLIC ACID

119

120 Oswaal CUET (UG) Chapterwise Question Bank R

(iv)

R'

CH—OH + [O] 2° Alcohol

R

(v)

(vi)

R'

2° Alcohol

R R

{ Friedel-Crafts

R'

R'

C = O + H 2O Ketone

R R'

C = O + H2 Ketone O

R

R' C=C

R

K2Cr2O7/H2SO4 or CrO3

Cu 573K

CH—OH

CHEMISTRY

+ O3

C

C

R

R' H2O, Zn – ZnO

R' O

O

RCOR + R'COR'

acylation reaction:

Ar/R

+ Ar/R

 Physical properties: {  The boiling points are higher than hydrocarbons and ethers of comparable molecular masses. {  Weak molecular association due to dipole-dipole interactions. { Solubility decreases with increase in carbon chain.

 Chemical properties: { Aldehydes are generally more reactive than ketones in nucleophilic addition reactions due to steric hindrance and electronic repulsions..

 Nucleophilic addition reactions: – ƒ HCN + OH– CN + H2O

ƒ

Tetrahedral intermediate

ƒ

C = O + NaHSO3

OSO 2H

C

ONa OMgX ƒ

C = O + RMgX

C

C

OSO 2Na OH

H 2O/H+

OH C

OH + Mg

R

R

X

OH ƒ

C = O + H2N—Z

C = N – Z + H2O

C NHZ

where Z = Alkyl, aryl, OH, NH2CONH2, etc. Aldehydes and ketones also react with ammonia derivatives like: ƒ NH2OH to give amines. ƒ NH2 – NH2 to give hydrazones. ƒ Phenyl hydrazine to give phenyl hydrazones.  Reduction: R R

C = O + H2

Ni or Pt

R R

CH—OH 2°Alcohol

 Ring substitution Aromatic aldehydes and ketones undergo electrophilic substitution reactions in which the carboxyl group acts as a deactivating and meta-directing group. COOH

COOH

Conc. HNO3+ Conc. H3SO4 NO2 m-Nitrobenzoic acid

121

ALDEHYDES, KETONES AND CARBOXYLIC ACID

 Name reactions: (i) Aldol condensation:



(ii) Cross aldol condensation:



(iii) Clemmensen reduction: C



HCl

CH2 + H2O

(iv) Wolff-Kishner reaction: C



Zn-Hg

NH2 NH2

O

–H2O

C

NNH2

KOH/ethyleneglycol

CH2 + N2

heat

Tollen's and Fehling test are used to distinguish between aldehydes and ketones. (v) Tollens' test: Ketone does not give this test. RCHO + 2[Ag ( NH 3 ) 2 ]+ + 3OH - ® RCOO - + 2Ag ¯ +2H 2O + 4 NH 3 (vi) Fehling test: ketone does not give this test.

Silver Mirror

R - CHO + 2Cu 2 + + 5OH - ® RCOO - + Cu 2O ¯ +3H 2O Red ppt

(vii) Cannizzaro reaction

H

—

O — — H—C—OH+H—C OK H Potassiumformate

—

H— H— — C=O+ — C = O + conc. KOH H H Formaldehyde

— —



O

Methylalcohol 2

— CHO + conc. KOH Benzaldehyde

— COONa

—CH2OH+ Benzyl alcohol

Sodiumbenzoate

 Test to distinguish: Test

Aldehydes

Ketones

1. Tollen’s test

Silver deposits

No reaction

2. Fehling’s test

Reddish brown deposits [except benzaldehyde]

No reaction

3. Schiff’s test

Pink colour is restored

No reaction

4. Iodoform test

Pale yellow crystals with those having keto methyl group

Pale yellow crystals with those having keto methyl group

CARBOXYLIC ACIDS:  Preparation: (i)

Scan to know more about this topic

Fehling's test Scan to know more about this topic

i ) Alk. KMnO 4 R - CH 2 - OH ¾(¾¾¾¾ ® R - COOH ( ii ) H O + 3

(ii)

R - CHO + [O] ¾¾¾¾¾¾ ® R - COOH O O (iii) H3O+ H3O+ R—C—NH2 R—C—OH R—C N O (iv) Dry ether H3O+ R—C—OMgX R—Mg— X + CO2 CH3 COOH K 2 Cr2 O7 / H 2 SO 4

(v)

Iodoform test O

Scan to know more about this topic

R—C—OH

alk.KMnO4 Toluene

Benzoic acid

Benzoic acid from toulene

122 Oswaal CUET (UG) Chapterwise Question Bank

CHEMISTRY

O C6H5—C O

H2O

CH3—C

C6H5COOH + CH3COOH Benzoic acid

Ethanoic acid

O



Benzoic ethanoic anhydride

CH3—CH2—CH2—COOCH2—CH3

(v)

NaOH

CH3 —CH2—CH2—COONa +

Ethyl butanoate

CH3CH2OH

+ H 3O

CH3CH2CH2COOH Butanoic acid

 Physical properties: ƒ Lower members are liquids with unpleasant odours. ƒ Higher acids are solids and are odourless, are less volatile. ƒ Have higher boiling point due to more association through intermolecular hydrogen bonding. ƒ Are miscible in water due to the formation of hydrogen bonds with water ƒ Solubility decreases with increasing number of carbon atoms. ƒ Benzoic acid is insoluble in cold water.  Chemical properties:

{ Esterification:

ƒ Carboxylic

acids are weak acids. liberate H+ ions on ionization ƒ Order of strength of Acids: HCOOH > CH3COOH > CH3CH2COOH > C3H7COOH CCl3COOH > CHCl2COOH > CH2ClCOOH > CH3COOH ƒ They

2R - COOH + 2 Na ® 2R - COONa + + H 2 Sodium carboxylate

R - COOH + NaOH ® R - COONa + + H 2O R - COOH + NaHCO3 ® R - COONa + + H 2O + CO 2 ­

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+ RCOOH + R ¢OH ¾H¾ ® RCOOR ¢ + H 2O

{ Reaction

with PCl5, PCl3 and SOCl2: RCOOH + PCl5 → RCOCl + POCl3 + HCl 3RCOOH + PCl3 → 3RCOCl + H3PO3 RCOOH + SOCl2 → RCOCl + SO2 ↑ + HCl ↑ { Reduction:

Esterification reaction

i ) LiAlH 4 /ether or B2 H 6 R - COOH ¾(¾¾¾¾¾¾ ¾ ® R - CH 2OH ( ii ) H O + 3

O COOH COOH

+ NH3

COO–NH4+

CONH2

COO–NH4

CONH2

– 2H2 O + ∆

C Strong heating

NH

–NH3

C Phthalamide

O Phthalimide

123

ALDEHYDES, KETONES AND CARBOXYLIC ACID { Hell Volhard

Zelinsky reaction:

Nitration COOH

Bromination COOH

COOH

Conc. HNO3 Conc. H2SO4

COOH

Br2/FeBr3 NO3

m-nitrobenzonic acid

Br m-bromobenzoic acid

{ Reduction

hydrocarbons when their sodium salts are heated with soda lime (NaOH and CaO in the ratio of 3 : 1). { Ring substitution in Benzoic acid: Aromatic carboxylic acids undergo electrophilic substitution reactions in which the carboxyl group acts as a deactivating and meta-directing group.

Carboxylic acids are reduced to primary alcohols by lithium aluminium hydride or better with diborane. { Decarboxylation Carboxylic acids lose carbon dioxide to form

OBJECTIVE TYPE QUESTIONS [A] MULTIPLE CHOICE QUESTIONS 1. Match List I with List II: List I

List II

(Nomenclature)

(Structure)

A. Acetophenone

CHO

I. B. Benzaldehyde

List I Given pair of organic compounds can be distinguished by

O

II.

C. Benzoic acid

C—CH3

O

III.

D. Benzophenone

C

COOH

IV.



3. Which of the following compound would undergo Aldol condensation? (1) Methanal (2) Benzaldehyde (3) 2, 2-Dimethylbutanal (4) Phenylacetaldehyde [CUET 2022, 10th Aug] 4. Match List I with List II:

Choose the correct answer from the options given below: (1) A – III, B – I, C – II, D – IV (2) A – II, B – I, C – IV, D – III (3) A – I, B – II, C – III, D – IV (4) A – IV, B – III, C – II, D – I [CUET 2022, 10th Aug] 2. Which simple chemical test is used to distinguish between ethanal and propanal? (1) Iodoform test (2) Tollen’s test (3) Fehling’s test (4) Lucas test [CUET 2022, 10th Aug]

List II

(A)

Ethanal/Propanal

I. Sodium Hydrogen carbonate test

(B)

Ethanol/Ethanoic acid

II. Fehling's Test

(C)

Butanal/Butan-2-one

III. Tollen's Test

(D) Benzaldehyde/Ethanal IV. Iodoform Test Choose the correct answer from the options given below: (1) A – IV, B – I, C – II, D – III (2) A – III, B – II, C – IV, D – I (3) A – IV, B – I, C – III, D – II (4) A – I, B – II, C – III, D – IV [CUET 2022, 17th Aug] 5. The major product of the reaction is: dil NaOH C6 H 5CHO + C6 H 5COCH 3 ¾(i)¾¾¾ ® __________ (ii) D

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

C6H5CH = CHC6H5 C6H5CH = CHCOC6H5 C6H5CH2OH + C6H5CH(OH)C6H5 C6H5CH2OH + C6H5COONa [CUET 2022, 18th Aug] 6. Benzaldehyde reacts with semi carbazide to give:

124 Oswaal CUET (UG) Chapterwise Question Bank (1)

+

–

N = NCl H

(2)

12. pKa value of

O

O

C = N – C – NHNH2 O

(4)

C – NH2

. BaSO 4 11. RCOCl + H 2 ¾Pd ¾¾¾ ® X + HCl

O

X can be: (A) CH3COCH3 (C)

(E)

CHO

(B)



is less acidic than

(3) O N 2

7. The presence of carbonyl group in glucose is confirmed by treating it with: (1) HI (2) Mild oxidising agent Br2 water (3) NH2OH (4) HNO3 [CUET 2022, 20th Aug] 8. Ethanal undergoes nucleophilic addition reaction at faster speed than acetone, this is because of  (1) Acetone has 6 alpha-hydrogen atoms. (2) Electrophilicity of carbon of carbonyl group is high in ethanal. (3) Electrophilicity of carbon of carbonyl group is less in ethanol. (4) Nucleophilicity of carbon of carbonyl group in ethanal is high. [CUET 2022, 20th Aug] 9. Aldehydes react with one equivalent of monohydric alcohols in the presence of dry hydrogen chloride to give: (1) An acetal (2) A hemiacetal (3) An oxime (4) A ketal [CUET 2022, 20th Aug] 10. The reagent which converts alkylbenzenes to carboxylic acids is: (1) CrO2Cl2 (2) CrO3 (3) Anhydrous AlCl3/CuCl (4) KMnO4 – KOH/Δ, HCl [CUET 2022, 21st Aug]

is less than of

because (1) NO2 is an electron donating group which increases electron- density on ring. (2) NO2 is an electron-withdrawing group which decreases electron- density on ring COOH

C = N – NH – C – NH2 H

(3)

CHEMISTRY

C – CH3

(D) CH3CH2COCH3

CHO

Choose the correct answer from the options given below: (1) A, B and C only (2) B, C and D only (3) C and E only (4) B and D only [CUET 2022, 23rd Aug]

COOH CH3N (4) Hyperconjugation is more in COOH NO2

[CUET 2022, 23rd Aug] 13. Write the correct IUPAC name of the compound given below CH3 – C– CH3 – CH3 – C – OH O O (1) 2-Oxopentanoic acid (2) 4-Formylpentanoic acid (3) 4-Oxopentanoic acid (4) 4-Onepentanoic acid 14. Which one of the following product is NOT formed when mixture of ethanal and propanol heated with dil. NaOH? (1) But-2-enal (2) Pent-3-enal (3) 2-Methylpent-2-enal (4) 2-Methylbut-2-enal [CUET 2022, 30th Aug] 15. Identify the product (A) formed in the following sequence of reaction:

O

(1)

OH

(2)

COOH COOH

(3)

COOH OCH3

OCOCH3

(4) COOH

[CUET 2022, 30th Aug] 16. Which of the following is the strongest acid? (1) Fluoro acetic acid (2) Trifluoroacetic acid (3) Chloroacetic acid (4) Difluoro acetic acid [CUET 2021, 23rd Sept] 17. The carboxylic acid that does not undergo HVZ reaction is: (1) CH3COOH (2) (CH3)2COOH (3) CH3CH2CH2CH2COOH (4) (CH3)3CCOOH

125

ALDEHYDES, KETONES AND CARBOXYLIC ACID

18. The reagent which does not react with both, acetaldehyde and benzaldehyde. (1) Sodium hydrogen sulphite (2) Phenyl hydrazine (3) Fehling's solution (4) Grignard reagent / I2 19. C6 H 5 - CO - CH 3 ¾NaOH ¾¾¾ ® ?+ ?

(1) C6H5COOH + CH4 (2) C6H5COONa + CHI3 (3) C6H6 + CH3COONa + HI (4) C6H5CH2COOH 20. Name the main compounds A and B formed in the following reaction: CH3 MgBr ( hG ) / conc. HCl CH 3CN ¾(i)¾¾¾¾ ® A ¾Zn ¾¾¾¾¾ ®B ( ii ) H O + 3

(1) CH3CH2COOH [A], CH3CH2CH3 [B] (2) CH3CH2CHO [A], C2H4 [B] (3) CH3COCH3 [A], CH3CH2CH3 [B] (4) CH3COCH3 [A], C2H6 [B] 21. Which of the following compounds is most reactive towards nucleophilic addition reactions? (1)

(2)

CH3—C—H

(3)

O C–H



(4)

CH3—C—CH3 O C – CH3

[NCERT Exemp. Q 2, Page 168] 22. The correct order of increasing acidic strength is : (1) Phenol < Ethanol < Chloroacetic acid < Acetic acid (2) Ethanol < Phenol < Chloroacetic acid < Acetic acid (3) Ethanol < Phenol < Acetic acid < Chloroacetic acid (4) Chloroacetic acid < Acetic acid < Phenol < Ethanol. [NCERT Exemp. Q 3, Page 168] 23. Compound can be prepared by the reaction of : (1) Phenol and benzoic acid in the presence of NaOH (2) Phenol and benzoyl chloride in the presence of pyridine (3) Phenol and benzoyl chloride in the presence of ZnCl2 (4) Phenol and benzaldehyde in the presence of palladium. [NCERT Exemp. Q 4, Page 169] 24. The reagent which does not react with both, acetone and benzaldehyde. (1) Sodium hydrogen sulphite (2) Phenyl hydrazine (3) Fehlings’ solution (4) Grignard reagent [NCERT Exemp. Q 5, Page 169] 25. CH3—C=CH

40%H2SO4 1%HgSO4

A

Isomersation

CH3—C=CH3



Structure of ‘A’ and type of isomerism in the above reaction are respectively : (1) Prop-1-en-2-ol, metamerism (2) Prop-1-en-1-ol, tautomerism (3) Prop-2-en-2-ol, geometrical isomerism (4) Prop-1-en-2-ol, tautomerism [NCERT Exemp. Q 8, Page 170] 26. Which of the following compounds will give butanone on oxidation with alkaline KMnO4 solution? (1) Butan-1-ol (2) Butan-2-ol (3) Both of these (4) None of these [NCERT Exemp. Q 11, Page 170] 27. In Clemmensen reduction carbonyl compound is treated with: (1) zinc amalgam + HCl (2) sodium amalgam + HCl (3) zinc amalgam + nitric acid (4) sodium amalgam + HNO3 [NCERT Exemp. Q 12, Page 170] 28. Through which of the following reactions number of carbon atoms can be increased in the chain? (1) Iodoform reaction (2) Cannizzaro reaction (3) Aldol condensation (4) HVZ reaction [NCERT Exemp. Q 16, Page 171] 29. Cannizzaro reaction is not given by: ____________ CHO (1) (3) HCHO



(2)

CHO

CH3

(4) CH3CHO [NCERT Exemp. Q 6, Page 169] 30. Compounds A and C in the following reaction are : Hydroboration

CH3 MgBr4 oxidation 2 SO 4 . A CH 3CHO ¾(i)¾¾¾¾ ®(A) ¾H¾¾¾ ®(B) ¾¾¾¾¾ ®(C) (ii) H2O

(1) identical (2) positional isomers (3) functional isomers (4) optical isomers [NCERT Exemp. Q 9, Page 170] [B] ASSERTION REASON QUESTIONS Directions: In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as: (1) Both A and R are correct and R is the correct explanation of A. (2) Both A and R are correct but R is NOT the correct explanation of A. (3) A is correct but R is not correct. (4) A is not correct but R is correct. 1. Assertion (A): Formaldehyde is a planar molecule. Reason (R): It contains sp2 hybridised carbon atom. [NCERT Exemp. Q 42, Page 175] 2. Assertion (A): Aromatic aldehydes and formaldehyde undergo Cannizzaro reaction.

126 Oswaal CUET (UG) Chapterwise Question Bank

Reason (R): Aromatic aldehydes are almost as reactive as formaldehyde. [NCERT Exemp. Q 45, Page 176] 3. Assertion (A): Aldehydes and ketones, both react with Tollen’s reagent to form silver mirror. Reason (R): Both, aldehydes and ketones contain a carbonyl group. [NCERT Exemp. Q 46, Page 176] 4. Assertion (A): Compounds containing –CHO group are easily oxidized to corresponding carboxylic acids. Reason (R): Carboxylic acids can be reduced to alcohols by treatment with LiAlH4. [NCERT Exemp. Q 43, Page 175] 5. Assertion (A): The α-hydrogen atom in carbonyl compounds is less acidic. Reason (R): The anion formed after the loss of α-hydrogen atom in carbonyl compounds is more acidic. [NCERT Exemp. Q 44, Page 176] 6. Assertion (A): Aromatic carboxylic acids undergo elec­ trophilic substitution reactions in the meta position. Reason (R): The carboxyl group exerts -R effect and acts as a deactivating group. 7. Assertion (A): All aldehydes and ketones undergo Aldol condensation in the presence of a base. Reason (R): Presence of alpha hydrogen is necessary for aldol condensation to occur. 8. Assertion (A): Electron withdrawing groups increase the acidity of carboxylic acids. Reason (R): Carboxylic acids are soluble in water. 9. Assertion (A): Aldehydes are more reactive than ketones. Reason (R): The more alkyl groups in aldehydes exhibit +I effect [electron donating effect]. 10. Assertion (A): Aldehydes and ketones are functional isomers. Reason (R): Both have same molecular formula but different functional groups. [C] COMPETENCY BASED QUESTIONS I. Based on following passage answer questions from 1-5: Compound ‘A’ was prepared by oxidation of compound ‘B’ with alkaline KMnO4. Compound A’ on reduction with lithium aluminium hydride gets converted back to compound ‘B’. When compound ‘A’ is heated with compound B in the presence of H2SO4 it produces fruity smell of compound C. [NCERT Exemp. Q 29, Page 173] 'A' is carboxylic acid (R–COOH), 'B' is an alcohol (R– CH2OH) and 'C' is an ester (RCH2–COOR) (KMnO4 ) R - CH 2OH ¾Oxidation ¾¾¾¾¾ ® R - COOH (B) (A) LiAlH 4 reduction Alcohol (B) ¬¾¾¾¾¾ ¾ Carboxylic acid (A)

O R—C—OH

O H2SO4 R—C—O—H2C—R H—OH3C—R (Eserification) Ester (C) (Fruity smell)

CHEMISTRY

1. The functional group present in compound A is: (1) Ester (2) Aldehyde (3) Ketone (4) Carboxylic acid 2. The compound formed when A reacts with B in the presence of H2SO4 is: (1) Ester (2) Aldehyde (3) Ketone (4) Ether 3. The role of H2SO4 in the reaction between A and B is that of : (1) An oxidising agent (2) A reducing agent (3) A dehydrating agent (4) A precipitating agent 4. Compound A reacts with NaHCO3. The correct statement regarding the reaction is: (1) White precipitate of CH3COONa is formed. (2) There is effervescence with the evolution of CO2. (3) Compound A gets reduced to an aldehyde. (4) Compound A gets reduced to an alcohol.. 5. Another reagent that can be used to oxidise compound B to A is: (1) Acidified K2Cr2O7 (2) Alkaline K2Cr2O7 (3) Zn-Hg/HCl (4) Soda lime II. Based on following passage answer questions from 6-10: An aromatic compound ‘P’ (Molecular formula C8H8O) gives positive 2, 4-DNP test forming a yellow precipitate of hydrazone, but does not give Tollen’s or Fehlings’ test for aldehydes Compound P also gives a yellow precipitate of Iodoform on treatment with iodine and sodium hydroxide solution also. On drastic oxidation with potassium per­ manganate compound P forms an aromatic carboxylic acid ‘R’ (Molecular formula C7H6O2), which gives a buff precipitate with neutral ferric chloride solution. 6. 2,4-DNP is a reagent used to detect the presence of the functional group: (1) Carboxylic acid (2) Ester (3) Carbonyl (4) Alcohol 7. The yellow compound formed when compound P reacts with I2/ NaOH is: (1) Iodoform (2) Picric acid (3) Aldol (4) Tri nitro toluene 8. Compound P is identified as: (1) Benzaldehyde (2) Benzoic acid (3) Benzophenone (4) Acetophenone 9. Tollen’s test is performed to detect the presence of: (1) Ketones (2) Aldehydes (3) Carboxylic acids (4) Esters 10. Compound R is identified as: (1) Phenol (2) Benzaldehyde (3) Benzoic acid (4) Toluene

127

ALDEHYDES, KETONES AND CARBOXYLIC ACID

ANSWER KEY [A] MULTIPLE CHOICE QUESTIONS 1. (2)

2. (1)

3. (4)

4. (3)

5. (2)

6. (2)

7. (2)

8. (2)

9. (2)

10. (4)

11. (3)

12. (1)

13. (3)

14. (2)

15. (3)

16. (2)

17. (4)

18. (3)

19. (2)

20. (3)

21. (1)

22. (3)

23. (2)

24. (3)

25. (4)

26. (2)

27. (1)

28. (3)

29. (4)

30. (2)

8. (2)

9. (3)

10. (1)

8. (4)

9. (2)

10. (3)

[B] ASSERTION REASON QUESTIONS 1. (1)

2. (3)

3. (4)

4. (2)

5. (4)

6. (1)

7. (4)

[C] COMPETENCY BASED QUESTIONS 1. (4)

2. (1)

3. (3)

4. (2)

5. (1)

6. (3)

7. (1)

ANSWERS WITH EXPLANATION 1. Option (2) is correct Explanation: Acetophenone: Ketone group:

CHO dil. Ph—CH2—CHO NaOH Phenylacetaldehyde A

Ph—CH2CH = C Ph

4. Option (3) is correct Benzaldehyde: Aldehyde group on benzene: CHO

Benzoic acid: Carboxylic acid on benzene: COOH

Benzophenone: Two benzene attached by ketone group O C

2. Option (1) is correct Explanation: Iodoform test is used to detect aldehydes and ketones having CH3CO− (methyl ketone) group. I2 CH 3CHO ¾NaOH ¾¾ ® CHl3 + HCOONa Å

Explanation: 1. Ethanal/Propanal – Iodoform test –Used to identify the presence of aldehyde or ketone in an unknown compound. 2. Ethanol/Ethanoic–acid - Sodium Hydrogen carbonate test – Used for the test of the carboxylic acid group. 3. Butanal/Butan-2-one Tollen’s Test- also known as the silver mirror test used to distinguish between Aldehyde and Ketone. 4. Benzaldehyde/Ethanal Fehling’s Test – For the detection of reducing and non-reducing sugar. 5. Option (2) is correct Explanation: Aldol condensation is an organic reaction in which an enolate ion reacts with a carboxyl compound in order to form a β–hydroxy aldehyde or β – hydroxy ketone. Hydroxide functions as a base and therefore moves the acidic α-hydrogen producing the reactive enolate ion. This reaction can be seen as an acid-base reaction.

Yellow Precipitate I2 CH 3CH 2CHO ¾NaOH ¾¾ ® No characteristic change

Tollen’s test, also known as silver-mirror test is used to distinguish between an aldehyde and a ketone. Fehling’s test is used for the estimation or detection of reducing sugars and non-reducing sugars. Lucas test is used to differentiate and categorise primary, secondary and tertiary alcohols. 3. Option (4) is correct Explanation: Aldehydes which contain at least 2 alpha hydrogen atoms, undergo Aldol condensation reaction Pheny­ lacetaldehyde contains 2 alpha hydrogen atoms and undergoes Aldol condensation reaction in presence of dilute alkali.

6. Option (2) is correct Explanation: When semi carbazide reacts with a ketone (or aldehyde) to form semi carbazone. Only one nitrogen atom of semi carbazide acts as a nucleophile and attacks the carbonyl carbon of the ketone. The product of the reaction consequently is R2C=N−NH−CONH2 rather than R2C=NCONH−NH2.

128 Oswaal CUET (UG) Chapterwise Question Bank 14. Option (2) is correct

C6H5CH=O + H2NNHCONH2 Benzaldehyde Semicarbazide C6H5

C6H5

H and

C

Explanation: H

C

N

N NHCONH2

NHCONH2

7. Option (2) is correct Explanation: Glucose reacts with hydroxylamine to give an oxime and adds a hydrogen cyanide molecule to give cyano­ hydrins. These reactions confirm the presence of a carbonyl group in glucose. Bromine water oxidises glucose to gluconic acid thus the reddish brown colour of bromine water is decolourised 8. Option (2) is correct Explanation: As the electrophilicity of carbon of carbonyl group is high in ethanal, it undergoes nucleophilic addition reaction at faster speed than acetone.

CH3CHO + CH3CH2CHO

NaOH ∆

15. Option (3) is correct

CH3CH = CH-CHO But-2-enal + CH3CH2CH = C(CH3)-CHO 2-Methylpent-2-enal + CH3-CH=CH(CH3)-CHO 2-methylbut-2-enal + CH3-CH=CH-CH2-CHO Pent-3-enal

Explanation: O (i) [Ag(NH3)2] + followed by H3O+ CHO

C—OH O

(ii) NABH4/H2O

R – CHO + R' – OH (1 eq)

R

dry HCl

OH

C—OH

C

O

R OR' (Hemiacetal)

10. Option (4) is correct.

O + (iii) (CH3CCO)2O/H3O

Explanation: 3

16. Option (2) is correct.

11. Option (3) is correct. Explanation: In the Rosenmund's reaction, acid chlorides are converted to corresponding aldehydes by catalytic reduction.

O

O

9. Option (2) is correct Explanation:

CHEMISTRY

O

C—CH3

COOH

Explanation: In CF3COOH, fluorine atom (an electron withdrawing group) is present on alpha position of the carboxyl group, withdraws electrons from the carbon of the carboxyl acid as well as from the oxygen of the O — H bond. This decreases electron density of the O — H bond. Thus, the O — H bond gets weakened and the release of H+ ion is favoured. Thus, trifluorocetic acid is the strongest acid. 17. Option (4) is correct.

CHO

H2 Cl Pd-BaSO4 Benzoyl chloride

Cl

+ H2

Benzaldehyde Pd-BaSO4 H + HCl

Explanation: Due to electron withdrawing nature of –NO2 which decreases electron density on the benzene ring. Lower is the pKa value, greater is the acid strength. The electron withdrawing nitro (−NO2) group increases the acidity of benzoic acid. 13. Option (3) is correct Explanation: 4

18. Option (3) is correct. Explanation: Aliphatic aldehydes(acetaldehyde) reduce the Fehling’s solution to red cuprous oxide. 19. Option (2) is correct.

12. Option (1) is correct

5

Explanation: The carboxylic acids having α-hydrogen atom undergo HVZ reaction. Since (CH3)3CCOOH does not contain α-H-atom; so, it does not undergo HVZ reaction.

3

2

1

CH3 – C – CH2 – CH2 – C – OH

IUPAC name 4-Oxopentanoic acid.

Explanation: / I2 C6 H 5COCH3 ¾NaOH ¾¾¾ ® C6 H 5COONa + CHl3

20. Option (3) is correct

129

ALDEHYDES, KETONES AND CARBOXYLIC ACID

21. Option (1) is correct.

28. Option (3) is correct

Explanation : CH3CHO is most reactive towards nucleophilic addition reactions. Carbonyl compounds are polar with positive charge on carbon atom which is attacked by nucleophiles. Two electron releasing alkyl groups in ketones make carbon less electron deficient than aldehydes. Benzene ring exhibits + R-effect which thereby decreases the ease of nucleophilic addition reaction in benzaldehyde and acetophenone. Hence, the reactivity order is: H H R

Explanation: Aldol condensation involves condensation of aldehydes and ketones with α- hydrogen







C=O > R

H





C=O >



C=O R

Explanation: Phenol is more stable than alcohol due to formation of more stable conjugate base after removal of H+ from phenol. CH3CH2OH