Preview 43 Years Chapterwise Topicwise Solved Papers (20211979) IIT JEE Chemistry by Ranjeet Shahi (2022)

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1 Some Basic Concepts of Chemistry 1-22

24 Alcohols and Ethers 364-377

25 Aldehydes and Ketones 378-396

26 Carboxylic Acids and

27 Aliphatic Compounds Containing Nitrogen 413-422

28 Benzene and Alkyl Benzene 423-440

29 Aromatic Compounds Containing Nitrogen 441-457

30 Aryl Halides and Phenols 458-470

31 Aromatic Aldehydes, Ketones

UNIT I 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 II 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, van der Waals' equation,

liquefaction of gases, critical constants

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, magnetic and dielectric

properties

UNIT III Atomic Structure

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, 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,

ψ and ψ , concept of atomic orbitals as one electron wave functions; 2

Variation of ψ and ψ with r for 1s and 2s orbitals; various quantum 2

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 –

aufbau principle, Pauli's exclusion principle and Hund's rule,

electronic configuration of elements, extra stability of half-filled and

completely filled orbitals

UNIT IV 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 V 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; ΔS of the

o

universe and ΔG of the system as criteria for spontaneity, ΔG

(Standard Gibb's energy change) and equilibrium constant

UNIT VI Solutions

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 point and osmotic pressure; Determination of molecular mass using colligative properties; Abnormal value of molar mass, van't Hoff factor and its significance

UNIT VII Equilibrium

Meaning of equilibrium, concept of dynamic equilibrium

Equilibria involving physical processes Solid -liquid, liquid - gas and solid - gas equilibria, Henry's law, general characteristics of equilibrium involving physical processes

Equilibria involving chemical processes Law of chemical equilibrium,

equilibrium constants (K and K) and their significance, significance of

o

ΔG and ΔG 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 VIII Redox Reactions and Electrochemistry

Electronic concepts of oxidation and reduction, redox reactions, oxidation number, rules for assigning oxidation number, balancing of redox reactions

JEE MAIN

SYLLABUS

Section A : PHYSICAL CHEMISTRY

Electrochemical cells - Electrolytic and Galvanic cells, different types of

electrodes, electrode 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; Corrosion and its prevention

UNIT IX 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

UNIT X Surface Chemistry

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

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

UNIT XI Classification of Elements and

Periodicity in Properties

Periodic Law and Present Form of the Periodic Table, s, p, d and f Block

Elements, Periodic Trends in Properties of Elementsatomic and Ionic Radii,

Ionization Enthalpy, Electron Gain Enthalpy, Valence, Oxidation States and

Chemical Reactivity

UNIT XII General Principles and Processes of Isolation of Metals

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 XIII 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 XIV s - Block Elements (Alkali and Alkaline Earth Metals)

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, sodium chloride, 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 XV p - Block Elements

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.Group wise 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,(PCl , PCl ); 3 5

Structures of oxides and oxoacids of nitrogen and phosphorus

Group 16 Preparation, properties, structures and uses of dioxygen and ozone; Allotropic forms of sulphur; Preparation, properties, structures and uses of sulphur dioxide, sulphuric acid (including its industrial preparation); Structures of oxoacids of sulphur

Group 17 Preparation, properties and uses of chlorine and hydrochloric acid; 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; Structures of fluorides and oxides of xenon

UNIT XVI 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

K Cr O and KMnO 2 2 7 4

Inner Transition Elements

Lanthanoids - Electronic configuration, oxidation states, chemical reactivity and lanthanoid contraction Actinoids - Electronic configuration and oxidation states

UNIT XVII Coordination Compounds

Introduction to coordination compounds, Werner's theory; ligands, coordination number, denticity, chelation; IUPAC nomenclature of mononuclear coordination 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 XVIII 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 environmental pollution

Section B : INORGANIC CHEMISTRY

UNIT XIX Purification & Characterisation of Organic Compounds

Purification Crystallisation, 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 XX 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: —C=C—,—C=C— 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 XXI Hydrocarbons

Classification, isomerism, IUPAC nomenclature, general methods of

preparation, properties and reactions

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 effect); Ozonolysis, oxidation, and

polymerization

Alkenes acidic character; addition of hydrogen, halogens, water and

hydrogen halides; polymerization

Aromatic hydrocarbons Nomenclature, benzene structure and

aromaticity; Mechanism of electrophilic substitution: halogenation,

nitration, Friedel – Craft's alkylation and acylation, directive influence of

functional group in mono-substituted benzene

UNIT XXII 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 XXIII 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 reactions such as - Nucleophilic addition

reactions (addition of HCN, NH and its derivatives), Grignard reagent; 3

oxidation; reduction (Wolff Kishner and Clemmensen); acidity of α -

hydrogen, aldol condensation, Cannizzaro reaction, Haloform reaction;

Chemical tests to distinguish between aldehydes and Ketones

Carboxylic Acids Acidic strength & factors affecting it

UNIT XXIV 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 XXV Polymers

General introduction and classification of polymers, general methods

of polymerization-addition 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 XXVI Biomolecules

General introduction and importance of biomolecules

Carbohydrates Classification aldoses and ketoses; monosaccharides (glucose and fructose), constituent monosaccharides of

oligosacchorides (sucrose, lactose, maltose) and polysaccharides (starch, cellulose, glycogen)

Proteins Elementary Idea of α-amino acids, peptide bond, polypeptides; proteins: primary, secondary, tertiary and quaternary structure (qualitative idea only), denaturation of proteins, enzymes Vitamins Classification and functions

Nucleic Acids Chemical constitution of DNA and RNA Biological functions of Nucleic acids

UNIT XXVII Chemistry in Everyday Life

Chemicals in medicines Analgesics, tranquilizers, antiseptics, disinfectants, antimicrobials, antifertility drugs, antibiotics, antacids, antihistamins - their meaning and common examples

Chemicals in food Preservatives, artificial sweetening agents - common examples

Cleansing agents Soaps and detergents, cleansing action

Unit XXVIII 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

— Inorganic compounds Mohr's salt, potash alum

— Organic compounds Acetanilide,

p-nitroacetan ilide, aniline yellow, iodoform.

— Chemistry involved in the titrimetric excercises - Acids bases and the use of indicators, oxali acid vs KMnO , Mohr's salt vs KMnO 4 4

— Chemical principles involved in the qualitative salt analysis

— Chemical principles involved in the following experiments

1 Enthalpy of solution 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

PHYSICAL CHEMISTRY

General Topics Concept of atoms and molecules,

Dalton's atomic theory, Mole concept, Chemical formulae,

Balanced chemical equations, Calculations (based on

mole concept) involving common oxidation-reduction,

neutralisation, and displacement reactions, Concentration

in terms of mole fraction, molarity, molality and normality

Gaseous and Liquid States Absolute scale of

temperature, ideal gas equation, Deviation from ideality,

van der Waals' equation, Kinetic theory of gases, average,

root mean square and most probable velocities and their

relation with temperature, Law of partial pressures,

Vapour pressure, Diffusion of gases

Atomic Structure and Chemical Bonding Bohr model,

spectrum of hydrogen atom, quantum numbers,

Wave-particle duality, de-Broglie hypothesis, Uncertainty

principle, Qualitative quantum mechanical picture of

hydrogen atom, shapes of s, p and d orbitals, Electronic

configurations of elements (up to atomic number 36),

Aufbau principle, Pauli's exclusion principle and Hund's

rule, Orbital overlap and covalent bond; Hybridisation

involving s, p and d orbitals only, Orbital energy diagrams

for homonuclear diatomic species, Hydrogen bond,

Polarity in molecules, dipole moment (qualitative aspects

only), VSEPR model and shapes of molecules (linear,

angular, triangular, square planar, pyramidal, square

pyramidal, trigonal bipyramidal, tetrahedral and

octahedral)

Energetics First law of thermodynamics, Internal energy,

work and heat, pressure-volume work, Enthalpy, Hess's

law, Heat of reaction, fusion and vaporization, Second law

of thermodynamics, Entropy, Free energy, Criterion of

spontaneity

Chemical Equilibrium Law of mass action, Equilibrium

constant, Le-Chatelier's principle (effect of concentration,

temperature and pressure), Significance of DG and DGo in

chemical equilibrium, Solubility product, common ion

effect, pH and buffer solutions, Acids and bases (Bronsted

and Lewis concepts), Hydrolysis of salts

Electrochemistry Electrochemical cells and cell reactions,

Standard electrode potentials, Nernst equation and its

relation to DG, Electrochemical series, emf of galvanic

cells, Faraday's laws of electrolysis, Electrolytic

conductance, specific, equivalent and molar conductivity,

Kohlrausch's law, Concentration cells

Chemical Kinetics Rates of chemical reactions, Order of

reactions, Rate constant, First order reactions, Temperature dependence of rate constant (Arrhenius equation)

Solid State Classification of solids, crystalline state, seven

crystal systems (cell parameters a, b, c), close packed structure of solids (cubic), packing in fcc, bcc and hcp lattices, Nearest neighbours, ionic radii, simple ionic compounds, point defects

Solutions Raoult's law, Molecular weight determination

from lowering of vapour pressure, elevation of boiling point and depression of freezing point

Surface Chemistry Elementary concepts of adsorption

(excluding adsorption isotherms), Colloids, types, methods of preparation and general properties, Elementary ideas of emulsions, surfactants and micelles (only definitions and examples)

Nuclear Chemistry Radioactivity, isotopes and isobars,

Properties of rays, Kinetics of radioactive decay (decay series excluded), carbon dating, Stability of nuclei with respect to proton-neutron ratio, Brief discussion on fission and fusion reactions

Transition Elements (3d series) Definition, general

characteristics, oxidation states and their stabilities, colour (excluding the details of electronic transitions) and

JEE ADVANCED

compounds, cis-trans and ionisation isomerisms,

hybridization and geometries of mononuclear

coordination compounds (linear, tetrahedral, square

planar and octahedral)

Preparation and Properties of the following

Compounds Oxides and chlorides of tin and lead, Oxides,

chlorides and sulphates of Fe , Cu and Zn , Potassium

permanganate, potassium dichromate, silver oxide, silver

nitrate, silver thiosulphate

Ores and Minerals Commonly occurring ores and

minerals of iron, copper, tin, lead, magnesium, aluminium,

zinc and silver

Extractive Metallurgy Chemical principles and reactions

only (industrial details excluded), Carbon reduction

method (iron and tin), Self reduction method (copper and

lead), Electrolytic reduction method (magnesium and

aluminium), Cyanide process (silver and gold)

Principles of Qualitative Analysis Groups I to V (only

Concepts Hybridisation of carbon, Sigma and pi-bonds,

Shapes of simple organic molecules, Structural and

geometrical isomerism, Optical isomerism of compounds

containing up to two asymmetric centres, (R,S and E,Z

nomenclature excluded), IUPAC nomenclature of simple

organic compounds (only hydrocarbons, mono-functional

and bi-functional compounds), Conformations of ethane

and butane (Newman projections), Resonance and

hyperconjugation, Keto-enol tautomerism, Determination

of empirical and molecular formulae of simple

compounds (only combustion method), Hydrogen bonds,

definition and their effects on physical properties of

alcohols and carboxylic acids, Inductive and resonance

effects on acidity and basicity of organic acids and bases,

Polarity and inductive effects in alkyl halides, Reactive

intermediates produced during homolytic and heterolytic

bond cleavage, Formation, structure and stability of

carbocations, carbanions and free radicals

Preparation, Properties and Reactions of Alkanes

Homologous series, physical properties of alkanes

(melting points, boiling points and density), Combustion

and halogenation of alkanes, Preparation of alkanes by

Wurtz reaction and decarboxylation reactions

Preparation, Properties and Reactions of Alkenes and

Alkynes Physical properties of alkenes and alkynes

(excluding the stereochemistry of addition and elimination), Reactions of alkenes with KMnO and ozone, 4

Reduction of alkenes and alkynes, Preparation of alkenes and alkynes by elimination reactions, Electrophilic addition reactions of alkenes with X , HX, HOX and H O 2 2

(X=halogen), Addition reactions of alkynes, Metal acetylides

Reactions of Benzene Structure and aromaticity,

Electrophilic substitution reactions, halogenation, nitration, sulphonation, Friedel-Crafts alkylation and acylation Effect of o-, m- and p-directing groups in monosubstituted benzenes

Phenols Acidity, electrophilic substitution reactions

(halogenation, nitration and sulphonation), Tiemann reaction, Kolbe reaction

Reimer-Characteristic Reactions of the following (including those mentioned above) Alkyl halides, rearrangement

reactions of alkyl carbocation, Grignard reactions, nucleophilic substitution reactions, Alcohols, esterification, dehydration and oxidation, reaction with sodium, phosphorus halides, ZnCl /concentrated HCl, 2conversion of alcohols into aldehydes and ketones, Ethers, Preparation by Williamson's Synthesis, Aldehydes and Ketones, oxidation, reduction, oxime and hydrazone formation, aldol condensation, Perkin reaction, Cannizzaro reaction, haloform reaction and nucleophilic addition reactions (Grignard addition), Carboxylic acids, formation

of esters, acid chlorides and amides, ester hydrolysis Amines, basicity of substituted anilines and aliphatic amines, preparation from nitro compounds, reaction with nitrous acid, azo coupling reaction of diazonium salts of aromatic amines, Sandmeyer and related reactions of diazonium salts, carbylamine reaction, Haloarenes, nucleophilic aromatic substitution in haloarenes and substituted haloarenes (excluding Benzyne mechanism and Cine substitution)

Carbohydrates Classification, mono and disaccharides

(glucose and sucrose), Oxidation, reduction, glycoside formation and hydrolysis of sucrose

Amino Acids and Peptides General structure (only

primary structure for peptides) and physical properties

Properties and Uses of Some Important Polymers

Natural rubber, cellulose, nylon, teflon and PVC

Practical Organic Chemistry Detection of elements (N, S,

halogens), Detection and identification of the following functional groups, hydroxyl (alcoholic and phenolic), carbonyl (aldehyde and ketone), carboxyl, amino and nitro, Chemical methods of separation of mono-functional organic compounds from binary mixtures

Topic 1 Mole Concept

Objective Questions I (Only one correct option)

1. 5 moles of AB2weight 125 10× −3kg and 10 moles of A B2 2

weight 300 10× −3kg The molar mass of A M( A)and molar

mass of B M( B)in kg mol−1are (2019 Main, 12 April I)

(a) M A =10 10× −3and M B = ×5 10−3

(b) M A =50 10× − 3and M B =25 10× − 3

(c) M A =25 10× −3and M B =50 10× −3

(d) M A = ×5 10− 3and M B =10 10× − 3

2. The minimum amount of O2( )g consumed per gram of

reactant is for the reaction (Given atomic mass : Fe=56,

O=16, Mg=24, P=31, C=12, H=1) (2019 Main, 10 April II)

(a) C H3 8( )g +5O2( )g → 3CO2( )g +4H O2 ( )l

(b) P4( )s +5O2( )g → P O4 10( )s

(c) 4Fe( )s +3O2( )g → 2Fe O2 3( )s

(d) 2Mg( )s +O2( )g → 2MgO( )s

3. At 300 K and 1 atmospheric pressure,

10 mL of a hydrocarbon required 55 mL of O2for complete

combustion and 40 mL of CO2is formed The formula of the

hydrocarbon is (2019 Main, 10 April I)

(a) C H Cl4 7 (b) C H4 6 (c) C H4 10 (d) C H4 8

4. 10 mL of 1 mM surfactant solution forms a monolayer

covering 0.24 cm2on a polar substrate If the polar head is

approximated as a cube, what is its edge length?

(2019 Main, 9 April II)

6. The percentage composition of carbon by mole in methane is

(2019 Main, 8 April II)

(a) 75% (b) 20% (c) 25% (d) 80%

7. 8 g of NaOH is dissolved in 18 g of H O.2 Mole fraction ofNaOH in solution and molality (in mol kg−1) of the solutionrespectively are (2019 Main, 12 Jan II)

(a) 0.2, 11.11 (b) 0.167, 22.20(c) 0.2, 22.20 (d) 0.167, 11.11

8. The volume strength of 1 M H O2 2is(Molar mass of H O2 2=34 g mol−1) (2019 Main, 12 Jan II)(a) 16.8 (b) 22.4 (c) 11.35 (d) 5.6

9. The amount of sugar (C H O )12 22 11 required to prepare 2 L ofits 0.1 M aqueous solution is (2019 Main, 10 Jan II)

(a) 17.1 g (b) 68.4 g (c) 136.8 g (d) 34.2 g

10. For the following reaction, the mass of water produced from

445 g of C H O57 110 6is :2C H O ( )57 110 6 s +163O ( )2 g →114CO ( )2 g +110 H O2 ( )l

(2019 Main, 9 Jan II)

(2017 JEE Main)

(a) 15 kg (b) 37.5 kg(c) 7.5 kg (d) 10 kg

13. 1 g of a carbonate (M2CO ) on treatment with excess HCl3produces 0.01186 mole of CO2 The molar mass of M2CO3

in g mol−1is (2017 JEE Main)

(a) 1186 (b) 84.3 (c) 118.6 (d) 11.86

Some Basic Concepts

of Chemistry

1

14. At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon

requires 375 mL air containing 20% O2 by volume for

complete combustion After combustion, the gases occupy

330 mL Assuming that the water formed is in liquid form

and the volumes were measured at the same temperature and

pressure, the formula of the hydrocarbon is (2016 Main)

(a) C H3 8 (b) C H4 8 (c) C H4 10 (d) C H3 6

15. The molecular formula of a commercial resin used for

exchanging ions in water softening is C H SO Na8 7 3

(molecular weight= 206) What would be the maximum

uptake of Ca2+ions by the resin when expressed in mole per

16. 3 g of activated charcoal was added to 50 mL of acetic acid

solution (0.06 N) in a flask After an hour it was filtered and

the strength of the filtrate was found to be 0.042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(2015 Main)

(a) 18 mg (b) 36 mg (c) 42 mg (d) 54 mg

17. The ratio mass of oxygen and nitrogen of a particular gaseous

mixture is 1 : 4 The ratio of number of their molecule is

(2014 Main)

(a) 1 : 4 (b) 7 : 32 (c) 1 : 8 (d) 3 : 16

18. The molarity of a solution obtained by mixing 750 mL of

0.5 M HCl with 250 mL of 2 M HCl will be (2013 Main)

(a) 0.875 M (b) 1.00 M (c) 1.75 M (d) 0.0975M

19. Dissolving 120 g of urea (mol wt 60) in 1000 g of water

gave a solution of density 1.15 g/mL The molarity of the

(a) 1.78 M (b) 2.00 M (c) 2.05 M (d) 2.22 M

20. Given that the abundances of isotopes54Fe,56Fe and 57Fe

are 5%, 90% and 5%, respectively, the atomic mass of Fe is

(2009)

(a) 55.85 (b) 55.95

(c) 55.75 (d) 56.05

21. Mixture X =0.02 mole of [Co(NH ) SO ]Br3 5 4 and 0.02 mole

of [Co(NH ) Br]SO3 5 4was prepared in 2 L solution

1 L of mixture X + excess of AgNO3solution→Y

1 L of mixture X + excess of BaCl2solution→Z

Number of moles of Y and Z are (2003, 1M)

(a) Molarity (b) Normality (c) Formality (d) Molality

26. A molal solution is one that contains one mole of solute in

(a) 1 : 1 (b) 1 : 2 (c) 2 : 1 (d) 9 : 4

30. The largest number of molecules is in (1979, 1M)

(a) 36 g of water(b) 28 g of CO(c) 46 g of ethyl alcohol(d) 54 g of nitrogen pentaoxide (N O2 5)

31. The total number of electrons in one molecule of carbon

(a) 22 (b) 44 (c) 66 (d) 88

32. A gaseous mixture contains oxygen and nitrogen in the ratio

of 1:4 by weight Therefore, the ratio of their number of

(a) 1 : 4 (b) 1 : 8 (c) 7 : 32 (d) 3 : 16

Numerical Answer Type Questions

33. A 100 mL solution was made by adding 1.43 g of

Na CO2 3⋅xH O2 The normality of the solution is 0.1 N The

value of x is ………

(The atomic mass of Na is 23 g/mol) (2020 Main, 4 Sep II)

34. Galena (an ore) is partially oxidised by passing air through it

at high temperature After some time, the passage of air isstopped, but the heating is continued in a closed furnace suchthat the content undergo self-reduction The weight (in kg) of

Pb produced per kg of O2consumed is ……… (Atomic weights in g mol−1: O =16,S=32,Pb=207)

(2018 Adv.)

35. To measure the quantity of MnCl2dissolved in an aqueoussolution, it was completely converted to KMnO4 using thereaction,

MnCl + K S O + H O2 2 2 8 2 →KMnO + H SO + HCl4 2 4(equation not balanced)

Few drops of concentrated HCl were added to this solution

and gently warmed Further, oxalic acid (225 mg) was

added in portions till the colour of the permanganate ion

disappeared The quantity of MnCl2(in mg) present in the

initial solution is ………

(Atomic weights in g mol−1: Mn 55, Cl 35.5)= =

(2018 Adv.)

36. In the following reaction sequence, the amount of D (in

gram) formed from 10 moles of acetophenone is ……

(Atomic weights in g mol−1 : H = 1, C = 12, N = 14,

O = 16, Br = 80 The yield (%) corresponding to the

product in each step is given in the parenthesis)

(2018 Adv.)

Fill in the Blanks

37. The weight of 1 10× 22 molecules of CuSO 5H O4⋅ 2 is

38. 3.0 g of a salt of molecular weight 30 is dissolved in 250 g

water The molarity of the solution is ……… (1983, 1M)

39. The total number of electrons present in 18 mL of water is

40. The modern atomic mass unit is based on the mass of

Integer Answer Type Questions

41. The mole fraction of a solute in a solution is 0.1 At 298 K,

molarity of this solution is the same as its molality Density

of this solution at 298 K is 2.0 g cm−3 The ratio of the

molecular weights of the solute and solvent, m

m

solute solvent

42. A compound H2X with molar weight of 80 g is dissolved

in a solvent having density of 0.4 g mL−1 Assuming no

change in volume upon dissolution, the molality of a 3.2

molar solution is (2014 Adv.)

43. 29.2% (w/W ) HCl stock solution has density of 1.25g mL

− 1 The molecular weight of HCl is 36.5 g mol−1 The

volume (mL) of stock solution required to prepare a 200

mL solution 0.4 M HCl is (2012)

Subjective Questions

44. 20% surface sites have adsorbed N2 On heating N2 gas

evolved from sites and were collected at 0.001 atm and 298

K in a container of volume is 2.46 cm3 Density of surface sites

is 6 023 10 × 14/cm2and surface area is 1000 cm2, find out thenumber of surface sites occupied per molecule of N2.(2005, 3M)

45. In a solution of 100 mL 0.5 M acetic acid, one gram of activecharcoal is added, which adsorbs acetic acid It is found that theconcentration of acetic acid becomes 0.49 M If surface area ofcharcoal is 3.01×102m , calculate the area occupied by single2acetic acid molecule on surface of charcoal (2003)

46. Find the molarity of water Given:ρ= 1000 kg/m3 (2003)

47. A plant virus is found to consist of uniform cylindrical particles

of 150 Å in diameter and 5000 Å long The specific volume ofthe virus is 0.75 cm3/g If the virus is considered to be a singleparticle, find its molar mass (1999, 3M)

48. 8.0575 ×10−2 kg of Glauber’s salt is dissolved in water toobtain 1 dm3of solution of density 1077.2 kg m−3 Calculatethe molality, molarity and mole fraction of Na SO2 4in solution

(1994, 3M)

49 A is a binary compound of a univalent metal 1.422 g of A reacts

completely with 0.321 g of sulphur in an evacuated and sealed

tube to give 1.743 g of a white crystalline solid B, that forms a hydrated double salt, C with Al (SO )2 4 3 Identify A, B and C.

(1994, 2M)

50. Upon mixing 45.0 mL 0.25 M lead nitrate solution with25.0 mL of a 0.10 M chromic sulphate solution, precipitation oflead sulphate takes place How many moles of lead sulphate areformed? Also calculate the molar concentrations of species leftbehind in the final solution Assume that lead sulphate iscompletely insoluble (1993, 3M)

51. Calculate the molality of 1.0 L solution of 93% H SO2 4,(weight/volume) The density of the solution is 1.84 g/mL

(1990, 1M)

52. A solid mixture (5.0 g) consisting of lead nitrate and sodiumnitrate was heated below 600°C until the weight of the residuewas constant If the loss in weight is 28.0 per cent, find theamount of lead nitrate and sodium nitrate in the mixture

(1990, 4M)

53 n-butane is produced by monobromination of ethane followed

by Wurtz’s reaction.Calculate volume of ethane at NTP

required to produce 55 g n-butane, if the bromination takes

place with 90% yield and the Wurtz’s reaction with 85% yield

(1989, 3M)

54. A sugar syrup of weight 214.2 g contains 34.2 g of sugar(C H O12 22 11) Calculate (i) molal concentration and (ii) molefraction of sugar in syrup (1988, 2M)

55. An unknown compound of carbon, hydrogen and oxygencontains 69.77% C and 11.63% H and has a molecular weight

of 86 It does not reduces Fehling’s solution but forms abisulphate addition compound and gives a positive iodoformtest What is the possible structure(s) of unknown compound?

(1987, 3M)

56. The density of a 3 M sodium thiosulphate solution (Na S O2 2 3)

is 1.25 g per mL Calculate (i) the percentage by weight of

(100%)

(50%) (50%)

(60%)

sodium thiosulphate (ii) the mole fraction of sodium

thiosulphate and (iii) the molalities of Na+ and S O2 32− ions

(1983, 5M)

57. (a) 1.0 L of a mixture of CO and CO2is taken This mixture

is passed through a tube containing red hot charcoal The

volume now becomes 1.6 L The volumes are measured

under the same conditions Find the composition of

mixture by volume

(b) A compound contains 28 per cent of nitrogen and

72 per cent of a metal by weight 3 atoms of metal

combine with 2 atoms of nitrogen Find the atomic

58. 5.00 mL of a gas containing only carbon and hydrogen were

mixed with an excess of oxygen (30 mL) and the mixture

exploded by means of electric spark After explosion, the

volume of the mixed gases remaining was 25 mL

On adding a concentrated solution of KOH, the volume

further diminished to 15 mL, the residual gas being pure

oxygen All volumes have been reduced to NTP Calculatethe molecular formula of the hydrocarbon gas (1979, 3M)

59. In the analysis of 0.5 g sample of feldspar, a mixture ofchlorides of sodium and potassium is obtained, which weighs0.1180 g Subsequent treatment of the mixed chlorides withsilver nitrate gives 0.2451 g of silver chloride What is thepercentage of sodium oxide and potassium oxide in the

60. The vapour density (hydrogen = 1) of a mixture consisting of

NO2 and N O2 4is 38.3 at 26.7°C Calculate the number ofmoles of NO2in 100 g of the mixture (1979, 5M)

61. Accounts for the following Limit your answer to twosentences, “Atomic weights of most of the elements are

62. Naturally occurring boron consists of two isotopes whoseatomic weights are 10.01 and 11.01 The atomic weight ofnatural boron is 10.81 Calculate the percentage of eachisotope in natural boron (1978, 2M)

Objective Questions I (Only one correct option)

1. An example of a disproportionation reaction is

2. In an acid-base titration, 0.1 M HCl solution was added to

the NaOH solution of unknown strength Which of the

following correctly shows the change of pH of the titration

mixture in this experiment? (2019 Main, 9 April II)

(2019 Main, 9 April II)

3. 0.27 g of a long chain fatty acid was dissolved in 100 cm3of

hexane 10 mL of this solution was added dropwise to the

surface of water in a round watch glass Hexane evaporatesand a monolayer is formed The distance from edge to centre

of the watch glass is 10 cm What is the height of themonolayer? [Density of fatty acid=0 9 g cm−3;π =3]

(2019 Main, 8 April II)

(a) 10−6m (b) 10−4m(c) 10−8m (d) 10−2m

4. In order to oxidise a mixture of one mole of each of FeC O2 4,

Fe (C O )2 2 4 3, FeSO4 and Fe (SO )2 4 3 in acidic medium, thenumber of moles of KMnO4required is (2019 Main, 8 April I)

5. 100 mL of a water sample contains 0.81 g of calciumbicarbonate and 0.73 g of magnesium bicarbonate Thehardness of this water sample expressed in terms ofequivalents of CaCO3is (molar mass of calcium bicarbonate

is 162 g mol−1and magnesium bicarbonate is 146 g mol−1)

(2019 Main, 8 April I)

(a) 5,000 ppm (b) 1,000 ppm(c) 100 ppm (d) 10,000 ppm

6. 50 mL of 0.5 M oxalic acid is needed to neutralise 25 mL ofsodium hydroxide solution The amount of NaOH in 50 mL

of the given sodium hydroxide solution is

(2019 Main, 12 Jan I)

(a) 40 g (b) 80 g (c) 20 g (d) 10 g

7. 25 mL of the given HCl solution requires 30 mL of 0.1 Msodium carbonate solution What is the volume of this HClsolution required to titrate 30 mL of 0.2 M aqueous NaOHsolution? (2019 Main, 11 Jan II)

8. In the reaction of oxalate with permanganate in acidic

medium, the number of electrons involved in producing one

molecule of CO2is (2019 Main, 10 Jan II)

9. The ratio of mass per cent of C and H of an organic

compound (C H O )x y z is 6 : 1 If one molecule of the above

compound (C H O )x y z contains half as much oxygen as

required to burn one molecule of compound C Hx y

completely to CO2 and H O2 The empirical formula of

compound C H Ox y zis (2018 Main)

(a) C H O3 6 3 (b) C H O2 4 (c) C H O3 4 2 (d) C H O2 4 3

10. An alkali is titrated against an acid with methyl orange as

indicator, which of the following is a correct combination?

(2018 Main)

11. From the following statements regarding H O2 2 choose the

incorrect statement (2015 Main)

(a) It can act only as an oxidising agent

(b) It decomposed on exposure to light

(c) It has to be stored in plastic or wax lined glass bottles in

dark

(d) It has to be kept away from dust

12. Consider a titration of potassium dichromate solution with

acidified Mohr’s salt solution using diphenylamine as

indicator The number of moles of Mohr's salt required per

mole of dichromate is (2007, 3M)

13. In the standardisation of Na S O2 2 3 using K Cr O2 2 7 by

iodometry, the equivalent weight of K Cr O2 2 7is (2001, 1M)

(a) (molecular weight)/2 (b) (molecular weight)/6

(c) (molecular weight)/3 (d) same as molecular weight

14. The reaction, 3ClO−(aq)→ClO3– (aq) + 2Cl−(aq) is an

15. An aqueous solution of 6.3 g oxalic acid dihydrate is made up

to 250 mL The volume of 0.1 N NaOH required to

completely neutralise 10 mL of this solution is (2001, 1M)

Objective Question II (More than one correct option)

24. For the reaction, I−+ClO−3+H SO2 4→ Cl + HSO− −4+I2

the correct statement(s) in the balanced equation is/are(a) stoichiometric coefficient of HSO4− is 6 (2014 Adv.)

(b) iodide is oxidised(c) sulphur is reduced(d) H O2 is one of the products

Numerical Answer Type Questions

25. 5.00 mL of 0.10 M oxalic acid solution taken in a conicalflask is titrated against NaOH from a burette usingphenolphthalein indicator The volume of NaOH required forthe appearance of permanent faint pink color is tabulatedbelow for five experiments What is the concentration, inmolarity, of the NaOH solution?

Exp No Vol of NaOH (mL)

26. Aluminium reacts with sulphuric acid to form aluminium

sulphate and hydrogen What is the volume of hydrogen gas

in litre (L) produced at 300 K and 1.0 atm pressure, when

5.4 g of aluminium and 50.0 mL of 5.0 M sulphuric acid are

combined for the reaction?

(Use molar mass of aluminium as 27.0 g mol−1, R=0 082

atm L mol−1K−1) (2020 Adv.)

27. A 20.0 mL solution containing 0.2 g impure H O2 2 reacts

completely with 0.316 g of KMnO4 in acid solution The

purity of H O2 2 (in%) is (molecular weight of

H O2 2=34; molecular weight of KMnO4 =158)

(2020 Main, 4 Sep I)

28. The ammonia prepared by treating ammonium sulphate with

calcium hydroxide is completely used by NiCl 6H O2⋅ 2 to

form a stable coordination compound Assume that both the

reactions are 100% complete If 1584 g of ammonium

sulphate and 952 g of NiCl 6H O2⋅ 2 are used in the

preparation, the combined weight (in grams) of gypsum

and the nickel-ammonia coordination compound thus

produced is

(Atomic weights in g mol− 1: H = 1, N = 14, O = 16, S = 32,

Cl = 35.5, Ca = 40, Ni = 59) (2018 Adv.)

Assertion and Reason

Read the following questions and answer as per the direction

given below :

(a) Statement I is true; Statement II is true; Statement II is the

correct explanation of Statement I

(b) Statement I is true; Statement II is true; Statement II is not

the correct explanation of Statement I

(c) Statement I is true; Statement II is false

(d) Statement I is false; Statement II is true

29 Statement I In the titration of Na CO2 3 with HCl using

methyl orange indicator, the volume required at the

equivalence point is twice that of the acid required using

phenolphthalein indicator

Statement II Two moles of HCl are required for the

complete neutralisation of one mole of Na CO2 3 (1991, 2M)

Fill in the Blanks

30. The compound YBa Cu O2 3 7, which shows super

conductivity, has copper in oxidation state ……… Assume

that the rare earth element yttrium is in its usual + 3 oxidation

Integer Answer Type Questions

31. The difference in the oxidation numbers of the two types ofsulphur atoms in Na S O2 4 6is (2011)

32. Among the following, the number of elements showing onlyone non-zero oxidation state is O, Cl, F, N, P, Sn, Tl, Na, Ti

(2010)

33. A student performs a titration with different burettes andfinds titrate values of 25.2 mL, 25.25 mL, and 25.0 mL Thenumber of significant figures in the average titrate value is

(2010)Subjective Questions

34. Calculate the amount of calcium oxide required when itreacts with 852 g of P O4 10 (2005, 2M)

35. Hydrogen peroxide solution (20 mL) reacts quantitativelywith a solution of KMnO4 (20 mL) acidified with dilute

H SO2 4 The same volume of the KMnO4 solution is justdecolourised by 10 mL of MnSO4 in neutral mediumsimultaneously forming a dark brown precipitate of hydratedMnO2 The brown precipitate is dissolved in 10 mL of 0.2 Msodium oxalate under boiling condition in the presence ofdilute H SO2 4 Write the balanced equations involved in thereactions and calculate the molarity of H O2 2 (2001)

36. How many millilitres of 0.5 M H SO2 4are needed to dissolve0.5 g of copper (II) carbonate? (1999, 3M)

37. An aqueous solution containing 0.10 g KIO3(formula weight = 214.0) was treated with an excess of KIsolution The solution was acidified with HCl The liberated

I2 consumed 45.0 mL of thiosulphate solution decolourisethe blue starch-iodine complex Calculate the molarity of thesodium thiosulphate solution (1998, 5M)

38. To a 25 mL H O2 2solution, excess of acidified solution ofpotassium iodide was added The iodine liberated required

20 mL of 0.3 N sodium thiosulphate solution Calculate thevolume strength of H O2 2solution (1997, 5M)

39. A 3.00 g sample containing Fe O Fe O3 4, 2 3 and an inertimpure substance, is treated with excess of KI solution inpresence of dilute H SO2 4 The entire iron is converted into

Fe2+ along with the liberation of iodine The resultingsolution is diluted to 100 mL A 20 mL of the dilutedsolution requires 11.0 mL of 0.5 M Na S O2 2 3 solution toreduce the iodine present A 50 mL of the dilute solution,after complete extraction of the iodine required 12.80 mL of0.25 M KMnO4 solution in dilute H SO2 4 medium for theoxidation of Fe2+ Calculate the percentage of Fe O2 3 and

Fe O3 4in the original sample (1996, 5M)

40. A 20.0 cm3mixture of CO, CH4and He gases is exploded by

an electric discharge at room temperature with excess ofoxygen The volume contraction is found to be 13.0 cm3

A further contraction of 14.0 cm3occurs when the residualgas is treated with KOH solution Find out the composition

of the gaseous mixture in terms of volume percentage

41. A 5.0 cm3solution of H O2 2liberates 0.508 g of iodine from

an acidified KI solution Calculate the strength of H O2 2

solution in terms of volume strength at STP (1995, 3M)

42. One gram of commercial AgNO3 is dissolved in 50 mL of

water It is treated with 50 mL of a KI solution The silver

iodide thus precipitated is filtered off Excess of KI in the

filtrate is titrated with (M/10) KIO3solution in presence of

6 M HCl till all I− ions are converted into ICl It requires

50 mL of (M/10) KIO3 solution, 20 mL of the same stock

solution of KI requires 30 mL of (M/10) KIO3under similar

conditions Calculate the percentage of AgNO3 in the

sample

Reaction KIO + 2KI + 6HCl3 → 3ICl + 3KCl + 3H O2

(1992, 4M)

43. A 2.0 g sample of a mixture containing sodium carbonate,

sodium bicarbonate and sodium sulphate is gently heated till

the evolution of CO2ceases The volume of CO2at 750 mm

Hg pressure and at 298 K is measured to be 123.9 mL A 1.5 g

of the same sample requires 150 mL of (M/10) HCl for

complete neutralisation Calculate the percentage

composition of the components of the mixture (1992, 5M)

44. A 1.0 g sample of Fe O2 3solid of 55.2% purity is dissolved in

acid and reduced by heating the solution with zinc dust The

resultant solution is cooled and made up to 100.0 mL An

aliquot of 25.0 mL of this solution requires for titration

Calculate the number of electrons taken up by the oxidant in

the reaction of the above titration (1991, 4M)

45. A solution of 0.2 g of a compound containing Cu2+ and

C O2 42− ions on titration with 0.02 M KMnO4in presence of

H SO2 4 consumes 22.6 mL of the oxidant The resultant

solution is neutralised with Na CO2 3, acidified with dilute

acetic acid and treated with excess KI The liberated iodine

requires 11.3 mL of 0.05 M Na S O2 2 3solution for complete

reduction Find out the mole ratio of Cu2+ to C O2 42− in the

compound Write down the balanced redox reactions

involved in the above titrations (1991, 5M)

46. A mixture of H C O2 2 4(oxalic acid) and NaHC O2 4weighing

2.02 g was dissolved in water and the solution made up to one

litre Ten millilitres of the solution required 3.0 mL of 0.1 N

sodium hydroxide solution for complete neutralisation In

another experiment, 10.0 mL of the same solution, in hot

dilute sulphuric acid medium, required 4.0 mL of 0.1 N

potassium permanganate solution for complete reaction

Calculate the amount of H C O2 2 4 and NaHC O2 4 in the

47. An organic compound X on analysis gives 24.24 per cent

carbon and 4.04 per cent hydrogen Further, sodium extract

of 1.0 g of X gives 2.90 g of silver chloride with acidified silver nitrate solution The compound X may be represented

by two isomeric structures Y and Z Y on treatment with

aqueous potassium hydroxide solution gives a dihydroxy

compound while Z on similar treatment gives ethanal Find out the molecular formula of X and gives the structure

48. An equal volume of a reducing agent is titrated separatelywith 1 M KMnO4in acid, neutral and alkaline medium Thevolumes of KMnO4required are 20 mL in acid, 33.3 mL inneutral and 100 mL in alkaline media Find out the oxidationstate of manganese in each reduction product Give thebalanced equations for all the three half reaction Find out thevolume of 1M K Cr O2 2 7consumed, if the same volume of thereducing agent is titrated in acid medium (1989, 5M)

49. A sample of hydrazine sulphate (N H SO2 6 4)was dissolved in

100 mL of water, 10 mL of this solution was reacted withexcess of ferric chloride solution and warmed to completethe reaction Ferrous ion formed was estimated and it,required 20 mL of M/50 potassium permanganate solution.Estimate the amount of hydrazine sulphate in one litre of thesolution

Reaction 4Fe3+ + N H2 4 → N + 4Fe2 2++ 4H+MnO + 5Fe4− 2+ + 8H+ → Mn2+ + 5Fe3++ 4H O2

(1988, 3M)

50. 5 mL of 8 N nitric acid, 4.8 mL of 5 N hydrochloric acid and

a certain volume of 17 M sulphuric acid are mixed togetherand made up to 2 L 30 mL of this acid mixture exactlyneutralise 42.9 mL of sodium carbonate solution containingone gram of Na CO 10H O2 3⋅ 2 in 100 mL of water Calculatethe amount in gram of the sulphate ions in solution

(1985, 4M)

51. 2.68×10−3moles of a solution containing an ion A n+require1.61×10− 3moles of MnO4−for the oxidation of A n+ to A O3−

in acidic medium What is the value of n ? (1984, 2M)

52. 4.08 g of a mixture of BaO and unknown carbonate MCO3

was heated strongly The residue weighed 3.64 g This wasdissolved in 100 mL of 1 N HCl The excess acid required

16 mL of 2.5 N NaOH solution for complete neutralisation

Identify the metal M. (1983, 4M)

Topic 1 Mole Concept

1 Key Idea To find the mass ofA and B in the given question,

mole concept is used

M

= given mass ( )molecular mass ( )

On solving the equation, we obtain

M A= ×5 10−3

So, the molar mass of A M( A) is

5 10× −3kg mol−1and B M( B) is 10 10× −3kg mol−1

124g

2 160g

∴Number of millimoles=10 mL×10−3M=10−2Number of moles=10−5

Now, number of molecules

=Number of moles×Avogadro’s number

=10− 5× ×6 1023= ×6 1018

Hints & Solutions

Surface area occupied by 6 10× 18molecules=0.24 cm2

∴Surface area occupied by 1 molecule

=

×

0.24

6 1018=0 04 ×10− 18 cm2

As it is given that polar head is approximated as cube Thus,

surface area of cube=a2, where

a=edge length

∴ a2= ×4 10−20cm2

a= ×2 10− 10cm=2pm

5 Key Idea The reactant which is present in the lesser amount,

i.e which limits the amount of product formed is called

limiting reagent

When 56 g of N2+10gof H2is taken as a combination then

dihydrogen (H2) act as a limiting reagent in the reaction

only 10 g of H2gas is present in option (a)

Hence, H2gas is the limiting reagent

gas is required for 35g of N2 and 8g of H2is present in reaction

mixture Mass of H2left unreacted= −8 7 5 g of H2

=0 5 gof H2.Similarly, in option (c) and (d), H2does not act as limiting

(by equation I)

6 Key Idea The percentage composition of a compound is given

by the formula

% composition=[Composition of a substance in a compound /

Total composition total of compound]×100

7. Mole fraction of solute

=number of moles of solute + number of moles solvent

number of moles of solute

χSolute Solute

Solute Solvent

=+

n

+

w Mw w Mw

w Mw

Solute Solute Solute Solute

Solvent Solvent

Given, wSolute=wNaOH =8 g

MwSolute=MwNaOH=40g mol− 1

1818

.

. =0167

Now, molality ( )m = Moles of solute

Mass of solvent (in kg)

w Mw w

Solute Solute Solvent(in g) 1000

840

Molarity (M)= x

112 or x=Molarity×11.2

where, x=volume strength

So, for 1 M H O2 2, x= ×1 112 112 = Among the given options, 11.35 is nearest to 11.2

9. Molarity=Number of moles of solute ( )

g)gmol1

∴ Molarity=w M

V

B/ B Given, w B=mass of solute ( )B in g

M B=Gram molar mass of B (C H O )12 22 11 =342 g mol−1Molarity=01 M

Volume ( )V =2 L

2 =w B/ ⇒ w B=01 342 × ×2g=68 4 g

10. 2 C H57 110 6O ( ) + 163 O ( )s 2g →110H O( ) + 114 CO ( )2 l 2g

Molecular mass of C H57 110O6

= ×2 (12×57+ ×1 110+16×6) g=1780 gMolecular mass of 110 H O2 =110 2 16( + ) = 1980 g

1780 g of C H57 110O6produced=1980 g of H O.2445g of C H57 110O6produced=1980×

1780 445 g of H O2

=495 of H O2

11. Molality ( )m = Number of moles of solute

Mass of solvent (in g) ×1000

= Mass of solute (in g) 1000×Molecular weight of solutemass of solvent (in g)

w

Na Na

10002

12. Given, abundance of elements by mass

oxygen=614 %, carbon=22 9 %, hydrogen=10% and

Inert part of air=80% of 375=300 mL

Total volume of gases =CO2+ Inert part of air

we have to find, mole per gram of resin

∴ 1g of C H SO8 7 3Na has number of mole

= weight of given resin

Molecular, weight of resin= 1

206molNow, reaction looks like

2C H SO Na8 7 3 +Ca2+ →(C H SO8 7 3)2Ca+ 2Na

Q 2 moles of C H SO8 7 3Na combines with 1 mol Ca2 +

∴1 mole of C H SO Na8 7 3 will combine with1

1206

m M m M

O N

O O N N

O N

2

2

28

32= ×1 =4

2832

732

18. From the formula,M M V M V

60 2Weight of solution=Weight of solvent+Weight of solute

21. 1.0 L of mixture X contain 0.01 mole of each [Co(NH ) SO ]Br3 5 4

and [Co(NH ) Br]SO3 5 4 Also, with AgNO3, only[Co(NH ) SO ]Br3 5 4 reacts to give AgBr precipitate as[Co(NH ) SO ]Br3 5 4 AgNO [Co(NH ) SO ]NO

1.0 mol

3 Excess

3 5 2

1 mol

Hence, moles of Y and Z are 0.01 each.

22. Number of atoms=Number of moles

× Avogadro’s number (N A)Number of atoms in 24 g C =24

12 ×N A = 2N A

only two replaceable hydrogens

Therefore, normality=molarity×basicity=0.3×2=0.60

25. Molality is defined in terms of weight, hence independent of

temperature Remaining three concentration units are defined in

terms of volume of solution, they depends on temperature

26. Molality of a solution is defined as number of moles of solute

present in 1.0 kg (1000 g) of solvent

27. The balanced chemical reaction is

3BaCl + 2Na PO2 3 4 → Ba (PO ) + 6NaCl3 4 2

In this reaction, 3 moles of BaCl2combines with 2 moles of

Na PO3 4 Hence, 0.5 mole of of BaCl2require

2

3×0.5 0.33 mole of Na PO= 3 4

Since, available Na PO3 4(0.2 mole) is less than required mole

(0.33), it is the limiting reactant and would determine the

amount of product Ba (PO )3 4 2

Q 2 moles of Na PO3 4gives 1 mole Ba (PO )3 4 2

∴0.2 mole of Na PO3 4would give1

2×0.2 = 0.1 mole Ba (PO )3 4 2

28. Unlike other metal carbonates that usually decomposes into

metal oxides liberating carbon dioxide, silver carbonate on

heating decomposes into elemental silver liberating mixture of

carbon dioxide and oxygen gas as :

Since, one mole of H2( )g is produced per mole of zinc with both

sulphuric acid and NaOH respectively, hydrogen gas is

produced in the molar ratio of 1:1 in the above reactions

30. Number of molecules present in 36 g of water

=36×

18 N A=2N A

Number of molecules present in 28 g of CO=28× =

28 N A N ANumber of molecules present in 46 g of C H OH2 5 =46×

46 N A = N A

Number of molecules present in 54 g of N O2 5= 54 ×

108 N A = 0.5 N A Here, N A is Avogadro’s number Hence, 36 g of water contain

the largest (2N A) number of molecules

31. In a neutral atom, atomic number represents the number ofprotons inside the nucleus and equal number of electrons around

it Therefore, the number of total electrons in molecule of CO2

= electrons present in one carbon atom

+ 2×electrons present in one oxygen atom

w w

( )( )

( )( )

ON

ON

2 2

2 2

32= ×1 =4

2832

732

33. Molar mass of Na CO2 3⋅xH O2 (Atomic mass of Na = 23, C=12,O=16)

=23× +2 12 48+ +18x

=46+12+48+18x

= (106+18x)Equivalent weight of Na CO2 3⋅xH O2

=Molar mass= = +factor

53 901

So, 10 143

53 9

2

− =+

In the presence of oxygen

2PbS + 3O2→2PbO + 2SO2 …(i)

By self reduction

2PbO + PbS→3Pb + SO2Thus 3 moles of O2produces 3 moles of Pb

K SO2 4+2MnSO4+8H O +10CO2 2Given, mass of oxalic acid added = 225mg

So, millimoles of oxalic acid added =225

90 =2 5.Now from equation 2

Millimoles of KMnO4used to react with oxalic acid=1 and

Millimoles of MnCl2required initially=1

∴Mass of MnCl2required initially = 1×(55 71+ ) = 126mg

Alternative Method

m moles of MnCl2=m moles of KMnO4=x (let)

and Meqof KMnO4=Meqof oxalic acid

Q 6.023×1023molecules of CuSO 5H O4⋅ 2 weigh 249.5 g

∴1022molecules will weigh 249.5

6 mol ( )A

Benzamide (50%) i.e.,

3 mol ( )B

Br /KOH2

NH2

Aniline (50%) i.e., 1.5 mol ( )C

Br (3-eqiv.) AcOH

2

NH2

2,4,6-tribromo aniline (100%) i.e., 1.5 mol ( )D

BrBr

Br

NH2BrBr

Br

NH2BrBr

Br

NH2BrBr

Br

38. Molarity=Number of moles of solute

Volume of solution in litre

= Weight of solute×

Molar mass

1000Volume in mL

= 3 ×

30

1000

250 =0.4 M

39. Considering density of water to be 1.0 g/mL, 18 mL of water is

18 g (1.0 mol) of water and it contain Avogadro number of

molecules Also one molecule of water contain

2×(one from each H-atom) + 8×(from oxygen atom)

= 10 electrons

⇒1.0 mole of H O2 contain=10×6.023×1023

=6.023×1024electrons

40. Carbon-12 isotope According to modern atomic mass unit, one

atomic mass unit (amu) is defined as one-twelfth of mass of an

atom of C-12 isotope, i.e

= ; Moles of solvent, n w

m

2 22

m w

Molarity= Solute (moles)

m

1 2

9(solute)(solvent)=

molarity into molality.

43. Mass of HCl in 1.0 mL stock solution

= 6.023×1014×1000 = 6.023×1017Surface sites used in adsorption = 20

×

2 6.023 106.023 10

⇒ millimol of CH COOH3 adsorbed = 50 – 49 = 1

⇒ number of molecules of CH COOH3 adsorbed

2 20

17 3

3 1 = 1.1773×10− 16g

⇒ Molar mass of virus

=Mass of one virus×Avogadro’s number

=1.1773×10− 16×6.023×1023g

= 70.91×106g

48. Molar mass of Glauber’s salt (Na SO 10H O)2 4⋅ 2

=23× +2 32+64+10 18× =322g

Mole of Na SOMole of Na SO Mole of water

2 4

2 4

=+

0.250.25 1041.718

= 4.3 10× − 3

49. Compound B forms hydrated crystals with Al SO2( 4) 3Also, B is

formed with univalent metal on heating with sulphur Hence,

compound B must has the molecular formula M2SO and4

compound A must be an oxide of M which reacts with sulphur to

give metal sulphate as

A+S→ M2SO4

B

Q 0.321 g sulphur gives 1.743 g of M2SO4

∴ 32.1 g S (one mole) will give 174.3 g M2SO4

Therefore, molar mass of M2SO = 174.3 g4

⇒ 174.3 = 2×Atomic weight of M+ 32.1 + 64

⇒ Atomic weight of M = 39, metal is potassium (K)

K SO2 4on treatment with aqueous Al SO2( 4)3gives potash-alum

K SO + Al (SO ) + 24H O2 4 2 4 3 2

B

→K SO Al (SO ) 24H O2 4 2 4 3⋅ 2

C

If the metal oxide A has molecular formula MO , two moles of itx

combine with one mole of sulphur to give one mole of metal

sulphate as

2KOx+ →S K SO2 4

⇒ x=2, i.e A is KO2

50. The reaction involved is

3Pb(NO ) + Cr (SO )3 2 2 4 3→3PbSO ( )4s ↓+ 2Cr(NO )3 3

millimol of Pb(NO )3 2taken=45×0.25 = 11.25

millimol of Cr (SO )2 4 3taken = 2.5

Here, chromic sulphate is the limiting reagent, it will determine

the amount of product

Q 1 mole Cr (SO )2 4 3produces 3 moles PbSO4

∴ 2.5 millimol Cr (SO )2 4 3will produce 7.5 millimol PbSO4

Hence, mole of PbSO4precipitate formed =7.5 10× −3

Also, millimol of Pb(NO )3 2remaining unreacted

= 2×millimol of Cr (SO )2 4 3reacted

⇒ Weight of residue left=5 – 1.4=3.6 g

Now, let the original mixture contain x g of Pb(NO )3 2

Q 330 g Pb(NO )3 2gives 222 g PbO

∴ x gPb(NO )3 2will give222

330

x

g PbOSimilarly, 85 g NaNO3gives 69 g NaNO2

⇒ (5 – x) g NaNO3will give69 5

+ ( − )= 3.6 g

Solving for x gives, x=3.3 g Pb(NO )3 2 ⇒ NaNO3=1.7 g

53. Reactions involved are

C H + Br2 6 2 →C H Br + HBr2 52C H Br + 2Na2 5 →C H + 2NaBr4 10Actual yield of C H4 10=55 gwhich is 85% of theoretical yield

⇒ Theoretical yield of C H4 10= 55 100×

85 = 64.70 gAlso, 2 moles (218 g) C H Br2 5 gives 58 g of butane

⇒ 64.70 g of butane would be obtained from

2

58×64 70 =2 23 moles C H Br2 5Also yield of bromination reaction is only 90%, in order to have2.23 moles of C H Br2 5 , theoretically

Weight of Solvent (g) 1000

= 0.1× =

180 1000 0.55(ii) Mole fraction of sugar= Mole of sugar

Mole of sugar + Mole of water

=+0.10.1 10=9.9 10× −3

55. From the given elemental composition, empirical formula can

Since, empirical formula weight and molecular weight both are

(86), empirical formula is the molecular formula also

Also, the compound does not reduce Fehling’s solution,

therefore it is not an aldehyde, but it forms bisulphite, it must be

O

— CH

3 3

3 3-methyl -2-butanone

56. (a) Let us consider 1.0 L solution for all the calculation

(i) Weight of 1 L solution=1250 g

Weight of Na S O2 2 3= ×3 158=474 g

⇒Weight percentage of Na S O2 2 3= 474 × =

1250 100 37.92(ii) Weight of H O2 in 1 L solution=1250 474− =776 g

Mole fraction of Na S O2 2 3=

3

3 77618

If the 1.0 L original mixture contain x litre of CO2, after

passing from tube containing red-hot charcoal, the new

volumes would be :

2x (volume of CO obtained from CO2)+1

– x(original CO)=1+x=1.6 (given)

Hence, original 1.0 L mixture has 0.4 L CO and 0.6 L of CO2,

i.e 40% CO and 60% CO2by volume

(b) According to the given information, molecular formula of

the compound is M3N Also, 1.0 mole of compound has 28 g2

of nitrogen If X is the molar mass of compound, then :

moles of (NaCl + KCl) = moles of AgCl(one mole of either NaCl or KCl gives one mole of AgCl)

Now, let the chloride mixture contain x g NaCl.

58.5

0.11874.5

0.2451143.5

Solving for x gives x=0.0338 g (mass of NaCl)

⇒ Mass of KCl=0.118 – 0.0338 = 0.0842 gAlso, moles of Na O =1

x mole ofNO2

⇒ 46x+92 1( −x)=76.6 ⇒ x=0.3348Also, in 100 g mixture, number of moles= 100

76.6

⇒Moles of NO2in mixture= 100 × =

76.6 0.3348 0.437

61. Most of the elements found in nature exist as a mixture of

isotopes whose atomic weights are different The atomic weight

of an element is the average of atomic weights of all its naturally

occurring isotopes

62. Average atomic weight

=ΣPercentage of an isotope×Atomic weight

Topic 2 Equivalent Concept, Neutralisation

and Redox Titration

1. In disproportionation reactions, same element undergoes

oxidation as well as reduction

e.g

Here, CuBr get oxidised to CuBr2and also it get reduced to Cu

Other given reactions and their types are given below

In the given reaction, MnO4− get oxidised to Mn2+and I−get

reduced to I2 It is an example of redox reaction The reaction

takes place in acidic medium

2KMnO4 →K MnO2 4+MnO2+O2

The given reaction is an example of decomposition reaction

Here, one compound split into two or more simpler compounds,

atleast one of which must be in elemental form

2NaBr+Cl2→2NaCl+Br2

The given reaction is an example of displacement reaction In

this reaction, an atom (or ion) replaces the ion (or atom) of

another element from a compound

2. The graph that shows the correct change of pH of the titration

mixture in the experiment is

In this case, both the titrants are completely ionised

HCl + NaOH-N a Cl + H O+ 2

−

As H⊕ is added to a basic solution, [OHÈ]

decreases and [H+]increases Therefore, pH goes on decreasing As the equivalence

point is reached,[OH ]È

is rapidly reduced After this point [OHÈ

]decreases rapidly and pH of the solution remains fairly constant

Thus, there is an inflexion point at the equivalence point

The difference in the volume of NaOH solution between the endpoint and the equivalence point is not significant for most of thecommonly used indicators as there is a large change in the pHvalue around the equivalence point Most of them change theircolour across this pH change

3. 100 mL (cm3) of hexane contains 0.27 g of fatty acid

In 10 mL solution, mass of the fatty acid,

m=0 27× =

100 10 0 027

Density of fatty acid, d=0 9 g cm−3

∴Volume of the fatty acid over the watch glass,

Let, height of the cylindrical monolayer=h cm

Q Volume of the cylinder=Volume of fatty acid

cmcm

100 10

6=10 000, ppm

+2 0 Reduction

6. The reaction takes place as follows,

Now, molarity= Number of moles

Volume of solution (in L)

= Weight / molecular massVolume of solution (in L)2

40

100050

=wNaOH ×

wNaOH=2 40 50

Thus, (*) none option is correct

7. The reaction of HCl with Na CO2 3is as follows:

× ×MHCl = ×0.1 2×

=30× =25

6

25MThe reaction of HCl with NaOH is as follows:

⇒5C O2 42–ions transfer 10e−to produce 10 molecules of CO2

So, number of electrons involved in producing 10 molecules of

CO2is 10 Thus, number of electrons involved in producing 1

molecule of CO2is 1

9. We can calculate the simplest whole number ratio of C and H

from the data given, as

Element Relative

mass

Molar mass

Relative mole

Simplest whole number ratio

12=0 5 0 5

0 5 1

=

0 5 =2Alternatively this ratio can also be calculated directly in the

to burn one molecule of compoundC Hx ycompletely to CO2and

H O2 We can calculate number of oxygen atoms from this asconsider the equation

C H Ox y z=C H O1 2 1.5

So, empirical formula will be [C H O ] 2 C H O1 2 1.5 × = 2 4 3

10. Methyl orange show Pinkish colour towards more acidicmedium and yellow orange colour towards basic or less acidicmedia Its working pH range is

Weak base have the pH range greater than 7 When methylorange is added to this weak base solution it shows yelloworange colour

Now when this solution is titrated againststrong acid the pH move towards more acidicrange and reaches to end point near 3.9 whereyellow orange colour of methyl orangechanges to Pinkish red resulting to similarchange in colour of solution as well

11. H O2 2acts as an oxidising as well as reducing agent, becauseoxidation number of oxygen in H O2 2is−1 So, it can be oxidised

to oxidation state 0 or reduced to oxidation state –2

H O2 2decomposes on exposure to light So, it has to be stored inplastic or wax lined glass bottles in dark for the prevention ofexposure It also has to be kept away from dust

Q 1 mole of dichromate = 6 equivalent of dichromate

∴6 equivalent of Mohr’s salt would be required

Since, n-factor of Mohr’s salt is 1, 6 equivalent of it would also

be equal to 6 moles

Hence, 1 mole of dichromate will oxidise 6 moles of Mohr’s salt

3.9 –4.5

Yellow orange Pinkish

13. The following reaction occur between S O2 32−and Cr O2 72−:

26H + 3S O+ 2 32−+ 4Cr O2 72− →6SO42− +8Cr3++ 13H O2

Change in oxidation number of Cr O2 7−per formula unit is 6 (it is

always fixed for Cr O2 7−)

Hence, equivalent weight of K Cr O2 2 7=Molecular weight

6

14. It is an example of disproportionation reaction because the same

species (ClO−)is being oxidised to ClO3−as well as reduced to Cl−

15. Oxalic acid dihydrate H C O 2H O2 2 4⋅ 2 : mw = 126

It is a dibasic acid, hence equivalent weight = 63

In NiF62−, Ni is in + 4 oxidation state

In CrO Cl2 2, oxidation state of Cr is +6

17. In S8, oxidation number of S is 0, elemental state

In S F2 2, F is in – 1 oxidation state, hence S is in + 1 oxidation

state

In H S,2 H is in +1 oxidation state, hence S is in – 2 oxidation

state

18. The balanced redox reaction is :

3MnO4−+5FeC O2 4+24H+ →3Mn2+ + 5Fe3+

+ 10CO2+12H O2

Q 5 moles of FeC O2 4require 3 moles of KMnO4

∴1 mole of FeC O2 4will require3

5mole of KMnO4.

19. The balanced chemical reaction is :

2MnO + 5SO4− 3− + 6H+ → 2Mn2+ +5SO4−+ 3H O2

Q 5 moles SO32−reacts with 2 moles of KMnO4

∴ 1 mole of SO3−will react with2

21. Volume strength of H O2 2=Normality×5.6=1.5×5.6=8.4 V

22. In Ba(H PO )2 2 2, oxidation number of Ba is +2 Therefore,

H PO : 22 2− ×(+1) + x+ 2× − = −( 2) 1

23. Equivalent weight in redox system is defined as :

E n-

= Molar massfactor

Here n-factor is the net change in oxidation number per formula unit of oxidising or reducing agent In the present case, n-factor

is 2 because equivalent weight is half of molecular weight Also,

MnSO4 → MnO2 2 (+ 2→+ 4)MnSO4 → MnO4− 5 (+ 2→+ 7)MnSO4 → MnO42− 4 (+ 2→+ 6)Therefore, MnSO4converts to MnO2

reaction consists of oxidation half-cell reaction and reduction half-cell reaction Write both half-cell reactions, i.e oxidation half-cell reaction and reduction half-cell reaction.Then balance both the equations.

Now determine the correct value of stoichiometry of H SO2 4.Oxidation half-reaction, 2I−→ +I2 2e− …(i)Here, I−is converted into I2 Oxidation number of I is increasingfrom –1 to 0 hence, this is a type of oxidation reaction

l Stoichiometric coefficient of HSO4−is 6

Hence, option (a), (b) and (d) are correct

25. Oxalic acid solution titrated with NaOH solution usingphenolphthalein as an indicator

H C O2 2 4+2NaOH→ Na C O2 2 4+2H O2Equivalent of H C O2 2 4reacted=Equivalent of NaOH reacted

= × ×5 2 01= × ×1000

1000 M(NaOH)

Impure H O2 2react with KMnO4(acidic)

KMnO + H O+7 4 2−12→Mn+2 +O02

KMnO4acts as an oxidising agent,

Mn + 5+7 e−→Mn+2 (valency factor=5)O

−

−

→

1 2

=26 86158.(weight)H2 O2 =017 g

(Purify)H2O2= (Pure) ×

(Impure)

H2 O2 H2O2

Now, in Eq (i)

if, 1584 g of ammonium sulphate is used

Further, as given in question,

24 moles of NH3produced in reaction (i) is completly utilised

by 952g or 4 moles of NiCl 6H O2⋅ 2 to produce 4 moles of

[Ni(NH ) ] Cl3 6 2

So, 4 moles of [Ni(NH ) ] Cl3 6 2= ×4 232=928gms

Hence, total mass of gypsum and nickel ammonia coordination

compound [Ni(NH ) ] Cl3 6 2=2064 928 2992+ =

29. Both assertion and reason are factually true but the reason does

not exactly explain the assertion The correct explanation is,

methyl orange and phenolphthalein changes their colour at

different pH

30. If x is the oxidation state of Cu then :

3+ × +2 2 3x+ × −7 ( 2)=0 ⇒ x=7 3/

31. Na2S4O6is a salt of H2S4O6which has the following structure

⇒Difference in oxidation number of two types of sulphur = 5

32. Only F and Na show only one non-zero oxidation state

34. The balanced reaction is

6CaO + P O4 10 → 2Ca (PO )3 4 2Moles of P O4 10= 852=

284 3Moles of CaO required= × =3 6 18Mass of CaO required =18×56 = 1008 g

35. Meq of oxalate=10×0.2× =2 4Meq of MnO2formed=Meq of oxalate=4

2KMnO + 5H O + 3H SO4 2 2 2 4→ 2MnSO + 5O4 2

+ K SO + 8H O2 4 2MnO + Na C O + 2H SO2 2 2 4 2 4→ MnSO + Na SO4 2 4

+ CO + 2H O2 2 2

36. The balanced chemical reaction is

CuCO3+ H SO2 4 → CuSO4+H O2 + CO2millimol of CuCO 0.5

millimol of KIO3used= 0.1

39. Let the original sample contains x millimol of Fe O3 4 and

y millimol ofFe O2 3 In the first phase of reaction,

Fe O3 4+I− → 3Fe2+ +I

2 (n-factor of Fe O3 4=2)

Fe O2 3+I− → 2Fe2+ +I

2(n-factor of Fe O2 3=2)

⇒Meq of I2formed=Meq (Fe O3 4+Fe O )2 3

=Meq of hypo required

⇒ 7.5 millimol KIO3would be required for original 50 mL KI

⇒ Original 50 mL KI solution contain 15 millimol of KI.After AgNO3treatment 5 millimol of KIO3is required, i.e 10millimol KI is remaining

⇒ 5 millimol KI reacted with 5 millimol of AgNO3

⇒ Mass of AgNO3= 5 ×

1000 170 = 0.85 g

⇒ Mass percentage of AgNO3= 85%

43. CO2is evolved due to following reaction :

2NaHCO3 → Na CO + H O + CO2 3 2 2Moles of CO2produced = pV

×

750760

1298

123.9

1000 0.082

= ×5 10−3

⇒ Moles of NaHCO3in 2 g sample= × ×2 5 10− 3= 0.01

⇒millimol of NaHCO3in 1.5 g sample

160 1000 = 3.45During treatment with Zn-dust, all Fe3+ is reduced to Fe2 +,hence

millimol of Fe2+(in 100 mL) = 3.45×2 = 6.90

⇒ In 25 mL aliquot,6.90

4 = 1.725 millimol Fe

2+ion

Finally Fe2+is oxidised to Fe3+, liberating one electron per Fe2+

ion Therefore, total electrons taken up by oxidant

Let there be x millimol of Cu2+

⇒ Meq of Cu2+=Meq of I2= meq of hypo

⇒Moles of Cu2+: moles of C O2 42−=0.565 : 1.13=1 : 2

46. Let us consider 10 mL of the stock solution contain x millimol

oxalic acid H C O2 2 4and y millimol of NaHC O2 4

When titrated against NaOH, basicity of oxalic acid is 2 while

that of NaHC O2 4is 1

When titrated against acidic KMnO4, n-factors of both oxalic

acid and NaHC O2 4would be 2

chlorine atoms per molecule

⇒X =C H Cl2 4 2with two of its structural isomers

Cl— CH — CH — Cl2 2

I

and CH — CHCl3 2

II

On treatment with KOH, I will give ethane-1, 2-diol, hence it is

Y Z on treatment with KOH will give ethanal as

OH

CHOH

48. Let the n-factor of KMnO4in acid, neutral and alkaline media

are N1, N2and N3respectively Also, same volumes of reducingagent is used everytime, same number of equivalents of KMnO4would be required every time

∴In acid medium MnO4− → Mn2+

In neutral medium MnO4− → Mn4+

In alkaline medium MnO4− → Mn6+

⇒ Meq of Fe2+ present in solution = 2

⇒ millimol of Fe2 +present in solution = 2 (n-factor = 1)

× = millimol N H2 4Therefore, molarity of hydrazine sulphate solution

= ×1 =2

110

120

In acid solution : Normality of HNO3= ×8 5

2000= 0.02Normality of HCl =5

2000

×4.8

=0.012Let normality of H SO2 4in final solution be N.

52. During heating MCO3is converted into MO liberating CO2

while BaO is remaining unreacted :

MCO ( )3s →Heat MO( ) + CO ( )s 2g ↑ 0.44 g = 0.01 mol

BaO( )4.08 g

s BaO( )3.64 g

s

From the decomposition information, it can be deduced that the

original mixture contained 0.01 mole of MCO3and the solidresidue, obtained after heating, contain 0.01 mole (10 millimol)

Also, molar mass of BaO = 138 + 16

⇒ 100 = (Atomic weight of metal) + (12 + 3×16)

⇒ Atomic weight of metal = 40, i.e Ca

Topic 1 Preliminary Developments and Bohr’s Model

Objective Questions I (Only one correct option)

1. 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) (2019 Main, 9 April II)

(a) The electron can be found at a distance 2a from0

the nucleus

(b) The magnitude of the potential energy is double

that of its kinetic energy on an average

(c) The probability density of finding the electron is

maximum at the nucleus

(d) The total energy of the electron is maximum when

it is at a distance a0from the nucleus

2. If p is the momentum of the fastest electron ejected

from a metal surface after the irradiation of light having

wavelength λ, then for 1.5 p momentum of the

photoelectron, the wavelength of the light should be

(Assume kinetic energy of ejected photoelectron to be

very high in comparison to work function)

(2019 Main, 8 April II)

3. What is the work function of the metal, if the light of

wavelength 4000 Å generates photoelectron of velocity

4. The ground state energy of hydrogen atom is−13 6 eV

The energy of second excited state of He+ion in eV is

(2019 Main, 10 Jan II)

(a)−54 4 (b)−3 4 (c)−6 04 (d)−27 2

5. Which of the graphs shown below does not represent therelationship between incident light and the electron ejected frommetal surface? (2019 Main, 10 Jan I)

6. A stream of electrons from a heated filament was passed between

two charged plates kept at a potential differenceV esu If e and m

are charge and mass of an electron, respectively, then the value of

h/λ (where, λ is wavelength associated with electron wave) is

(a) 2 meV (b) meV (c) 2meV (d) meV

7. Rutherford’s experiment, which established the nuclear model ofthe atom, used a beam of (2002, 3M)

(a)β-particles, which impinged on a metal foil and got absorbed(b)γ-rays, which impinged on a metal foil and got scattered(c) helium atoms, which impinged on a metal foil and gotscattered

(d) helium nuclei, which impinged on a metal foil and gotscattered

8. Rutherford’s alpha particle scattering experiment eventually led

to the conclusion that (1986, 1M)

(a) mass and energy are related(b) electrons occupy space around the nucleus(c) neutrons are burried deep in the nucleus(d) the point of impact with matter can be precisely determined

K.E of s

es

Intensity of light 0

K.E of s

es

Frequency of light 0

K.E of s

es

Frequency of light 0

Number

of e ss

9. The radius of an atomic nucleus is of the order of (1985, 1M)

(a) 10−10cm (b) 10−13cm (c) 10−15cm (d) 10 cm−8

10. Bohr’s model can explain (1985, 1M)

(a) the spectrum of hydrogen atom only

(b) spectrum of an atom or ion containing one electron only

(c) the spectrum of hydrogen molecule

(d) the solar spectrum

11. The increasing order (lowest first) for the values of e/m

(charge/mass) for electron ( ),e proton (p), neutron (n) and

(a) nucleus (b) atom (c) electron (d) neutron

13. Rutherford’s experiment on scattering ofα-particles showed

for the first time that the atom has (1981, 1M)

(a) electrons (b) protons

(c) nucleus (d) neutrons

Objective Questions II

(One or more than one correct option)

14. The energy of an electron in the first Bohr orbit of H-atom is

–13.6 eV The possible energy value(s) of the excited state(s)

for electrons in Bohr orbits of hydrogen is (are) (1988)

(a)−3.4 eV (b)−4.2 eV (c)−6.8 eV (d)+6.8 eV

15. The atomic nucleus contains (1988, 1M)

(a) protons (b) neutrons (c) electrons (d) photons

16. The sum of the number of neutrons and proton in the isotope

17. When alpha particles are sent through a thin metal foil, most

of them go straight through the foil, because (1984, 1M)

(a) alpha particles are much heavier than electrons

(b) alpha particles are positively charged

(c) most part of the atom is empty space

(d) alpha particles move with high velocity

18. Many elements have non-integral atomic masses, because

(a) they have isotopes (1984, 1M)

(b) their isotopes have non-integral masses

(c) their isotopes have different masses

(d) the constituents, neutrons, protons and electrons,

combine to give fractional masses

Match the Columns

19. Consider the Bohr’s model of a one-electron atom where the

electron moves around the nucleus In the following List-I

contains some quantities for the nth orbit of the atom and

List-II contains options showing how they depend on n.

(II) Angular momentum of the electron

in the nth orbit

(Q) ∝n− 1(III) Kinetic energy of the electron in the

(II) Angular momentum of the electron in

the nth orbit

(Q) ∝n− 1(III) Kinetic energy of the electron in the

21. According to Bohr’s theory,

E n=Total energy K n=Kinetic energy

V n =Potential energy r n =Radius of nth orbit

Fill in the Blanks

22. The light radiations with discrete quantities of energy arecalled (1993, 1M)

23. The mass of a hydrogen is …… kg (1982, 1M)

24. Isotopes of an element differ in the number of …… in their

nuclei

(1982, 1M)

25. Elements of the same mass number but of different atomic

numbers are known as …… (1983, 1M)

Numerical Answer Type Questions

26. The figure below is the plot of potential energy versus

internuclear distance ( )d of H2molecule in the electronic

ground state What is the value of the net potential energy

E0(as indicated in the figure) in kJ mol−1, for d =d0 at

which the electron-electron repulsion and the

nucleus-nucleus repulsion energies are absent? As

reference, the potential energy of H atom is taken as zero

when its electron and the nucleus are infinitely far apart

Use Avogardo constant as 6 023 10 × 23mol−1 (2020 Adv.)

Subjective Questions

27. With what velocity should anα-particle travel towards thenucleus of a copper atom so as to arrive at a distance 10−13mfrom the nucleus of the copper atom ? (1997 (C), 3M)

Electronic Configuration and Quantum Number

Objective Questions I (Only one correct option)

1. The figure that is not a direct manifestation of the quantum

nature of atoms is (2020 Main, 2 Sep I)

2. The number of orbitals associated with quantum numbers

n=5, m s= +1

(a) 25 (b) 50 (c) 15 (d) 11

3. Among the following, the energy of 2s-orbital is lowest in

(2019 Main, 12 April II)

(a) Lyman and Paschen (b) Brackett and Pfund(c) Paschen and Pfund (d) Balmer and Brackett

6. The graph between |ψ|2and r (radial distance) is shown below.

This represents (2019 Main, 10 April I)

(a) 1s-orbital (b) 2 p-orbital (c) 3s-orbital (d) 2s-orbital

7. For any given series of spectral lines of atomic hydrogen,

let∆ν ν= max−νminbe the difference in maximum and

minimum frequencies in cm−1 The ratio

∆νLyman/∆νBalmer is (2019 Main, 9 April I)

9. If the de-Broglie wavelength of the electron in nthBohr orbit

in a hydrogenic atom is equal to 15 πa (a0 0is Bohr radius),

then the value of n Z/ is (2019 Main, 12 Jan II)

(a) 1.0 (b) 0.75 (c) 0.40 (d) 1.50

10. The de-Broglie wavelength ( )λ associated with a photoelectron

varies with the frequency ( )ν of the incident radiation as, [ν0is

threshold frequency] (2019 Main, 11 Jan II)

11. Which of the following combination of statements is true

regarding the interpretation of the atomic orbitals?

(2019 Main, 9 Jan 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 is inversely proportional to the azimuthal

quantum number

III According to wave mechanics, the ground state angular

momentum is equal to h

2π.

IV The plot of ψvs r for various azimuthal quantum

numbers, shows peak shifting towards higher r value.

(a) I, III (b) II, III (c) I, II (d) I, IV

12. 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 10 cm5 –1,

(a) non linear (b) linear with slope−RH

(c) linear with slope RH (d) linear with intercept−RH

14. The radius of the second Bohr orbit for hydrogen atom is(Planck’s constant ( )h =6 6262 10 × − 34Js; mass of electron

=91091 10 × −31 kg ; charge of electron( )e =160210 10 × −19C; permitivity of vacuum

( )∈ = × −

0 8 854185 10 12kg− 1m A− 3 2) (2017 Main)(a) 1.65 Å (b) 4.76 Å

(c) 0.529 Å (d) 2.12 Å

15 P is the probability of finding the 1s electron of hydrogen

atom in a spherical shell of infinitesimal thickness, dr,

at a distance r from the nucleus The volume of this shell is

4πr dr The qualitative sketch of the dependence of P on r is2

(2016 Adv.)

16. Which of the following is the energy of a possible excitedstate of hydrogen? (2015 Main)

(a) + 13.6 eV (b) – 6.8 eV(c) –3.4 eV (d) + 6.8 eV

17. The correct set of four quantum numbers for the valenceelectrons of rubidium atom (Z=37 is) (2013 Main)

(a) 5 0 0 1

2

2, , ,+

(c) 5 1 1 1

2

, , , ,+ (d) 5 0 1 1

2, , ,+

18. Energy of an electron is given by

n

2.178 10 18J

2 2

Wavelength of light required to excite an electron in an

hydrogen atom from level n=1to n=2 will be

(h=6.62 10× −34Js and c=3.0 10 ms× 8 −1)(a) 1.214×10−7m (b) 2.816×10−7m(c) 6.500×10−7m (d) 8.500×10−7m

r

0

(d)(c)

19. The kinetic energy of an electron in the second Bohr orbit of a

hydrogen atom is [a0is Bohr radius] (2012)

2 2 0

16π (c)

h ma

2 0

32π2 (d) h

ma

2 2 0

64π

20. The number of radial nodes in 3s and 2p respectively are

21. Which hydrogen like species will have same radius as that of

Bohr orbit of hydrogen atom? (2004, 1M)

(a) n=2, Li2+ (b) n=2, Be3+

(c) n=2, He+ (d) n=3, Li2+

22. If the nitrogen atom had electronic configuration 1s , it7

would have energy lower than that of the normal ground

state configuration 1 2s2 s22p3,because the electrons would

be closer to the nucleus, yet 1s is not observed, because it7

(a) Heisenberg uncertainty principle

(b) Hund’s rule

(c) Pauli exclusion principle

(d) Bohr postulate of stationary orbits

23. The quantum numbers + 1

24. The wavelength associated with a golf ball weighing 200 g

and moving at a speed of 5 m/h is of the order (2001, 1M)

(a) 10− 10m (b) 10− 20m

(c) 10− 30m (d) 10− 40m

25. The number of nodal planes in a p xorbital is (2001, 1M)

(a) one (b) two (c) three (d) zero

26. The electronic configuration of an element is

1s2,2 2s2 p6,3s23p63d5, 4s1 This represents its (2000, 1M)

(a) excited state (b) ground state

(c) cationic form (d) anionic form

27. The electrons, identified by quantum numbers n and l,

(i) n=4, l=1, (ii) n=4, l=0, (iii) n=3, l=2, (iv) n=3, l=1

can be placed in order of increasing energy, from the lowest to

(a) (iv) < (ii) < (iii) < (i) (b) (ii) < (iv) < (i) < (iii)

(c) (i) < (iii) < (ii) < (iv) (d) (iii) < (i) < (iv) < (ii)

28. The energy of an electron in the first Bohr orbit of H-atom is

–13.6 eV The possible energy value(s) of the excited state(s)

for electrons in Bohr orbits of hydrogen is (are) (1998, 2M)

29. For a d-electron, the orbital angular momentum is(1997, 1M)

(a) 62

30. The first use of quantum theory to explain the structure of

34. Which of the following does not characterise X-rays ?(a) The radiation can ionise gases (1992, 1M)

(b) It causes ZnS to fluoresce(c) Deflected by electric and magnetic fields(d) Have wavelengths shorter than ultraviolet rays

35. The correct set of quantum numbers for the unpairedelectron of chlorine atom is (1989, 1M)

40. The ratio of the energy of a photon of 200 Å wavelength

radiation to that of 4000 Å radiation is (1986, 1M)

(a)1

1

41. Which one of the following sets of quantum numbers

represents an impossible arrangement? (1986, 1M)

42. Electromagnetic radiation with maximum wavelength is

(a) ultraviolet (b) radio wave (1985, 1M)

43. Which electronic level would allow the hydrogen atom to

absorb a photon but not to emit a photon? (1984, 1M)

44. Correct set of four quantum numbers for the valence

(outermost) electron of rubidium (Z=37 is) (1984, 1M)

(a) 5, 0, 0,+ 1

2(c) 5, 1, 1,+ 1

2

45. The principal quantum number of an atom is related to the

(b) spin angular momentum

(c) orientation of the orbital in space

(d) orbital angular momentum

46. Any p-orbital can accommodate upto (1983, 1M)

(a) four electrons

(b) six electrons

(c) two electrons with parallel spins

(d) two electrons with opposite spins

Objective Questions II

(One or more than one correct option)

47. The ground state energy of hydrogen atom is

−13 6 eV Consider an electronic state Ψ 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

(a) It is a 4 d state

(b) The nuclear charge experienced by the electron in this state

is less than 2e, where e is the magnitude of the electronic

charge

(c) It has 2 angular nodes

(d) It has 3 radial nodes

48. The ground state electronic configuration of nitrogen atomcan be represented by (1999, 3M)

49. Which of the following statement (s) is (are) correct ?

(1998, 2M)

(a) The electronic configuration of Cr is [Ar] 3d54s (atomic1

number of Cr = 24)(b) The magnetic quantum number may have a negative value(c) In silver atom, 23 electrons have a spin of one type and 24 ofthe opposite type (atomic number of Ag = 47)

(d) The oxidation state of nitrogen in HN3is – 3

50. An isotone of3276Ge is (1984, 1M)(a)3277Ge (b)3377As

(c)3477Se (d)3478Se

Assertion and Reason

Read the following questions and answer as per the direction given below :

(a) Both Statement I and Statement II are correct; Statement

II is the correct explanation of Statement I(b) Both Statement I and Statement II are correct; Statement

II is not the correct explanation of Statement I(c) Statement I is correct; Statement II is incorrect(d) Statement I is incorrect; Statement II is correct

51 Statement I The first ionisation energy of Be is greater than

that of B

Statement II 2p-orbital is lower in energy than 2s. (2000)

Passage Based Questions

The hydrogen-like species Li2+is in a spherically symmetric

state S1with one radial node Upon absorbing light the ion

undergoes transition to a state S2 The state S2has one radialnode and its energy is equal to the ground state energy of thehydrogen atom

55. For He+ion, the only INCORRECT combination is

(a) (I) (i) (S) (b) (II) (ii) (Q)

(c) (I) (iii) (R) (d) (I) (i) (R)

56 For the given orbital in Column 1, the Only CORRECT

combination for any hydrogen-like species is

(a) (II) (ii) (P) (b) (I) (ii) (S)

(c) (IV) (iv) (R) (d) (III) (iii) (P)

57 For hydrogen atom, the only CORRECT combination is

(a) (I) (i) (P) (b) (I) (iv) (R)

(c) (II) (i) (Q) (d) (I) (i) (S)

58. Match the entries in Column I with the correctly related

quantum number(s) in Column II (2008, 6M)

B A hydrogen-like one-electron

wave function obeying

Pauli’s principle

q Azimuthalquantumnumber

C Shape, size and orientation

of hydrogen-like atomic

orbitals

r Magneticquantumnumber

D Probability density of

electron at the nucleus in

hydrogen-like atom

s Electron spinquantumnumber

Fill in the Blanks

59. The outermost electronic configuration of Cr is (1994, 1M)

60. 8 g each of oxygen and hydrogen at 27°C will have the totalkinetic energy in the ratio of (1989, 1M)

61. The uncertainty principle and the concept of wave nature ofmatter were proposed by and respectively

65. In a given electric field,β-particles are deflected more than

α-particles in spite ofα-particles having larger charge

Match the Columns

Answer Q 55, Q 55 and Q 56 by appropriately matching the information given in the three columns of the following table.

The wave function,ψn, ,l m lis a mathematical function whose value depends upon spherical polar coordinates ( , , )rθ φ of the electron

and characterised by the quantum number n l , and m l Here r is distance from nucleus,θis colatitude andφis azimuth In the

mathematical functions given in the Table, Z is atomic number and a0is Bohr radius (2017 Adv.)

0

(P)

0 3

0

5 2

Integer Answer Type Questions

70. Not considering the electronic spin, the degeneracy of the

second excited state (n=3 of H-atom is 9, while the)

degeneracy of the second excited state of H−is (2015 Adv.)

71. In an atom, the total number of electrons having quantum

n=4, |m l |=1 and m s= −1

2is

72. The atomic masses of He and Ne are 4 and 20 amu,

respectively The value of the de-Broglie wavelength of He

gas at−73°C is ‘M’ times that of the de-Broglie wavelength

of Ne at 727°C M is (2013 Adv.)

73. The work function (φ) of some metals is listed below The

number of metals which will show photoelectric effect when

light of 300 nm wavelength falls on the metal is (2011)

Φ(eV) 2.4 2.3 2.2 3.7 4.8 4.3 4.7 6.3 4.75

74. The maximum number of electrons that can have principal

quantum number, n=3 and spin quantum number,

where, a0is Bohr’s radius Let the radial node in 2s be at r0

Then, find r in terms of a0

(b) A base ball having mass 100 g moves with velocity

100 m/s Find out the value of wavelength of base ball

(2004, 2M)

77. The wavelength corresponding to maximum energy for

hydrogen is 91.2 nm Find the corresponding wavelength for

78. Calculate the energy required to excite 1 L of hydrogen gas

at 1 atm and 298 K to the first excited state of atomic

hydrogen The energy for the dissociation of H—H bond is

79. An electron beam can undergo diffraction by crystals

Through what potential should a beam of electrons be

accelerated so that its wavelength becomes equal to 1.54 Å

(1997 (C), 2M)

80. Consider the hydrogen atom to be proton embedded in a

cavity of radius a0 (Bohr’s radius) whose charge is

neutralised by the addition of an electron to the cavity in

vacuum, infinitely slowly Estimate the average total energy

of an electron in its ground state in a hydrogen atom as the

work done in the above neutralisation process Also, if themagnitude of the average kinetic energy is half themagnitude of the average potential energy, find the average

81. Calculate the wave number for the shortest wavelengthtransition in the Balmer series of atomic hydrogen.(1996, 1M)

82. Iodine molecule dissociates into atoms after absorbing light

to 4500Å If one quantum of radiation is absorbed by eachmolecule, calculate the kinetic energy of iodine atoms.(Bond energy of I = 240 kJ mol2 –1) (1995, 2M)

83. Find out the number of waves made by a Bohr’s electron inone complete revolution in its 3rd orbit (1994, 3M)

84. What transition in the hydrogen spectrum would have the

same wavelength as the Balmer transition n = 4 to n = 2 of

86. According to Bohr’s theory, the electronic energy of

hydrogen atom in the nth Bohr’s orbit is given by :

E

n

n= −21.7 10×2 −19JCalculate the longest wavelength of electron from the thirdBohr’s orbit of the He+ion (1990, 3M)

87. What is the maximum number of electrons that may bepresent in all the atomic orbitals with principal quantumnumber 3 and azimuthal quantum number 2 ? (1985, 2M)

88. Give reason why the ground state outermost electronicconfiguration of silicon is (1985, 2M)

89. The electron energy in hydrogen atom is given by

E

n

n= − 21.7 10×2 −12 erg Calculate the energy required to

remove an electron completely from the n = 2 orbit What is

the longest wavelength (in cm) of light that can be used tocause this transition? (1984, 3M)

90. Calculate the wavelength in Angstroms of the photon that is

emitted when an electron in the Bohr’s orbit, n = 2 returns to the orbit, n = 1in the hydrogen atom The ionisation potential

of the ground state hydrogen atom is 2.17×10− 11 erg per

Topic 1 Preliminary Developments

and Bohr’s Model

1. Statement (d) is incorrect For 1s-orbital radial probability

density (R2)against r is given as:

For 1s-orbital, probability density decreases sharply as we move

away from the nucleus

The radial distribution curves obtained by plotting radial

probability functions vs r for 1s-orbital is

The graph initially increases and then decreases It reaches a

maximum at a distance very close to the nucleus and then

decreases The maximum in the curve corresponds to thedistance at which the probability of finding the electron inmaximum

2. The expression of kinetic energy of photo electrons,

2 2

λ energy of incident light.

E0=threshold energy or work functions,

12

12

p m

Q p= momentum=mv

As per the given condition,

λλ

2 1 1 2

49