Oswaal IIT JEE Main Solved Papers Chemistry Chapterwise and Topicwise 2019 and 2020 All shifts 32 Papers

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CHEMISTRY

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1ST EDITION, YEAR 2020-21

ISBN SYLLABUS COVERED

“978-93-5423-008-0”

JEE (MAIN)

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PREFACE Logical thinkers, problem solvers, creators, engineers are definitely the most recognisable and valued people around the world. But many students preparing for JEE still ask; how good is engineering as a career option? Engineering has become the most lucrative profession now more than ever, with the demand for engineers exceeding their supply. Engineers bring ideas to life. In every field, engineers play a very important role in devising and executing new and better technologies. Thus, to take up engineering as a profession cracking the JEE (Main) exam is the first and the most important step. Every aspiring engineer in the country works diligently for this exam. This makes JEE a highly competitive exam. With the core objective of “Learning Made Simple” Oswaal Editorial Board has designed Oswaal JEE (Main) Solved Papers to help these aspiring candidates crack one of the most difficult examinations in the country. The book includes Previous Years’ Papers; to tell the students what to expect and familiarise them with the exam pattern that has been followed year on year along with the Scheme of Examination issued by the NTA on 16th Dec 2020.

Benefits of studying from this book are:

1. Based on the Scheme of Examination issued by the NTA on 16th Dec 2020 2. Fully Solved Papers of 2019 & 2020 JEE (Main) 3. Chapter-wise & Topic-wise presentation for systematic learning 4. Subjective (Integer Types) Questions for extensive practice 5. Concept Revision (Video Based) for hybrid learning 6. Commonly Made Errors to polish concepts 7. Mind Maps for better retention

8. Mnemonics for memorise the concepts.

This book aims to make students exam-ready, boost their confidence and help them achieve better results. We hope that OSWAAL JEE (Main) Solved Papers 2019 & 2020 will help you at every step as you move closer to your educational career goal. We wish you all great success ahead!

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TEAM OSWAAL

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CONTENTS g

Latest Syllabus for Academic Year 2021

g

10 Tips & Tricks to Crack JEE (Main)

13 - 13

g

Trend Analysis JEE (Main) 2019 & 2020

14 - 14

9 - 12

g Mnemonics

15 - 24

1. Some Basic Concepts in Chemistry

1 - 8

8. Redox Reactions and Electrochemistry 67 - 77

2. States of Matter

9 - 17



Topic 1. Redox Reactions



Topic 2. Electrochemistry



Topic 1. Gases and Liquid State



Topic 2. Solid State

3. Atomic Structure

9. Chemical Kinetics and Surface

18 - 27

Chemistry

78 - 94



Topic 1. Bohr’s Atomic Model, Sommerfeld Concept



Topic 1. Chemical Kinetics



Topic 2. Quantum Approach



Topic 2. Surface Chemistry



Topic 3. Photoelectric Effect

10. Classification of Elements and Periodicity

4. Chemical Bonding and Molecular Structure



in Properties

11. General Principles and Processes of

28 - 35



Topic 1. Development of Chemical Bonding



Isolation of Metals



Topic 2. Molecular Orbital Theory



Topic 1. Metallurgy



Topic 2. Ellingham Diagram

5. Chemical Thermodynamics

36 - 45



Topic 1. First Law of Thermodynamics, Reversible and Irreversible processes



Topic 2. Chemical Thermodynamics, Hess’s Law



Topic 3. 2nd Law of Thermodynamics

6. Solutions

46 - 55



Topic 1. Different Methods for Expressing Concentration of Solution



Topic 2. Colligative Properties of dilute



solutions, Van’t Hoff ’s Factor and



its significance

7. Equilibrium

56 - 66



Topic 1. Equilibria Involving Physical Process



Topic 2. Concept of Acid, Base, Ionic Equilibria, Hydrolysis of Salts, Buffer Solution

95 - 100

12. Hydrogen, s & p Block Elements

Topic 1. Hydrogen



Topic 2. s-block Elements



Topic 3. p-block Elements

101 - 107

108 - 118

13. d & f - Block Elements, and Coordination Compounds

Topic 1. d and f Block Element



Topic 2. Co-ordination Compounds

119 - 134

14. Environmental Chemistry

135 - 139

15. Purification, Basic Principles and Characteristics of organic compounds

140 - 148



Topic 1. Purification and Characterisation



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of Organic Compounds

... contd.

Topic 2. Some Basic Principles of Organic Chemistry

16. Hydrocarbons and their Halogen Derivatives

Topic 1. Alkanes, Alkenes, Alkynes



Topic 2. Halogen and their derivatives



Topic 1. Amines



Topic 2. Diazonium Salts

19. Polymers and Biomolecules 149 - 165



Topic 1. Polymers



Topic 2. Biomolecules

20. Analytical Chemistry and Chemistry in

17. Organic Compounds Containing Oxygen

Everyday life 166 - 185



Topic 1. Alcohols, Phenols and Ethers





Topic 2. Aldehydes and Ketones





Topic 3. Carboxylic Acids and its Derivatives



18. Organic Compounds Containing

186 - 199

What some preparation tips ? There's an ocean of knowledge waiting for you. Dive in by scanning the QR code.

SCAN THE CODE

Chemistry

Topic 2. Practical chemistry in inorganic chemistry

Topic 3. Chemistry in everyday life

220 - 232 qq

l Appendix



6 Tips to Prepare for JEE (Main) and Class 12th Board Exams Simultaneously

SCAN

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211 - 219

Topic 1. Practical Chemistry in Organic



Nitrogen

200 - 210

JEE (Main) Latest Syllabus for Academic Year (2021) CHEMISTRY

SECTION - A PHYSICAL CHEMISTRY aufbau principle, Pauli’s exclusion principle and Hund’s rule, electronic configuration of elements, extra stability of half- filled and completely filled orbitals.

Unit 1 : Some Basic Concepts in Chemistry



Matter and its nature, Dalton’s atomic theory; Concept of atom, molecule, element and compound; Physical quantities and their measurements in Chemistry, precision and accuracy, significant figures, S.I. units, dimensional analysis; Laws of chemical combination; Atomic and molecular masses, mole concept, molar mass, percentage composition, empirical and molecular formulae; Chemical equations and stoichiometry.

Unit 2 : States Of Matter Classification of matter into solid, liquid and gaseous states. Gaseous State: Measurable properties of gases; Gas laws- Boyle’s law, Charle’s law, Graham’s law of diffusion, Avogadro’s law, Dalton’s law of partial pressure; Concept of Absolute scale of temperature; Ideal gas equation; Kinetic theory of gases (only postulates); Concept of average, root mean square and most probable velocities; Real gases, deviation from Ideal behaviour, compressibility factor and van der Waals equaation. Liquid State: Properties of liquids- vapour pressure, viscosity and surface tension and effect of temperature on them (qualitative treatment only). Solid State: Classification of solids: molecular, ionic, covalent and metallic solids, amorphous and crystalline solids (elementary idea); Bragg’s Law and its applications; Unit cell and lattices, packing in solids (fcc, bcc and hcp lattices), voids, calculations involving unit cell parameters, imperfection in solids; Electrical and magnetic properties.



Unit 3 : Atomic Structure



Thomson and Rutherford atomic models and their limitations; Nature of electromagnetic radiation, photoelectric effect; Spectrum of hydrogen atom, Bohr model of hydrogen atomits postulates, derivation of the relations for energy of the electron and radii of the different orbits, limitations of Bohr’s model; Dual nature of matter, de- Broglie’s relationship, Heisenberg uncertainty principle. Elementary ideas of quantum mechanics, quantum mechanical model of atom, its important features. Concept of atomic orbitals as one electron wave functions; Variation of Y and Y2 with r for 1s and 2s orbitals; various quantum numbers (principal, angular momentum and magnetic quantum numbers) and their significance; shapes of s, p and d-orbitals, electron spin and spin quantum number; Rules for filling electrons in orbitals-

Unit 4 : Chemical Bonding and Molecular Structure Kossel - Lewis approach to chemical bond formation, concept of ionic and covalent bonds. Ionic Bonding: Formation of ionic bonds, factors affecting the formation of ionic bonds; calculation of lattice enthalpy. Covalent Bonding: Concept of electronegativity, Fajan’s rule, dipole moment; Valence Shell Electron Pair Repulsion (VSEPR) theory and shapes of simple molecules. Quantum mechanical approach to covalent Bonding: Valence bond theory - Its important features, concept of hybridization involving s, p and d orbitals; Resonance. Molecular Orbital Theory: Its important features, LCAOs, types of molecular orbitals (bonding, antibonding), sigma and pi-bonds, molecular orbital electronic configurations of homonuclear diatomic molecules, concept of bond order, bond length and bond energy. Elementary idea of metallic bonding. Hydrogen bonding and its applications.



Unit 5 : Chemical Thermodynamics





Fundamentals of thermodynamics: System and surroundings, extensive and intensive properties, state functions, types of processes First law of thermodynamics: Concept of work, heat internal energy and enthalpy, heat capacity, molar heat capacity; Hess’s law of constant heat summation; Enthalpies of bond dissociation, combustion, formation, atomization, sublimation, phase transition, hydration, ionization and solution. Second law of thermodynamics; Spontaneity of processes; ∆S of the universe and ∆G of the system as criteria for spontancity, ∆G0 (Standard Gibbs energy change) and equilibrium constant.

Unit 6 : Solutions

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Different methods for expressing concentration of solution- molality, molarity, mole fraction, percentage (by volume and mass both), vapour pressure of solutions and Raoult’s Law- Ideal and non-ideal solutions, vapour pressure composition, plots for ideal and non-ideal solutions; Colligative properties of dilute solutions - relative lowering of vapour pressure, depression of freezing point, elevation of boiling and osmotic pressure; Determination of molecular mass using colligative properties; Abnormal value of molar mass, van’t Hoff factor and its significance.

...contd. Unit 7 : Equilibrium Meaning of equilibrium, concept of dynamic

equilibrium. Equilibria involving physical processes: Solidliquid, liquid- gas and solid - gas equilibria, Henry’s law, general characteries of equilibrium involving physical processes. Equilibria involving chemical processes: Law of chemical equilibrium, equilibrium constants (Kp and Kc) and their significance, significance of ∆G and ∆G0 in chemical equilibria, factors affecting equilibrium concentration, pressure, temperature, effect of catalyst; Le Chatelier’s principle. Ionic equilibrium : Weak and strong electrolytes, ionization of electrolytes, various concepts of acids and bases (Arrhenius , Bronsted - Lowry and Lewis) and their ionization, acid- base equilibria (including multistage ionization) and ionization constants, ionization of water, pH scale, common ion effect, hydrolysis of salts and pH of their solutions, solubility of sparingly soluble salts and solubility products, buffer solutions.

Unit 8 : Redox Reactions And Electro Chemistry







Electronic concepts of oxidation and reduction, redox reactions, oxidation number, rules for assigning oxidation number, balancing of redox reactions. Electrolytic and metallic conduction, conductance in electrolytic solutions, molar conductivities and their variation with concentraton: Kohlrausch’s law and its application. Electrochemical cells- Electrolytic and Galvanic cells, different types of electrodes, electrodes potentials including standard electrode potential, half- cell and cell reactions, emf of a Galvanic cell and its measurement; Nernst equation and its applications; Relationship between cell potential and Gibbs’ energy change; Dry cell and lead accumulator; Fuel cells.

Unit 9 : Chemical Kinetics



Rate of a chemical reaction, factors affecting the rate of reactions; concentration, temperature, pressure and catalyst; elementary and complex reactions, order and molecularity of reactions, rate law, rate constant and its units, differential and integral forms of zero and first order reactions, their characteristics and half lives, effect of temperature on rate of reactions- Arrhenius theory, activation energy and its calculation, collision theory of bimolecular gaseous reactions (no derivation).

Colloidal state- distinction among true solutions, colloids and suspensions, classification of colloids - lyophilic, lyophobic; multimolecular, macromolecular and associated colloids (micelles), preparation and properties of colloids - Tyndall effect, Brownian movement, electrophoresis, dialysis, coagulation and flocculation; Emulsions and their characteristics.

SECTION - B INORGANIC CHEMISTRY Unit 11 : Classification Of Elements And Periodicity In Properties

Unit 12 : General Principles And Processes Of Isolation Of Metals

Adsorption- Physisorption and chemisorption and their characteristics, factors affecting adsorption of gases on solids- Freundlich and Langmuir adsorption isotherms, adsorption from solutions. Catalysis- Homogeneous and heterogeneous, activity and selectivity of solid catalysts, enzyme catalysis and its mechanism.



Modes of occurrence of elements in nature, minerals, ores; Steps involved in the extraction of metals- concentration, reduction(chemical and electrolytic methods) and refining with special reference to the extraction of Al, Cu, Zn and Fe; Thermodynamic and electrochemical principles involved in the extraction of metals.

Unit 13 : Hydrogen



Position of hydrogen in periodic table, isotopes, preparation, properties and uses of hydrogen; Physical and chemical properties of water and heavy water; Structure, preparation, reactions and uses of hydrogen peroxide; Classification of hydrides- ionic, covalent and interstitial; Hydrogen as a fuel.

Unit 14 : S-Block Elements (Alkali And Alkaline earth Metals)







Unit 10 : Surface Chemistry

Modern periodic law and present form of the periodic table, s, p, d and f block elements, periodic trends in properties of elements atomic and ionic radii, ionization enthalpy, electron gain enthalpy, valence, oxidation states and chemical reactivity.

Group - 1 and 2 Elements General introduction, electronic configuration and general trends in physical and chemical properties of elements, anomalous properties of the first element of each group, diagonal relationships. Preparation and properties of some important compounds- sodium carbonate and sodium hydroxide and sodium hydrogen carbonate; Industrial uses of lime, limestone, Plaster of Paris and cement; Biological significance of Na, K, Mg and Ca,

Unit 15 : P-Block Elements

( 10 )

Group - 13 to Group 18 Elements General Introduction : Electronic configuration and general trends in physical and chemical properties of elements across the periods and down the groups; unique behaviour of the first element in each group.

...contd. Groupwise study of the p- block elements Group - 13 Preparation, properties and uses of boron and aluminium; Structure, properties and uses of borax, boric acid, diborane, boron trifluoride, aluminium chloride and alums. Group - 14 Tendency for catenation; Structure, properties and uses of Allotropes and oxides of carbon, silicon tetrachloride, silicates, zeolites and silicones. Group - 15 Properties and uses of nitrogen and phosphorus; Allotrophic forms of phosphorus; Preparation, properties, structure and uses of ammonia, nitric acid, phosphine and phosphorus halides, (PCI3, PCI5); Structures of oxides and oxoacids of nitrogen and phosphorus. Group - 16 Preparation, properties, structures and uses of ozone; Allotropic forms of sulphur; Preparation, Properties, structures and uses of sulphuric acid (including its industrial preparation); Structures of oxoacids of sulphur. Group - 17 Preparation, properties and uses of hydrochloric acidl; Trends in the acidic nature of hydrogen halides; Structures of Interhalogen compounds and oxides and oxoacids of halogens. Group - 18 Occurrence and uses of noble gases; Structure of fluorides and oxides of xenon.

Unit 16 : d-and f - Block Elements Transition Elements

General introduction, electronic configuration, occurrence and characteristics, general trends in properties of the first row transition elements - physical properties, ionization enthalpy, oxidation states, atomic radii, colour, catalytic behaviour, magnetic properties, complex formation, interstitial compounds, alloy formation; Preparation, properties and uses of K2, Cr2, O2 and KMnO4 Inner Transition Elements Lanthanoids - Electronic configuration, oxidation states and lanthanoid contraction. Actinoids - Electronic configuration and oxidation states.

Unit 17 : Co-Ordination Compounds Introduction to co-ordination

compounds, Werner's theory; ligands, co-ordination number, denticity, chelation; IUPAC nomenclature of mononuclear co-ordination compounds, isomerism; Bonding-Valence bond approach and basic ideas of Crystal field theory, colour and magnetic properties; Importance of coordination compounds (in qualitative analysis, extraction of metals and in biological systems).

Unit 18 : Environmental Chemistry Environmental Pollution- Atmospheric, water and soil.

Atmospheric Pollution- Tropospheric and Stratospheric. Tropospheric pollutants- Gaseous pollutants: Oxides of carbon, nitrogen and sulphur, hydrocarbons; their sources, harmful effects and prevention; Green house effect and Global warming: Acid rain; Particulate pollutants- Smoke, dust, smog, fumes, mist; their sources, harmful effects and prevention. Stratospheric pollution- Formation and breakdown of ozone, depletion of ozone layer- its mechanism and effects. Water pollution- Major pollutants such as, pathogens, organic wastes and chemical pollutants; their harmful effects and prevention. Soil pollution- Major pollutants such as: Pesticides (insecticides, herbicides and fungicides), their harmful effects and prevention. Strategies to control enviromental pollution.

SECTION- C ORGANIC CHEMISTRY Unit 19 : Purification And Characterisation Of Organic Compounds

Purification- Crystallization, sublimation, distillation, differential extraction and chromatography- principles and their applications. Qualitative analysis- Detection of nitrogen, sulphur, phosphorus and halogens. Quantitative analysis (basic principles only)Estimation of carbon, hydrogen, nitrogen, halogens, sulphur, phosphorus. Calculations of empirical formulae and molecular formulae; Numerical problems in organic quantitative analysis.

Unit 20 : Some Basic Principles Of Organic Chemistry

Tetravalency of carbon; Shapes of simple molecules- hybridization (s and p); Classification of organic compounds based on functional groups: and those containing halogens, oxygen, nitrogen and sulphur; Homologous series; Isomerism structural and stereoisomerism. Nomenclature (Trivial and IUPAC) Covalent bond fission - Homolytic and heterolytic; free radicals, carbocations and carbanions; stability of carbocations and free radicals, electrophiles and nucleophiles. Electronic displacement in a covalent bond - Inductive effect, electromeric effect, resonance and hyperconjugation. Common types of organic reactions- Substitution, addition, elimination and rearrangement.

Unit 21 : Hydrocarbons

( 11 )



Classification, isomerism, IUPAC nomenclature, general methods of preparation, properties and reactions.

...contd. Alkanes - Conformations; Sawhorse and Newman projections (of ethane); Mechanism of halogenation of alkanes. Alkenes - Geometrical isomerism; Mechanism of electrophilic addition: addition of hydrogen, halogens, water, hydrogen halides ( Markownikoff ’s and peroxide effects); Ozonolysis and polymerization. Alkynes- Acidic character; Addition of hydrogen, halogens, water and hydrogen halides; Polymerization. Aromatic hydrocarbonsNomenclature, benzene- structure and aromaticity; Mechanism of electrophilic substitution; halogenation, nitration, Friedel- Craft’s alkylation and acylation, directive influence of functional group in monosubstituted benzene.

Unit 22 : Organic Compounds Containing Halogens

General methods of preparation, properties and reactions; Nature of C-X bond; Mechanisms of substitution reactions. Uses; Environmental effects of chloroform, iodoform freons and DDT.

Unit 23 : Organic Compounds Containing Oxygen

General methods of preparation, properties reactions and uses. ALCOHOLS, PHENOLS AND ETHERS Alcohols : Identification of primary, secondary and tertiary alcohols; mechanism of dehydration. Phenols : Acidic nature, electrophilic substitution reactions; halogenation, nitration and sulphonation, Reimer- Tiemann reaction. Ethers : Structure Aldehyde and Ketones: Nature of carbonyl group; Nucleophilic addition to >C=O group, relative reactivities of aldehydes and ketones; Important reaction such as - Nucleophilic addition reactions (addition of HCN, NH3 and its derivatives), Grignard reagent; oxidation; reduction (Wolff Kishner and Clemmensen); activity of α-hydrogen, aldol condensation, Cannizzaro reaction, Haloform reaction; Chemical tests to distinguish between aldehydes and Ketones. Carboxylic Acids Acidic strength and factors affecting it.

Unit 24 : Organic Compounds Containing Nitrogen

General methods of preparation, properties, reactions and uses. Amines : Nomenclature, classification, structure, basic character and identification of primary, secondary and tertiary amines and their basic character. Diazonium Salts : Importance in synthetic organic chemistry

Unit 25 : Polymers General introduction and classification of

polymers, general methods of polymerizationaddition and condensation, copolymerization. Natural and synthetic rubber and vulcanization:

some important polymers with emphasis on their monomers and uses- polythene, nylon, polyester and bakelite.

Unit 26 : Biomolecules General introduction and importance of biomol-

ecules. Carbohydrates - Classification: Aldoses and ketoses; monosaccharides (glucose and fructose) and constituent monosaccharides of oligosaccharides (sucrose, lactose and maltose). Proteins - Elementary Idea of -amino acids, peptide bond, polypeptides: Proteins: primary, secondary, tertiary and quaternary of proteins, eanzymes. Vitamins - Classification and functions. Nucleic acids - Chemical constitution of DNA and RNA. Biological functions of nucleic acids.

Unit 27 : Chemistry In Everyday Life

Chemical in medicines - Analgesics, tranquilizers, antiseptics, disinfectants, antimicrobials, antifertility drugs, antibiotics, antacids, antihistamins - their meaning and common examples. Chemical in food - Preservatives, artificial sweetening agents- common examples. Cleansing agents - Soaps and detergents, cleansing action.

Unit 28 : Principles Related To Practical Chemistry  Detection of extra elements (N, S, halogens) in

organic compounds; Detection of the following functional groups: hydroxyl (alcoholic and phenolic). carbonyl (aldehyde and ketone). carboxyl and amino groups in organic compounds.  Chemistry involved in the preparation of the following: Inoganic compounds: Mohr’s salt, potash alum. Organic compounds: Acetanilide, p-nitroacetanilide, aniline yellow, iodoform.  Chemistry involved in the titrimetric excercisesAcids bases and the use of indicators, oxalic- acid vs KMnO4, Mohr’s salt vs KMnO4.  Chemical principles involved in the qualitative salt analysis: Cations- Pb2+, Cu2+, Ai3+, Fe3+, Zn2+, Ni2+, Ca2+, Ba2+, Mg2+, NH +4 Anions- CO32 − S2- SO2 − NO3− NO − Cl–, Br–, I– 4 , 2 , , (Insoluble salts excluded)  Chemical principles involved in the following experiments: 1. Enthalpy of solutions of CuSO4 2. Enthalpy of neutralization of strong acid and strong base. 3. Preparation of lyophilic and lyophobic sols. 4. Kinetic study of reaction of iodide ion with hydrogen peroxide at room temperature. qq



( 12 )

10 TIPS & tricks TO CRACK JEE (main) EXAM IN THE FIRST ATTEMPT

Joint Entrance Examination or JEE (Main) is conducted by NTA. After clearing JEE (Main), candidates will be eligible to apply for admission to IITs, NITs, CFTIs and almost all prestigious engineering colleges. Cracking the JEE (Main) Exam in the very first attempt; given the difficulty level, can be a tough task. But it is quite attainable if done diligently as well as smartly. Here we are giving you 10 tips that you must follow by heart to crack the exam in the very first attempt:

1. Start Studying from The Beginning

All the aspirants are aware of how vast, comprehensive and detailed the syllabus of the JEE (Main) exam is. To crack the exam in the first attempt you have to start preparing for the exam from the beginning of Class 12th. It is only then that you will be able to complete the entire syllabus. Following this approach will also leave you with plenty of time to revise.

2. Prioritize & Plan

Devise a study plan which is realistic rather than being idealistic. Try to cover a certain number of to ics ever day.

3. Formulas Are Very Important

For JEE (Main) 2021, some of the questions will be based on direct formulas and thus students should remember them. A simple trick to remember all the formulas are that write all the formulas on a paper and paste them in front of study table, it will be helpful to revise anytime.

4. Get the Right Tools and Study Material

Collecting and preparing from the appropriate study material is something that you can’t ignore. You should refer to Oswaal JEE (Main) 15 Mock Test papers to boost your preparation. The books you choose to study from should be on the lines of the current syllabus and the ones that could be trusted on before examination.

5. Practice Important Topics & Strengthen Your Weak Areas

You must analyse the detailed syllabus of all the sections and then start preparing first for the important topics, which are frequently questioned upon in the previous years’ exams. Try to cover those topics which are your weak areas and then cover the ones which are your strength.

6. Understand the Concepts

No one can crack the JEE (Main) exam just by mugging up all the concepts and topics. The syllabus of the exam is in-depth and to achieve good scores you need to understand every concept.

7. Revise Whenever You Get Time

Make sure you revise as much as possible. The revision will help you in keeping the concepts fresh in your mind until the day of the final examinations. You may refer to a few good Question Banks, Sample Papers and your self-made notes for this purpose.

8. Keep A Track on Time

While you are solving papers, make sure you keep a track on time i.e., how much time does it take to solve one sections and the type of questions which take minimum and maximum time.

9. Exam Day Strategy

First & foremost, try to be in time at the exam centre, it will help you keep yourself calm. Scoring good marks is all about identifying the questions which you should be attempting first and the ones which are to be solved in the second round. Always try to attempt those questions first which seems familiar, less time consuming and easy.

10. Keep Yourself Motivated & Healthy Don’t be Anxious, keep yourself Calm! Taking care of your thought process and keeping it positive is the first and the best course of action that one is required to take. This time is very important for you, so is everything you are eating or thinking. Eat healthy and easy to digest food and take proper sleep. Always remember that to achieve good scores you will need consistent efforts and calm mind. Trust on your honest efforts, if in case at any point of time you feel stressed, don’t be hesitant in taking help from counsellors or family members. Your focus should only be on clearing the JEE exam and to give your best shot.

All the Best!!



( 13 )

TREND ANALYSIS JEE (MAIN) 2019 & 2020 CHEMISTRY Total Papers Chap. No.

Chapter Name

8

8

6

10

Jan 2019

Apr 2019

Jan 2020

Sep 2020

Phase - 1

Phase - 2

Phase - 1

Phase - 2

1

Some Basic Concepts in Chemistry

4

4

5

11

2

States of Matter

10

8

4

8

3

Atomic Structure

9

10

4

9

4

Chemical Bonding & Molecular Structure

4

6

4

8

5

Chemical Thermodynamics

12

9

6

7

6

Solutions

11

10

6

7

7

Equilibrium

10

10

7

13

8

Redox Reactions and Electrochemistry

9

10

8

14

9

Chemical Kinetics and Surface Chemistry

15

15

10

21

10

Classification of Elements and Periodicity in Properties

5

4

6

11

11

General Principles and Processes of Isolation of Metals

8

9

4

7

12

Hydrogen, s & p - Block Elements

21

20

15

23

13

d & f - Blocks Elements and Coordination Compounds

19

25

19

23

14

Environmental Chemistry

13

9

2

6

15

Purification, Basic Principles and Characteristics of Organic Compounds

11

18

8

13

16

Hydrocarbons and their Halogen Derivatives

8

16

12

12

17

Organic Compound Containing Oxygen

31

19

10

25

18

Organic Compound Containing Nitrogen

12

14

5

7

19

Polymers and Biomolecules

12

16

7

13

20

Analytical Chemistry and Chemistry in Everyday life

16

8

8

12

240

240

150

250

Total Questions



( 14 )

M → Motion (Gas Particle are in Random Motion) S → Size (negligible size of particle to total volume)

Chapter - 1 Some Basic Concepts in Chemistry 1.

P → Pressure (Pressure exerted due to Collision with walls of container) E → Elastic Collision A → Attractive forces are not present K → K.E ∝ Temp S → Speed (Distribution of speed of particles remain const.)

Metric System

The Great Morning King Henry Doesn't Usually Drink chocolate Milk Mixed with Natural Powder The → Tera (×1012) Great → Giga (×109) Morning → Mega (×106) King → Kilo (×103) Henry → Hecto (×102) Doesn't → Deca (×10) Usually → Unit (×1) Drink → Deci (×10–1) Chocolate → Centi (×10–2) Milk → Milli (×10–3) Mixed with → Micro (×10–6) Natural → Nano (×10–9) Powder → Pico (×10–12)

5.

Cu Te MOTHe R 3224 Unit Cell - Cubic, Tetragonal, Monoclinic, Orthorhombic, Triclinic, Hexagonal, Rhombohetral Edge Length - a=b=c, a=b≠c, a≠b≠c, a≠b≠c, a≠b≠c, a=b≠c, a=b=c Axial Length - a=β=g, a=β=g, a=β≠g, ∝=β=g, a≠β≠g, a=β≠g, a=β=g No. of Bravias Lattice - 3, 2, 2, 4, 1, 1, 1 6.

Triclinic, Orthorhombic, Monoclinic (a≠b≠c) Hexagonal, Tetragonal (a=b≠c) Cubic, Rhombohetral (a=b=c)

Gas Law's

PTV (letters that touches are directly proportional & letter don't are indirectly proportional)

7.

Tetragonal, Orthorhombic, Rhombohedral, Cubic (a=β=g) Hexagonal, Monoclinic (∝=β≠g) Triclinic (∝≠β≠g)

Const terms in Gas Laws

Chapter - 3 Atomic Structure

Paid TV Can Be Good 1.

Const terms → Pressure (P) Temp (T) Volume (V) Gas Law → Boyle's (Gay-Lussac's) 3.

Atomic No & Mass No

APEMAN

Ideal Gas Behavior

Atomic No. = No. of Protons = No. of Electrons Mass No. = Atomic No. + No. of neutrons

PLIGHT High temp & Low pressure to achieve ideal Gas behavior PL → Pressure Low IG → Inert Gas HT → High Temp 4.

Axial Angles

TORC Has More (HM) Twists (T)

1  [P∝T], [V∝T], P ∝  V  

2.

Edge Length

TOM Handpicked Tag (HT) of Class Representative (CR)

Chapter - 2 States of Matter 1.

Crystal System

Kinetic Theory of Gas

Mother SPEAKS ( 16 )

2.

Isotopes, Isobars & Isotones

Bring Top Talented MAN (BTT MAN) Atoms having same Isobars → Mass Number Isotopes → Atomic Number Isotones → Neutrons

3.

Electromagnetic Spectrum

8.

Roman Men Invented Very Unusual X-ray Gun

SPAM S → Spin Quantum no. (mS)

Roman → Radiowaves Men → Microwaves Invented → IR waves Very → Visible rays Unusual → UV waves X-ray → X-rays

P → Principal Quantum no. (n) A → Azimuthal Quantum no. (l) M → Magnetic Quantum no. (m2) 9.

Sequence of orbitals

Sober Physicists Don't Find Giraffes Hiding In Kitchen s,p,d,f,g,h,i,k

Gun → g -rays (Gamma rays) 4.

Quantum Numbers

Visible Region of EMR

VIBGYOR Chapter - 4 Chemical Bonding & Molecular Structure

Visible (a) Violet (b) Indigo (c) Blue (d) Green (e) Yellow (f) Orange (g) Red 5.

1.

For Very Lovely Son! [Formal Charge (F.C) = Valence e– in free state (V.E) – Lone pair (l.p) –1/2 × Shared e– (S.E)]

Planck's Quantum theory

Employee's Provident Fund (EPF) 2.

Energy = Planck's constant × Frequency [E = hr] 6.

H-atom spectral lines

Fluorine, Oxygen, Nitrogen 3.

H2, N2, F2, O2, I2, Cl2, Br2 4.

Electronic video Recording (EVR)

Ionic > Covalent > H–bonds > Dipole > Vanderwaal 5.

Z2 Energy (E) ∝ 2 n z n

Radius

∝

n2 z2

Chemical Bond Strength

I can't Handle Dirty Vans

Bohr Model of an atom

∝

Diatomic Molecules

Have No Fear of Ice Cold Beer

Lyman (n1=1) Balmer (n1=2) Paschen (n1=3) Brackett (n1=4) Pfund (n1=5)

Velocity

H–bonding

H-bonding is FON (Fun)!

Myan Mer Pasta Bread Fund

7.

Formal Charge

Bond Polarity

SNAP Symmetrical → Non Polar Asymmetrical → Polar 6.

Hybridisation

(VMCA)

( 17 )

Chapter - 6 Solutions

Steric No. = 1/2 [V+M–C+A] V → Valence e– of central atom M → Monovalent atoms (H/X) C → Cationic Charge A → Anionic Charge

Ideal & Non ideal Solutions

HIV

Chapter - 5 Chemical Thermodynamics 1.

Ideal Enthalpy (∆H) ∆H=0 Intermolecular A–A & B–B is Forces same as A–B Volume (∆V) ∆V=0

Process Boring ACT

Peer's Hard Verified Test

Chapter - 7 Equilibrium

Process ISO Bar Adiabatic ISO Choric ISO Therm Const → Pressure (P) Heat (q) Volume (V) Temp (T) 2.

State Function

Bronsted Acid-Base Concept

PVT HUGS

Strong Army, Lost to Carelessly Weak Bandits

Pressure, Volume, Temp, Enthalpy (H), Internal Energy (U), Gibbs free energy (G) Entropy (S) 3.

4.

Strong Acid gives Weak Conjugate Base

I Work Hard

Chapter - 8 Redox Reactions and Electrochemistry

Change in internal energy (U) = Work (w)+Heat (q)

1.

First law of Thermodynamics

Loss of e– is oxidation Gain of e– is reduction

CP–CV = R

2.

Criteria of Spontaneity

Reduction at Cathode

(dH)S,P Calcium > Magnesium > Aluminium > Zinc > Chromium > Iron > Lead > Copper > Mercury (Hg) > Silver > Gold

T

Gibb's Free Energy

5.

Metal activity series

Get High Test Scores

FAT CAT

DG = DH – TDS

Flow of e– from anode to cathode

( 18 )

6.

Metal activity series Amount of Hundred Ceins Balancing Half Cell Steps : (1) Atoms (2) Oxygen (3) Hydrogen (4) Charge

7.

Medium

Solid Sol

Solid

Solid

Solid Sol

Solid

Liquid

Sol

Solid

Gas

Aerosol

Solid

Gel Emulsion

Liquid

Gas

Aerosol

Electro Chemical Series

Gas

Solid

Solid Sol

Priyanka Chopra Sees Movie About Zebra In The Libya Hiring Cobra Studying Algebra

Gas

Liquid

Foam

Chapter - 10 Classification of Elements and Periodicity in Properties 1.

Elements of Atomic No (1-18)

Happy Harry Listen BBC Network Over French Network. Native Magpies Always Sit Peacefully Searching Clear Areas

For Galvanic Cell

H, He, Li, Be, B, C, N, O, F, Ne, Na, Al, Si, P, S, Cl, Ar 2.

Electrolytic Cell

Group I Elements

Little Nasty Kids Rub Cats Fur

LOAP

Lithium (Li), Sodium (Na), Potassium (K), Rubidium (Rb) Caesium (Cs), Francium (Fr)

Loss of e Oxidation Anode Positive –

3.

Group II Elements

Beer Mugs Can Snap Bar's Reputation

Chapter - 9 Chemical Kinetics and Surface Chemistry

Beryllium (Be), Magnesium (Mg), Calcium (Ca), Strontium (Sr), Barium (Ba), Radium (Ra) 4.

Mechanism of Heterogeneous Catalysis

Group III Elements

BAGIT

RAID Program

Boron (B), Aluminium (Al), Gallium (Ga), Indium (In), Thallium (Tl)

(a) Reactant diffusion on surface (b) Adsorption of Reactant (c) Intermediate formation (d) Desorption of product (e) Product leaves the surface 2.

Phase

Liquid

Loss of e– Oxidation Anode Negative

1.

Type of Colloids

Liquid

LOAN

9.

Dispersion

Liquid

Potassium < Calcium < Sodium < Magnesium < Aluminium < Zinc < Iron < Tin < Lead < Hydrogen < Copper < Silver < Gold (Au) 8.

Dispersed

5.

Group IV B Elements

Can Simple Germans Surprise Public Carbon (C), Silicon (Si), Germanium (Ge), Tin (Sn), Lead (Pb)

Types of Colloids

Soft SAGE And Shredded Face (SSAGEASF)

6.

Group V B Elements

New Police Assign Subordinate Bikram on Duty ( 19 )

Nitrogen (N), Phosphorus (P), Arsenic (As), Antimony (Sb), Bismuth (Bi) 7.

(a) Concentration of Ore (b) Isolation (c) Purification

Group VI B Elements

2.

Old Sahranpur Seems Terribly Polluted

Honest Man Feeling Low (HMFL)

Oxygen (O), Sulphur (S), Selenium (Se), Tellurium (Te), Polonium (Po) 8.

(a) Hydraulic Washing (b) Magnetic Separation (c) Froath Floatation Method (d) Leaching

Group VII B Elements

First Class Biryani In Australia

9.

3.

Conversion to Oxide

Fluorine (F), Chlorine (Cl), Bromine (Br), Iodine (I), Astatine (Al)

CRAP

Group VIII B/18 Elements

Calcination → Absence of O2 Roasting → Presence of O2

He never Arrived; Karan exited with Rohan

4.

Ores

MISH Prime Minister Going China

Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Ex), Radon (Rn)

Iron ores → Magnetite, Iron pyrites, Siderite, Haematite Copper ores → Copper pyrites, Malachite, Copper Glance, Cuprite

10. 3d-Series

Scary Tiny Vicious Creatures are Mean females come to Night Club Zen

Chapter - 12 Hydrogen, s & p-Block Elements Hydrogen

Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn, 11. 4d-Series

Yes S(z)sir Nob. Most Technicians Rub Rod's Pale Silver Cadillac

1.

Isotopes of Hydrogen

Pro-Diabetic Treatment PDT)

Y, Zr, Nb>Mo, Tc, Ru, Rh, Pd, Ag, Cd 12. 5d-Series

Protium

Late Harry Took Walk, Reached Office In Pajamas After an Hour

1   H 1 

2  Deuterium  H  1 

La......., Hf, Ta, W, Re, OS, Ir, Pt, Au, Hg

Tritium

13. Lanthanides

Ladies Can't Put Needles Properly in Slot-machnies. Every Girl Tries Daily However, Every Time You'd be lose

2.

3   H 1 

H–Bonding

iso FON ! (Say Fun)

La, Ce, Pr, Nd, Pm, Sm, Eu Gd, Tb, Dy, Ho, Er, Tm, Yb, Lu

Fluorine, Oxygen, Nitrogen 3.

Chapter - 11 General Principles and Process of Isolation of Metal 1.

Concentration of Ore

Hardness of Water

CM is temporarily hard with Head Clerks (HC) but permanently Temporary hardness due to Mg(HCO3)2, Ca(HCO3)2

Process of Metallurgy

Permanent hardness due to MgCl2, CaCl2, MgSO4, – 2– CaSO4 Hard with civil servants (CS) Cl , SO

CIP (Read opp PIC) ( 20 )

hydrogen Carbonate (HCO3–)

s-block 4.

elements

Ion

Group I Elements

Mn2+

Little Nasty Kids Ruts Cats Far

Fe2+/Fe3+

Lithium (Li), Sodium (Na), Potassium (K), Rubidium (Rb), Caesium (Cs), Francium (Fr) 5.

Cement Modified Soil (CMS) Oxidised

Green

→

FYG

Grey

→

NBG

Cu2+

Blue

Red

→

CBR

→

CDD

Deep Blue Deep Blue

elements

New Police Assigns Subordinate Bikram on duty Nitrogen (N) Phosphorus (P) Arsenic (As) Antimony (Sb) Bismuth (Bi)

Old Sahranpur Polluted

Sodium ion(Na+) is oxidised ACC → reduced→Anode of carbon on which Cl is reduced –

(Na/Li + liq.NH3) – (Reducing in nature, Paramagnetic, conducting, Coloured)

Seems

Terribly

Oxygen (O) Sulphur (S) Selenium (Se) Tellurium (Te) Polonium (Pu)

Properties of Birch Reagent

Roman People Can Commute (RPCC)

9.

MPC

12. Group 16 Elements

Cathode Mercury (Hg) on which

8.

Yellow

→

11. Group 15 Elements

Castner Kellnar Cell

p-block

Colour less

Brown

Co2+

Beryllium (Be), Magnesium (Mg), Calcium (Ca), Strontium (Sr), Barium (Ba), Radium (Ra)

7.

Pink

Ni2+

p-block

Group II Elements

Beer Mug Can Snape Bar's Reputation

6.

Oxidising Flame Reducing Flame

13. Group 17 Elements

First Class Biryani In Australia

elements

Fluorine (F)

Group 13 Elements

Chlorine (Cl)

BAGIT

Bronine (Br)

Boron (B), Aluminium (Al), Gallium (Ga), Indium (In), Thallium (Tl)

Astatine (At)

Iodine (I)

14. Group 18 Elements

Group 14 Elements

He Never Arrived; Karan exited with Rohan

Can Simple Germans Surprise Public Carbon (C), Silicon (Si), Germanium (Ge), Tin (Sn), Lead (Pb),

Helium (He) Neon (Ne) Argon (Ar)

10. Borax bead Test

Krypton (Kr)

Multiple Program Combined (MPC) for Your Growth (FYG). New Boys Get (NBG) Common Boys Room (CBR) for Combining Desktop Drawing (CDD)

Xenon (Xe)

Chapter - 13 d & f block elements and Coordination Compounds

( 21 )

1.

3d-Series

I Bought Some Copies to Study Fundamental of Chemistry He Nutured Excellence in Necessary Coordination Compounds

Scary Tiny Vicious Creatures are Mean; Females Come to Night Club Zen Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn 2.

I– < Br– < SCN– < Cl– < S2– < OH– < C2O42– < H2O < NCS– < EDTA4– < NH3 < CN– < CO

4d-Series

I– = I

Yes S(z)ir, Nob Most Technicians Rub Rod's Pale Silver Cadillac

Br– = Brought SCN– = Some

Y, Zr, Nb, Mo,Tc, Ru, Rh, Pd, Ag, Cd 3.

Cl– = Copies to

5d-Series

S2– = Study

Late Harry Took Walk, Reached Office In Pajamas After an Hour

F = Fundamental OH– = Of C2O42– = Chemistry

La....... Hf, Ta, W, Re, Os, Ir, Pt, Au, Hg 4.

H2O = He

Lanthanides

NCS– = Nutured

Ladies Can't Put Needles Properly is Slot-machines. Every Girl Tries Daily, However, Every Time You'd be Lose

EDTA4– = Excellence in NH3 = Necessary CN– = Coordination CO = Compounds

La, Ce, Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy, Ho, Er, Tm, Yb, Lu 5.

9.

Common League People win Hearts Vice-Versa

Spectrochemical Series

I Bought Some Copies to Study Fundamental of Chemistry. He Nurtured Excellence in Necessary Coordination Compound

(i) CFSE (∆0) < Pairing Energy (P.E) Ligand → Weak field ligand Type of complex → High spin Complex Pairing of e– in t2g orbital

I < Br– < SCN– < Cl– < S2– < OH– < C2O42– < H2O < NC5– < EDTA4– < NH3 < CN– < CO 6.

Pairing of e– Octahedral Complexes

(ii) CFSE (∆0) < Pairing Energy (P.E) 10. Werner's theory

PIcturesque SNow

Common League People win Hearts

Primary valency → Ionisable i.e., Charge on (PIcturesque) Ionisation sphere

CFSE (∆0) < Pairing Energy Ligand → Weak field Type of complex → High spin Pairing of e– in t2g orbital 7.

Secondary valency → Non (SNow) Coordination number

Werner's theory

Ionisable

i.e.,

Chapter - 14 Environmental Chemistry

PIcturesque SNow Primary valency → Ionisable (Charge on Ionisation sphere) Secondary valency → Non Ionisable (Coordination Number) 8.

Pairing of e– Octahedral Complexes

Spectrochemical series

( 22 )

1.

Gases air Pollutants

HOSCN Hydrocarbons, Oxides of Sulphur (SO2, SO3), Carbon (CO, CO2), Nitrogen (NO, NO2)

2.

Chapter - 16 Hydrocarbons and their Halogen Derivatives

Components of Photochemical Smog

O FAN PAN Ozone, Formaldehyde, Acrolein, Nitric oxide, PAN

1.

Queen Elizabeth Second's Navy Commands, Controls

Chapter - 15 Purification, Basic Principles and Characteristics of Organic Compounds 1.

Qiatonary,

Functional group preference order

2.

ASEHA NAKAA Delhi Training Camp

Alkyl (–R) Halogen (–X) Alkoxyl (–OR) Amino (–NH2, NHR, NR2) Hydroxyl (–OH) Amide (–NHCl) Phenyl (C6H5)

No Preference Functional Group

NAHE Nitro, Alkyl / Aryl, Halo, Ethers 3.

4.

3.

Carbon Chain

CURT-I

Meth, Eth, Prop, But

Carbocation Intermediate Unimolecular Reaction Racemic mixture is obtained Two step process Ist order kinetics

3-D Representation

Solid → Towards observer (

6.

4.

)

Chirality

Dashed → Away from observer (IIIII)

CANS

Types of Organic Reaction

Chiral → Non-Super imposable mirror Images

EARS

Achiral → Super imposable Mirror Images

(a) Elimination (b) Addition (c) Rearrangement (d) Substitution

Chapter - 17 Organic Compound Containing Oxygen 1.

Structural Isomerism

TASte → Tollen's test, Aldehyde group, Silver Mirror

(a) Position (b) Functional Group (c) Metamerism (d) Chair

FAAR → Fehling's test, Aliphatic Aldehyde, RedBrown ppt IMLY → Iodoform test, Methyl group linked to O , Yellow ppt –C–

Optical Isomerism

=

GO (a) Geometrical (b) Optical

Detection test

TASte FAAR IMLy

Poor Farmer Managing Crops (PFMC)

7.

SN1 reaction

Monkey Eat Peeled Bananas

So towards Do away

5.

o, p–directing

AHA AHA P

Carboxylic Acid > Sulphonic Acid > Ester > Acid Halides > Acid Amides >Nitrile > Aldehyde > Ketone > Alcohol > Amnes = > = 2.

m-directing Group

( 23 )

2.

Common Names of Carboxylic Acid

Sucrose → Glucose + Fructose

Frog Are Polite, Being Very Courteous

Non-Reducing Sugar 2.

Formic, Acetic, Propionic, Butyric, Valeric, Caproic 3.

PVT TIM HALL

Dicarboxylic Acid

(Phenylalanine, Valine, Threonine, Tryptophan, Isoleucine, Methionine, Histidine, Arginine, Leucine, Lysine)

Oh My, Such Good Apple Pie, Sweet As Sugar 3.

Oxalic, Malonic, Succinic, Glutaric, Adipic, Pimelie, Subric, Azelaic, Sebacic 4.

Rest all Vitamins are water Soluble 4.

Reaction to convert –C– to alkane Clemmen → son → Zn–Hg/HCl

DNA A=T, G≡C

Wolf → Reaction → NH2–NH2/OH–

(2 H–bonds b/w Adenine & Thymine 3 H–bonds b/w Guanine & Cytosine)

Chapter - 18 Organic Compounds Containing Nitrogen

G≡C A=T Also, GCAT

Chapter - 20 Analytical Chemistry and Chemistry in Everyday life

Carbylamine test

PAFSI (Say Pepsi) Primary amine gives Foul smell of Isocyanicle with CHCl3+KOH Amine Smell RNH2+CHCl3+KOH

1.

Artificial Sweetening Agents

ASSA

RNC+KCl+H2O

Coupling Reaction

Aspartame, Saccharin, Sucrolose, Alitame

DSPO DAY (Say, DeSPO DAY)

Also, Assac Sue Ali 2.

Diazonium Salt + Phenol → Orange dye + N2Cl– +

OH

OH

N=N

OH

(orange dye) + N2Cl– +

NH2

H+

Antiseptic & Disinfectants

Bitter Chlor

Diazonium Salt + Aniline → Yellow dye

N=N

NH2 (yellow dye)

Chapter - 19 Polymers and Biomolecules 1.

DNA & RNA

G3Cinema AT 2PM

(to remember regents of reaction)

2.

Fatsoluble Vitamins → Vitamin K, E, D, A

KEDA

Clemmenson and wolf Reaction

Can Zebra Woo Nightingale

1.

Essential Amino Acids

Disaccharides

Non-reducing SGF ( 24 )

Bithionol , Terpineol, Chloroxylenol 3.

Antacids

His Interaction Presented by lime Ran (Say Simran) Interaction of Histamine prevented by limetidine, Ranitidine

SOME BASIC CONCEPTS IN CHEMISTRY

1

Some Basic Concepts in Chemistry

Chapter 1 Syllabus

Matter and its nature, Dalton's atomic theory, Concept of atom, molecule, element and compound, physical quantities and their measurements in chemistry, precision an accuracy, significant figures, S.I. units, dimensional analysis, Laws of chemical combination, atomic and molecular mass, mole concept, molar mass, percentage composition, empirical and molecular formulae, chemical equations and stoichiometry.



Concept Revision (Video Based) For details, scan the code

Part-1



For details, scan the code Mole Concept

Stoichiometry Part-2



JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions Q.1. The average molar mass of chlorine is 35.5 g mol–1. The ratio of 35Cl to 37Cl in naturally occurring chlorine is close to  [JEE (Main)-6th Sept -2020-(Shift-II)) (b) 4 : 1 (a) 3 : 1 (c) 2 : 1 (d) 1 : 1 Q.2. Amongst the following which is not a postulate of Dalton’s atomic theory  [JEE (Main)-7th Jan-2020-(Shift-I)] (a) Matter is formed of indivisible atoms. (b) Under identical conditions of pressure and temperature gases combines and give gaseous products in simple volume ratio. (c) During chemical reaction atoms remains conserved and only pass through rearrangement (d) Some atoms have same properties including atomic mass Q.3. 0.6 g of area of strong heating with NaOH evolves NH3 liberated NH3 will combine completely with which of the following HCl solution ?  [JEE (Main)-7th Jan-2020-(Shift-II)] (a) 100 ml of 0.2 N– HCl (b) 400 ml of 0.2 N– HCl (c) 200 ml of 0.1 N– HCl (d) 200 ml of 0.2 N– HCl Q.4. 5 g of Zn reacts with (I) Excess of NaOH



(II)  Dilute HCl, then volume ratio of H2 gas evolved in (I) and II is  [JEE (Main)-4th Jan-2020-(Shift-II)] (a) 2 : 1 (b) 1 : 2 (c) 1 : 1 (d) 3 : 1 Q.5. 25 g of an unknown hydrocarbon upon burning produce 88 g of CO2 and 9 g of H2O. This unknown hydrocarbon contains. [JEE (Main)-12th April-2019-(Shift-I)] (a) 20 g of carbon and 5 g of hydrogen (b) 22 g of carbon and 3 g of hydrogen (c) 24 g or carbon and 1 g of hydrogen (d) 18 g of carbon and 7 g of hydrogen Q.6. At 300 K and 1 atmosphere pressure 10 ml of a hydrocarbon required 55 mL of O2 for complete combustion and 40 mL of CO2 is formed. The formula of hydrocarbon is [JEE (Main)-10th April-2019-(Shift-I)] (b) C4H6 (a) C4H2Cl (c) C4H10 (d) C4H8 Q.7. The minimum amount of O2(g) consumed per gram of reactant is for the reaction (Given atomic mass Fe = 56, O = 16, Mg = 24, D = 31, C = 12, H = 1) [JEE (Main)-10th April-2019-(Shift-II)] (a) C3H8(g) + 5O2(g) ® 3CO2(g) + 4H2O(l) (b) P4(s) + 502(g) ® P4O10(s)

3

SOME BASIC CONCEPTS IN CHEMISTRY

CO2 is 25.0 L under such condition, what is the percentage of sodium bicarbonate in each tablet? Molar mass of NaHCO3 = 84 g mol–1 [JEE (Main)-11th Jan-2019-(Shift-I)] (b) 0.84 (a) 8.4 (d) 33.6 (c) 16.8 Q.16. In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in producing one molecule of CO2 is [JEE (Main)-10th Jan-2019-(Shift-I)] (b) 5 (a) 2 (d) 10 (c) 1 Q.17. For the following reaction, the mass of matter produced from 445 g of C57 H110O6 is 2C57H110O6 (s) + 163O2 ® 114CO3(g) + 110 H2O (l) [JEE (Main)-8th Jan-2019-(Shift-I)] (b) 495 g (a) 490 g (d) 890 g (c) 445 g

ANSWERS – KEY

1. (a) 2. (b) 3. (c) 5. (c) 6. (b) 7. (c) 9. (d) 10. (a) 11. (a) 13. (a) 14. (d) 15. (a) 17. (b)

4. (c) 8. (a) 12. (d) 16. (c)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (a) is correct. Let the percentage of 35Cl = x and the percentage of 37Cl = (100 – x) Þ 35 × 5 =

35 ´ x + 37 ( 100 - x )

=

100 Þ 2x = 150 Þ x = 75 Therefore 35Cl : 37Cl = 75 : (100 – 75) = 3 : 1 Ans. 2. Option (b) is correct. Under identical condition of pressure and temperature gases combines and give gaseous products in simple volume ratio. This postulate is the statement of French chemist Gay-Lussac in 1808. Ans. 3. Option (c) is correct. – O O OH C NH 2 C H2N NH2 H2N O—OH O H–OH H N—C—OH +N H =

=

(c) 4 Fe(s) + 302(g) ® 2 Fe2O3(s) (d) 2 Mg(s) + O2(g) ® 2 MgO(s) Q.8. In order to oxidise a mixture of one mole of each of FeC2O4, Fe2(C2O4)3, FeSO4 and Fe2(SO4)3 in acidic medium, the number of moles of KMnO4 required in [JEE (Main)-8th April-2019-(Shift-I)] (b) 1 (a) 2 (d) 1.5 (c) 3 Q.9. 100 ml of a water sample contains 0.81 g of calcium bicarbonate and 0.75 g of magnesium bicarbonate. The hardness of this water sample expressed in terms of equivalent of CaCO3 is [molar mass of calcium bicarbonate is 162 g mol–1 and magnesium bicarbonate is 146 g mol–1] [JEE (Main)-8th April-2019-(Shift-I)] (b) 1,000 ppm (a) 5,000 ppm (d) 10,000 ppm (c) 100 ppm Q.10. 0.27 g of long chain fatty acid wad dissolved in 100 cm3 of hexane. 10 mL of this solution was added drop wise to the surface of water in a round watch glass. Hexane evaporates and a monolayer if formed the distance from edge to centre of the watch glass if 10 cm. What is the height of monolayer? [Density of fatty acid = 0.9 g cm–3 p = 2] [JEE (Main)-8th April-2019-(Shift-I)] –6 (b) 10–4 m (a) 10 m –8 (d) 10–2 m (c) 10 m Q.11. For a reaction N2(g) + 3 H2(g) ® 2 NH3(g) identify dihydrogen (H2) as a limiting reagent on the following reaction mixture [JEE (Main)-9th April-2019-(Shift-I)] (a) 56 g of N2 + 10 g of H2 (b) 35 g of N2 + 8 g of H2 (c) 14 g of N2 + 4 g of H2 (d) 28 g of N2 + 6 g of H2 Q.12. 5 moles of AB2 weight 125 × 10–3 kg and 10 moles of A2B2 weight 300 × 10–3 kg. The molar mass of A (MA) and molar mass of B (MB) in kg mol–1 is [JEE (Main)-2nd April-2019-(Shift-I)] (a) MA = 10 × 10–3 and MB = 5 × 10–3 (b) MA = 50 × 10–3 and MB = 25 × 10–3 (c) MA = 25 × 10–3 and MB = 5 × 10–3 (d) MA = 5 × 10–3 and MB = 10 × 10–3 Q.13. The hardness of water sample in terms of equivalents of CaCO3 containing 10–3 M CaSO4 is (Molar mass of CaSO4 = 136 g mol–1) [JEE (Main)-12th Jan-2019-(Shift-I)] (b) 10 ppm (a) 100 ppm (d) 90 ppm (c) 50 ppm Q.14. Decomposition of X exhibits a rate constant of 0.05 mg/year. How many years are required for the decomposition of 0.5 mg × into 2.5 mg? [JEE (Main)-12th Jan-2019-(Shift-I)] (b) 25 (a) 20 (d) 50 (c) 40 Q.15. A 10 mg effervescent tablet containing sodium bicarbonate and oxalic acid releases 0.25 ml of CO2 at T = 298.15 K and p = 1 bar. If molar volume of

2

NH 3 + CO2 + CO2

3

–OH similarly

O ||

® H 2 N - C - O- + NH 3 CO ( NH 2 )2 + OH ¾¾ -

4 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Ans. 7. Option (c) is correct.

O ||

-

+

H 2 N - C - O Na + H 2O ® NH 3 + CO2 + NaOH

CO(NH2)2 + H2O ® 2NH3 + CO2 60 34 60 g urea evolves ammonia 34 g 0.6 g urea evolves ammonia 0.34 g   17 g NH3 º 1000 ml 1 N HCl   0.34 g NH3 º 20 ml 1 N HCl º 200 ml 0.1 N HCl Ans. 4. Option (c) is correct. When Zn reacts with NaOH, Zn + 2NaOH ® Na2 ZnO2 + H2 65.38 22.4 Lit at N.T.P Volume of H2 at N.T.P when reacts with 5g of Zn 22 × 4 = ´ 5 Lit 65 × 38 When Zn reacts with dilute HCl Zn + 2HCl ® ZnCl2 + H2 65.38 22.4 Lit at N.T.P Volume of H2 at N.T.P when reacts 5g of Zn 22 × 4 = ´ 5 Lit 65 × 38 Therefore the ratio of produced H2 22 × 4 ´ 5 22 × 4 ´ 5 = = 65 × 38 65 × 38 = 1 : 1 Ans. 5. Option (c) is correct. Let the formula of hydrocarbon is Cx Hy y yˆ Ê CxHy(s) + Á x + ˜ O2(g) = xCO2(g) + H2O (l) Ë 2 4¯ 12 Mass of carbon = ¥ 88 = 24 g 44 2 ¥ 9 = 1g Mass of hydrogen = 18 Ans. 6. Option (b) is correct. Let the formula of hydrocarbon is Cx Hy. The combustion reaction for Cx Hy is yˆ y Ê 300 K Æ xCO2 ( g ) + H 2 O(l ) C x H y ( g ) + Á x + ˜ O2 ( g ) æDæææ atom Ë 4¯ 2 \ 10 ml

yˆ Ê 10 Á x + ˜ ml Ë 4¯

10 x ml

Given volume of CO2 = 40 mL \ 10x = 40 fi x = 4 The volume of O2 used = 55 ml yˆ Ê 10 Á x + ˜ = 55 or, Ë 4¯ yˆ Ê 10 Á 4 + ˜ = 55 Ë 4¯ 5y fi = 15 2 fi y = 6 Therefore hydrocarbon is C4H6. fi

CHEMISTRY

(a) C3 H 8 ( g ) + 5O2 ( g ) Æ 3CO2 ( g ) + 4 H 2 O(l ) 5¥16 ¥ 2 44 160 160 \ 1 g of C3 H8 consumed O2 = = 3.64 g 44 P4 ( s) + 5O2 ( g ) Æ P4 O10 ( g ) (b) 4 ¥ 31 = 124 160 160 \ 1 g of P4 consumed O2 = = 1.29 g 124 (c) 4 Fe( s)+ 3O2 ( g ) ® 2 Fe2O3 ( s) 4 ´ 56 3 ´ 16 ´ 2 = 96 = 224 \ 1 g of Fe consumed O2 = (d) 2 Mg( s)+ O2 ( g ) Æ 2 MgO( s) 2 ¥ 24 32 = 48 1 g of Mg consumed O2 =

96 = 0.43 g 224

32 = 0.67 g. 48

Ans. 8. Option (a) is correct. The oxidation of a mixture of one mole of each of FeC2O4, Fe2(C2O4)3, FeSO4 and Fe2(SO4)3 with KMnO4 in acidic medium is as follows. FeC 2O4 + MnO-4 Æ Mn 2 + + CO2 + Fe3 +

...( i )

Fe2 (C 2O4 )3 + MnO-4 Æ Mn 2 + + CO2 + Fe3 +

...( ii )

FeSO4 + MnO-4 Æ Mn 2 + + CO2 + Fe3 + ...( iii ) Change in oxidation number of Mn is 5. Total change in oxidation in the above (i), (ii), (iii) equations are +3, +6, +1 respectively. \ No. of equivalent KMnO4 = Total change in oxidation numbers fi n × 5 = 1 × 3 + 1 × 6 + 1 × 1 fi n = 2 Ans. 9. Option (d) is correct. Molecular weight of calcium bicarbonate = 162 Molecular weight of magnesium bicarbonate = 146 Molecular weight of calcium carbonate = 100 162 g of Ca(HCO3)2 = 100 g CaCO3 100 ¥ 0.81 g CaCO3 0.81 g of Ca(HCO3)2 = 162 = 0.5 g CaCO3 146 g of Mg(HCO3)2 = 100 g of CaCO3 100 ¥ 0.73 g of CaCO3 0.73 g of Mg (HCO3)2 = 146 = 0.5 g CaCO3 100 ml of sample water contains (0.5 + 0.5) g = 1 g of CaCO3 Thus hardness of water sample 1 ¥ 10 6 100 = 10,000 ppm Ans.10. Option (a) is correct. 100 mL of hexane contains 0.27 g of fatty acid. =

5

SOME BASIC CONCEPTS IN CHEMISTRY

0.27 ¥ 10 g of fatty acid 100 = 0.027 g of fatty acid Density of fatty acid = 0.9 g cm–3 Volume of fatty acid (V) 0.027 g M = = = 0.03 cm3 d 0.9 g cm -3 Let the height of cylindrical monolayer = h \ The distance from edge of centre of watch glass (r) = 10 cm \ V = pr2h V fi h = 2 pr 10 mL of hexane contains

=

0.03 cm

3

3.14 ¥ (10 )2 cm 2

Ans. 15. Option (a) is correct. 2 NaHCO3 + H 2 C 2 O4 Æ 2CO2 + Na2 C 2 O4 + H 2 O 2 mole

Therefore number of moles of CO2 produced 0.25 n = 25000 = 10–5 mol = 10–5 mol of NaHCO3 = 10–5 × 84 g Percentage of sodium bicarbonate in each tablet =

-3

6

¥ 10 ¥ 100 g 1000 = 100 ppm Ans. 14. Option (d) is correct. The unit of rate constant follows zero order reaction. Therefore for zero order reaction half life period a t1/2 = 2K a = initial concentration K = rate constant 5mg \ t1/2 = = 50 years 2 ´ 0.05mg / year =

10

84 ´ 10 -5 10 ´ 10 -3

´ 100 = 8.4%

Ans. 16. Option (c) is correct. The reaction of oxalate with permanganate in acidic medium occurs as 5C 2 O24 - + 2 MnO4- + 16 H + Æ 10CO2 + 2 Mn 2 + + 8 H 2 O

1 × 10–4 cm = 1 × 10–6 m Ans. 11. Option (a) is correct. According to stoichiometry of balanced equation 28 g of N2 reacts with 6 g of hydrogen 6 ¥ 56 \ 56 g of N2 reacts with g of hydrogen 28 = 12 g of hydrogen But only 10 g of H2 is present in option (a) Therefore here H2 is the limiting reagent for the synthesis of NH3. Ans. 12. Option (d) is correct. Given, molar mass of A and B are MA, MB respectively. By the problem, 5(MA + 2MB) = 125 × 10–3 ...(i) [as n × M = W n = no. of moles M = Molecular weight] and 10(2MA + 2MB) = 300 × 10–3 ...(ii) Solving equations (i) and (ii), MA = 5 × 10–3, MB = 10 × 10–3 Ans. 13. Option (a) is correct. Given : Strength of solution = 10–3 M \ 1000 ml of solution contains CaCO3 = 10–3 M = 10–3 × 100 g of CaCO3 \ 106 parts of solution contains CaCO3

2 mole

2 MnO-4 + 16 H + Æ 2 Mn 2 + + 8 H 2 O 5C 2 O24 - Æ 10CO2 + 10e2 MnO-4 + 5C 2 O24 - 16 H + Æ 2 Mn 2 + + 10CO2 + 8 H 2 O \ To produce CO2, 10 e– are involved for the above reaction. 10 mole electrons involve in 10 mole CO2. Therefore 1 mole electron involve in 1 mole CO2. Ans. 17. Option (b) is correct. 2C57 H110O6(s) + 163O2 ® 114CO2(g) + 110H2O(l) 2 (12×57+1×110+6×16)   110 × 18 = 1780           = 1980 From reaction, 1780 g C57H110O6 pound produces = 1980 g H2O \ 445 g C57H110O6 compound produces 1980 ¥ 445 = g 1780 = 495 g of H2O

Integer Type Questions (Chapter Based) Q1. The minimum number of moles of O2 required for complete combustion of 1 mole of propane and 2 moles of butane is …….  [JEE (Main) 2020, 5th Sept (Shift - I)] Sol. The combustion reaction react on for propane

C3H8 + 5O2 = 3 CO2 + 4 H2O The combustion reaction for two moles of butane



2C4H10 + 13O2 = 8CO2 + 10 H2O



Required no. of moles of q2 = (5 + 13) = 18 moles

Q2. The volume in mL of 0.02M K2Cr2O7 solution required to react with 0.288 g of ferrous oxalate in acidic medium ….. (Molar mass of Fe = 56 g mol–1)  [JEE (Main) 2020 5th Sept (Shift - II)] Sol. Molecular as of ferrous oxalate (FeC2O4)  = 56 + 24 + 64 = 144 0 ⋅ 288 \ 0.288 g of ferrous oxalate = moles 144

6 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

 = 0.002 mole milliequivalent of Fe C2O4 º milliequivalent of K2Cr2O7 0.002 × 3 × 1000 º 6 × 0.02 × Volume of K2Cr2O7

\ Volume of K2Cr2 O7 = 100 ml Q3. The mass of ammonia in grams produced when 2.8kg of dinitrogen quantitatively reacts with 1 kg of dihydrogen is [JEE (Main) 2020 4th Sept (Shift - I)] Sol. Ammonia is produced as follows.



N2 + 3H2 = 2NH3



Initially no. moles of N2 =





2800 = 100 moles 28

1000 No. moles of H 2 = = 500 moles 2 \ N2 + 3H2 = 2NH3

Initial mole  Final mole

100

Here limiting reagent is N2. 500 – 300 = 200

0

\ Mass of NH3 formed 200 moles = 200 × 17 gram = 3400 grams. Q4. A 100 ml solution was made by adding 1.43 g of Na2CO3 x H2O. The normality of the solution is 0.1 N.The value of x is ………. (The atomic mass of Na is 23gm −1) Sol. Equivalent of solute = 0.1 × 0.1 1 Mole of (Na2 CO3, x H2O) = (0.1 × 0.1) 2 1 (0.1 × 0.1) × (106 + 18x) = 1.43 2 Þ   106 + 18x = 286 Þ 18x = 180 Þ x = 10 Q.5. In a saturated acylic compound the mass ratio of C : H is 4 : 1 and C : 0 is 3 : 4. Find the no. of moles of O2 required to react with 2 moles compound to give CO2 and water.  [JEE (Main) 2020 2nd Sept (Shift - II)] H 1 3 3 = ´ = Sol. C 4 3 12





C 3 4 12 = ´ = O 3 4 16 \ H : C : O = 3 : 12 : 16 Elements

Atomic weight

Mass ratio

No. of moles

H

1

3

3 =3 1

C

12

12

12 =1 12

O



16

16



Þ 2C2 H6 O2 + 5O2 ® 4CO2(g) + 6H2O(g)



So, required moles of O2 is 5.

Q6. Given a solution of HNO3 of density 1.4 g/mL and 63% w/w. Determine Molarity of HNO3 solution.

Empirical formula

C H3 O

16 =1 16

[JEE (Main) 2020, 9th Jan (Shift - I)]



Sol. 1 ml solution of HNO3 laving weight 1.4 g Q7.

500

CHEMISTRY



1000 ml solution of HNO3 having weight 1400 g 63 HNO3 present in 1000 ml solution = 1400 ´ 100 1400 ´ 63 \ Molarity of HNO3 solution = M 63 ´ 100 = 14.00 M 0.3g [ML6]Cl3 of molar mass 267.46 g/mol is reacted with 0.125 M AgNO3 (aq) solution, calculate volume of AgNO3 required in ml. [JEE (Main) 2020, 8th Jan (Shift - I)]

Sol. [ML6]Cl3 + 3 AgNO3 ® 3 AgCl 0.3 g Vm 0.125 M 0×3 \ ´ 3 = 0 × 125 ´ V ´ 10 -3 267 × 46 0 × 3 ´ 3 ´ 1000 = 26 × 92 mL 267 × 46 ´ 0 × 125 Q8. Calculated the mass of FeSO4. 7 H2O which must be added in 100 kg of wheat to get 10 ppm of Fe.



Þ V=

[JEE (Main) 2020, 8th Jan (Shift - I)]



Sol. 100 kg of wheat = 100 × 1000 g = 105 g Let Fe present = x g 105 part of wheat contains Fe x g x ´ 10 6 g = 10 xg 106 part of wheat contains Fe 10 5 Therefore 10 x = 10 Þ x=1 Molecular out of FeSO4, 7H2O = 56 + 96 + 7 × 18  = 278 56 g = 1 mole Fe 1 1 mole Fe = ´ 278 g Fe SO4.7 H2O 1g = 56 56 = 4.96 g of FeSO4 7 H2O Q9. Given following reaction, NaClO3 + Fe ® O2 + FeO + NaCl In the above reaction 492L of O2 is obtained at 1 atm & 300 K temperature. Determine mass of NaClO3 required (in kg) [R = 0.082 L atm mol−1 K−1)  [JEE (Main) 2020, 8th Jan (Shift - I)] Sol. NaClO3 + Fe ® O2 + FeO + NaCl



As the given organic compound is acyclic saturated



So molecular formula is C2 H6 O2 5 C 2 H 6 O2 + O (g) ® 2CO2(g) + 3 H2O(g) 2 2







23 + 35.5 + 48 106.5

V2 =

22.44 L at N.T.P

P1V1 ´ T2 1atm ×492L×273K =  448Lit T1 ´ P2 300K ×1atm

22.4 Litre of O2 is produced from 106.5g NaClO3 106 × 5 ´ 448 \ 4480 Litre of O2 is produced from g 22.4 = 2,128.66875 g NaClO3  02.13 kg

7

SOME BASIC CONCEPTS IN CHEMISTRY

Q10. Each of solution A and B of 100 L containing 4 g NaOH and 9.8 g H2SO4. Find pH of solution which obtain by mixing 40 L solution of A and 10 L solution of B  [JEE (Main) 2020, 7th Jan (Shift - I)] Sol. 1 litre solution contains NaOH = 4 × 10−2 g



1 litre solution contains H2SO4 = 98 × 10−3 g 4 \ Strength of NaOH solution ´ 10 -2 = 1 ´ 10 -3 N 40 98 And strength of H2SO4 Solution = ´ 10 -3 49 = 2 × 10−3 40 Lit 10 -3 N - NaOH º 40 ´ 10 -3 Lit 1 N - NaOH 10 Lit 10 -3 N - H 2SO4 º 20 ´ 10 -3 Lit 1 N H 2SO4 50 Lit x N - H 2SO4 º 20 ´ 10 -3 Lit 1N NaOH



\



Þ





20 ´ 10 -3 N 50 2 -3   x = ´ 10 N NaOH 5

   x =

2 \ éëOH - ùû = ´ 10 -3 N 5 [OH−] = 4 × 10−4 N−





By the problem we get, 36.5x 73(1 - x ) + = 0.5475 84 106 fi x = 0.5555 g % NaHCO3 = 55.55%

Q.12. Calculate molarity of water, if its density is 1000 kg/m3. Sol. Volume of water 10–3 m3 = 1 litre = 1000 ml 1 m3 H2O weigh 103 kg H2O = 106 g H2O 10–3 m3 H2O weight = 106 × 10–3 g H2O = 103g H2O





\

10 3 18 1000 Molarity = = 55.6 18 ¥ 1

Mole of water =

Q.13. On heating 1.763 g of hydrated BaCl2 to dryness 1.505 g of anhydrous salt remained. Calculate the number of H2O molecules present in hydrated BaCl2? Sol. By applying heat on hydrated BaCl2 the reaction takes place as follows,

D BaCl 2 . nH 2 O ææÆ BaCl 2 + n H 2 O Molecular weight 208 ( 208 + 18n)

Þ − log [OH−] = − log 4 − log10−4 Þ pOH = − 2 × 0.301 + 4 = 3.398 (208 + 18n)g BaCl2 . nH2O gives \ pH = 14 − 3.398 = 10.602 = 208 g of BaCl2 Q.11. 1.0 g of a mixture of Na2CO3 and NaHCO3 is dissolved in water to make 250 ml solution. \ 1.763 g. BaCl2 . nH2O gives 25 ml of the solution required 15 ml 0.1 N HCl 208 ´ 1.763 = g BaCl2 for neutralization. Calculate the percentage of ( 208 + 18n) NaHCO3 in the mixture. \ By the problem, Sol. Let the weight of NaHCO3 = x g 208 ´ 1.763 \ Weight of Na2CO3 = (1 – x) g = 1.505 208 + 18n ∵ 25 mL of the solution requires 15 ml 0.1 N HCl for \ n = 1.92 » 2 neutralisation. Q .14. An organic compound contains 69.4 % C and 5.8 \ 250 mL of the solution requires 150 ml 0.1 N HCl % H and 13.25 % of O. A sample of 0.303 g of this for neutralisation. compound was analysed for N2 by Kjeldahl’s ∫ 15 ml 1 N HCl method. The NH3 evolved was absorbed in 50 ml We know that, of 0.05 M H2SO4. The excess acid required 25 ml 1000 ml 1 N HCl ∫ 36.5 g HCl 0.1 M NaOH for neutralisation. Determine the total number of atoms present in the compound if 36.5 ¥ 15 g HCl 15 ml 1 N HCl ∫ its molecular wt. is 121. 1000 Sol. 1 M H2SO4 = 2N H2SO4, ∫ 0.5475 g HCl 0.05 M H2SO4 = 2 × 0.05 N H2SO4 NaHCO3 + HCl = NaCl + CO2 + H 2 O The amount of NH3 evolved 36.5 g 84 g ∫ (50 × 0.05 × 2 – 25 × 0.1) ml of N H2SO4 84 g of NaHCO3 reacts with 36.5 g HCl ∫ 2.5 ml of 1 N H2SO4 36.5 ¥ x g HCl Again 1 ml 1 N H2SO4 x g of NaHCO3 reacts with 84 ∫ 0.017 g of NH3 Again, ∫ 0.014 g of nitrogen Na2 CO3 + 2 HCl = 2 NaCl + CO2 + H 2 O \ 2.5 ml 1 N H2SO4 2 ¥ 36.5 g 106 g ∫ 2.5 × 0.014 g of nitrogen = 73 g ∫ 0.035 g of nitrogen \ 0.303 g of the compound contains 0.033 g of N ∵ 106 g of Na2CO3 combine with 73 g of Na2CO3 0.035 ¥ 100 73(1 - x ) = 11.55 % % of nitrogen = \ ( 1 – x) of Na2CO3 combine with g of Na2CO3 0.303 106

8 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) \

C : H : O : N 69.4 5.8 13.25 11.55 : : : 12 1 16 14 = 5.78 : 5.78 : 5.8 : 0.828 = : 7 : : 7 1 1 Therefore empirical formula of the compound is C7 H7NO Let the molecular formula of the compound is (C7H7NO)n Molecular wt 121 = =1 \ n = emipirical wt 121 \ Molecular formula of the compound is C7H7NO Total no. of atoms present in the compound is = 16 Q.15. 6.02 × 1020 molecules of urea are present in 100 mL of its solution. Calculate the concentration of the solution. Sol. Molarity of the solution 6.02 ¥ 10 20 ¥ 1000

= 0.01 M 100 ¥ 6.02 ¥ 10 23 Q.16. Calculate the amount of water produced ( in g) in the oxidation of 1 mole of rhombic sulphur by conc. HNO3 to a compound with highest oxidation state of sulphur is ........... [Molar mass of water = 18 g mol–1] =

Sol. S 8 + Conc. HNO3 Æ H 2SO4 + NO2 + H 2O

On balancing the above equation,



1 mole of S8 produces 16 moles of H2O Mass of H2O produced = 16 × 18 g = 288 g

S 8 + 48 HNO3 Æ 8 H 2 SO4 + 48 NO2 + 16 H 2 O

CHEMISTRY

Q.17. The mole fraction of urea in an aqueous urea solution containing 900 g of water is 0.05. If the density of the solution is 1.2 g cm–3, the molarity of urea solution then calculate [Molar masses of urea and water are 60 g mol–1 and 18 g mol–1 respectively] Sol. Given mole fraction of urea = 0.05 900 Number of moles of H2O = = 50 18 nurea \ = 0.05 nurea + nH 2O nurea fi = 0.05 nurea + 50 fi nurea = 2.63 Mass of urea = 2.63 × 60 g = 157.8 g Total mass of solution = (157.8 + 900) g = 1057.8 g 1057.8 ml Volume of solution = 1.2 = 881.5 cm3 = 0.8815 litre No. of moles Molarity = Vol. in litres 2.63 = 0.8815 = 2.9835 M » 2.98 M (approx.)



8 RT 8 PV   m m

8P d

3P d

2 RT 2 PV   m m

2P D

ula r

aw

Ve loc ity

m ’s L

dRT P

•Volume of one mole of gas at critical temperature is called critical volume (Vc ) and pressure at this temperature is called critical pressure (Pc )

Mo lec

Gr ah a

M=

States of Matter

is the force acting per unit length perpendicular to the line drawn on the surface of liquid.

Denoted by Gamma () Unit: Nm-1

sc Vi os Intermolecular forces of liquid state are stronger than gaseous state. • Boiling is the condition of free vapourisation throughout the liquid. • Normal b.p. is boiling point at 1 atm. •Standard b.p.is boiling point at 1 bar.

ity

• Real gases show deviations from ideal gas law because molecules interact with each other. •Pressure exerted by the gas is lower than the pressure exerted by the ideal gas. an2 Pideal = Preal + V2 Observed Correction pressure term • Compressibility factor Z = PV nRT

• A gas contains a large number of small particles called molecules. Size and mass of all molecules of each gas are identical. •There is no force of attraction between the particles of a gas at ordinary temperature and pressure. • Particles of a gas are always in constant and random motion. • Particles of a gas move in all possible directions in straight lines. • Collisions of gas molecules are perfectly elastic. • At any particular time, different particles in the gas have different speeds and hence different kinetic energies. • If a molecule has variable speed, then it must have a variable kinetic energy.

cp 

• Most probable velocity (Cp)

c

Crms 

3 RT 3 PV   m m • Average velocity (c– )

• Root mean square velocity (Crms)

Under the same conditions of temperature and pressure the rates of diffusion of different gases are inversely proportional to the square roots of their densities

fo ar

Gas Liquid Solid Predominance of intermolecular interactions

Predominance of thermal energy

Gas  Liquid Solid

dx F=A– dz - Viscosity coefficient () is the force when velocity gradient is unity and the area of contact is unit area. - SI unit of viscosity coefficient is Newton second per square metre -2 (N s m ) = Pa s (Pascal second)

R= 8.314 JK-1 mol-1

pV R = nT , R is gas constant / universal gas constant

•Boyle's Law (Pressure – Volume relationship): At constant temperature, the pressure of a fixed amount of gas varies inversely with its volume. V P P1V1 = P2V2 = Constant  1 = 1 P2 V2 •Charle's Law (Temperature – Volume relationship) : At constant pressure, the volume of a fixed mass of a gas is directly proportional to the absolute temperature V1 = V2 V =Constant T1 T2 T •Gay Lussac's Law (Pressure – Temperature relationship): At constant volume, pressure of a fixed amount of a gas varies directly with temperature.P = constant T •Avogadro’s Law (Volume – Amount relationship): Equal volume of all gases under the same conditions of temperature and pressure Mass of gas contain equal number of molecules. V = kn; n= Molar Mass

Physical Properties: • Gases are highly compressible. • Gases exert pressure equally in all directions. • Gases have much lower density than solids and liquids. • Volume and shape of gases are not fixed. • They mix evenly and completely in all proportions.

ion act ter n i al rm

•London dispersion forces are the weakest intermolecular forces present among non-polar atoms and molecules. •Dipole – dipole forces act between the molecules possessing permanent dipole. •Dipole – induced dipole forces are the attractive forces operate between polar molecules having permanent dipole. •Hydrogen bond is found in the molecules in which highly polar N–H, H–O and H–F bonds are present.

tate

us S

eo Gas

l cu le

he st sv e rc

Gas Law s

er m o

In t

Total pressure exerted by the mixture of non- reactive gases is equal to sum of the partial pressures of individual gases. Ptotal = p1 +p2 +p3 ......... (At constant T,V)

STATES OF MATTER

9

States of Matter

Chapter 2 Syllabus

States of Matter Classification of matter into solid, liquid and gaseous states. Gaseous State: Measurable properties of gases; Gas laws- Boyle’s law, Charle’s law, Graham’s law of diffusion, Avogadro’s law, Dalton’s law of partial pressure; Concept of Absolute scale of temperature; Ideal gas equation; Kinetic theory of gases (only postulates); Concept of average, root mean square and most probable velocities; Real gases, deviation from Ideal behaviour, compressibility factor and van der Waals equation. Liquid State: Properties of liquids- vapour pressure, viscosity and surface tension and effect of temperature on them (qualitative treatment only). Solid State: Classification of solids: molecular, ionic, covalent and metallic solids, amorphous and crystalline solids (elementary idea); Bragg’s Law and its applications; Unit cell and lattices, packing in solids (fcc, bcc and hcp lattices), voids, calculations involving unit cell parameters, imperfection in Solids; Electrical and magnetic properties.

Topic-1

Gases and Liquid State Concept Revision (Video Based) For details, scan the code

Kinetic theory of gases

For details, scan the code

Van der Waals gas equation

LIST OF TOPICS : Topic-1 : Gases and Liquid State 

.... P. 10

Topic-2 :  Solid State

.... P. 13

For details, scan the code Liquefaction of gases,  Andrew curve, critical constant

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions

Q.1. A mixture of one mole each of H2, He and O2 each are enclosed in a cylinder of volume V at temperature T. If the partial pressure of H2 is 2 atm, the total pressure of the gases in the cylinder is  [JEE (Main)- 3rd Sept-2020-(Shift - II)] (a) 22 atm (b) 14 atm (c) 6 atm (d) 38 atm.

Q.2.

A B No. of Molecules

Speed

C

11

STATES OF MATTER



Select the correct options :  [JEE (Main)-7th Jan-2020-(Shift - II)] (a) A = CMPS, B = CAverage, C = CRMS (b) A = Caverage, B = CMPS, C = CRMS (c) A = CRMS, B = CAverage, C = CMPS (d) A = CAverage, = B = CMPS, C = CRMS Q.3. Which of the following graphs is not correct for ideal gas ? [JEE (Main)-2nd Jan-2020-(Shift - II)] (a)



(b) d

d

T

T

(c)



(d)

d

d

1 P T [d = density, P = pressure, T = Temperature] Q.4. Consider the following table

Gas

a(atm dm3 mol–1)

b/(dm3 mol–1)

A

642.32

0.05196

B

155.21

0.04136

C

431.91

0.05196

D

155.21

0.4382

a and b are van der Waals constants. The correct statement about the gases is [JEE (Main)-10th April-2019-(Shift-I)] (a) gas C will occupy lesser volume than gas A, gas B will be lesser compressible than gas D. (b) gas C will occupy more volume than gas A, gas B will be more compressible than gas D. (c) gas C will occupy more volume than gas A, gas B will be lesser compressible than gas D. (d) gas C will occupy more volume than gas A, gas B will be lesser compressible than gas D. Q.5. Points I, II and III in the following plot respectively correspond to (vmp : most probable velocity) [JEE (Main)-10th April-2019-(Shift-II)] f (v)→ Distribution function



Gas 6

–2

a/(atom dm mol ) –2

3

–1

b/(10 dm mol )

Ar

Ne

Kr

Xe

1.3

0.2

5.1

4.1

3.2

1.7

1.0

5.0



Which gas is expected to have the highest critical temperature?  [JEE (Main)-9th April-2019-(Shift-I)] (a) Kr (b) Xe (c) Ar (d) Ne Q.7. At a given temperature T, gases Ne, Ar, Xe and Kr are found to deviate from ideal gas behavior. Their RT equation of state is given as P = at T. Here b V-b the van der Waal’s constant. Which gas will exhibit steepest increase in the plot of Z (compression factor) vs P2. [JEE (Main)-9th April-2019-(Shift-II)] (a) Xe (b) Ar (c) Kr (d) Ne Q.8. 10 mL of 1 mM surfactant solution forms a monolayer covering 0.24 cm2 on polar substrate. Consider the surfactant is adsorbed only on one face of the cube. If the polar head is approximated as a cube, what is its edge length? [JEE (Main)-9th April-2019-(Shift-II) (a) 2.0 pm (b) 0.1 nm (c) 1.0 pm (d) 2.0 nm Q.9. The volume of gas A in twice than that of gas B. The compressibility factor of gas A in thrice than that of gas B at same temperature. The pressures of gases for equal number of moles are [JEE (Main)-12th Jan-2019-(Shift-I)] (a) PA = 2PB (b) 2PA = 3PB (c) PA = 3PB (d) 3PA = 2PB Q.10. An open vessel at 27°C is heated until two fifth of the air (assumed as an ideal gas) in it has escaped from the vessel. Assuming that the volume of the vessel remains constant, the temperature at which the vessel has been heated is [JEE (Main)-12th Jan-2019-(Shift-II)] (a) 750 K (b) 500 K (c) 750°C (d) 500°C

ANSWERS – KEY

I



(d) vmp of N2 (300 K); Vmp of H2 (300 K); vmp of O2 (400 K) Q.6. Consider the van der Waal’s constants a and b for the following gases,

II Speed → v

III

(a) vmp of H2 (300 K); vmp of N2 (300 K); vmp of O2 (400 K) (b) vmp of O2 (400 K); vmp of N2 (300 K); vmp of H2 (300 K) (c) vmp of N2 (300 K); vmp of O2 (400 K); vmp of H2 (300 K)



1. (c)

2. (a)

3. (b)

4. (b)



5. (c)

6. (a)

7. (d)

8. (a)



9. (b)

10. (b)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (c) is correct. For n mole ideal gas, nRT P= V When n, T, V are constant

12 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Þ

y = c+x



The plot of = P is found to be

→

→

∴P∝n P= P= 2 atm (As same number of and P= H2 O2 H2 moles are present in a cylinder) ∴ PTotal = 2 + 2 + 2 = 6 atm. Ans. 2 Option (a) is correct. As we know that C rms : C av : C m.p.s

3RT 8RT 2RT : : = = 1.224 : 1.128 : 1.000 M πM M The curve passing though a maximum indicating that a maximum indicating that a maximum number of molecules is moving is small represented by A and C indicating the order distribution of the curve. Ans. 3 Option (b) is correct. From ideal gas equation we get, PM = dRT  PM  1 ⇒ d=   R T So graph d vs T is not straight line. Ans. 4. Option (b) is correct. Gas A and C have same value of b but different values of ‘a’ so gas having higher value of ‘a’ occupy more strong force of attraction by which they come closer to each other hence have less volume. But ‘B’ and D have same value of ‘a’ but different values of ‘b’ so gas having lesser value of ‘b’ will be more compressible. Ans. 5. Option (c) is correct.

2RT vmp = M

fi

vmp =

T M

For N2, O2, H2

300 400 < < 28 32

300 2

or, vmp of N2 (300 K) < vmp of O2 (400 K) < vmp of H2 (300 K) Ans. 6. Option (a) is correct. 8a Critical temperature (TC) = 27 Rb Since, value of a is highest for Kr therefore Kr gas has highest value of critical temperature. Ans. 7. Option (d) is correct. RT P = ( V - b) Þ P( V - b) = RT Þ PV - Pb = RT Þ Þ

PV Pb = 1+ RT RT z = 1+

Pb RT

have, Z =

PV RT

CHEMISTRY

b Slope =—– RT

z1

P → At constant T, slope of z vs P group µ C The gas with high value of b will be steepest as slope is directly proportional to b. b is the van der Waal’s constant and it is equal to the four time of actual 4 volume of the gas molecules (b = 4Nb¢ and b¢ = 3 pr3 ) Xe has the maximum radius, maximum b and hence the group will be steepest. Ans. 8. Option (a) is correct. Total area = Area occupied by one particle × no. of particles 2 fi 0.24 cm2 = a ´

10 -3 ´ 10 10 3

´ NA

fi 0.24 cm2 = a2 × 10–5 × 6 × 1023 fi 0.24 cm2 = a2 × 6 × 1018 fi

a2 =

0.04 1018

cm 2

fi a2 = 4 × 10–20 cm2 fi a = 2 × 10–10 cm = 2 × 10–12m = 2pm Ans. 9. Option (b) is correct. Compressibility factor of gas A = 3 × Compressibility factor of gas B fi ZA = 3ZB ...(i) Again compressibility PV Z = [for real gases] ...(ii) nRT \ Substituting the value of Z in eqn. (i) we get 3PBVB PAVA = h h A RTA B RTB [Given nA = nB, TA = TB VA = 2VB] fi

3PB ´ VB PA ´ 2VB = n A RTA n A RTA

fi 2PA = 3PB Ans. 10. Option (b) is correct. Here the volume is constant, as the vessel is open, the pressure is also constant. Let at the initial state n1, moles are present in the vessel and in the final state n2 moles remains in vessel. Let initial and final temperatures are T1, T2 respectively. So the gas equation for two cases becomes,

13

STATES OF MATTER

PV = n1RT1 and PV = n2RT2 \

gas, the final no. of moles

n1RT1 = n2RT2

fi

= 1 -

n1T1 = n2T2 ...(i)

2 = 0.6 5

From eqn. (i) we get fi 1 × 300 = 0.6 × T2 fi T2 = 500

Let in initial state the number of moles = 1 As no. of moles is proportional to the volume of the

Topic-2

Solid State Concept Revision (Video Based)



For details, scan the code Bravais Lattices

For details, scan the code

For details, scan the code Close Packing in Crystals

HCP Crystal Classification of Ionic Structures

Defects in Crystals

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions Q.1. A crystal is made up of metal ions, M1 and M2 and oxide ions. Oxide ions form a ccp lattice structure. The cation M1 occupies 50% of octahedral voids and the cation M2 occupies 12.5% of tetrahedral voids of oxide lattice. The oxidation number of M1 and M2 are respectively.  [JEE (Main)-6th Sept-2020-(Shift - II)] (a) +1, + 3 (b) +3, +2 (c) +2, +4 (d) +4, +2 Q.2. A diatomic molecule X2 has a body centered cubics (bcc) structure with a cell edge of 300 ppm. The density of the molecule is 6.47 gcm–3. The number of molecules present in 200 g of X2 is  [JEE (Main)-5th Sept-2020-(Shift - I)] (a) 4NA (b) 2NA (c) 4ONA (d) 8NA Q.3. An element crystallises in face–centered cubic (fcc) unit cell edge a. The distance between the centers of two nearest octahedral voids in crystal lattice is  [JEE (Main)-5th Sept-2020-(Shift - II)] a (a) (b) a 2 (c)

a 2

(d) 2 a

Q.4. An alkaline earth metal ‘M’ reacted forms water soluble sulphate and water insoluble hydroxide, its oxide MO is very stable to heat and does not have rock salt structure M is.  [JEE (Main)-4th Sept-2020-(Shift - II)]

(a) Ca (b) Mg (c) Sr (d) Be Q.5. An element has a face centred cubic (fcc) structure with a cell edge of a. The distance between the centres of the two nearest tetrahedral words in the lattice is  [JEE (Main)-2019-12th April-(Shift-I)] (a)

2a

(b) a

3 (c) a (d) a 2 2 Q.6. The ratio of number of atoms present in a simple cubic, body centred cubic and face centred cubic structure are respectively, [JEE (Main)-2019-12th April-(Shift-II)] (a) 8 : 1 : 6 (b) 1 : 2 : 4 (c) 4 : 2 : 1 (d) 4 : 2 : 3 Q.7. Element ‘B’ forms ccp structures and A occupies half of the octahedral voids, while oxygen atoms occupy all the tetrahedral voids. The structure of bimetallic oxide is : 

[JEE (Main)-2019-8th April-(Shift-I)]

(a) A2BO4 (b) AB2O4 (c) A2B2O (d) A4B2O Q.8. Consider the bcc unit cells of the solids 1 and 2 with the position of atoms as shown below. The radius of atom B is twice that of atom A. The unit cell edge length is 50% more in solid 2 than in 1. What is approximate packing efficiency in solid 2?

14 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) A

A

•

A

•

•

•A

A

A

•

•

B

•A •

•

A

•A A

A

•

•A •A

•A •

•A

A

(b) hcp lattice A,

1 tetrahedral voids B 3

(c) hcp lattice B,

1 tetrahedral voids A 3

(d) hcp lattice B,

2 tetrahedral voids A 3

(d) Eav is doubled when its temperature is increased by four times. Q.14. At 100°C, copper (Cu) has FCC unit cell structure with cell edge length of x Å. What is the approximate density of Cu (in g cm–3) at this temperature? (Atomic mass of Cu = 63.55u] [JEE (Main)-2019-9th Jan-(Shift-I)]

•

A Solid 2 th [JEE (Main)-2019-8 April-(Shift-II)] (a) 65% (b) 90% (d) 45% (c) 75% Q.9. A solid having density of 9 × 103 kg m–3 forms face centred cubic crystals of edge length 200 2 pm. What is the molar mass of the solid? [Avogadro's constant = 6 × 1023 mol–3, p = 3] [JEE (Main)-2019-12th Jan-(Shift-I)] (b) 0.4320 kg mol–1 (a) 0.03050 kg mol–1 (c) 0.0432 kg mol–1 (d) 0.0216 kg mol–1 Q.10. The radius of the largest sphere which fits properly at the centre of the edge of a body centred cubic unit cell is ( Edge length is represented by a). [JEE (Main)-2019-11th Jan-(Shift-II)] (b) 0.027a (a) 0.134a (d) 0.067a (c) 0.047a Q.11. Which primitive unit cell has unequal edge lengths (a ¹ b ¹ c) and all axial angles different from 90°? [JEE (Main)-2019-11th Jan-(Shift-I)] (a) Hexagonal (b) Monoclinic (c) Tetragonal (d) Triclinic Q.12. A compound of formula A2B3, has the hcp lattice. Which atom forms the hcp lattice and what fraction of tetrahedral voids in occupied by the other atoms? [JEE (Main)-2019-10th Jan-(Shift-II)] 2 (a) hcp lattice A, tetrahedral voids B 3 Solid 1

Q.13. Which of the following statement(s) is are correct regarding the root mean square speed (Urms) and average translational kinetic energy (Eav) of a molecule in a gas at equilibrium? [JEE (Main)-2019-9th Jan-(Shift-I)] (a) Urms is double when its temperature is increased four times. (b) Eav at a given temperature does not depend on its molecular mass. (c) Urms is inversely proportional to the square root of its molecular mass.

CHEMISTRY

(a) (c)

211 x3 105 x3

205 x3

(b)

422

(d)



x3

Q.15. Which of the exhibit both Frenkel & Schottky defect ? (b) KCl (a) Ag Br (c) CsCl (d) Zns

ANSWERS – KEY

1. (b)

2. (a)

3. (a)

4. (d)



5. (c)

6. (b)

7. (b)

8. (b)



9. (a)

10. (d)

11. (d)

12. (c)



13. (a, b, c)

14. (d)

15.(a)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (b) is correct. In ccp lattice structure number of O2– ions = 1 1 8´ + 6´ = 4 8 2  In ccp number of O2 void = 4 In ccp number of Td void = 8

50 =2 100 12 × 5 =1 Number of M2 metal atoms = 8 ´ 100 Therefore formula of the compound is (M1)2 (M2) O4 i.e. it has spinel structure (Like Zn Al2 O4) Therefore, oxidation state of M1 is + 3 and oxidation state of M2 is + 2 Ans. 2. Option (a) is correct. In bcc number of particle (z) = 2 Z´M Again density (d) = N A ´ volume 2´M ⇒ 6 × 47 g cm -3 = 3 6 × 022 ´ 10 23 3 ´ 10 -8 cm Number of M1 metals atoms = 4 ´

(

)

200 g = 4 moles 50 gmol -1 = 4 NA

Therefore number of moles =´

Ans. 3. Option (a) is correct. In fcc crystal voids are located at the edge centers and body center. \ Minimum distance between the two nearest O2 voids in the crystal

15

STATES OF MATTER 2

Lattice =

2

a a a 2 +2 = 2     a/2

Ans. 4. Option (d) is correct. The only water soluble salt of alkaline earth metal is BeSO4 but its hydroxide Be(OH)2 is insoluble in water. Its oxide MO is a BeO have hexagonal wurtzite structure. Beryllium forms highly soluble (in water) beryllium sulfate (BeSO4). It forms water insoluble beryllium hydroxide (Be(OH)2. On heating, beryllium oxide becomes inert to heating due to increase in the degree of polymerization.. Beryllium hydroxide dissolves in sodium hydroxide solution to form sodium beryllate (Na2BeO2). Ans. 5. Option (c) is correct.

•

• •

•

•



1 No. of atoms of A = × No. of O2 voids 2

3 a = 2r + 2(2r) = 6r 6r =2 3r fi a = 3 4 3 4 p r + p ( 2r )3 3 3 \ Packing fraction = a3

a

•

No. of atoms of B = 4

1 = × 4 = 2 2 No. of ‘O’ atoms = No. of all Td voids = 8 =A : B : O = 2 : 4 : 8 = 1 : 2 : 4 The formula of the compound is AB2O4. Ans. 8. Option (b) is correct According to problem

a/2

•

\

•

4 p ( r 3 + 8r 3 ) = 3 3 ´ 8 ´ 3r 3 =

% of packing efficiency = 0.706 × 100 = 90.6 % Ans. 9. Option (a) is correct. Density of the unit cell Z´M Z´M = r = 3 N A ( apm ) N A a3 pm 3 fi 9 × 103 kgm–3 =

•

In fcc, tetrahedral voids are located on the body 3 diagonal at a distance of a from the corner. 4 1 æ ö cos–1 ç ÷ = angle between body diagonal and an è 3ø edge. Therefore, the projection of the line on an edge to a 4 a So the other tetrahedral void will also be 4 away. Hence distance between these two is é aù a a = ê a - 4 ú - 4 = 2 . ë û Ans. 6. Option (b) is correct. In simple cubic the no. of atoms = 8 × 8 = 1 In body centred cubic the no. of atoms 1 = 8 × +1=2 8 In face centered cubic the no. of atoms 1 1 = 8 × +6× =4 8 2 Hence the ratio is 1 : 2 : 4 Ans. 7. Option (b) is correct No. of lattice points = No. of O2 voids

4p ´ 9 p = = 0.906 3´3´8´ 3 2 3

4´M 6 ´ 10 23 mol -1 ( 200 2 ´ 10 -12 m)3 ( 9 ´ 10 3 )kg m -3 ´ ( 6 ´ 10 23 )mol -1

´ ( 200 2 ´ 10 -12 )3 m3 4 M = 0.03050 kg mol–1 Ans. 10. Option (d) is correct. fi

M =

R

R

2r

T 2r

a

R

In body centred cubic lattice \ fi

Body diagonal = 2a2 + a2 = a 3 Again Body diagonal = 4R 4R = a 3 R =

According to the problem 2(R + r) = a

a 3 ...(i) 4

16 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) fi

R + r =

a a 3 + r = 2 4 a a 3 fi r = 2 4 fi

2a - a 3 r = 4

fi

r =

fi fi

é a 3ù êSubstituting R = ú 4 úû êë

=

0.268 a 4

1 × Tv : hcp 3

1 × 12 : 6 3 fi 4 : 6 fi 2 : 3 fi

Therefore hcp lattice = B, 1 A= Tetrahedral voids. 3 Ans. 13. Option (a, b, c) are correct Hence Urms is double when its temperature is increased four times. Hence option (a) is correct. Urms =

3RT fi Urms µ M

1 M

3R ´ 4 T fi Urms 2 × M Hence option (c) is correct Urms =

3RT M

3 KT fi Eav does not depend on 2 molecular mass at given temperature. Hence option (b) is correct Ans. 14. Option (d) is correct For fcc no. of atoms present per unit cell (Z) = 4 Atomic mass of Cu = 63.55 u

Eav =

r =

=

fi r = 0.067a Ans. 11. Option (d) is correct Among the seven crystal system triclinic have primitive crystalline system in which intercepts (a ¹ b ¹ c) as well as interfacial angles (a ¹ b ¹ g ¹ 90°) are unequal. Ans. 12. Option (c) is correct In hcp unit lattice total effective number of atoms = Number of Octahedral voids =6 No. of Tetrahedral voids = 2 times of Octahedral voids = 2 × 6 = 12 Since formula of the compound is A2B3 Let A : B



Mass of the unit cell Volume of the unit cell Z´

a( 2 - 3 ) 4 a( 2 - 1.732 ) r = 4 r =

Edge Length (a) = xÅ = x × 10– 8cm Volume of the unit cell = (x × 10–8cm)3 = x3 × 10–24cm3 Density of the unit cell

a 2

fi

CHEMISTRY

M NA

x 3 ´ 10 -24 cm 3 4 ´ 63.55 g mol -1 6.023 ´ 10 23 mol -1 ´ x 3 ´ 10 -24 cm 3

422.048 g cm -3 = x3 Ans. 15. Option (a) is correct Ag Br is a compounds in which both schottky and Frenkel defects are found because Ag Br is highly ionic but there is a great difference in the size of positive and negative ions.

Integer Type Questions (Chapter Based) Q.1. A spherical balloon of radius 3 cm containing helium gas has a pressure of 48 × 10–3 bar. At the same temperature, the pressure of a spherical billon of radius 12 cm containing the same amount of gas will be ……. × 10–6 bar  [JEE Main 2020, 6th Sept (Shift - I)] Sol. For gas when temperature and number of moles are constant then we can apply Boyle's law. According to Boyle's law, P1V1 = P2V2 Given, P1 = 48 × 10–3 bar V1 = 4/3π(3)3 V2 = 4/3π(12)3 P2 = ? \ P1 V1 = P2 V2 P2 = {48 × 10–3 × (3)3 } / (12)3 = 7.5 × 10–4 = 750 × 10–6 bar Q.2. In a solid substances edge length of unit cell is 405 pm, density of solid is 10.9 g/cm3 and molar mass of substances is 109 g, then find radius of atom which form this units cell in pm.  [JEE Main 2020, 3rd sept (Shelf – I)] ZM Sol. We know that density(r) = N A ´ a3 Z ´ 109 gmol -1



⇒ 10.9 g cm–3 =



⇒ Z = 4 = fcc unit cell In fcc unit cell



4r = a 2 a

6 × 02 ´ 10 23 mol -1 éë 4 × 05 ´ 10 -8 cm ùû

3

1 × 414 ´ 405pm ⇒ r= = 143 × 17 pm . 4 Q.3. At 400 K, the root mean square (rms) speed of a gas A (molecular weight = 40) is equal to the most



17

STATES OF MATTER

probable speed of gas B at 60 K. The molecular weight of the gas B is Sol. Given, T1 = 400 K, T2 = 60 K Molecular weight of A (M1) = 40 Let the molecular weight of B be M2 3RT1 M1



\

Cr.m.s. =



give,

Cr.m.s. = Cmp



\



fi

3R ¥ 400 = 40 30 =

Cmp =

2 RT2 M2

2 R ¥ 60 M2 120 M2

fi M2 = 4 Q.4. If pd v/s, P (where P denotes pressure in atom and d denotes density in g/L) is plotted for He gas (assume ideal) at a definite temperature. If é d( pd ) ù = 5 . Find out the temperature? ê ú ë dp û P =8.21 Sol. From ‘n’ mole ideal gas, we get PV = hRT g fi PV = RT M

fi

P =

g RT V M



fi

P =

dRT M



fi

d =

PM RT



fi

Pd =



fi



fi



fi

P2 M RT

Q. 5. The density of CaO is 3.35 g/cm3. The oxide crystallizes in one of the cubic system with an edge of 4.80 Å. How many Ca2+ ions an O2– ions are present in each unit cell? ZM Sol. r = 3 a NA

ρa3 N A M 3.35 ( 4.8 ´ 10 -8 )3 ´ 6.023 ´ 10 23 fi Z = 56 fi Z = 3.98 fi Z ≈ 4.0 Therefore each unit cell contains 4 molecules of CaO. So no. of Ca2+ ions = 4 and O2– ions = 4 Q.6. The ionic radii of X+ and Y– ions are 0.98 × 10–10 m and 1.81 × 10–10 m. Find out the co-ordination number of each ion in XY? g X + 0.98 ´ 10 -10 = = 0.54 Sol. g Y - 1.81 ´ 10 -10

fi



When the value of radius ratio lies in between 0.414 and 0.732, the coordination number of each ion will be 6.

Q.7. In a solid AB having the NaCl structure A atoms occupy the corner of the cubic unit cell. If all the face-centered atoms along one of axes are removed, find out the total number of atoms present in unit cell. – Sol. Cl  ion is forming fcc unit cell as A and Na+ ion is in the octahedral voids as B. In NaCl crystal eight Cl – (A) ions are present in the corners and 6 Cl –(A) ions on the faces so that Cl –(A) ions per unit cell 1 1 =8× +6× =4 2 8

2 PM d( Pd) =5 = dp RT 2 ´ 8.21 ´ 4 = 50 0.0821 ´ T

T =

2 ´ 8.21 ´ 4 = 16 0.0821 ´ 50

Z =



No. of Cl – or A – ion after removing atoms along one 1 1 axes = 8 × +4× =3 2 8 1 +1=4 4 \ Total number of atoms in unit cell = 4 + 3 = 7 No. of B+ or Na+ = 12 ×



Fundamental Equation was developed by

• Heisenberg's Uncertainty Principle: It is impossible to determine simultaneously, the exact position and momentum of an electron. h Δx. Δpx > –= 4 ⇒ Δv. Δx = h 4 m

• Dual behaviour of atom i.e., particle and wavelike. De Broglie equation: λ= h = h mv p

Postulates: •Electron in H atom can move around the nucleus in a circular path of fixed radius and energy called as orbits. These orbits are arranged concentrically around the nucleus. •Each of these orbits has a definite energy known as energy levels or stationary states. •When an electron jumps from a lower energy level to higher one, some energy is absorbed. ΔE E2-E1 •v= n = n Bohr's frequency rule. Angular momentum of electron: mevr = n h n=1,2,3..... 2 Limitations: •Unable to account for finer details of H atom. Spectrum observed by sophisticated spectroscopic techniques. •Could not explain the ability of atoms to form molecules by chemical bonds.

Ty p

2

2

es :

s

p

d

(i)s p d ......... notation (ii)Orbital diagram

a b c

rog en Ato m

Tritium 31 T).

•Aufbau Principle : In the ground state of atoms, the orbitals are filled according to increasing energies. •Pauli Exclusion Principle : No two electrons in an atom can have same set of four quantum numbers. •Hund's rule : Pairing of electrons in the orbitals belonging to same subshell does not take place until each electron belonging to the subshell is singly occupied.

Atomic Structure

tron of Neu

It is the phenomenon of ejection of electrons from the surface of a metal when light of suitable frequency strikes on it.

1s

z

x x

2s

y

y

z

s Orbitals: Spherically symmetric Size increases with increase in size in n, ie, 4s>3s>2s>1s

y

2px

Z

2pz

x y x Size.4p>3p>2p

y

Z

Z

2py

x dxy

y

dx2–y2

y

x

x

dyz

z y

dz2

Z

dxz

z

x

d Orbitals – ‘’Clover leaf ’’ distribution. – Two angular nodes.

x

Dual nature i.e. wave like and particle like of the electromagnetic radiation. • Experimental results regarding atomic spectra. • Wave nature of electromagnetic radiation. It was given by James Maxwell. Frequency (ν): Number of waves that pass a given point in one second. Unit – Hertz (Hz), Velocity of light = Frequency × Wavelength Wave number (–ν): Number of wavelengths per unit length. Unit – m-1 • Particle nature of electromagnetic radiation: Planck's quantum theory: E= hν Planck's constant (h) = 6.626 × 10–34 Js p Orbitals: Each p Orbital consists of two sections called lobes on either side of plane passing through the nucleus.

Lower the value of (n+l) for an orbital, lower is its energy.

Positive space

Electron

Postulates: • Positive charge and most of the mass of atoms was densely concentrated in extremely small region i.e. nucleus. • Nucleus is surrounded by electrons that move around the nucleus with high speed in circular path called orbits. • Electrons and nucleus are held together by electrostatic forces of attraction. Drawbacks: • It cannot explain the stability of an atom. • It does not say anything about the electronic structure of atoms.

Atom possesses a spherical shape in which the positive charge is uniformly distributed.

By James Chadwick Charge on neutron = 0 Mass of neutron= 1.675 × 10–27 Kg

By Ernest Rutherford Charge on proton = +1.6022 × 10–19C Mass of Proton= 1.672 × 10–27 Kg

By J.J.Thomson Charge to mass ratio of electron 11 -1 = 1.758820 × 10 C kg Charge electron = 1.6022 × 10–19C Mass of electron = 9.1094 × 10–31Kg

: (Watermelon M odel of Atom odel) so n M Thom Rutherford's Nuclear Model of A tom

y over Disc

Atoms of different elements with different atomic number but same mass number. 40 (40 20Ca and 18Ar)

Atoms of same element having same atomic number but different mass number. (Isotopes of hydrogen: 1 Protium 1 H, Deuterium 21 D and

rs

Number of protons (Z) + number of neutrons (n)

ba Iso

Number of protons in nucleus of an atom or Number of electrons in a neutral atom

(i) Principal quantum number (n) :n = 1,2,3,4………….. Shell = K, L, M, N....... (ii) Azimuthal quantum number (l) : For given value of n, l = 0 to n-1 (iii) Magnetic quantum number (m) : for subshell with ‘l' value ml=2l+1 (iv) Spin quantum Number (ms): +1/ (↑), –1/ (↓)

Schrodinger as Hψ = E= where H= Hamiltonian

>

5

>



Bohr's Model for Hyd

• Emission Spectra : Spectrum of radiation emitted by a substance that has absorbed energy. • Absorption Spectra : It is like photographic negative of an emission spectra. • Line / Atomic Spectra : Emission spectra which do not show a continuous spread of wavelength from red to violet, rather they emit light only at specific wavelength with dark space between them. – = 109677 ( 1 1 ) cm–1 where v n1 2 n 2 2 n1=1,2………….n2=n1+1, n1+2……… Series n1 n2 Spectral Region Ultraviolet Lyman 2,3......... 1 Balmer Visible 2 3,4......... Paschen 3 Infrared 4,5......... Brackett 4 Infrared 5,6......... Pfund 5 6,7......... Infrared

18 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) CHEMISTRY

Atomic Structure

Chapter 3 Syllabus

Discovery of sub-atomic particles (electron, proton and neutron); Thomson and Rutherford atomic models and their limitations; Nature of electromagnetic radiation, photoelectric effect, spectrum of hydrogen atom, Bohr model of hydrogen atom - its postulates, deviation of the relations for energy of the electron and radii of the different orbits. Limitation of Bohr's model, Dual nature of matter, deBroglie's relation, heisenberg's uncertainty principle. Elementary idea of quantum mechanics, quantum mechanical model of an atom, Its important features Y1 and Y2, concept of atomic orbitals as one electron wave functions; variation of Y1 and Y2 with r for 1s and 2s orbitals; various quantum number (principal, angular momentum and magnetic quantum numbers) and their significance, shapes of s, p and d-orbitals, electron spin and spin quantum numbers; rules for filling electrons in orbital – Aufbau principle, Pauli exclusion principle and Hund's rule, electronic configuration of elements, extra stability of half filled and completely filled orbitals.

Topic-1

Bohr's Atomic Model, Sommerfeld Concept Concept Revision (Video Based) For more details, scan the code Bohr's Model for Hydrogen Atom

For more details, scan the code

LIST OF TOPICS : Topic-1 : Bohr's Atomic Model, Sommerfeld Concept .... P. 19 Topic-2 :  Quantum Approach .... P. 22 Topic-3 :  Photoelectric Effect .... P. 25

For more details, scan the code Sommerfeld atom model

Application of Bohr's theory Part-1

Part-2

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions

Q.1. The difference between the radii of 3rd and 4th orbits of Li2+ is ∆R1. The difference between the radii of 3rd and 4th orbits of He+ is ∆R2. Ratio ∆R1 : ∆R2 is  [JEE (Main)-5th Sept-2020-(Shift-I)] (a) 2 : 3 (b) 3 : 8 (c) 3 : 2 (d) 8 : 3 Q.2. The region of electromagnetic spectrum in which the Balmer series lines appear is  [JEE (Main)-4th Sept-2020-(Shift-I)] (a) Infrared (b) Microwave (c) Ultraviolet (d) Visible

Q.3. The shortest wavelength of H atom in the Lyman series is l1. The longest wavelength in the Balmer series of He+ is  [JEE (Main)-4th Sept-2020-(Shift-II)] 36 l 9l1 (b) (a) 5 5 27 l 1 (c) 5l 1 (d) 5 9 Q.4. No of subshells having n = 4 & m = −2 are  [JEE (Main)-2nd Sept-2020-(Shift-II)] (a) 2 (b) 4 (c) 8 (d) 16

20 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Q.5. Determine wavelength of electron in 4th Bohr’s orbit of H-atom?[JEE (Main)-9th Jan-2020-(Shift-I)] (b) 2pa0 (a) 4pa0 (c) 8pa0 (d) 6pa0 Q.6. Given for H – atom é 1 1ù u = RH ê 2 - 2 ú ë n1 n2 û Select the correct option regarding this formula for Balmer series (A) n1 = 2 (B) Ionisation energy of H atom can be calculated from above formula (C) lmaximum is for n = 3 (D) If l decreases then spectrum lines will converse. [JEE (Main)-8th Jan-2020-(Shift-I)] (b) C, D (a) A, B (c) A & C (d) A, B, C & D Q.7. Determine Bohr’s radius of Li2+ ion for n = 2. Given (Bohr’s radius of H atom for n = 1 is a0)  [JEE (Main)-8th Jan-2020-(Shift-II)] 4a 3a (b) 0 (a) 0 3 4

a0 (d) 16 a0 3 9 Q.8. The ratio of the shortest wavelength of two spectral series of hydrogen spectrum is found to be about 9. The spectral series are [JEE (Main)-10th April-2019-(Shift-I)] (a) Lyman and Paschen (b) Brakett and Pfund (d) Balmer and Brackett (c) Paschen an Pfund Q.9. For any given series of spectral lines of atomic hydrogen. Let Dn = nmax – nmin will be the difference in maximum and minimum frequencies Dn Lyman is in cm–1. The ratio Dn Balmer (c)

[JEE (Main)-9th April-2019-(Shift-I)] (a) 27 : 5 (b) 5 : 4 (d) 4 : 1 (c) 9 : 4 Q.10. The ground state energy of hydrogen atom is –13.6 ev. The energy of second excited state of He+ ion in eV is [JEE (Main)-10th Jan-2019-(Shift-II)] (a) –54.4 (b) –3.4 (c) –6.04 (d) –27.2 Q.11. Heat treatment of muscular pain involves radiation of wavelength of about 900 nm. Which spectral line of H-atom is suitable for this purpose? [RH = 1 × 105 cm–1, h = 6.6 × 10–34 Js, c = 3 × [JEE (Main)-9th Jan-2019-(Shift-I)] 108 ms–1] (a) Paschen 5 ® 3 (b) Paschen • ® 3 (c) Lyman • ® 1 (d) Balmer • ® 2 Q.12. For emission line of atomic hydrogen from xi = 8 1 to xf = x, the plot of wave number (n) against æç ö÷ è n2 ø will be The Rydberg constant, (RH) is in wave number unit) [JEE (Main)-9th Jan-2019-(Shift-I)]

CHEMISTRY

(a) Non linear (b) Linear with slope – RH (c) Linear with slope RH (d) linear with intercept – RH Q.13. Which of the following combination of statements is true regarding the interpretation of the atomic orbitals? [JEE (Main)-9th Jan-2019-(Shift-II)] I. An electron in an orbital of high angular momentum stays away from the nucleus than an electron in the orbital of lower angular momentum. II. For a given value of the principal quantum number, the size of the orbit in inversely proportional to the azimuthal quantum number. III. According to wave mechanics, the ground sate h angular momentum is equal to 2≠ IV. The plot of Y vs r for various azimuthal quantum numbers, shows peak shifting towards higher r value. (b) II, III (a) I, III (d) I, IV (c) I, II

ANSWERS – KEY

1. (a)

2. (d)

3. (a)

4. (a)



5. (c)

6. (c)

7. (b)

8. (a)



9. (c)

10. (c)

11. (b)

12. (d)



13. (a)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (a) is correct. we know that rn = 0 × 529 \   

(r ) Li 2 +

n=4

( )

- rLi 2+

n2 A° Z

n=3

And for He+   

(r ) He+

n=4

\

(

- rHe+  

\

)

n=3

=

=

0 × 529 (16 - 9 ) = DR1 3

0 × 529 (16 - 9 ) = DR 2 2

DR1 2 = DR 2 3

∆R1 : ∆R2 = 2 : 3

Ans. 2. Option (d) is correct. In hydrogen spectrum Balmer series lines are appeared in visible region. Ans. 3. Option (a) is correct. As Energy µ

1 Wavelength

We know that

Þ

æ 1 1 1ö = RH ç 2 - 2 ÷ l1 è h1 h2 ø

1 1 ö æ1 = RH ç 2 - 2 ÷ l1 è1 ¥ ø

21

ATOMIC STRUCTURE

1 λ1 For He+ ion in Balmer series 1 1  1 RH × 2 2  2 − 2  λ =   2 3 He + Þ RH =

Now if x1 = 3 and x2 = 1 Therefore the spectral series Paschen and Lyman. Ans. 9. Option (c) are correct. From Rydberg equation

1 5 2 Þ λ =RH × ( 2 ) × 36 He +

Þ

1 5 = RH × λ He + 9

Þ

1 1 5 = × λ He + λ 1 9

x1 3 = x2 1

\

Ê 1 1ˆ n = RH Á 2 - 2 ˜ Ë x1 x 2 ¯ For Lyman series

1 ö æ 1 n max = RH ç 2 - 2 ÷ ¥ ø è1



9λ Þ λ He + =1 5 Ans. 4. Option (a) is correct. when n = 4, l = 0, 1, 2, 3, ml For l = 2, ml = +2, +1, 0, −1, (−2) And again l = 3, ml = +3, +2, +1, 0, −1, (−2), 3 Therefore 2 sub shells are possible

1 ö æ = RH ç 1 - 2 ÷ ¥ ø è

Ans. 5. Option (c) is correct. We know that 2prn = nl n2 Þ 2 π × a0 = nλ Z

For Balmer series,

na Þ 2π × 0 = λ Z Þ

 λ = 2 × π ×

...(iii)



æ1 1 ö n max = RH ç - ÷ è4 ¥ø

...(iv)



Ê 1 1ˆ n min = RH Á - ˜ Ë 4 9¯

...(v)



(A) n1 = 2 for Balmer series is correct (C) Energy required to transfer e– from 2nd orbit to 3rd orbit in Balmer series is lowest. Hence l is maximum for this transition Ans. 7. Option (b) is correct. a x2 rx = o Z

ao × ( 2 ) 4 ao = Þ r2 = 3 3 2

\

1 4

...(ii)

1 9

...(vi)

From (iii) and (iv)

Ans. 6. Option (c) is correct.

RH l1 x2 9 x2 = 1 = 12 = RH x l2 1 2 x 22

\ n max - n min = RH ¥

\ n max - n min = RH ¥

4 a0 = 8 πa0 1

Ans. 8. Option (a) are correct. From Rydberg equation æ 1 1 1 ö = RH çç 2 - 2 ÷÷ l1 è x1 ¥ 2 ø æ 1 1 ö 1 RH ç 2 - 2 ÷ and = ÷ ç l2 è x1 ¥ 2 ø as for shortest wavelengths

1ˆ Ê n (min) = RH Á 1 - ˜ Ë 4¯



...(i)

Dn Lyman Dn Balmer

=

9 4

Ans. 10. Option (c) is correct. The ground state energy of hydrogen atom is –13.6 eV. We know that

E = – 13.6 ×

z2 x2

For He, 22 E (He+) = –13.6 × 2 [Here z = 2 3 n = 2 + 1 = 3]

...(i) ...(ii)

= –6.04 eV Ans. 11. Option (b) is correct. We know that Rydberg equation, fi

fi

Ê 1 1ˆ 2 1 = RH Á 2 - 2 ˜ z l Ë x1 x 2 ¯ Ê 1 1 1ˆ Á 2 - 2 ˜ = 2 R ¥ Ë x1 x 2 ¯ H l¥z 1 Ê 1 1ˆ Á 2 - 2 ˜ = (1 ¥ 107 m -1 ) ¥ (900 ¥ 10 -9 m) ¥ (1)2 Ë x1 x 2 ¯ [For H, z = 1]

22 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) fi

1 x12

-

1 x 22

=

1 9

It is only possible when x1 = 3, x2 = • Ans. 12. Option (d) is correct. According to Rydberg equation,

Ê 1 1ˆ 2 Wave number n = - RH Á 2 - 2 ˜ z Ë x 2 x1 ¯ [Where z = 1] 1ˆ Ê 1 n = - RH Á 2 - 2 ˜ Ëx 8 ¯ [∵x2 = x, x1 = 8] RH RH n = 64 - 2 x

fi

fi

CHEMISTRY

µ distance from the nucleus II. Size of orbit µ Azimuthal quantum number(l) [When x = constant] So statement II is wrong nh III. mvr = 2p For Ground state, n = 1 Angular momentum = mvr =

h 2p

So, statement III is correct IV. The plot of Y vs r for various azimuthal quantum number as shown below. L = 0 (n = 1) l = 1 (n = 2)

Computing with equation of straight line y = mx + L, we get 1 x = 2 Slope (x) = – RH x Ans. 13. Option (a) is correct. xh I. Angular momentum mvr = 2l fi mvr µ x

l = 2 (n = 3) 

l = 3 (n = 4)

Topic-2

Quantum Approach Concept Revision (Video Based) For more details, scan the code Quantum Concept

   

  

For more details, scan the code Aufbau Principle            

Heisenberg uncertainty principle

Pauli's Exclusion Principle

   

Hund's Rule              

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions Q.1. The correct statement about probability density (except at infinite distance from nucleus) is  [JEE (Main)-5th Sept-2020-(Shift-I)] (a) It can be zero for 1s orbital (b) It can never be zero for 2s orbital (c) It can be negative for 2p orbital (d) It can be zero for 3p orbital Q.2. Consider the hypothetical situation where the azimuthal quantum number takes values

0, 1, 2……(x + 1) where x is the principal quantum number. Then, the element with atomic number 

[JEE (Main)-3rd Sept-2020-(Shift-II)] (a) 8 is the first noble gas (b) 9 is the first alkali metal (c) 6 has a 2p valence subshell (d) 13 has a half filled valence subshell

23

ATOMIC STRUCTURE

Q.3. The electrons are more likely to be found a (x) x b x c

[JEE (Main)-12th April-2019-(Shift-I)] (a) In the region a and c (b) In the region a and b (c) Only in the region a (d) Only in the region c Q.4. The graph between |Y|2 and r (radial distance) is shown below. This represents [JEE (Main)-10th April-2019-(Shift-I)]

Intensity

||2

II. n = 3, l = 2 ml = 4 ms = +

1 2

III. n = 4, l = 1 ml = 0 ms = +

1 2

1 2 The correct order of their increasing energies will be : [JEE (Main)-8th April-2019-(Shift-I)] (a) IV < III < II < I (b) I < II < III < IV (c) IV < II < III < I (d) I < III < II < IV



Wavlength Increasing wavelength

(b) Adsorption spectrum

(d)

Internal Energy of Ar

(c)

K. E of photoelectron

r (a) 1s-orbital (b) 2p-orbital (c) 3s-orbital (d) 2s-orbital Q.5. Which one of the following about an electron occupying the 1s-orbital in a hydrogen atom is incorrect? (The Bohr radius is represented by a0) [JEE (Main)-9th April-2019-(Shift-II)] (a) The total energy of an electron is maximum when it is at a distance a0 from the nucleus. (b) The magnitude of the potential energy is double that of its kinetic energy of an average. (c) The probability density of finding the electron is maximum at the nucleus (d) The electron can be found at a distance 2a0 from the nucleus. Q.6. The quantum number of four electrons are given below: 1 I. n = 4, l = 2 ml = – 2 ms = 2

IV. n = 3, l = 1 ml = 1 ms = -

Q.7. If the de-Broglie wavelength of the electron in nth Bohr orbit in a hydrogen atom in equal to 1.5 pa0 x (a0 is Bohr radius), than the value of is z [JEE (Main)-12th Jan-2019-(Shift-II)] (a) 1.0 (b) 0.75 (c) 0.40 (d) 1.5 Q.8. The 71st electron of an element X with an atomic number of 71 enters into orbital [JEE (Main)-10th Jan-2019-(Shift-II)] (a) 4f (b) 6p (c) 5d (d) 6s Q.9. The de-Broglie wavelength (l) associated with a photoelectron varies with the frequency (n) of the incident radiation as [n0 is threshold frequency] [JEE (Main)-12th Jan-2019-(Shift-II)] 1 1 (a) l µ (b) l µ 3/2 1/4 n n ( (n 0 - n 0 ) 0) 1 1 (c) l µ (d) l µ 1/2 n n ( (n - n 0 ) 0) Q.10. The figure that is not a direct manifestation of the quantum nature of atom is T2>T1 (a) Black T2 body radiation T1

Pb K Na

Frequency of incident radiation

300 400 500 600 700

Temperature (K)

Q.11. The ground state energy of hydrogen atom is –136 eV. Consider an electronic state Y of He+ whose energy, azimuthal quantum number and magnetic quantum number are –3.4 eV, 2 and 0, respectively. Which of the following statement(s) is (are) true for the state (Y) ? (a) It is a 4d state (b) It has 2 angular nodes (c) It has 3–radial nodes (d) The nuclear charge experienced by the electron in this state in less than 2e– where e is the magnitude of the electronic charge.

24 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) ANSWERS – KEY

1. (d)

2. (a, b)

3. (a)

4. (d)



5. (a)

6. (c)

7. (b)

8. (c)



9. (d)

10. (d)

11. (d)



Kinetic energy (K.E.) =

\

CHEMISTRY

1 KZe 2 2 r

|(P.E.)| = 2 |(K.E.)|

2

(c) Y  (Probability density) is maximum at nucleus

ANSWERS WITH EXPLANATIONS Ans. 1. Option (d) is correct. In Schrodinger wave equation y has no physical significance but y2 has physical significance. y2 represent the probability of finding electron around nucleus in space. The variation of y2 at different distance as shown in below. 1s Ψ

(d) Electron can be found at any distance from nucleus P = Probability function

2s Ψ

²

²

Node

r

r

Ψ

²

Ψ

²

r

Potential energy (P.E.) = -

2 (Probability density)

2



node r

Ans. 2. Options (a, b) are correct. When n = 1, l = 0, 1, 2 and when n = 2, l = 0, 1, 2, 3 So, according to (n + l) selection rule subshells have the following order of increasing energy levels 1s, 1p, 2s, 1d 2p, 3s, 2d 3p 4s (a) For the 1st noble gas configuration is 1s2 1p6 having atomic number 8 (b) For the first alkali metals 1s2 1p6 2s1, having atomic number 9 (c) When z =6, the configuration is 1s2 1p4 (a) When z =13, the configuration = 1s2 1p6 2s2 1d3 where d – orbital in not half filled. Ans. 3. Option (a) is correct. Radial probability function P(x) = 4px2 × Yxz Therefore probability will be max at a and c. Ans. 4. Option (d) is correct. Graph of |Y|2 vs r, touches at 1 point so it has one radial node and since at r = 0, it has some value so it should be for ‘s’-orbital \ n – l – 1 = 1  [Where l = 0, it has one radial node] fi n – 1 = 1 fi n = 2 fi ‘2s’ orbital Ans. 5. Option (a) is correct. (a) Total energy of electron is minimum in first orbit i.e. at a0 distance from nucleus. Hence, (a) option is correct. (b)

s

K Ze 2 r

r

Ans. 6. Option (c) is correct. According to selection rule, smaller the value of (n + l), lower is the energy level or between the two subshell having same value of (n + l), the subshell which have lower value of n will be lower level. n

l

n+l

I

4

2

4+2=6

II

3

2

3+2=5

III

4

1

4+1=5

IV

3

1

3+1=1

Therefore the correct order of their increasing energies will be IV < II < III < I Ans. 7. Option (b) is correct. 2pr = xl 2 πr 2 πx 2 a0 x = = 2 π a0 fi l = x x×z z l = 1.5 pa0 x \ 2p a0 = 1.5 pa0 z fi

1.5 x = = 0.75 2 z

Ans. 8. Option (c) is correct. The element which have atomic number 71 belong in the Lanthanide series, from La(58) to Lu(71). lanthanide elements have general outermost electronic configuration (n – 2) f 1–14 (n – 1) d0–1 ns2 Y6 (70) has electronic configuration [Xe] f 14 6s2. So

25

ATOMIC STRUCTURE

71st electron of Lu(71) has electronic configuration [Xe] f 14 5d1 6s2. Ans. 9. Option (d) is correct. de-Broglie wave length in terms of kinetic energy we get, h l = ...(i) 2mKE Again, according to photoelectric effect, KE = hn – hn0 ...(ii) h l= 2m( hn - hn 0 ) l=

h 2m( n - n 0 )

Þlµ

1 (n - n0 )

Ans. 10. Option (d) is correct. a, b, c are according to quantum theory and d is according to kinetic theory of gases. From (i) and (ii) we get, h l = 2m ¥ ( hn - hn 0 )

\

l =

1 ( n - n 0 )n 2

Ans. 11. Option (d) is correct. En for He+ = -13.6 ¥ \ –13.6 ×

4 x2

x2

= –3.4

x2 =

fi

z2

13.6 4 ¥4 3.4

fi x = 4 Azimuthal quantum number (l) = 2 Magnetic quantum number (x) = 0 \ It's 4d orbital Radial nodes = n – l – 1 = 4 – 2 – 1 = 1 Angular nodes L = 2 As He­ion is a single electron species, i.e. nuclear charge experienced in 2e–.

Topic-3

Photo Electric Effect Concept Revision (Video Based)



For more details, scan the code Photo Electric Effect



For more details, scan the code Schrodinger Equation

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions

(b)

O

Number of e– (s) O

(d)

Energy of Light

K.E. of electron

O

(c)

O

Intensity of Light

Frequency of Light

K.E. of electron

(a)

K.E. of electron

Q.1. Which of the graph shown below does not represent the relationship between incident light and the electron ejected from metal surface? [JEE (Main)-9th Jan-2019-(Shift-I)]

Frequency of Light

Q.2. What is the work function of the metal, if the light of wavelength 4000 Å generates photoelectron of velocity 6 × 105 ms–1 from it? [Mass of electron

26 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

(c)

2 l 3

(d)

2. (b)

3. (a)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (d) is correct. The kinetic energy of the ejected electron in photoelectric effect is given by

hn = hn0 + KE

fi

KE = hn – hn0

[Where hn0 = photoelectric work function = E0 fi

KE = E – E0 n = Frequency of incident light n0 = Threshold energy



KE = Kinetic energy of ejected electron]

(a) Slope = 1 Intercept = E0

KE O

No. of ejected e's vs frequency of light This is correct option. (d) Slope = +h Intercept = –h0

KE O

0



Therefore option (d) is wrong Ans. 2. Option (b) is correct. Work function of metal (f) = hn0 [n0 = threshold frequency] Also 1 m n 2 = hn – hn0 2 e fi 1 men 2 = hn – f 2 1 fi × 9 × 10–31 kg (6 × 105 ms–1)2 – 2

hn = Energy of incident light = E

O Frequency of Light (ν )

1 l 2

ANSWERS – KEY 1. (d)

(c) Number of electrons

= 9 × 10–31 kg, Velocity of light = 3 × 108 ms–1, Planck's constant = 6.626 × 10–34 Js, Charge of electron = 1.6 × 10–19 J] [JEE (Main)-12th Jan-2019-(Shift-I)] (a) 4.0 eV (b) 2.1 eV (c) 0.9 eV (d) 3.1 eV Q.3. If p is the momentum of fastest electron ejected from a metal surface after the irradiation of light having wavelength l, then for 1.5 p momentum of the photo electron, the wavelength of the light should be (Assume kinetic energy of ejected photo electron to be very high comparison to work function) [JEE (Main)-8th April-2019-(Shift-II)] 3 4 (a) l (b) 4 l 9

E0

4000 ¥ 10 -10 m

m 2n 2 p 2 = 2m 2m [p = mn = momentum]

For new wave length KE

Intensity of Light

KE is independent of intensity it does not depends on frequency of light.

-f

Ans. 3. Option (a) are correct. In photoelectric effect hc = w + KE of electron l Given that KE of ejected photoelectron is very high in comparison to work function w 1 hc = KE = mn2 2 l

Graph (a) is correct (b)

6.626 ¥ 10 -34 3s ¥ 3 ¥ 10 8 ms-1

fi f = 1.62 × 10–21 kg m2s–2 – 4.96 × 1019J = 3.36 × 10–19 J [1 kg m2 s2 = 1 J] = 2.1 eV

=

O

CHEMISTRY



(1.5 p )2 225 ¥ p 2 lc = = 2m 200 ¥ m l1

fi

lc 9 hc = l1 4 l

fi

l1 =

4 l 9

27

ATOMIC STRUCTURE

Integer Type Questions (Chapter Based)



For He+,

Q.1. A metal having work function = 4.41 × 10−19 J is subjected to a light having wavelength 300 nm, then maximum kinetic energy of the emitted photoelectron is …………. × 10−21 J (Given h = 6.63 × 10−34 Js & C = 8 × 108 m/s) [JEE (Main)-3rd Sept.-2020-(Shift-I)] Sol. From Einstein photoelectric equation we get, 1 hu = KE + hu0 = mv 2 + hu0 2







E3 (for He+)



= -



Þ KE = hu − W hc Þ KE = -W

l

6 × 63 ´ 10 -34 JS ´ 3 ´ 10 8 ´ ms -1 Þ KE = - 4 × 41 ´ 10 -19 J 300 ´ 10 -9 = 2.22 × 10−19 J = 222 × 10−21 J Q.2. How many electrons in an atom with atomic number 105 can have (n + l) = 8? Sol. 17 e’s 5f14, 6d3 5f14 = (n + l) = 5 + 3 = 8 6d3 = (n + l) = 6 + 2 = 8

Q.3. An ion Mn+ has magnetic moment equal to 4.9 BM. Find out the value of x? Sol. We know that magnetic moment = x( x + 2 ) Therefore x( x + 2 ) = 4.9 fi x(x + 2) = (4.9)2 fi x2 + 2x = 24.01 fi x2 + 2x – 24 = 0 fi x2 + 6x – 4x – 24 = 0 fi x(x + 6) – 4(x + 6) = 0 fi (x + 6) 0 or x – 4 = 0 \ x = 4 Q. 4. According to Bohr's theory, the electronic energy of hydrogen atom in the nth Bohr's orbit is given by En =

-21.76 ¥ 10

-19

n2

J

Calculate the longest wavelength of electron from the third Bohr's orbit of the He+ ion. Sol.



En = -

21.76 ¥ 10 x2

-19

J



En = -

21.76 ¥ 10 -19 x2

z2

21.76 ¥ 10 -19 ¥ 2 2

32 = – 9.68 × 10–19 J Energy required to remove an e– from the third bohr orbit of He+ DE = 0 – E3 = 0 – (– 9.68 × 10–19) J = 9.68 × 10–19 J hc l = DE =

6.626 ´ 10 -34 Js ´ 3 ´ 10 8 ms-1 9.68 ´ 10 -19 J

= 2050 × 10–10 m = 2050 Å Q.5. In a sample of H-atom, electrons make transition from 6th excited state upto ground state, producing all possible type of photons. Find out total number of lines which are in infrared region? Sol. Infrared lines = Total lines – visible lines – UV lines 7(7 - 1) – 5 – 6 = 21 – 11 = 10 = 2 Visible lines = 5, 6 ® 2, 6 ® 2, 5 ® 2, 4 ® 2, 3 ® 2 UV lines = 6;7 ® 3, 6 ® 1, 5 ® 1, 4 ® 1, 3 ® 1, 2 ® 1 Q.6. In an atom, the total number of electrons having quantum numbers 1 n = 4, |ml| = 1 and ms = is. 2

Sol. When n = 4, the total possible orbitals are 4s

4p

0

–1

0

4d +1

+2 +1

0

–1

–2

4f +3

+2

+1

0

–1

–2

–3

Again |ml| = 1, total possible values are ml i.e., +1 and –1 and one orbital contains maximum two 1 or ml = electrons having spin either ms = + 2 1 . Therefore total no. of electrons = 6 (When 2 1 ms = – ). 2 –



(i) Bond Length : Equilibrium distance between the nuclei of two bonded atoms in molecule. (ii) Bond Angle : Angle between the orbitals containing bonding electron pairs around central atom in a molecule complex ion. (iii) Bond Enthalpy : Amount of energy required to break one mole of bonds of particular type between two atoms. (iv) Bond Order : Number of bonds between the two atoms of a molecule. (v) Resonance Structures : A set of two or more Lewis structures that collectively describe the electronic bonding a single polyatomic species. (vi) Dipole Moment : Product of the magnitude of the charge and distance between centres of positive and negative charge. µ= Q x r

Postulates : • Shape of molecule depends upon the number of valence shell electron pairs around central atom. • Pairs of electrons in the valence shell repel one another. • These pairs of electrons tend to occupy such positions in space that minimize repulsion. • The valence shell is taken as a sphere with electron pairs localising on spherical surface at maximum distance from one another. • A multiple bond is treated as if it is a single electron pair and the two or three electron pairs of a multiple bond are treated as a single super pair. • When one or more resonance structures can represent a molecule, VSEPR model is applicable. • Decreasing order of repulsive interaction : lp – lp > lp – bp > bp – bp Valence Bond Theory : Given by L Pauling. It explains that a covalent bond is formed between two atoms by overlap of their half-filled valance orbitals, each of which contains one unpaired electron. Orbital Overlap Concept : Formation of a covalent bond results by pairing of electrons in valence shell with opposite spins. Types of Overlapping : (i) Sigma (σ) bond – end to end. (ii) Pi (π) bond – axis remain parallel to each other. Hybridisation : Process of intermixing of orbitals of different energies resulting in formation of new set of orbitals of equivalent energies and shape. Types of Hybridisation –(i) sp (ii) sp2 (iii) sp3 Bonding Molecular Orbitals : Addition of atomic orbitals. Antibonding Molecular Orbitals : Subtraction of atomic orbitals.

Energy required to completely separate one mole of a solid ionic compound into gaseous constituent ions.

.B

A

A .. B

.B

120°

AB4 B

.. B

A

5'

9°

10

B Tetrahedral

BF3

A

B CH4,NH4+

.B.

B:

AB6 90° :B

.. B

:B PCl5

Trigonal bipyramidal

120°

AB5 :B

90°

.B.

.B. Octahedral

A

.. B

B

AB2E

Chemical Bonding and Molecular Structure

.B.

B:

90°

B

SO2,O3

SF6

Bent

A

..

E

Kossel Lewis approach to chemical bonding :

B

NH3

H2 O :E B B Bent (Tetrahedral)

A

..

E

B Trigonal pyramidal (Tetrahedral)

B

A

E

..

A B

B SF4

A E:

E: CIF3 B T–shape(Trigonalbipyramidal)

AB3E2 B

B

B See saw (Trigonalbipyramidal)

AB4E :E

B

B

..

A

B

B

B BrF5

AB4E2

..

A

E B XeF4 B E .. Square planar (Octahedral)

B

B

E Square Pyramidal (Octahedral)

AB5E

B

* Lewis pictured the atom as a positively charged 'kernel' and the outer shell accommodates a maximum of eight electrons. • Lewis postulated that atoms achieve the stable octet when linked by chemical bonds. • Kossel gave following facts: * In the periodic table, highly electronegative halogens and highly electropositive alkali separated by noble gases. * Formation of a negative ion from a halogen atom and a positive ion from an alkali metal atom is associated with gain and loss of electron by respective atoms. * Negative and positive ions formed attain noble gas electronic configuration. • Negative and positive ions are stabilized by electrostatic attraction. Octet Rule : Atoms can combine either by transfer of valence electrons from one atom to another or by sharing of valence electrons to complete octet in their valence shells. Lewis dot Structure provides a picture of bonding in molecules and ions in terms of the shared pairs of electrons and the octet rule. How To Write A Lewis Dot Structure: Step 1 : Add the valence electrons of the combining atoms to obtain total number of electrons. Step 2 : For anions, each negative charge means addition of one electron. For cations, each positive charge means subtraction of one electron from total number of valence electrons. Step 3 : Write chemical symbols of combining atoms. Step 4 : Least electronegative atom occupies central position. Step 5 : After accounting for shared pairs of electrons, remaining are either utilized for multiple bonding or remain as lone pairs. Formal Charge= (Total number of valence electrons in free atom) – [(Total number of non-bonding electrons) + 1/2(Total number of bonding electrons)] Limitations Of Octet Rule : • Shows three types of exceptions i.e. incomplete octet of central atom, odd-electron molecules and expanded octet. • Does not account for the shape of molecules. • Fails to explain stability of molecules. Hydrogen Bond: Formed when the negative end of one molecule attracts the positive end of other. Types: (i) Intermolecular : Between two different molecules of same or different substances. (ii) Intramolecular : Between two highly electronegative atoms in the same molecule.

AB3E

AB2E2

Postulates : • Electrons in a molecule are present in various molecular orbitals as electrons are present in atomic orbitals. • Atomic orbitals of comparable energies and proper symmetry combine. •Atomic orbitals is monocentric while a molecular orbital is polycentric. • Number of molecular orbitals formed is equal to number of combining atomic orbitals. • Bonding molecular orbitals has low energy and high stability. Types of MO : σ(Sigma), (Pi), δ (Delta)

180° :B BeCl2,HgCl2

Trigonal planar

AB3

AB2 :B

Types : (i) Covalent Bond : A chemical bond formed between two atoms by mutual sharing of electrons between them to complete their octet. (ii) Ionic Bond : A chemical bond formed by complete transfer of electrons from one atom to another acquire the stable nearest noble gas configuration. (iii) Coordinate bond : A chemical bond formed by donation of two electrons from one atom to another to complete their octct.

28 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) CHEMISTRY

Chapter

Chemical 4 Bonding and Molecular Structure

Syllabus Kossel Lewis approach to chemical bond formation, concept of ionic and covalent bonds. Ionic Bonding formation of ionic bonds, factors affecting the formation of ionic bonds. Calculation of lattice enthalpy. Covalent bonding concept of electronegativity, Fajan's rule, dipole moment, valence shell electron pair repulsion (VSEPR) theory and shapes of simple molecules. Quantum mechanical approach to covalent bonding : Valence bond theory - its important features, concept of hybridization involving, s, p and d-orbitals, resonance. Molecular orbital theory : Its important features, LCAOs, types of molecular orbitals (bonding, antibonding), sigma and pi-bonds, molecular orbital, electronic configuration of homonuclear diatomic molecules, concept of bond order, bond length and bond energy. Elementary idea of metallic bonding hydrogen bonding and its applications.

Topic-1

LIST OF TOPICS : Topic-1 : Development of Chemical Bonding .... P. 29

Development of Chemical Bonding Concept Revision (Video Based) For details, scan the code Covalent & Ionic Bonding



Topic-2 :  Molecular Orbital Theory 

.... P. 33

For details, scan the code Fajan's Rule Concept of Polarisation  

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions

(a) Pyramidal

Q.1. If AB4 molecule is a polar molecule, a possible

(b) Square planar (c) Trigonal or bipyramidal

geometry of AB4 is 

[JEE (Main)-2

nd

Sept-2020-(Shift-I)]

(a) Square pyramidal

(d) Trigonal planar Q.3. Boiling point of water is 373 K, then boiling point of H2S is

(b) Rectangular planar (c) Square planar (d) Tetrahedral Q.2. The structure of SF6 is octahedral. What is the structure of SF4 (including lone pairs if any) 

[JEE (Main)-2nd Sept-2020-(Shift-II)]



[JEE (Main)-2nd Sept-2020-(Shift-I)] (a) Less than 300 K (b More than 300 K but less than 373 K (c) More than 373 K (d) More than 300 K

30 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

C

O

O

H

O [JEE (Main)-3rd Sept-2020-(Shift-II)]



(1) (B) is more likely to be crystalline than (A)



(2) (B) has higher boiling point than (A)



(3) (B) dissolves more readily than (A) in water



Identify the correct option from below [JEE (Main)-3rd Sept-2020-(Shift-II)]



(a) only (1) is true (b) (1) and (2) are true (c) (2) and (3) are true (d) (1) and (3) are true Q.5. The intermolecular potential energy for the molecules A, B, C and D given below suggests that Potential Energy (KJmol–1 )

0



50

100

150

–200 –300

A–D

–400

–600

A–C

A –A A –B

Inter atomic distance(pm)

[JEE (Main)-4th Sept-2020-(Shift-I)]



(d) Energy Internuclear distance

Internuclear

distance Q.8. Lattice enthalpy and enthalpy of solution of NaCl are 788 kJ mol−1 and 4 kJ mol−1 respectively. The hydration enthalpy of NaCl is  [JEE (Main)-5th Sept-2020- (Shift-II)]

(a) −784 kJ mol−1

(b) −780 kJ mol−1

(d) 784 kJ mol−1 (c) 780 kJ mol−1 Q.9. Correct order of intermolecular forces. [JEE (Main)-7th Jan-2020-(Shift-I)] (a) Dipole – Dipole > ion – ion > Dipole – ion (b) ion – ion > Dipole – Dipole > Dipole – ion (c) Dipole – ion > Dipole – Dipole > ion – ion (d) ion – ion > Dipole – ion > Dipole – Dipole Q.10. Decreasing order of dipole moment in CHCl3, CCl4 & CH4 is  [JEE (Main)-7th Jan-2020-(Shift-I)] (a) CHCl3 > CCl4 = CH4 (b) CHCl3 > CCl4 > CH4

–100

–500

(c) Energy

Q.4. Consider the following molecules and statements related to them O–H O H (A) = (B) = .. O .. C H

CHEMISTRY

(a) A−D has the shortest bond length (b) D is more electronegative than other atoms (c) A − A has largest bond enthalpy

(c) CCl4 > CHCl3 > CH4 (d) CCl4 = CH4 > CHCl3 Q.11. Following van der Waal forces are present in ethyl acetate liquid  [JEE (Main)-8th Jan-2020-(Shift-I)] (a) H – bond, London forces (b) dipole – dipole interaction, H – bond (c) dipole – dipole interaction, London forces (d) H – bond, dipole – dipole maturation, London forces Q.12. There are two compounds A and B of molecular formula C9H18O3, A has higher boiling point than B. what are the possible structures of A and B ?  [JEE (Main)-8th Jan-2020-(Shift-II)]

(a)

CH3O

(d) A − B has the stiffest bond Q.6. The structure of PCl5 in the solid state is

(a) Tetrahedral [PCl4+] and octahedral [PCl6−]

(b) Trigonal bipyramidal

(a)

(b)

HO A=

(c)

CH3O

OCH3

OCH3

A=

(d)

HO B=

Internuclear distance

OCH3

OH

OH OCH3

HO A=

OCH3

B=

OCH3

Energy

Energy



OH

CH3O

OH

OH

OH

(b)

Internuclear distance

HO B=

OCH3

(c) Square planar [PCl4]+ and octahedral [PCl6−] (d) Square pyramidal Q.7. The potential energy curve for the H2 molecule as a function of internuclear distance is :  [JEE (Main)-5th Sept-2020-(Shift-I)]

OH

A=

HO B=

OH

OH

31

CHEMICAL BONDING AND MOLECULAR STRUCTURE

Q.13. A substance X having low melting point, does not conduct electricity in both solid and liquid state ‘X’ can be  [JEE (Main)-9th Jan-2020-(Shift-I)] (a) Hg (b) ZnS (c) SiC (d) CCl4 Q.14. Select the correct statements among the followings  [JEE (Main)-9th Jan-2020-(Shift-II)] (A) LiCl does not dissolve in pyridine (B) Li does not react ethyne to form ethynide (C) Li and Mg react slowly with water (D) Among alkali metal Li has highest hydration tendency (a) B, C, D (b) A, B, C, D (c) A, B, C (d) C, D Q.15. The ion that has sp3d2 hybridization for the central atom is  [JEE (Main)-8th April-2020-(Shift-II)] (a) [ICl2]– (b) [BrF2]– – (c) [ICl4] (d) [IF6]– Q.16. The correct statement among I to III are I. Valence bond theory can not explain the colour exhibited by transition metal complexes. II. Valence bond theory can predict qualitatively the magnetic properties of transition metal complexes. III. Valence bond theory cannot distinguish ligands as weak and strong field ones. [JEE (Main)-9th April-2019-(Shift-II)] (a) II and III only (b) I, II and III (c) I and II only (d) I and III only.

1 [ VE + MA − C + A ] 2

For SF4, Hybridization =

1 [6 + 4 − 0 + 0=] 5 → sp3d. 2 lone pair of electron = H⋅O −M A − D    =       

        (monovalent  (Divalent         atom)   atom)

   = 5 − 4 = 1 Therefore the molecule is AB4L type i.e having trigonal bipyramidal shape. (SF4 have Sea-Saw structure) Ans. 3. Option (a) is correct. In H2S; S in a larger - atom having lower electronegativity though it has two pair of electron It cannot form hydrogen bond whereas oxygen atom in H2O can form hydrogen bond. H2S remains as discrete molecule and molecular attraction i.e. van der Waals force of attraction is low. So it has boiling point less than 300 K. Ans. 4. Option (b) is correct. O H  Intramolecular hydrogen .. bond. (Remains as discrete O .. molecule, so boiling point C H is low.) O H

H

C—O : H—O

: :

:

C—O :

=

:

=

O

ANSWER – KEY

:

H—O

: :

Q.17. The isoelectronic set of ions is [JEE (Main)-10th April-2019-(Shift-I)] – – + (a) F , Li , Na and Mg2+ (b) N3–, Li+, Mg2+ and O2– (c) Li+, Na+, O2ª and F– (d) N3–, O2–, F– and Na+

Square pyramidal is polar because the structure is B B A: B B The dipole moment of A is never cancelled by others. Ans. 2. Option (c) is correct.

O

H

4. (b)



5. (d)

6. (a)

7. (b)

8. (a)



9. (d)

10. (a)

11. (c)

12. (b)

13. (d)

14. (a)

15. (c)

16. (d)

17. (d)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (a) is correct. For AB4 molecule the possible geometry are : b·p l·p

Nature of molecule

H—O

C—O : H—O

: :

3. (a)

:

2. (c)

:

1. (a)

=



O

Intermolecular hydrogen bond (Remains as a polymeric molecules, apparent molecular weight becomes high, Due to molecular association, boiling point is high.) Ans. 5. Option (d) is correct. Greater the extent of overlapping region greater will be bond strength i.e. why bond enthalpy for A − B bond becomes highest and B is more electronegative atom than others Order of bond length is A−B dipole – dipole. Ans. 10. Option (a) is correct. Dipole moment (µ) is zero for symmetrical molecules i.e. µ CH4 = µ CCl 4 = 0 but µ CHCl3 > 0 . Ans. 11. Option (c) is correct. Ethyl acetate is polar molecule, so dipole – dipole interaction will be present. :O–: O CH3— C — O — C2H5

CHEMISTRY

+

CH3— C =OC 2 H5

Ans. 12. Option (b) is correct. Compound (A) can form extensive intermolecular hydrogen bonding. Consequently it can forms molecular association and remains as a polymeric molecule, apparent molecular weight become high. So boiling point of A is very high whereas smaller oxygen atom in compound is attached with carbon not with hydrogen. So it can not form hydrogen bond and remains as discrete molecule and B has low boiling point. Ans. 13. Option (d) is correct. CCl4 is a covalent compound. Therefore it cannot conduct electricity is both solid and liquid state. Again it has low melting point due to non-polar nature.

M+ ions :  Li+ > Na+ > K+ > Rb+ > Cs+ Ans. 15. Option (c) is correct. We know that hybridization (H) 1 = [ V.E.+ M.A. - C+ A] 2 V. E. = No. of valance electrons M.A. = No. of monovalent atom C = No. of cationic charge A = No. of anionic charge For [ICl2]– and [BrF2]– 1 3 H = [7 + 2 - 0 + 1] = 5 ( sp d ) 2 For [ICl4]–

H =

For [IF6]–

1 [7 + 4 - 0 + 1] = 6 ( sp 3 d 2 ) 2

1 [7 + 6 - 0 + 1] = 7 ( sp 3 d 3 ) 2 Ans. 16. Option (d) is correct. These (I & III) are limitation of V.B.T. V.B.T cannot distinguish between weak field ligands and strong field ligands and can not explain the colour of transition metal complexes. Ans. 17. Option (d) is correct. Two or more species which contains same no. of electron(s) are called iso electronic species.

H =

Species (ions) F – Li+

At. No. (Z)

No. of electrons

9

9 + 1 = 10*

3

3–1=2

+

11

11 – 1 = 10*

2+

Mg

12

12 – 2 = 10

N3–

7

7 + 3 = 10*

Na

2–

O 8 8 + 2 = 10* Therefore option (d) is the correct answer.

33

CHEMICAL BONDING AND MOLECULAR STRUCTURE

Topic-2

Molecular Orbital Theory

Concept Revision (Video Based) For details, scan the code

Hydrogen Bonding

     

For details, scan the code Hybridization        

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions Q.1. Which of the following has weakest bond ?  [JEE (Main)-3rd Sept-2020-(Shift-I)] (a) NO+

(b) NO2+

(c) NO–

(d) NO

Q.2. Bond order and magnetic nature of CN– are respectively,  [JEE (Main)-7th Jan-2020-(Shift-II)] (a) 3, diamagnetic (b) 3, paramagnetic (c) 2, 5, paramagnetic (d) 2, 5, diamagnetic Q.3. Which of the following species have one unpaired electron each ?  [JEE (Main)-8th Jan-2020-(Shift-I)] (a) O2, O-2

(b) O2, O+2

(c) O+2 , O-2

(d) O2, O22 -

Q.4. According to molecular orbital theory, which of the following is true with respect to Li2+ and Li2– ? [JEE (Main)-9th Jan-2019-(Shift-I)] (a) Both are unstable (b) Li2+ is unstable and Li2– is stable (c) Both are stable (d) Li2+ is stable and Li2– is unstable Q.5. In which of the following processes, the bond order has increased and paramagnetic character has changes to diamagnetic ? [JEE (Main)-9th Jan-2019-(Shift-II)] + (a) O2 ® O2 (b) N2 ® N2+ 2– (c) O2 ® O2 (d) NO ® NO+ Q.6. Two pi and half sigma bonds are present in [JEE (Main)-10th Jan-2019-(Shift-I)] + (a) O2 (b) N2 (c) N2+ (d) O2 Q.7. Among the following molecules/ions C22–, N22–, O22–, O2 Which one is diamagnetic and has the shortest bond length ? [JEE (Main)-8th April-2019-(Shift-II)] (a) C22– (b) O2 (c) O22– (d) N22– Q.8. Among the following the molecule expected to be stabilised by an ion formation is

C2, O2, NO, F2 [JEE (Main)-9th April-2019-(Shift-I)] (a) C2 (b) F2 (c) NO (d) O2 Q.9. Among the following species, the diamagnetic molecule is  [JEE (Main)-9th April-2019-(Shift-II)] (a) CO (b) B2 (c) NO (d) O2 Q.10. During the change of O2 to O2–, the incoming electron goes to the orbital [JEE (Main)-10th April-2019-(Shift-I)] (a) p 2px (b) p*zpx (c) p 2py (d) s*zpz

ANSWER – KEY

1. (a)

2. (a)

3. (c)



4. (d)

5. (d)

6. (c)

7. (a) 8. (a)

9. (a)

10. (b)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (a) is correct. Molecule Bond order 10 - 4 NO+ = 3 × 0 2 9-4 NO2+ = 2 × 5 2 10 - 6 NO– = 2×0 2 NO

10 - 5 = 2×5 2

Ans. 2. Option (a) is correct. 10 - 4 =3 2 – CN is a 14 electron system having electronic CN–; B. O. =

2 2 = p 22´ y < d 2p configuration d 1s2 < d 1s*2 < d 2s2 < d 2s* 2 < p 2px 2

CN– is diamagnetic in character due to presence of paired electron.

34 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Ans. 3. Option (c) is correct. O2 ⟶  

{

}{

σ1s2 , σ*1s2 , σ 2s2 , σ*2s2 , σ 2p2 , π 2p2 = π 2p2 , π*2p1 = π*2p1 z

z

x

y

x

y

}

′ ′ → 3 unpaired electrons π= λ2py 2× y O+2 ⟶ 1, unpaired electron O−2 ⟶ 1, unpaired electron

O22 ⟶ no, unpaired electron Therefore option ‘c’ is correct. Ans. 4. Option (d) is correct. Electronic configuration of Li2+ ; s1s2 s*1s2 s 2s1 and electronic configuration of Li2– ; s 1s2 s*1s2 s 2s2 s*2s1 Now, N - Na 3 - 2 = Bond order of Li2+ = b 2 2 =

Bond order of Li2– =

1 2 4-3 1 = 2 2

Both Li2+ and Li2– ions have same value of positive bond order but the species which have greater number of anti bonding electron, will be less stable. So, their order to stability will be Li2+ > Li2–. Ans. 5. Option (d) is correct. If the total electrons lie between 8 to 14 N-8 Bond order (B.O.) = 2 And if the total electrons lie between 15 to 20,

20 - N Bond order (B.O.) = 2 Therefore for 20 - 16 = 2.0 O2 (B.O.) = 2 O2+ (B.O.) =

20 - 15 = 2.5 2

For N2 (B.O.) =

14 - 8 = 3.0 2

N2

+

13 - 8 = 2.5 (B.O.) = 2

20 - 16 = 2.0 For O2 (B.O.) = 2 O2 For

2–

20 - 18 = 1.0 (B.O.) = 2

20 - 15 = 2.5 NO (B.O.) = 2

NO+ (B.O.) =

14 - 8 = 3.0 2

CHEMISTRY

Therefore B.O. increases for the pair of species O2, O2+ and NO, NO+ But O2 and O2+ both these species are para magnetic in nature For NO :- s 1s2 s 1*s2 s 2s2 s* 2s2 p2px2 = p 2py2 s 2pz2 p*2p'x = p*2p'y –Paramagnetic For NO+ :- s 1s2 s*1s2 s 2s2 s*2s2 p 2px2 = p 2py2 s 2pz2 –Diamagnetic Ans. 6. Option (c) is correct. For 20 - 15 = 2.5 O2+ B.O. = 2 Thus for O2+ : s1s2 s*1s2­ s 2s2 s*2s2­s 2p2z p 2px2 = p2 2py {p 2p1y = p 2py} 14 - 8 = 3.0 (2p and 1 s bond) For N2 B.O. = 2 13 - 8 = 2.5 For N2+ B.O. = 2 For N2+ – s 1s2 s*1s2 s 2s2 s*2s2 s2p2z {p2p2x = p 2p1y} (i.e. N2+ ions have bond one s bond present 20 - 16 =2 But O2 have B.O. = 2 Ans. 7. Option (a) is correct. We know that 1 Bond length µ B.O. For C22–,

B.O. =

14 - 8 = 3.0 2

For N22–,

B.O. =

20 - 16 = 2.0 2

For O22–,

B.O. =

20 - 18 = 1.0 2

20 - 16 = 2.0 2 Therefore the shortest bond length is present for the ion C22–. Ans. 8. Option (a) is correct. Stability µ B.O. 13 - 8 = 2.5 (a) For C2–; B.O. = 2 For O2,

(b) F2–, B.O. =

B.O. =

20 - 18 = 0.5 2

(c) NO–; B.O. = (d) O2–, B.O. =

20 - 16 = 2.0 2 20 - 17 = 1.5 2

Ans. 9. Option (a) is correct. Substances which are weakly repelled in magnetic field is known as diamagnetic. A substance shows diamagnetism due to presence of paired electrons. (a) CO (No. of electrons = 14)

35

CHEMICAL BONDING AND MOLECULAR STRUCTURE

m r = m 12 + m 22 + 2m 1m 2 cos 120° {m 1 = m 2 }



= 2m 12 - m 2 = m 2 = m = 1.5D

Q. 2. Final out the percentage of d character of SF6. Sol. Hybridization of SF6 is (H) 1 = [6 + 6 - 0 + 0] 2

= 6 ® sp3d2 % d character



=

2 (no. of d - orbital) ¥ 100 6 (total no. of hybrid orbital)

= 33 % Q.3. Among B2H6, B3N3H60 N2O, N2O4, H2S2O3 and H2S2O8, the total number of molecules containing covalent bond between two atoms of the same kind is ............. Sol. In the following formulae covalent bond is present between two atoms of the same kind = 4 O O

S

— —



— — —



O

— — H2S2O8 :

Cl The bond moments of para position cancel out as they are in opposite direction. The remain two Cl

HO OH O

O

Cl



S

H2S2O3 :

Cl Cl

– + —O N — — N—

Cl compound ?

Cl

Sol. 1.5

O

—

Cl

N2O:

O

— —

the dipole moment of Cl

is 1.5 D. What will be

S— O OH

— —

Q.1. The dipole moment of

N — N

— —

N2O4 :

Cl

— —

Integer Type Questions (Chapter Based)

atoms are at meta position having bond angle 120°. Therefore resultant dipole moment

O

O

S— — OH O —

Electronic configuration : s 1s2 s* 1s2 s 2s2 s* 2s2 s 2pz2 {p zpx2 = p zpy2} – diamagnetic (b) B2 (No. of electrons = 10) Electronic configuration : s 1s2 s* 1s2 s 2s2 s* 2s2 {p1 2px = p1 2py} – para magnetic (c) NO (No. of electrons = 15) Electronic configuration, s 1s2 s*1s2 s 2s2 s*2s2 s2p2z {p 2px2 = p 2py2} {p*2px 1 = p*zpy0 (d) O2 (No. of electron = 16) Electronic configuration s 1s2 s*1s2 s 2s2 s*2s2 s 2p2z {p2p2x = p2p2y} {p2p'x' = p2p'y} Ans. 10. Option (b) is correct. The electronic configuration of O2– is [s(2s)2] [s*(2s)]2 [s(2pz)]2 [p(2px)]2 [p(2py)]2 [p*(2px)]1 [p*(2py)]1 So to form O2– ion from O2 the incoming electron goes to the p*2px molecular orbital.



rH= rH1+ rH2+ rH3............ rH1

B

rH2

D

rH3

is the enthalpy change when one mole of an ionic compound dissociates into its ions in gaseous state.

Applicatio ns

∑bond enthalpies reactants – ∑bond enthalpies products

∆aH is the enthalpy change on breaking one mole of bonds completely to obtain atoms in gas phase.

ΔG = ΔH – TΔS ΔG 0, process is non-spontaneous

Chemical Thermodynamics

° sol H is the enthalpy change when one mole of a substance dissolves in a specified amount of solvent.

If a reaction takes place in several steps then its standard reaction enthalpy is the sum of standard enthalplies of international reactions into which the overall reactions may be divided at the same temperature. rH1 A C

–∑ b H i reactants rH = ∑aiH products i (a) Standard Enthalpy of reactions is the enthalpy change for a reaction when all the participating substances are in their standard states. (b) Enthalpy changes during phase transformations: Standard enthalpy of fusion / molar enthalpy of fusion, fusHθ is the enthalpy change that accompanies melting of one mole of a solid substance in standard state. Standard enthalpy of vaporization or molar enthalpy of vaporization. vapHθ is the amount of heat required to vaporize one mole of a liquid at constant temperature and under standard pressure. (c) Standard molar enthalpy of formation rHθ is the standard enthalpy change for the formation of one mole of a compound from its elements in their most stable state of aggregation.

First Law of Thermodynomics ; Energy can neither be created nor be destroyed but it can be converted from one form to another. Second Law of Thermodynomics: All spontaneous processes are accompanied by a net increase of entropy. Third Law of Thermodynomics: The entropy of a pure and perfectly crystalline substance at the absolute zero temperature is zero.

In calorimetry, the process is carried out in a vessel called calorimeter, immersed in a known volume of a liquid. (a) U measurements: The energy changes are measured at constant volume. No work is done. (b) H measurements: In exothermic reaction, heat is evolved, so qp and rH will be negative. In endothermic reaction, heat is absorbed, so qp and rH will be positive.

Cp - C v = R

∆cH is the enthalpy change per mole of a substance when it undergoes combustion and all reactants and products are in standard state.

Δr H o

Property whose value does not depends on the quantity or size of matter present. (Temperature, density, Pressure etc).

m Te r

y og ol in

Property whose value depends on the quantity or size of matter present in the system. eg. mass Volume Internal energy etc.

ΔSTotal = ΔSsystems + ΔSsurroundings q rev S= T

q=Cxmx T=C T

ΔH is negative: Exothermic reactions. ΔH is positive: Endothermic reactions.

Reversible Process: The process which can be reversed at any moment by an infinitesimal change. Irreversible Process: Processes other than reversible process. At constant temperature Wrev = 2.303 nRT log Vf Vi For adiabatic change, q=0, U = Wad

Vi

Work= –Pex ∆V=–Pex (Vf –Vi) = Pex dV

Vf

•System and Surroundings: A system refers to that part of universe in which observations are made and remaining universe constitutes the surroundings. •Types of System: 1. Open System: There is exchange of energy and matter between system and surroundings. 2. Closed System: There is no exchange of matter but exchange of energy is possible. 3. Isolated System: There is no exchange of energy or matter between the system and surroundings. 4. State of system : Described by its measurable or macroscopic properties. 5. Internal Energy : as a state function: It is the sum of chemical, electrical, mechanical or any other type of energy. 6. (a) Work: Adiabatic process is in which there is no transfer of heat between system and surroundings. ΔU = U2 – U1 = Wad (b) Heat is change in internal energy of a system by transfer of heat from the surroundings to the system or vice-versa without expenditure of work. q is positive: Heat is transferred from surroundings to system. q is negative: Heat is transferred from system to surroundings. (c) General Case First law of Thermodynamics: U = q+W the energy of an isolated system is constant.

36 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) CHEMISTRY

En tro py

Chapter

Chemical 5 Thermodynamics

Syllabus Fundamentals of thermodynamics : System and surroundings, extensive and intensive properties, state functions, types of processes. First law of thermodynamics, concept of work, heat internal energy and enthalpy, heat capacity, molar heat capacity, Hess’s law of constant heat summation, enthalpies of bond dissociation, combustion, formation, atomization, sublimation, phase transition, hydration, ionization and solution. Second law of thermodynamics spontaneity of processes, DS of the universe and DG of the system as criteria for spontaneity, DG° (Standard Gibb’s energy change) and equilibrium constant.

Topic-1



First Law of Thermodynamics, Reversible and Irreversible Processes Concept Revision (Video Based) For details, scan the code

Part - 1

Topic-2 :  Chemical Thermodynamics, Hess’s Law Topic-3 : 2

nd

Law of Thermodynamics

.... P. 40 .... P. 41

For details, scan the code

First Law of Thermodynamics

  

LIST OF TOPICS : Topic-1 : First Law of Thermodynamics, Reversible and Irreversible processes .... P. 37

Part - 2

    

Reversible and Irreversible Process

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions

Q.1. When helium gas balloon explodes, this process is  [JEE (Main)- 3rd Sept-2020-(Shift - I)] (a) Isothermal reversible (b) Isothermal irreversible (c) Adiabatic reversible (d) Adiabatic irreversible Q.2. For one mole of an ideal gas, which of these statements must be true ? (1) U and H each depends only on temperature (2) Compressibility factor Z is not equal to 1 (3) Cp, m − Cv, m = R (4) Cvdt for any process  [JEE (Main)- 4th Sept-2020-(Shift-I)]

(a) 2, 3 and 4 (b) 1 and 3 (c) 1, 3 and 4 (d) 3 and 4 Q.3. Five moles of an ideal gas of 1 bar and 298 K is expanded into vaccum to double the volume. The work done its [JEE (Main)- 4th Sept-2020-(Shift - II)] (a) −RT (V2 − V1) (b) zero V2 (d) C2 (T2 − T1) (c) - RT ln V1 Q.4. For Br2(l) Enthalpy of atomization = x KJ / mole Bond dissociation enthalpy of bromine = y KJ / mole then [JEE (Main)- 9th Sept-2020-(Shift - I)] (a) x > y (b) x < y (c) x = y (d) Relation does not exist

38 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Q.5. 0.5 moles of gas A and x moles of gas B exert a pressure of 200 Pa in a container of volume 10 m3 at 1000 K. Given R is the gas constant in JK–1 mol–1, x is  [JEE (Main)-3rd Jan-2020-(Shift-I)] 4-R (a) 2 R (b) 2R 4-R 2R 4+R (c) (d) 4+R 2R Q.6. Consider the reversible isothermal expansion of an ideal gas in a closed system at two different temperatures T1 and T2 (T1 < T2). The correct graphical depiction of the dependence of work done (w) on the final volume (v) is [JEE (Main)- 9th Jan-2019-(Shift-I)] (a)

T2

T1

(b)

W

O

(c)

T2

T1

W

lnv

O

(d)

T2

T1

lnv T2

T1

W

W

O O lnv lnv Q.7. An ideal gas undergoes isothermal compression from 5m3 to 1m3 against a constant external pressure of 4 Nm–2. Heat released in this process is used to increase the temperature of 1 mole of Al. If molar heat capacity of Al is 24 J mol–1 K–1 the temperature of Al increases by [JEE (Main)-3rd Jan-2019-(Shift-II)] 3 (a) K (b) 1 K 2 2 (c) 2 K (d) K 3 Q.8. For a diatomic ideal gas in a closed system which of the following plots does not correctly describe the relation between various thermodynamic quantities ? [JEE (Main)- 12th Jan-2019-(Shift-I)]

(a)

(b) Cv

Cp

P

(c)

T

(d) U

Cv

T V Q.9. The combination of plots which does not represent isothermal expansion on an ideal gas is [JEE (Main)- 12th Jan-2019-(Shift-I)]

(A)

CHEMISTRY

(B) P

P O

1/

O

Vn

(C)

Vm

(D) PVn

U

O

T

O

V (a) (A) and (C) (b) (A) and (D) (c) (B) and (C) (d) (B) and (D) Q.10. Which one of the following equations does not correctly represent the first law of thermodynamics for the given processes involving an ideal gas? (Assume non-expansion work is zero) [JEE (Main)- 8th April-2019-(Shift-I)] (a) Cyclic process : q = – W (b) Adiabatic process : DU = – W (c) Isochoric process : DU = q (d) Isothermal process : q = – W Q.11. 5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If CV = 28 JK–1 mol–1, calculate DU and Dpv for this process (R = 8 JK–1 mol–1) [JEE (Main)- 8th April-2019-(Shift-II)] (a) DU = 2.8 KJ; D (pv) = 0.8 KJ (b) DU = 14 J; D (pv) = 0.8 KJ (c) DU = 14 KJ; D (pv) = 4 KJ (d) DU = 14 KJ; D (pv) = 18 KJ Q.12. Among the following the set of parameters that represents path function is [JEE (Main)- 9th April-2019-(Shift-II)] (A) q + W (B) q (C) W (D) H – TS (a) A and D (b) A, B and C (c) B, C and D (d) B and C Q.13. During compression of a spring the work done is 10 KJ and 2 KJ escaped to the surroundings as heat The change in internal energy, DU (in KJ) is [JEE (Main)-9th Jan-2019-(Shift-I)] (a) 8 (b) 12 (c) –12 (d) –8 Q.14. An ideal gas is allowed to expand from 1 L to 10 L against a constant external pressure of 1 bar. The work done in KJ is [JEE (Main)- 12th April-2019-(Shift-I)] (a) –9.0 (b) +100 (c) –0.9 (d) –2.0

ANSWERS – KEY 1. (d) 5. (b) 9. (d) 13. (a)

2. (c) 6. (c) 10. (b) 14. (c)

3. (b) 7. (d) 11. (c)

4. (a) 8. (a) 12. (d)

39

CHEMICAL THERMODYNAMICS

ANSWERS WITH EXPLANATIONS Ans. 1. Option (d) is correct. When He gas balloon explode, the process occurs, so rapidly to be treated as adiabatic and equilibrium is disturbed after the completion of process i.e it is also an irreversible process. Ans. 2. Option (c) is correct. (a) A as du = CvdT and dH = CpdT i.e Both u and H depends on temperature (c) Again Cp − Cv = R (d) ∆u = CvdT for all process Ans. 3. Option (b) is correct. When an ideal gas is expanded into vaccum, though the expansion is spontaneous but no energy change takes place, This is because energy change can take place as heat on work. fi Work done = −Pext ∆V As Pext = 0, therefore w= 0 Ans. 4. Option (a) is correct. ∆Hatomization=xKJ/mole

Br2 (l) ∆HVap

Br2 (g)

Bond

y= energ

2Br(g)

ole

yKJ/m

\ ∆ H atomization = ∆Hvap + Bond energy Therefore x>y Ans. 5. Option (b) is correct. Given : From the ideal gas equation, PV = hRT fi 200 Pa × 10 m3 = (0.5 + x) mole × R × JK–1 mol–1 1000 K –2 3 fi 200 Nm × 10 m = (0.5 + x) mole × R × NmK–1 mol–1 1000 K fi 200 × 10 = (0.5 + x)R × 1000 200 ¥ 10 1 + x = R ¥ 1000 2 2 1 fi x = R 2 4-R fi x = 2R Ans. 6. Option (c) is correct. For reversible isothermal expansion we get,

V2

W = –nRT ln V 1

fi |W| = nRT [ln V2 – V1] fi |W| = –nRT ln V2 – nRT ln V1 Comparing with y = mx + c Thus for two different temperature we get the curve of straight line seems to be touching the origin and y intercept becomes zero which is not possible. Slope of curve 2 is more than the curve 1 and curve is also move negative than curve 1. Ans. 7. Option (d) is correct. From 1st law of thermodynamics (when compression occurs) du = d q + dw

fi fi

du = d q + P2 (V2 – V1) O = d q + P2 (V2 – V1) [When is other – mol, du = q] fi d q = –4 Nm–2 (1 – 5) m3 = +16 Nm = +16 J Again d q = nCV DT fi 16 J = 1 mol × 24 J mol–1 K–1 × DT 16 2 K= K fi DT = 24 3 Ans. 8. Option (a) is correct. We know that 7 CP = R [Independent of P] 2 Graph (1) is not the correct graph when variation of CP vs P 5 And also CV = R (does not depend on volume) 2 Ans. 9. Option (d) is correct. (B) and (D) are not correct representation for isothermal expansion of ideal gas. Ans. 10. Option (b) is correct. 1st law of thermodynamics we get, DU = q + W For cyclic process DU = O; q = –W (a) is correct For adiabatic process q = 0 DU = W (b) is incorrect For isochoric process; DV = 0 Again W = PDV \ W = 0 \ DU = q (c) is correct For isothermal process, D = 0 \ q = –W (d) is correct Ans. 11. Option (C) is correct. DU = nCV DT = 5 mol × 28 JK–1 mol–1 × (200 – 100) K = 14000 J = 14 kJ D(PV) = nR (T2 – T1) = 5 mol × 8 JK–1 mol–1 × (200 – 100) K = 4000 J = 4 kJ Ans. 12. Option (d) is correct. State function is a function which does not depend upon the change of path but depend on final state and initial state. But path function always depend upon the change of path. q and W depends on the change of path but their summation equals to internal energy ‘u’ which is independent of path i.e., state function. Again H – TS = G; ‘G’ free energy depends final state and initial state. So it is a state function. Ans. 13. Option (a) is correct. From 1st law of thermodynamics we get DU = q + W DU = 1 – 2 kJ + 10 kJ \ DU = 8 kJ Ans. 14. Option (c) is correct. Work done (W) = – Pext (V2 – V1) = – 1 bar (10 – 1) Lit =– 9 bar – Lit = – 900 J [ 1 L bar = 100 J]

40 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

CHEMISTRY

Topic-2

Chemical Thermodynamics, Hess's Law

For details, scan the code

For details, scan the code

Thermochemistry Problems

For details, scan the code

Chemical Thermodynamics

Hess's Law Part - 1

Part - 2

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions Q.1. Given (i) C (graphite) + O2(g) ® CO2 (g); DrH° = x kJ mol–1 1 (ii) C (graphite) + O2(g) ® CO (g); 2 DrH° = y kJ mol–1 1 (iii) CO (g) + O2(g) ® CO2 (g); 2 DrH° = z kJ mol–1 Based on the above thermodynamical equations, find out which one of the following algeberic relationships is correct ? [JEE (Main)- 12th Jan-2019-(Shift-II)] (b) x = y – z (a) y = 2z – x (d) x = y + z (c) z = x + y Q.2. For silver CP (JK–1 mol–1) = 23 + 0.01 T. If the temperature (T) of 3 moles of silver is raised from 300 K to 1000 K at 1 atom pressure, the value of DH will be close to [JEE (Main)- 8th April-2019-(Shift-I)] (a) 62 kJ (b) 16 kJ (c) 21 kJ (d) 13 kJ Q.3. The difference between DH and DU (DH – DU); when the combustion of one mole of heptane (l) is carried out at a temperature T, is equal to [JEE (Main)- 10th April-2019-(Shift-II)] (a) – 4RT (b) 3RT (c) 4RT (d) –3RT Q.4. Enthalpy of sublimation of iodine is 24 cal g–1 at 200°C. If specific heat of I2(s) and I2 (vap.) are 0.055 and 0.031 cal g–1 K–1 respectively, then enthalpy of sublimation of iodine at 250°C in cal g–1 is [JEE (Main)- 12th April-2019-(Shift-I)] (a) 2.85 (b) 5.7 (c) 22.8 (d) 11.4

ANSWER – KEY

1. (d) 2. (a)

3. (a)

4. (c)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (d) is correct. C(graphite) + O2(g) ® CO2(g); DrH° = x kJ mol–1 (i) 1 C(graphite) + O2(g) ® CO(g); DrH° = y kJ mol–1 2 (ii)

1 CO(g) + O2(g) ® CO2(g); DrxH° = z kJ mol–1 (iii) 2 Now adding equation (i) & (ii) we get 1 C (graphite) + O2(g) ® CO(g) DxH° = y kJ mol–1 2 1 CO (g) + O2(g) ® CO2 (g) DrH° = z kJ mol–1 2 C (graphite) + O2 = CO2 (g) DrH° = (y + z) kJ mol–1 If the equals to the equation (i) \ (x = y + z) Ans. 2. Option (a) is correct. T2



DH =

Ú nCpdT

T1

1000

= n

Ú

300

( 23 + 0.01 T ) dT

[Here CP = (23 + 0.01 T)] 1000

È 0.01 T 2 ˘ ˙ = 3 Í23T + 2 ˙˚ ÍÎ 300

È 2 ¥ 23 ¥ 700 + 9100 ˘ = 3 Í ˙ 2 Î ˚ = 61950 J/mol = 62 kJ/mol Ans. 3. Option (a) is correct. The general combustion reaction for hydrocarbon is yˆ y Ê CxHy(g) + Á x + ˜ O2(g) ® x CO2(g) + H2O (l) ¯ Ë 4 2 Therefore combustion reaction for heptane (l) is an follows C7H16(l) + 11 O2 (g) ® 7 CO2(g) + 8 H2O (l) \ Dng = np + nr = 7 – 11 = – 4 Again DH = DU + DngRT \ DH – DU = – 4RT Ans. 4. Option (c) is correct. DH2 = CPDT fi H2 – H1 = CP (T2 – T1)

41

CHEMICAL THERMODYNAMICS

fi H2 = H1 + CPr × n (T2 – T1) fi H2 = 24 cal g–1 + (0.031 – 0.055) cal g–1 K –1 × 50 K [\ T2 – T1 =(523 – 473)K

= 50 K] fi H2 = 24 cal g–1 – 1.2 cal g–1 fi H2 = 22.8 cal g–1 Therefore, the enthalpy of sublimation of iodine at 250°C = 22.8 cal g–1

Topic-3

2nd Law of Thermodynamics Concept Revision (Video Based) For details, scan the code

  2nd Law of Thermodynamics                     Entropy Concept 

  

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions Q.1. Select the correct option



th

[JEE (Main)- 9 Jan-2020-(Shift - II)] (a) Entropy is a function of temperature and also entropy change is a function of temperature (b) Entropy is a function of temperature & entropy change is not a function if temperature (c) Entropy is not a function of temperature & entropy change is a function of temperature (d) Both entropy & entropy change are not a function of temperature Q.2. The entropy change associated with the conversion of 1 kg of ice at 273 K to water vapours at 383 K is (specific heat of water liquid and water vapour are 4.2 kJ K–1 Kg–1 and 2.0 kJ K–1 Kg–1 heat of liquid fusion and vapourisation of water are 334 kJ Kg–1 and 2491 kJ Kg–1 respectively) (log 273 = 2.436, log 373 = 2.572, log 383 = 2.583) [JEE (Main)- 9th Jan-2020-(Shift-II)] –1 –1 (a) 9.23 kJ Kg K (b) 8.49 kJ Kg–1 K–1 –1 –1 (c) 7.90 kJ Kg K (d) 2.64 kJ Kg–1 K–1 Q.3 A process has DH = 200 J mol–1 and DS = 40 JK–1 mol–1. Out of the values given below choose the minimum temperature above which the process will be spontaneous 

[JEE (Main)- 10th Jan-2020-(Shift-II)] (a) 20 K (b) 4 K (c) 5 K (d) 12 K Q.4. The process with negative entropy is [JEE (Main)- 10th Jan-2020-(Shift-II)] (a) Synthesis of ammonia from N2 and H2 (b) Dissociation of CaSO4(s) to CaO (s) and SO2(g) (c) Dissolution of iodine in water (d) Sublimation of dry ice Q.5. For the chemical reaction X  Y, the standard reaction Gibb's energy depends on temperature T(in K) as



3 T 8 The major component of the reaction mixture at T. DTG° (in kJ mol–1) = 120 –

[JEE (Main)- 11th Jan-2019-(Shift-I)] (a) Y if T = 280 K (b) X if T = 350 K (c) X if T = 315 K (d) Y if T = 300 K Q.6. Two blocks of the same metal having same mass and at temperature T1 and T2 respectively are brought in contact with each other and allowed to attain thermal equilibrium at constant pressure. The change in entropy, DS, for this process is [JEE (Main)- 11th Jan-2019-(Shift-I)] 1˘ È Í (T1 + T2 ) 2 ˙ (a) 2 CP ln Í ˙ Í T1 T2 ˙ Î ˚

È T1 + T2 ˘ ˙ (b) 2 CP ln Í Î 4 T1 T2 ˚

È T1 + T2 ˘ È (T1 + T2 )2 ˘ ˙ (c) CP ln Í (d) 2 CP ln Í 2 T T ˙ Î 1 2˚ ÍÎ 4 T1 T2 ˙˚ Q.7. The reaction MgO (s) + C(s) ® Mg(s) + CO(g), for which DrH° = + 491 kJ mol–1 and DrS° = 198. 0 kJ–1 mol–1, is not feasible at 298 K. Temperature above which reaction will be feasible is

[JEE (Main)- 11th Jan-2019-(Shift-I)] (a) 2040.5 K (b) 1890.0 K (c) 2380.5 K (d) 2480.3 K Q.8. The standard reaction Gibb's energy for a chemical reaction at an absolute temperature T is given by DrG° = A – BT Where A and B are non zero constants which of the following is true about the reaction?

[JEE (Main)- 11th Jan-2019-(Shift-II)] (a) Endothermic of A< 0 and B > 0 (b) Exothermic if B < 0

42 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) (c) Exothermic of A > 0 and B < 0 (d) Endothermic if A > 0 Q.9. A process will be spontaneous at all temperature if [JEE (Main)- 10th April-2019-(Shift-I)] (a) D H > 0 and D S < 0 (b) D H < 0 and D S > 0 (c) D H < 0 and D S < 0 (d) D H > 0 and D S > 0 Q.10. The incorrect match in the following is [JEE (Main)- 12th April-2019-(Shift-II)] (a) DG° < 0, K > 1 (b) DG° = 0, K = 1 (c) DG° > 0 K < 1 (d) DG° < 0, K < 1 Q.11. Choose the reaction (s) from the following options for which the standard enthalpy of reaction is equal to the standard enthalpy of formation. (a) 2 H2(g) + O2(g) ® 2 H2O (l) (b) 2 C(g) + 3 H2(g) ® C2H6(g) 3 (c) O2(g) ® O3(g) 2 1 (d) S8(s) + O2 (g) ® SO2(g) 8

ANSWERS – KEY

1. (a) 5. (c)

9. (b)

2. (a) 6. (c)

3. (c) 7. (d)

10. (d)

11. (c, d)

4. (a) 8. (d)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (a) is correct. Entropy is defined as follows δ qrev dS = T du + δw = T C v dT + Rdv = T  P R C v dT R  T = V  = + dv    T V Hence entropy is a function of temperature and also entropy change is a function of temperature. Ans. 2. Option (a) is correct. The conversion of 1 kg ice at 273 into vapour at 373 K takes place as follows H2O(s) 273 K

DS1

H2O(l) 273 K

DS2

H2O(l) 373 K

DS3

H2O(g) 373 H2O(g) 383 K

\ DS1 =

DH 273 K

=

334 kJ Kg -1 273K

= 1.22 kJ Kg–1 K–1

T2

DS2 = m C ln T 1

CHEMISTRY

373K 273K = 4.2 × 2.303 [log 373 – log 273] kJ = 1.31 kJ K–1 Kg–1 DH vap DS3 = 373 K = 4.2 kJ K–1 kg–1 ln

2491 kJ Kg -1 373 K = 6.67 kJ Kg–1 K–1 =



T2

DS4 = m C ln T 1

383 373 = 2 × 2.303 (log 383 – log 373) kJ K–1 Kg–1 = 0.05 kJ K–1 Kg–1 \ DStotal = DS1 + DS2 + DS3 + DS4 = 9.26 kJ Kg–1 K–1 Ans. 3. Option (c) is correct. From Gibbs free energy we get, DG = DH – T D S A process would be spontaneous when DG < 0. It is po sible when T D S > D H = 2 × ln

\ T >

200 J mol -1 DH = =5K DS 40 JK -1mol -1

Therefore the minimum temperature is 5 K above which the process will be spontaneous. Ans. 4. Option (a) is correct. Synthesis of ammonia, N2(g) + 3 H2(g) ® 2 NH3(g) \ Dn = 2 – (1 + 3) = – 2 Therefore DS = – ve Dissociation of CaSO4, CaSO4 (s) = CaO(s) + SO3(g) Dn = 1 + 0 – 0 = 1 \ DS = +ve Dissolution of I2 in water Water I2(s) ææææÆ I2(aq) KI Randomness increases \ DS = +ve Sublimation of dry ice CO2(s) ® CO2(g) Randomness increases, \ DS = + ve Ans. 5. Option (c) is correct. XY (i) When DrG° < 0, spontaneous process and Y will be major component while X will be minor component. (ii) But when DrG° > 0, non-spontaneous process, and X will be major component while Y will be minor component 3 (a) DrG° = 120 – × 280 = 15 8

43

CHEMICAL THERMODYNAMICS

i.e., DrG° > 0; major component X 3 (b) DrG° = 120 – × 350 = –11.25 8 i.e. DrG° < 0, major component Y 3 (c) DrG° = 120 – × 315 8 = 1.875 i.e., DrG° > 0, major component X 3 (d) DrG° = 120 – × 300 = 7.5 8 i.e., DrG° > 0, major component = X Ans. 6. Option (c) is correct. T + T2 At thermal equilibrium Tf = 1 2 For the 1st block DS1 = CP ln

Tf T1 Tf

For the 2nd block DS2 = CP ln T 2 \ When they brought in contact with each other, DS = DS1 + DS2 = CP ln

Tf

Tf

+ CP ln T T1 2

Ê Tf Tf ˆ = CP ln Á ¥ ˜ Ë T1 T2 ¯ 2

Tf = CP ln T T 1 2

ÈÊ T + T ˆ 2 ˘ 2 ÍÁ 1 ˜ ˙ ÍË 2 ¯ ˙ = CP ln Í ˙ T1 T2 Í ˙ ÍÎ ˙˚

È ( T1 + T2 )2 ˘ = CP ln Í 4 T T ˙ ÍÎ 1 2 ˙ ˚

Ans. 7. Option (d) is correct. Given DH° = 491.1 kJ mol–1 DS° = 198.0 kJ mol–1 From Gibbs Helmholtz equation, DG° = DH° – TDS° ...(i) At equilibrium DG° = 0 From eqn (i) DH° = TDS° DH∞ \ T = DS∞ fi

T =

491.1 ¥ 10 3 J mol -1

198.0 JK -1 mol -1 \ T = 2480.3 K Therefore above 2480.3 K reaction will be feasible Ans. 8. Option (d) is correct. From Gibbs Helmholtz equation, DG° = DH° – TDS° ...(i) Given equation DG° = A – BT ...(ii) Comparing above two equation, we get at absolute temperature the equation (ii) reduced to

DG° = A Therefore if A > 0, reaction is endothermic. Ans. 9. Option (b) is correct. Gibbs Helmholtz equation, DG = DH – TS For spontaneous process at all temperature DG < 0 and it is possible when DH < 0 and DS > 0 Ans. 10. Option (d) is correct. We know that DG° = –RT ln K \ If K > 1 then DG° < 0 If K < 1 then DG > 0 If K = 1 then DG° = 0 Therefore the incorrect match is, DG° < 0, K < 1 Ans 11. Option (c, d) is correct. The standard enthalpy of formation of a compound is defined as the change in enthalpy for the reaction in which one mole of a compound in its standard state (i.e., at particular temperature and 1 atm pressure) is produced from its constituent elements in their standard elemental state 1 ¸ S 8 ( s) + O2 ( g ) Æ SO2 ( g )Ô Ô 8 ˝ formation reactions 3 O 2 ( g ) Æ O3 ( g ) Ô Ô˛ 2

Integer Type Questions (Chapter Based) Q.1. Heat of combustion of ethanol to give CO2 and water at constant pressure and 27°C is − 327 kCal. How much heat is evolved in (Cal) in combustion at constant volume at 27°C ?  [JEE Main 2020, 2nd Sept (Shift - I)] Sol. For the heat of combustion of ethanol the required equation is C2H5OH (l) + 3O2(g) = 2 CO2(g) + 3H2O(l)  ∆H = −327 Kcal ∆ng = 2 − 3 = − 1 \ ∆H = ∆U + ∆ng RT Þ − 327000 cal = ∆U − 1 mol × 2 cal mol−1 K−1  × 300 K Þ ∆U = (−327000 + 600) cal = − 326400 cal Q.2. The internal energy (in J) when 90 g of water under goes complete evaporation at 100°C is…….. (Given ∆Hvap for water at 373 K = 41 kJ/mol R = 8.314 JK mol−1)  [JEE Main 2020 2nd Sept (Shift - I)] Sol. No. of moles of H 2O =

90g = 5 mol 18 g mol −1

∆H = ∆U + ∆ngRT Þ 41000 × 5 J = ∆U + 5 mole × 8.3145 mol−1 K−1  × 373 K Þ 205000 J = ∆U + 15505.61 J Þ ∆U = 189494.39 J = 189494 J Q.3. For a dimerization reaction 2A(g) ® A2 (g)

44 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) at 298K, ∆U = −20 KJ mol−1 ∆S = −30 JK−1 mol−1, then the ∆G will be………  [JEE (Main) 2020, 5th Sept (Shift - II)] Sol. We know that ∆H = ∆U + ∆ngRT Þ ∆H = −20,000 J mol−1 + (1 − 2) 8.314 J mol−1K−1  × 298 K Þ ∆H = −22477.572 J \ ∆G =∆H − T∆S = −22477.572 J − 298 K × −30 JK−1 mol−1 = −13537.572 J = − 13538 J Q.4. A(l) ® 2B (g) ∆U = 2.1 KCal, ∆S = 20 cal/K, T = 300 K Find ∆G(in KCal)  [JEE (Main) 2020, 7th Jan (Shift - I)] Sol. A(l) ® 2B (g) We know that ∆H = ∆U + ∆ngRT  Here ∆ng = 2 mole = 2.1 K Cal + 2 mol × 2 cal K−1 × 300 K = 2100 Cal + 1200 Cal = 3300 cal Again ∆G = ∆H − T∆S = 3300 cal − 300 K  × 20 cal K−1 = −2700 cal = −2.70 K cal Q.5. Calculate ∆fH° (ln kJ / mol) for C2H6 (g) if ∆cH° [C(Graphite)] = −393.5 kJ/mol ∆cH° [H2(g)] = −286 kJ/mol and ∆cH° [C2H6(g)] = − 1560 kJ/mol  [JEE (Main) 2020, 7th Jan (Shift - II)] Sol. 2C (Graphite) + 3H2(g) ® C2H6(g) C(Graphite) + O2(g) ® CO2(g)…(i)  ∆H = −393.5 kJ/mol 1 …(ii) H 2 ( g ) + O2 ( g ) → H 2O ( l )  2  ∆H = −286 kJ/mol 7 C 2 H 6 ( g ) + O2 ( g ) → 2CO2 ( g ) + 3H 2O ( l ) …(iii) 2  ∆H = −1560 kJ/mol \ ∆fH° for C2H6 = [Equn(i) × 2 + Equn(ii) × 3 − Equn (iii)] = [2 × − 393.5 + 3 × (−286) − (−1560)] kJ/mol = – 85.00 kJ/mol Q.6. A gas undergoes expansion according to the following graph. Calculate work done by the gas.

P(Paocal) P(Pascal) 8

8

(2, 2)



12 v(m³)

[JEE (Main) 2020, 8th Jan (Shift - I)]

Sol. [W]= 1 ( 6 + 10 ) × 6= 48J 2

CHEMISTRY

Q.7. Temperature of 4 moles of gas increases from 300 K to 500 K find Cv of ΔU = 5000 J.  [JEE (Main) 2020, 8th Jan (Shift - II)] Sol. DU = nCv DT ⇒5000 J = 4 mol × Cv (500 – 300) K ⇒ Cv = 06⋅25 JK–1 mol–1 Q.8. Find out the efficiency of the reversible cycle as shown in the following figure.

B

250 T(K) 150

C

A

1000 (Entropy)

1500 (JK–1)

Sol. Efficiency of cycle =

Area of the closed cycle ¥ 100 Area under the curve

=

1 (1500 - 1000 )( 250 - 150 ) 2

1 (1500 - 1000 )( 250 - 150 ) + (1500 - 100 )(150 - 0 ) 2 1 ¥ 500 ¥ 100 2 ¥ 100 = 1 ¥ 500 ¥ 100 + 500 ¥ 150 2

¥ 100

= 25 %

Q.9. The molar entropy of vaporization of acetic acid is 14.4 Cal K–1 mol–1 at its boiling point 118°C. What will be the latent heat of vaporization of acetic acid? δq δqp ∆H = = Sol. ds = T T T \ DHv = TdS = 391 K × 14.4 cal K–1 mol–1 391 × 14.4 cal = 60 g = 93.84 cal g–1  94 cal g–1 Q.10. One mole of a non-ideal gas undergoes a change of state from (2.0 atm, 3.02 L, 250 K) to (5.0 atm, 6 L, 300 K) with a change in internal energy (DU) = 40 L – atm. Find out the change of enthalpy of the process in Lit-atm ? Sol. P and V both changes in a process. \ DH = DU + D(PV) = 40 L.atm + (P2V1 – P1V2) = 40 L.atm + (30 – 6) L.atm = 40 L.atm + 24 L. atm = 64 L.atm

45

CHEMICAL THERMODYNAMICS

Q.11. Enthalpy of neutralization of HCl by NaOH is –13.7 kcal mol–1 and by NH4OH is – 12.7 kcal mol–1. Determine the enthalpy of dissociation of NH4OH. Sol. Enthalpy of dissociation of NH4OH = – 12.7 + (13.7) Kcal mol–1 = 1.0 Kcal mol–1 Q.12. 4.48 L of an ideal gas at STP requires 12.0 calories to raise its temperature by 15°C at constant volume. What will be the value of CP (in calories)? Sol. From n mole ideologies equation we get



n =

=

1 atm×4.48 L 0.082 mol -1 K -1 ¥ 273 K

= 0.2 mole

12 cal dQ = = 4 cal 0 . 2 mole ¥ 15 K nDT \ CP = CV + R = 4 + 2 = 6 cal



PV RT

CV =



Mass of solute × 100 Volume of solution

Volume of component × 100 Total volume of solution

s

Molality: Number of moles of solute per kilogram of the solvent

Molecular mass Valency

No. of moles of solute×100 Volume of solution

Molarity : Number of moles of solute in 1L solution

Gas – Solid → O2 in Pd Liquid – Solid → Amalgam of Hg with Na Solid – Solid → Cu dissolved in gold

Gas – Liquid → O2 dissolved in water Liquid – Liquid → Ethanol dissolved in water Solid – Liquid → Glucose dissolved in water

No. of moles of solute×100 Mass of solvent

Gram Equivalents of solute Mass of solute = Equivalent weight Equivalent weight =

Exothermic ∆sol H < 0, Solubility decreases

Endothermic ∆sol H > 0, Solubility increases

Not significant

Increases with decrease in temperature

Gas – Gas → Mixture of O2 and N 2 Liquid – Gas → Chloroform with N2 Solid – Gas → Camphor in N2

No. of gram equivalentof solute×100 Volume of solution

No. of moles of component Total no. of moles of all components

No. of parts of components×10 6 Total no. of parts of components of solution

Mass of component in solution × 100 Total mass of solution

ou se

Normality: Number of gram equivalents of the solute dissolved in one litre of solution

Solutions

Mole fraction

Parts per million : For trace quantities

Mass percentage w/w

Volume percentage v/v

Mass by volume p ercentage (w/v)

Maximum boiling azeotrope

∆H mix = negative ∆Vmix = negative

∆Vmix = positive ∆H mix = positive

For any solution, the partial vapour pressure of each volatile component is directly proportional to its mole fraction.

Non-ideal solution → (Mixture of chloroform and acetone)

Ideal solution → (n-hexane and n-heptane)

Minimum boiling azeotrope

W2 ×M 1 P°1 - P1 = M 2 ×W1 P°1

K b ×1000 × W2 M 2 × W1

K f × W2 ×1000 M 2 ×W1

• Relative lowering of vapour pressure →

• Elevation of boiling point → ∆Tb =

• Depression in freezing point → ∆Tf =

• Osmotic pressure → π = CRT

Increases with increase in pressure

Gas in L iq u id

Normal molar mass = Abnormal molar mass

G a

Partial pressure of gas in vapour phase is proportional to the mole fraction of gas in the solution. p =KHx

46 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) CHEMISTRY

Solutions

Chapter 6 Syllabus

Different methods for expressing concentration of solution - molality, molarity, mole fractions percentage (by volume and mass both), vapour pressure of solutions and Raoult’s Law - ideal and non-ideal solutions, vapour pressure - composition plots for ideal and non-ideal solutions. Colligative properties of dilute solutions-relative lowering of vapour pressure, depressing of freezing point, elevation of boiling point and osmotic pressure, determination of molecular mass using colligative properties, abnormal value of molar mass, Van’t Hoff's factor and its significance.

Topic-1

LIST OF TOPICS : Topic-1 : Different Methods for Expressing Concentration of Solution  .... P. 47 Topic-2 :  Colligative Properties of dilute solutions, Van’t Hoff's Factor and its significance .... P. 50

Different Methods for Expressing Concentration of Solution Concept Revision (Video Based) For details, scan the code

For details, scan the code

Normality, Molarity, Molality, Formality

For details, scan the code

Raoult's Law

 

Henry's Law

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions

Q.1. An open beaker of water in equilibrium with water vapour is in sealed container. When a few grams of glucose are added to the beaker of water, the rate at which water molecules.  [JEE (Main)-2nd Sept-2020-(Shift - I)] (a) Leaves the vapour decreases (b) Leaves the solution decreases (c) Leaves the vapour increases (d) Leaves the solution increases. Q.2. Four gases α, β, γ and d have KH values 50 K bar, 20 K bar, 2 × 10–3 K bar and 2 K bar respectively, then  [JEE (Main)-3rd Sept-2020-(Shift - I)] (a) α is more soluble in water (b) Pressure of γ in 55.5 molal solution is 1 bar (c) Pressure of d in 55.5 molal solution is 2.50 bar (d) Pressure of b in 55.5 molal solution 50 bar. Q.3. A solution of two components containing n1 moles of the 1st component and n2 moles of the 2nd component is present. M1 and M2 are the molecular



weights of component 1 and 2 respectively. It d is the density of the solution in g mol–1. C2 is the molarity and x2 is the mole fraction of the 2nd component, then C2 can be expressed as [JEE (Main)-6th Sept-2020-(Shift – I)] dx1 (a) C 2 = M 2 + x 2 ( M 2 - M1 ) (b) C 2 =

1000 x 2 M1 + ( M 2 - M1 ) x 2

(c) C 2 =

1000 dx 2 M1 + x 2 ( M 2 - M1 )

(d) C 2 =

dx 2 M 2 + ( M 2 - M1 ) x 2

Q.4. A set of solution is prepared using 180g of water as a solvent and 10g of different non-volatile solutes A, B and C. The relative lowering of vapour pressure in the presence of these solutes are in water.



[Given : Molar mass A = 100 g mol–1 B = 200 g mol–1 = 10,000 g mol–1] [JEE (Main)-6th Sept-2020-(Shift - II)]

(a) A > B > C (b) B > C > A (c) C > B > A (d) A > C > B Q.5. A solution of sodium sulphate contains 92 g of Na+ ions per kilogram of water. The molality of Na+ ions in that solution in mol kg–1 is [JEE (Main)-2nd Sept-2020-(Shift-I)] (a) 16 (b) 4 (c) 132 (d) 8 Q.6. Which one of the following statements regarding Henry's Law is not correct ? [JEE (Main)-9th Jan-2019-(Shift-I)] (a) Different gases have different KH (Henry's law constant) value at the same temperature (b) Higher the value of KH at a given pressure higher is the solubility of the gas in the liquids. (c) The value of KH increases with increase of temperature and KH is function of the nature of the gas. (d) The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution. Q.7. Liquids A and B form an ideal solution in the entire composition range. At 350 K, the vapour pressure of pure A and pure B are 7 × 103 Pa and 12 × 303 Pa, respectively. The composition of the vapour in equilibrium with a solution containing 40 mole percent of A at this temperature is [JEE (Main)-10th Jan-2019-(Shift-I)] (a) xA = 0.76; xB = 0.24 (b) xA = 0.28; xB = 0.72 (c) xA = 0.4; xB = 0.6 (d) xA = 0.37; xB = 0.63 Q.8. The amount of sugar (C12H22O11) required to prepare 2 L of its 0.1 M aqueous solution is [JEE (Main)-10th Jan-2019-(Shift-I)] (a) 17.1 g (b) 68.4 g (c) 136.8 g (d) 34.2 g Q.9. 8 g of NaOH is dissolved in 18 g of H2O. Mole fraction of NaOH is solution and molality (in mole Kg–1) of the solution respectively are [JEE (Main)-12th Jan-2019-(Shift-I)]

z y x w

(0,0) Mole fraction of water

(b)

Partial pressure

(a)

Partial pressure

(a) 0.2, 11.11 (b) 0.67, 22.20 (c) 0.2, 22.20 (d) 0.167, 11.11 Q.10. For the solution of gases w, x, y and z in water at 298 K in water at 298 K, the Henry's law constants (KH) are 0.5, 2, 35 and 40 K bar respectively. The correct plot for the given data is [JEE (Main)-8th April-2019-(Shift-II)]

z y

x w (0,0) Mole fraction of water

(c)

(d) z y x

w

CHEMISTRY

Partial pressure



Partial pressure

48 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

z y x w

(0,0) Mole fraction (0,0) Mole fraction of water of water Q.11. Molal depression constant for a solvent in 4.0 K kg mol–1. The depression in the freezing point of the solvent for 0.03 mol kg–1 solution of K2SO4 is (Assume complete dissociation of the electrolyte) [JEE (Main)-9th April-2019-(Shift-II)] (a) 0.18 K (b) 0.36 K (c) 0.12 K (d) 0.24 K Q.12. What would be the molality of 20% (mass/mass) aqueous solution of KI ? (Molar mass of KI = 166 g mol–1) [JEE (Main)-9th April-2019-(Shift-II)] (a) 1.48 (b) 1.51 (c) 1.35 (d) 1.08 Q.13. The mole fracton of a solvent in aqueous solution of solute in 0.8. The molality (in mol kg–1) of the aqueous solution is [JEE (Main)-12th April-2019-(Shift-II)] –2 (a) 1388 × 10 (b) 13.88 × 10–1 (c) 13.88 (d) 13.88 × 10–3

ANSWERS – KEY 1. (c) 2. (b) 5. (b) 6. (b) 9. (d) 10. (b) 13. (c)

3. (c) 7. (b) 11. (b)

4. (a) 8. (b) 12. (b)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (c) is correct. The vapour pressure of pure solvent is greater than the vapour pressure of solution. So when few grams of glucose are added to the beaker in water, vapour pressure of solution will be decreased than the vapour pressure pure solvent. In order to maintain the new equilibrium some vapour molecule will be convert in liquid water. Ans. 2. Option (b) is correct. (a) From Henry’s law we get p = KH x; where p = partial pressure, x is the mole traction and KH = Henry’s law constant. At a given pressure higher the value of KH for the gas lower will be its solubility in given solvent. So α gas is the least soluble gas. (b) For γ pγ = (KH)γ x

(c) For d (d) For β

55 ⋅ 5 = 2× = 1 bar 1000 55 ⋅ 5 + 18   pd = (KH)d x 1   = 2 × 103 × = 1000 bar 2 pβ = (KH)β x   =2 ×

1 = 1 bar 2

49

SOLUTIONS

Ans. 3. Option (c) is correct.

n1 Mole fraction of first component (X1) = n1 + n2 Mole fraction of second component (X2) =

n2 n1 + n2

Mass of solution = n1M1 + n2M2

∴ Volume of the solution =

n1M1 + n2 M 2 d

w1 M1 n1 M1 = w1 n1 =

No.of molesof solute × 1000 Molarity (C2) = Volumeof thesolution

      

1000   n M + n M 2 2 = n2  1 1 d  

    

n2 n2 × d × 1000 = x2 = n1 + n2 n1M1 + n2 M 2   

  

x 2 × d × 1000 n2= x 2 ( n1 + n2 )= n1 M1 + n2 M2 n1 + n2

x 2 × d × 1000 x 1 M1 + x 2 M 2    =

  =

x 2 × d × 1000

(1 − x2 ) M1 + x2 M 2 [As x1 + x2 = 1]

x 2 × d × 1000 = M1 + x 2 ( M 2 − M1 )

  Ans. 4. Option (a) is correct.

10 g = 0 ⋅ 1mole 100 g mol −1 10 g No. of moles of B= = 0 ⋅ 05mole 200 g mol −1 10 g No. of moles of C= = 0 ⋅ 001mole 10, 000 g mol −1 Relative lowering of vapour = mole fraction of the non-volatile solute. Therefore the relative lowering vapour follow as A > B > C. Ans.5. Option (b) is correct. Solute persent in 1000g of solvent Molality (m) = Molecular weight of solute No. of moles of A=

92 mol kg −1 23 = 4 mol kg–1 Ans.6. Option (b) is correct. According to Henry's Law Pgas = KH × xgas Therefore with increase of KH value the solubility of the gas will decrease in solution. Ans. 7. Option (b) is correct. =

For ideal solution, P = PA0 xA + PB0 xB fi P = 7 × 103 × 0.4 + 12 × 103 × 0.6 fi P = 2.8 × 103 + 7.2 × 103 fi P = 10 × 3 Pa = 1 × 104 Pa In vapour phase,

xA =

=

PA PA0 x A = P P 7 ¥ 10 3 ¥ 0.4 1 ¥ 10 4

= 0.28

xB = 1 – 0.28 = 0.72

Ans.8. Option (b) is correct. Molecular weight of C12H22O11 = (12 × 12 + 1 × 22 + 16 × 11) g mol–1 = 342 g mol–1 We know that, 1000 ml 1 M C12H22O11 ∫ 342 g or 2000 ml 0.1 M C12H22O11

∫

342 ¥ 2000 ¥ 0.1 g ∫ 68.4 g 1000

Ans. 9. Option (d) is correct. xNaOH = xNaOH

8 40 ∫ 8 18 + 40 18 =

0.2 0.2 1 = = = 0.167 0.2 + 1 1.2 6

Molality (m) =

Solute present in 1000 g of solvent Molecular wegiht of solute

=

8 × 1000 mol kg −1 18 × 40

= 11.11 mol kg–1 Ans. 10. Option (b) is correct. According to Henry's Law, Pgas = KH xgas

x gas + K H2 O = v fi Pgas = K H (1 - x H O ) x gas = 1 - x H2 O fi Pgas = K H - K H x H 2O Pgas = Partial pressure of the gas above the solution with a liquid (solvent) say water. xgas = Mole fraction of the gas (solute) in the solution. x H 2 O = Mole fraction of water (solvent) 2

Comparing with straight line equation, y = mx + c Slope = –KH (negative) Intercept = +KH Higher the value of KH more negative will be slope and lower the value of KH less negative will be slope. Moreover the higher the value fo KH, higher will be the intercept.

50 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Ans. 11. Option (b) is correct. K2SO4  2K+ + SO42– \ i = 3 From DTf = i Kf × m = 3 × 4 × 0.03 = 0.36 K Ans. 12. Option (b) is correct. Wt of solvent (H2O) = (100 – 20) g = 80 g 80 g of H2O contain KI, 20 g by weight 20 ¥ 1000 1000 g of H2O contain KI g by weight 80 20 ¥ 1000 Molality = = 1.51 mol/kg 80 ¥ 166

CHEMISTRY

Ans. 13. Option (c) is correct. Mole fraction of solvent (Xsolvent) = 0.8 But XSolvent + XSolute = 1 fi XSolute = 1 – 0.8 fi XSolute = 0.2 By the definition of molality, w2 1000 × m = w1 M2 =

n2 × 1000 n1 × M1

0.2 ¥ 1000 0.8 ¥ 18 = 13.88 mol kg–1

=

Topic-2

Colligative Properties of dilute solutions, Van’t Hoff’s Factor and its significance Concept Revision (Video Based) For details, scan the code Colligative Properties of Solution

For details, scan the code Van't hoff Factor

 

JEE (Main) Questions – 2019 and 2020 Multiple Choice Questions Q.1. If a mango shrinks when kept in concentrated salt solution then which of the following process take place ? [JEE (Main)-2nd Sept-2020-(Shift - II)] (a) Diffusion (b) Dialysis (c) Osmosis (d) Reverse osmosis Q.2. Vapour pressure of pure CS2 and CH3COCH3 are 512 mm of Hg and 312 mm of Hg respectively. Total vapour pressure of mixtures is 600 mm of Hg than find incorrect statement.  [JEE (Main)-7th Jan-2020-(Shift - I)] éA - - - - - Aù (a) ê B - - - - - B ú > A ------- B ë û (b) Does not obey Raoult’s laws (c) Endothermic solution (d) After adding 100 ml of each, then volume is less than 200 ml Q.3. There are two beakers (I) having pure volatile solvent and (II) having volatile solvent and non volatile solute. If both beakers are placed together in a closed container then :  [JEE (Main)-7th Jan-2020-(Shift - II)] (a) Volume of solvent beaker will decrease and solution beaker will increase. (b) Volume of solvent beaker will increase and solution beaker will also increase. (c) Volume of solvent beaker will decrease and solution beaker will also decrease.

(d) Volume of solvent beaker will increase and solution beaker will decrease. X Y Z Q.4. V.P

Temperature



(1) Intermolecular force of attraction of x > y (2) Intermolecular force of attraction of x < y (3) Intermolecular force of attraction of z < x Select correct options : [JEE (Main)-8th Jan-2020-(Shift - I)] (a) 1 & 3 (b) 1 & 2 (c) 2 only (d) 2 & 3 Q.5. A solution contain 62 g of ethylene glycol in 250 g of water in cooled upto –10°C. If Kg for water is 1.86 K kg mol–1, then amount of water (in g) separated as ice is [JEE (Main)-9th Jan-2019-(Shift-II)] (a) 32 (b) 48 (c) 64 (d) 16 Q.6. Elevation in the boiling point for 1 molal solution of glucose in 2 K. The depression in the freezing

51

SOLUTIONS

point for 2 molal solution of glucose in the same solvent is 2 K. The relation between Kb and Kf is [JEE (Main)-10th Jan-2019-(Shift-II)] (a) Kb = 1.5 Kf (b) Kb = 0.5 Kf (c) Kb = Kf (d) Kb = 2 Kf Q.7. The freezing point of a dilute milk sample is found to be –0.2°C while it should have been –0.5°C for pure milk. How much water has been added to pure milk to make the dilute sample? [JEE (Main)-11th Jan-2019- (Shift-I)] (a) 2 cup of water to 3 cups of pure milk (b) 1 cups of water to 3 cups of pure milk (c) 3 cups of water to 2 cups of pure milk (d) 1 cups of water to 2 cups of pure milk Q.8. K2 HgI4 is 40% ionised in aqueous solution. The value of its Van't Hoff factor (i) is [JEE Main 2019, 11th Jan (Shift-II)] (a) 1.6 (b) 1.8 (d) 2.0 (c) 2.2 Q.9. Freezing point of a 4% aqueous solution of X is equal to freezing point of 12% aqueous solution of Y. If molecular weight of X is A, then molecular weight of Y is [JEE (Main)-12th Jan-2019-(Shift-I)] (a) 4A (b) 2A (c) 3A (d) A Q.10. Molecules of benzoic acid (C6H5COOH) dimerise in benzene ‘w’ of the acid dissolved in 30 g of benzene shows a depression in freezing point equal to 2 K. If the percentage association of the acid to form dimer in the solution is 80, then w is (Given that Kf = 5 K of kg mol–1, molar mass of benzoic acid = 122 g mol–1) [JEE (Main)-12th Jan-2019-(Shift-II)] (a) 1.8 g (b) 1.0 g (c) 2.4 g (d) 1.5 g Q.11. The vapour pressures of pure liquids A and B are 400 and 600 mm Hg, respectively at 298 K. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquids B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components A and B in vapour phase, respectively are [JEE (Main)-8th April-2019-(Shift-I)] (a) 450 mm Hg, 0.4, 0.6 (b) 500 mm Hg 0.5, 0.5 (c) 450 mm Hg, 0.5, .05 (d) 500 mm Hg, 0.4, 0.6 Q.12. The osmatic pressure of a dilute solution of an ionic compound XY in water is four times that of a solution of 0.01 M BaCl2 in water. Assuming complete dissociation of the given ionic compounds in water, the concentration of XY (in mol–1) in solution is [JEE (Main)-9th April-2019-(Shift-I)] –2 (a) 4 × 10 (b) 16 × 10–4 –4 (c) 4 × 10 (d) 6 × 10–2 Q.13. Liquid M and Liquid N form an ideal solution. The vapour pressures of pure liquids M and N are 450 and 700 mm Hg, respectively at the same temperature. The correct statement is xM = mole fraction of M in solution xN = mole fraction of N in solution



YM = mole fraction of M in vapour phase YN = mole fraction of N in vapour phase [JEE (Main)-9th April-2019-(Shift-I)] x Y x Y (b) M = M (a) M > M x N YN x N YN

x M YM (d) (xM – YM) < (xN – YN) < x N YN Q.14. At room temperature, a dilute solution of urea is prepared by dissolving 0.60 g of urea in 360 g of water. If vapour pressure of pure water at this temperature is 35 mm Hg, lowering of vapour pressure will be (Molar mass of urea = 60 g mol–1) [JEE (Main)-10th April-2019-(Shift-I)] (a) 0.027 mm Hg (b) 0.031 mm Hg (c) 0.017 mm Hg (d) 0.028 mm Hg Q.15. 1 g of a non-volatile, non-electrolyte solute is dissolved in 100 g of two different solvents A and B. Whose ebullisocopic constant are in the ratio of 1 : 5. The ratio of the elevation in their boiling DTb ( A) is points, DTb (B) [JEE (Main)-10th April-2019-(Shift-II)] (a) 5 : 1 (b) 10 : 1 (c) 1 : 5 (d) 1 : 0.2 Q.16. A solution is prepared by dissolving 0.6 g of urea (molar mass = 60 g mol–1 and 1.8 g of glucose (molar mass = 180 g mol–1) in 100 mL of water at 27°C. The osmotic pressure of the is (R = 0.08206 L atm K–1 mol–1) [JEE (Main)-12th April-2019-(Shift-II)] (a) 8.2 atm (b) 2.46 atm (d) 1.64 atm (c) 4.92 atm (c)

ANSWERS – KEY

1. (c)

2. (d)

3. (a)

4. (c)



5. (c)

6. (d) 7. (c)

8. (b)



9. (c)

10. (c)

11. (d)

12. (d)

14. (c)

15. (c)

16. (c)

13. (a)

ANSWERS WITH EXPLANATIONS Ans. 1. Option (c) is correct. When mango kept in concentrated salt solution, the solvent (water) from mango move to concentrated salt solution. Consequently mango shrinks in solution and this phenomenon is known as osmosis. Ans. 2. Option (d) is correct. CS2 is a non-polar solvent but CH3COCH3 possess dipole moment. Therefore interaction between CS2 and acetone in solution is less than A – A and B – B interactions. It shows positive deviation i.e. does not obey the Raoult’s law. In such solution ΔH = +ve, as the energy is required to break the bonds, after adding 100 ml of each, then net volume is greater than 200 ml. Hence the incorrect statement is (d). Ans. 3. Option (a) is correct. Beaker (I) containing pure solvent which volatile in nature. Hence its volume decreases gradually. Beaker (II) having volatile solvent and non-volatile

52 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise)

Vapour Pressure

solute shows lowering of vapour pressure. Ans. 4. Option (c) is correct. X Y Z

Temperature At a definite temperature as intermolecular force of attraction increases vapour pressure decreases. Hence the option (c) is correct. Ans.5. Option (c) is correct. We know that DTf = Kf × m W ¥ 1000 fi Tf¢ – Tg¢ = Kf × 2 M2 ¥ W1

fi 0 – (–10) = 1.86 × fi

W1 =

62 ¥ 1000 62 ¥ W1

1.86 ¥ 62 ¥ 1000 62 ¥ 10

fi W1 = 186 g Therefore, the amount of water separated as ice is (DW) = (250 – 186) g = 64 Ans.6. Option (d) is correct. The relation between elevation in boiling point and depression in freezing point is DTb Kb ¥ m = DTf Kf ¥ m fi fi

Kb ¥ 1 2 = Kf ¥ 2 2

fi

Freezing point of pure milk

Freezing point of diluted milk

Tf = –0.5°C

Tf = –0.2°C

DTf = 0.5°C

DTf = 0.2°C

M

DTf = Kf cup ( DTf )1 ( DTf )2

=

K f ¥ x ¥ 1000 ¥ W2 K f ¥ x ¥ W1 ¥ 1000

fi

0.5 W2 = W1 0.2

fi

W2 =

5 W1 2

5 3 W1 - W1 = W1 2 2 Therefore 3 cups of water will be added with 2 cups of pure milk. Ans.8. Option (b) is correct. \

Wwater =

fi i = 1.8 Ans.9. Option (c) is correct. For the same freezing point of solution the molality of the solution of X and Y will be equal. \ mx = my fi

12 ¥ 1000 4 ¥ 1000 = 88 ¥ Y 96 ¥ A

or, Y = 3.27A or, Y = 3A Ans.10. Option (c) is correct. Degree of association for C6H5COOH is given as follows i -1 a = 1 -1 n [As 2C6H5COOH  (C6H5COOH)2] i -1 80 fi = [Give, n = 2 1 100 - 1 a = 80%] 2 \ i = 0.6 Again DTf = iKf × m W ¥ 1000 fi 2 = 0.6 × 5 × 122 ¥ 30 W =

2 ¥ 122 ¥ 30 3 ¥ 1000

fi W = 2.44 \ W  2.4 g

Ans.7. Option (c) is correct.



K2[Hg I4]  2K+ + [Hg I4]2– Van’t Hoff factor for ionisation reaction is given as follows, i = 1 + a (n – 1) i -1 \ a = [Where n = no of ions n-1 a = degree of dissociation] i -1 fi 0.4 = 3-1

fi

Kb = 2 K f

CHEMISTRY

Ans. 11. Option (d) is correct. For ideal solution, Ptotal = PA° × A + PB° × B \ Ptotal = {400 × 0.5 + 600 (1 – 0.5)} mm Hg = (200 + 300)mm Hg = 500 mm Hg Therefore mole fraction of A in vapour phase YA =

PA P° ×A = A Ptotal Ptotal

=

0.5 ¥ 400 = 0.4 500

Mole fraction of B in vapour phase, YA + YB = 1 fi YB = 1 – YA fi YB = 1 – 0.4 fi YB = 0.6 Ans. 12. Option (d) is correct.

53

SOLUTIONS

From osmotic pressure of solution we get, p = i CRT Given concentration of BaCl2 = 0.01 M (Given) pxy = 4p BaCl 2

fi i Cxy RT = 4i × CBaCl 2 ¥ RT ...(i) Again XY  X+ + Y– [Here i = 2] And BaCl2  Ba2+ + 2Cl– [Here i = 3] From (i) we get 2 × Cxy = 4 × 3 × 0.01 0.12 fi Cxy = 2 fi Cxy = 0.06 mol L–1 = 6 × 10–2 mol L–1 Ans. 13. Option (a) is correct. P°M = 450 mm Hg, P°N = 700 mm Hg \ PM = P°M × M = YM ×PT Y fi P°M = M ×PT XM

Ans. 15. Option (c) are correct. DTb = Kb × m Kb (A) DTb (A) = [Given mA = mB] Kb ( B) DTb ( B) \

Ans. 16. Option (c) are correct. From Van't Hoff's law p = CRT n RT = V =

Ans. 14. Option (c) is correct. We know that, Relative Cowering of vapour pressure = mole fraction of solute. DP fi = X2 [X2 = mole fraction of solute P0 n2 = ] n1 + n2

fi



Integer Type Questions (Chapter Based)

XM YM < XN YN

n2 DP = n1 + n2 P0



V

= (0.01 + 0.01) × 10 × 0.0821 × 300 p = 4.92 atm

Given, P°N < P°N YN YM fi < XN XM

fi

( nurea + nglucose ) RT

Ê 0.6 18 ˆ ÁË 60 + 180 ˜¯ ¥ 0.0821 ¥ 300 = 100 1000

Similarly we get, YN ×PT P°N = XN

fi

DTb (A) 1 = DTb ( B) 5

[n1 = no. of moles of solvent n2 = no. of moles of solute]

W2 M DP 2 = W W P0 1 + 2 M1 M 2



fi

0.60 DP 60 = 0.60 360 P0 + 60 18

fi

DP = 0.0005 P0

 È Given wt ˘ Íno. of moles ˙ Mol. wt ˚ Î

fi DP = 0.0005 × P0 = 0.0005 × 35 mm Hg = 0.0175 mm Hg

Q.1. In 0.1 mole fraction glucose solution, find % (w/w) of water in the sample (in nearest integer).  [JEE (Main) 2020, 3rd Sept (Shift – I)] Sol. Let total mole fraction of solution (x1 + x2) = 1 So, mole fraction of glucose (x1) = 0.1 mole fraction of H2O = 0.9 0 ⋅ 9 × 18   = %(w/w) of H 2O   × 100 ⋅ × + ⋅ × 0 9 18 0 1 180    = 47.368    47.37 Q.2. 6.023 × 1022 molecules are present in 10 g of substance ‘X’. The molarity of a solution containing 5 g of substance ‘X’ in 2L solution is ……… × 10–3.  [JEE (Main) 2020, 3rd Sept (Shift - II)] Sol. 6.023 × 1022 molecules of X having weight 10 g 6.023 × 1023 molecules of X having weight 10 × 6.023 × 10 23 = = 100g 6.023 × 10 22 

∴ Molecular weight of X substance is 100 g 2L i.e. 2000 ml solution contains X by 5.g 1000 ml solution contains X by 2.5 g 2⋅5 ∴ Molarity = M = 25 × 10–3 100

Q.3. At 300 K, the vapour pressure of a solution containing 1 mole of n – hexane and 3 moles of n – heptane is 550 mm of Hg. At the same temperature. If one more mole of n – heptane is added to this solution, the vapour pressure of the solution increases by 10 mm of Hg. What is the vapour pressure in mm Hg of n – heptane in its pure state ……. ? (JEE (Main) 2020, 4th Sept (Shift - I))

54 Oswaal JEE (MAIN) Solved Papers (Chapter-wise & Topic-wise) Sol. We know that PT = PA° X A + P°B X B Here, PA° = Partial pressure of hexane, XA = Mole fraction of Hexane, PB° = Partial pressure of heptane, XB = Mole fraction of Hepane

Now,



⇒ 550 = P°A ×

1 3 + P°B × 4 4

⇒ 2200 = PA0 + 3PB0

or= 550 0.26PA0 + 0.75PB0  After adding 1 mole of n–heptane, 1 4 560 = P°A × + P°B × 5 5

….(i)

560 0.2PA0 + 0.8PB0 …(ii) ⇒= Subtracting from (ii) to (i) P°B =600

∴ Vapour pressure of n – heptane is = 600 mm of Hg Q.4. If 250 cm3 of an aqueous solution containing 0.73 g of a protein A is isotonic with one litre of another aqueous solution containing 1.65 g of a protein B at 298 K, the ratio of the molecular masses of A and B is …… × 10–2 (to the nearest integer).  [JEE (Main) 2020, 3rd Sept (Shift - II)] Sol. As the solution is isotonic ∴ i2 C1 = i2 C2 (Here i1 = i2 = 1 for protein) ⇒ C1 = C2 0 ⋅ 73 × 1000 1 ⋅ 65 ⇒ = 250 × M A MB

⇒

M A 0 ⋅ 73 × 4 = =1 ⋅ 77 =177 × 10 −2 MB 1 ⋅ 65

Q.5. The osmotic pressure of a solution of NaCl is 0.10 atm and that of a glucose solution is 0.20 atm. The osmotic pressure of a solution formed by mixing 1L of the sodium chloride solution with 2L of the glucose solution is x × 10–3 atom x is …. (The atomic mass of Na is 23g mol–1)  [JEE (Main) 2020, 4th Sept (Shift - II)] Sol. We know that   π = ic RT n ⇒    π =i RT V inRT ⇒ πV = Therefore from Boyle – Van’t hoff law we get Therefore πfinal (V1 + V2) = π1V1 + π2V2

∴ πfinal = 0 ⋅ 1 × 1 + 0 ⋅ 2 × 2 3  [When temperature and mass i.e. number moles remains constant]

0 ⋅ 5 500 × 10 −3 = = 167 × 10 −3 atm 3 3 ∴   x = 167 Q.6. The elevation of boiling point of 0.10 m aqueous CrCl3 × NH3 solutions is two terms that of 0.05 m aqueous CaCl2 solution. The value of x is ……. [Assume 100% ionization of complex and CaCl3

CHEMISTRY

coordination number of Cr as 6 and that all NH3 molecules are present inside the coordination sphere] [JEE (Main) 2020, 4th Sept(Shift – II)] 2+ Sol. CaCl  Ca + 2Cl–

0.05 m

0.05 m

0.10 m

∴ 0.05m aqueous CaCl2 solution = 0.15m solution of non-electrolyte. By the problem [CrCl3.x.NH3] have concentration = 0.15 × 2 = 0.30 m. Therefore CrCl3.x.NH3 must produce three ions in aqueous solution maintaining coordination number of central metal ion 6 and the formula of the given compound will be [Cr(NH3)5Cl]Cl2. Therefore x = 5.00. Q.7. The strength of 5.6 volume of H2O2 (of density 1 g/mL) in terms of mass percentage and molarity(M) respectively are.  [JEE (Main) 2020, 3rd Sept(Shift - II)] Sol. 2H2O2 = 2H2O + O2 68 22400 ml at NTP At N.T.P. 5.6 ml of O2 is produced from 1 mL of H2O2 solution 22400 ∴ 22400ml of O2 is produced from mL of 5⋅6 H O solution



2

2

= 4000 mL of H2O2 solution 400 ml H2O2 solution contains 68 g of H2O2 68 × 100 100 ml H2O2 solution contains of H2O2 4000 = 1.7%



Again 4000 ml H2O2 solution contains 68 g of H2O2 ∴ 100 ml H2O2 solution contains 17g of H2O2 17 ∴ Molarity = M = 0.500 M 34 Q.8. Determine the amount of NaCl to be dissolved in 600g H2O to decrease the freezing point by 0.2°C. Given Kf of H2O = 2K kg mol–1 density of H2O(l) = 1 g/ml [JEE (Main) 2020, 9th Jan(Shift – I)] w2 × 1000 Sol. ∆Tf = i K f × M 2 × w1

−1 ⇒ 0 ⋅ 2K =2 × 2K kg mol ×



⇒ w= 2

w2 × 1000 58 ⋅ 5g mol −1 × 600 g

0 ⋅ 2 × 58 ⋅ 5 × 600 g = 01 ⋅ 76 g 1000 × 4

Q.9. 0.2 ml of an ideal gas has volume 1 dm3 in locked box with friction less piston. The gas is in thermal equilibrium with excess of 0.5 m aqueous ethylene glycol at its freezing point. If piston is released all of a sudden at 1 atm then determine the final volume of gas in dm3 (R = 0.8 atm mol–1 K–1, Kf = 2.0 K mol–1)  [JEE (Main) 2020, 9th Jan(Shift – II)] Ideal gas

Sol.

=

0.5 m ethylene glycerol in water V1 =1 dm³

55

SOLUTIONS

ΔTf = Kf × m ⇒ ΔTf = 2K kg mol–1 × 0⋅5 mol kg–1 = 1 ⇒ Tf = (273 – 1)K = 272 K For n mole ideal gas equation we get

= ρgas

nRT 0 ⋅ 1mol × 0 ⋅ 08 atm mol −1k −1 × 272 K = V 1L

= 2⋅176 atm After releasing the piston, the final volume will be P1V1 2 ⋅ 176 atm × 1dm 3 V2 = = P2 1atm

= 2 ⋅ 176 dm 3  2 ⋅ 18 dm 3 Q.10. What will be the mass urea that should be dissolved in 50 g of water in order to produce the same lowering of vapour pressure of produced by dissolving 3 g of glucose in the same quantity of water ? n2 W2 M1 P 0 - PS = Sol. = X2 = n1 W1 M2 P0

As





P 0 - PS 0

P c Ê W2 M1 ˆ ÁË W M ˜¯ 1

2 urea

=

n2 same in the two cases. n1

Ê W2 M1 ˆ = Á Ë W1 M2 ˜¯ glucose





= 113.77 Van't Hoff factor (i) =  Normal molecular weight  O observedmolecular weight



=



= 0.53 2 CH3 COOH  (CH3 COO)2 a 1– a 2



Tb =

This indicates that complex compound dissociation produces three ions as

=

60 113.77

1-a +

7353 21 fi Tb = 350.1 K fi Tb = 77°C Q.12. If the freezing point of a 0.01 molal aqueous solution of a cobalt (III) chloride - ammonia complex (which behave as a strong electrolyte) is –0.0558°C, the number of chloride(s) in the coordination sphere of the complex is......... [Kf of water = 1.86 K Kg mol–1] Sol. DTf = 0 – (0.0558 °C) = 0.0558°C Again DTf = i Kfm (Van't Hoff factor) D Tf 0.0558 i = = =3 1.86 ¥ 0.01 Kfm fi

5.12 ¥ 0.2 ¥ 1000 0.45 ¥ 20





3 ¥ 18 W2 ¥ 18 fi = 50 ¥ 180 50 ¥ 60 fi W2 = 1.00 g Q.11. The heat of vaporisation of benzene is 7353 cal mol–1. What will be the approximate boiling point of benzene? Sol. According to Trouton's rule D H vap = 21 cal K–1 mol–1 Tb

Q.13. The freezing point of a solution containing 0.2 g of ethanoic acid in 20 g of benzene is lowered by 0.45°C. Calculate the degree of association of ethanoic acid in benzene. Assume ethanoic acid dimerises in the solvent benzene. [Kf of benzene = 5.12 K mol–1 kg] K f ¥ W ¥ 1000 Sol. Molecular weight = D Tf ¥ w

on

[Co(NH 3 )5 Cl] Cl 2  [Co(NH 3 )5 Cl] 2 + + 2 Cl Therefore only one chloride ion is present in the coordination sphere is 1.00.



Van't Hoff factor i =

1

a 2 = 0.53

a fi 1 - = 0.53 2 fi a = 1.06 – 2 fi a = 0.94 \ Degree of association a = 0.94  94% Q.14. What weight (in kg) of solute (molecular weight = 60 g mol–1) is required to dissolved it in 1800 g 3 of water of reduce the vapour pressure to th of 5 pure water? Sol. From lowering of vapour pressure are gas



P 0 - PT w¥M = PT m¥W



3 P0 - P0 w ¥ 18 5 fi = 3 0 1800 ¥ 60 P 3 5 [Here PT = P 5 0 m = 1800 W = 60 g mol–1 M = 18 g mol–1]



fi



fi

w = 6000 ¥



fi

w = 4000 g = 4 kg



w 2P0 5 ¥ 0 = 1000 5 3P 2000

2 3



o

Arrhenius Concept: Arrhenius acid: gives H+ ions Arrhenius base: gives OH – ions Bronsted–Lowry Concept: Bronsted –Lowry acid: proton donor Bronsted –Lowry base: proton acceptor form conjngate acid - base pair Lewis Concept: Lewis acid: Electron pair acceptor Lewis base: Electron pair donor

ΔG is negative, reaction is spontaneous and proceeds in forward direction . ΔG is positive, reaction is non-spontaneous and proceeds in backward direction . ΔG is 0, reaction has achieved equilibrium. ° K = e– G /RT

Step 1: Write the balanced equation for the reaction. Step 2: Make a table that lists for each substance involved in reaction; Initial concentration, change in concentration, equilibrium concentration. Step 3: Substitute the equilibrium concentrations into the equilibrium equation for the reaction and solve. Step 4: Calculate equilibrium concentrations from the calculated value of concentration of substances. Step 5: Check results by substituting them into the equilibrium equation. Buffer Solutions: These are the solutions which reside change in pH on dilution or addition of acid or alkali. Solubility Product (Ksp): For a sparingly soluble salt, it is the product of molar concentration of ions raised to power equal to numbers of times each ion.

• If Qc < Kc, net reaction goes from left to right. • If Qc > Kc, net reaction goes from right to left. • If Qc = Kc, no net reaction occurs

• Kc < 10-3 reaction proceeds rarely. • Kc > 103 reaction proceeds nearly to completion • Kc → 10–3 to 103, reaction is at equilibrium

a Reactants and products are in same phase.

Ba

m

Equilibrium Law o r

Law of

pH of a solution is negative logarithm to base 10 of the activity of hydrogen ion Size increases HF