Chapterwise Topicwise
Solved Papers
2021-1979
IITJEE
JEE Main & Advanced
Chemistry
Ranjeet Shahi
Arihant Prakashan (Series), Meerut
Arihant Prakashan (Series), Meerut
All Rights Reserved
© Author
Administrative & Production Offices
Regd. Office
‘Ramchhaya’ 4577/15, Agarwal Road, Darya Ganj, New Delhi -110002
Tele: 011- 47630600, 43518550
Head Office
Kalindi, TP Nagar, Meerut (UP) - 250002, Tel: 0121-7156203, 7156204
Sales & Support Offices
Agra, Ahmedabad, Bengaluru, Bareilly, Chennai, Delhi, Guwahati,
Hyderabad, Jaipur, Jhansi, Kolkata, Lucknow, Nagpur & Pune.
ISBN 978-93-25796-14-0
PO No : TXT-XX-XXXXXXX-X-XX
Published by Arihant Publications (India) Ltd.
For further information about the books published by Arihant, log on to
www.arihantbooks.com or e-mail at info@arihantbooks.com
Follow us on
CONTENTS
1-22
19. Extraction of Metals
282-293
2. Atomic Structure
23-40
20. Qualitative Analysis
294-306
3. Periodic Classification and
Periodic Properties
21. Organic Chemistry Basics
307-331
41-47
22. Hydrocarbons
332-349
4. Chemical Bonding
48-66
23. Alkyl Halides
350-363
5. States of Matter
67-83
24. Alcohols and Ethers
364-377
25. Aldehydes and Ketones
378-396
26. Carboxylic Acids and
their Derivatives
397-412
1. Some Basic Concepts of Chemistry
6. Chemical and Ionic Equilibrium
84-108
7. Thermodynamics and
Thermochemistry
109-129
8. Solid State
130-139
9. Solutions and Colligative Properties
140-155
27. Aliphatic Compounds
Containing Nitrogen
413-422
10. Electrochemistry
156-177
28. Benzene and Alkyl Benzene
423-440
11. Chemical Kinetics
178-195
12. Nuclear Chemistry
196-199
29. Aromatic Compounds
Containing Nitrogen
441-457
13. Surface Chemistry
200-206
30. Aryl Halides and Phenols
458-470
14. s-Block Elements
207-217
31. Aromatic Aldehydes, Ketones
and Acids
471-484
15. p-Block Elements-I
218-227
16. p-Block Elements-II
228-248
32. Biomolecules and Chemistry
in Everyday Life
485-501
33. Environmental Chemistry
502-504
17. Transition and
Inner-Transition Elements
249-258
18. Coordination Compounds
259-281
JEE Advanced Solved Paper 2021
1-16
SYLLABUS
JEE MAIN
Section A : PHYSICAL CHEMISTRY
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 ψ2, concept of atomic orbitals as one electron wave functions;
Variation of ψ and ψ2 with r for 1s and 2s orbitals; various quantum
numbers (principal, angular momentum and magnetic quantum
numbers) and their significance; shapes of s, p and d - orbitals, electron
spin and spin quantum number; rules for filling electrons in orbitals –
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
universe and ΔG of the system as criteria for spontaneity, ΔGo
(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
ΔG and ΔGo 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.
Eectrolytic and metallic conduction, conductance in electrolytic
solutions, specific and molar conductivities and their variation with
concentration: Kohlrausch's law and its applications.
temperature on rate of reactions - Arrhenius theory, activation energy
and its calculation, collision theory of bimolecular gaseous reactions
(no derivation).
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 X Surface Chemistry
Adsorption - Physisorption and chemisorption and their
characteristics, factors affecting adsorption of gases on solidsFreundlich 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 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
Section B : INORGANIC CHEMISTRY
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,(PCl3, PCl5);
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
K2 Cr2 O7 and KMnO4.
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 C : ORGANIC 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 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.
UNIT XXII Organic Compounds Containing Halogens
General methods of preparation, properties and reactions; Nature of
C—X bond; Mechanisms of substitution reactions.
—
Inorganic compounds Mohr's salt, potash alum.
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
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 KMnO4, Mohr's salt vs KMnO4.
Alcohols Identification of primary, secondary and tertiary alcohols;
mechanism of dehydration.
Chemistry involved in the preparation of the following
—
Chemical principles involved in the qualitative salt analysis
—
Cations — Pb2+ , Cu2+, Al3+, Fe3+, Zn2+, Ni2+, Ca2+, Ba2+ , Mg2+ NH4+.
Anions – CO32-, S2-, SO42-, NO2, NO3, Cl -, Br-, I- (Insoluble salts
excluded).
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, NH3 and its derivatives), Grignard reagent;
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.
—
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.
JEE ADVANCED
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.
INORGANIC CHEMISTRY
Isolation/Preparation and Properties of the following
Non-metals Boron, silicon, nitrogen, phosphorus, oxygen,
sulphur and halogens, Properties of allotropes of carbon
(only diamond and graphite), phosphorus and sulphur.
Preparation and Properties of the following
Compounds Oxides, peroxides, hydroxides, carbonates,
bicarbonates, chlorides and sulphates of sodium,
potassium, magnesium and calcium, Boron, diborane,
boric acid and borax, Aluminium, alumina, aluminium
chloride and alums, Carbon, oxides and oxyacid (carbonic
acid), Silicon, silicones, silicates and silicon carbide,
Nitrogen, oxides, oxyacids and ammonia, Phosphorus,
oxides, oxyacids (phosphorus acid, phosphoric acid) and
phosphine, Oxygen, ozone and hydrogen peroxide,
Sulphur, hydrogen sulphide, oxides, sulphurous acid,
sulphuric acid and sodium thiosulphate, Halogens,
hydrohalic acids, oxides and oxyacids of chlorine,
bleaching powder, Xenon fluorides.
Transition Elements (3d series) Definition, general
characteristics, oxidation states and their stabilities,
colour (excluding the details of electronic transitions) and
calculation of spin-only magnetic moment; Coordination
compounds: nomenclature of mononuclear coordination
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 Fe2+, Cu2+ and Zn2+, 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
Ag+, Hg2+, Cu2+, Pb2+, Bi3+, Fe3+, Cr3+, Al3+, Ca2+, Ba2+, Zn2+, Mn2+
and Mg2+), Nitrate, halides (excluding fluoride), sulphate
and sulphide.
ORGANIC CHEMISTRY
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
(boiling points, density and dipole moments), Acidity of
alkynes, Acid catalysed hydration of alkenes and alkynes
(excluding the stereochemistry of addition and
elimination), Reactions of alkenes with KMnO4 and ozone,
Reduction of alkenes and alkynes, Preparation of alkenes
and alkynes by elimination reactions, Electrophilic
addition reactions of alkenes with X2, HX, HOX and H2O
(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), ReimerTiemann reaction, Kolbe reaction.
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, ZnCl2/concentrated HCl,
conversion 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.
1
Some Basic Concepts
of Chemistry
6. The percentage composition of carbon by mole in methane is
Topic 1 Mole Concept
(2019 Main, 8 April II)
Objective Questions I (Only one correct option)
1. 5 moles of AB2 weight 125 × 10−3 kg and 10 moles of A2 B2
weight 300 × 10−3 kg. The molar mass of A ( M A ) and molar
mass of B ( M B ) in kg mol −1 are
(2019 Main, 12 April I)
(a) M A = 10 × 10−3 and M B = 5 × 10−3
(b) M A = 50 × 10−3 and M B = 25 × 10−3
(d) M A = 5 × 10
and M B = 10 × 10
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)
C3 H8 ( g ) + 5O2 ( g ) → 3CO2 ( g ) + 4H2 O( l )
P4 ( s ) + 5O2 ( g ) → P4 O10 ( s )
4Fe( s ) + 3O2 ( g ) → 2Fe2 O3 ( s )
2Mg ( s ) + O2 ( g ) → 2MgO( s )
10 mL of a hydrocarbon required 55 mL of O2 for complete
combustion and 40 mL of CO2 is formed. The formula of the
hydrocarbon is
(2019 Main, 10 April I)
(c) C4H10
(d) C4H 8
(a) C4H7Cl (b) C4H 6
4. 10 mL of 1 mM surfactant solution forms a monolayer
covering 0.24 cm 2 on a polar substrate. If the polar head is
approximated as a cube, what is its edge length?
(2019 Main, 9 April II)
(b) 0.1 nm
(c) 1.0 pm
(d) 80%
7. 8 g of NaOH is dissolved in 18 g of H2 O. Mole fraction of
NaOH in solution and molality (in mol kg− 1 ) of the solution
respectively 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
(d) 2.0 nm
5. For a reaction,
N 2 ( g ) + 3H 2 ( g ) → 2NH 3 ( g ), identify dihydrogen (H 2 )
as a limiting reagent in the following reaction mixtures.
(2019 Main, 9 April I)
(a) 56 g of N 2 + 10 g of H 2 (b) 35 g of N 2 + 8 g of H 2
(c) 14 g of N 2 + 4 g of H 2 (d) 28 g of N 2 + 6 g of H 2
(2019 Main, 12 Jan II)
(d) 5.6
9. The amount of sugar (C12 H22 O11 ) required to prepare 2 L of
its 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 C57 H110 O6 is :
2C57 H110 O6 ( s ) + 163O2 ( g ) → 114CO2 ( g ) + 110 H2 O ( l )
(2019 Main, 9 Jan II)
(a) 490 g
3. At 300 K and 1 atmospheric pressure,
(a) 2.0 pm
(c) 25%
(Molar mass of H2 O2 = 34 g mol −1 )
(a) 16.8
(b) 22.4
(c) 11.35
−3
2. The minimum amount of O2 ( g ) consumed per gram of
(a)
(b)
(c)
(d)
(b) 20%
8. The volume strength of 1 M H2 O2 is
(c) M A = 25 × 10−3 and M B = 50 × 10−3
−3
(a) 75%
(b) 495 g
(c) 445 g
(d) 890 g
11. A solution of sodium sulphate contains 92 g of Na + ions per
kilogram of water. The molality of Na + ions in that solution
in mol kg−1 is
(2019 Main, 9 Jan I)
(a) 16
(b) 4
(c) 132
(d) 8
12. The most abundant elements by mass in the body of a healthy
human adult are oxygen (61.4%), carbon (22.9%), hydrogen
(10.0 %), and nitrogen (2.6%). The weight which a 75 kg
person would gain if all 1 Hatoms are replaced by 2 Hatoms is
(2017 JEE Main)
(a) 15 kg
(c) 7.5 kg
(b) 37.5 kg
(d) 10 kg
13. 1 g of a carbonate (M 2 CO3 ) on treatment with excess HCl
produces 0.01186 mole of CO2 . The molar mass of M 2 CO3
in g mol −1 is
(2017 JEE Main)
(a) 1186
(b) 84.3
(c) 118.6
(d) 11.86
2 Some Basic Concepts of Chemistry
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)
(b) C4 H8
(c) C4 H10
(d) C3 H6
(a) C3 H8
15. The molecular formula of a commercial resin used for
24. The normality of 0.3 M phosphorus acid (H3PO3) is
(1999, 2M)
(a) 0.1
(b) 0.9
(c) 0.3
(d) 0.6
25. In which mode of expression, the concentration of a solution
remains independent of temperature?
(a) Molarity (b) Normality (c) Formality
(1988, 1M)
(d) Molality
26. A molal solution is one that contains one mole of solute in
(1986, 1M)
exchanging ions in water softening is C8 H7 SO3 Na
(molecular weight = 206). What would be the maximum
uptake of Ca 2+ ions by the resin when expressed in mole per
gram resin?
(2015 Main)
1
1
2
1
(b)
(c)
(d)
(a)
103
206
309
412
(a) 1000 g of solvent
(b) 1.0 L of solvent
(c) 1.0 L of solution
(d) 22.4 L of solution
27. If 0.50 mole of BaCl 2 is mixed with 0.20 mole of Na 3 PO4 ,
the maximum number of moles of Ba 3 (PO4 )2 that can be
formed is
(1981, 1M)
(a) 0.70
(b) 0.50
(c) 0.20
(d) 0.10
16. 3 g of activated charcoal was added to 50 mL of acetic acid
28. 2.76 g of silver carbonate on being strongly heated yields a
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
solution is
(2011)
(a) 1.78 M (b) 2.00 M (c) 2.05 M
(d) 2.22 M
20. Given that the abundances of isotopes
54 Fe, 56 Fe and 57 Fe
are 5%, 90% and 5%, respectively, the atomic mass of Fe is
residue weighing
(a) 2.16 g
(b) 2.48 g
(1979, 1M)
(c) 2.32 g
(d) 2.64 g
29. When the same amount of zinc is treated separately with
excess of sulphuric acid and excess of sodium hydroxide, the
ratio of volumes of hydrogen evolved is
(1979, 1M)
(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 (N2 O5 )
31. The total number of electrons in one molecule of carbon
dioxide is
(1979, 1M)
(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
molecules is
(1979, 1M)
(a) 1 : 4
(b) 1 : 8
(c) 7 : 32
(d) 3 : 16
(2009)
(a) 55.85
(c) 55.75
(b) 55.95
(d) 56.05
Numerical Answer Type Questions
21. Mixture X = 0.02 mole of [Co(NH3 )5 SO4 ]Br and 0.02 mole
of [Co(NH3 )5 Br]SO4 was prepared in 2 L solution.
1 L of mixture X + excess of AgNO3 solution → Y
1 L of mixture X + excess of BaCl 2 solution → Z
Number of moles of Y and Z are
(a) 0.01, 0.01
(b) 0.02, 0.01
(c) 0.01, 0.02
(d) 0.02, 0.02
22. Which has maximum number of atoms?
(a) 24 g of C (12)
(c) 27 g of Al (27)
(2003, 1M)
(2003, 1M)
(b) 56 g of Fe (56)
(d) 108 g of Ag (108)
23. How many moles of electron weighs 1 kg?
23
(a) 6.023 × 10
(c)
6.023
× 1054
9.108
1
(2002, 3M)
(b)
× 1031
9.108
1
(d)
× 108
9.108 × 6.023
33. A 100 mL solution was made by adding 1.43 g of
Na 2CO3 ⋅ xH 2O. 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 is
stopped, but the heating is continued in a closed furnace such
that the content undergo self-reduction. The weight (in kg) of
Pb produced per kg of O 2 consumed is ……… .
(Atomic weights in g mol −1 : O = 16, S = 32, Pb = 207)
(2018 Adv.)
35. To measure the quantity of MnCl 2 dissolved in an aqueous
solution, it was completely converted to KMnO4 using the
reaction,
MnCl 2 + K 2 S2 O8 + H2 O → KMnO4 + H2 SO4 + HCl
(equation not balanced).
Some Basic Concepts of Chemistry 3
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 MnCl 2 (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)
O
NaOBr
H3O+
A
NH3, ∆
(60%)
B
(2005, 3M)
45. In a solution of 100 mL 0.5 M acetic acid, one gram of active
charcoal is added, which adsorbs acetic acid. It is found that the
concentration of acetic acid becomes 0.49 M. If surface area of
charcoal is 3.01 × 102 m2 , calculate the area occupied by single
acetic 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 of
the virus is 0.75 cm 3 /g. If the virus is considered to be a single
particle, find its molar mass.
(1999, 3M)
obtain 1 dm 3 of solution of density 1077.2 kg m −3 . Calculate
the molality, molarity and mole fraction of Na 2 SO4 in solution.
C
(50%)
Br2(3 equivalent )
AcOH
number of surface sites occupied per molecule of N2 .
48. 8.0575 × 10−2 kg of Glauber’s salt is dissolved in water to
Br2/KOH
(50%)
K in a container of volume is 2.46 cm 3 . Density of surface sites
is 6.023 × 1014 /cm 2 and surface area is 1000 cm 2 , find out the
(1994, 3M)
D
(100%)
(2018 Adv.)
Fill in the Blanks
37. The weight of 1 × 1022 molecules of CuSO4 ⋅ 5H2 O is
…………. .
(1991, 1M)
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 2 (SO4 )3 . Identify A, B and C.
(1994, 2M)
50. Upon mixing 45.0 mL 0.25 M lead nitrate solution with
(1980, 1M)
25.0 mL of a 0.10 M chromic sulphate solution, precipitation of
lead sulphate takes place. How many moles of lead sulphate are
formed? Also calculate the molar concentrations of species left
behind in the final solution. Assume that lead sulphate is
completely insoluble.
(1993, 3M)
40. The modern atomic mass unit is based on the mass of
51. Calculate the molality of 1.0 L solution of 93% H2 SO4 ,
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
…………. .
…………. .
(1980, 1M)
(weight/volume). The density of the solution is 1.84 g/mL.
(1990, 1M)
Integer Answer Type Questions
52. A solid mixture (5.0 g) consisting of lead nitrate and sodium
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

m
molecular weights of the solute and solvent,  solute  is ...
 msolvent 
.
(2016 Adv.)
42. A compound H2 X 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
nitrate was heated below 600°C until the weight of the residue
was constant. If the loss in weight is 28.0 per cent, find the
amount 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
(C12 H22 O11 ). Calculate (i) molal concentration and (ii) mole
fraction of sugar in syrup.
(1988, 2M)
55. An unknown compound of carbon, hydrogen and oxygen
contains 69.77% C and 11.63% H and has a molecular weight
of 86. It does not reduces Fehling’s solution but forms a
bisulphate addition compound and gives a positive iodoform
test. What is the possible structure(s) of unknown compound?
(1987, 3M)
44. 20% surface sites have adsorbed N2 . On heating N2 gas
56. The density of a 3 M sodium thiosulphate solution ( Na 2 S2 O3 )
evolved from sites and were collected at 0.001 atm and 298
is 1.25 g per mL. Calculate (i) the percentage by weight of
4 Some Basic Concepts of Chemistry
sodium thiosulphate (ii) the mole fraction of sodium
thiosulphate and (iii) the molalities of Na + and S2 O2−
3 ions.
oxygen. All volumes have been reduced to NTP. Calculate
the molecular formula of the hydrocarbon gas. (1979, 3M)
(1983, 5M)
59. In the analysis of 0.5 g sample of feldspar, a mixture of
57. (a) 1.0 L of a mixture of CO and CO2 is taken. This mixture
chlorides of sodium and potassium is obtained, which weighs
0.1180 g. Subsequent treatment of the mixed chlorides with
silver nitrate gives 0.2451 g of silver chloride. What is the
percentage of sodium oxide and potassium oxide in the
sample?
(1979, 5M)
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
weight of metal.
(1980, 5M)
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
60. The vapour density (hydrogen = 1) of a mixture consisting of
NO2 and N2 O4 is 38.3 at 26.7°C. Calculate the number of
moles of NO2 in 100 g of the mixture.
(1979, 5M)
61. Accounts for the following. Limit your answer to two
sentences, “Atomic weights of most of the elements are
fractional”.
(1979, 1M)
62. Naturally occurring boron consists of two isotopes whose
atomic weights are 10.01 and 11.01. The atomic weight of
natural boron is 10.81. Calculate the percentage of each
isotope in natural boron.
(1978, 2M)
Topic 2 Equivalent Concept, Neutralisation and Redox Titration
Objective Questions I (Only one correct option)
1. An example of a disproportionation reaction is
(2019 Main, 12 April I)
(a)
2MnO−4
−
+
+ 10I + 16H → 2Mn2 + +5I2 + 8H 2O
(b) 2NaBr + Cl2 → 2NaCl + Br2
(c) 2KMnO4 → K 2MnO4 + MnO2 + O2
(d) 2CuBr → CuBr2 + Cu
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)
pH
surface of water in a round watch glass. Hexane evaporates
and a monolayer is formed. The distance from edge to centre
of the watch glass is 10 cm. What is the height of the
monolayer? [Density of fatty acid = 0.9 g cm −3 ; π = 3]
(2019 Main, 8 April II)
(a) 10−6 m
(c) 10−8 m
(b) 10−4 m
(d) 10−2 m
4. In order to oxidise a mixture of one mole of each of FeC2 O4 ,
Fe2 (C2 O4 )3 , FeSO4 and Fe2 (SO4 )3 in acidic medium, the
number of moles of KMnO4 required is (2019 Main, 8 April I)
(a) 2
(b) 1
(c) 3
(d) 1.5
5. 100 mL of a water sample contains 0.81 g of calcium
bicarbonate and 0.73 g of magnesium bicarbonate. The
hardness of this water sample expressed in terms of
equivalents of CaCO3 is (molar mass of calcium bicarbonate
is 162 g mol−1 and magnesium bicarbonate is 146 g mol−1 )
pH
V(mL)
V(mL)
(A)
(B)
(2019 Main, 8 April I)
(a) 5,000 ppm
(c) 100 ppm
(b) 1,000 ppm
(d) 10,000 ppm
6. 50 mL of 0.5 M oxalic acid is needed to neutralise 25 mL of
pH
sodium hydroxide solution. The amount of NaOH in 50 mL
of the given sodium hydroxide solution is
pH
(2019 Main, 12 Jan I)
(a) (D)
(c) (B)
V(mL)
V(mL)
(C)
(D)
(a) 40 g
(2019 Main, 9 April II)
(b) (A)
(d) (C)
3. 0.27 g of a long chain fatty acid was dissolved in 100 cm 3 of
hexane. 10 mL of this solution was added dropwise to the
(b) 80 g
(c) 20 g
(d) 10 g
7. 25 mL of the given HCl solution requires 30 mL of 0.1 M
sodium carbonate solution. What is the volume of this HCl
solution required to titrate 30 mL of 0.2 M aqueous NaOH
solution?
(2019 Main, 11 Jan II)
(a) 75 mL
(b) 25 mL
(c) 12.5 mL
(d) 50 mL
Some Basic Concepts of Chemistry 5
8. In the reaction of oxalate with permanganate in acidic
medium, the number of electrons involved in producing one
(2019 Main, 10 Jan II)
molecule of CO2 is
(a) 2
(b) 5
(c) 1
(d) 10
9. The ratio of mass per cent of C and H of an organic
compound (Cx H y O z ) is 6 : 1. If one molecule of the above
compound (Cx H y O z ) contains half as much oxygen as
required to burn one molecule of compound Cx H y
completely to CO2 and H2 O. The empirical formula of
compound Cx H y O z is
(2018 Main)
(c) C3 H4 O2
(d) C2 H4 O3
(a) C3 H6 O3 (b) C2 H4 O
10. An alkali is titrated against an acid with methyl orange as
indicator, which of the following is a correct combination?
(2018 Main)
Base
Acid
End point
(a)
Weak
Strong
Colourless to pink
(b)
Strong
Strong
Pinkish red to yellow
(c)
Weak
Strong
Yellow to pinkish red
(d)
Strong
Strong
Pink to colourless
17. The oxidation number of sulphur in S 8 , S 2 F 2 , H 2 S
respectively, are
(a) 0, +1 and –2
(c) 0, +1 and +2
18. The number of moles of KMnO 4 that will be needed to react
completely with one mole of ferrous oxalate in acidic
medium is
(1997)
2
3
4
(b)
(c)
(d) 1
(a)
5
5
5
19. The number of moles of KMnO 4 that will be needed to react
with one mole of sulphite ion in acidic solution is
(1997)
2
(a)
5
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)
(a) 3
(b) 4
(c) 5
(d) 6
13. In the standardisation of Na 2 S2 O3 using K 2 Cr2 O7 by
iodometry, the equivalent weight of K 2 Cr2 O7 is (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
(2001)
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)
(a) 40 mL
(b) 20 mL
(c) 10 mL
(d) 4 mL
16. Among the following, the species in which the oxidation
number of an element is + 6
(a) MnO−4
(b) Cr(CN)3−
6
(c) NiF62−
(d) CrO2 Cl 2
3
(b)
5
4
(c)
5
(d) 1
20. For the redox reaction
MnO4− + C2 O24 − + H+ → Mn 2+ + CO2 + H2 O
The correct coefficients of the reactants for the balanced
reaction are
11. From the following statements regarding H2 O2 choose the
example of
(a) oxidation reaction
(b) reduction reaction
(c) disproportionation reaction
(d) decomposition reaction
(1999)
(b) +2, +1 and –2
(d) –2, +1 and –2
(2000)
(a)
(b)
(c)
(d)
MnO−4
2
16
5
2
C2 O42 −
5
5
16
16
H+
16
2
2
5
21. The volume strength of 1.5 N H2 O2 is
(a) 4.8
(b) 8.4
(c) 3.0
(1992)
(1990, 1M)
(d) 8.0
22. The oxidation number of phosphorus in Ba(H2 PO2 )2 is
(a) +3
(c) +1
(b) +2
(d) –1
(1988)
23. The equivalent weight of MnSO 4 is half of its molecular
weight, when it converts to
(a) Mn 2 O3
(b) MnO2
(c) MnO−4
(1988, 1M)
(d) MnO2−
4
Objective Question II (More than one correct option)
24. For the reaction, I− + ClO3− + H2 SO4 → Cl − + HSO4− + I2
the correct statement(s) in the balanced equation is/are
(a) stoichiometric coefficient of HSO−4 is 6
(2014 Adv.)
(b) iodide is oxidised
(c) sulphur is reduced
(d) H2 O is one of the products
Numerical Answer Type Questions
25. 5.00 mL of 0.10 M oxalic acid solution taken in a conical
flask is titrated against NaOH from a burette using
phenolphthalein indicator. The volume of NaOH required for
the appearance of permanent faint pink color is tabulated
below for five experiments. What is the concentration, in
molarity, of the NaOH solution?
6 Some Basic Concepts of Chemistry
Exp. No.
1
2
3
4
5
Integer Answer Type Questions
Vol. of NaOH (mL)
12.5
10.5
9.0
9.0
9.0
31. The difference in the oxidation numbers of the two types of
sulphur atoms in Na 2 S4 O6 is
32. Among the following, the number of elements showing only
one non-zero oxidation state is O, Cl, F, N, P, Sn, Tl, Na, Ti
(2010)
(2020 Adv.)
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 −1 K −1 )
(2020 Adv.)
27. A 20.0 mL solution containing 0.2 g impure H 2O2 reacts
completely with 0.316 g of KMnO4 in acid solution. The
purity of H 2O2 (in%) is ............. (molecular weight of
H 2O2 = 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 2 ⋅ 6H2 O 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 2 ⋅ 6H2 O 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,
(2018 Adv.)
Cl = 35.5, Ca = 40, Ni = 59)
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 2 CO3 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 2 CO3 . (1991, 2M)
Fill in the Blanks
30. The
(2011)
compound YBa 2 Cu 3 O7 , which shows super
conductivity, has copper in oxidation state ………. Assume
that the rare earth element yttrium is in its usual + 3 oxidation
state.
(1994, 1M)
33. A student performs a titration with different burettes and
finds titrate values of 25.2 mL, 25.25 mL, and 25.0 mL. The
number of significant figures in the average titrate value is
(2010)
Subjective Questions
34. Calculate the amount of calcium oxide required when it
reacts with 852 g of P4 O10 .
(2005, 2M)
35. Hydrogen peroxide solution (20 mL) reacts quantitatively
with a solution of KMnO4 (20 mL) acidified with dilute
H2 SO4 . The same volume of the KMnO4 solution is just
decolourised by 10 mL of MnSO4 in neutral medium
simultaneously forming a dark brown precipitate of hydrated
MnO2 . The brown precipitate is dissolved in 10 mL of 0.2 M
sodium oxalate under boiling condition in the presence of
dilute H2 SO4 . Write the balanced equations involved in the
reactions and calculate the molarity of H2 O2 .
(2001)
36. How many millilitres of 0.5 M H2 SO4 are needed to dissolve
0.5 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 KI
solution. The solution was acidified with HCl. The liberated
I2 consumed 45.0 mL of thiosulphate solution decolourise
the blue starch-iodine complex. Calculate the molarity of the
sodium thiosulphate solution.
(1998, 5M)
38. To a 25 mL H2 O2 solution, excess of acidified solution of
potassium iodide was added. The iodine liberated required
20 mL of 0.3 N sodium thiosulphate solution. Calculate the
volume strength of H2 O2 solution.
(1997, 5M)
39. A 3.00 g sample containing Fe3 O4 , Fe2 O3 and an inert
impure substance, is treated with excess of KI solution in
presence of dilute H2 SO4 . The entire iron is converted into
Fe2+ along with the liberation of iodine. The resulting
solution is diluted to 100 mL . A 20 mL of the diluted
solution requires 11.0 mL of 0.5 M Na 2 S2 O3 solution to
reduce the iodine present. A 50 mL of the dilute solution,
after complete extraction of the iodine required 12.80 mL of
0.25 M KMnO4 solution in dilute H2 SO4 medium for the
oxidation of Fe2+ . Calculate the percentage of Fe2 O3 and
(1996, 5M)
Fe3 O4 in the original sample.
40. A 20.0 cm 3 mixture of CO, CH4 and He gases is exploded by
an electric discharge at room temperature with excess of
oxygen. The volume contraction is found to be 13.0 cm 3 .
A further contraction of 14.0 cm 3 occurs when the residual
gas is treated with KOH solution. Find out the composition
of the gaseous mixture in terms of volume percentage.
(1995, 4M)
Some Basic Concepts of Chemistry 7
41. A 5.0 cm 3 solution of H2 O2 liberates 0.508 g of iodine from
an acidified KI solution. Calculate the strength of H2 O2
solution in terms of volume strength at STP.
(1995, 3M)
Calculate the amount of H2 C2 O4 and NaHC2 O4 in the
mixture.
(1990, 5M)
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
of Y and Z.
(1989, 5M)
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) KIO 3 solution in presence of
6 M HCl till all I− ions are converted into ICl. It requires
50 mL of (M/10) KIO 3 solution, 20 mL of the same stock
solution of KI requires 30 mL of (M/10) KIO3 under similar
conditions. Calculate the percentage of AgNO3 in the
sample.
Reaction KIO3 + 2KI + 6HCl → 3ICl + 3KCl + 3H2 O
(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 CO2 ceases. The volume of CO2 at 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 Fe2 O3 solid 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)
48. An equal volume of a reducing agent is titrated separately
with 1 M KMnO4 in acid, neutral and alkaline medium. The
volumes of KMnO4 required are 20 mL in acid, 33.3 mL in
neutral and 100 mL in alkaline media. Find out the oxidation
state of manganese in each reduction product. Give the
balanced equations for all the three half reaction. Find out the
volume of 1M K 2 Cr2 O7 consumed, if the same volume of the
reducing agent is titrated in acid medium.
(1989, 5M)
49. A sample of hydrazine sulphate ( N2 H 6 SO4 ) was dissolved in
100 mL of water, 10 mL of this solution was reacted with
excess of ferric chloride solution and warmed to complete
the 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 the
solution.
Reaction 4Fe3+ + N2 H4 → N2 + 4Fe2+ + 4H+
MnO−4 + 5Fe2+ + 8H+ → Mn 2+ + 5Fe3+ + 4H2 O
45. A solution of 0.2 g of a compound containing Cu 2+ and
C2 O2−
4 ions on titration with 0.02 M KMnO4 in presence of
H2 SO4 consumes 22.6 mL of the oxidant. The resultant
solution is neutralised with Na 2 CO3 , acidified with dilute
acetic acid and treated with excess KI. The liberated iodine
requires 11.3 mL of 0.05 M Na 2 S2 O3 solution for complete
reduction. Find out the mole ratio of Cu 2+ to C2 O2−
4 in the
compound. Write down the balanced redox reactions
involved in the above titrations.
(1991, 5M)
(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 together
and made up to 2 L. 30 mL of this acid mixture exactly
neutralise 42.9 mL of sodium carbonate solution containing
one gram of Na 2 CO3 ⋅ 10H2 O in 100 mL of water. Calculate
the amount in gram of the sulphate ions in solution.
(1985, 4M)
51. 2.68 × 10−3 moles of a solution containing an ion A n+ require
1.61 × 10−3 moles of MnO−4 for the oxidation of A n+ to A O−3
in acidic medium. What is the value of n ?
(1984, 2M)
46. A mixture of H2 C2 O4 (oxalic acid) and NaHC2 O4 weighing
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.
52. 4.08 g of a mixture of BaO and unknown carbonate MCO3
was heated strongly. The residue weighed 3.64 g. This was
dissolved 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)
Answers
Topic 1
1. (d)
5. (d)
9. (b)
2. (c)
6. (b)
10. (b)
3. (b)
7. (d)
11. (b)
4. (a)
8. (c)
12. (c)
13.
17.
21.
25.
29.
(b)
(b)
(a)
(d)
(a)
14.
18.
22.
26.
30.
(*)
(a)
(a)
(a)
(a)
15.
19.
23.
27.
31.
(d)
(c)
(d)
(d)
(a)
16.
20.
24.
28.
32.
(d)
(b)
(d)
(a)
(c)
8 Some Basic Concepts of Chemistry
33. (10.00)
34. (6.47kg)
35.
37. (4.14 g)
38. (0.4)
39.
isotope
41. (9)
42. (8)
43.
45. (5 × 10 −19 m 2 ) 46. (55.56 mol
L −1) 47. (70.91 × 10 6g) 48. (4.3 × 10 −3)
51. (10.42)
52. (1.7 g)
53.
56. (i) (37.92), (ii) (0.065), (iii) (7.73m)
59. (i) (0.0179 g), (ii) (10.6 %)
60.
(126 mg) 36. (495 g)
(6.023×10 24 ) 40. C-12
(8 mL)
44. (2)
(55.55 L) 54. (9.9 × 10 −3)
57. (a) (0.6), (b) (24)
(0.437)
62. (20 %)
Topic 2
1. (d)
2. (b)
3. (a)
4. (a)
5.
9.
13.
17.
21.
25.
28.
32.
37.
45.
51.
(d)
(d)
(b)
(a)
(b)
(0.11)
(2992)
(2)
(0.062 M)
(1:2)
(2)
6.
10.
14.
18.
22.
26.
29.
33.
38.
48.
52.
(*)
(c)
(c)
(b)
(c)
(6.15)
(b)
(3)
(1.334 V)
(16.67 mL)
(Ca)
7.
11.
15.
19.
23.
27.
30.
34.
42.
49.
(b)
(a)
(a)
(a)
(b)
(85)
7/3
(1008 g)
(85%)
(6.5gL −1)
8.
12.
16.
20.
24.
(c)
(d)
(d)
(a)
(a,b,d)
31.
36.
44.
50.
(5)
(8.096 mL)
(1.04)
(6.5376 g)
Hints & Solutions
Topic 1 Mole Concept
1.
Key Idea To find the mass of A and B in the given question,
mole concept is used.
given mass (w)
Number of moles( n) =
molecular mass (M )
(c) 4Fe(s) + 3O2(g ) → 2Fe2O3 (s)
244g
96g
96
g of O2 consumed = 0.43 g
⇒ 1 g of reactant =
224
(d) 2Mg(s) + O2(g ) → 2MgO(s)
48 g
32 g
32
g of O2 consumed = 0.67 g
48
So, minimum amount of O2 is consumed per gram of reactant
(Fe) in reaction (c).
⇒ 1 g of reactant =
Compound
Mass of A (g)
Mass of B (g)
AB2
MA
2 MB
A 2B 2
2 MA
2 MB
3. In eudiometry,
300 K
y
y

CxH y +  x +  O2 → x CO2 + H2O

1 atm
4
2
We know that,
Number of moles (n) =
given mass (w)
molecular mass (M )
n×M = w
Using equation (A), it can be concluded that
1 mol
…(A)
5(M A + 2M B ) = 125 × 10−3 kg
…(i)
10(2M A + 2M B ) = 300 × 10−3 kg
…(ii)
From equation (i) and (ii)
1 (M A + 2M B )  125 
=

2 (2M A + 2M B )  300
On solving the equation, we obtain
M A = 5 × 10−3
M B = 10 × 10−3
and
So, the molar mass of A (M A ) is
5 × 10−3 kg mol −1 and B (M B ) is 10 × 10−3 kg mol −1.
2. (a) C3H8 (g ) + 5O2 (g ) → 3CO2 (g ) + 4H2O(l )
44g
160g
⇒ 1g of reactant =
160
g of O2 consumed = 3.64 g
44
(b) P4 (s) + 5O2(g ) → P4O10 (s)
124g
160g
⇒1 g of reactant =
160
g of O2 consumed = 129
. g
124
1 mL
10 mL
y

 x +  mol

4
y

 x +  mL

4
y

 x +  × 10 mL

4
x mol
x mL
10x mL
Given, (i) VCO2 = 10x = 40 mL ⇒ x = 4
y

(ii) VO2 = 10  x +  mL = 55 mL

4
y

10  4 +  = 55

4
y × 10
⇒
40 +
= 55
4
10
4
⇒
y×
= 15 ⇒ y = 15 ×
=6
4
10
So, the hydrocarbon (CxH y ) is C4H6.
⇒
[Qx = 4]
4. Given, volume = 10 mL
Molarity = 1mM = 10−3 M
∴Number of millimoles = 10 mL × 10−3 M = 10−2
Number of moles = 10−5
Now, number of molecules
= Number of moles × Avogadro’s number
= 10−5 × 6 × 1023 = 6 × 1018
Some Basic Concepts of Chemistry 9
Surface area occupied by 6 × 1018 molecules = 0.24 cm2
∴Surface area occupied by 1 molecule
0.24
=
= 0.04 × 10−18 cm2
6 × 1018
As it is given that polar head is approximated as cube. Thus,
surface area of cube = a2, where
a =edge length
2
∴
a = 4 × 10−20 cm2
a = 2 × 10−10 cm = 2 pm
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 + 10 g of H2 is taken as a combination then
dihydrogen (H2 ) act as a limiting reagent in the reaction.
…(I)
N2 (g ) + 3H2 (g ) → 2NH3 (g )
2 × 14 g
28g
3 × 2g
6g
2(14 + 3) g
34g
28g N2 requires 6g H2 gas.
6g
× 56 g = 12g of H2
56g of N2 requires
28 g
12g of H2 gas is required for 56g of N2 gas but
only 10 g of H2 gas is present in option (a).
Hence, H2 gas is the limiting reagent.
In option (b), i.e. 35g of N2 + 8 g of H2.
As 28 g N2 requires 6g of H2.
6g
× 35 g H2 ⇒ 7.5 g of H2.
35g N2 requires
28 g
Here, H2 gas does not act as limiting reagent since 7.5 g of H2
gas is required for 35g of N2 and 8g of H2 is present in reaction
mixture. Mass of H2 left unreacted = 8 − 7.5 g of H2.
= 0.5 gof H2.
Similarly, in option (c) and (d), H2 does not act as limiting
reagent.
For 14 g of N2 + 4 g of H2.
As we know 28g of N2 reacts with 6g of H2.
6
14 g of N2 reacts with
× 14 g of H2 ⇒ 3g of H2.
28
For 28g of N2 + 6 g of H2, i.e. 28g of N2 reacts with 6g of H2
(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
In CH4 ,
mole of carbon = 1
mole of hydrogen = 4
∴ % of carbon by mole in CH4 =
1
× 100 = 20%
1+ 4
7. Mole fraction of solute
=
number of moles of solute + number of moles solvent
number of moles of solute
χ Solute
Given,
w Solute
n Solute
Mw Solute
=
=
w Solute
w
n Solute + nSolvent
+ Solvent
Mw Solute MwSolvent
wSolute = wNaOH = 8 g
Mw Solute = MwNaOH = 40g mol − 1
w Solvent = w H2 O = 18g
Mw Solvent = 18 g mol− 1
0.2
0.2
8 / 40
=
=
= 0167
.
∴ χ Solute = χ NaOH =
8 18 0.2 + 1
12
.
+
40 18
Moles of solute
Now, molality (m) =
Mass of solvent (in kg)
w Solute
8
Mw Solute
40
=
× 1000 =
× 1000
wSolvent (in g )
18
0.2
=
× 1000 = 11.11mol kg − 1
18
Thus, mole fraction of NaOH in solution and molality of the
solution respectively are 0.167 and 11.11 mol kg − 1 .
8. Concentration of H2O2 is expressed in terms of volume strength,
i.e. “volume of O2 liberated by H2O2 at NTP”. Molarity is
connected to volume strength as:
x
or x = Molarity × 11.2
Molarity (M) =
112
.
where, x = volume strength
So, for 1 M H2O2, x = 1 × 112
. = 112
.
Among the given options, 11.35 is nearest to 11.2.
Number of moles of solute (n)
9. Molarity =
Volume of solution (in L)
wB (g)
Also, n =
M B (gmol−1 )
w / MB
Molarity = B
∴
V
Given, wB = mass of solute ( B ) in g
M B = Gram molar mass of B (C12H22O11 ) = 342 g mol −1
Molarity = 01
. M
Volume (V ) = 2 L
w / 342
01
. = B
⇒ wB = 01
. × 342 × 2 g = 68.4 g
⇒
2
10. 2 C57 H110O6 (s) + 163 O2 (g ) → 110H2O(l ) + 114 CO2 (g )
Molecular mass of C57H110O6
= 2 × (12 × 57 + 1 × 110 + 16 × 6) g = 1780 g
Molecular mass of 110 H2O = 110 (2 + 16) = 1980 g
1780 g of C57H110O6 produced = 1980 g of H2O.
1980
445g of C57H110O6 produced =
× 445 g of H2O
1780
= 495 of H2O
Number of moles of solute
11. Molality (m) =
× 1000
Mass of solvent (in g)
10 Some Basic Concepts of Chemistry
Mass of solute (in g) × 1000
Molecular weight of solute 
× mass of solvent (in g)

wNa + × 1000 92 × 1000
=
=
= 4 mol kg − 1
M Na + × wH 2 O 23 × 1000
=
16. Given, initial strength of acetic acid = 0.06 N
Final strength = 0.042 N;
∴Initial millimoles of CH3COOH = 0.06 × 50 = 3
Final millimoles of CH3COOH = 0.042 × 50 = 2.1
∴ Millimoles of CH3COOH adsorbed = 3 − 2.1= 0.9 mmol
12. Given, abundance of elements by mass
oxygen = 614
. %, carbon = 22.9%, hydrogen = 10% and
nitrogen = 2.6%
Total weight of person = 75 kg
75 × 10
Mass due to 1 H =
= 7.5 kg
100
1
2
H atoms are replaced by H atoms,
Mass due to 2 H = (7.5 × 2) kg
∴ Mass gain by person = 7.5 kg
13. M 2CO3 + 2HCl → 2M Cl + H2O + CO2
1g
0.01186
mole
Number of moles of M 2CO3 reacted = Number of moles of
CO2 evolved
1
[M = molar mass of M 2CO3]
= 0.01186
M
1
M =
= 84.3 g mol − 1
0.01186
y
y
14. CxH y (g ) +  x +  O2 (g ) → xCO2 (g ) + H2O(l )

4
75 mL
30 mL
2
= 20% of 375 = 75 mL
O 2 used
Inert part of air = 80% of 375 = 300 mL
Total volume of gases = CO2 + Inert part of air
= 30 + 300 = 330 mL
x 30
=
⇒x=2
1 15
y
x+
4 = 75 ⇒ x + y = 5
1
15
4
⇒
x = 2, y = 12 ⇒ C2 H12
= 0.9 × 60 mg = 54 mg
(mO 2 )
17.
nO 2
nN 2
=
(M O 2 )
(mN 2 )
(M N 2 )
where, mO 2 = given mass of O2 , mN 2 = given mass of N2 ,
M O 2 = molecular mass of O2 , M N 2 = molecular mass of N2 , nO 2 =
number of moles of O2 , nN 2 = number of moles of N2
 mO  28 1 28 7
= 2
= ×
=
 mN 2  32 4 32 32
18. From the formula, M f =
M 1V1 + M 2V2
V1 + V2
Given, V1 = 750 mL, M 1 = 0.5 M
V2 = 250 mL, M 2 = 2 M
750 × 0.5 + 250 × 2 875
=
=
= 0.875 M
750 + 250
1000
19. Molarity =
Moles of solute
Volume of solution (L)
120
=2
60
Weight of solution = Weight of solvent + Weight of solute
= 1000 + 120 = 1120 g
1120 g
1
⇒
Volume =
×
= 0.973 L
1.15 g / mL 1000 mL / L
2.000
Molarity =
= 2.05M
⇒
0.973
Moles of urea =
20. From the given relative abundance, the average weight of Fe can be
calculated as
15. We know the molecular weight of C8 H7 SO3Na
= 12 × 8 + 1 × 7 + 32 + 16 × 3 + 23 = 206
we have to find, mole per gram of resin.
Volume = 50 mL
A=
54 × 5 + 56 × 90 + 57 × 5
= 55.95
100
21. 1.0 L of mixture X contain 0.01 mole of each [Co(NH3 )5 SO4 ]Br
∴ 1g of C8 H7 SO3Na has number of mole
weight of given resin
1
=
=
mol
Molecular, weight of resin 206
and [Co(NH3 )5 Br]SO4. Also, with AgNO3, only
[Co(NH3 )5 SO4 ]Br reacts to give AgBr precipitate as
Now, reaction looks like
With BaCl 2, only [Co(NH3 )5 Br]SO4 reacts giving BaSO4
precipitate as
[Co(NH3 )5 Br]SO4 + BaCl 2 → [Co(NH3 )5 Br]Cl 2 + BaSO4
2C8 H7 SO3Na + Ca 2+ → (C8 H7 SO3 )2 Ca + 2Na
Q 2 moles of C8 H7 SO3Na combines with 1 mol Ca 2+
1
∴1 mole of C8 H7 SO3Na will combine with mol Ca 2+
2
1
mole of C8 H7 SO3 Na will combine with
∴
206
1
1
1
mol Ca 2+ =
mol Ca 2+
×
412
2 206
[Co(NH3 )5 SO4 ]Br + AgNO3 → [Co(NH3 )5 SO4 ]NO3 + AgBr
1.0 mol
1.0 mol
1.0 mol
Excess
1 mol
Excess
Hence, moles of Y and Z are 0.01 each.
22. Number of atoms = Number of moles
Number of atoms in 24 g C =
× Avogadro’s number (N A)
24
× NA = 2NA
12
Some Basic Concepts of Chemistry 11
56
NA = NA
56
27
Number of atoms in 27 g of Al =
NA = NA
27
108
Number of atoms in 108 g of Ag =
NA = NA
108
Hence, 24 g of carbon has the maximum number of atoms.
Number of atoms in 56 g of Fe =
30. Number of molecules present in 36 g of water
36
× N A = 2N A
18
28
Number of molecules present in 28 g of CO =
× NA = NA
28
46
Number of molecules present in 46 g of C2H5OH =
× NA = NA
46
54
Number of molecules present in 54 g of N2O5 =
× N A = 0.5 N A
108
Here, NA is Avogadro’s number. Hence, 36 g of water contain
the largest (2NA ) number of molecules.
=
23. Mass of an electron = 9.108 × 10−31 kg
Q 9.108 × 10−31 kg = 1.0 electron
1
1031
1
electrons
=
×
∴ 1 kg =
−31
9.108 6.023 × 1023
9.108 × 10
1
=
× 108 mole of electrons
9.108 × 6.023
31. In a neutral atom, atomic number represents the number of
protons 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
= 6 + 2 × 8 = 22.
24. Phosphorus acid is a dibasic acid as :
O

H—P — OH only two replaceable hydrogens

OH
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 2 + 2Na 3PO4 → Ba 3 (PO4 )2 + 6NaCl
In this reaction, 3 moles of BaCl 2 combines with 2 moles of
Na 3PO4. Hence, 0.5 mole of of BaCl 2 require
2
× 0.5 = 0.33 mole of Na 3PO4.
3
Since, available Na 3PO4 (0.2 mole) is less than required mole
(0.33), it is the limiting reactant and would determine the
amount of product Ba 3 (PO4 )2.
Q 2 moles of Na 3PO4 gives 1 mole Ba 3 (PO4 )2
1
∴0.2 mole of Na 3PO4 would give × 0.2 = 0.1 mole Ba 3 (PO4 )2
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 :
Heat
Ag2CO3 (s) → 2Ag (s) + CO2 (g ) +
1
O (g )
2 2
MW = 276 g
2 × 108 = 216 g
Hence, 2.76 g of Ag2CO3 on heating will give
216
× 2.76 = 2.16g Ag as residue.
276
29. The balanced chemical reaction of zinc with sulphuric acid and
NaOH are
Zn + H2SO4 → ZnSO4 + H2 (g ) ↑
Zn + 2NaOH + 2H2O → Na 2[ Zn(OH)4 ] + H2 (g ) ↑
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.
32.
Weight of a compound in gram (w)
= Number of moles (n)
Molar mass (M )
Number of molecules (N )
=
Avogadro number (NA )
w (O2 ) N (O2 )
…(i)
⇒
=
32
NA
w (N2 ) N (N2 )
And
…(ii)
=
28
NA
Dividing Eq. (i) by Eq. (ii) gives
N (O2 ) w (O2 ) 28 1 28 7
= ×
=
=
×
N (N2 ) w (N2 ) 32 4 32 32
33. Molar mass of Na 2CO3⋅ xH2O .
(Atomic mass of Na = 23, C = 12, O = 16)
= 23 × 2 + 12 + 48 + 18x
= 46 + 12 + 48 + 18x
= (106 + 18x )
Equivalent weight of Na 2CO3⋅ xH2O
Molar mass M
=
=
= (53 + 9x )
2
n factor
[ Here, m = molar mass and n factor = 2]
Weight
Gram equivalent =
Equivalent weight
[Given, weight of Na 2CO3 ⋅ xH2O = 143
. g]
Hence, gram equivalent of
1.43
Na 2CO3 ⋅ xH2O =
53 + 9x
Gmeq
Normality =
Vlitre
143
.
01
. =
53 + 9x
.
01
As, volume = 100 mL = 0.1 L
12 Some Basic Concepts of Chemistry
So,
10−2 =
143
.
53 + 9x
36. Given,
O
53 + 9x = 143
9x = 90
x = 10.00
NH3/∆
Br2/KOH
Br2(3-eqiv.)
NaOBr
C
D
A
B
AcOH
H3O+ (60%)
(50%)
(100%)
(50%)
34. The equations of chemical reactions occurring during the process
are
In the presence of oxygen
HO
O
2PbS + 3O2 → 2PbO + 2SO2
By self reduction
207 g
So, 32 g of O 2 gives 207 g of Pb
207
g of Pb
1 g of O 2 will give
32
207
1000g of O 2 will give
× 1000 = 6468.75 g
32
= 6.46875 kg ≈ 6.47kg
Benzoic acid
(60%)
i.e.,
6 mol
(A)
Acetophenone
10 mol
(1) 2MnCl 2 + 5K 2S2O 8 + 8H2O →
2KMnO 4 + 4 K 2SO 4 + 6H2SO 4 + 4HCl
(2) 2KMnO 4 + 5H2C 2O 4 + 3H2SO 4 →
K 2SO 4 + 2MnSO 4 + 8H2O +10CO 2
Given, mass of oxalic acid added = 225mg
225
So, millimoles of oxalic acid added =
= 2.5
90
Now from equation 2
Millimoles of KMnO 4 used to react with oxalic acid=1 and
Millimoles of MnCl 2 required initially=1
∴ Mass of MnCl 2 required initially = 1 × (55 + 71) = 126mg
Alternative Method
O
NH3/∆
NaOBr
H3O+
Benzamide
(50%)
i.e.,
3 mol
(B)
NH2
Br2/KOH
NH2
Br
Br
Br2(3-eqiv.)
AcOH
Br
2,4,6-tribromo aniline
(100%)
i.e.,
1.5 mol
(D)
Aniline
(50%)
i.e.,
1.5 mol
(C)
NH2
Br
Br
So, 1.5 mol of
are produced from
35. The balanced equations are
m moles of MnCl 2 = m moles of KMnO 4 = x (let)
and M eq of KMnO 4 = M eq of oxalic acid
225
So,
x×5=
×2
90
Hence, x = 1
∴ m moles of MnCl 2 = 1
Hence mass of MnCl 2 = (55 + 71) × 1 = 126 mg.
H2N
O
…(i)
2PbO + PbS → 3Pb + SO2
Thus 3 moles of O2 produces 3 moles of Pb
i.e. 32 × 3 = 96 g of O 2 produces 3 × 207 = 621 g of Pb
So 1000 g (1kg) of oxygen will produce
621
× 1000 = 6468.75 g
96
= 6.4687 kg ≈ 6.47 kg
Alternative Method
From the direct equation,
PbS + O2 → Pb + SO2
32 g
The products formed are
Br
10 moles of acetophenone.
NH2
Br
Br
Molar mass of
= 240 + 14 + 4 + 72 = 330
Br
NH2
Br
Br
Hence, amount of
produced is 330 × 1.5 = 495 g
Br
37. Molar mass of CuSO4 ⋅ 5H2O
= 63.5 + 32 + 4 × 16 + 5 × 18
= 249.5 g
Also, molar mass represents mass of Avogadro number of
molecules in gram unit, therefore
Q 6.023 × 1023 molecules of CuSO4 ⋅ 5H2O weigh 249.5 g
249.5
× 1022 = 4.14 g
∴ 1022 molecules will weigh
6.023 × 1023
Some Basic Concepts of Chemistry 13
Number of moles of solute
Volume of solution in litre
Weight of solute
1000
=
×
Molar mass
Volume in mL
3 1000
=
×
= 0.4 M
30 250
38. Molarity =
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 H2O contain = 10 × 6.023 × 1023
= 6.023 × 1024 electrons.
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.
1
1 amu (u) =
× weight of an atom of C-12 isotope.
12
w
w
41. Moles of solute, n1 = 1 ; Moles of solvent, n2 = 2
m1
m2
χ 1 (solute) = 0.1and χ 2 (solvent) = 0.9
χ 1 n1 w1 m2 1
= =
⋅
=
χ 2 n2 m1 w2 9
Solute (moles) w1 × 1000 × 2
Molarity =
=
Volume (L)
m1 (w1 + w2 )
∴
Total mass of solution  w1 + w2 
Note Volume =
=
 mL
 2 
Density
Molality =
Given,
hence,
∴
∴
42.
Solute (moles) w1 × 1000
=
Solvent (kg)
m1 × w2
molarity = molality
2000 w1
1000 w1
=
m1 (w1 + w2 )
m1 w2
w2
1
=
⇒ w1 = w2 = 1
w1+ w2 2
w1 m2 1
m (solute)
= ⇒ 1
=9
m1 w2 9
m2 (solvent)
PLAN This problem can be solved by using concept of conversion of
molarity into molality.
Molarity = 3.2 M
Let volume of solution = 1000 mL = Volume of solvent
Mass of solvent = 1000 × 0.4 = 400 g
Since, molarity of solution is 3.2 molar
∴
n solute = 3.2 mol
3.2
Molality (m) =
=8
400 / 1000
Hence, correct integer is (8).
43. Mass of HCl in 1.0 mL stock solution
= 1.25 ×
29.2
= 0.365 g
100
Mass of HCl required for 200 mL 0.4 M HCl
200
=
× 0.4 × 36.5 = 0.08 × 36.5 g
1000
∴ 0.365 g of HCl is present in 1.0 mL stock solution.
0.08 × 36.5
= 8.0 mL
0.08 × 36.5 g HCl will be present in
0.365
44. Partial pressure of N2 = 0.001 atm,
T = 298 K, V = 2.46 dm 3.
From ideal gas law : pV = nRT
pV 0.001 × 2.46
n(N2 ) =
=
= 10−7
RT
0.082 × 298
⇒ Number of molecules of N2 = 6.023 × 1023 × 10−7
= 6.023 × 1016
Now, total surface sites available
= 6.023 × 1014 × 1000 = 6.023 × 1017
20
Surface sites used in adsorption =
× 6.023 × 1017
100
= 2 × 6.023 × 1016
⇒ Sites occupied per molecules
2 × 6.023 × 1016
Number of sites
=
= 2
=
Number of molecules
6.023 × 1016
45. Initial millimol of CH3COOH = 100 × 0.5 = 50
millimol of CH3COOH remaining after adsorption
= 100 × 0.49 = 49
⇒ millimol of CH3COOH adsorbed = 50 – 49 = 1
⇒ number of molecules of CH3COOH adsorbed
1
=
× 6.023 × 1023 = 6.023 × 1020
1000
3.01 × 102
⇒ Area covered up by one molecule =
6.02 × 1020
= 5 × 10−19 m 2
46. Mass of 1.0 L water = 1000 g
⇒
Molarity =
1000
= 55.56 mol L−1
18
47. Volume of one cylinderical plant virus = πr2l
= 3.14 (75 × 10−8 )2 × 5000 × 10−8 cm 3 = 8.83 × 10−17 cm 3
Volume of a virus
⇒ Mass of one virus =
Specific volume
=
8.83 × 10−17 cm 3
= 1.1773 × 10−16 g
0.75 cm 3 g−1
⇒ Molar mass of virus
= Mass of one virus × Avogadro’s number
= 1.1773 × 10−16 × 6.023 × 1023 g
= 70.91 × 106 g
48. Molar mass of Glauber’s salt (Na 2SO4 ⋅ 10H2O)
= 23 × 2 + 32 + 64 + 10 × 18 = 322g
14 Some Basic Concepts of Chemistry
⇒ Mole of Na 2SO4 ⋅ 10H2O in 1.0 L solution =
80.575
= 0.25
322
⇒ Molarity of solution = 0.25 M
Also, weight of 1.0 L solution = 1077.2 g
weight of Na 2SO4 in 1.0 L solution = 0.25 × 142 = 35.5 g
⇒ Weight of water in 1.0 L solution = 1077.2 – 35.5 = 1041.7 g
0.25
× 1000 = 0.24 m
⇒ Molality =
1041.7
Mole of Na 2SO4
Mole fraction of Na 2SO4 =
Mole of Na 2SO4 + Mole of water
0.25
= 4.3 × 10−3.
=
1041.7
0.25 +
18
49. Compound B forms hydrated crystals with Al 2 (SO4 )3. Also, B is
formed with univalent metal on heating with sulphur. Hence,
compound B must has the molecular formula M 2SO4 and
compound A must be an oxide of M which reacts with sulphur to
give metal sulphate as
A + S → M 2SO4
B
Q 0.321 g sulphur gives 1.743 g of M 2SO4
∴ 32.1 g S (one mole) will give 174.3 g M 2SO4
Therefore, molar mass of M 2SO4 = 174.3 g
⇒ 174.3 = 2 × Atomic weight of M + 32.1 + 64
⇒ Atomic weight of M = 39, metal is potassium (K)
K2SO4 on treatment with aqueous Al 2 (SO4 )3 gives potash-alum.
K2SO4 + Al 2 (SO4)3 + 24H2O → K2SO4Al 2 (SO4)3 ⋅ 24H2O
B
C
If the metal oxide A has molecular formula MOx, two moles of it
combine with one mole of sulphur to give one mole of metal
sulphate as
2KOx + S → K2SO4
⇒
x = 2, i.e. A is KO2.
50. The reaction involved is
3Pb(NO3 )2 + Cr2 (SO4 )3 → 3PbSO4 (s) ↓ + 2Cr(NO3 )3
millimol of Pb(NO3 )2 taken = 45 × 0.25 = 11.25
millimol of Cr2 (SO4 )3 taken = 2.5
Here, chromic sulphate is the limiting reagent, it will determine
the amount of product.
Q 1 mole Cr2 (SO4 )3 produces 3 moles PbSO4.
∴ 2.5 millimol Cr2 (SO4 )3 will produce 7.5 millimol PbSO4.
Hence, mole of PbSO4 precipitate formed = 7.5 × 10−3
Also, millimol of Pb(NO3 )2 remaining unreacted
11.25 – 7.50 = 3.75
⇒ Molarity of Pb(NO3 )2 in final solution
millimol of Pb(NO3 )2 3.75
= 0.054 M
=
=
Total volume
70
Also, millimol of Cr(NO3 )2 formed
= 2 × millimol of Cr2 (SO4 )3 reacted
5
= 0.071 M
⇒ Molarity of Cr(NO3 )2 =
70
51. 93% H2SO4 solution weight by volume indicates that there is
93 g H2SO4 in 100 mL of solution.
If we consider 100 mL solution, weight of solution = 184 g
Weight of H2O in 100 mL solution = 184 – 93 = 91 g
Moles of solute
⇒ Molality =
× 1000
Weight of solvent (g)
93 1000
= 10.42
=
×
98
91
52. Heating below 600°C converts Pb(NO3 )2 into PbO but to
NaNO3 into NaNO2 as
∆
Pb(NO3 )2 → PbO(s) + 2NO2 ↑ +
MW :
330
∆
222
NaNO3 → NaNO2 (s) +
MW :
85
1
O ↑
2 2
1
O ↑
2 2
69
28
Weight loss = 5 ×
= 1.4 g
100
⇒ Weight of residue left = 5 – 1.4 = 3.6 g
Now, let the original mixture contain x g of Pb(NO3 )2.
Q 330 g Pb(NO3 )2 gives 222 g PbO
222 x
g PbO
∴ x g Pb(NO3 )2 will give
330
Similarly, 85 g NaNO3 gives 69 g NaNO2
69 (5 − x )
g NaNO2
⇒ (5 – x) g NaNO3 will give
85
222 x 69 (5 − x )
= 3.6 g
⇒ Residue :
+
330
85
Solving for x gives, x = 3.3 g Pb(NO3 )2 ⇒ NaNO3 = 1.7 g.
53. Reactions involved are
C2H6 + Br2 → C2H5Br + HBr
2C2H5Br + 2Na → C4H10 + 2NaBr
Actual yield of C4H10 = 55 g which is 85% of theoretical yield.
55 × 100
= 64.70 g
⇒ Theoretical yield of C4H10 =
85
Also, 2 moles (218 g) C2H5Br gives 58 g of butane.
⇒ 64.70 g of butane would be obtained from
2
× 64.70 = 2.23 moles C2H5Br
58
Also yield of bromination reaction is only 90%, in order to have
2.23 moles of C2H5Br, theoretically
2.23 × 100
= 2.48 moles of C2H5Br required.
90
Therefore, moles of C2H6 required = 2.48
⇒ Volume of C2H6 (NTP) required = 2.48 × 22.4 = 55.55 L.
34.2
54. Moles of sugar =
= 0.1
342
Moles of water in syrup = 214.2 – 34.2 = 180 g
Moles of solute
Therefore, (i) Molality =
× 1000
Weight of Solvent (g)
0.1
=
× 1000 = 0.55
180
Mole of sugar
(ii) Mole fraction of sugar =
Mole of sugar + Mole of water
0.1
=
= 9.9 × 10−3
0.1 + 10
Some Basic Concepts of Chemistry 15
55. From the given elemental composition, empirical formula can
be derived as :
Element
Weight %
Mole %
C
69.77
5.81
H
11.63
11.63
Simple ratio
5
10
O
18.60
1.1625 (obtained by
dividing from M )
1
Hence, empirical formula is C5H10O and empirical formula
weight is 86.
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
a ketone.
Also, it gives positive iodoform test, it must be a methyl ketone.
O

C3H7 — C — CH3
Based on the above information, the compound may be one of
the following :
O
CH3 O



CH3CH2CH2— C — CH3 or CH3 — CH— C — CH3
2-pentanone
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 2S2O3 = 3 × 158 = 474 g
474
⇒ Weight percentage of Na 2S2O3=
× 100 = 37.92
1250
(ii) Weight of H2O in 1 L solution = 1250 − 474 = 776 g
3
Mole fraction of Na 2S2O3 =
= 0.065
776
3+
18
3×2
(iii) Molality of Na + =
× 100 = 7.73 m
776
57. (a) After passing through red-hot charcoal, following reaction
occurs
C(s) + CO2 (g ) → 2CO(g )
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)
⇒
x = 0.6
Hence, original 1.0 L mixture has 0.4 L CO and 0.6 L of CO2 ,
i.e. 40% CO and 60% CO2 by volume.
(b) According to the given information, molecular formula of
the compound is M 3N2. Also, 1.0 mole of compound has 28 g
of nitrogen. If X is the molar mass of compound, then :
28
X ×
= 28
100
⇒
X = 100 = 3 × Atomic weight of M + 28
72
⇒ Atomic weight of M =
= 24
3
58. In the present case, V ∝ n (Q all the volumes are measured
under identical conditions of temperature and pressure) Hence,
the reaction stoichiometry can be solved using volumes as :
y
y

CxH y (g ) +  x +  O2 (g ) → x CO2 (g ) + H2O (l )

4
2
volume of CO2 (g ) + O2 (g ) (remaining unreacted) = 25
⇒ Volume of CO2 (g ) produced
= 10 mL (15 mL O2 remaining)
Q 1 mL CxH y produces x mL of CO2
∴ 5 mL CxH y will produce 5x mL of CO2 = 10 mL ⇒ x = 2
y

Also, 1 mL CxH y combines with  x +  mL of O2

4
y

5 mL CxH y will combine with 5  x +  mL of O2

4
y

⇒ 5  x +  = 15 (15 mL of O2 out of 30 mL)

4
(remaining unreacted)
⇒ y = 4, hence hydrocarbon is C2H4.
59. Oxides of sodium and potassium are converted into chlorides
according to following reactions :
Na 2O + 2HCl → 2NaCl + H2O
K2O + 2HCl → 2KCl + H2O
Finally all the chlorides of NaCl and KCl are converted into
AgCl, hence
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.
x
0.118 − x 0.2451
⇒
+
=
58.5
74.5
143.5
Solving for x gives x = 0.0338 g (mass of NaCl)
⇒
Mass of KCl = 0.118 – 0.0338 = 0.0842 g
1
Also, moles of Na 2O = × moles of NaCl
2
1 0.0338
× 62 = 0.0179 g
⇒ Mass of Na 2O = ×
2
58.5
1 0.0842
Similarly, mass of K2O = ×
× 94 = 0.053 g
2
74.5
0.0179
Mass % of Na 2O =
⇒
× 100 = 3.58 %
0.5
0.053
Mass % of K2O =
× 100 = 10.6 %
0.5
60. From the vapour density information
Molar mass = Vapour density × 2 (Q Molar mass of H2 = 2)
= 38.3 × 2 = 76.6
Now, let us consider 1.0 mole of mixture and it contains
x mole of NO2.
⇒
46 x + 92 (1 − x ) = 76.6 ⇒ x = 0.3348
100
Also, in 100 g mixture, number of moles =
76.6
100
⇒ Moles of NO2 in mixture =
× 0.3348 = 0.437
76.6
16 Some Basic Concepts of Chemistry
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
Σ Percentage of an isotope × Atomic weight
100
10.01x + 11.01 (100 − x)
⇒ x = 20%
⇒ 10.81 =
100
=
Therefore, natural boron contains 20% (10.01) isotope and 80%
other isotope.
Topic 2 Equivalent Concept, Neutralisation
and Redox Titration
1. In disproportionation reactions, same element undergoes
The difference in the volume of NaOH solution between the end
point and the equivalence point is not significant for most of the
commonly used indicators as there is a large change in the pH
value around the equivalence point. Most of them change their
colour 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,
0.27
× 10 = 0.027 g
100
Density of fatty acid, d = 0.9 g cm−3
m=
∴Volume of the fatty acid over the watch glass,
m 0.027
V = =
= 0.03 cm3
d
0.9
Let, height of the cylindrical monolayer = h cm
Q Volume of the cylinder = Volume of fatty acid
oxidation as well as reduction.
Reduction
e.g.
+2
+1
2CuBr
0
h cm
CuBr2 +Cu
Oxidation
Here, CuBr get oxidised to CuBr2 and also it get reduced to Cu.
Other given reactions and their types are given below.
10 cm
⇒
V = πr 2 × h
⇒
h=
Reduction
+7
–
2 MnO4
–
+10I + 16 H
+
+2
2Mn
+ 5I2 + 8H2O
= 1 × 10−4cm
Oxidation
In the given reaction, MnO−4 get oxidised to Mn 2+ and I− get
reduced to I2. It is an example of redox reaction. The reaction
takes place in acidic medium.
2KMnO4 → K 2MnO4 + 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
FeC2O4 , Fe2 (C2O4 )3 FeSO4 and
Fe2 (SO4 )3 in acidic medium with KMnO4 is as follows :
…(i)
FeC2O4 + KMnO4 → Fe3 + + CO2 + Mn 2+
Fe2(C2O4 )3 + KMnO4 → Fe3 + + CO2 + Mn 2+
FeSO4 + KMnO4 → Fe
3+
+ SO24−
+ Mn
2+
…(ii)
…(iii)
Change in oxidation number of Mn is 5. Change in oxidation
number of Fe in (i), (ii) and (iii) are +3, + 6, + 1, respectively.
neq KMnO 4 = neq [ FeC 2O 4 + Fe 2 ( C 2O 4 ) 3 + FeSO 4 ]
∴
n × 5 = 1× 3 + 1× 6 + 1× 1
n=2
5. Given, W Ca (HCO 3) 2 = 0.81 g
W Mg (HCO ) = 0.73 g
3 2
M Ca (HCO ) = 162 g mol −1,
3 2
M Mg(HCO 3 ) 2 = 146 mol −1
V H 2 O = 100mL
V(mL)
Now,
In this case, both the titrants are completely ionised.
⊕
= 1 × 10−6 m
4. The oxidation of a mixture of one mole of each of
pH
HCl + NaOH
0.03 cm3
V
=
πr2 3 × (10)2 cm2
+ −
- 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,[OHÈ ] is rapidly reduced. After this point [OH È]
decreases rapidly and pH of the solution remains fairly constant.
Thus, there is an inflexion point at the equivalence point.
neq (CaCO3) = neq [Ca(HCO3 )2 ]+ neq [Mg(HCO3 )2 ]
W
0.81
0.73
×2=
×2+
×2
100
162
146
W
= 0.005 + 0.005
∴
100
W = 0.01 × 100 = 1
1
Thus, hardness of water sample =
× 106 = 10,000 ppm
100
Some Basic Concepts of Chemistry 17
12x 6
= (given and molar mass of C = 12, H = 1)
y
1
6. The reaction takes place as follows,
H2C2O4 + 2NaOH → Na 2C2O4 + 2H2O
Now, 50 mL of 0.5 M H2C2O4 is needed to neutralize 25 mL of
NaOH.
∴ Meq of H2C2O4 = Meq of NaOH
50 × 0.5 × 2 = 25 × M NaOH× 1
M NaOH = 2M
Number of moles
Now,
molarity =
Volume of solution (in L)
Weight / molecular mass
=
Volume of solution (in L)
1000
w
2 = NaOH ×
40
50
2 × 40 × 50
wNaOH =
= 4g
1000
Thus, (*) none option is correct.
y 
y

Number of oxygen atoms required = 2 × x +
= 2x +

4  
2 
1
y 
y
Now given, z = 2x +
= x+
2 
2  
4 
Here we consider x and y as simplest ratios for C and H so now
putting the values of x and y in the above equation.
y 
2

z= x+
= 1+
= 1.5

4  
4 
Thus, the simplest ratio figures for x , y and z are x = 1, y = 2 and
z = 15
.
Now, put these values in the formula given i.e.
CxH yOz = C1H2O1.5
So, empirical formula will be [C1H2O1.5 ] × 2 = C2H4O3
7. The reaction of HCl with Na 2CO3 is as follows:
2HCl + Na 2CO3 → 2NaCl + H2O + CO2
We know that, M eq of HCl = M eq of Na 2CO3
25
30
× 1 × M HCl =
× 0.1 × 2
1000
1000
30 × 0.2 6
M
M HCl =
=
25
25
The reaction of HCl with NaOH is as follows:
10. Methyl orange show Pinkish colour towards more acidic
medium and yellow orange colour towards basic or less acidic
media. Its working pH range is
Pinkish
Red
NaOH + HCl → NaCl + H2O
Also, M eq of HCl = M eq of NaOH
6
30
V
×1×
=
× 0.2 × 1
25
1000 1000
V = 25mL
8. Reaction of oxalate with permanganate in acidic medium.
5C2O24− + 2MnO−4 → 10CO2 + 2Mn 2+ + 8H2O
n-factor :
Number of mole
⇒ 5C2O42–
(4 − 3 ) × 2 = 2
5
(7 − 2 ) = 5
2
10
−
ions transfer 10e to produce 10 molecules of CO2.
So, number of electrons involved in producing 10 molecules of
CO2 is 10. Thus, number of electrons involved in producing 1
molecule of CO2 is 1.
9. We can calculate the simplest whole number ratio of C and H
from the data given, as
Element
Now, after calculating this ratio look for condition 2 given in the
question i.e. quantity of oxygen is half of the quantity required
to burn one molecule of compoundC xH y completely to CO2 and
H2O. We can calculate number of oxygen atoms from this as
consider the equation.
y
y

CxH y + x + O2 → xCO2 + H2O

4 
2
3.9 –4.5
Yellow
orange
Weak base have the pH range greater than 7. When methyl
orange is added to this weak base solution it shows yellow
orange colour.
Now when this solution is titrated against
strong acid the pH move towards more acidic
1
range and reaches to end point near 3.9 where
NH2
yellow orange colour of methyl orange HN
In conjugation
changes to Pinkish red resulting to similar
change in colour of solution as well.
11. H2O2 acts as an oxidising as well as reducing agent, because
oxidation number of oxygen in H2O2 is −1. So, it can be oxidised
to oxidation state 0 or reduced to oxidation state –2.
H2O2 decomposes on exposure to light. So, it has to be stored in
plastic or wax lined glass bottles in dark for the prevention of
exposure. It also has to be kept away from dust.
12. n-factor of dichromate is 6.
C
Relative
mass
6
Molar
mass
12
H
1
1
Relative
mole
6
= 0.5
12
1
=1
1
Simplest whole
number ratio
0.5
=1
0.5
1
=2
0.5
Alternatively this ratio can also be calculated directly in the
terms of x and y as
Also, n-factor of Mohr’s salt is 1 as :
O. A
FeSO4 (NH4 )2 SO4⋅ 6H2O → Fe3+
Mohr’s salt
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.
18 Some Basic Concepts of Chemistry
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,
1
n-factor MnSO4 → Mn 2O3 1 (+ 2 → + 3)
2
2 (+ 2 → + 4)
MnSO4 → MnO2
−
5 (+ 2 → + 7)
MnSO4 → MnO4
13. The following reaction occur between S2O32− and Cr2O2−
7 :
26H +
+
→ 6SO24− + 8Cr 3+ + 13H2O
Change in oxidation number of Cr2O72− per formula unit is 6 (it is
always fixed for Cr2O2−
7 ).
+
3S2O32−
4Cr2O72−
Molecular weight
6
14. It is an example of disproportionation reaction because the same
species (ClO− ) is being oxidised to ClO−3 as well as reduced to Cl − .
Hence, equivalent weight of K2Cr2O7 =
15. Oxalic acid dihydrate H2C2O4 ⋅ 2H2O : mw = 126
It is a dibasic acid, hence equivalent weight = 63
6.3 1000
= 0.4 N
Normality =
×
⇒
63
250
⇒
N 1V1 = N 2V2
⇒
Hence,
0.1 × V1 = 0.4 × 10
V1 = 40 mL
16. In MnO−4 , oxidation state of Mn is +7
In Cr(CN)63− , oxidation state of Cr is +3
In NiF62− , Ni is in + 4 oxidation state.
MnSO4 → MnO42−
24.
PLAN This problem includes concept of redox reaction. A redox
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 H2SO4.
Oxidation half-reaction, 2 I − → I2 + 2 e −
…(i)
Here, I − is converted into I2. Oxidation number of I is increasing
from –1 to 0 hence, this is a type of oxidation reaction.
Reduction half-reaction
6H+ + ClO3− + 6e− → Cl − + 3H2O
l
In CrO2Cl 2, oxidation state of Cr is +6.
17. In S8 , oxidation number of S is 0, elemental state.
In S2F2, F is in – 1 oxidation state, hence S is in + 1 oxidation
state.
In H2S, H is in +1 oxidation state, hence S is in – 2 oxidation
state.
19. The balanced chemical reaction is :
2MnO−4 + 5SO23− + 6H+ → 2Mn 2+ + 5SO24− + 3H2O
Q 5 moles SO2−
3 reacts with 2 moles of KMnO 4
2
mole KMnO4.
∴ 1 mole of SO2−
3 will react with
5
Hence, the coefficients of reactants in balanced reaction are 2, 5
and 16 respectively.
21. Volume strength of H2O2 = Normality × 5.6 = 1.5 × 5.6 = 8.4 V
l
⇒
23. Equivalent weight in redox system is defined as :
Molar mass
E=
n-factor
Stoichiometric coefficient of HSO−4 is 6.
Hence, option (a), (b) and (d) are correct.
25. Oxalic acid solution titrated with NaOH solution using
phenolphthalein as an indicator.
H2 C2 O4 + 2NaOH → Na 2 C2 O4 + 2H2 O
Equivalent of H2C2O4 reacted = Equivalent of NaOH reacted
9 × M(NaOH) × 1
5 × 2 × 01
.
=
=
1000
1000
1
M (NaOH) = = 0.11
9
26. Aluminium reacts with sulphuric acid to form aluminium
sulphate and hydrogen.
2Al

 5. 4
= 0. 2 mol 


 27
+
3H2 SO4
 50 × 5

= 0. 25 mol 

 1000

→ Al 2 (SO4 )3 + 3H2
H2SO 4 is limiting reagent and moles of H2 (g ) produced
= 0.25 mol
Using ideal gas equation,
22. In Ba(H2PO2 )2, oxidation number of Ba is +2. Therefore,
H2PO2− : 2 × (+1) + x + 2 × (−2) = − 1
x=+1
Here, H2 O releases as a product. Hence, option (d) is
correct.
6I− + ClO−3 + 6H2SO4 → Cl − + 3I2 + 3H2O + 6HSO−4
20. The balanced redox reaction is :
2MnO4− + 5C2O24− + 16H + → 2Mn 2+ + 10CO2 + 16H2O
…(ii)
Multiplying equation (i) by 3 and adding in equation (ii)
6I− + ClO3− + 6H+ → Cl − + 3I2 + 3H2O
18. The balanced redox reaction is :
3MnO−4 + 5FeC2O4 + 24H+ → 3Mn 2+ + 5Fe3+
+ 10CO2 + 12H2O
Q 5 moles of FeC2O4 require 3 moles of KMnO4
3
∴ 1 mole of FeC2O4 will require mole of KMnO4.
5
4 (+ 2 → + 6)
Therefore, MnSO4 converts to MnO2.
⇒
pV = nRT
0.25 × 0.082 × 300
V =
⇒ 6.15 L
1 atm
27. Given, volume of solution = 20.0 mL
Impure sample of H2O2 = 0.2 g
Mass of KMnO4 = 0.316 g
Some Basic Concepts of Chemistry 19
Impure H2O2 react with KMnO4 (acidic)
+7
−1
+2
KMnO4 + H2O2 → Mn
30. If x is the oxidation state of Cu then :
3 + 2 × 2 + 3x + 7 × (− 2) = 0 ⇒ x = 7/ 3
0
+ O2
31. Na2S4O6 is a salt of H2S4O6 which has the following structure
KMnO 4 acts as an oxidising agent,
+7
+2
Mn + 5e− → Mn
−1
O2
0
−
→ O2 + 2e
(valency factor = 2)
We have to compare both KMnO4 and H2O2.
(Mass equivalent) H O2 = (Mass equivalent) KMnO4
2
Weight
× 1000
molecular weight / valence factor
weight
=
× 1000
molecular weight / valence factor
(weight)H 2 O 2
34 / 2
2O2
=
=
O
(v)
S
OH
O
⇒ Difference in oxidation number of two types of sulphur = 5
figure cannot be greater than the same in either of them being
manipulated.
34. The balanced reaction is
(Pure)H2 O2
(Impure)H2 O2
× 100
017
.
× 100 = 85%
0.2
(i) ( NH4)2SO 4 + Ca(OH)2 → CaSO 4 ⋅ 2H2O +
1 mol
172 g
2NH3
2 mol
(2 × 17) = 34 g
(ii) NiCl 2 ⋅ 6H2O + 6NH3 → [Ni(NH3)6] Cl 2 + 6H2O
6 mol
102 g
S
33. Average titrate value is 25.15, but the number of significant
28. Balanced equations of reactions used in the problem are as follows
1 mol
238 g
S
Cl = − 1 to + 7
N = − 3 to + 5
P = − 3 to + 5
Sn = + 2, + 4
Tl = + 1, + 3 (rare but does exist)
Ti = + 2, + 3, + 4
0.316
× 1000
158 / 5
0.316
34
=
×5×
158
2
26.86
=
158
1 mol
132 g
S
O = O− , O2− , O2+ ;
(weight)H2 O2 = 017
. g
(Purify) H
HO
O
(0)
32. Only F and Na show only one non-zero oxidation state.
× 1000 =
(weight)H2 O2
O
(valency factor = 5)
1 mol
232 g
Now, in Eq. (i)
if, 1584 g of ammonium sulphate is used.
1584
i.e., 1584 g (NH4 )2 SO4 =
= 12 mol
132
So, according to the Eq. (i) given above 12 moles of (NH4 )2 SO4
produces
(a) 12 moles of gypsum
(b) 24 moles of ammonia
Here, 12 moles of gypsum = 12 × 172 = 2064 g
and
24 moles of NH3 = 24 × 17 = 408g
Further, as given in question,
24 moles of NH3 produced in reaction (i) is completly utilised
by 952g or 4 moles of NiCl 2 ⋅ 6H2O to produce 4 moles of
[Ni(NH3 )6 ] Cl 2.
So, 4 moles of [Ni(NH3 )6 ] Cl 2 = 4 × 232 = 928gms
Hence, total mass of gypsum and nickel ammonia coordination
compound [Ni(NH3 )6 ] Cl 2 = 2064 + 928 = 2992
6CaO + P4O10 → 2Ca 3 (PO4 )2
852
Moles of P4O10 =
=3
284
Moles of CaO required = 3 × 6 = 18
Mass of CaO required = 18 × 56 = 1008 g
35. Meq of oxalate = 10 × 0.2 × 2 = 4
Meq of MnO2 formed = Meq of oxalate = 4
Meq of KMnO4 in 20 mL = 4
Normality of H2O2 × 20 = 4
⇒
Normality of H2O2 = 0.20 N
⇒
0.20
Molarity of H2O2 =
⇒
= 0.10 M
2
The balanced reactions are
2KMnO4 + 5H2O2 + 3H2SO4 → 2MnSO4 + 5O2
+ K2SO4 + 8H2O
MnO2 + Na 2C2O4 + 2H2SO4 → MnSO4 + Na 2SO4
+ 2CO2 + 2H2O
36. The balanced chemical reaction is
CuCO3 + H2SO4 → CuSO4 + H2O + CO2
0.5 × 1000
= 4.048
millimol of CuCO3 =
123.5
⇒ Millimol of H2SO4 required = 4.048
Millimol = Molarity × Volume (in mL)
Q
4.048
Volume =
⇒
= 8.096 mL
0.50
37. The redox reaction involved are
29. Both assertion and reason are factually true but the reason does
IO−3 + 5I− + 6H+ → 3I2 + 3H2O
not exactly explain the assertion. The correct explanation is,
methyl orange and phenolphthalein changes their colour at
different pH.
I2 + 2S2O32− → 2I− + S4O62−
20 Some Basic Concepts of Chemistry
0.1
× 1000 = 0.467
214
⇒ millimol of I2 formed = 3 × 0.467 = 1.4
⇒ millimol of Na 2S2O3 consumed = 2 × 1.4 = 2.8
2.8
⇒ Molarity of Na 2S2O3 =
= 0.062 M
45
4
× 100= 20%
20
Vol % of He = 30%
millimol of KIO3 used =
Vol % of CH4 =
41. The redox reaction involved is :
H2O2 + 2I− + 2H+ → 2H2O + I2
38. Meq of H2O2 = Meq of I2 = Meq of Na 2S2O3
If N is normality of H2O2, then
N × 25 = 0.3 × 20 ⇒ N = 0.24
Volume
strength
⇒
= N × 5.6 = 1.334 V
39. Let the original sample contains x millimol of Fe3O4 and
y millimol of Fe2O3. In the first phase of reaction,
Fe3O4 + I− → 3Fe2+ + I2 (n-factor of Fe3O4 = 2)
−
2+
Fe2O3 + I → 2Fe
+ I2 (n-factor of Fe2O3 = 2)
⇒ Meq of I2 formed = Meq (Fe3O4 + Fe2O3 )
= Meq of hypo required
…(i)
⇒
2x + 2 y = 11 × 0.5 × 5 = 27.5
Now, total millimol of Fe2+ formed = 3x + 2 y. In the reaction
Fe2+ + MnO−4 + H+ → Fe3+ + Mn 2+
n-factor of Fe2+ = 1
⇒
Meq of
MnO−4
2+
= Meq of Fe
⇒
3x + 2 y = 12.8 × 0.25 × 5 × 2 = 32
Solving Eqs. (i) and (ii), we get
x = 4.5
and
y = 9.25
4.5
Mass of Fe3O4 =
⇒
× 232 = 1.044 g
1000
1.044
% mass of Fe3O4 =
× 100 = 34.80%
3
9.25
Mass of Fe2O3 =
× 160 = 1.48 g
1000
1.48
% mass of Fe2O3 =
× 100 = 49.33%
3
…(ii)
40. The reaction involved in the explosion process is
CO(g ) +
x mL
1
O (g ) → CO2 (g )
2 x2
2
mL
x mL
CH4 (g ) + 2O2 (g ) → CO2 (g ) + 2H2O(l )
y mL
2 y mL
y mL
The first step volume contraction can be calculated as :
x


 x + + y + 2 y − (x + y) = 13


2
…(i)
⇒
x + 4 y = 26
The second volume contraction is due to absorption of CO2.
Hence,
…(ii)
x + y = 14
If M is molarity of H2O2 solution, then
0.508 × 1000
5M =
(Q 1 mole H2O2 ≡≡ 1 mole I2)
254
⇒
M = 0.4
Also, n-factor of H2O2 is 2, therefore normality of H2O2 solution
is 0.8 N.
⇒ Volume strength = Normality × 5.6 = 0.8 × 5.6 = 4.48 V
42. The reaction is
KIO3 + 2KI + 6HCl → 3ICl + 3KCl + 3H2O
KIO3 required for 20 mL original KI solution = 3 millimol.
⇒ 7.5 millimol KIO3 would be required for original 50 mL KI.
⇒ Original 50 mL KI solution contain 15 millimol of KI.
After AgNO3 treatment 5 millimol of KIO3 is required, i.e. 10
millimol KI is remaining.
⇒ 5 millimol KI reacted with 5 millimol of AgNO3.
5
⇒ Mass of AgNO3 =
× 170 = 0.85 g
1000
⇒ Mass percentage of AgNO3 = 85%
43. CO2 is evolved due to following reaction :
2NaHCO3 → Na 2CO3 + H2O + CO2
750 123.9
1
pV
Moles of CO2 produced =
=
×
×
760 1000 0.082 × 298
RT
= 5 × 10−3
⇒ Moles of NaHCO3 in 2 g sample = 2 × 5 × 10−3 = 0.01
⇒ millimol of NaHCO3 in 1.5 g sample
0.01
=
× 1.5 × 1000= 7.5
2
Let the 1.5 g sample contain x millimol Na 2CO3, then
2x + 7.5 = millimol of HCl = 15
x = 3.75
7.5 × 84
= 0.63 g
⇒ Mass of NaHCO3 =
1000
3.75 × 106
Mass of Na 2CO3 =
= 0.3975 g
1000
0.63
× 100 = 42 %
⇒ % mass of NaHCO3 =
1.50
0.3975
% mass of Na 2CO3 =
× 100 = 26.5%
1.5
⇒
44. Mass of Fe2O3 = 0.552 g
Now, solving equations (i) and (ii),
x = 10 mL, y = 4 mL and volume of He = 20 – 14 = 6 mL
10
Vol % of CO =
× 100 = 50%
⇒
20
0.552
× 1000 = 3.45
160
During treatment with Zn-dust, all Fe3+ is reduced to Fe2 + ,
hence
millimol of Fe2O3 =
Some Basic Concepts of Chemistry 21
millimol of Fe2 + (in 100 mL) = 3.45 × 2 = 6.90
6.90
= 1.725 millimol Fe2+ ion.
⇒ In 25 mL aliquot,
4
Finally Fe2+ is oxidised to Fe3+ , liberating one electron per Fe2+
ion. Therefore, total electrons taken up by oxidant.
= 1.725 × 10−3 × 6.023 × 1023
= 1.04 × 10
5C2O42− + 2MnO−4 + 16H+ → 2Mn 2+ + 10CO2 + 8H2O
Let, in the given mass of compound, x millimol of C2O2−
4 ion is
present, then
Meq of C2O42− = Meq of MnO−4
⇒
2x = 0.02 × 5 × 22.6 ⇒ x = 1.13
At the later stage, with I − , Cu 2+ is reduced as :
2Cu 2+ + 4I − → 2CuI + I2
and
I2 + 2S2O32− → 2I− + S4O62−
Let there be x millimol of Cu 2+ .
Meq of Cu 2+ = Meq of I2 = meq of hypo
⇒
x = 11.3 × 0.05 = 0.565
⇒ Moles of Cu
2+
:
moles of C2O42− = 0.565 : 1.13 = 1 : 2
46. Let us consider 10 mL of the stock solution contain x millimol
oxalic acid H2C2O4 and y millimol of NaHC2O4.
When titrated against NaOH, basicity of oxalic acid is 2 while
that of NaHC2O4 is 1.
… (i)
⇒
2x + y = 3 × 0.1 = 0.3
When titrated against acidic KMnO4, n-factors of both oxalic
acid and NaHC2O4 would be 2.
… (ii)
⇒
2x + 2 y = 4 × 0.1 = 0.4
Solving equations (i) and (ii) gives
y = 0.1, x = 0.1
⇒ In 1.0 L solution, mole of H2C2O4 =
Mole of NaHC2O4 =
⇒ X = C2H4Cl 2 with two of its structural isomers.
Cl— CH2— CH2 — Cl
and
I
CH3— CHCl 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
20
45. With KMnO4, oxalate ion is oxidised only as :
⇒
Because X can be represented by two formula of which one
gives a dihydroxy compound with KOH indicates that X has two
chlorine atoms per molecule.
0.1
× 100 = 0.01
1000
0.1
× 100 = 0.01
1000
⇒ Mass of H2C2O4 = 90 × 0.01 = 0.9 g
Mass of NaHC2O4 = 112 × 0.01= 1.12 g
35.5
47. Mass of chlorine in 1.0 g X =
× 2.9 = 0.717 g
143.5
Now, the empirical formula can be derived as :
ClCH2CH2Cl + OH− → CH2 — CH2


OH
OH
(Y )
–H2 O
CH3CHCl 2 + KOH → CH3CH(OH)2 → CH3CHO
Unstable
(Z )
48. Let the n-factor of KMnO4 in acid, neutral and alkaline media
are N 1, N 2 and N 3 respectively. Also, same volumes of reducing
agent is used everytime, same number of equivalents of KMnO4
would be required every time.
100
5
N 2 = 100N 3 ⇒ N 1 = N 2 = 5N 3
⇒ 20N 1 =
3
3
Also, n-factors are all integer and greater than or equal to one but
less than six, N 3 must be 1.
⇒
N 1 = 5, N 2 = 3
∴ In acid medium
MnO−4 → Mn 2+
In neutral medium
MnO−4 → Mn 4+
In alkaline medium MnO−4 → Mn6+
⇒ meq of K2Cr2O7 required = 100
100 = 1 × 6 × V (n-factor = 6)
⇒
⇒
V = 100/ 6 = 16.67 mL
1
49. Meq of MnO−4 required = 20 ×
×5=2
50
⇒ Meq of Fe2+ present in solution = 2
⇒ millimol of Fe2+ present in solution = 2 (n-factor = 1)
Also,
Q 4 millimol of Fe2+ are formed from 1 millimol N2H4
1
1
∴ 2 millimol Fe2+ from × 2 = millimol N2H4
4
2
Therefore, molarity of hydrazine sulphate solution
1 1
1
= ×
=
2 10 20
1
mol N2H6SO4 is present.
⇒ In 1 L solution
20
1
⇒ Amount of N2H6SO4 =
× 130 = 6.5 gL−1
20
C
H
Cl
% wt :
24.24
4.04
71.72
Mole :
2
4
2
⇒ Molarity of carbonate solution =
Simple ratio :
1
2
1
⇒ Normality of carbonate solution = 2 × 0.035 = 0.07 N
⇒ Empirical formula = CH2Cl.
50. Molecular weight of Na 2CO3 ⋅10H2O = 286
1
1000
= 0.035
×
286 100
22 Some Basic Concepts of Chemistry
In acid solution : Normality of HNO3 =
Normality of HCl =
8×5
= 0.02
2000
5 × 4.8
= 0.012
2000
Let normality of H2SO4 in final solution be N.
⇒ (N + 0.02 + 0.012) × 30 = 0.07 × 42.9
⇒
N = 0.0681
⇒ Gram equivalent of
⇒ Mass of
SO2−
4
SO2−
4
in 2 L solution = 2 × 0.0681
= 0.1362
96
in solution = 0.1362 ×
= 6.5376 g
2
51. For the oxidation of A n+ as :
A n+ → AO−3 n-factor = 5 – n
⇒ Gram equivalent of A n+ = 2.68 × 10−3 (5 − n)
Now equating the above gram equivalent with gram equivalent
of KMnO4 :
2.68 × 10−3 (5 − n) = 1.61 × 10−3 × 5
⇒
n=+2
52. During heating MCO3 is converted into MO liberating CO2
while BaO is remaining unreacted :
Heat
MCO3 (s) → MO(s) + CO2 (g ) ↑ 0.44 g = 0.01 mol
BaO(s)
4.08 g
BaO(s)
3.64 g
From the decomposition information, it can be deduced that the
original mixture contained 0.01 mole of MCO3 and the solid
residue, obtained after heating, contain 0.01 mole (10 millimol)
of MO.
Also, millimol of HCl taken initially = 100
millimol of NaOH used in back-titration = 16 × 2.5 = 40
⇒ millimol of HCl reacted with oxide residue = 60
HCl reacts with oxides as :
MO
+ 2HCl
→ MCl 2 + H2O
10 millimol
20 millimol
BaO
+
2HCl
→ BaCl 2 + H2O
60 – 20 = 40 millimol
Therefore, the residue contain 20 millimol of BaO.
Also,
molar mass of BaO = 138 + 16
= 154
154 × 20
= 3.08 g
Mass of BaO =
⇒
1000
⇒ Mass of MCO3 = 4.08 – 3.08 = 1.0 g
Q 0.01 mole of MCO3 weight 1.0 g
∴ 1 mole of MCO3 = 100 g
⇒ 100 = (Atomic weight of metal) + (12 + 3 × 16)
⇒ Atomic weight of metal = 40, i.e. Ca
2
Atomic Structure
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 2a0 from
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 a0 from 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)
4
λ
9
2
(c) λ
3
3
λ
4
1
(d) λ
2
(b)
(a)
3. What is the work function of the metal, if the light of
wavelength 4000 Å generates photoelectron of velocity
6 × 105 ms −1 from it?
−31
(Mass of electron = 9 × 10
kg
Velocity of light = 3 × 10 ms −1
8
Planck’s constant = 6.626 × 10−34 Js
Charge of electron = 1.6 × 10−19 JeV−1 )
(2019 Main, 12 Jan I)
(a) 4.0 eV
(c) 0.9 eV
(b) 2.1 eV
(d) 3.1 eV
The energy of second excited state of He+ ion in eV is
(2019 Main, 10 Jan II)
(b) −3.4
(c) −6.04
relationship between incident light and the electron ejected from
metal surface?
(2019 Main, 10 Jan I)
K.E. of
e ss
K.E. of
es s
(a)
(b)
0
0
Energy of light
Number
of ess
Intensity of light
K.E. of
ess
(c)
(d)
0
Frequency of light
0
Frequency of light
6. A stream of electrons from a heated filament was passed between
two charged plates kept at a potential difference V 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
given by
(2016 Main)
(a) 2 meV
(b) meV
(c) 2meV
(d) meV
7. Rutherford’s experiment, which established the nuclear model of
the 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 got
scattered
(d) helium nuclei, which impinged on a metal foil and got
scattered
8. Rutherford’s alpha particle scattering experiment eventually led
4. The ground state energy of hydrogen atom is −13.6 eV.
(a) −54.4
5. Which of the graphs shown below does not represent the
(d) −27.2
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
24 Atomic Structure
9. The radius of an atomic nucleus is of the order of
−10
(a) 10
−13
cm (b) 10
−15
cm (c) 10
cm
−8
(d) 10 cm
10. Bohr’s model can explain
List-I
(1985, 1M)
(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
List-II
(I)
Radius of the n th orbit
(P)
∝ n −2
(II)
Angular momentum of the electron
in the nth orbit
(Q)
∝ n −1
(III)
Kinetic energy of the electron in the
nth orbit
(R)
∝ n0
(IV)
Potential energy of the electron in
the nth orbit
(S)
∝ n1
(T)
∝ n2
(U)
∝ n 1/ 2
11. The increasing order (lowest first) for the values of e/m
(charge/mass) for electron ( e ), proton (p), neutron (n) and
alpha particle (α) is
(1984, 1M)
(a) e, p, n, α
(b) n, p, e, α
(c) n , p , α , e
(d) n , α, p , e
12. Rutherford’s scattering experiment is related to the size of
the
(a) nucleus (b) atom
(1983, 1M)
(c) electron
(d) neutron
13. Rutherford’s experiment on scattering of α-particles showed
for the first time that the atom has
(a) electrons
(b) protons
(c) nucleus
(d) neutrons
(1981, 1M)
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
(b) neutrons
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.
List-II
(I)
Radius of the nth orbit
(P)
∝ n −2
(II)
Angular momentum of the electron in
the nth orbit
(Q)
∝ n −1
(III) Kinetic energy of the electron in the
nth orbit
(R)
∝ n0
(IV) Potential energy of the electron in the
nth orbit
(S)
∝ n1
(T)
∝ n2
(U)
∝ n 1/ 2
(1988, 1M)
(c) electrons
(d) photons
16. The sum of the number of neutrons and proton in the isotope
of hydrogen is
(a) 6
(b) 5
20. Consider the Bohr’s model of a one-electron atom where the
List-I
Objective Questions II
(One or more than one correct option)
(a) protons
Which of the following options has the correct combination
considering List-I and List-II?
(2019 Adv.)
(a) (III), (P)
(b) (III), (S)
(c) (IV), (U)
(d) (IV), (Q)
(1986, 1M)
(c) 4
(d) 3
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
Which of the following options has the correct combination
considering List-I and List-II?
(2019 Adv.)
(a) (II), (R)
(b) (I), (P)
(c) (I), (T)
(d) (II), (Q)
21. According to Bohr’s theory,
En = Total energy
Vn = Potential energy
Match the following :
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.
K n = Kinetic energy
rn = Radius of nth orbit
(2006, 6M)
Column I
Column II
A. V / K = ?
n
n
B.
If radius of nth orbit ∝
C.
Angular momentum in lowest
orbital
D.
1
r
n
∝ Zy, y = ?
E xn
,x= ?
p.
0
q.
–1
r.
–2
s.
1
Fill in the Blanks
22. The light radiations with discrete quantities of energy are
called ................ .
(1993, 1M)
Atomic Structure 25
23. The mass of a hydrogen is …… kg.
(1982, 1M)
Use Avogardo constant as 6.023 × 1023 mol −1 .
(2020 Adv.)
24. Isotopes of an element differ in the number of …… in their
Potential energy
(kJ mol–1)
nuclei.
(1982, 1M)
25. Elements of the same mass number but of different atomic
numbers are known as …… .
(1983, 1M)
Numerical Answer Type Questions
H
H
E0
d0
26. The figure below is the plot of potential energy versus
internuclear distance ( d ) of H2 molecule 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 = d 0 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.
d
Internuclear distance (d)
Subjective Questions
27. With what velocity should an α-particle travel towards the
nucleus of a copper atom so as to arrive at a distance 10−13 m
from the nucleus of the copper atom ?
(1997 (C), 3M)
Topic 2 Advanced Concept (Quantum Mechanical Theory)
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)
3. Among the following, the energy of 2s-orbital is lowest in
(2019 Main, 12 April II)
(a) K
(b) H
(c) Li
(d) Na
4. The electrons are more likely to be found
Increasing wavelength
a
Ψ (x)
(a)
b
Rb
(b)
(c)
x
–x
Absorption spectrum
K
c
Na
(2019 Main, 12 April I)
Kinetic energy of
photoelectrons
(a) in the region a and c
(c) only in the region a
(b) in the region a and b
(d) only in the region c
Frequency of incident radiation
5. The ratio of the shortest wavelength of two spectral series of
300
6. The graph between | ψ |2 and r (radial distance) is shown below.
hydrogen spectrum is found to be about 9. The spectral series
are
(2019 Main, 10 April II)
(a) Lyman and Paschen
(b) Brackett and Pfund
(c) Paschen and Pfund
(d) Balmer and Brackett
Internal
energy of
Ar
400
500
600
Temperature (K)
This represents
(2019 Main, 10 April I)
T 2 >T 1
(d)
Intensity
of black body
radiation
2
|Ψ|
T1
Wavelength
2. The number of orbitals associated with quantum numbers
1
n = 5, ms = + is
2
(a) 25
(b) 50
(2020 Main, 7 Jan I)
(c) 15
(d) 11
r
(a) 1s-orbital
(c) 3s-orbital
(b) 2 p-orbital
(d) 2s-orbital
26 Atomic Structure
7. For any given series of spectral lines of atomic hydrogen,
let ∆ν = ν max − ν min be the difference in maximum and
minimum frequencies in cm −1 . The ratio
∆ νLyman / ∆ νBalmer is
(2019 Main, 9 April I)
(a) 27 : 5
(b) 5 : 4
(c) 9 : 4
(d) 4 : 1
8. The quantum number of four electrons are given below:
I. n = 4 , l = 2, ml = − 2, ms = −
1
2
(2019 Main, 8 April I)
(b) I < II < III < IV
(d) I < III < II < IV
th
9. If the de-Broglie wavelength of the electron in n Bohr orbit
in a hydrogenic atom is equal to 15
. πa0 (a0 is 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, [ν 0 is
threshold frequency]
(2019 Main, 11 Jan II)
1
1
(b) λ ∝
(a) λ ∝
1
3
( ν − ν0 )4
1
(c) λ ∝
( ν − ν0 )
(d) λ ∝
1
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
h
momentum is equal to .
2π
IV. The plot of ψ vs 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 × 105 cm–1 ,
(a) Paschen, 5 →3
(c) Lyman, ∞ → 1
Rydberg constant, RH is in wave number unit)
(2019 Main, 9 Jan I)
(a) non linear
(c) linear with slope RH
(b) linear with slope −RH
(d) linear with intercept −RH
(Planck’s constant ( h ) = 6.6262 × 10− 34 Js; mass of electron
= 91091
× 10− 31
.
(2019 Main, 11 Jan I)
(b) Paschen, ∞ → 3
(d) Balmer, ∞ → 2
kg
;
charge
of
electron
C; permitivity of vacuum
( e ) = 160210
.
× 10
(∈0 ) = 8.854185 × 10− 12kg − 1 m − 3A 2 )
(2017 Main)
− 19
(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
4 πr2 dr. The qualitative sketch of the dependence of P on r is
(2016 Adv.)
P
P
(a)
(b)
0
( ν − ν0 )2
1
( ν − ν0 )2
h = 6.6 × 10−34 Js, c = 3 × 108 ms −1 ]
 1
the plot of wave number ( ν ) against  2  will be (The
n 
14. The radius of the second Bohr orbit for hydrogen atom is
1
II. n = 3, l = 2, ml = 1, ms = +
2
1
III. n = 4 , l = 1, ml = 0, ms = +
2
1
IV. n = 3, l = 1, ml = 1, ms = −
2
The correct order of their increasing energies will be
(a) IV < III < II < I
(c) IV < II < III < I
13. For emission line of atomic hydrogen from ni = 8 to n f = n,
0
r
P
r
P
(c)
(d)
0
r
0
r
16. Which of the following is the energy of a possible excited
state of hydrogen?
(a) + 13.6 eV
(c) –3.4 eV
(2015 Main)
(b) – 6.8 eV
(d) + 6.8 eV
17. The correct set of four quantum numbers for the valence
electrons of rubidium atom ( Z = 37 ) is
1
(b) 5, 1, 0, +
(a) 5, 0, 0, +
2
1
(c) 5, 1, ,1, +
(d) 5, 0, 1, +
2
18. Energy of an electron is given by
 Z2 
E = − 2.178 × 10−18 J  2 
n 
(2013 Main)
1
2
1
2
(2013 Main)
Wavelength of light required to excite an electron in an
hydrogen atom from level n = 1to n = 2 will be
(h = 6.62 × 10−34 Js and c = 3.0 × 108 ms −1 )
(a) 1.214 × 10−7 m
(c) 6.500 × 10
−7
m
(b) 2.816 × 10−7 m
(d) 8.500 × 10−7 m
Atomic Structure 27
19. The kinetic energy of an electron in the second Bohr orbit of a
hydrogen atom is [a0 is Bohr radius]
(2012)
h2
h2
h2
h2
(d)
(b)
(c)
(a)
4π 2 ma02
16π 2 ma02
32π 2 ma02
64π 2 ma02
20. The number of radial nodes in 3s and 2p respectively are
(a) 2 and 0
(c) 1 and 2
(b) 0 and 2
(d) 2 and 1
(2005, 1M)
21. Which hydrogen like species will have same radius as that of
Bohr orbit of hydrogen atom?
(a) n = 2, Li
2+
(2004, 1M)
3+
(b) n = 2, Be
(c) n = 2, He+
(d) n = 3, Li 2+
22. If the nitrogen atom had electronic configuration 1s , it
would have energy lower than that of the normal ground
state configuration 1s2 2s2 2 p 3 , because the electrons would
be closer to the nucleus, yet 1s7 is not observed, because it
violates
(2002, 3M)
(a) Heisenberg uncertainty principle
(b) Hund’s rule
(c) Pauli exclusion principle
(d) Bohr postulate of stationary orbits
24. The wavelength associated with a golf ball weighing 200 g
and moving at a speed of 5 m/h is of the order
(d) 10−40 m
(b) two
(c) three
(2001, 1M)
(2001, 1M)
(d) zero
26. The
electronic configuration of an element is
2
2
1s , 2s 2 p 6 , 3s2 3 p 6 3d 5 , 4 s1 . This represents its (2000, 1M)
(a) excited state
(c) cationic form
atom was made by
(1997, 1M)
(a) Heisenberg
(c) Planck
(b) Bohr
(d) Einstein
31. Which of the following has the maximum number of
unpaired electrons ?
(1996, 1M)
(d) Fe2+
(c) V3+
(b) Ti 3+
(a) Mg2+
(1996, 1M)
1 h
(a) + ⋅
2 2π
h
(c)
2π
(b) zero
(d) 2 ⋅
h
2π
33. Which of the following relates to photons both as wave
motion
and
as
a
stream
of
particles
?
(1992, 1M)
(b) E = mc2
(d) E = hν
(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 unpaired
electron of chlorine atom is
(a)
(c)
n
2
3
l
1
1
m
0
1
(1989, 1M)
n
(b) 2
(d) 3
l
1
0
m
1
0
36. The correct ground state electronic configuration of
chromium atom is
(1989, 1M)
(a) [ Ar ] 3d 5 4 s1
(b) [ Ar ] 3d 4 4 s2
(c) [ Ar ] 3d 6 4 s0
(d) [ Ar ] 4 d 5 4 s1
37. The outermost electronic configuration of the most
25. The number of nodal planes in a px orbital is
(a) one
30. The first use of quantum theory to explain the structure of
34. Which of the following does not characterise X-rays ?
(a) rotation of the electron in clockwise and anti-clockwise
direction respectively
(b) rotation of the electron in anti-clockwise and clockwise
direction respectively
(c) magnetic moment of the electron pointing up and down
respectively
(d) two quantum mechanical spin states which have no classical
analogue
(c) 10−30 m
 h
(d) 2  
 2π 
 h
(c)  
 2π 
(a) Interference
(c) Diffraction
1
1
23. The quantum numbers + and − for the electron spin
2
2
represent
(2001, 1M)
(b) 10−20 m
 h
 h
(a) 6   (b) 2  
 2π 
 2π 
32. The orbital angular momentum of an electron in 2s-orbital is
7
(a) 10−10 m
29. For a d-electron, the orbital angular momentum is(1997, 1M)
(b) ground state
(d) anionic form
electronegative element is
(1988, 90, 1M)
(a) ns2 np3
(b) ns2 np4
(c) ns2 np5
(d) ns2 np6
38. The orbital diagram in which the Aufbau principle is
violated
(1988, 1M)
(a)
(b)
(c)
(d)
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
highest, as
(1999, 2M)
(a) (iv) < (ii) < (iii) < (i)
(c) (i) < (iii) < (ii) < (iv)
(b) (ii) < (iv) < (i) < (iii)
(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)
(a) –3.4 eV
(c) – 6.8 eV
(b) – 4.2 eV
(d) + 6.8 eV
39. The wavelength of a spectral line for an electronic transition
is inversely related to
(1988, 1M)
(a) the number of electrons undergoing the transition
(b) the nuclear charge of the atom
(c) the difference in the energy of the energy levels involved in
the transition
(d) the velocity of the electron undergoing the transition
28 Atomic Structure
40. The ratio of the energy of a photon of 200 Å wavelength
radiation to that of 4000 Å radiation is
1
(a)
4
(b) 4
(1986, 1M)
1
(c)
2
41. Which one of the following sets of quantum numbers
(d)
49. Which of the following statement (s) is (are) correct ?
(1998, 2M)
(b) radio wave
(d) infrared
(1985, 1M)
absorb a photon but not to emit a photon?
(b) 2p
(c) 2s
(1984, 1M)
(d) 1s
44. Correct set of four quantum numbers for the valence
(outermost) electron of rubidium ( Z = 37 ) is
1
(a) 5, 0, 0, +
2
1
(c) 5, 1, 1, +
2
(b)
(c)
43. Which electronic level would allow the hydrogen atom to
(a) 3s
(1999, 3M)
(1986, 1M)
42. Electromagnetic radiation with maximum wavelength is
(a) ultraviolet
(c) X-ray
can be represented by
(a)
(d) 2.
represents an impossible arrangement?
n
l
m
s
1
(a) 3
2
–2
2
1
(b) 4
0
0
2
1
(c) 3
2
–3
2
1
(d) 5
3
0
−
2
48. The ground state electronic configuration of nitrogen atom
(1984, 1M)
1
(b) 5, 1, 0 , +
2
1
(d) 6, 0, 0, +
2
45. The principal quantum number of an atom is related to the
(a) size of the orbital
(b) spin angular momentum
(c) orientation of the orbital in space
(d) orbital angular momentum
46. Any p-orbital can accommodate upto
(1983, 1M)
(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.
(a) The electronic configuration of Cr is [Ar] 3d 5 4 s1 (atomic
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 of
the opposite type. (atomic number of Ag = 47)
(d) The oxidation state of nitrogen in HN3 is – 3
76
32 Ge
50. An isotone of
(a)
(c)
is
77
32 Ge
77
34 Se
(1984, 1M)
(b)
(d)
77
33 As
78
34 Se
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 S 1 with one radial node. Upon absorbing light the ion
undergoes transition to a state S 2 . The state S 2 has one radial
node and its energy is equal to the ground state energy of the
hydrogen atom.
52. The state S 1 is
(a) 1s
(b) 2s
(2010)
(c) 2p
(d) 3s
Which of the following statement(s) is(are) true for the
state Ψ?
(2019 Adv.)
(a) It is a 4 d state
53. Energy of the state S 1 in units of the hydrogen atom ground
(b) The nuclear charge experienced by the electron in this state
is less than 2e, where e is the magnitude of the electronic
charge
54. The orbital angular momentum quantum number of the state
(c) It has 2 angular nodes
(d) It has 3 radial nodes
state energy is
(a) 0.75
(b) 1.50
(2010)
(c) 2.25
(d) 4.50
S 2 is
(a) 0
(2010)
(b) 1
(c) 2
(d) 3
Atomic Structure 29
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, ψ n , l , ml is a mathematical function whose value depends upon spherical polar coordinates ( r, θ , φ ) of the electron
and characterised by the quantum number n , l and ml . Here r is distance from nucleus, θ is colatitude and φ is azimuth. In the
mathematical functions given in the Table, Z is atomic number and a0 is Bohr radius.
(2017 Adv.)
Column 1
(I)
1s-orbital
Column 2
Column 3
3  Zr 
 2 −  a 0 
e
(i)
Z
ψn, l , ml ∝  
 a0 
(P)
ψn, I, ml (r)
0
r/a0
(II) 2s-orbital
(ii)
One radial node
(III) 2 pz-orbital
(iii)
(IV) 3 dz2-orbital
(iv)
5
ψn, l ml
(Q)
 Zr 
 Z  2 − a 
∝   re  0  cosθ
 a0 
xy-plane is a nodal plane
55. For He+ ion, the only INCORRECT combination is
(a) (I) (i) (S)
(c) (I) (iii) (R)
(b) (II) (ii) (Q)
(d) (I) (i) (R)
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
(b) (I) (iv) (R)
(d) (I) (i) (S)
(2008, 6M)
Column I
A.
B.
Column II
Orbital angular momentum
of the electron in a
hydrogen-like atomic orbital.
p.
A hydrogen-like one-electron
wave function obeying
Pauli’s principle.
q.
a03
(R)
Probability density is maximum at nucleus
(S)
Energy needed to excite electron from n = 2 state to n = 4 state
27
is
times the energy needed to excite electron from n = 2
32
state to n = 6 state
Fill in the Blanks
outermost electronic configuration of Cr is
.......................... .
(1994, 1M)
60. 8 g each of oxygen and hydrogen at 27°C will have the total
kinetic energy in the ratio of .......... .
(1989, 1M)
61. The uncertainty principle and the concept of wave nature of
matter were proposed by ............ and .............respectively.
(1988, 1M)
62. Wave functions of electrons in atoms and molecules are
58. Match the entries in Column I with the correctly related
quantum number(s) in Column II.
1
59. The
56. For the given orbital in Column 1, the Only CORRECT
(a) (I) (i) (P)
(c) (II) (i) (Q)
Probability density at nucleus ∝
Principal
quantum
number
Azimuthal
quantum
number
C.
Shape, size and orientation
of hydrogen-like atomic
orbitals.
r.
Magnetic
quantum
number
D.
Probability density of
electron at the nucleus in
hydrogen-like atom.
s.
Electron spin
quantum
number
called .............. .
(1993, 1M)
63. The 2 px, 2 p y and 2 p z orbitals of atom have identical shapes
but differ in their ........... .
(1993, 1M)
64. When there are two electrons in the same orbital, they have
…… spins.
(1983, 1M)
True/False
65. In a given electric field, β -particles are deflected more than
α-particles in spite of α-particles having larger charge.
(1993, 1M)
orbital is
y
zero.
(1986, 1M)
67. The energy of the electron in the 3d-orbital is less than that in
the 4s-orbital in the hydrogen atom.
(1983, 1M)
68. Gamma rays are electromagnetic radiations of wavelengths
of 10−6 to 10−5 cm.
(1983, 1M)
69. The outer electronic configuration of the ground state
chromium atom is 3d 4 4 s2 .
(1982, 1M)
66. The electron density in the XY-plane in 3d x 2 −
2
30 Atomic Structure
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
numbers
(2014 Adv.)
1
n = 4, | ml | = 1 and ms = − is
2
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)
Metal
Li
Na
K
Mg
Cu
Ag
Fe
Pt
W
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,
(2011)
ms = − 1 / 2 , is
Subjective Questions
75. (a) Calculate velocity of electron in first Bohr orbit of hydrogen
atom (Given, r = a0 ).
(b) Find de-Broglie wavelength of the electron in first Bohr
orbit.
(c) Find the orbital angular momentum of 2p-orbital in terms of
(2005, 2M)
h / 2π units.
76. (a) The Schrodinger wave equation for hydrogen atom is
ψ 2s =
 1
1
 
4 (2π )1/ 2  a0 
3/ 2

r  − r / 2a 0
2 −  e
a0 

where, a0 is 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
He+ ion.
(2003, 2M)
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
436 kJ mol −1 .
(2000)
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 the
magnitude of the average kinetic energy is half the
magnitude of the average potential energy, find the average
potential energy.
(1996, 2M)
81. Calculate the wave number for the shortest wavelength
transition 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 each
molecule, calculate the kinetic energy of iodine atoms.
(Bond energy of I2 = 240 kJ mol –1 )
(1995, 2M)
83. Find out the number of waves made by a Bohr’s electron in
one 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
(1993, 3M)
He+ spectrum?
85. Estimate the difference in energy between 1st and 2nd
Bohr’s orbit for a hydrogen atom. At what minimum atomic
number, a transition from n = 2 to n = 1 energy level would
result in the emission of X-rays with l = 3.0 × 10–8 m ?
Which hydrogen atom-like species does this atomic number
correspond to?
(1993, 5M)
86. According to Bohr’s theory, the electronic energy of
hydrogen atom in the nth Bohr’s orbit is given by :
En =
−21.7 × 10−19
J
n2
Calculate the longest wavelength of electron from the third
Bohr’s orbit of the He+ ion.
(1990, 3M)
87. What is the maximum number of electrons that may be
present in all the atomic orbitals with principal quantum
number 3 and azimuthal quantum number 2 ?
(1985, 2M)
88. Give reason why the ground state outermost electronic
configuration of silicon is
(1985, 2M)
3s
3p
3s
3p
and not
89. The electron energy in hydrogen atom is given by
En = −
21.7 × 10−12
erg. Calculate the energy required to
n2
remove an electron completely from the n = 2 orbit. What is
the longest wavelength (in cm) of light that can be used to
cause 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
atom.
(1982, 4M)
91. The energy of the electron in the second and third Bohr’s
orbits of the hydrogen atom is − 5.42 × 10−12 erg and
− 2.41 × 10−12 erg respectively. Calculate the wavelength of
the emitted light when the electron drops from the third to the
second orbit.
(1981, 3M)
Answers
Topic 1
1.
5.
9.
13.
17.
21.
(d)
2. (a)
3. (b)
(d)
6. (c)
7. (d)
(b)
10. (b)
11. (d)
(c)
14. (a)
15. (a,b)
(a,c)
18. (a,c)
19. (a)
A → r; B → q; C → p, D → s
22. (photons)
23. (1 . 66 × 10 –27 kg)
25.
(isobars) 26. (–5242.41) 27.
4.
8.
12.
16.
20.
(c)
(b)
(a)
(d)
(c)
24. (neutrons)
6.3 × 10 6
Topic 2
1. (c)
2. (a)
3. (a)
4. (a)
5. (a)
6. (d)
7. (c)
8. (c)
9. (b)
10. (d)
11. (d)
12. (b)
13. (c)
14. (d)
15. (c)
16. (c)
17. (a)
18. (a)
19. (c)
20. (a)
21. (b)
22. (c)
23. (d)
24. (c)
25. (a)
26. (b)
27. (a)
28. (a)
29. (a)
30. (b)
31. (d)
32. (b)
33. (a)
34. (c)
35. (c)
36. (a)
37. (c)
38. (b)
39. (c)
40. (d)
41. (c)
42. (b)
43. (d)
44. (a)
45. (a)
46. (d)
47. (a,c)
48. (a,d)
49. (a,b,c)
50. (b,d)
51. (c)
52. (b)
53. (c)
54. (b)
55. (c)
56. (a)
58. A → q;
57. (d)
B → p, q, r, s
5
59. Cr = [Ar] 3d , 4s
C → p, q, r
1
D → p, q, r
60. 1 : 16
61. Heisenberg, de-Broglie.
62. orbital
63. Orientation in space
64. opposite
65. True
69. False
73. (4.14 eV)
66. False
70. (3)
74. (9)
67. True
71. (6)
77. (22.8 nm)
79. (63.56 V)
81. (2.725× 10 6 M −1)
82. (2.16 × 10 20 J/atom)
89. (3.66 × 10
−5
86. (471 nm)
68. False
72. (5)
78. (98.44 kJ)
87. (10)
cm) 90. (1220 Å) 91. (660 nm)
Hints & Solutions
Topic 1 Preliminary Developments
and Bohr’s Model
1. Statement (d) is incorrect. For 1s-orbital radial probability
density (R 2 ) against r is given as:
decreases. The maximum in the curve corresponds to the
distance at which the probability of finding the electron in
maximum.
2. The expression of kinetic energy of photo electrons,
1
mv 2 = E − E0
2
When, KE > > E0, the equation becomes,
1
KE = mv 2 = E
2
1 2 hc
p2
hc
⇒
=
mv =
⇒
2
λ
2
λ
2m
1
1
2
⇒ λ = hc × 2m × 2 ⇒ λ ∝ 2
p
p
hc
E=
= energy of incident light.
λ
E0 = threshold energy or work functions,
1 2 1 (mv )2 1 p2
mv = ×
= × 2
2
2
2 m
m2
Q p = momentum = mv
As per the given condition,
KE =
R2
1s
r
4πrR2
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
λ 2  p1 
= 
λ 1  p2 
2
2
r
The graph initially increases and then decreases. It reaches a
maximum at a distance very close to the nucleus and then
⇒
2
λ2  p 
4
 2
=
 =  =
 3
. × p
λ  15
9
32 Atomic Structure
⇒
4
λ
9
λ2 =
Q λ 1 = λ 
 p1 = p 
So, option (a) is correct.
(ii) KE of ejected electrons does not depend on the intensity of
incident light.
3. Work function of metal (φ ) = hν 0
where, ν 0 = threshold frequency
KE
1
mev 2 = hν − hν 0
2
1
or
mev 2 = hν − φ
2
1
hc
−φ
mev 2 =
λ
2
Given : λ = 4000 Å = 4000 × 10−10m
Also,
…(i)
0
…(ii)
Intensity of light
So, option (b) is correct.
(iii) When, number of ejected electrons is plotted with frequency
of light, we get
v = 6 × 105 ms −1,
me = 9 × 10−31 kg, c = 3 × 108 ms −1
Number
of e–’s
h = 6.626 × 10−34 Js
Thus, on substituting all the given values in Eq. (i), we get
1
× 9 × 10−31 kg × (6 × 105 ms −1 )2
2
6.626 × 10−34 J s × 3 × 108 ms −1
−φ
=
4000 × 10−10 m
0
So, option (c) is also correct.
(iv) KE = hν − hν 0
φ = 162
. × 10−21 kgm 2s −2 − 4.96 × 10−19 J
∴
= 3.36 × 10−19 J
[1 kgm2s −2 = 1J]
KE
= 21
. eV
0
4. The ground state energy of H-atom is + 13.6 eV.
For second excited state, n = 2 + 1 = 3
Z2
[Qfor He+ , Z = 2]
∴ E3 (He+ ) = − 13.6 × 2 eV
n
22
= − 13.6 × 2 eV = − 6.04 eV
3
KE = E − E0
Slope = + h, intercept = − hν 0. So, option (d) is not correct.
6.
Plan As you can see in options, energy term is mentioned hence, we
have to find out relation between h / λ and energy. For this, we
shall use de-Broglie wavelength and kinetic energy term in eV.
⇒
|Ψ|2
For 1s-orbital
number of radial node = 1–0–1=0
r
KE = Kinetic energy of ejected electrons.
E = Energy of incident light = hν
E0 = Threshold energy = hν0
ν = Frequency of incident light
ν0 = Threshold frequency
KE
E0
Slope = ± 1, intercept = − E0
E
h
p
h
λ
p=
…(i)
Kinetic energy of an electron = eV
As we know that, KE =
∴
where,
0
n
n0
de-Broglie wavelength for an electron (λ ) =
5. For photoelectric effect,
(i)
Frequency of light (n)
p2
2m
eV =
p2
2m
or p = 2 meV
…(ii)
h
= 2 meV
λ
nuclei) in his experiment.
From equations (i) and (ii), we get
7. Rutherford used α-particle (He2+
8. According to Rutherford’s model, there is a heavily positively
charged nucleus and negatively charged electrons occupies space
around it in order to maintain electro-neutrality.
9. Radius of a nucleus is in the order of 10−13 cm, a fact.
10. Bohr’s model is applicable to one-electron system only.
11. Neutron has no charge, hence e/ m is zero for neutron. Next,
α-particle (He2+ ) has very high mass compared to proton and
electron, therefore very small e/ m ratio. Proton and electron
Atomic Structure 33
have same charge (magnitude) but former is heavier, hence has
smaller value of e/ m.
e
: n<α < p< e
m
21. A. Vn = −
Kn =
12. The negligibly small size of nucleus compared to the size of
atom was first established in Rutherford’s experiment.
13. The most important findings of Rutherford’s experiment is
13.6
where,
eV
n2
13.6
In excited states, E2 = −
= − 3.4 eV
4
13.6
E3 = −
= − 1.51 eV etc.
9
En = −
n =1, 2, 3, ....
1  Ze2 


8πε 0  r 
⇒
Vn
= − 2  (r )
Kn
B.
En = −
discovery of nucleus.
14. Energy of electron in H-atom is determined by the expression:
1  Ze2 


4 πε 0  r 
⇒
Ze2
∝ r −1
8πε 0r
x = − 1  (q )
C. Angular momentum = l (l + 1)
D.
rn =
a0n2
⇒
Z
1
∝ Z  (s)
rn
15. Nucleus is composed of neutrons and protons.
22. Photons have quantised energy.
16. The isotopes of hydrogen are 1 H2 and 1H3.
23. Mass of one H-atom =
17. Alpha particles passes mostly undeflected when sent through
h
= 0 in 1s-orbital — (p)
2π
10−3
kg = 1.66 × 10−27 kg
6.023 × 1023
thin metal foil mainly, because
24. Isotopes have different number of neutrons.
(i) it is much heavier than electrons.
(ii) most part of atom is empty space.
25. Isobars have same mass number but different atomic numbers.
18. Many elements have several isotopes. For such elements,
atomic mass is average of the atomic masses of different
isotopes, which is usually non-integral.
19. (III) Kinetic energy of the electron in nth orbit,
K.E. = + 13.6 ×
Z2
n2
1
or K.E. ∝ n−2
n2
From list-II, correct match is (III P).
(IV) Potential energy of the electron in the nth orbit,
Z2
P.E. = − 2 × 13.6 × 2
n
1
P.E. ∝ 2
n
P.E. ∝ n−2
or
K.E. ∝
26. Given that, electrons and nucleus are at infinite distance, so
potential energy of H-atom is taken as zero.
Therefore, according to Bohr’s model, potential energy of a
H-atom with electron in its ground state = − 27.2 eV
At d = d0, nucleus-nucleus and electron-electron repulsion is
absent.
Hence, potential energy will be calculated for 2H atoms
= − 2 × 27.2 eV= − 54.4 eV
Potential enery of 1 mol H atoms in kJ
54.4 × 6.02 × 1023 × 16
. × 10−19
= − 5242. 4192 kJ/mol
=
1000
27. When α-particle stop at 10−13m from nucleus, kinetic energy is
zero, i.e. whole of its kinetic energy at the starting point is now
converted into potential energy.
Potential energy of this α-particle can be determined as
PE = −
From List II, correct match is (IV P).
Hence, correct matching from list-I and list-II on the basis of
given option is (III, P).
20. (I) Radius of the nth orbit,
r = 0.529 ×
Here,
n2
Z
⇒
(Z1 = + 2, Z2 = + 29, ε 0 = 8.85 × 10−12 J−1 C2 m −1,
r = 10−13 m)
−19 2
2 × 29 × (1.6 × 10 )
J
| PE | =
4 × 3.14 × 8.85 × 10−12 × 10−13
= 1.33 × 10−13 J
r ∝ n2
From list-II, correct match is (I, T )
(II) Angular momentum of the electron,
nh
or mvr ∝ n
mvr =
2π
From list-II, correct match (II, S)
Hence, correct matching from list-I and list-II on the basis of
given option is (I, T ).
Z1 × Z2e2
(4 πε 0 ) r
⇒
⇒
= kinetic energy of α-particle at t = 0
1
KE = mv 2 = 1.33 × 10−13
2
v=
2 × 1.33 × 10−13
= 6.3 × 106 ms−1
4 × 1.66 × 10−27
34 Atomic Structure
Topic 2 Advanced Concept (Quantum Mechanical
Theory) Electronic Configuration and
Quantum Number
The graph between wavefunction (ψ ) and distance (r) from the
nucleus helps in determining the shape of orbital.
5. According to Rydberg’s equation,
1 RH  1
1
=
 − 
hc  n12 n22 
λ
1. Quantum nature of atoms are associated with following
phenomena or processes :
(i) Absorption spectrum (ii) Emission spectrum
(iii) Black body radiation (iv)
Photo-electricity
Internal energy (U ) of particles like atoms depends upon the
thermodynamic variables ( p, V , T ) of the system. Thus,
U = f ( p,T ) or U = f1 (T ,V ) orU = f2 ( p,V )
So, quantum nature of atom is not associated with its internal
energy.
Graph (c) is not a direct manifestation of the quantum nature of
atoms.
2. According to quantum mechanical atom model, for each value
of n (principal quantum number), there are ‘n’ different values of
l (azimuthal quantum number), i.e. l = 0, 1, 2, …, (n − 1). And,
for each value of l, there are 2l + 1 different values of ml
(magnetic quantum number), i.e. ml = 0, ±1, ±2 … ± l.
∴Total number of possible combinations of n, l and ml , for a
given value of n is n2, and each such combination is associated
with an orbital. Each orbital can occupy a maximum of two
electrons, having a different value of spin quantum number (ms ),
1
1
which are + or − .
2
2
∴ Number of orbitals associated with n = 5 is n2 = 25. Each of
1
as well as
those orbitals can be associated with ms = +
2
1
ms = − .
2
∴ Answer = 25
3. The energy of 2s-orbital is lowest in K(potassium). An orbital
Principal quantum
number
gets larger as the principal quantum number n increases.
Correspondingly, the energy of the electron in such an orbital
becomes less negative, meaning that the electron is less strongly
bound and has less energy. The graph of principal quantum
number with atomic number is
or
1  1
1
∝ − 
λ  n12 n22 
For shortest wavelength, i.e. highest energy spectral line, n2 will
be (∞ ).
For the given spectral series, ratio of the shortest wavelength of
two spectral series can be calculated as follows :
1
1
1
−
−0
λ L 32 ∞ 2 9
1
(a)
=
=
=
1
1
λP
1
−
0
9
−
12 ∞ 2
1
1
− 2
2
λ Bk
1 16 16
5
∞
(b)
=
=
×
=
1
1
λ Pf
25
1 25
−
4 2 ∞2
1
1
− 2
2
λP
1 9 9
∞
5
(c)
=
=
× =
1
1
λ Pf
25
1 25
−
32 ∞ 2
1
1
− 2
2
λB
1 4 1
∞
4
(d)
=
=
× =
1
1
λ Bk
16
1 4
−
22 ∞ 2
Note Lyman = L (n1 = 1), Balmer = B (n1 = 2)
Paschen = P (n1 = 3), Brackett = Bk (n1 = 4 )
Pfund = Pf (n1 = 5)
6. The graphs between | ψ |2 and r are radial density plots having
(n − l − 1) number of radial nodes. For 1s, 2s, 3s and 2 p-orbitals
these are respectively.
|Ψ|2
2
For 2s-orbital
number of radial node = 2–0–1=1
r
1
2s
1
25
50
75
|Ψ|2
For 3s-orbital
number of radial node = 3–0–1=2
100
Atomic number
r
4. The electrons are more likely to be found in the region aand c. At
b, wave function becomes zero and is called radial nodal surface
or simply node.
a
|Ψ|2
Y (x)
For 2p-orbital
number of radial node = 2–1–1=0
+x →
b
← –x
r
c
Thus, the given graph between | ψ |2 and r represents 2s-orbital.
Atomic Structure 35
Thus, Eq. (i) becomes
7. For any given series of spectral lines of atomic hydrogen.
Let ∆ν = ν max − ν min be the difference in maximum and
minimum frequencies in cm−1.
For Lyman series,
2πa0
2πa0
∴
∆ν = ν max − ν min
General formula:
1
1
ν = 109677  2 − 2 
nf 
 ni
For Lyman n1 = 1, n2 = 2, 3, K
1 
1 1
ν max = 109,677 −  = 109,677  − 0
1 
 1 ∞
ν min
n2
= n (1.5 πa0) [Given, λ = 1 .5 πa0]
Z
n 15
. πa0 1.5
=
= 0.75
=
2
Z
2πa0
λ=
λ∝
∴
For Balmer series,
...(i)
1
(ν − ν 0 )1/ 2
11. (I) Angular momentum, mvr =
109677
4
nh
2π
⇒
mvr ∝ n
∝ distance from the nucleus
(II) This statement is incorrect as size of an orbit
∝ Azimuthal quantum number (l )
(Q n = constant)
(III) This statement is incorrect as at ground state,
n = 1, l = 0
⇒ Orbital angular momentum (wave mechanics)
h
= l (l + 1)
=0
[Q l = 0 ]
2π
(IV) The given plot is
 1
109677 × 5
1 
ν min = 109,677 2 − 2  ⇒
36
 (2) (3) 
∆ν = ν max − ν min
109,677 109,677 
 1
∆ν Balmer =
−
× 5 = 109,677  
 9
4
36


∆ν Lyman 109,677 / 4
=
∆ν Balmer 109,677 / 9
∆ν Lyman 9
=
∆ν Balmer 4
∆ν Lyman
is 9 : 4.
∴The ratio of
∆ν Balmer
l=0 (n=1)
l=1 (n=2)
8. Smaller the value of (n + l ), smaller the energy. If two or more
sub-orbits have same values of (n + l ), sub-orbits with lower
values of n has lower energy. The (n + l ) values of the given
options are as follows :
I. n = 4, l = 2 ; n + l = 6
II. n = 3, l = 2; n + l = 5
III. n = 4 , l = 1, n + l = 5
IV. n = 3, l = 1, n + l = 4
Among II and III, n = 3 has lower value of energy. Thus, the
correct order of their increasing energies will be
IV < II < III < I
Circumference
2πr
9. Number of waves =
⇒ n=
Wavelength
λ
∴
2πr = nλ
Also, we know that radius (r) of an atom is given by
a n2
r= 0
Z
h
2 m K.E
Also, according to photoelectric effect
KE = hν − hν 0
On substituting the value of KE in Eq (i), we get
h
λ=
2m × (hν − hν 0 )
∆ν Lyman = ν max − ν min
109,677 × 3  109,677
= 109,677 − 
 =
4
4

⇒
...(ii)
10. de-Broglie wavelength (λ) for electron is given by
= 109,677
1
1 
= 109,677 − 2 
 1 (2) 
 1
1
ν max = 109,677 2 − 
 (2) ∞ 
n2
= nλ
Z
...(i)
l=2 (n=3)
ψ
l=3 (n=4)
r
12. ∆E = hc ×

  1
1
1
= hc × RH  2 − 2  × Z 2 
λ

  n1 n2 
1
1
hc
[for H, atom Z = 1]
− 2=
2
n1 n2 RH × λ × Z 2 × hc
1
1
1
=
=
×
RH × λ (1 × 10 7 m−1 ) (900 × 10−9 m)
1
1 1
⇒
−
=
n12 n22 9
⇒
So, in option (b)
1
1
1
1
−
= −0=
9
32 ∞ 2 9
∴ n1 = 3,
 n2 = ∞ 
36 Atomic Structure
13. According to Rydberg’s formula,
1
1
wave number (ν) = RH Z  2 − 2 
nf 
 ni
Given, ni = n, nf = 8 [Q it is the case of emission]
1 1
ν = RH × (1)2  2 − 2 
n 8 
1 R
R
1
ν = RH  2 −  = H2 − H
64
 n 64  n
2
On comparing with equation of straight line, y = mx + c, we get
−RH
Slope = RH , intercept =
.
64
1
Thus, plot of wave number (ν) against 2 will be linear with
n
slope (+ RH ).
n2h2
4π 2me2kZ
1
k=
4 π ∈0
rn =
rn =
Thus, its electronic configuration is [ Kr ]5s1. Since, the last
electron or valence electron enter in 5s subshell.
So, the quantum numbers are n = 5, l = 0, (for s-orbital) m = 0
(Q m = + l to −l), s = + 1 / 2 or − 1 / 2.
Z2 
18. Given, in the question E = − 2.178 × 10−18 J  2 
n 
For hydrogen
Z = 1,
1
So,
E1 = − 2.178 × 10−18 J  2 
1 
1
−18
E2 = − 2.178 × 10 J  2 
2 
Now, E1 − E2
1  hc
1
i.e.
∆E = 2.178 × 10−18  2 − 2  =
1
2  λ
1
6.62 × 10−34 × 3.0 × 108
1
2.178 × 10−18  2 − 2  =
1
λ
2 
∴
λ ≈ 1.21 × 10−7 m
14. Bohr radius (rn ) = ∈0 n2h2
∴
17. Given, atomic number of Rb, Z = 37
19. According to Bohr’s model,
mvr =
2 2
n h ∈0
a
= n2 0
Z
πme2Z
nh
2π
⇒ (mv )2 =
n2h2
4 π 2r2
e = charge of electron
n2h2
1
mv 2 =
2
8π 2r2m
Also, Bohr’s radius for H-atom is, r = n2a0
h = Planck’s constant
Substituting ‘r’ in Eq. (i) gives
⇒
where, m = mass of electron
k = Coulomb constant
rn =
KE =
n2 × 0.53
Å
Z
Radius of n
th
KE =
h2
8π 2n2a02m
when n = 2 , KE =
…(i)
h2
32π 2 a02 m
20. The number of radial nodes is given by expression (n − l − 1).
Bohr orbit for H-atom
2
= 0.53 n Å
[Z = 1for H-atom]
∴Radius of 2nd Bohr orbit for H-atom
For 3s, number of nodes = 3 − 0 − 1 = 2
For 2p, number of nodes = 2 − 1 −1 = 0
21. Expression for Bohr’s orbit is, rn =
= 0.53 × (2)2 = 212
. Å
15. This graph shows the probability of finding the electron within
shell at various distances from the nucleus (radial probability).
The curve shows the maximum, which means that the radial
probability is greatest for a given distance from the nucleus.
This distance is equal to Bohr’s radius = a0
a0n2
= a0
Z
when n = 2, Z = 4.
22. 1s7 violate Pauli exclusion principle, according to which an
orbital cannot have more than two electrons.
1
1
23. + and − just represents two quantum mechanical spin states
2
2
which have no classical analogue.
24. Using the de-Broglie’s relationship :
P
λ=
1s
r
(a)
(b)
(c)
(d)
It is for 2s-orbital.
It is radial wave function for 1s.
Correct
Probability cannot be zero at a certain distance from nucleus.
13.6
eV
n2
−13.6
In excited states, E2 =
= −3.4 eV
4
16. ∴
En = −
where, n = 1, 2, 3 ...
h
6.625 × 10−34
=
= 2.3 × 10−30 m
5
mv 0.2 ×
60 × 60
25. Nodal plane is an imaginary plane on which probability of
finding an electron is minimum. Every p-orbital has one nodal
plane :
px
YZ-plane, a nodal plane
Atomic Structure 37
26. 1s2 2 s2 2 p6 3s2 3 p6 3d 5 4 s1 is ground state electronic configuration
38. Option (b) is wrong representation according to aufbau
principle. A high energy atomic orbital (2p) cannot be filled
unless the low energy orbital (2s) is completely occupied.
of Cr.
27.
(i)
(ii)
(iii)
(iv)
n = 4, l = 1
n = 4, l = 0
n = 3, l = 2
n = 3, l = 1
⇒
⇒
⇒
⇒
4 p-orbital
4s-orbital
3d-orbital
3d-orbital
 1
1  hc
− 2 =
2
 n1 n2  λ
39. Transition energy (∆E ) = kZ 2 
(iv) 3p < (ii) 4s < (iii) 3d < (i) 4 p
28. The energy of an electron in a Bohr atom is expressed as
where, k = Constant,
Z = Atomic number,
n = Orbit number
= − 13.6 eV for H (n = 1)
− 13.6
when n = 2 , E2 =
eV = − 3.40 eV
22
kZ 2
En = − 2
n
(n can have only integral value 1, 2, 3,…… ∞)
h
29. The orbital angular momentum (L) = l (l + 1)
2π
h
= 6
(l = 2 for d - orbital )
2π
30. Bohr first made use of quantum theory to explain the structure of
atom and proposed that energy of electron in an atom is
quantised.
31. Mg2+ = 1s2 2s2 2 p6 no unpaired electron
Ti
3+
V
3+
2
2
6
2
6
1
2
2
6
2
6
2
= 1s 2 s 2 p 3s 3 p 3d one unpaired electron
= 1s 2 s 2 p 3s 3 p 3d two unpaired electrons
Fe2+ = 1s2 2 s2 2 p6 3s2 3 p6 3d6 four unpaired electrons
32. Expression for orbital angular momentum (L) is
h
L = l (l + 1)
= 0 for 2s-electrons
2π
Q For s-orbital, l = 0.
33. Diffraction is property of wave, E = mc2 determine energy of
particle and E = hν determine energy of photon. Interference
phenomena is exhibited by both matter and waves.
34. X-rays is electrically neutral, not deflected in electric or
magnetic fields.
35. Cl (17) = 1s2 2 s2 2 p6 3s2 3 p5
∆E ∝
i.e.
According to Aufbau principle, energies of above mentioned
orbitals are in the order of
40. E =
hc
E
λ
⇒ 1 = 2 =2
λ
E2 λ 1
41.
n
l
1
λ
m
s
1
2
3
−3
2
This is the wrong set of quantum number because | m | cannot be
greater than l.
42. The wavelength order is
X-ray < ultraviolet < infrared < radio wave
43. When electron jumps to lower orbit photons are emitted while
photons are absorbed when electron jumps to higher orbit.
1s-orbital is the lower most, electron in this orbital can absorb
photons but cannot emit.
44. The valence shell configuration of Rubidium (Rb) is
[ Kr ] 5s1
n = 5, l = 0, m = 0, s = +
1
or
2
−
1
2
45. The principal quantum number ‘n’ represents orbit number
hence, determine the size of orbitals.
46. According to Pauli exclusion principle, an atomic orbital can
accommodate at the most, two electrons, with opposite spins.
47. Given, ground state energy of hydrogen atom = − 13.6 eV
Energy of He + = − 3.4 eV, Z = 2
13.6 × Z 2
eV
n2
− 13.6 × (2 )2
13.6 × 4
− 3.4 eV =
⇒ n=
⇒ n=4
3.4
n2
Given, azimuthal quantum number (l ) = 2 (d – subshell
Magnetic quantum number (m) = 0
∴ Angular nodes (l ) = 2
Radial node = n − l − 1= 4 − 2 − 1= 1
nl = 4 d state
Hence, options (a), (c) are correct.
Energy of He + , E = −
48. Both (a) and (d) are correct. The three electrons in the
2p-orbitals must have same spin, no matter up spin or down spin.
The last, unpaired electron has, n = 3, l = 1( p) and m can have
any of the three value (− 1, 0, + 1).
36. Cr (24) = 1s2 2 s2 2 p6 3s2 3 p6 3d 5 4 s1
1442443
Ar
The above configuration is exception to Aufbau’s principle.
37. Fluorine, a halogen, is the most electronegative atom, has the
2
5
electronic configuration 2s 2 p (valence shell).
49. (a) Cr = [Ar] 3d 5 4 s1 , an exception to aufbau principle.
(b) For a given value of l, m can have any value from
(−l to + l), so can have negative value.
(c) Ag is in copper group with d 10s1 configuration,
i.e. 46 electrons are spin paired.
50. Isotones have same number of neutrons.
76
77
32 Ge , 33As
and 34Se78 have same number (44) of neutrons,
hence they are isotones.
38 Atomic Structure
51. Assertion is correct Be(1s2 , 2s2 ) has stable electronic
configuration, removing an electron require more energy than
the same for B(2 p1 ). Reason is incorrect
(Aufbau principle).
52. S 1 is spherically symmetrical state, i.e. it correspond to a
s-orbital. Also, it has one radial node.
Number of radial nodes = n − l − l
⇒
n − 0 − 1 = 1 ⇒ n = 2 i.e. S 1 = 2s-orbital.
66. 3dx2 − y2 orbital lies in XY-plane.
67. Aufbau principle.
68. This is the wavelength of infrared radiation.
69. Cr = 3d 5 4 s1.
70. In an one electron (hydrogenic) system, all orbitals of a shell
remains degenerate, hence in second excited state, the
degeneracy of H-atom is nine
EH =
3p
3s
53. Ground state energy of electron in H-atom (EH )
H(1s1)
Ground state
kZ 2
= k (Z = 1, n = 1)
n2
For S 1 state of Li 2+ ,
3d
Second excited
state of H-atom
All are degenerate
degeneracy = 9
In case of many electrons system, different orbitals of a shell are
non-degenerate. Hence,
k (3)2 9
E = 2 = k = 2.25 k
4
2
1s
H
2+
54. In S 2 state, E (Li ) = K (given)
Ground state
qk
K = 2 ⇒ n=3
n
Since, S 2 has one radial node.
3 − l −1 = 1 ⇒ l = 1
2p
1s 2s
1s
2s
2p
–
Second excited
state only three p-orbitals
(2px, 2py, 2pz) are
First excited state
degenerate.
71.
+
55. In the wave function (ψ ) expression for 1s-orbital of He , there
should be no angular part. Hence (iii) can’t be true for ψ 1s of
He+ .
PLAN This problem is based on concept of quantum number. Follow
the following steps to solve this problem.
Write all possible orbitals having combination of same principal,
azimuthal, magnetic and spin quantum number.
Then count the all possible electrons having given set of
quantum numbers.
56. Correct : 2s orbital has one radial node.
No of radial node = n − l − 1 = 2 − 0 − 1 = 1
For n = 4, the total number of possible orbitals are
Also, when radial part of wave function (ψ ) is plotted against
‘‘r’’, wave function changes its sign at node.
4s
4p
0
–1 0 +1
57. i is the correct expression of wave function for 1s-orbital of
4d
–2 –1 0 +1 +2
4f
–3 –2 –1 0 +1 +2 +3
i.e. L depends on azimuthal quantum number only.
According to question | m l | = 1, i.e. there are two possible
values of m l , i.e. +1 and –1 and one orbital can contain
1
maximum two electrons one having s = + and other having
2
s = − 1/ 2 .
So, total number of orbitals having {| m l | = 1} = 6
B. To describe a one electron wave function, three quantum
numbers n, l and m are needed. Further to abide by Pauli
exclusion principle, spin quantum number(s) is also needed.
Total number of electrons having
1
{| m l | = 1and ms = − } = 6
2
hydrogenic system.
58. A. Orbital angular momentum
(L) = l (l + 1)
h
2π
C. For shape, size and orientation, only n, l and m are needed.
D. Probability density (ψ 2 ) can be determined if n, l and m are
known.
72.
PLAN KE =
λ (wavelength) =
65. Very large mass of alpha particles than beta particles is
responsible for less deflection in former case.
h
∝
2mKE
h
2m(T )
T = Temperature in Kelvin
h
λ (He at −73° C = 200 K) =
2 × 4 × 200
proposed wave nature of electron.
64. Two electrons in same orbital must have opposite spin.
h
=
mv
where,
61. Heisenberg proposed uncertainty principle and de-Broglie
62. orbital
63. 2 px ,2 py and 2pz have different orientation in space.
m2v 2 = 2mKE ∴ mv = 2mKE
∴
59. Cr = [Ar] 3d 5 4 s1
60. 1 : 16
1
3
mv 2 = RT
2
2
λ (Ne at 727°C = 1000 K) =
∴
Thus,
h
2 × 20 × 1000
λ (He)
=M =
λ (Ne)
M =5
2 × 20 × 1000
=5
2 × 4 × 200
Atomic Structure 39
73. Energy of photon
=
–34
8
hc
hc
6.625 × 10 × 3 × 10
= 4 . 14 eV
J=
eV =
λ
eλ
300 × 10–9 × 1.602 × 10–19
For photoelectric effect to occur, energy of incident photons
must be greater than work function of metal. Hence, only Li, Na,
K and Mg have work functions less than 4.14 V.
74. When n = 3, l = 0, 1, 2 i.e. there are 3s, 3p and 3d-orbitals. If all
these orbitals are completely occupied as
1
and 9 with
2
1
s=– .
2
Total 18 electrons, 9 electrons with s = +
Alternatively In any nth orbit, there can be a maximum of 2n2
electrons. Hence, when n = 3, number of maximum electrons =
1
18. Out of these 18 electrons, 9 can have spin –
2
1
and remaining nine with spin + .
2
nh
75. (a) mvr =
2π
⇒
6.625 × 10−34
nh
v=
=
2πmr 2 × 3.14 × 9.1 × 10−31 × 0.529 × 10−10
= 2.18 × 106 ms − 1
(b) λ =
h
6.625 × 10−34
= 0.33 × 10−9 m
=
mv 9.1 × 10−31 × 2.18 × 106
(c) Orbital angular momentum
h
 h
(L) = l (l + 1)
= 2 
 2π 
2π
78. Moles of H2 =
⇒ Bond energy = 0.0409 × 436 = 17.84 kJ
Number of H-atoms produced after dissociation
= 2 × 0.0409 × 6.023 × 1023 = 4.93 × 1022
1

Transition energy/atom = 2.18 × 10−18 1 −  J

4
3
= × 2.18 × 10−18 J
4
⇒ Total transition energy
3
= × 2.18 × 10−18 × 4.93 × 1022 J
4
= 80.60 × 103 J = 80.60 kJ
Therefore, total energy required
= dissociation energy + transition energy
= (17.84 + 80.60) kJ = 98.44 kJ
79. If accelerated by potential difference of V volt, then
1
mv 2 = eV
2
p2
= eV , here p = momentum (mv )
2m
h
h
Using de-Broglie equation, λ = =
p
2meV
⇒
⇒
1.54 × 10−10 =
80. The work done in the given neutralisation process is
W =−∫
2
2
⇒
(b)
r0
2
r0  − a 0
 1  
ψ 22s = 0 = 
 2 −  e
a0 
 4 2π  
r0
2−
= 0 ⇒ r0 = 2a0
a0
λ=
h
6.625 × 10−34
= 6.625 × 10−35 m
=
mv 100 × 10−3 × 100
= 6.625 × 10−25 Å (negligibly small)
77. The general Rydberg’s equation is
ν=
⇒
1
= R (Z )2
λ
 1
1
 2 − 2
n
n
 1
2
1
∝ Z2
λ
⇒
1
Z (H)2
λ (He+ )
=
=
λ (H)
Z (He+ )2 4
⇒
λ (He+ ) =
λ (H) 91.2
=
nm = 22.8 nm
4
4
6.625 × 10−34
(2 × 9.1 × 10−31 × 1.6 × 10−19 V )1/ 2
Solving for V gives : V = 63.56 V.
[Q For p-orbital, l = 1]
76. (a) At radial node, ψ must vanishes, i.e.
pV
1×1
=
= 0.0409
RT 0.082 × 298
⇒
W =
∞
a0
F dr and F =
e2
4πε 0r2
∞
e2  1 
e2
=
−
= Total energy (E )
4 πε 0  r  a 0
4 πε 0r
Now, if ‘V’ is magnitude of potential energy, then according to
given information, kinetic energy (Ek ) is V / 2. Therefore,
V
(PE is always negative)
E = −V +
2
V
=−
2
⇒
V = − 2E =
− e2
2πε 0r
81. The Rydberg’s equation for H-atom is
 1
1
1
= ν (wave number) = RH  2 − 2 
λ
 n1 n2 
For Balmer series, n1 = 2 and n2 = 3, 4 , 5, ..., ∞
For shortest λ , n2 has to be maximum, i.e. infinity. Then
1.09 × 107
 1 1 R
ν = RH  −  = H =
= 2.725 × 106 m −1
 4 ∞
4
4
40 Atomic Structure
82. After breaking of the bond of I2 molecule, the remaining energy
would be distributed uniformly to iodine atoms as their kinetic
energy, i.e.
E (energy of photon) = Bond energy + 2 × kinetic energy
⇒
240 × 103
6.625 × 10−34 × 3 × 108
+ 2 × Ek
=
4500 × 10−10
6.023 × 1023
Ek = 2.16 × 1020 J/atom
⇒
For longest wavelength transition from 3rd orbit, electron must
jump to 4th orbit and the transition energy can be determined as
1 1
∆E = + 4 × 21.7 × 10−19  −  J = 4.22 × 10−19 J
 9 16
hc
λ
∆E =
Also, Q
λ=
∴
83. The Bohr de-Broglie relationship is
hc 6.625 × 10−34 × 3 × 108
m
=
∆E
4.22 × 10−19
= 471 × 10−9 m = 471 nm
2πr = nλ = circumference of Bohr’s orbit.
i.e. number of complete waves formed in one complete
revolution of electron in any Bohr orbit is equal to orbit number,
hence three.
87. Ten, the given value of n and l correspond to 3d-orbital which
84. The expression for transition wavelength is given by Rydberg’s
multiplicity which states that the singly occupied degenerate
atomic orbitals must have electrons of like spins.
equation :
 1
1
1
= RH Z 2  2 − 2 
λ
 n1 n2 
has five fold degeneracy level.
88. The 2nd configuration is against Hund’s rule of maximum
89. The required transition is n1 = 2 to n2 = ∞ and corresponding
transition energy is
 1
1
∆E = 21.7 × 10−12  2 − 2  erg
 n1 n2 
+
Equating the transition wavelengths of H-atom and He ion,
 1
1
4
4
RH  2 − 2  = RH  2 − 2 
2
4 
 n1 n2 
=
Equating termwise on left to right of the above equation gives
n1 = 1 and n2 = 2
85. For H-atom, the energy of a stationary orbit is determined as
k
En = − 2
n
where, k = constant (2.18 × 10
−18
⇒
The longest wavelength that can cause above transition can be
determined as :
λ=
J)
1 3

⇒ ∆E (n = 2 to n = 1) = k 1 −  = k

4 4
= 1.635 × 10−18 J
For a H-like species, energy of stationary orbit is determined as
kZ 2
En = − 2
n
where, Z = atomic number
1 ∆E k 2  1
Z  −
=
=
1
hc hc
λ
⇒
Z2 =
90. Ionisation potential of H-like species
= E1 = 2.17 × 10−11 erg
⇒
= 1.6275 × 10−18 J ⇒ λ =
3
1
2
 = RH Z ×

4
4
6.625 × 10−34 × 3 × 108
m
1.6275 × 10−18
91. Transition energy = [ − 2.41 − (− 5.42)] × 10−12 erg
= 3.01 × 10−12 erg
= 3.01 × 10−19 J
86. For H-like species, the energy of stationary orbit is expressed as
E ( X ) = Z 2 × E (H )
Also,
⇒
+
⇒ For He (Z = 2)
E=−
hc
∆E
= 122 × 10−9 m = 1220 Å
Z = 2 (He+ )
⇒
1

∆E = 2.17 × 10−11 1 − 2  × 10−7 J

2 
=
4
4
=
= 4.05
3RH λ 3 × 1.097 × 107 × 3 × 10−8
hc 6.625 × 10−34 × 3 × 108
=
∆E
5.425 × 10−12 × 10−7
= 3.66 × 10−7 m = 3.66 × 10−5 cm
 1
1
∆E = kZ 2  2 − 2 
 n1 n2 
⇒
21.7
× 10−12 erg = 5.425 × 10−12 erg
4
4 × 21.7 × 10−19
J
n2
∆E =
λ=
[Q 1 erg = 10−7 J ]
hc
λ
6.625 × 10−34 × 3 × 108
m
3.01 × 10−19
= 660 × 10−9 m = 660 nm
3
Periodic Classification
and Periodic Properties
Topic 1 History and Periodic Classification
Objective Questions I (Only one correct option)
1. The IUPAC symbol for the element with atomic number 119
would be
(a) unh
(c) uun
(2019 Main, 8 April II)
(b) uue
(d) une
2. The element with Z = 120 (not yet discovered) will be an/a
(2019 Main, 12 Jan I)
(a) transition metal
(c) alkaline earth metal
(b) inner-transition metal
(d) alkali metal
Objective Question II
(One or more than one correct option)
4. The statements that is/are true for the long form of the
periodic table is/are
3. The statement that is not correct for the periodic
classification of elements, is
(c) the first ionisation energies of elements along a period do not
vary in a regular manner with increase in atomic number
(d) for transition elements the d-subshells are filled with electrons
monotonically with increase in atomic number
(1992, 1M)
(a) the properties of elements are the periodic functions of their
atomic numbers
(b) non-metallic elements are lesser in number than metallic
elements
(1988, 1M)
(a) it reflects the sequence of filling the electrons in the order of
sub-energy level s, p, d and f
(b) it helps to predict the stable valency states of the elements
(c) it reflects tends in physical and chemical properties of the
elements
(d) it helps to predict the relative ionicity of the bond between any
two elements
Topic 2 Periodic Properties
Objective Questions I (Only one correct option)
2−
3−
−
3. The group number, number of valence electrons and valency
2+
1. The correct order of the ionic radii of O , N , F , Mg ,
+
3+
Na and Al is
(2020 Main, 5 Sep II)
(a) N3 − < O2 − < F− < Na + < Mg 2+ < Al 3+
(b) Al 3+ < Na + < Mg 2+ < O2 − < F− < N3 −
(c) Al 3+ < Mg 2+ < Na + < F− < O2 − < N3 −
3−
(d) N
−
2−
<F <O
< Mg
2+
+
< Na < Al
3+
2. Within each pair of elements F and Cl, S and Se, and Li and
Na, respectively, the elements that release more energy upon
an electron gain are
(2020 Main, 7 Jan II)
(a) F, Se and Na
(b) F, S and Li
(c) Cl, S and Li
(d) Cl, Se and Na
of an element with atomic number 15, respectively, are
(2019 Main, 12 April I)
(a) 16, 5 and 2
(c) 16, 6 and 3
(b) 15, 5 and 3
(d) 15, 6 and 2
4. The element having greatest difference between its first and
second ionisation energy, is
(a) Ca
(b) Sc
(c) Ba
(d) K
(2019 Main, 9 April I)
5. The
correct option with respect to
electronegativity values of the elements is
the
Pauling
(2019 Main, 11 Jan II)
(a) P > S
(c) Te > Se
(b) Si < Al
(d) Ga < Ge
42 Periodic Classification and Periodic Properties
6. The correct order of the atomic radii of C, Cs, Al and S is
(2019 Main, 11 Jan I)
(a) C < S < Al < Cs
(c) S < C < Cs < Al
(b) C < S < Cs < Al
(d) S < C < Al < Cs
7. In general, the properties that decrease and increase down a
group in the periodic table, respectively are
(2019 Main, 9 Jan I)
(a) electronegativity and atomic radius
(b) electronegativity and electron gain enthalpy
(c) electron gain enthalpy and electronegativity
(d) atomic radius and electronegativity
3−
2−
8. The ionic radii (in Å) of N ,O
and F respectively are
(b) 1.36, 1.71 and 1.40
(d) 1.71, 1.36 and 1.40
(2015 Main)
(c) Kr
(d) Xe
12. The first ionisation potential of Na is 5.1 eV. The value of
electron gain enthalpy of Na + will be
(2013 Main)
(b) − 5.1 eV
(d) + 2.55 eV
(b) [Ne] 3s2 3 p3
(d) [Ar] 3d 10 4s2 4 p3
(1989, 1M)
(a) N3−
(c) F −
(b) O2−
(d) Na +
21. The first ionisation potential of Na, Mg, Al and Si are in the
order
(a) Na < Mg >Al < Si
(c) Na < Mg <Al >Si
(1988, 1M)
(b) Na > Mg > Al > Si
(d) Na > Mg > Al <Si
(2013 Main)
(b) S < Se < Ca < Ba < Ar
(d) Ca < Ba < S < Se < Ar
14. Identify the least stable ion amongst the following.
(b) Be−
(d) C −
(2001, 1M)
(d) Ge > Si > C
(1987, 1M)
(b) 1.60, 1.60
(d) None of these
(a) 14.6, 13.6
(c) 13.6, 13.6
(1987, 1M)
(b) 13.6, 14.6
(d) 14.6, 14.6
25. The hydration energy of Mg 2+ is larger than that of
(a) Al 3+
(c) Be2+
(1984, 1M)
(b) Na +
(d) Mg3+
(1982, 1M)
(a) boron
(c) nitrogen
(b) carbon
(d) oxygen
27. The correct order of second ionisation potential of carbon,
nitrogen, oxygen and fluorine is
16. The correct order of radii is
(2000, 1M)
(a) N < Be < B
(b) F − < O2− < N3−
(c) Na < Li < K
(d) Fe3+ < Fe 2+ < Fe4+
17. The incorrect statement among the following.
respectively given by
(a) 0.72, 1.60
(c) 0.72, 0.72
26. The element with the highest first ionisation potential is
potential is
(b) Be > Mg > Ca
23. Atomic radii of fluorine and neon in Angstrom units are
(2002, 3M)
15. The set representing the correct order of first ionisation
(c) B > C > N
(1987, 1M)
(b) N, Si, C, P
(d) P, Si, N, C
oxygen atoms are respectively given by
increasing first ionisation enthalpy for Ca, Ba, S, Se and Ar?
(a) K > Na > Li
(a) C, N, Si, P
(c) Si, P, C, N
24. The first ionisation potential in electron volts of nitrogen and
13. Which of the following represents the correct order of
(a) Li +
(c) B−
2
the order
(2015 Main)
11. Which one has the highest boiling point?
(a) Ca < S < Ba < Se < Ar
(c) Ba < Ca < Se < S < Ar
configurations are given below), the one having the highest
ionisation energy is
(1990, 1M)
22. The electronegativity of the following elements increases in
(b) Br2
(d) ICl
(a) − 2.55 eV
(c) − 10.2 eV
19. Amongst the following elements (whose electronic
20. Which one of the following is the smallest in size?
(b) BeSO4
(d) SrSO4
(b) Ne
(d) Fe2+
−
10. Which among the following is the most reactive?
(a) He
(c) V 3+
(c) [Ne] 3s 3 p
has its hydration enthalpy greater than its lattice enthalpy?
(a) Cl 2
(c) I2
(1996, 1M)
(b) Ti 3+
2
9. Which one of the following alkaline earth metal sulphates
(a) CaSO4
(c) BaSO4
unpaired electrons ?
(a) Mg2+
(a) [Ne] 3s2 3 p1
(2015 Main)
(a) 1.36, 1.40 and 1.71
(c) 1.71, 1.40 and 1.36
18. Which of the following has the maximum number of
(1997(C), 1M)
(a) The first ionisation potential of Al is less than the first
ionisation potential of Mg
(b) The second ionisation potential of Mg is greater than the
second ionisation potential of Na
(c) The first ionisation potential of Na is less than the first
ionisation potential of Mg
(d) The third ionisation potential of Mg is greater than third
ionisation potential of Na
(a) C > N > O > F
(c) O > F > N > C
(1981, 1M)
(b) O > N > F > C
(d) F > O > N > C
Objective Questions II
(One or more than one correct option)
28. The option(s) with only amphoteric oxides is(are)
(2017 Adv.)
(a) NO, B2O3 , PbO, SnO2
(c) Cr2O3 , BeO, SnO, SnO2
(b) Cr2O3 , CrO, SnO, PbO
(d) ZnO, Al 2O3 , PbO, PbO2
29. Ionic radii of
(1999, 3M)
35
Cl − <
(a) Ti 4+ < Mn 7 +
(b)
37
(c) K + > Cl −
(d) P 3+ > P 5+
Cl −
Periodic Classification and Periodic Properties 43
30. The first ionisation potential of nitrogen and oxygen atoms
are related as follows.
(1989, 1M)
(a) The ionisation potential of oxygen is less than the ionisation
potential of nitrogen
(b) The ionisation potential of nitrogen is greater than the
ionisation potential of oxygen
(c) The two ionisation potential values are comparable
(d) The difference between the two ionisation potential is too large
31. Sodium sulphate is soluble in water whereas barium sulphate
is sparingly soluble because
(1989, 1M)
(a) the hydration energy of sodium sulphate is more than its lattice
energy
(b) the lattice energy of barium sulphate is more than its hydration
energy
(c) the lattice energy has no role to play in solubility
(d) the hydration energy of sodium sulphate is less than its lattice
energy
Atomic
number
n
n+1
n+ 2
n+ 3
Ionisation enthalpy (kJ/mol)
I1
I2
I3
1681
3374
6050
2081
3952
6122
496
4562
6910
738
1451
7733
Fill in the Blanks
38. Compounds that formally contain Pb 4+ are easily reduced to
Pb 2+ . The stability of the lower oxidation state is due to
…… .
(1997, 1M)
39. Ca 2+ has a smaller ionic radius than K + because it has
............
(1993, 1M
40. On Mulliken scale, the average of ionisation potential and
electron affinity is known as ................
(1985, 1M)
41. The energy released when an electron is added to a neutral
Assertion and Reason
gaseous atom is called …… .
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.
32. Statement I Nitrogen and oxygen are the main components
in the atmosphere but these do not react to form oxides of
nitrogen.
Statement II The reaction between nitrogen and oxygen
requires high temperature.
(2015 Main)
33. Statement I Pb 4+ compounds are stronger oxidising agents
4+
than Sn compounds.
Statement II The higher oxidation states for the group 14
elements are more stable for the heavier members of the
group due to ‘inert pair effect’.
(2008, 3M)
34. Statement I Band gap in germanium is small.
Statement II The energy spread of each germanium atomic
energy level is infinitesimally small.
(2007, 3M)
35. Statement I The first ionisation energy of Be is greater than
True/False
42. The basic nature of the hydroxides of group 13 (III B)
decreases progressively down the group.
(2000, (S), 1M)
36. Statement I F-atom has a less negative electron affinity than
Cl-atom.
Statement II Additional electrons are repelled more
effectively by 3 p-electrons in Cl-atom than by 2 p-electrons in
F-atom.
(1998, 2M)
(1993, 1M)
43. The decreasing order of electron affinity of F, Cl, Br is
F > Cl > Br.
(1993, 1M)
44. In group IA of alkali metals, the ionisation potential
decreases down the group. Therefore, lithium is a poor
reducing agent.
(1987, 1M)
45. The softness of group IA metals increases down the group
with increasing atomic number.
(1986, 1M)
Subjective Questions
46. Arrange the following ions in order of their increasing radii
Li + , Mg 2+ , K + , Al 3+.
(1997, 1M)
47. Compare qualitatively the first and second ionisation
potentials of copper and zinc. Explain the observation.
(1996, 2M
48. Arrange the following as stated :
“Increasing order of ionic size’’ N3– ,Na + , F− , O2 − , Mg 2+
(1991, 1M)
that of B.
Statement II 2p-orbital is lower in energy than 2s.
(1982, 1M)
49. Explain the following :
“The first ionisation energy of carbon atom is greater than
that of boron atom whereas, the reverse is true for the second
ionisation energy.’’
(1989, 2M)
50. Arrange the following in the order of their increasing size:
Cl − , S2 − , Ca 2+ , Ar
(1986, 1M)
51. Arrange the following in order of their
(i) decreasing ionic size Mg 2+ , O2 − , Na + ,F−
Numerical Answer Type Questions
37. The 1st, 2nd and 3rd ionisation enthalpies, I1 , I 2 , and I 3 , of
four atoms with atomic numbers n , n + 1, n + 2, and n + 3,
where n < 10, are tabulated below. What is the value of n ?
(2020 Adv.)
(ii) increasing first ionisation energy Mg, Al, Si, Na
(iii) increasing bond length F2 , N2 , Cl 2 , O2
(1985, 3M)
44 Periodic Classification and Periodic Properties
Answers
Topic 1
1. (b)
2. (c)
3. (d)
4. (b,c,d)
Topic 2
1.
5.
9.
13.
17.
(c)
(d)
(b)
(c)
(b)
2.
6.
10.
14.
18.
3.
7.
11.
15.
19.
(c)
(a)
(d)
(b)
(d)
(b)
(a)
(d)
(b)
(b)
4.
8.
12.
16.
20.
(d)
(c)
(b)
(b)
(d)
21.
25.
29.
33.
37.
39.
(a)
22. (c)
23. (a)
(b)
26. (c)
27. (c)
(d)
30. (a,b,c)
31. (a,b)
(c)
34. (c)
35. (c)
(9)
38. (inert pair effect)
(higher effective nuclear charge)
24.
28.
32.
36.
(a)
(a,b)
(a)
(c)
40. (electronegativity)
41. (electron affinity)
42. F
44. F
43. F
45. T
Hints & Solutions
Topic 1 History and Periodic Classification
1.
Atomic number (119) = 1
1
9
un un en
So, symbol of the element = uue
Name of the element = ununennium
It is expected to be s-block element an alkali metal and the first
element in eighth period. It is the lightest element that has not
yet been synthesised.
2. The element with Z = 120 will be an alkaline earth metal.
Recently, oganesson (Og) with atomic number 118 is named by
IUPAC is a noble gas and placed just two place before 120. So,
the general electronic configuration is represented as [noble gas]
ns2 and element with Z = 120 exist as an alkaline earth metal.
3. (a) Correct statement According to Moseley’s law, the
properties of elements are the periodic function of their
atomic numbers.
(b) Correct statement The whole s-block, d-block, f -block
and heavier p-block elements are metal.
(c) Correct statement Trend is not regular, Be has higher first
ionisation energy than B, nitrogen has higher first ionisation
energy than oxygen.
(d) Inccorrect statement d-subshells
are
not
filled
monotonically, regularity break at chromium and copper.
4. (a) Incorrect Electrons are not filled in sub-energy levels
s, p, d and f in the same sequence.
(b) Correct Number of valence shell electrons usually
determine the stable valency state of an element.
(c) Correct Physical and chemical properties of elements are
periodic function of atomic number which is the basis of
modern, long form of periodic table.
(d) Correct Relative ionicity of the bond between any two
elements is function of electronegativity difference of the
bonded atoms which in turn has periodic trend in long form
of periodic table.
Topic 2 Periodic Properties
1. Size of species ∝
1
Nuclear charge
Iso-electronic species are those atoms or ions which has the
same number of electrons.
Size of species decreases with increasing protons.
More is effective nuclear charge (Zeff) lesser will be ionic size.
Correct order of ionic radii
Al 3+ < Mg2+ < Na + < F− < O2− < N3−
2. The first electron gain enthalpy is exothermic (or negative).
Generally, electron gain enthalpy becomes less exothermic (or
less negative) when comparing elements of a group from top to
bottom.
Therefore, electron gain enthalpy of S > Se and Li > Na.
But there are some exceptions to this.
One of them is the case of a group 17 elements where electron
gain is most negative for Cl instead of F, due to extra small size
of fluorine.
∴ Upon an electron gain, energy releases in the order :
Cl > F, S > Se and Li > Na
3. The group number, number of valence electrons and valency of
an element with atomic number 15 are 15, 5 and 3 respectively.
Modern periodic table is based on the atomic number. Number
of valence electrons present in an atom decides the group
number. Electronic configuration of element having atomic
number 15 = 1s2 2s2 2 p6 3s2 3 p3
Valence electrons
As five electrons are present in valence shell, its group number
is 15. Valency of element having atomic number 15 is +3
(8 − 5 = 3).
4. The electronic configuration of given elements are as follows :
K(19) = 1s2 2s2 2 p6 3s2 3 p6 4 s1
Periodic Classification and Periodic Properties 45
Mg(12) = 1s2 2s2 2 p6 3s2
Sr(38) = 1s2 2s2 2 p6 3s2 3 p6 4 s2 3d 10 4 p6 5s2
Sc(21) = 1s2 2s2 2 p6 3s2 3 p6 4 s2 3d 1
First ionisation enthalpy (IE) of K is lowest among the given
options. Here, the energy required to remove an electron from
4 s1 is least as only one electron is present in the outermost shell.
IE (I) is comparatively high for Mg and Sr and two electrons
(fully-filled) are placed in s-orbital. Second ionisation enthalpy
of K is highest among the given options.
Now, removal of an electron occur from p6 (fully-filled). So,
high energy is required to remove the electron. From the above
discussion, it can be concluded that (IE 2 − IE 1 ) value is
maximum for K (potassium).
5. The electronegativity values of given elements on the Pauling
scale can be shown as follows:
Period No.
3
4
5
Group 13
Al (1.5)
Ga (1.6)
Group 14
Si (1.8)
Ge (1.8)
Group 15
P (2.1)
Group 16
S (2.5)
Se (2.4)
Te (2.01)
On moving from left to right across a period, i.e. from Ga to Se,
the effective nuclear charge increases and size decreases.
As a result, the value of electronegativity increases due to
increase in the attraction between the outer electrons and the
nucleus. Whereas on moving down the group, (i.e. from Se to
Te), the atomic size increases.
As a result, the force of attraction between the outer electron and
the nucleus decreases. Hence, the electronegativity decreases.
6. Element
C
Al
S
Cs
Period
No.
2nd
3rd
3rd
6th
Group No.
14
13
16
1
Along the period atomic radius
decreases, so, radii : Al > S.
With the addition of a new shell, period number as well as atomic
radius increases. It is because of the successive addition of one
extra shell of electrons. So, the order of the atomic radii of the
given elements will be: C < S < Al < Cs
7. The summary of variation of periodic properties is given in table
below:
S.No.
Periodic property
1.
2.
Atomic radius
Electron gain
enthalpy
Electronegativity
3.
Variation
Along a period Along a group
Decreases
Increases
Increases
Decreases
Increases
Decreases
Thus, electronegativity decreases and atomic radius increases
down a group in the periodic table.
8. Number of electrons in N3−, = 7 + 3 = 10
Number of electrons in O2− = 8 + 2 = 10
Number of electrons in F− = 9 + 1 = 10
Since, all the three species have each 10 electrons, hence they
are isoelectronic species.
It is considered that, in case of isoelectronic species as the
negative charge increases, ionic radii increases and therefore the
value of ionic radii are
(highest among the three)
N3− = 1.71
O2− = 1.40
F− = 136
. (lowest among the three)
Time Saving Technique There is no need to mug up the radius
values for different ions. This particular question can be solved
through following time saving.
Trick The charges on the ions indicate the size as
N3− > O2− > F− . Thus, you have to look for the option in which
the above trend is followed. Option(c) is the only one in which
this trend is followed. Hence, it is the correct answer.
9. As we move down the group, size of metal increases. Be has
lower size while SO2−
4 has bigger size, that’s why BeSO 4 breaks
easily and lattice energy becomes smaller but due to lower size
of Be, water molecules are gathered around and hence hydration
energy increases.
On the other hand, rest of the metals, i.e Ca, Ba, Sr have bigger
size and that’s why lattice energy is greater than hydration energy.
Time Saving Technique In the question of finding hydration
energy only check the size of atom. Smaller sized atom has more
hydration energy. Thus, in this question Be is placed upper most
in the group has lesser size and not comparable with the size of
sulphates. Hence, BeSO4 is the right response.
10. Cl 2, Br2 and I2 are homonuclear diatomic molecule in which
electronegativity of the combining atoms is same, so they are
more stable and less reactive, whereas, I and Cl have different
electronegativities and bond between them are polarised and
reactive. Therefore, interhalogen compounds are more reactive.
Time Saving Technique In this type of question of halogen,
only go through the polarity of the molecules. As we know,
diatomic molecule does not have polarity but molecules with
dissimilar sizes have polarity resulting in more reactivity.
11. As we move down the group of noble gases, molecular mass
increases by which dipole produced for a moment and hence
London forces increases from He to Xe.
Therefore, more amount of energy is required to break these
forces, thus boiling point also increases from He and Xe.
12.
Na → Na + + e− First IE
+
Na + e− → Na
Electron gain enthalpy of Na + is reverse of (IE)
Because reaction is reverse so ∆H (eq) = − 5.1 eV
13. Ionisation energy increases along a period from left to right and
decreases down a group. The position of given elements in the
periodic table is as
Group No. 2
Ca
Ba
16
S
Se
18
Ar
Thus, the order of increasing ∆H IE1 is Ba < Ca < Se < S < Ar
14. Be− is the least stable ion, Be (1s2 2s2 ) has stable electronic
configuration, addition of electron decreases stability.
15. In a group, ionisation energy decreases down the group
Be > Mg > Ca
16. Among isoelectronic species, greater the negative charge,
greater the ionic size, hence F− < O2− < N3− .
46 Periodic Classification and Periodic Properties
17. (a) Correct statement In a period, element of 2nd group has
higher first ionisation potential than element of group 13.
(b) Incorrect statement Mg+ require less energy for further
ionisation than Na + because of noble gas configuration of Na + .
(c) Correct statement Ionisation energy increases from left
to right in a period.
18. Mg2+ = 1s2 2 s2 2 p6 = no unpaired electron
Ti 3+ = 1s2 2 s2 2 p6 3s2 3 p6 3d 1 = one unpaired electron
V3+ = 1s2 2 s2 2 p6 3s2 3 p6 3d 2 = two unpaired electrons
2+
Fe
2
2
6
2
6
6
= 1s 2 s 2 p 3s 3 p 3d = four unpaired electrons
19. [Ne] 3s2 3 p3 has highest ionisation energy, periodic trend.
20. Among isoelectronic species, the relation in size is
21. Ionisation energy increases from left to right in a period.
However, exception occur between group 2 and group 13
elements on account of stability of electronic configuration of
valence shell.
IE
Group 2 =
> Group 13 =
ns2
np1
⇒ The desired order is Na < Mg > Al < Si
22. Electronegativity increases from left to right in a period and
decreases from top to bottom in a group. Variation is more rapid
in group than in a period, hence the desired order is
Electronegativity : Si < P < C < N
23. Atomic radius of noble gases are greater than halogens of same
period, hence (a) is the correct answer.
24. First ionisation energy of oxygen is less than that of nitrogen on
the ground of stability of valence shell configuration, hence (a)
is the correct answer.
25. Hydration energy depends on charge of ion and ionic radius.
Higher the charge, greater the hydration energy. On the other
hand, smaller the size, greater the hydration energy. Charge is
considered first for comparison. Hence, Mg2+ has higher
hydration energy than Na + .
26. Nitrogen has highest ionisation potential due to exceptional
stability of its valence shell configuration mentioned in
question 21.
27. For second ionisation potential, electron will have to be
removed from valence shell of the following ions:
C+ (5e–) = 1s2
2s2
2p
+
–
N (6e ) =
1s2
2s2
2p
O+ (7e–) = 1s2
(d) is incorrect because CrO is basic oxide.
29. (a) Ti 4+ > Mn 7+ is the correct order of size due to lower positive
charge on Ti 4+ .
Cl − = 37 Cl − : Isotopes with same charge have same size
because isotopes differ in compositions of nuclei which do
not affect the atomic/ionic radius.
(c) K+ < Cl − is the correct order. Among isoelectronic species,
anion has greater size than cation.
(d) P3+ > P5+ is the correct order. For the same elements, lower
the positive charge, larger the ions.
37
30. (a) and (b) are infact the same statements and both are correct. N
has slightly greater ionisation energy than oxygen which is
against periodic trend. This exception is due to completely
half-filled (2 p3 ) orbital in nitrogen that makes ionisation
slightly difficult than oxygen.
(c) Also correct : Although N has greater first ionisation
potential than oxygen, two values of ionisation potentials are
comparable since they are adjacent in a period, i.e. electrons
are removed from same orbit during ionisation.
(d) Incorrect – opposite to (c). of the bonded atoms which in
turn has periodic trend in long form of periodic table.
31. (a) Correct For greater solubility, hydration energy must be
greater than lattice energy.
(b) Correct Greater lattice energy discourage dissolution of a salt.
(c) Incorrect When a salt dissolve, energy is required to break
the lattice, which comes from hydration process.
(d) Incorrect Explained in (A).
32. Statement I and II are true and Statement II is the correct
explanation of statement I.
33. Statement I is true. Stronger oxidising agent is one which itself
can easily be reduced. Pb4+ is unstable, due to inert pair effect,
can easily be reduced to stable Pb2+ , hence a stronger oxidising
agent than Sn 4+ .
Statement II is false. Due to inert pair effect, the higher
oxidation states of group 14 elements becomes less stable for
heavier member.
34. Both statements I and II are true and Statement II is the correct
explanation of statement I.
2s2
2p
F+ (8e–) = 1s2
28. (c) is incorrect because NO is neutral oxide.
(b)
cation < neutral < anion
Hence, Na + has smallest size.
ns2
In general, ionisation energy increases from left to right in a
period. However, exception occur between adjacent atoms in a
period, greater amount energy is required for removal of
electron from completely half-filled or completely filled orbital
than the same for adjacent atom with either less than completely
half-filled or less than completely filled orbital. Therefore,
ionisation potential of O+ is greater than that of F+ . Also ionisation
potential of N+ is greater than C+ but less than both O+ and F+
(periodic trend). Hence, overall order is 2nd IP : O > F > N > C.
2s2
2p
35. Statement I is true Be has higher first ionisation energy than B
which is against periodic trend.
Statement II is false 2s-orbital is lower in energy than 2p,
Aufbau’s principle.
Periodic Classification and Periodic Properties 47
36. Statement I is true; Statement II is false.
F atom has slightly lower affinity for the electron than chlorine.
It is due to the reason that additional electrons are repelled more
effectively by 2p-electrons in F than by 3p-electrons in Cl-atom.
37. By observing the values of different ionisation energies, I 1 , I 2
and I 3 for atomic number (n + 2 ), it is observed that there is very
large difference between the second ionisation energy and first
ionisation energy (I 2 >> I 1 ).
This indicates that number of valence shell electrons is 1 and
atomic number (n + 2 ) should be an alkali metal.
Also for atomic number (n + 3 ), I 3 >> I 2.
This indicates that it will be an alkaline earth metal which
suggests that atomic number (n + 1) should be a noble gas and
atomic number (n) should belong to halogen family. Since,
n < 10; hence, n = 9 (F atom)
38. Inert pair effect-favours lower oxidation state.
39. Higher effective nuclear charge due to greater p/e ratio.
40. Electronegativity =
IP + EA
2
(Mulliken formula)
41. Electron affinity–definition.
42. Basic nature of hydroxides increases down a group.
43. Cl has maximum electron affinity, hence the correct order is
Cl > F > Br
44. Ionisation potential decreases down the group but this is not the
only criteria of reducing power.
45. In a group, size increases from top to bottom.
46. Li + < Al 3+ < Mg2+ < K+ . Size decreases from left to right in a
period and it increases from top to bottom in a group. Variation
is more pronounced in group than in period.
47. Zn = 3d 10 4 s2 , Cu = 3d 10 4 s1
The first ionisation energy is greater for Zn but reverse is true for
2nd ionisation energy.
48. Ionic size Mg2+ < Na + < F− < O2− < N3−
49. The first ionisation energy of carbon is greater than the same of
boron as predicted from periodic trend. However, for 2nd
; more stable than C+ =1s2 2s2
B+ = 1s2
2s2
2p1
ionisation trend is reversed due to stability of completely filled
2s-orbital of B+ :
50. Size Ca 2+ < Ar < Cl − < S2− .
51. (i) Mg2+ , O2− , Na + and F −
are all isoelectronic, has
10 electrons each. Among isoelectronic species, the order of
size is cation < neutral < anion.
Also, between cations, higher the charge, smaller the size
and between anions, greater the negative charge, larger the
size. Therefore, the decreasing order of ionic radii :
O2− > F− > Na + > Mg2+
(ii) First ionisation energy increases from left to right in a period.
However, exception occur between group 2 and 13 and group 15
and 16 where trend is reversed on the grounds of stability of
completely filled and completely half-filled orbitals. Therefore,
Ionisation energy (1st) : Na < Al < Mg < Si
(iii) If the atoms are from same period, bond length is inversely
proportional to bond order. In a group, bond length is related
directly to atomic radius.
Therefore,
bond length N2 < O2 < F2 < Cl 2
4
Chemical Bonding
Topic 1 Preliminary Concepts of Electrovalent
and Covalent Bonding
Objective Questions I (Only one correct option)
1. The isoelectronic set of ions is
−
+
+
(a) F , Li , Na and Mg
(2019 Main, 10 April I)
2+
(b) N3 − , Li + , Mg 2+ and O2 −
(c) Li + , Na + , O2 − and F−
(d) N3 − , O2 − , F− and Na +
2. Which of the following compounds contain(s) no covalent
bond(s)?
KCl, PH3 , O2 , B2 H6 , H2 SO4
(a) KCl, B2 H6 , PH3
(c) KCl
(2018 Main)
(b) KCl, H2 SO4
(d) KCl, B2 H6
3. The intermolecular interaction that is dependent on the
inverse cube of distance between the molecules is (2015 Main)
(a) ion-ion interaction
(b) ion-dipole interaction
(c) London force
(d) hydrogen bond
4. The nodal plane in the π-bond of ethene is located in
(a) the molecular plane
(2002, 3M)
(b) a plane parallel to the molecular plane
(c) a plane perpendicular to the molecular plane which
bisects the carbon-carbon σ-bond at right angle
(d) a plane perpendicular to the molecular plane which
contains the carbon-carbon σ-bond
5. Amongst H2 O, H2 S, H2 Se and H2 Te, the one with the highest
boiling point is
(2000, 1M)
(a) H2 O because of hydrogen bonding
(b) H2 Te because of higher molecular weight
(c) H2 S because of hydrogen bonding
(d) H2 Se because of lower molecular weight
6. Arrange the following compounds in order of increasing
dipole moment, toluene (I), m-dichlorobenzene (II),
o-dichlorobenzene (III), p-dichlorobenzene (IV) (1996, 1M)
(a) I < IV < II < III
(b) IV < I < II < III
(c) IV < I < III < II
(d) IV < II < I < III
7. The number and type of bonds between two carbon atoms in
CaC2 are
(1996, 1M)
(a) one sigma (σ ) and one pi ( π ) bonds
(b) one sigma (σ ) and two pi ( π ) bonds
(c) one sigma (σ ) and one half pi ( π ) bonds
(d) one sigma (σ ) bond
8. The molecule which has zero dipole moment is (1989, 1M)
(d) ClO2
(c) NF3
(a) CH2 Cl 2 (b) BF3
9. Element X is strongly electropositive and element Y is
strongly electronegative. Both are univalent. The compound
formed would be
(1980, 1M)
(b) X −Y +
(c) X −− Y
(d) X → Y
(a) X +Y −
10. Which of the following compound is covalent?
(a) H2
(c) KCl
(1980, 1M)
(b) CaO
(d) Na 2 S
11. The total number of electrons that take part in forming the
bonds in N2 is
(a) 2
(b) 4
(1980, 1M)
(c) 6
(d) 10
12. The compound which contains both ionic and covalent bonds
is
(a) CH4
(1979, 1M)
(b) H2
(c) KCN
(d) KCl
Objective Questions II
(One or more than one correct option)
13. Each of the following options contains a set of four
molecules. Identify the option(s) where all four molecules
posses permanent dipole moment at room temperature.
(2019 Adv.)
(a) SO2 , C6 H5 Cl, H2 Se, BrF5
(b) BeCl 2 , CO2 , BCl 3 , CHCl 3
(c) NO2 , NH3 , POCl 3 , CH3 Cl
(d) BF3 , O3 , SF6 , XeF6
14. Dipole moment is shown by
(1986, 1M)
(a) 1, 4-dichlorobenzene
(b) cis-1, 2-dichloroethene
(c) trans-1, 2-dichloroethene (d) trans-1, 2-dichloro-2- pentene
Chemical Bonding 49
Numerical Answer Type Questions
15. Consider the following compounds in the liquid form :
O 2 ,HF,H 2O,NH 3 ,H 2O 2 ,CCl 4 ,CHCl 3 , C6H 6 ,C6H 5Cl
When a charged comb is brought near their flowing stream,
how many of them show deflection as per the following
figure?
(2020 Adv.)
17. Statement I LiCl is predominantly a covalent compound.
Statement II Electronegativity difference between Li and Cl
is too small.
(1998, 2M)
Fill in the Blank
18. There are …… π -bonds in a nitrogen molecule.
(1982, 1M)
True/False
19. All molecules with polar bonds have dipole moment.
(1985, 1/2 M)
20. Linear overlapping of two atomic p-orbitals leads to a sigma
bond.
(1983, 1M)
Subjective Questions
21. Arrange the following ions in order of their increasing radii:
Li + , Mg 2+ , K + , Al 3+ .
+
(1997, 1M)
+
22. Between Na and Ag , which is stronger Lewis acid and
why?
(1997, 3M)
−
23. In the reaction, I + I2 →
I3− ,
which is the Lewis acid?
(1997, 1M)
16. Among the species given below, the total number of
diamagnetic species is____
H atom, NO2 monomer, O−2 (superoxide), dimeric sulphur in
vapour phase, Mn 3 O4 ,( NH4 )2 [ FeCl 4 ], ( NH4 )2 [ NiCl 4 ],
(2018 Adv.)
K 2 MnO4 , K 2 CrO4
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 correct; Statement II is incorrect
(d) Statement I is incorrect; Statement II is correct
24. Explain the difference in the nature of bonding in LiF and LiI.
(1996, 2M)
25. The dipole moment of KCl is 3.336 × 10−29 C-m which
indicates that it is a highly polar molecule. The interatomic
distance between K + and Cl − in this molecule is
2.6 × 10−10 m. Calculate the dipole moment of KCl molecule
if there were opposite charges of one fundamental unit
located at each nucleus. Calculate the percentage ionic
character of KCl.
(1993, 2M)
26. Give reasons in two or three sentences only for the following :
“Hydrogen peroxide acts as an oxidising as well as a reducing
agent.’’
(1992, 1M)
27. State four major physical properties that can be used to
distinguish between covalent and ionic compounds. Mention
the distinguishing features in each case.
(1978, 2M)
Topic 2 VBT, Hybridisation and VSEPR Theory
Objective Questions I (Only one correct option)
1. The correct statements among I to III are :
I. Valence bond theory cannot explain the color exhibited
by transition metal complexes.
II. Valence bond theory can predict quantitatively the
magnetic properties of transition metal complexes.
III. Valence bond theory cannot distinguish ligands as weak
and strong field ones.
(2019 Main, 9 April II)
(a) II and III only
(b) I, II and III
(c) I and II only
(d) I and III only
2. The correct statement about ICl 5 and ICl −4 is
(2019 Main, 8 April II)
(a) ICl 5 is square pyramidal and ICl −4 is tetrahedral
(b) ICl 5 is square pyramidal and ICl −4 is square planar
(c) Both are isostructural
(d) ICl 5 is trigonal bipyramidal and ICl −4 is tetrahedral
3. The ion that has sp 3 d 2 -hybridisation for the central atom, is
(2019 Main, 8 April II)
(a) [ICl 2 ]−
(b) [BrF2 ]−
(c) [ICl 4 ]−
(d) [IF6 ]−
50 Chemical Bonding
4. The size of the iso-electronic species Cl − , Ar and Ca 2+ is
affected by
(2019 Main, 8 April I)
(a) azimuthal quantum number of valence shell
(b) electron-electron interaction in the outer orbitals
(c) principal quantum number of valence shell
(d) nuclear charge
5. In which of the following processes, the bond order has
increased and paramagnetic character has changed to
diamagnetic?
(2019 Main, 9 Jan II)
(a) O2 → O2+
(b) N2 → N+2
(c) O2 → O22 −
(d) NO → NO+
(2018 Main)
(b) 6
(d) 12
(2017 Main)
(a) O2 − , F− , Na + , Mg 2 +
(b) O− , F− , Na, Mg +
(c) O2 − , F− , Na, Mg 2 +
(d) O− , F− , Na + , Mg 2 +
8. The correct statement for the molecule, CsI3 is
(2014 Main)
it is a covalent molecule
it contains Cs + and I−3 ions
it contains Cs 3+ and I− ions
it contains Cs + , I− and lattice I2 molecule
(a) SO3
(b) BrF3
(c)
(2010)
(d) OSF2
magnetic nature of the diatomic molecule B2 is
(a) 1 and diamagnetic
(b) 0 and diamagnetic
(c) 1 and paramagnetic
(d) 0 and paramagnetic
(2010)
11. The species having bond order different from that in CO is
(b) NO+
(a) NO−
(2007, 3M)
(d) N2
12. Among the following, the paramagnetic compound is
(2007, 3M)
(a) Na 2 O2
(b) O3
(c) N2 O
(d) KO2
13. Which of the following contains maximum number of lone
pairs on the central atom?
(b) XeF 4
(a) ClO −3
(c) SF 4
(d) I −3
14. Number of lone pair(s) in XeOF4 is/are
(a) 0
(b) 1
(c) 2
(2005, 1M)
(2004, 1M)
(d) 3
15. Which of the following are isoelectronic and isostructural ?
(a)
(c)
NO–3 ,
–
NO3 , CO2−
3
ClO–3 , CO2–
3
–
CO2–
3 , ClO3 , SO3
(2003, 1M)
(b) SO3 , NO–3
(d) CO2–
3 , SO3
(2003, 1M)
(b) CH 2 Cl 2
(d) CCl 4
of N and B atoms in a 1 : 1 complex of BF3 and NH3 .
(a) N : tetrahedral, sp 3 ; B: tetrahedral, sp 3
(2002, 3M)
(b) N : pyramidal, sp 3 ; B: pyramidal, sp 3
(c) N: pyramidal, sp 3 ; B: planar, sp 2
(d) N: pyramidal, sp 3 ; B: tetrahedral, sp 3
(2001, 1M)
(a) dsp 2 , dsp 3 , sp 2 and sp 3 (b) sp 3 , dsp 2 , sp 3 d and sp 2
(c) dsp 2 , sp 2 , sp 3 and dsp 3 (d) dsp 2 , sp 3 , sp 2 and dsp 3
20. The common features among the species CN – , CO
and NO+ are
(a) bond order three and isoelectronic
(b) bond order three and weak field ligands
(c) bond order two and acceptors
(d) isoelectronic and weak field ligands
(2001, 1M)
(2000, 1M)
(a) sp, sp 3 and sp 2 respectively
(b) sp, sp 2 and sp 3 respectively
(c) sp 2 , sp and sp 3 respectively
(d) sp 2 , sp 3 and sp respectively
22. In the compound CH2 == CH  CH2  CH2  C ≡≡ CH, the
C2  C3 bonds is of
(a) sp - sp 2
(c) sp - sp 3
(1999, 2M)
(b) sp 3 - sp 3
(d) sp 2 - sp 3
23. The geometry of H2 S and its dipole moment are
(a) angular and non-zero
(c) linear and non-zero
(1999, 2M)
(b) angular and zero
(d) linear and zero
24. The geometry and the type of hybrid orbital present about the
central atom in BF3 is
(1998, 2M)
(a) linear, sp
(b) trigonal planar, sp 2
(c) tetrahedral, sp 3
(d) pyramidal, sp 3
25. Which one of the following compounds has
sp 2 - hybridisation?
(1997, 1M)
(a) CO2
(d) CO
(b) SO2
(c) N2 O
26. Among KO2 , AlO−2 , BaO2 and NO+2 , unpaired electron is
present in
(a) NO+2 and BaO2
(b) KO2 and AlO−2
(c) Only KO2
(d) Only BaO2
(1997 C, 1M)
27. The cyanide ion CN− and N2 are isoelectronic, but in contrast
16. Among the following, the molecule with the highest dipole
moment is
(a) CH 3 Cl
(c) CHCl 3
18. Specify the coordination geometry around and hybridisation
and NH+4 are
10. Assuming that Hund’s rule is violated, the bond order and
(c) CN−
(d) O2–
2
21. The hybridisation of atomic orbitals of nitrogen in NO+2 , NO−3
9. The species having pyramidal shape is
SiO2−
3
(2002, 3M)
(c) O–2
following species NH3 , [PtCl 4 ]2 − , PCl 5 and BCl 3 is
7. The group having isoelectronic species is
(a)
(b)
(c)
(d)
electron (s)?
(b) F2
(a) N2
19. The correct order of hybridisation of the central atom in the
6. Total number of lone pair of electron in I−3 ion is
(a) 3
(c) 9
17. Which of the following molecular species has unpaired
to CN− , N2 is chemically inert because of
(1997 C, 1M)
(a) low bond energy
(b) absence of bond polarity
(c) unsymmetrical electron distribution
(d) presence of more number of electron in bonding orbitals
Chemical Bonding 51
28. Among the following species, identify the isostructural
pairs.
NF3 , NO3− ,
+
BF3 , H3 O , N3 H
(1996, 1M)
(a) [NF3 ,NO−3 ] and [BF3 ,H3 O+ ]
(b) [NF3 , N3 H] and [NO−3 ,BF3 ]
(c) [ NF3 , H3 O+ ] and [NO–3 , BF3 ]
(d) [NF3 , H3 O+ ] and [N3 H, BF3 ]
(b) NCl 3
(c) PH3
(1996, 1M)
(d) BF3
30. The maximum possible number of hydrogen bonds a water
molecule can form is
(a) 2
(b) 4
(1992, 1M)
(c) 3
ClO−2 is
(a) sp
(1992, 1M)
(b) sp
(c) sp
2
(d) None of these
32. The molecule which has pyramidal shape is
(a) PCl 3
(b) SO3
(c) CO2–
3
(1989, 1M)
(d) NO–3
33. Which of the following is paramagnetic?
(a) O–2
(b) CN–
(c) CO
(1989, 1M)
(d) NO+
34. The Cl—C—Cl angle in 1, 1, 2, 2-tetrachloroethene and
tetrachloromethane respectively will be about
(a) 120° and 109.5°
(b) 90° and 109.5°
(c) 109° and 90°
(d) 109.5° and 120°
35. The molecule that has linear structure is
(a) CO2
(b) NO2
(c) SO2
(1988, 1M)
(1988, 1M)
(d) SiO2
36. The species in which the central atom uses sp 2 -hybrid
orbitals in its bonding is
(a) PH3
(b) NH3
(1984, 1M)
(1988, 1M)
(c) CH+3
(d) SbH3
37. Of the following compounds, which will have a zero dipole
moment ?
(a) 1, 1-dichloroethylene
(b) cis-1, 2-dichloroethylene
(c) trans-1, 2-dichloroethylene
(d) None of the above
(a) its planar structure
(1983, 1M)
(b) its regular tetrahedral structure
(c) similar sizes of carbon and chlorine atoms
(d) similar electron affinities of carbon and chlorine
42. The ion that is isoelectronic with CO is
(a) CN−
(1987, 1M)
(b) O+2
(c) O−2
(1982, 1M)
(d) N+2
43. Among the following, the linear molecule is
(a) CO2
(d) 1
31. The type of hybrid orbitals used by the chlorine atom in
3
(a) two mutually perpendicular orbitals
(b) two orbitals at 180°
(c) four orbitals directed tetrahedrally
(d) three orbitals in a plane
41. Carbon tetrachloride has no net dipole moment because of
29. Which one of the following molecules is planar?
(a) NF3
40. On hybridisation of one s and one p-orbital we get
(b) NO2
(c) SO2
(1982, 1M)
(d) ClO2
44. If a molecule MX 3 has zero dipole moment, the sigma
bonding orbitals used by M (atomic number < 21) are
(a) pure p
(b) sp-hybridised
(1981, 1M)
(d) sp 3 -hybridised
(c) sp 2 -hybridised
Objective Questions II
(One or more than one correct option)
45. The molecules that will have dipole moment are
(a) 2, 2-dimethyl propane
(c) cis-3-hexene
(1992, 1M)
(b) trans-2-pentene
(d) 2,2,3,3-tetramethyl butane
46. Which of the following have identical bond order?
(a) CN–
(b) O–2
(c) NO+
(d) CN+
(1992, 1M)
47. The linear structure assumed by
(a) SnCl 2
(b) CS2
(c) NO+2
(1991, 1M)
(d) NCO–
48. CO2 is isostructural with
(a) HgCl 2
(b) C2 H2
(1986, 1M)
(c) SnCl 2
(d) NO2
Match the Columns
49. Match the orbital overlap figures shown in Column I with the
description given in Column II and select the correct answer
using the codes given below the Columns.
(2014 Adv.)
Column I
Column II
p-d π antibonding
A.
1.
B.
2.
d-d σ bonding
C.
3.
p-dπ bonding
D.
4.
d-d σ antibonding
38. The hybridisation of sulphur in sulphur dioxide is (1986, 1M)
(a) sp
(b) sp 3
(c) sp 2
(d) dsp 2
39. The bond between two identical non-metal atoms has a pair
of electrons
(a) unequally shared between the two
(b) transferred fully from one atom to another
(c) with identical spins
(d) equally shared between them
(1986, 1M)
52 Chemical Bonding
64. In benzene, carbon uses all the three p-orbitals for
Codes
A
(a) 4
(c) 2
B
3
3
C
2
1
D
1
4
A
1
4
(b)
(d)
B
2
1
C
3
2
D
4
3
50. Match each of the diatomic molecules in Column I with its
property/properties in Column II.
(2009)
A.
p.
Paramagnetic
B.
N2
q.
Undergoes oxidation
C.
O−2
r.
Undergoes reduction
D.
O2
s.
Bond order ≥ 2
t.
Mixing of ‘s’ and ‘p’ orbitals
ColumnII
B
C
p, r, t, s q, r, t
q, r, s, t p, q, r, t
p, q, r
r, s, t
p, q, s
p, t
65. SnCl 2 is a non-linear molecule.
1
(1985, M)
2
Integer Answer Type Questions
central atom in the following species is
[TeBr6 ]2 − , [BrF2 ]+ , SNF3 and [XeF3 ]−
Codes
A
q, r, s
p, q, r, t
q, r, s, t
p, q, s, t
(1987, 1M)
66. The sum of the number of lone pairs of electrons on each
Column I
B2
(a)
(b)
(c)
(d)
hybridisation.
D
p, q, t
p, r, s, t
p, q, r, t
q, r, t
(Atomic numbers : N = 7, F = 9, S = 16, Br = 35,
Te = 52, Xe = 54)
(2017 Adv.)
N−3 ,
N2 O, NO+2 ,
67. Among the triatomic molecules/ions BeCl 2 ,
O3 , SCl 2 , ICl −2 , I−3 and XeF2 , the total number of linear
molecules(s)/ion(s) where the hybridisation of the central
atom does not have contribution from the d-orbital(s) is
[atomic number of S = 16 , Cl = 17 , I = 53 and Xe = 54]
(2015 adv.)
68. A list of species having the formula XZ4 is given below
(2014 Adv.)
XeF4 , SF4 , SiF4 , BF4− , BrF4− , [Cu(NH3 )4 ] 2+ , [FeCl 4 ] 2− ,
Fill in the Blanks
51. Among N2 O, SO2 , I+3 and I–3 , the linear species are …… and
……
52. When N2 goes to
(1997 C, 1M)
N+2 ,
the N  N bond distance … , and when
O2 goes to O+2 the O  O bond distance ……
[CoCl 4 ] 2− and [PtCl 4 ] 2−
Defining shape on the basis of the location of X and Z atoms,
the total number of species having a square planar shape is
69. The total number of lone-pair of electrons in melamine is
(1996, 1M)
(2013 Adv.)
53. The two types of bonds present in B2 H6 are covalent and ……
70. Based on VSEPR theory, the number of 90° F—Br—F angles
(1994, 1M)
(2010)
54. The kind of delocalisation involving sigma bond orbitals is
called.................
(1994, 1M)
55. The valence atomic orbitals on C in silver acetylide is
.............hybridised.
56. The shape of CH3 + is …… .
(1990, 1M)
(1990, 1M)
57. …… hybrid orbitals of nitrogen atom are involved in the
in BrF5 is
Subjective Questions
71. Predict whether the following molecules are isostructural or
not. Justify your answer.
(ii) N(SiMe3 )3
(i) NMe3
(2005, 2M)
72. On the basis of ground state electronic configuration, arrange
(1982, 1M)
the following molecules in increasing O—O bond length
(2004, 2M)
order. KO2 , O2 , O2 [AsF6 ]
58. Pair of molecules which forms strongest intermolecular
73. Draw the shape of XeF4 and OSF4 according to VSEPR
formation of ammonium ion.
hydrogen bonds is ……… . (SiH4 and SiF4 , acetone and
(1981, 1M)
CHCl 3 , formic acid and acetic acid)
59. The angle between two covalent bonds is maximum in …… .
(CH4 , H2 O, CO2 )
(1981, 1M)
60. The dipole moment of CH3 F is greater than that of CH3 Cl.
(1993, 1M)
(1993, 1M)
62. The presence of polar bonds in a polyatomic molecule
suggests that the molecule has non-zero dipole moment.
(1990, 1M)
63. sp 3 hybrid orbitals have equal s and p character.
(2004, Main, 2M)
74. Using VSEPR theory, draw the shape of PCl 5 and BrF5 .
(2003, 2M)
75. Draw the molecular structures of XeF 2 , XeF 4 and XeO2 F2 ,
True/False
61. H2 O molecule is linear.
theory. Show the lone pair of electrons on the central atom.
(1987, 1M)
indicating the location of lone pair(s) of electrons.
(2000, 3M)
76. Interpret the non-linear shape of H2 S molecule and
non-planar shape of PCl 3 using valence shell electron pair
repulsion (VSEPR) theory. (Atomic number : H = 1, P = 15,
S = 16, Cl = 17)
(1998, 4M)
77. Using the VSEPR theory, identify the type of hybridisation
and draw the structure of OF2 . What are the oxidation states of
O and F ?
(1997, 3M)
Chemical Bonding 53
78. Write the Lewis dot structural formula for each of
 H 
 •••• 
H • •O• • H




the following. Give also, the formula of a neutral molecule,
which has the same geometry and the same arrangement of
the bonding electrons as in each of the following. An
example is given below in the case of H3 O+ and NH3 .
+
Lewis dot. structure
(i)
O2−
2
(ii)
CO2−
3
 H 
 •••• 
H • •N• • H




Neutral .molecule
(iii) CN
(1983, 4M)
(iv) NCS−
−
Topic 3 Resonance, LCAO, MOT, Other Bonding Types
Objective Questions I (Only one correct option)
1. The intermolecular potential energy for the molecules A, B, C
and D given below suggests that:
(2020 Main, 4 Sep I)
0
100
–100
A–D
following will not be a viable molecule?
(b) He+2
(a) He2+
2
A–B
–600
(c) H−2
(a) A-B has the stiffest bond
(b) D is more electronegative than other atoms
(c) A-A has the largest bond enthalpy
(d) A-D has the shortest bond length
(b) π 2 px
(c) π2 p y
*
(d) σ 2 p z
3. HF has highest boiling point among hydrogen halides,
because it has
(2019 Main, 9 April II)
(a) lowest ionic character
(b) strongest van der Waals’ interactions
(c) strongest hydrogen bonding
(d) lowest dissociation enthalpy
4. Among the following species, the diamagnetic molecule is
(2019 Main, 9 April II)
(a) CO
(b) B2
(c) NO
(d) O2
5. Among the following, the molecule expected to be stabilised
by anion formation is C2 , O2 , NO, F2 .
(a) C2
(b) F2
(c) NO
(d) O2
(2019 Main, 9 April I)
2−
2−
6. Among the following molecules/ions, C2−
2 , N2 , O2 , O2
Which one is diamagnetic and has the shortest bond length?
(2019 Main, 8 April II)
(a) C2−
2
(b) O2
(c) O2−
2
(2019 Main, 9Jan I)
(2018 Main)
(d) H2−
2
(2017 Main)
(2019 Main, 10 April I)
*
(d) O2
10. Which of the following species is not paramagnetic?
2. During the change of O2 to O−2 , the incoming electron goes to
the orbital.
(a) π2 px
(c) N+2
9. According to molecular orbital theory, which of the
A–C
A–A
–500
(b) N2
is true with respect to Li +2 and Li −2 ?
(a) Both are unstable
(b) Li+2 is unstable and Li−2 is stable
(c) Both are stable
(d) Li+2 is stable and Li−2 is unstable
150
–200
Potential
–300
Energy
(kJ mol–1)
–400
(2019 Main, 10 Jan I)
(a) O+2
8. According to molecular orbital theory, which of the following
Interatomic distance (pm)
50
7. Two pi and half sigma bonds are present in
(d) N 2−
2
(a) NO
(c) O2
(b) CO
(d) B2
11. Assuming 2s-2p mixing is not operative, the paramagnetic
species among the following is
(2014 Adv.)
(a) Be2
(b) B2
(c) C2
(d) N2
12. Stability of the species Li 2 , Li −2 and Li +2 increases in the order
of
(2013 Main)
(b) Li –2 < Li +2 < Li 2
(a) Li 2 < Li 2+ < Li 2−
(c) Li 2 < Li 2− < Li 2+
(d) Li −2 < Li 2 < Li +2
13. In which of the following pairs of molecules/ions both the
species are not likely to exist?
(a) H+2 , He2−
(b) H −2 , He22 −
2
(c) H22 + , He2
(2013 Main)
(d) H−2 , He2+
2
14. Hyperconjugation involves overlap of which of the following
orbitals?
(a) σ - σ
(c) p - p
15. According to MO theory,
(2008, 3M)
(b) σ - p
(d) π - π
(2004, 1M)
(a) O+2 is paramagnetic and bond order greater than O2
(b) O+2 is paramagnetic and bond order less than O2
(c) O+2 is diamagnetic and bond order is less than O2
(d) O+2 is diamagnetic and bond order is more than O2
54 Chemical Bonding
16. Molecular shape of SF4 , CF4 and XeF4 are
(2000, 1M)
The electronic structure of O 3 is
24. Statement I
+
(a) the same, with 2, 0 and 1 lone pair of electrons respectively
(b) the same, with 1, 1 and 1 lone pair of electrons respectively
(c) different, with 0, 1 and 2 lone pair of electrons respectively
(d) different, with 1, 0 and 2 lone pair of electrons respectively
17. In compounds of type ECl 3 , where E = B, P, As or Bi, the
angles Cl—E—Cl is in order
(a) B > P = As = Bi
(b) B > P > As > Bi
(c) B < P = As = Bi
(d) B < P < As < Bi
(1999, 2M)
(1999, 2M)
(b) CO2 < CO2–
3 < CO
(d) CO < CO2 < CO2–
3
19. Which contains both polar and non-polar bonds?
(a) NH4 Cl
(c) H2 O2
(b) HCN
(d) CH4
(b) Liquid NH3
(d) HCl
••
−
O •• .
••
•
•
O
••
O ••
••
••
structure is not allowed
(2016 adv.)
(b) O2+
2 is expected to have a longer bond length than O2
(c) N+2 and N−2 have the same bond order
has the same energy as two isolated He atoms
22. Hydrogen bonding plays a central role in which of the
following phenomena?
(2014 Adv.)
(a) Ice floats in water
(b) Higher Lewis basicity of primary amines than tertiary
amines in aqueous solutions
(c) Formic acid is more acidic than acetic acid
(d) Dimerisation of acetic acid in benzene
23. Which one of the following molecules is expected to exhibit
diamagnetic behaviour?
(2013 Main)
(a) C 2
(b) N 2
(c) O 2
(d) S 2
Assertion and Reason
Read the following questions and answer as per the direction given
below:
(a) Statement I is correct; Statement II is correct; Statement II is
the correct explanation of Statement I.
(b) Statement I is correct; Statement II is 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.
(1998, 2M)
25. Match the reactions in Column I with nature of the
reactions/type of the products in Column II.
(2007, 6M)
Column I
A.
O−2
B.
C.
Column II
→ O2 +
O22 −
1.
Redox reaction
CrO24 − + H+ →
2.
One of the products has
trigonal planar structure
MnO−4 + NO−2
3.
Dimeric bridged
tetrahedral metal ion
4.
Disproportionation
+
+ H →
NO−3 + H2 SO4
+ Fe
21. According to molecular orbital theory, which of the
(d)
•
•
Match the Columns
D.
Objective Questions II
(One or more than one correct option)
He+2
Statement II
(1983, 1M)
following statements is(are) correct?
(a) C2−
2 is expected to be diamagnetic
••
O
O
(1997, 1M)
20. Which one among the following does not have the hydrogen
bond?
(a) Phenol
(c) Water
•
•
because octet around O cannot be expanded.
18. The correct order of increasing CO bond length of
CO, CO23 − , CO2 is
(a) CO23 − < CO2 < CO
(c) CO < CO2–
3 < CO2
O
Codes
A
(a) 2
(c) 2
2+
→
B
1, 4
3
C
3
1
D
4
4
A
(b) 1, 4
(d) 3
B
3
4
C
1, 2
2, 3
D
1
1
Integer Answer Type Questions
26. Chlorine reacts with hot and concentrated NaOH and produces
compounds ( X ) and (Y ). Compound ( X ) gives white
precipitate with silver nitrate solution. The average bond
order between Cl and O atoms in (Y ) is …… .
(2020 Main, 7 Jan I)
27. Among H2 ,He+2 , Li 2 , Be2 , B2 , C2 , N2 , O−2 and F2 , the
number of diamagnetic species is
(Atomic numbers : H = 1, He = 2, Li = 3, Be = 4, B = 5, C = 6,
N = 7, O = 8, F = 9)
(2017 Adv.)
Subjective Questions
28. Write the MO electron distribution of O2 . Specify its bond
order and magnetic property.
(2000, 3M)
29. Arrange the following as stated.
“Increasing strength of hydrogen bonding ( X  H  X ).”
O, S, F, Cl, N
(1991, 1M)
30. What effect should the following resonance of vinyl chloride
have on its dipole moment?
CH2 ==CH  Cl ←→
CH2–
(1987, 1M)
+
 C HCl
Answers
Topic 1
1.
5.
9.
13.
2.
6.
10.
14.
(d)
(a)
(a)
(a, c)
17. (c)
25. (80.2%)
3.
7.
11.
15.
(c)
(b)
(a)
(a)
18. (2)
4.
8.
12.
16.
(b)
(b)
(c)
(6)
19. F
(a)
(b)
(c)
(1)
20. T
Topic 2
1.
5.
9.
13.
17.
21.
25.
29.
33.
37.
41.
45.
2.
6.
10.
14.
18.
22.
26.
30.
34.
38.
42.
46.
(d)
(d)
(d)
(d)
(c)
(b)
(b)
(d)
(a)
(c)
(b)
(b, c)
3.
7.
11.
15.
19.
23.
27.
31.
35.
39.
43.
47.
(b)
(c)
(a)
(b)
(a)
(d)
(c)
(b)
(a)
(c)
(a)
(a, c)
(c)
(a)
(a)
(a)
(b)
(a)
(b)
(a)
(a)
(d)
(a)
(b, c, d)
4.
8.
12.
16.
20.
24.
28.
32.
36.
40.
44.
48.
(d)
(d)
(d)
(a)
(a)
(b)
(c)
(a)
(c)
(b)
(c)
(a, b)
49. (c) A → 2; B → 3; C →1; D →4
50. (b) A → p, q, r, t; B → q, r, s, t; C → p, q, r, t; D → p, r, s, t
52. increases, decreases
51. N 2O, I 3−
53. three centre bond-two electrons
54. hyperconjugation
55. sp
56. Triangular planar
57. sp 3
58. HCOOH and CH3COOH
59. CO2
60. F
61. F
65. T
63. F
68. (4)
64. F
69. (6)
62. F
66. (6)
Topic 3
1.
5.
9.
13.
17.
21.
25.
27.
(a)
(a)
(d)
(c)
(b)
(a, c)
(b) A → 1, 4;
(6)
2. (b)
3. (c)
6. (a)
7. (c)
10. (b)
11. (c)
14. (b)
15. (a)
18. (a)
19. (c)
22. (a, b, d)
23. (a,b)
B → 3; C →1, 2; D →1
28. (2)
4.
8.
12.
16.
20.
24.
26.
(a)
(d)
(b)
(d)
(d)
(a)
(1.67)
Hints & Solutions
Topic 1 Preliminary Concepts of Electrovalent
and Covalent Bonding
H
1. Key Idea Isoelectronic species contains same number of
electrons.
O
H
H
O
(O2)
(PH3)
The species with its atomic number and number of electrons are
as follows :
Species (ions)
At. no. ( Z )
No. of electrons
N3−
7
7 + 3 = 10
O2−
8
8 + 2 = 10
F−
9
9 + 1 = 10
Na +
11
11 − 1 = 10
3
3 −1 = 2
12
12 − 2 = 10
Li
Covalent
bond
P
+
Mg
2+
Thus, option (d) contains isoelectronic set of ions.
2. KCl is the only ionic compound. The structure of PH3, O2, B2H6
and H2SO4 are given below
H
120°
H
B
97°
H 1.33Å
Å H
1.19
B
H
H
H
H
H
H
B
H
Covalent
Bonds
B
H (B2H6) H
O
H
H
B
B
H
H
H
Banana bond
S
HO OH O
Sulphuric acid
(H2SO4)
All bond between S and O atom are covalent bonds.
56 Chemical Bonding
3. Ion-ion interaction is dependent on the square of distance, i.e.
ion-ion interaction ∝
1
r2
F
Similarly, ion-dipole interaction ∝
London force ∝
1
r6
F
Three identical vectors acting in outward direction at equal
angles in a plane cancel each other giving zero resultant, hence
non-polar.
r3
(In case of dipole-dipole interaction)
From the above, it is clear that the ion-dipole interaction is the
better answer as compared to dipole-dipole interaction, i.e.
hydrogen bonding.
4.
sp2
1
Superficially it seems as both ion-dipole interaction and
hydrogen bonding vary with the inverse cube of distance between
the molecules but when we look at the exact expressions of field
(force) created in two situations,
it comes as
2| P |
(In case of ion-dipole interaction)
| E | or | F | =
4 π ∈ r3
H
9. Strongly electropositive, univalent X will form an 1 : 1 ionic
compound with strongly electronegative, univalent Y.
X + Y → X +Y −
10. H2 is a covalent, diatomic molecule with a sigma covalent bond
between two hydrogen atoms.
11. N2 has triple bond and each covalent bond is associated with
one pair of electrons, therefore, six electrons are involved in
forming bonds in N2.
12. In KCN, the bonding between potassium ion and cyanide ion is
ionic while carbon and nitrogen are covalently bonded in
cyanide ion as:
Covalent bonds
H
+ –
[K] [ C ≡≡N]
C—–C
H
120°
F—B
1
r3
and dipole-dipole interaction ∝
2q2r – 4 q2a
and F =
4 π ∈0 r3
8. BF3 has triangular planar arrangement.
H
Ionic bond
Pi bond is formed by the p-orbitals whose lobes have minima in
the plane of molecule, hence molecular plane is the nodal plane of
pi-bond.
Key Idea Dipole moment of a bond depends on the
difference in the electronegativities of bonded atoms.
More is the difference in the electronegativities, greater
will be the dipole moment. Also,
For symmetrical molecule, µ = 0
For unsymmetrical molecule, µ ≠ 0
13.
5. H-bond is the strongest intermolecular force.
All are different with 1, 0 and 2 lone pairs of electrons at central
atom.
The molecules which gives permanent dipole moment are
polar in nature.
6. p-dichlorobenzene is non-polar.
Cl—
Cl
—Cl
S
Se
p-dichlorobenzene
,
The two dipole vectors cancelling each other giving zero
resultant dipole moment. o-dichlorobenzene has greater
dipole moment than meta-isomer.
Cl
O
µ1
Cl
F
F
µ2
F
(m-dichlorobenzene)
dipole vectors are at 120° angle
p-dichlorobenzene (IV) < toluene (I) < m-dichlorobenzene (II)
< o-dichlorobenzene (III)
Cl
–
m1
m¹0
(polar)
Be
Cl
,
F
m2
Cl
mnet» 0
(non-polar)
Cl
m2
,
O
: C ≡≡ C : one sigma and two pi-bonds.
m¹0
(polar)
F
m1
Toluene is less polar than both ortho and para
dichlorobenzene. Therefore, the increasing order of dipole
moment is
7. The carbide (C2−
2 ) ion has the following bonding pattern:
H
H
m¹0
(polar)
Br
Cl
(o-dichlorobenzene)
dipole vectors are at 60° angle
–
O
m¹0
(polar)
µ1 > µ2
Cl
,
C
H
C
m¹0
(polar)
Cl
Cl
N
,
O
Cl
m 0
(non-polar)
Cl
N
Cl ,
B
O
mnet 0
(non-polar)
O
m¹0
(polar)
,
H H H
m¹0
(polar)
O
P
m¹0
(polar)
Cl ,
Cl
Chemical Bonding 57
F
C
H
F
F
H
,
F
H m¹0
O
B
Cl ,
O
(vi) (NH4)2FeCl 4 has Fe as central metal atom with +2 oxidation
state. The electronic configuration of Fe 2+ in the complex is
Xe
,
F
m=0
(non-polar)
(polar)
F
3d
4s
4p
Cl–
Cl– Cl– Cl–
O
m¹0
(polar)
F
sp3
F
F
4 unpaired
electrons
m»0
(non-polar)
Thus, options (a, c) are correct.
14. 1,4-dichlorobenzene is non-polar, individual dipole vectors
(vii) (NH4)2NiCl 4 has Ni as central metal atom with +2 oxidation
state. The electronic configuration of Ni 2+ in the complex is
cancel each other.
3d
4s
4p
Cl–
Cl– Cl– Cl–
sp3
µ≠0
Cl H
Cl
Cl ClCH2
C==C
H
H Cl
Polar
µ=0
Non-polar
Cl
H
C2 H 5
Polar
(viii) In K2MnO4 central metal atom Mn has +6 oxidation state
with following structure
O–
15. Only polar liquid will be attracted towards charged comb due to
the formation of electrically charged droplets in the polar liquid
stream, induced by a nearby charged object. Hence, liquid
showing deflection are HF,H2O,NH3 ,H2O2 ,CHCl 3 , C6H5Cl.
16. Among the given species only K2CrO4 is diamagnetic as central
0
metal atom Cr in it has [ Ar ]3d electronic configuration i.e., all
paired electrons. The structure and oxidation state of central
metal atom of this compound are as follows
O
, Oxidation state Cr6+
–
O– O
Rest all the compounds are paramagnetic. Reasons for their
paramagnetism are given below
(i) H-atom have 1s1 electronic configuration, i.e. 1 unpaired
electron.
Electronic configuration of Mn6+ is
3d
4s
17. Statement I is correct but Statement II is incorrect. The
covalency in LiCl is due to small size of Li + ion which brings
about large amount of polarisation in bond.
18. These are 2π-bonds in a nitrogen molecule.
19. The resultant of individual bond dipoles may or may not be
non-zero.
20. Linear overlapping of p-orbitals form sigma bond while sidewise
N
O
O
–
O O
one unpaired
electron
Cr
O
Mn
2K+
O
Structure K+
hybridisation
2 unpaired
electrons
C==C
C==C
H
(ii) NO2 , i.e.
hybridisation
in itself is an odd electron species.
(iii) O−2 (Superoxide) has one unpaired electron in π * molecular
orbital.
(iv) S2 in vapour phase has O2 like electronic configuration i.e.,
have 2 unpaired electrons in π * molecular orbitals.
(v) Mn 3O4 has following structure
+2
O
Mn
+4
O
Mn
O
+2
Mn
O
Thus, Mn is showing +2 and +4 oxidation states. The outermost
electronic configuration of elemental Mn is 3d 5 4 s2. Hence, in
both the above oxidation states it has unpaired electrons as
3d
overlapping of two p-orbitals forms a pi bond.
21. Li + < Al 3+ < Mg2+ < K+
22. Ag+ is stronger Lewis acid because it can easily accommodate
lone pair of electrons from Lewis base. On the other hand, Na +
has noble gas configuration, cannot accept lone pair of electron,
not at all a Lewis acid.
23. I2 is Lewis acid because I − coordinate its one lone pair to I2.
24. Both LiF and LiI are expected to be ionic compounds. However ,
LiI is predominantly covalent because of small size of Li + and
large size of iodide ion. A smaller cation and a larger anion
introduces covalency in ionic compound.
25. Dipole moment is calculated theoretically as
µ = q⋅d
4s
Mn2+
Here,
µ Theo = 1.6 × 10−19 × 2.6 × 10−10 = 4.16 × 10−29 cm
5 unpaired electrons
3d
Mn4+
3 unpaired electrons
q = 1.6 × 10−19 C and d = 2.6 × 10−10 m
4s
% ionic character =
3.336 × 10−29
µ obs
× 100
× 100 =
µ Theo
4.16 × 10−29
= 80.2%
58 Chemical Bonding
The hybridisation of given species are as follows :
26. In hydrogen peroxide (H2O2 ), oxygen is in –1 oxidation state,
can be oxidised to O2 (zero oxidation state) or can be reduced to
H2O (–2 oxidation state of oxygen).
l
H =
Hence, H2O2 can act as both oxidising agent and reducing agent.
With strong oxidising agent like KMnO4, H2O2 acts as a
reducing agent while with strong reducing agent like H2C2O4 , it
acts as an oxidising agent.
l
For [ ICl4 ]− ,
H =
27. (i) Melting points Ionic compounds have higher melting points
l
than covalent compounds.
(ii) Boiling points Ionic compounds have higher boiling points
than covalent compounds.
(iii) Solubility Ionic compounds have greater solubility in water
than a covalent compound.
(iv) Conductivity in aqueous solution Ionic compounds have
greater electrical conductivity in aqueous solution while
covalent compounds are usually non-conducting.
For [ IF6 ]− ,
5.
Bond Order Paramagnetic/
 N b − N a  Diamagnetic
 Nature



2
Species
Valence MOs
NO(15e− )
[ 8 e− ] π 2 px2 = π 2 py2σ 2 pz2
π * 2 p1x = π * 2 py0σ * 2 pz 0
–e−
NO + (14e− )
[ 8 e− ] π 2 px2 = π 2 py2σ 2 pz2
6−1
= 2.5
2
6−0
=3
2
Paramagnetic
Diamagnetic
π * 2 px0 = π * 2 py0σ * 2 pz0
••
2. For ICl5
1
(7 + 5 − 0 + 0) = 6 (sp3d 2 )
2
–e
Cl
3
N 2 (14 e− )
[ 8 e− ] π 2 px2 = π 2 py2σ2 pz2
N 2+ (13 e− )
π * 2 px0 = π * 2 py0 , σ 2 pz0
−
Cl
[ 8e− ] π 2 px2 = π 2 p2y σ 2 pz 1
2
sp d hybridised
Cl
Geometry : Octahedral
Shape / Structure : Square pyramidal
I
Cl
Cl
Cl
s
Geometry : Octahedral
Shape/Structure : Square planar
I
Cl
So, ICl5 and
sp 3 d 2 hybridised
Cl
ICl−4
are isolobal but not isostructural.
Key Idea The hybridisation for a central atom in a species
can be calculated using formula
1
H = (V + M − C + A )
2
where, H = No. of hybridised orbitals used by central atoms.
V = No. of valence electrons of the central atom.
M = No. of mono-valent atoms (bonded).
C = No. of cationic (positive) charge.
A = No. of anionic (negative) charge.
6−0
=3
2
5−0
= 2.5
2
Diamagnetic
6−2
=2
2
Paramagnetic
Paramagnetic
π * 2 px0 = π * 2 p0y σ * 2 pz0
Cl
For &&ICl−4
1
H = (7 + 4 − 0 + 1) = 6 (sp3d 2 )
2
3.
1
(7 + 4 − 0 + 1) = 6 (sp3d 2 )
2
1
(7 + 6 − 0 + 1) = 7 (sp3d 3 )
2
4. The radius of isoelectronic species is inversely proportional to
their nuclear charge or atomic number (Z). Thus, greater the
value of Z, lesser the radii of isoelectronic species.
only. Valence bond theory (VBT) cannot explain the colour
exhibited by transition metal complexes. This theory cannot
distinguish ligands as weak and strong field ones.
H=
1
(7 + 2 − 0 + 1) = 5 (sp3d )
2
H =
Topic 2 VBT, Hybridisation and
VSEPR Theory
1. Among the given statements, correct statements are I and III
For [ ICl2 ]− and [ BrF2 ]−
O2 (16 e− )
–e−
[ 8 e− ] σ 2 pz2 π 2 px2 = π 2 py2
π * 2 p1x = π * 2 p1y σ * 2 pz0
O2+ (15 e− )
[ 8 e− ] σ 2 pz2 π 2 px2 = π 2 py2
π * 2 p1x = π * 2 py0σ * 2 pz0
+2e−
O22− (18 e− )
6−1
= 2.5
2
Paramagnetic
[ 8 e− ] σ 2 pz2 π 2 px2 = π 2 py2
π * 2 px2 = π * 2 py2σ * 2 pz0 6 − 4 = 1
2
Diamagnetic
So, only in the conversion of NO → NO+, the bond order has
increased (2.5 → 3) and paramagnetic character has changed to
diamagnetic.
6. The structure of I−3 ion is
–
I
I
I
Hence, 9 is the correct answer.
Chemical Bonding 59
SF4 : F
7. Isoelectronic species are those which contains same number of
electrons.
S
Species
Atomic number
Number of electrons
O2 −
8
10
F−
9
10
11
10
Na
+
Mg
2+
O−
Na
Mg
one lone pair at S.
F
+
12
10
8
9
11
11
12
11
∴ Option (a) is correct which contains isoelectronic species
O2 − , F − , Na + , Mg 2 + .
8. I −3 is an ion made up of I 2 and I − which has linear shape.
While Cs + is an alkali metal cation.
S
—
F
F
S is sp3 hybridised
F
••
I3− :
F
14.
F
I I
••
O

Xe
•
•
O
←
•
•
•• −
I ••
••
three lone pairs at central iodine.
F
At central atom (Xe), there is one lone pair.
• •
F
15. NO−3 and CO2−
3 both have 32 electrons, central atom
sp2 hybridised, triangular planar.
16. CH3Cl has the highest dipole moment.
17. O−2 has odd number(17) of electrons, therefore it must contain
at least one unpaired electron.
F



18. F  B− ←



F

9.
F—S==O
F
F
H



N+  H Both ‘B’ and ‘N’ sp3 tetrahedral.



H

19. NH3 = sp3 ,[ PtCl 4 ]2− = dsp2 , PCl 5 = sp3d , BCl 3 = sp2
Pyramidal
SO3 is planar (S is sp2 hybridised), BrF3 is T-shaped and SiO2−
3 is
planar (Si is sp2 hybridised).
20. All three have 14 electrons (iso electronic) with bond order of
three.
10. For molecules lighter than O2, the increasing order of energies of
+
molecular orbitals is
*
π 2 p y 
*
π 2 p y .....
σ1s σ* 1sσ 2s σ* 2s
σ
σ
2
2
p
p
x
x

π
2
p
*

z
π 2 p z
where, π2 p y and π2 p z are degenerate molecular orbitals, first
singly occupied and then pairing starts if Hund’s rule is obeyed.
If Hund’s rule is violated in B2 , electronic arrangement would
be
π 2 p 2y
σ1 s2 σ* 1s2 σ 2s2 σ* 2s2 
...
π 2 p z
21.
H
O—N==O,
N
–
O==N==O,
H
sp2
sp
1
O
2
3
4
5
with CO, have the same bond order as CO. NO− (16e− ) has bond
order of 2.
12. O−2 in KO2 has 17 electrons, species with odd electrons are always
paramagnetic.
••
13. ClO−3 : − O  Cl == O one lone pair at Cl.

O
XeF4 : F
F
Xe
F
two lone pairs at Xe.
F
6
23. H2S has sp3 hybridised sulphur, therefore, angular in shape
with non-zero dipole moment.
H
S
H
bonding electrons − antibonding electrons
2
6− 4
=
=1
2
11. The bond order of CO = 3. NO , CN and N2 are isoelectronic
sp3 H
Hybridisation at C2 = sp2 and at C3 = sp3.
Bond order =
−
H
22. CH2 == CH  CH2  CH2  C ≡≡ CH
No unpaired electron-diamagnetic.
+
+
(Non-linear, polar molecule)
F
24.
F—B
F
sp2
(Trigonal planar)
25. Sulphur in SO2 is sp2-hybridised.
S
O
O
Electron pair = 2 (σ-bonds) + 1 (lone pair) = 3
Hybridisation = sp 2
Carbon in CO2 is sp-hybridised, N in N2 O is sp-hybridised,
carbon in CO is sp-hybridised.
60 Chemical Bonding
32. PCl 3 has sp3-hybridised phosphorus, with one lone pair. Therefore,
26. Molecular orbital electronic configuration are
KO2 (O−2 ) : σ1s2 σ* 1s2 σ 2s2 σ* 2s2 σ 2 px2
molecule has pyramidal shape like ammonia.
π 2 p 2y π* 2 p 2y
π 2 p 2z
σ* 2 px0
33. O−2 has odd number of electrons, hence it is paramagnetic.
*
π 2 p1z
Cl
Has one unpaired electron in π* 2 p orbital.
AlO−2 has both oxygen in O2− state, therefore, no unpaired
electron is present. BaO2 (O2−
2 )
C==C
34.
Cl
C
Cl
Cl
sp2-hybridised
π 2 p 2y
σ1s2 σ* 1s2 σ 2s2 σ* 2s2 σ 2 px2
π 2 p 2z
Has no unpaired electron.
35. CO2 is linear because carbon is sp-hybridised.
π* 2 p 2z
has [O== N==O] bonding, hence no unpaired
electron.
36. In CH+3 , there are only three electron pairs around carbon atom
giving sp2-hybridisation state.
+
27. N2 is a neutral, non-polar, inert molecule while CN − is a
H
28.
F
N
F
F
:
O==N
–
O
+
H 3O :
O
H
37. Dipole vectors in trans-1, 2-dichloroethylene are at 180° and
directed in opposite direction, cancelling each other.
Cl
H
C==C
+
H
H
Pyramidal
(O-sp3)
Triangular planar
(N-sp2)
N3H ••
F
Triangular planar
(B-sp2)
O
sp2-hybridised
F
F—B
BF3 :
Pyramidal
(N-sp3)
–
NO3
+
••
••
• N == N == N  H
•
Central nitrogen is sp -hybridised
29. BF3 has triangular planar arrangement.
F
F—B 120°
F
sp2-hybridised
There identical vectors acting in outward direction, at
equal angles in a plane, cancel each other giving zero
resultant, hence non-polar.
30. A water molecule can form at the most four H-bonds.
38. In SO2, the Lewis-dot structure is
• •
O == S == O
NOTE
π-bonded electrons are not present in hybrid orbitals, therefore not
counted in electron pairs. Rather π bonds are formed by lateral
overlapping of pure p-orbitals.
39. Bonds between identical non-metal is purely covalent due to same
electronegativities of the bonded atoms. Hence, the bonded atoms
have equal holds on the shared pair of electrons.
40. Hybridisation of one ‘s’ and one ‘p’ orbitals gives two
sp hybrid orbitals oriented linearly at 180°.
s + p → 2 sp hybrid orbitals
41. CCl 4 has a regular tetrahedral shape.
Cl
H
µ
O
H
Four sites of H-bonding
31.
• •
O  Cl == O
• •
electron pairs at Cl = 2 (σ-bonds) + 2 (lone-pairs) = 4
Hybridisation at Cl = sp
3
dipole moment = 0
H
Cl
Electron pairs at S = 2 (σ-bonds) + 1 (lone-pair) = 3
sp2 hybridised.
Therefore, NF3 , H3O+ and BF3 , NO3− pairs have same shape.
−
H
H—C
highly polar, highly active ion.
NF3 :
Cl
Cl
π 2 p 2y *
σ 2 px0
109°
sp3-hybridised
*
+
NO+2
Cl
120°
Cl
Cl
C
µ
Cl
Net dipole = 0
Cl
42. CO has a total of 14 electrons and CN− also has 14 electrons.
C (6e− ) + N (7e− ) + e− → CN− (14 e− )
43. CO2 is a linear molecule because of sp-hybridisation around
carbon atom.
44. For non-polar MX 3, it must have triangular planar arrangement,
i.e. there should be sp2-hybridisation around M.
Chemical Bonding 61
45.
CH3

H3C  C  CH3

CH3
H
H3C
H
Polar
Symmetric, non-polar
CH3H2C
CH2CH3
C == C
H
H
Polar
CH2CH3
CH3 CH3
 
H3C  C  C  CH3
 
CH3 CH3
Symmetric, non-polar
46. CN− and NO+ are isoelectronic, have the same bond order of 3.
+
47. S== C == S
Linear
••
Sn
Cl
−
O== N == O
Linear
O  C ≡≡ N
Linear
••
Cl
O
Bent
S
d-d σ antibonding
D.
∴ A → 2, B → 3, C → 1, D → 4
Hence, (c) is the correct option.
50. (A) B2 : σ1s2 σ* 1s2 σ 2 s2 σ* 2s2
π 2 p1y
π 2 pz1
paramagnetic.
6− 4
=1
2
Bond is formed by mixing of s and p orbitals.
B2 undergoes both oxidation and reduction as
Bond order =
O
Bent
48. CO2 , HgCl 2 , C2H2 are all linear.
49.
p-d π antibonding
C.
C == C
Heat
PLAN This problem includes basic concept of bonding. It
can be solved by using the concept of molecular orbital theory.
B2 + O2 → B2 O3
(Oxidation)
B2 + H2 → B2 H6
(Reduction)
(B) N2 : σ1s2 σ* 1s2 σ 2 s2 σ* 2 s2σ 2 px2
+ ve phase
π 2 p 2y
π 2 p 2z
diamagnetic.
10 − 4
= 3> 2
2
N2 undergoes both oxidation and reduction as
Bond order =
– ve phase
∆
N2 + O2 →
NO
Any orbital has two phase +ve and –ve. In the following
diagram, +ve phase is shown by darkening the lobes and –ve by
without darkening the lobes.
Catalyst
N2 + 3H2 → NH3
In N2 , bonds are formed by mixing of s and p orbitals.
*
*
(C) O2− : σ1s2 σ
1s 2 σ 2 s 2 σ
2 s2 σ 2 px2
Bonding MO
Antibonding MO
When two same phase overlap with each other, it forms bonding
molecular orbital otherwise antibonding.
B.
d-d σ bonding
p-d π bonding
π 2 p2y π* 2 p1y
π 2 pz2 π* 2 p1
z
*
σ
2 px0
Paramagnetic with bond order = 2.
O2 undergoes reduction and the bond involves mixing of
s and p-orbitals.
51. N2O and I−3 are linear species.
52. Bond order in N2 is 3 while same in N+2 is 2.5, hence bond
distance
A.
*
σ
2 px0
−
*
*
(D) O2 : σ1s2 σ
1s2 σ 2 s2 σ
2 s2 σ 2 px2
On the basis of above two concepts, correct matching can be
done as shown below:
π 2 pz2 π* 2 pz1
Paramagnetic with bond order = 1.5. O2 undergoes both
oxidation and reduction and bond involves mixing of s and
p-orbitals.
σ-bond
π-bond
π 2 p2y π* 2 p2
y
increases
as
N2
goes
to
N+2 .
Bond order in O2 is 2 while same in O+2 is 2.5, hence bond
distance decreases as O2 goes to O+2 .
53. Three centred-2 electrons.
54. Hyperconjugation involves delocalisation of σ-electrons.
55. sp-hybridised.
56. Triangular planar. Carbon in CH+3 is sp2 hybridised.
62 Chemical Bonding
57. sp3-hybrid orbital holding the lone pair is involved in formation
H3N
of ammonium ion.
O
O


58. H  C  OH and CH3  C  OH . Both are capable of forming
H-bonds.
2+
Cu
H3N
NH3
Cl
NH3
Cl
Cl
2–
Pt
Cl
SF4 (See-saw) as shown below:
59. CO2, it is 180°.
F
60. Dipole moment (µ ) = q.d
S
Since electronegativity of F and Cl are very close, it is the
internuclear distance (d) that decides dipole moment here.
Hence, C  Cl bond has greater dipole moment the C-F bond.
F
F
F
SiF4 , BF4− , [FeCl 4 ]2− , [CoCl 4 ]2− are tetrahedral as shown below:
61. H2O is V-shaped molecule.
O
H
H
V-shaped
62. False
63. In sp3-hybrid orbital, there is 25 % s-character and 75 % p-character.
64. Carbon in benzene is sp2-hybridised, i.e. uses only two of its
p-orbitals in hybridisation.
65. Sn in SnCl 2 has sp2-hybridisation.
66.
S.N.
No. of σ-bonds
with central atom
6
Species
No. of L.P at
central atom
1
(i)
In [ TeBr6 ]2−
(ii)
In [ BrF2 ]+
2
2
(iii)
(iv)
In SNF3
In [ XeF3 ]−
4
3
0
3
••
N ≡≡ N → N ••
••
↑
sp
67. Cl  Be  Cl
↑
sp
69.
PLAN Melamine is a heterocyclic compound.
N
O
S
O
sp2-bent
Cl
70.
sp3-V-shaped
71. No, (i) NMe3 is pyramidal while (ii) N(SiMe3 )3 is planar. In the
latter case, pπ - dπ back bonding between N and Si makes N
sp2-hybridised.
sp3d-linear
XeF4 , BrF4− , [Cu(NH3 )4 ]2+ , [PtCl 4 ]2− are square planar as shown
below:
72. Bond order : O2− = 1.5, O2 = 2, O2+ = 2.5
Bond length : O+2 < O2 < O−2
Xe
F
F
O
Xe
F
F
F
F
F
F
Br
F
73.
s
F
F
Lone pair would push the Br—F bond pairs in upward direction
and all Br—F bond angles will contract.
PLAN This problem includes concept of hybridisation using VBT,
VSEPR theory, etc.,
F
F
Br
sp3d-linear
–
[Although ICl 2 , I–3 and XeF2 all also are linear but in them
d-orbital contribute in hybridisation.]
F
F
F
F
[I  I  I]– F  Xe  F
68.
N
Each nitrogen atom has one pair of lone pair. Thus, in all six
lone pairs.
[Cl  I  Cl]
Cl
sp3d-linear
NH2
NH2
All the above mentioned molecules/ions have sp-hybridised
central atom and no one pair at central atom, hence linear also.
Others are :
O
N
H2N
–
+
O == N == O
↑
sp
••
N≡≡ N → O••
••
↑
sp
Hence, correct integer is 4.
F
Square planar
F
S
F
F
Trigonal bipyramidal
Chemical Bonding 63
74.
Cl
F
F
P
Cl
(d) A - D has shortest bond length, it is incorrect because inter
molecular distance in between A - D more than 150 (pm)
which is height in all.
F
Cl
2. The change of O2 to O−2 can be as follows:
Br
Cl
Cl
[Dioxygen]
Energy
[8e– ] σ2p2z π2px2
= π2py2 π* 2px1 = π* 2p1y
Square pyramidal
(Br is sp3d 2 -hybridised)
Trigonal bipyramidal
(P is sp3d-hybridised)
75.
F
F
F
F
Xe
Xe
F
F
Square planar
Linear
O
Xe
F
F
O
See-saw shaped
76. In H2S, S is sp -hybridised with two lone pairs of electrons on it
giving V-shaped (water like) shape. In PCl 3 , P is sp3-hybridised
with one lone pair of electrons on it. Therefore, PCl 3 is
pyramidal in shape.
+2
×
×
F
××
××
••
××
ו
O
••
•×
F
××
×
×
−1
O
F
F −1
V-shaped
Electron pair = P = 2 + 2 = 4
Hybridisation = sp3

••
••

•
•
78. (i) O2−
2 :  • O •• O • 
••
••
2−


and


•
•
•O•


••
••


2−
(ii) CO3 : 
C

 • • • • 
• O • • O •
 • • • • • • • • 
••

(iv) NCS− :  •• S
 ••
••
••C ••
••
*
*
π 2px or π 2py molecular orbital (anti-bonding) which is
half-filled in O2.
3. HF has highest boiling point among hydrogen halides because it
has strongest hydrogen bonding. Here, the hydrogen bond exists
between hydrogen of one molecule and fluorine atom of another
molecule as shown below.
δ+
δ−
δ+
δ−
δ+
δ−
…H F…H F…H F
In this molecule, hydrogen bond behaves like a bridge between
two atoms that holds one atom by covalent bond and the other by
hydrogen bond.
••
••
•
•
• Cl •• Cl •
••
••
(Cl 2 )
Among the given options, CO is a diamagnetic molecule. It can
be proved by molecular orbital (MO) theory. The electronic
configuration of given diatomic molecules are given below.
•
•
Since, there is no unpaired electron in the CO molecule, so it
is diamagnetic.
F
••
••
C •• •• O ••
CO (Number of electrons = 14)
Electronic configuration = σ1s2 , σ* 1s2 , σ 2s2 , σ* 2s2,
σ2 pz2 , π 2 px2 ≈ π 2 p2y
F • • B • • F (BF3 )
l
(CO)
σ* 2s2 , σ 2 pz2 , π 2 px2 ≈ π 2 p2y, π * 2 px1 ≈ π * 2 p0y
••

••
N and •• Cl • • C • • N (ClCN)
••
•
•

Since, NO has one unpaired electron in π * 2 px1 orbital, so it is
paramagnetic.
l
(a) A - B has stiffest bond.
This is correct because the potential energy of A - B has
maximum negative value [between −500 to −600]. More
negative is the potential energy means high energy is
released. Higher the released energy higher is the stability
and hence the stiffness bond.
(b) D is more electronegative than other atoms, it is incorrect.
(c) A - A has the largest bond enthalpy, it is incorrect because
value of potential energy is more negative in A - B.
NO (Number of electrons = 15)
Electronic configuration = σ1s2 , σ* 1s2 , σ 2s2,
Topic 3 Resonance, LCAO, MOT,
Other Bonding Types
1.
= π* 2p1y
theory. Presence of unpaired electrons means paramagnetic and
absence of unpaired electrons means diamagnetic in nature.
and
−
[8e– ] σ2p2z π2p2x = π2p2y π* 2px2
Half-filled
anti-bonding
π*-MOs
l
−
[Super-oxide]
Energy
4. Key Idea Magnetic nature can be detected by molecular orbital
2−
• •

••
(iii) CN − :  •• C • • N  and
••


–
O2 (17 e–)
So, in the formation of O−2 from O2, the 17th electron goes to the
3
77.
+e –
O2 (16 e –)
F
F
B2 (Number of electrons = 10)
Electronic configuration
π 2 px1 ≈ πp1y
= σ1s2 , σ* 1s2 ,
σ2s2,
σ* 2s2,
Since, two unpaired electrons are present in π 2 px1 and π 2 p1y
orbital. So, it is paramagnetic.
l
O2 (Number of electrons = 16)
Electronic configuration
= σ1s2 , σ* 1s2 , σ 2s2 ,
σ 2 pz2 , π 2 px2 ≈ π 2 p2y, π * 2 px1 ≈ π * 2 p1y
σ* 2s2,
Since, two unpaired electrons are present in π * 2 px1 and π * 2 p1y
orbital. So, it is also paramagnetic.
64 Chemical Bonding
5. C2 will be stabilised after forming anion. The electronic configuration of carbon is1s2 2s2 2 p2. There are twelve electrons inC2. After forming
anion (i.e. C–2 ), the electronic configuration is
l
C−2 : (σ1s) 2 (σ *1s) 2 (σ 2 s) 2 (σ * 2 s) 2 ( π 2 px2 = π 2 p 2y ) (σ2 pz1 ) or KK (σ 2 s) 2 (σ * 2 s) 2 ( π 2 px2 = π 2 p 2y ). σ 2 pz1
Bond order =
1
1
(N b − N a ) = (9 − 4 ) = 2.5
2
2
For other options such as F2− , O2− , NO− , the electronic configurations are as follows :
l
F2− : (σ1s)2 (σ* 1s)2 (σ 2s)2 (σ* 2s)2 (σ 2 pz )2 (π 2 px2 = π 2 py2 ) (π * 2 px2 = π * 2 p2y )(σ* 2 pz1 )
Bond order = 1 / 2(N b − N a ) = 1 / 2(10 − 9) = 0.5
l
l
O−2 : (σ1s)2. (σ* 1s)2 (σ 2s)2 (σ* 2s)2 (σ 2 pz )2(π 2 px2 = π 2 p2y ) (π * 2 px2 = π * 2 p1y )
1
1
Bond order = (N b − N a ) = (10 − 7) = 15
.
2
2
−
2
*
2
2
2
NO : (σ1s) (σ 1s) (σ 2s) (σ * 2s) (σ 2 pz )2 (π 2 px2 = π 2 p2y ) (π * 2 px1 = π * 2 p1y )
1
1
(N b − N a ) = (10 − 6) = 2
2
2
The value of bond order of C−2 is highest among the given options. Bond order between two atoms in a molecule may be taken as an
approximate measure of the bond length.
The bond length decreases as bond order increases. As a result, stability of a molecule increases.
Bond order =
6.
Species
Bond order (BO)
6−0
=3
2
6−2
=2
2
6−4
=1
2
6−2
=2
2
MO energy order
C22− (14e− )
[ 8e ] π
O2 (16e− )
[ 8 e ]σ
O22− (18e− )
[ 8 e ]σ
N22− (16e− )
[ 8e ] π
2p x2
2p z2
2p z2
2p x2
= π
π
π
2p x2
2p x2
= π
2p y2
σ
= π
= π
2p y2
σ
2p z2
2p y2
2p y2
π*
π*
2p1x
2p x2
*
2p z2
π
2p1x
= π*
*
= π
2p1y
2p y2
= *π 2p1y
1
. So order of bond length
BO (Bond order)
= N22− < O22−
Bond length ∝
C22−
(BO = 3)
< O2
(BO = 2)
(BO = 1)
The diamagnetic species with shortest bond length is C2−
2
(option-a).
7. The energy order of MOs of the given species are as follows:
O2 (16 e− ’s) = σ1s2 , σ* 1s2 , σ 2s2 , σ* 2s2 , σ 2 p2z ,
π 2 p2x = π 2 p2y , π * 2 px1 = π * 2 p1y ,
O+2 (15e− ’s ) = σ1s2, σ* 1s2 , σ2s2, σ* 2s2 , σ2 p2z ,
π 2 p2x
=
π 2 p2y , π * 2 px1
≈π
*
=
N+2 (13e− ’s) = σ1s σ 1s σ2 s σ* 2s2
π 2 p2x = π 2 pz2 σ 2 pz1
Thus, in case of N+2 , two π-bonds and half σ-bond are present in
the bonding MOs.
8. Considering molecular orbital theory (MOT) :
The electronic configuration of
Diamagnetic
2
Paramagnetic
0
Diamagnetic
2
Paramagnetic
Nb − Na 3 − 2 1
=
=
2
2
2
The electronic configuration of
Li−2 (Z = 7) = σ1s2 , σ* 1s2 , σ2 s2 , σ ∗ 2 s1
N − Na 4 − 3 1
Bond order (BO) = b
=
=
2
2
2
For the species having the same value of BO, the specie having
lesser number of antibonding electrons[ N a ] will be more stable.
Here, N a of Li+2 (2) < N a of Li −2 (3) . So, their order of stability
will be Li+2 > Li2− .
Bond order (BO) =
Electronic
, σ 2 pz2
2 * 2
2
2
0
be judged through the calculation of bond order.
π 2 p2y
Li+2
Magnetic character
9. Key Idea According to M.O.T, the viability of any molecule can
2 p0y
N2 (14 e− ’s ) = σ 1s2 σ* 1s2 , σ 2s2 , σ* 2s2
π 2 p2x
n, number of unpaired e−
*
2
(Z = 5) = σ1s , σ 1s , σ 2s
1
Configuration
He+2
σ 2 σ* 1
1s
1s
H−2
σ
H2−
2
σ
He2+
2
σ
1s 2
1s 2
1s 2
Bond order
2−1
= 0.5
2
σ*
2−1
= 0.5
2
σ*
2−2
=0
2
1s1
1s 2
2−0
=1
2
Chemical Bonding 65
Stability order is Li −2 < Li 2+ < Li 2 (because Li −2 has more number
The molecule having zero bond order will not be viable hence,
H2−
2 (option d) is the correct answer.
of electrons in antibonding orbitals which destabilises the
species).
10. To identify the magnetic nature we need to check the molecular
orbital configuration. If all orbitals are fully occupied, species is
diamagnetic while when one or more molecular orbitals is/are
singly occupied, species is paramagnetic.
(a) NO (7 + 8 = 15) − σ1s2 , σ* 1s2 , σ 2s2 , σ* 2s2 ,
13. Species having zero or negative bond order do not exist.
0
H2+
2 (1 + 1 − 2 = 0) = σ1s
Bond order = 0
He2 (2 + 2 = 4 ) = σ1s2 , σ* 1s2
N − Na 2 − 2
Bond order = b
=
=0
2
2
So, both H2+
2 and He2 do not exist.
π 2 px2 = π 2 p2y , π 2 pz2 , π * 2 px1 = π * 2 p0y
One unpaired electron is present. Hence, it is paramagnetic.
(b) CO (6 + 8 = 14 ) − σ1s2 , σ* 1s2 , σ 2s2 , σ* 2s2 ,
π 2 px2 = π 2 p2y, σ 2 pz2
No unpaired electron is present. Hence, it is diamagnetic.
(c) O2 (8 + 8 = 16) − σ1s2 , σ* 1s2 , σ 2s2 , σ* 2s2 , σ 2 pz2 ,
π 2 px2 = π 2 px2 , π * 2 px1 = π * 2 p1y
14.
Two unpaired electrons are present.
Hence, it is paramagnetic.
(d) B2(5 + 5) − σ1s2 , σ* 1s2 , σ 2s2 , σ* 2s2 , π 2 px1 = π 2 p1y
PLAN This problem can be solved by using the concept involved in
molecular orbital theory. Write the molecular orbital electronic
configuration keeping in mind that there is no 2s-2p mixing,
then if highest occupied molecular orbital contain unpaired
electron then molecule is paramagnetic otherwise diamagnetic.
15. O+2 (15e− ) : σ1s2 σ* 1s2 σ 2s2 σ* 2s2 σ 2 px2
x
y
x
y
z
* 1s2 , σ 2s2 , σ
* 2s2 (diamagnetic)
(a) Be2 → σ1s , σ
2
* 1s2 , σ 2s2 , σ
* 2s2 , σ 2 p2 ,
(b) B2 → σ1s , σ
z
2
π 2 px0
π 2 py0
F
F
F
* 1s2 , σ 2s2 , σ
* 2s2 , σ 2 p2 , π 2 px ,
(c) C2 → σ1s2 , σ
z
π 2 p1y
π* 2 px0 * 0
, σ 2 p (paramagnetic)
S
16.
z
2
* 1s2 , σ 2s2 , σ
* 2s2 , σ 2 p2 , π 2 px ,
(d) N2 → σ1s2 , σ
z
π 2 py2
π* 2 px0
, σ * 2 pz0 (diamagnetic)
*
0
π 2p
y
Hence, (c) is the correct choice.
12. Li 2 (3 + 3 = 6) = σ1s2 , σ* 1s2 , σ 2s2
Nb − Na 4 − 2
=
=1
2
2
+
2 * 2
Li 2 (3 + 3 − 1 = 5) = σ1s , σ 1s , σ 2s1
3−2 1
Bond order =
= = 0.5
2
2
Li −2 (3 + 3 + 1 = 7) = σ1s2 , σ*1s2 , σ 2s2 σ* 2s1
4−3 1
Bond order =
= = 0.5
2
2
Bond order =
Xe
C
F
F
See-saw shape
molecule
F
F
F
F
F
(diamagnetic)
1
π* 2 py0
π 2 p2y π* 2 p1y
*
σ
2 px0
π 2 pz2 π* 2 p0
z
10 − 5
Bond order =
= 2.5; paramagnetic.
2
π 2 p2y π* 2 p1y *
−
2*
*
O2 (16e ) : σ1s σ 1s2 σ 2s2 σ
2s2 σ 2 px2
σ 2 px
π 2 pz2 π* 2 p1
z
10 − 6
Bond order =
=2
2
Hence, (a) is the correct answer.
Assuming that no 2s-2p mixing takes place the molecular orbital
electronic configuration can be written in the following sequence
of energy levels of molecular orbitals
* 1s, σ 2s, σ
* 2s, σ 2 p , π 2 p ≡ π 2 p , π* 2 p ≡ π* 2 p , σ
* 2p
σ1s, σ
z
(II)
(I)
I and II are hyperconjugation structures of propene and involves
σ-electrons of C—H bond and p-orbitals of pi bond in
delocalisation.
Two unpaired electrons are present.
Hence, it is paramagnetic.
11.
H
H


−
H  C  CH == CH2 ←→ H  C == CH  CH2

+
H
H
Tetrahedral
F
F
Square planar
17. When E = Bin BCl 3 , bond angle is 120°. When E = P, As or Bi in
ECl 3 , hybridisation at E will be sp3. Also, if central atoms are
from same group, bond angle decreases down the group provided
all other things are similar. Hence, the order of bond angles is
BCl 3 > PCl 3 > AsCl 3 > BiCl 3
1
18. Bond length ∝
Bond order
1 4
Bond order : CO2 = 2, CO = 3 , CO23− = 1 + =
3 3
Therefore, order of bond length is CO23− < CO2 < CO
19. H2O2
H
O—O
H
polar bond
Non-polar bond
66 Chemical Bonding
20. HCl does not form hydrogen bond. For formation of hydrogen
bond, atleast one hydrogen atom must be bonded to one of the
three most electronegative atom O , N and F.
21.
Species Electrons MOEC
C2−
2
N B N A BO
σ1s2 , σ * 1s2 ,
14
10
4
3
Magnetic
character
Diamagnetic
state, i.e. reduced. Hence, in the above reaction, oxygen
( O−1 / 2 ) is simultaneously oxidised and reduced
disproportionated.
2−
(B) In acidic medium, CrO2−
4 is converted into Cr2 O7 which
is a dimeric, bridged tetrahedral.
O–
O–
Cr
σ 2s2 , σ * 2s2 ,
O
π 2 px2 ≈ − π 2 p2y ,
14
As above
according to
number of
electrons
10
4
3
Diamagnetic
10
6
9
4
2.5 Paramagnetic
2
Paramagnetic
O2
16
N+2
13
N −2
15
10
5
2.5 Paramagnetic
He+2
3
2
1
0.5 Paramagnetic
The above is a redox reaction and a product NO−3 has
trigonal planar structure.
(D) NO−3 + H2SO4 + Fe2+ → Fe+ + NO
The above is a redox reaction.
26. The reaction is,
3Cl2 + 6 NaOH
5 NaCl + NaClO3 + 3H2O
(Y)
(X)
Thus, (a) is correct.
(b) Bond order O2+
2 > O 2 thus,
+ AgNO3
Bond length of O2+
2 < O 2 thus, incorrect.
(c)
(d)
N+2 and N−2 have same bond order thus correct.
He+2 with bond order = 0.5 is more stable thus, less energy
AgCl
(white ppt.)
than
isolated He atoms. Thus, (d) is incorrect.
22.
+
23. C2 (6 + 6 = 12) = σ1s , σ* 1s , σ 2s , σ* 2s ,
2
2
2
–
The structure of ClO3 (chlorate ion) is,
π 2 px2
N2 ( 7 + 7 = 14 ) = σ1s , σ* 1s2, σ 2s2 ,
σ* 2s2 , π 2 px2 ≈ π 2 p 2y , σ 2 p 2z
2
It is also a diamagnetic species because of the absence of
unpaired electrons.
O2 (8 + 8 = 16) or S2 = σ1s2 , σ* 1s2 , σ 2s2 , σ∗ 2s2 ,
σ 2 p 2 , π 2 p 2 ≈ π 2 p 2 π* 2 p1 ≈ π* 2 p1
x
y
x
Cl
O
≈ π 2 p2y
Since, all the electrons are paired, it is a diamagnetic species.
z
structure of ozone.
O
O
O
−O
== Cl
O O
O
O
O
Cl
O
O
27. H2 , Li 2 , Be2 , C2 , N2 and F2 are diamagnetic according to
molecular orbital theory.
28. O2 : σ1s2 σ* 1s2 σ 2s2 σ* 2s2 σ 2 px2
Bond order =
+
←→
O−
O
‘ClO’ bond order in the hybrid
Number of bonds between Cl and O
=
Total number of O (surrounding atoms)
5
= = 166
. or 1.67
3
24. Statement I is correct, given structure is one of the resonance
+
O
Cl
O
y
Due to the presence of two unpaired electrons, O2 and S2 both are
paramagnetic molecules.
O
–
Y = NaClO3 (Na and ClO3 )
PLAN This problem can be solved by using concept of H-bonding and
applications of H-bonding.
2
O
O
(C) MnO−4 + NO−2 + H+ → Mn 2+ + NO−3
σ 2 pz2
O2+
2
O
Cr
O
π 2 p2y π* 2 p1y
π 2 pz2 π* 2 p1
z
10 − 6
= 2, paramagnetic.
2
29. Strength of hydrogen bonding in X—H—X depends on
O
Statement II is also correct because oxygen cannot expand its
octet. It is also the explanation for the given structure of ozone.
25. (A) In the reaction : O−2 → O2 + O22−
Oxygen on reactant side is in − 1/ 2 oxidation state. In
product side, one of the oxygen is in zero oxidation state,
i.e. oxidised while the other oxygen is in –1 oxidation
electronegativity as well as size of X . X with higher
electronegativity and smaller size forms stronger H-bond.
Hence, increasing order of strength of H-bond is
S < Cl < N < O < F
30. Resonance in vinyl chloride increases polar character of the
molecule.
5
States of Matter
Topic 1 Gaseous State
Objective Questions I (Only one correct option)
1. If the distribution of molecular speeds of a gas is as per the
figure shown below, then the ratio of the most probable, the
average, and the root mean square speeds, respectively, is
(2020 Adv.)
(a) vmp of H 2 ( 300 K ); vmp of N 2(300 K); vmp of
O2(400 K)
(b) vmp of O2 ( 400 K ); vmp of N 2(300 K); vmp of
H 2(300 K)
(c) vmp of N 2(300 K); vmp of O2 ( 400 K ); vmp of
H 2(300 K)
(d) vmp of N 2(300 K); vmp of H 2(300 K); vmp of
O2(400 K)
Fraction of
molecules
4. Consider the following table.
Speed
(a) 1 : 1 : 1
(c) 1 : 1.128 : 1.224
(b) 1 : 1 : 1.224
(d) 1 : 1.128 : 1
2. For one mole of an ideal gas, which of these statements must be
true?
(2020 Main, 4 Sep I)
(A) U and H each depends only on temperature.
(B) Compressibility factor Z is not equal to 1.
(C) C p,m − CV,m = R
(D) dU = CV dT for any process.
(a) (B), (C) and (D)
(b) (A) and (C)
(c) (A), (C) and (D)
(d) (C) and (D)
3. Points I, II and III in the following plot respectively correspond
to (vmp : most probable velocity)
Distribution function f( v)
(2019 Main, 10 April II)
Gas
a/(k Pa dm6mol −1)
b/(dm 3mol −1 )
A
642.32
0.05196
B
155.21
0.04136
C
431.91
0.05196
D
155.21
0.4382
a and b are van der Waals’ constants. The correct
statement about the gases is
(2019 Main, 10 April I)
(a) gas C will occupy lesser volume than gas A; gas B
will be lesser compressible than gas D
(b) gas C will occupy more volume than gas A; gas B
will be more compressible than gas D
(c) gas C will occupy more volume than gas A; gas B
will be lesser compressible than gas D
(d) gas C will occupy more volume than gas A; gas B
will be lesser compressible than gas D
5. At a given temperature T, gases Ne, Ar, Xe and Kr are
found to deviate from ideal gas behaviour. Their equation
RT
of state is given as, p =
at T.
V −b
Here, b is the van der Waals’ constant. Which gas will
exhibit steepest increase in the plot of Z (compression
factor) vs p?
(2019 Main, 9 April II)
(a) Xe
(b) Ar
(c) Kr
(d) Ne
6. Consider the van der Waals’ constants, a and b, for the
following gases.
Gas
I
II
Speed, v
III
6
−2
3
−1
a/(atm dm mol )
−2
b/(10
dm mol )
Ar
Ne
Kr
Xe
1.3
0.2
5.1
4.1
3.2
1.7
1.0
5.0
68 States of Matter
Which gas is expected to have the highest critical
temperature ?
(2019 Main, 9 April I)
figure below. The temperature of one of the bulbs is then
(2016 Main)
raised to T2 . The final pressure p f is
(a) Kr
 T1 
(a) 2 pi 

 T1 + T2 
 T 
(b) 2 pi  2 
 T1 +T2 
 TT 
(c) 2 pi  1 2 
 T1 +T2 
 TT 
(d) pi  1 2 
 T1 +T2 
(b) Xe
(c) Ar
(d) Ne
7. The combination of plots which does not represent isothermal
expansion of an ideal gas is
(2019 Main, 12 Jan II)
p
13. If Z is a compressibility factor, van der Waals’ equation at
p
O
O
1/Vm
(A )
low pressure can be written as
(2014 Main)
RT
a
(a) Z = 1 +
(b) Z = 1 −
pb
VRT
Vm
(B)
pb
pb
(d) Z = 1 +
RT
RT
14. For gaseous state, if most probable speed is denoted by C *,
average speed by C and root square speed by C, then for a
large number of molecules, the ratios of these speeds are
(a) C * : C : C = 1.225 : 1.128 : 1
(2013 Main)
(b) C * : C : C = 1.128 : 1.225 : 1
(c) C * : C : C = 1 : 1.128 : 1.225
(c) Z = 1 −
pVm
U
O
O
p
(C)
(a) ( A ) and (C )
(c) ( B ) and ( D )
Vm
(D)
(b) ( B ) and (C )
(d) ( A ) and ( D )
(d) C * : C : C = 1 : 1.225 : 1.128
8. An open vessel at 27ºC is heated until two fifth of the air
15. For one mole of a van der Waals’ gas when b = 0 and
(assumed as an ideal gas) in it has escaped from the vessel.
Assuming that the volume of the vessel remains constant, the
temperature at which the vessel has been heated is
T = 300 K, the pV vs 1/V plot is shown below. The value of
the van der Waals’ constant a (atm L mol − 2 )
(2012)
(2019 Main, 12 Jan II)
(b) 500 K
(d) 500ºC
9. The volume of gas A is twice than that of gas B. The
compressibility factor of gas A is thrice than that of gas B at
same temperature. The pressures of the gases for equal
number of moles are
(2019 Main, 12 Jan I)
(a) p A = 2 pB
(b) 2 p A = 3 pB
(d) 3 p A = 2 pB
(c) p A = 3 pB
10. A 10 mg effervescent tablet containing sodium bicarbonate
and oxalic acid releases 0.25 mL of CO2 at T = 29815
. K and
p = 1 bar. If molar volume of CO2 is 25.0 L under such
condition, what is the percentage of sodium bicarbonate in
each tablet?
[Molar mass of NaHCO 3 = 84 g mol −1 ] (2019 Main, 11 Jan I)
(a) 8.4
(b) 0.84
(c) 16.8
(d) 33.6
11. 0.5 moles of gas A and x moles of gas B exert a pressure of 200
3
Pa in a container of volume 10m at 1000 K. Given R is the
gas constant in JK −1 mol−1 , x is
(2019 Main, 9 Jan I)
(a)
2R
4−R
(b)
4 −R
2R
(c)
4 +R
2R
(d)
2R
4+R
12. Two closed bulbs of equal volume (V) containing an ideal gas
initially at pressure pi and temperature T1 are connected
through a narrow tube of negligible volume as shown in the
24.6
pV (L atm mol–1)
(a) 750 K
(c) 750ºC
23.1
21.6
20.1
0
(a) 1.0
2.0
3.0
1/ V (mol L–1)
(b) 4.5
(c) 1.5
(d) 3.0
16. The term that corrects for the attractive forces present in a
real gas in the van der Waals’ equation is
(a) nb
(b) n 2 a / V 2
2
2
(c) − ( n a / V )
(d) − nb
(2009)
17. The given graph represent the variations of Z
(compressibility factor ( Z ) = pV / nRT ) ) versus p, for three
real gases A, B and C. Identify the only incorrect statement.
(2006, 5M)
C
A
1
B
Z
0 p (atm)
A
Ideal gas
B
C
States of Matter 69
(a) For the gas A, a = 0 and its dependence on p is linear at all
25. A gas will approach ideal behaviour at
pressure
(b) For the gas B, b = 0 and its dependence on p is linear at all
pressure
(c) For the gas C, which is typical real gas for which neither a
nor b = 0 . By knowing the minima and the point of
intersection, with Z = 1, a and b can be calculated
(d) At high pressure, the slope is positive for all real gases
(a) low temperature and low pressure
(b) low temperature and high pressure
(c) high temperature and low pressure
(d) high temperature and high pressure
18. If helium and methane are allowed to diffuse out of the
container under the similar conditions of temperature and
pressure, then the ratio of rate of diffusion of helium to
methane is
(2005)
(a) 2.0
(b) 1.0
(c) 0.5
(d) 4.0
19. For a monatomic gas kinetic energy = E. The relation with
rms velocity is
 2E 
(a) u =  
 m
1/ 2
 E
(c) u =  
 2m
1/ 2
(2004, 1M)
 3E 
(b) u =  
 2m
1/ 2
 E
(d) u =  
 3m
1/ 2
26. According to Graham’s law, at a given temperature the ratio
rA
of gases A and B is given by
rB
(where, p and M are pressures and molecular weights of gases
A and B respectively)
of the rates of diffusion
1
 p   M 2
(a)  A   A 
 pB   M B 
1
 p   M 2
(c)  A   B 
 pB   M A 
1
1
 M   p 2
(d)  A   B 
 M B   pA 
27. The compressibility factor for an ideal gas is
(a) 1.5
(b) 1.0
(c) 2.0
and that of O2 at 800 K is
(a) 4
21. Which of the following volume (V ) temperature (T ) plots
represents the behaviour of one mole of an ideal gas at the
atmospheric pressure?
(2002, 3M)
(38.8 L,
373 K)
(22.4 L,
273 K)
(b)
V(L)
(20.4 L,
273 K)
(1997, 1M)
(d) ∞
28. The ratio between the root mean square speed of H2 at 50 K
of
(2003, 1M)
(a) molecular interaction between atom and pV / nRT > 1
(b) molecular interaction between atom and pV / nRT < 1
(c) finite size of atoms and pV / nRT > 1
(d) finite size of atoms and pV / nRT < 1
V(L)
(1998, 2M)
 M   p 2
(b)  A   A 
 M B   pB 
(b) 2
(1996, 1M)
(c) 1
20. Positive deviation from ideal behaviour takes place because
(a)
(1999, 2M)
(28.6 L,
373 K)
1
(d)
4
29. Equal weights of ethane and hydrogen are mixed in an empty
container at 25° C. The fraction of the total pressure exerted
by hydrogen is
(1993, 1M)
(a) 1 : 2
(b) 1 : 1
(c) 1 : 16
(d) 15 : 16
30. At constant volume, for a fixed number of moles of a gas the
pressure of the gas increases with rise of temperature due to
(a) increase in average molecular speed
(1992, 1M)
(b) increase rate of collisions amongst molecules
(c) increase in molecular attraction
(d) decrease in mean free path
31. According to kinetic theory of gases, for a diatomic molecule
(1991, 1M)
T(K)
(c)
V(L)
(30.6 L,
373 K)
(22.4 L,
273 K)
T(K)
(d)
(a) the pressure exerted by the gas is proportional to mean
velocity of the molecule
V(L)
(b) the pressure exerted by the gas is proportional to the root
(22.4 L,
273 K)
(14.2 L,
373 K)
mean velocity of the molecule
(c) the root mean square velocity of the molecule is inversely
proportional to the temperature
T(K)
T(K)
22. The root mean square velocity of an ideal gas at constant
pressure varies with density (d ) as
(a) d 2
(b) d
(c) d
(2001, S, 1M)
(d) 1 / d
23. The compressibility of a gas is less than unity at STP.
Therefore,
(a) Vm > 22.4 L
(c) Vm = 22.4 L
(b) Vm < 22.4 L
(d) Vm = 44.8 L
(2000, 1M)
24. The rms velocity of hydrogen is 7 times the rms velocity of
nitrogen. If T is the temperature of the gas
(a) T (H2 ) = T (N2 )
(c) T (H2 ) < T (N2 )
(b) T (H2 ) > T (N2 )
(d) T (H2 ) = 7 T (N2 )
(2000, 1M)
(d) the mean translational kinetic energy of the molecule is
proportional to the absolute temperature
32. The rate of diffusion of methane at a given temperature is
twice that of a gas X. The molecular weight of X is
(1990, 1M)
(a) 64.0
(b) 32.0
(c) 4.0
(d) 8.0
33. The density of neon will be highest at
(a) STP
(c) 273° C, 1 atm
(1990, 1M)
(b) 0° C, 2 atm
(d) 273° C, 2 atm
34. The value of van der Waals’ constant a for the gases O2 , N2 ,
NH3 and CH4 are 1.360, 1.390, 4.170 and 2.253L2 atm mol −2
respectively. The gas which can most easily be liquefied is
(a) O2
(b) N2
(c) NH3
(d) CH4
70 States of Matter
35. A bottle of dry ammonia and a bottle of dry hydrogen
chloride connected through a long tube are opened
simultaneously at both ends the white ammonium chloride
ring first formed will be
(1988, 1M)
(a) at the centre of the tube
(b) near the hydrogen chloride bottle
(c) near the ammonia bottle
(d) throughout the length of the tube
Objective Questions II
(One or more than one correct option)
45. Which of the following statement(s) is(are) correct regarding
the root mean square speed (U rms ) and average translational
kinetic energy ( Eav ) of a molecule in a gas at equilibrium?
(2019 Adv.)
term that accounts for intermolecular forces is (1988, 1M)
a

(a) (V − b ) (b) RT
(c)  p + 2  (d) ( RT )−1

V 
(a) Urms is inversely proportional to the square root of its
molecular mass
(b) Urms is doubled when its temperature is increased four times
(c) Eav is doubled when its temperature is increased four times
(d) Eav at a given temperature does not depend on its
molecular mass
37. The average velocity of an ideal gas molecule at 27° C is
46. One mole of a monoatomic real gas satisfies the equation
36. In van der Waals’ equation of state for a non-ideal gas, the
0.3 m/s. The average velocity at 927° C will be (1986, 1M)
(a) 0.6 m/s (b) 0.3 m/s
(c) 0.9 m/s
(d) 3.0 m/s
38. Rate of diffusion of a gas is
(1985, 1M)
(a) directly proportional to its density
(b) directly proportional to its molecular weight
(c) directly proportional to the square root of its molecular weight
(d) inversely proportional to the square root of its
molecular weight
39. Equal weights of methane and hydrogen are mixed in an
empty container at 25° C. The fraction of the total pressure
exerted by hydrogen is
(1984, 1M)
1
8
1
16
(b)
(c)
(d)
(a)
2
9
9
17
40. When an ideal gas undergoes unrestrained expansion, no
cooling occurs because the molecules
(a) are above the inversion temperature
(b) exert no attractive forces on each other
(c) do work equal to loss in kinetic energy
(d) collide without loss of energy
(1984, 1M)
41. Helium atom is two times heavier than a hydrogen molecule.
At 298 K, the average kinetic energy of a helium atom is
(a) two times that of a hydrogen molecule
(1982, 1M)
(b) same as that of a hydrogen molecule
(c) four times that of a hydrogen molecule
(d) half that of a hydrogen molecule
42. Equal weights of methane and oxygen are mixed in an empty
container at 25°C. The fraction of the total pressure exerted
by oxygen is
(1981, 1M)
1
1
2
1 273
(b)
(c)
(d) ×
(a)
3
2
3
3 298
43. The temperature at which a real gas obeys the ideal gas laws
over a wide range of pressure is
(1981, 1M)
(a) critical temperature
(b) Boyle temperature
(c) inversion temperature (d) reduced temperature
44. The ratio of root mean square velocity to average velocity of
a gas molecule at a particular temperature is
(1981, 1M)
(a) 1.085 : 1 (b) 1 : 1.086 (c) 2 : 1.086 (d) 1.086 : 2
p(V − b)= RT where, b is a constant. The relationship of
interatomic potential V(r) and interatomic distance r for gas is
given by
(2015 Adv.)
V(r)
V(r)
(a) 0
r
r
V(r)
V(r)
(c) 0
(b) 0
r
(d) 0
47. According to kinetic theory of gases
r
(2011)
(a) collisions are always elastic
(b) heavier molecules transfer more momentum to the wall
of the container
(c) only a small number of molecules have very high velocity
(d) between collisions, the molecules move in straight lines
with constant velocities
48. A gas described by van der Waals’ equation
(2008, 4M)
(a) behaves similar to an ideal gas in the limit of large molar
volumes
(b) behaves similar to an ideal gas in the limit of large pressures
(c) is characterised by van der Waals’ coefficients that are
dependent on the identity of the gas but are independent
of the temperature
(d) has the pressure that is lower than the pressure exerted
by the same gas behaving ideally
49. If a gas is expanded at constant temperature
(1986, 1M)
(a) the pressure decreases
(b) the kinetic energy of the molecules remains the same
(c) the kinetic energy of the molecules decreases
(d) the number of molecules of the gas increases
Numerical Answer Type Questions
50. A spherical balloon of radius 3 cm containing helium gas has a
pressure of 48 × 10−3 bar. At the same temperature, the
pressure, of a spherical balloon of radius 12 cm containing the
same amount of gas will be ............ × 10−6 bar.
(2020 Main, 6 Sep I)
States of Matter 71
51. A closed tank has two compartments A and B, both filled with
54. The experimental value of d is found to be smaller than the
oxygen (assumed to be ideal gas). The partition separating the
two compartments is fixed and is a perfect heat insulator (Fig.
1). If the old partition is replaced by a new partition which can
slide and conduct heat but does not allow the gas to leak
across (Fig. 2), the volume (in m3 ) of the compartment A after
the system attains equilibrium is ____.
estimate obtained using Graham’s law. This is due to
(a) larger mean free path for X as a compared of that of Y
(b) larger mean free path for Y as compared to that of X
(c) increased collision frequency of Y with the inert gas as
compared to that of X with the inert gas
(d) increased collision frequency of X with the inert gas as
compared to that of Y with the inert gas
55. The value of d in cm (shown in the figure), as estimated from
Graham’s law, is
(a) 8
(b) 12
(c) 16
(d) 20
1 m3, 5 bar
400 K
A
3 m3, 1 bar, 300 K
B
Fig. 1
Match the Column
56. Match the gases under specified conditions listed in
Column I with their properties/laws in Column II.
B
A
Fig. 2
(2018 Adv.)
Assertion and Reason
Read the following questions and answer as per the direction given
below :
(a) Statement I is correct; Statement II is correct; Statement
II is the correct explanation of Statement I
(b) Statement I is correct; Statement II is 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
52. Statement I The pressure of a fixed amount of an ideal gas is
proportional to its temperature.
Statement II Frequency of collisions and their impact both
increase in proportion to the square root of temperature.
(2000)
53. Statement I The value of van der Waals’ constant ‘a’ is
larger for ammonia than for nitrogen.
Statement II Hydrogen bonding is present in ammonia.
(1998)
Passage Based Questions
L = 24 cm
X and Y are two
volatile liquids with
molar weights of
10 g mol −1 and 40 g
Cotton wool
Initial formation
mol −1 respectively. Cotton wool d
soaked in Y
of the product
soaked
in
X
Two cotton plugs, one
soaked in X and the
other soaked in Y, are simultaneously placed at the ends of a tube of
length L = 24 cm, as shown in the figure.
The tube is filled with an inert gas at 1 atm pressure and a
temperature of 300 K. Vapours of X and Y react to form a product
which is first observed at a distance d cm from the plug soaked in X.
Take X and Y to have equal molecular diameters and assume ideal
behaviour for the inert gas and the two vapours.
(2014 Adv.)
Column I
A. Hydrogen gas ( p = 200 atm,
T = 273 K)
B. Hydrogen gas ( p ~ 0, T = 273 K)
C. CO2 ( p = 1 atm, T = 273 K)
D. Real gas with very large molar
volume
Column II
p. compressibility
factor ≠ 1
q. attractive forces
are dominant
r. pV = nRT
s.
p (V − nb ) = nRT
(2007, 6M)
Fill in the Blanks
57. The absolute temperature of an ideal gas is …… to/than the
average kinetic energy of the gas molecules.
(1997, 1M)
58. 8 g each of oxygen and hydrogen at 27°C will have the total
kinetic energy in the ratio of …… .
(1989, 1M)
59. The value of pV for 5.6 L of an ideal gas is ......... RT, at NTP.
(1987, 1M)
60. The rate of diffusion of a gas is .......... proportional to both
.......... and square root of molecular mass.
61. C p − CV for an ideal gas is ……
(1986, 1M)
(1984, 1M)
62. The total energy of one mole of an ideal monoatomic gas at
27°C is …… cal.
(1984, 1M)
True / False
63. A mixture of ideal gases is cooled up to liquid helium
temperature (4.22 K) to form an ideal solution.
(1996, 1M)

n2a
64. In the van der Waals’ equation,  p + 2  (V − nb ) = nRT
V 

the constant ‘a’ reflects the actual volume of the gas
molecules.
(1993, 1M)
65. A gas in a closed container will exert much higher pressure
due to gravity at the bottom than at the top.
66. Kinetic energy of a molecule is zero at 0°C.
(1985, 1/2 M)
(1985, 1/2 M)
Integer Answer Type Questions
67. The diffusion coefficient of an ideal gas is proportional to its
mean free path and mean speed. The absolute temperature of
72 States of Matter
an ideal gas is increased 4 times and its pressure is increased 2
times. As a result, the diffusion coefficient of this gas
increases x times. The value of x is ...
(2016 Adv.)
68. A closed vessel with rigid walls contains 1 mole of 238
92 Uand 1
mole of air at 298 K. Considering complete decay of 238
92 U to
206
the
ratio
of
the
final
pressure
to
the
initial
pressure
of
Pb,
82
the system at 298 K is
(2015 Adv.)
mol −1 and
the value of Boltzmann constant is 1.380 × 10−23 JK −1 , then
the number of significant digits in the calculated value of the
universal gas constant is
(2014 Adv.)
69. If the value of Avogadro number is 6.023 × 10
23
70. To an evacuated vessel with movable piston under external
pressure of 1 atm, 0.1 mole of He and 1.0 mole of an unknown
compound (vapour pressure 0.68 atm at 0° C ) are introduced.
Considering the ideal gas behaviour, the total volume (in
litre) of the gases at 0° C is close to
(2011)
Subjective Questions
71. At 400 K, the root mean square (rms) speed of a gas X
(molecular weight = 40) is equal to the most probable speed of
gas Y at 60 K. The molecular weight of the gas Y is (2009)
72. The average velocity of gas molecules is 400 m s − 1 , find the
rms velocity of the gas.
(2003, 2M)
73. The density of the vapour of a substance at 1 atm pressure and
500 K is 0.36 kg m–3 . The vapour effuses through a small hole
at a rate of 1.33 times faster than oxygen under the same
condition.
(i) Determine, (a) molecular weight (b) molar volume
(c) compression factor (Z) of the vapour and (d) which
forces among the gas molecules are dominating, the
attractive or the repulsive?
(ii) If the vapour behaves ideally at 1000 K, determine the
average translational kinetic energy of a molecule.
(2002, 5M)
74. The compression factor (compressibility factor) for one mole
of a van der Waals’ gas at 0° C and 100 atm pressure is found to
be 0.5. Assuming that the volume of a gas molecule is negligible,
calculate the van der Waals’ constant ‘a’.
(2001, 5M)
75. Calculate the pressure exerted by one mole of CO2 gas at
273 K if the van der Waals’ constant a = 3.592 dm 6 atm
mol −2 . Assume that the volume occupied by CO2 molecules is
negligible.
(2000)
76.
(i) One mole of nitrogen gas at 0.8 atm takes 38 s to diffuse
through a pin-hole, whereas one mole of an unknown
compound of xenon with fluorine at 1.6 atm takes 57 s to
diffuse through the same hole. Calculate the molecular
formula of the compound.
(ii) The pressure exerted by 12 g of an ideal gas at
temperature t ° C in a vessel of volume V litre is one atm.
When the temperature is increased by 10°C at the same
volume, the pressure increases by 10%. Calculate the
temperature t and volume V.
(Molecular weight of the gas = 120)
(1999, 5M)
77. Using van der Waals’ equation, calculate the constant a when
two moles of a gas confined in a four litre flask exert a
pressure of 11.0 atm at a temperature of 300 K. The value of b
is 0.05 L mol –1 .
(1998, 4M)
78. An evacuated glass vessel weighs 50.0 g when empty 148.0 g
when filled with a liquid of density 0.98 g mL–1 and 50.5 g
when filled with an ideal gas at 760 mm Hg at 300 K.
Determine the molar mass of the gas.
(1998, 3M)
79. A mixture of ideal gases is cooled up to liquid helium
temperature (4.22 K) to form an ideal solution. Is this
statement true or false ? Justify your answer in not more than
two lines.
(1996, 1M)
80. The composition of the equilibrium mixture (Cl 2 s 2Cl)
which is attained at 1200° C, is determined by measuring the
rate of effusion through a pin-hole. It is observed that at
1.80 mm Hg pressure, the mixture effuses 1.16 times as fast
as krypton effuses under the same conditions. Calculate the
fraction of chlorine molecules dissociated into atoms (atomic
weight of Kr = 84)
(1995, 4M)
81. A mixture of ethane (C2 H6 ) and ethene (C2 H4 ) occupies 40 L
at 1.00 atm and at 400 K. The mixture reacts completely with
130 g of O2 to produce CO2 and H2 O. Assuming ideal gas
behaviour, calculate the mole fractions of C2 H4 and C2 H6 in
the mixture.
(1995, 4M)
82. An LPG (liquefied petroleum gas) cylinder weighs 14.8 kg
when empty. When full it weighs 29.0 kg and shows a pressure
of 2.5 atm. In the course of use at 27° C, the weight of the full
cylinder reduces to 23.2 kg. Find out the volume of the gas in
cubic metres used up at the normal usage conditions, and the
final pressure inside the cylinder. Assume LPG to the n-butane
with normal boiling point of 0° C.
(1994, 3M)
83. A 4 : 1 molar mixture of He and CH4 is contained in a vessel
at 20 bar pressure. Due to a hole in the vessel, the gas mixture
leaks out. What is the composition of the mixture effusing out
initially?
(1994, 2M)
84. A gas bulb of 1 L capacity contains 2.0 × 1021 molecules of
nitrogen exerting a pressure of 7.57 × 103 Nm–2 . Calculate
the root mean square (rms) speed and the temperature of the
gas molecules. If the ratio of the most probable speed to root
mean square speed is 0.82, calculate the most probable speed
for these molecules at this temperature.
(1993, 4M)
85. At room temperature, the following reaction proceed nearly
to completion. 2NO + O2 → 2NO2 → N2 O4
The dimer, N2 O4 , solidifies at 262 K. A 250 mL flask and a
100 mL flask are separated by a stopcock. At 300 K, the nitric
oxide in the larger flask exerts a pressure of 1.053 atm and the
smaller one contains oxygen at 0.789 atm.
The gases are mixed by opening the stopcock and after the
end of the reaction the flasks are cooled to 220 K. Neglecting
the vapour pressure of the dimer, find out the pressure and
composition of the gas remaining at 220 K. (Assume the
gases to behave ideally).
(1992, 4M)
States of Matter 73
86. At 27° C, hydrogen is leaked through a tiny hole into a vessel
for 20 min. Another unknown gas at the same temperature
and pressure as that of hydrogen is leaked through same hole
for 20 min. After the effusion of the gases the mixture exerts a
pressure of 6 atm. The hydrogen content of the mixture is 0.7
mole. If the volume of the container is 3 L. What is the
molecular weight of the unknown gas?
(1992, 3M)
87. Calculate the volume occupied by 5.0 g of acetylene gas at
50° C and 740 mm pressure.
(1991, 2M)
88. The average velocity at T1 K and the most probable at T2 K of
CO2 gas is 9.0 × 104 cm s –1 . Calculate the value of T1 and T2
(1990, 4M)
89. A spherical balloon of 21 cm diameter is to be filled up with
hydrogen at NTP from a cylinder containing the gas at 20 atm
at 27° C. If the cylinder can hold 2.82 L of water, calculate the
number of balloons that can be filled up.
(1987, 5M)
90. Calculate the root mean square velocity of ozone kept in a closed
vessel at 20° C and 82 cm mercury pressure.
(1985, 2M)
91. Give reasons for the following in one or two sentences.
(i) Equal volumes of gases contain equal number of moles.
(1984, 1M)
(ii) A bottle of liquor ammonia should be cooled before
opening the stopper.
(1983, 1M)
92. Oxygen is present in one litre flask at a pressure of 7.6 × 10−10
mm Hg. Calculate the number of oxygen molecules in the
flask at 0° C.
(1983, 2M)
93. When 2 g of a gas A is introduced into an evacuated flask kept
at 25° C, the pressure is found to be one atmosphere. If 3 g of
another gas B is then added to the same flask, the total
pressure becomes 1.5 atm. Assuming ideal gas behaviour,
calculate the ratio of the molecular weights M A : M B .
(1983, 2M)
94. At room temperature, ammonia gas at 1 atm pressure and
hydrogen chloride gas at p atm pressure are allowed to effuse
through identical pin holes from opposite ends of a glass tube
of one metre length and of uniform cross-section.
Ammonium chloride is first formed at a distance of 60 cm
from the end through which HCl gas is sent in. What is the
value of p ?
(1982, 4M)
95. Calculate the average kinetic energy, in joule per molecule in
8.0 g of methane at 27°C.
(1982, 2M)
96. The pressure in a bulb dropped from 2000 to 1500 mm of
mercury in 47 min when the contained oxygen leaked through
a small hole. The bulb was then evacuated. A mixture of
oxygen and another gas of molecular weight 79 in the molar
ratio of 1 : 1 at a total pressure of 4000 mm of mercury was
introduced. Find the molar ratio of the two gases remaining in
the bulb after a period of 74 min.
(1981, 3M)
97. A hydrocarbon contains 10.5 g of carbon per gram of
hydrogen. 1 L of the vapour of the hydrocarbon at 127°C and
1 atm pressure weighs 2.8 g. Find the molecular formula of
the hydrocarbon.
(1980, 3M)
98. 3.7 g of a gas at 25°C occupied the same volume as 0.184 g of
hydrogen at 17°C and at the same pressure. What is the
molecular weight of the gas ?
(1979, 2M)
99. 4.215 g of a metallic carbonate was heated in a hard glass
tube, the CO2 evolved was found to measure 1336 mL at
27°C and 700 mm of Hg pressure. What is the equivalent
weight of the metal ?
(1979, 3M)
100. Calculate the density of NH3 at 30°C and 5 atm pressure.
(1978, 2M)
Topic 2 Liquid State
Objective Questions I (Only one correct option)
2. At 100°C and 1 atm if the density of the liquid water is
1. The qualitative sketches I, II and III given below show the
Concentration
Surface tension
I
Surface tension
Surface tension
variation of surface tension with molar concentration of three
and
different aqueous solutions of KCl, CH3 OH
−
+
CH3 (CH2 )11 OSO3 Na at room temperature.
(2016 Adv.)
II
Concentration
The correct assignment of the sketches is
I
II
(a) KCl
(b)
CH3 (CH2 )11 OSO3−
CH3OH
III
Concentration
III
KCl
Na CH3OH
(c) KCl
CH3 (CH2 )11 OSO−3 Na +
(d) CH3OH
KCl
(a) 6 cm 3
(b) 60 cm 3
(c) 0.6 cm 3
(d) 0.06 cm 3
3. The critical temperature of water is higher than that of O2
because the H2O molecule has
CH3 (CH2 )11 OSO−3 Na +
+
1.0 g cm–3 and that of water vapour is 0.0006 g cm−3 , then the
volume occupied by water molecules in 1 L of steam at this
temperature is
(2000, 1M)
CH3OH
CH3 (CH2 )11 OSO3− Na +
(1997)
(a) fewer electrons than O2
(b) two covalent bonds
(c) V-shape
(d) dipole moment
4. A liquid is in equilibrium with its vapour at it’s boiling point.
On the average, the molecules in the two phases have equal
(a) inter-molecular forces
(b) potential energy
(c) kinetic energy
(d) total energy
Answers
Topic 1
1.
5.
9.
13.
17.
21.
25.
29.
33.
37.
41.
45.
49.
53.
(b)
(a)
(b)
(b)
(b)
(c)
(c)
(d)
(b)
(a)
(b)
(a, b, d)
(a,b)
(a)
2.
6.
10.
14.
18.
22.
26.
30.
34.
38.
42.
46.
50.
54.
3.
7.
11.
15.
19.
23.
27.
31.
35.
39.
43.
47.
51.
55.
(c)
(a)
(a)
(c)
(a)
(d)
(c)
(a)
(c)
(d)
(a)
(c)
(750)
(d)
(c)
(c)
(b)
(c)
(a)
(b)
(b)
(d)
(b)
(b)
(b)
(a)
(2.22)
(c)
4.
8.
12.
16.
20.
24.
28.
32.
36.
40.
44.
48.
52.
(b)
(b)
(b)
(b)
(a)
(c)
(c)
(a)
(c)
(b)
(a)
(a,c)
(d)
56. (A → p,s; B → r; C→ p, q; D → r)
57. (less)
58. (1 : 16)
60. (inversely, time)
63.
67.
71.
77.
82.
86.
90.
(F)
64.
(4)
68.
(4)
72.
(6.46)
78.
(2.46 m 3)
83.
(1020 g mol −1) 87.
(390.2 ms −1) 92.
59. (0.25)
61. (R)
(F)
65. (F)
(9)
69. (4)
435 ms −1 74. (1.25)
(123 g mol −1)
(8:1)
84. (407 ms −1)
(5.23 L)
89. (10)
(2.7 × 1010 g mol −1)
62. (900)
66.
70.
75.
80.
85.
(F)
(7 L)
(0.99 atm)
(0.14)
(0.221 atm)
93. (1 : 3)
94. (2.20 atm) 95. (6.2 × 10 −21 J/molecule)
98. (41.32 g)
99. (12.15)
100. (3.42 gL −1)
Topic 2
1. (d)
2. (c)
3. (d)
4. (c)
Hints & Solutions
Topic 1 Gaseous State
1. Fraction of molecules vs velocity graph is Maxwell distribution
curve.
This curve is slightly unsymmetrical as shown below.
Most probable speed (vp)
Fraction of molecules
Average speed (vavg)
Root-mean-square speed (vrms)
vp vavg vrms
Speed v
The ratio of most probable, the average and the root mean square
speeds for this graph is 1 : 1.128 : 1.224
But the graph in question is completly symmetrical. Therefore,
the most probable and the average speed will be same here but
root mean square speed will be greater than average speed. So,
the correct ratio is 1 : 1: 1224
. .
2. Statements (A), (C) and (D) are true whereas (B) is false.
(A) For 1 mole, U = internal energy and H = enthalpy.
Heat capacity at constant volume.
dU
dt
(n = 1 mol)
dU = nCV dT
(QCV = constant)
dU = 1 × CV × dT
(Function of temperature)
U = f (T )
U depends on temperature.
nCV =
Heat capacity at constant volume,
dH
nCp =
dT
[n = 1mol]
dH = nC p dT
dH = 1 × C p × dT
In mathematically term,
[QC p = constant]
H = f (T )
Hence, H depends on function of temperature
(b) Compressibility factor (Z ) describe the deviation of real gas
from ideal gas behaviour.
pV
…(i)
Z=
RT
For ideal gas,
[n = 1mol]
pV = nRT
…(ii)
pV = RT
Put in the value of pV from Eq. (ii) to Eq. (i)
RT
Z=
=1
RT
Compressibility factor is 1 for ideal gas then option (b) is
incorrect.
(c) For ideal gas,
(n = 1mol)
C p − CV = nR
C p − CV = R
It is correct statement.
(d)
∆U = CV dT
dU
nCV =
dT
(n = 1mol)
dU = nCV dT
dU = CV dT
It is also correct.
States of Matter 75
3.
Key Idea From kinetic gas equation,
2RT
Most probable velocity (vmp ) =
M
where, R = gas constant, T = temperature,
M = molecular mass
vmp =
Z=
M
T ( K)
T/M
H2
2
300
300 / 2 = 150 …III (Highest)
N2
28
300
300 / 28 = 10.71 …I (Lowest)
O2
32
400
400 / 32 = 12.5 … II
So,
I. corresponds to vmp of N2 (300 K)
II. corresponds to vmp of O2 (400 K)
III. corresponds to vmp of H2 (300 K)
4. For 1 mole of a real gas, the van der Waals’ equation is
a

 p + 2  (V − b) = RT

V 
The constant ‘a’ measures the intermolecular force of attraction
of gas molecules and the constant ‘b’ measures the volume
correction by gas molecules after a perfectly inelastic binary
collision of gas molecules.
For gas A and gas C given value of ‘b’ is
0.05196 dm3 mol −1. Here,
a ∝ intermolecular force of attraction
∝ compressibility ∝ real nature
1
∝
volume occupied
Value of a/(kPa dm6 mol −1) for gas A (642.32) > gas C (431.91)
So, gas C will occupy more volume than gas A. Similarly, for a
given value of a say 155.21 kPa dm6 mol −1 for gas B and gas D
1
∝ intermolecular force of attraction
b
∝ compressibility ∝ real nature
1
∝
volume accupied
b/(dm 3 mol −1) for gas B (0.04136) < Gas D (0.4382)
So, gas B will be more compressible than gas D.
5. Noble gases such as Ne, Ar, Xe and Kr found to deviate from
ideal gas behaviour.
Xe gas will exhibit steepest increase in plot of Z vs p.
Equation of state is given as:
RT
p=
⇒ p(V − b) = RT
(V − b)
⇒ pV = RT + pb
pV
RT
pb
⇒ y = c + mx
RT
The plot of z vs p is found to be
so,
Z =1+
2RT
T
, i.e. vmp ∝
M
M
Gas
pV − pb = RT
pV
pb
=1+
RT
RT
As,
slope =
b
RT
Z
p
The gas with high value of b will be steepest as slope is directly
proportional to b. b is the van der Waals’ constant and is equal to
four times the actual volume of the gas molecules. Xe gas
possess the largest atomic volume among the given noble gases
(Ne, Kr, Ar). Hence, it gives the steepest increase in the plot of Z
(compression factor) vsp.
6. Critical temperature is the temperature of a gas above which it
cannot be liquefied what ever high the pressure may be. The
kinetic energy of gas molecules above this temperature is
sufficient enough to overcome the attractive forces. It is
represented as Tc.
8a
Tc =
27Rb
8 × 13
.
For Ar,
Tc =
= 0.0144
27 × 8.314 × 3.2
8 × 0.2
For Ne,
Tc =
= 0.0041
27 × 8.314 × 17
.
8 × 51
.
For Kr,
Tc =
= 018
.
27 × 8.314 × 1.0
8 × 4.1
For Xe,
Tc =
= 0.02
27 × 8.314 × 5.0
The value of Tc is highest for Kr (Krypton).
7. In isothermal expansion, pVm = K (constant)
This relation is plotted in graph ‘C’
K
Likewise,
p=
Vm
This relation is plotted in graph “ A”.
Thus, graph B and D are incorrect.
For them the correct graphs are:
for graph B and
p
Vm
for graph D
U
Vm
8. Given, temperature (T1 ) = 27° C = 273 + 27 = 300 K
Volume of vessel = constant
Pressure in vessel = constant
2
3
Volume of air reduced by so the remaining volume of air is .
5
5
76 States of Matter
Let at T1 the volume of air inside the vessel is n so at T2 the
3
volume of air will be n.
5
Now, as p and V are constant, so
3
...(i)
n ⋅T1 = n T2
5
Putting the value of T1 in equation (i) we get,
3
n × 300 = n × T2
5
5
or
T2 = 300 × = 500 K
3
After mixing, number of moles in left chamber =
Number of moles in right chamber =
Total number of moles =
pV
[for real gases]
nRT
On substituting in equation (i), we get
pAVA
3p V
= B B
nARTA nB RTB
Compressibility factor (Z) =
…(ii)
14.
= 1.02 × 10 mol
Weight of NaHCO3
Molecular mass of NaHCO3
wNaHCO3 = 102
. × 10−5 × 84 × 103 mg
= 0.856 mg
0.856
× 100 = 8.56%
⇒ NaHCO 3 % =
10
pV = ΣnRT
Given: p = 200 Pa, V = 10m3 , T = 1000 K
...(i)
2RT
M
8RT
πM
C = Root square speed corrected as root means square speed, i.e.
rms =
3RT
and as we know C* < C < C
M
*
C : C : C =1 :
11. From the ideal gas equation,
4
3
:
= 1 : 1.128 : 1.225
p
2
NOTE
As no option correspond to root square speed, it is understood as
misprint. It should be root mean square speed.
15. The van der Waals’ equation of state is

n2a
 p + 2  (V − nb) = nRT
V 

nA = 0.5 moles, nB = x moles
On substituting the given values in equation (i), we get
200 × 10 = (nA + nB ) × R × 1000
200 × 10
0.5 + x =
R × 1000
1
2 2 1 4−R
+x= = − =
2R
R R 2
2
For one mole and when b = 0, the above equation condenses to
a

 p + 2  V = RT

V 
⇒
12. Initially,
pV = RT −
a
V
piV
RT1
…(i)
1
whose slope is ‘
V
− a’. Equating with slope of the straight line given in the graph.
20.1 − 21.6
−a=
= − 1.5 ⇒ a = 1.5
3− 2
Eq. (i) is a straight equation between pV and
pV
Number of moles of gases in each container = i
RT1
Total number of moles of gases in both containers = 2
C * = Most probable speed =
C = Average speed =
−5
∴
PLAN To solve this problem, the stepwise approach required, i.e.
(i) Write the van der Waals’ equation, then apply the condition that
at low pressure, volume become high,
i.e.
V − b~
−V
2 mol
⇒ In the reaction, number of mole of CO2 produced.
pV
1 bar × 0.25 × 10−3 L
n=
=
RT 0.082 L atm K−1mol −1 × 298.15 K
Number of mole of NaHCO3 =
pf V  1
1
 + 
R  T1 T2 
a

At low pressure,  p + 2  V = RT

V 
a
a
pV +
= RT or pV = RT −
⇒
V
V
pV
a
Divide both side by RT,
=1−
RT
RTV
10. 2NaHCO3 + H2C2O4 → 2CO2 + Na 2C4O4 + H2O
1 mol
RT2
=
According to van der Waals’ equation,
a

 p + 2  (V − b) = RT

V 
2 pA = 3 pB
2 mol
RT2
pf V
(ii) Now calculate the value of compressibility factor (Z ).
[ Z = pV / RT ]
Also, it is given that
VA = 2VB , nA = nB and TA = TB
∴ Eq. (ii) becomes
pA × 2VB 3 pBVB
=
nB RTB
nB RTB
⇒
RT1
+
pf V
As total number of moles remains constant.
pf V
pf V
 T2 
2 piV
Hence,
=
+
⇒ pf = 2 pi 

 T1 + T2 
RT1
RT1
RT2
13.
9. Given, ZA = 3ZB
pf V
pf V
RT1
States of Matter 77
16. In the van der Waals’ equation
V
=1
Videal
For ideal gas V = Videal
27. Compressibility factor (Z ) =

n a
 p + 2  (V − nb) = nRT
V 

2
n2a
corrects for
V2
intermolecular force while b corrects for molecular volume.
The additional factor in pressure, i.e.
Q
28. Expression of rms is, urms =
urms (H2 at 50 K)
=
urms (O2 at 800 K)
⇒
17. Option (b) is incorrect statement because at high pressure slope
of the line will change from negative to positive.
18.
r(He)
r(CH4 )
=
16
=2 :1
4
=
3
2
19. Kinetic energy (E ) = kT
RMS speed (u) =
x
30
x
Mole of hydrogen =
2
pV
pV
, for positive deviation,
> 1.
Q Z=
nRT
nRT
21. Option (b) and (d) are ruled out on the basis that at the initial
point of 273 K, 1 atm, for 1.0 mole volume must be 22.4 L, and it
should increase with rise in temperature.
Option (a) is ruled out on the basis that initial and final points are
not connected by the ideal gas equation V ∝ T , i.e. V /T do not
have the same value at the two points.
In option (c), at the initial point, the volume is 22.4 L as required
by ideal gas equation and (V /T ) have the same value at both
initial and final points.
22. Root mean square velocity (urms ) =
3RT
M
V
< 1(given)
Vid
V < 22.4 L
Vid (1 mol ) = 22.4 L at STP
23. Compressibility factor (Z ) =
24. Root mean square speed urms =
⇒
Mole fraction of hydrogen =
3RT
M
14T (H2 )
urms (H2 )
T (H2 )
28
⇒ 7=
= 7=
×
urms (N2 )
2
T (N2 )
T (N2 )
T (N2 ) = 2T (H2 ) i.e. T (H2 ) < T (N2 )
25. At high temperature and low pressure, the gas volume is
infinitely large and both intermolecular force as well as
molecular volume can be ignored. Under this condition
postulates of kinetic theory applies appropriately and gas
approaches ideal behaviour.
26. Rate of effusion ∝ pi ; pi = Partial pressure of ith component
∝
1
M
x
2
x
x
+
2 30
=
15
16
Partial pressure of H2
= Mole fraction of hydrogen
Total pressure
= 15 : 16
8RT
30. Average speed =
πM
⇒
i.e. at constant volume, for a fixed mass, increasing temperature
increases average speeds and molecules collide more frequently
to the wall of container leading to increase in gas pressure.
31. The mean translational kinetic energy (∈) of an ideal gas is
p ⋅ M = dRT
Substituting for RT / M in urms expression gives,
3p
1
urms =
⇒ urms ∝
d
d
⇒
50 32
×
=1
2 800
Mole of ethane =
20. Positive deviation corresponds to Z > 1
⇒
Q
3R × 50
2
3R × 800
32
29. Let x g of each gas is mixed.
3kT
2E
⇒ u=
m
m
Also,
3RT
M
∈=
32.
3
k BT ; T = Absolute temperature, i.e. ∈ ∝ T
2
rCH 4
rX
=2=
MX
16
⇒ M X = 64
33. The ideal gas equation, pV = nRT =
⇒
w
RT
M
 w
pM =   RT = dRT
V 
(d = density)
pM
RT
i.e. density will be greater at low temperature and high pressure.
⇒
d=
34. The ease of liquefication of a gas depends on their intermolecular
force of attraction which in turn is measured in terms of van der
Waals’ constant a. Hence, higher the value of a, greater the
intermolecular force of attraction, easier the liquefication.
In the present case, NH3 has highest a, can most easily be
liquefied.
35. HCl will diffuse at slower rate than ammonia because rate of
effusion ∝
1
.
M
Therefore, ammonia will travel more distance than HCl in the
same time interval and the two gas will first meet nearer to HCl
end.
78 States of Matter
36. In van der Waals’ equation of state
44. The two types of speeds are defined as;
a

 p + 2  (V − b) = RT

V 
(For 1 mole)
Root mean square speed (urms ) =
3RT
M
The first factor ( p + a/V 2 ) correct for intermolecular force while
the second term (V − b) correct for molecular volume.
Average speed (uav ) =
8RT
πM
37. Expression for average velocity is uav =
8RT
πM
For the same gas but at different temperature
uavg (T1 )
uavg (T2 )
⇒
=
T1
300
1
=
=
T2
1200 2
uav (927° C) = 2 × uav (27° C) = 0.6 ms−1
1
38. Rate of effusion ∝
,
M
x
2
x
Moles of CH4 =
16
Moles of H2 =
⇒
x
2
8
Mole fraction of H2 =
=
x
x
9
+
2 16
8
Partial pressure of H2
= Mole fraction of H2 =
9
Total pressure
40. According to postulates of kinetic theory, there is no
intermolecular attractions or repulsions between the molecules
of ideal gases.
3
41. According to kinetic theory, average kinetic energy (E ) = k BT
2
where, k B is Boltzmann’s constant. Since, it is independent of
molar mass, it will be same for He and H2 at a given temperature.
42. If x g of both oxygen and methane are mixed then :
x
32
x
Mole of methane =
16
x
1
32
Mole fraction of oxygen =
⇒
=
x
x
3
+
32 16
According to law of partial pressure
Partial pressure of oxygen ( pO 2 ) = Mole fraction × Total pressure
pO 2 1
=
⇒
3
p
Mole of oxygen =
43. It is the Boyle temperature TB . At Boyle temperature, the first
virial coefficient ( B ) vanishes and real gas approaches ideal
behaviour.
TB =
= 3:
8
= 3 : 2.54 = 1.085 : 1
π
45. The explanation of given statements are as follows:
(a) Urms is inversely proportional to the square root of its
molecular mass.
39. Let x grams of each hydrogen and methane are mixed,
⇒
For the same gas, at a given temperature, M and T are same,
therefore
urms
3RT
8RT
=
:
πM
uav
M
a
Rb
Here, a and b are van der Waals’ constants.
3RT
M
Hence, option (a) is correct.
(b) When temperature is increased four times then Urms become
doubled.
Urms =
Urms =
3R
× 4T
M
Urms = 2 ×
3RT
M
Hence, option (b) is correct.
(c) and (d) Eav is directly proportional to temperature but
does not depends on its molecular mass at a given
3
temperature as Eav = KT . If temperature raised four times
2
than Eav becomes four time multiple.
Thus, option (c) is incorrect and option (d) is correct.
46. Equation of state p(V − b) = RT indicates absence of
intermolecular attraction or repulsion, hence interatomic
potential remains constant on increasing ‘π’ in the beginning. As
the molecules come very close, their electronic and nuclear
repulsion increases abruptly.
47. (a) According to a postulate of kinetic theory of gases, collision
between the molecules as well as with the wall of container is
perfectly elastic in nature.
(b) If a gas molecule of mass m moving with speed u collide to
the wall of container, the change in momentum is
∆p = – 2mu. Therefore, heavier molecule will transfer more
momentum to the wall as there will be greater change in
momentum of the colliding gas molecule. However, this is
not postulated in kinetic theory.
(c) According to Maxwell-Boltzmann distribution of molecular
speed, very few molecules have either very high or very low
speeds. Most of the molecules moves in a specific,
intermediate speed range.
(d) According to kinetic theory of gases, a gas molecule moves
in straight line unless it collide with another molecule or to
the wall of container and change in momentum is observed
only after collision.
States of Matter 79
48. Option (a) is correct because in the limit of large volume, both
or
intermolecular force and molecular volume becomes negligible
in comparison to volume of gas.
Option (d) is wrong statement because Z can be either less or
greater than unity, hence real pressure can be less or greater than
ideal pressure.
51. Given p1 = 5 bar, V1 = 1 m 3, T1 = 400 K
So,
n1 =
5
400 R
Similarly, p2 = 1 bar, V2 = 3 m 3, T2 = 300 K, n2 =
11 20
= or
9 9
52. Assertion is incorrect because besides amount, pressure also
depends on volume. However, reason is correct because both
frequency of collisions and impact are directly proportional to
root mean square speed which is proportional to square root of
absolute temperature
53. a is the measure of intermolecular force of attraction. Greater the
intermolecular force of attraction (H-bond in the present case)
higher the value of a.
54. X is a lighter gas than Y, hence X has greater molecular speed. Due
to greater molecular speed of X, it will have smaller mean free path
and greater collision frequency with the incrt gas molecules. As a
result X will take more time to travel a given distance along a
straight line. Hence X and Y will meet at a distance smaller than one
calculated from Graham’s law.
50. Initial pressure ( p1 ) = 48 × 10−3 bar
Final pressure ( p2 ) = …… × 10−6 bar
4
Initial volume (V1 ) = π (3)3
3
4
Final volume (V2 ) = π (12)3
3
According to Boyle’s law p1V1 = p2V2
pV
p2 = 1 1
V2
4
48 × 10−3 × π (3)3
48 × 10−3 × (3)3
3
p2 =
=
4
(12)3
π (12)3
3
48 × 10−3 × 27
=
= 0.0277 × 27 × 10−3 = 750 × 10−6 bar
1728
Hence, the correct answer is 750.
11
9
2.22.
49. Pressure is inversely proportional to volume at constant
temperature, hence (a) is correct.
Average kinetic energy of a gas is directly proportional to
absolute temperature, hence (b) is correct.
Expansion at constant temperature cannot change the number of
molecules, hence (d) is incorrect.
4 + 4 x = 15 − 5x or x =
Hence, new volume of A i.e., (1+ x ) will comes as 1 +
Option (b) is wrong statement because in the limit of large
pressure Z > 1.
Option (c) is correct statement. For a van der Waals’ gas,
van der Waals’ constants a and b are characteristic of a gas,
independent of temperature.
4 (1+ x ) = 15 − 5x or
Hence, (d) is the correct choice.
55.
PLAN This problem can be solved by using the concept of Graham’s law
of diffusion according to which rate of diffusion of non-reactive
gases under similar conditions of temperature and pressure are
inversely proportional to square root of their density.
1
Rate of diffusion ∝
molar weight of gas
Let distance covered by X is d, then distance covered by Y is
24 – d.
If rX and rY are the rate of diffusion of gases X and Y,
40
rX
d
=
=
=2
10
rY 24 − d
[Q Rate of diffusion ∝ distance travelled]
d = 48 − 2d
⇒
3d = 48 ⇒ d = 16 cm
Hence, (c) is the correct choice.
56. A. At p = 200 atm, very high pressure, Z > 1. Also, at such a high
(from pV = nRT )
3
300 R
Let at equilibrium the new volume of A will be (1+ x )
So, the new volume of B will be (3− x )
Now, from the ideal gas equation.
p1V1
pV
= 2 2
n1RT1 n2RT2
and at equilibrium (due to conduction of heat)
p1 p2
=
T1 T2
V1 V2
So,
= or V1n2 = V2n1
n1 n2
After putting the values
3
(3 − x ) 5
5
or (1+ x ) =
(1+ x ) ×
= (3 − x ) ×
300 R
4
400 R
 n2a
pressure, the pressure correction factor  2  can be ignored
V 
in comparison to p.
B. At p ~ 0, gas will behave like an ideal gas, pV = nRT .
C. CO2 (p = 1atm, T = 273 K), Z < 1.
D. At very large molar volume, real gas behaves like an
ideal gas.
3
2
57. Less; E = RT
3
nRT . At same temperature, KE (total) ∝ n.
2
1
59. 0.25 RT because at NTP, 5.6 L = mole.
4
60. Inversely, time.
58. 1 : 16, KE =
61. For an ideal gas, Cp − CV = R
3
2
3
2
62. At 27°C, E = RT = × 2 × 300 = 900 cal
80 States of Matter
63. An ideal gas cannot be liquefied because there exist no
urms = umps
71. Given,
intermolecular attraction between the molecules of ideal gas.
3RT
=
M (X )
⇒
64. a is the measure of intermolecular force.
65. In a close container, gas exert uniform pressure everywhere in
the container.
3
66. KE = RT where, T is absolute temperature (in Kelvin).
2
3R × 400 2R × 60
=
40
M (Y )
⇒
72.
67. (DC) Diffusion coefficient ∝ λ (mean free path) ∝ U mean
⇒
Thus (DC) ∝ λ Umean
T
RT
⇒ λ∝
p
2 N0 σp
λ=
But,
rgas
rO 2
8RT
πM
U mean =
and
73.
DC ∝
 T2 
 
 T1 
3/ 2
 p   4T 
=  1   1
 2 p1   T1 
92 U
238
→
82Pb
206
+ 8 2He4 (g ) + 6− 1β 0
n(gas)[Initial] = 1 (air)
n(gas)[Final] = 8 (He) + 1(air) = 9
⇒ At constant temperature and volume;
p ∝ n.
pf nf 9
So,
=
= =9
pi
ni 1
69.
urms =
3π
uav =
8
= 1.33 =
32
Mgas
(ii) Ek =
a

 p + 2  V = RT

V 
pV
a
+
=1
RT VRT
a
Z+
=1
 ZRT 

 RT
 p 
⇒
⇒
a=
∴
⇒ pV +
a
= RT
V
 pV

Q
=Z
 RT

⇒
Z+
ap
=1
ZR 2T 2
ZR 2T 2 (1 − Z ) 0.5 (0.082 × 273)2 (1− 0.5)
=
p
100
a = 1.25 atm L2 mol −2
k = Boltzmann constant
75. In case of negligible molecular volume, b = 0 and
N A = Avogadro’s number
−23
3
3
k BT = × 1.38 × 10−23 × 1000 J = 2.07 × 10−20 J
2
2
74. In case of negligible molecular volume, b = 0. For 1 mole of gas
Universal gas constant, R = kN A
and
3 × 3.14
× 400 = 434 ms−1
8
(d) Q Z > 1, repulsive force is dominating.
⇒
PLAN This problem can be solved by using the concept involved in
calculation of significant figure.
where,
8
3π
18
= 50 L mol −1
0.36
pV
1 × 50
(c) Z =
=
= 1.22
RT 0.082 × 500
3/ 2
 1
=   (8) = 4
 2
68.
8RT
3RT
=
:
M
πM
(b) Vm =
(T )3/ 2
p
p
(DC)2
(x ) =  1 
 p2 
(DC)1
uav
=
urms
⇒ M (Y ) = 4
(i) (a) Mgas = 18 g mol −1
U mean ∝ T
∴
2RT
M (Y )
23
R = 1.380 × 10 × 6.023 × 10 J/Kmol
~ 8.312
= 8.31174 =
Since, k and N A both have four significant figures, so the value of
R is also rounded off upto 4 significant figures.
[When number is rounded off, the number of significant figure is
reduced, the last digit is increased by 1 if following digits ≥ 5 and
is left as such if following digits is ≤ 4.]
Hence, correct integer is (4).
70. Since, the external pressure is 1.0 atm, the gas pressure is also
1.0 atm as piston is movable. Out of this 1.0 atm partial pressure
due to unknown compound is 0.68 atm.
Therefore, partial pressure of He
= 1.00 – 0.68 = 0.32 atm.
n(He)RT 0.1 × 0.082 × 273
Volume =
=
=7L
⇒
p(He)
0.32
⇒ Volume of container = Volume of He.
van der Waals’ equation reduces to

n2a
 p + 2  V = nRT
V 

RT
a
− 2
(n =1 mole)
V
V
0.082 × 273
3.592
= 0.99 atm
=
−
22.4
(22.4)2
⇒
p=
76. (i) For the same amount of gas being effused
r1 t2 p1
= =
r2 t1 p2
⇒
M2
57 0.8
=
⇒
M1
38 1.6
M2
28
M 2 = 252 g mol −1
Also, one molecule of unknown xenon-fluoride contain only
one Xe atom [M (Xe) = 131], formula of the unknown gas can
be considered to be XeFn.
⇒ 131 + 19n = 252; n = 6.3, hence the unknown gas is XeF6.
States of Matter 81
(ii) For a fixed amount and volume, p ∝ T
1
T
where, T = Kelvin temperature
⇒
=
1.1 T + 10
⇒
⇒
82. Weight of butane gas in filled cylinder = 29 − 14.8 kg = 14.2 kg
⇒ During the course of use, weight of cylinder reduces to 23.2 kg
⇒ Weight of butane gas remaining now = 23.2 − 14.8 = 8.4 kg
Also, during use, V (cylinder) and T remains same.
p1 n1
Therefore,
=
p2 n2

n 
n
w 
 8.4 
p2 =  2  p1 = 
⇒
 × 2.5 Here, 2 = 2 
 14.2
 n1 
n
w1 

1
T = 100 K = t + 273
t = − 173° C
nRT  12  0.082 × 100
Volume =
=
= 0.82 L
 ×
 120
p
1
77. The van der Waals’ equation is
= 1.48 atm
Also, pressure of gas outside the cylinder is 1.0 atm.
⇒
pV = nRT
nRT (14.2 − 8.4 ) × 103 0.082 × 30
⇒
V =
=
×
L
p
58
1

n2a
 p + 2  (V − nb) = nRT
V 

⇒
a=
 (4 )2  2 × 0.082 × 300

V 2  nRT
− p =
− 11
2 
2 
4
−
2
(
0.05
)
n V − nb
2
(
)



= 6.46 atm L2 mol −2
78. Mass of liquid = 148 − 50 = 98 g
98
= 100 mL = volume of flask
⇒ Volume of liquid =
0.98
mass of gas = 50.5 − 50 = 0.50 g
 w
Now applying ideal gas equation : pV =   RT
M
wRT 0.5 × 0.082 × 300
⇒
M =
=
= 123 g mol −1
pV
1 × 0.1
79. False, ideal gas cannot be liquefied as there is no intermolecular
attraction between the molecules of ideal gas. Hence, there is no
point of forming ideal solution by cooling ideal gas mixture.
80. If ‘α’ is the degree of dissociation, then at equilibrium
Cl 2 r 2Cl
Moles
1 −α
2α
From diffusion information
r(mix)
= 1.16 =
r(Kr)
⇒
⇒
84
M (mix)
pV
1 × 40
=
RT 0.082 × 400
= 1.22
81. The total moles of gaseous mixture =
Let the mixture contain x mole of ethane. Therefore,
7
C2H6 + O2 → 2CO2 + 3H2O
2
x
C2H4 + 3O2 → 2CO2 + 2H2O
7
x
x + 3 (1.22 − x ) = + 3.66
2
2
130 x
⇒
= + 3.66
32 2
⇒ x = 0.805 mole ethane and 0.415 mole ethene.
0.805
= 0.66
⇒ Mole fraction of ethane =
1.22
Mole fraction of ethene = 1 − 0.66 = 0.34
Total moles of O2 required =
rCH4
=
M CH 4
nHe
n CH4
M He
=
4
1
16
=8
4
Initial ratio of rates of effusion gives the initial composition of
mixture effusing out. Therefore, n (He) : n (CH4 ) = 8 : 1
84. Number of moles =
2 × 1021
= 0.33 × 10−2
6 × 1023
p = 7.57 × 103 Nm −2
pV = nRT
7.57 × 103 × 10−3
pV
T =
= 276 K
=
nR 0.33 × 10−2 × 8.314
Now,
⇒
⇒
Also,
α = 0.14
1.22− x
rHe
83.
Total = 1 + α
M (mix) = 62.4
71
M (mix) =
= 62.4
1+ α
⇒
= 2460 L = 2.46 m 3
⇒
urms =
3RT
=
M
3 × 8.314 × 276
m s−1 = 496 ms−1
28 × 10−3
umps
= 0.82
urms
umps = 0.82 × urms = 0.82 × 496 ms−1 = 407 ms−1
85. First we calculate partial pressure of NO and O2 in the combined
system when no reaction taken place.
pV = constant ⇒ p1V1 = p2V2
1.053 × 250
⇒
p2 (NO) =
= 0.752 atm
350
0.789 × 100
p2 (O2 ) =
= 0.225 atm
350
Now the reaction stoichiometry can be worked out using partial
pressure because in a mixture.
pi ∝ ni
2NO + O2
→ 2NO2 → N2O4
Initial
Final
0.752 atm
0.302
0.225 atm
0
0
0
0
0.225 atm
Now, on cooling to 220 K, N2O4 will solidify and only unreacted
NO will be remaining in the flask.
Q
p∝T
p1 T1
∴
=
p2 T2
0.302 300
=
⇒
p2
220
⇒
p2 (NO) = 0.221 atm
82 States of Matter
86. Total moles of gas in final mixture =
pV
6×3
=
= 0.731
RT 0.082 × 300
Q Mole of H2 in the mixture = 0.70
∴ Mole of unknown gas ( X ) = 0.031
Because both gases have been diffused for same time
r (H2 ) 0.70
M
=
=
⇒ M = 1020 g mol −1
2
r ( X ) 0.031
87.
pressure inside the bottle. When the bottle is opened, there is
chances of bumping of stopper. To avoid bumping, bottle
should be cooled that lowers the pressure inside.
(ii) According to Avogadro’s hypothesis, “Under identical
conditions of pressure and temperature, equal volume of ideal
gases contain equal number of molecules.”
and
nRT
p
5
For acetylene gas, 5 g =
mol
26
740
p = 740 mm =
atm
760
V =
⇒

 7.6 × 10−10
1
=
×
 × 6.023 × 1023
760
0.082 × 273

= 2.7 × 1010 molecules
T = 50° C = 323 K
93. From the given information, it can be easily deduced that in the
Substituting in ideal gas equation
5 0.082 × 323
V =
×
× 76 = 5.23 L
26
74
88.
uav (average velocity ) =
4
9 × 10
ms−1 =
100
⇒
Hence,
8 × 8.314 T1
⇒
3.14 × 44 × 10−3
⇒
8T1
1
×
=
2T2
π
4T1
πT2
4T
4 × 1682.5
= 2142 K
T2 = 1 =
π
3.14
T1 = 1682.5 K, T2 = 2142 K
3
89. Volume of balloon =
4 3 4
 21
πr = × 3.14 ×   cm 3
 2
3
3
= 4847 cm 3 ≈ 4.85 L
Now, when volume of H2 (g ) in cylinder is converted into NTP
volume, then
p1V1 p2V2
20 × 2.82 1 × V2
=
=
⇒
T1
T2
300
273
V2 = NTP volume ⇒ V2 = 51.324 L
Also, the cylinder will not empty completely, it will hold 2.82 L
of H2 (g ) when equilibrium with balloon will be established.
Hence, available volume of H2 (g ) for filling into balloon is
51.324 − 2.82 = 48.504 L
48.504
= 10
⇒ Number of balloons that can be filled =
4.85
90. urms =
partial pressure of A = 1.0 atm
partial pressure of B = 0.5 atm
pV
V
nA = A =
RT
RT
pBV 0.5 V
nB =
=
RT
RT
3 M 
nB 1 wB
M
= =
× A = ×  A
nA 2 M B
wA 2  M B 
Also
4T1
πT2
1=
⇒
final mixture,
8RT1
πM
T1 = 1682.5 K
Also, for the same gas
8RT1
2RT2
uav
=
:
=
umps
M
πM
⇒
pV
RT
N (Number of molecules)
n=
N A (Avogadro number)
 pV 
N = nNA = 
 N
 RT  A
92. Number of moles (n) =
3 × 8.314 × 293
3RT
= 390.2 ms−1
=
M
48 × 10−3
91. (i) NH3 (l ) is highly volatile, a closed bottle of NH3 (l ) contains
large number of molecules in vapour phase maintaining high
⇒
M A :M B = 1 : 3
p
94. Rate of effusion (r) ∝
M
⇒
r (NH3 )
1
=
×
r (HCl )
17
⇒
p=
3
2
36.5
⇒
p
40 1
=
60 p
36.5
17
36.5
= 2.20 atm
17
3
2
3
= × 1.38 × 10−23 × 300 J = 6.21 × 10−21 J/molecule
2
dp
kp
96. Rate of effusion is expressed as −
=
dt
M
k = constant, p = instantaneous pressure
dp k dt
⇒
−
=
p
M
p
kt
Integration of above equation gives ln  0  =
 p
M
k
2000
47


Using first information : ln 
 =
 1500 
32
95. KE = k BT : k B = Boltzmann’s constant
⇒
k=
32
 4
ln  
 3
47
…(i)
Now in mixture, initially gases are taken in equal mole ratio, hence
they have same initial partial pressure of 2000 mm of Hg each.
After 74 min :
States of Matter 83
 2000 74 k
 =
ln 
32
 pO 2 
Also, the decomposition reaction is :
MCO3 → MO + CO2
0.05 mol
Substituting k from Eq. (i) gives
 2000
74
32
 4
 =
×
ln 
ln  
 3
p
47
32
 O2 
Q
∴
⇒
⇒
Q
 2000 74
 4
 =
ln 
ln  
 3
p
47
 O2 
Solving gives p (O2 ) at 74 min = 1271.5 mm
 2000 74 k
 =
For unknown gas : ln 
79
 pg 
100. The ideal gas equation :
pV = nRT =
Substituting k from (i) gives
 2000
74
32
 4
 =
ln 
ln  
×
 3
p
47
79
 g 
97. First we determine empirical formula as
C
H
Weight
10.5
1
Mole
10.5
= 0.875
12
1
Simple ratio
1
Whole no.
7
pM =
⇒
Topic 2 Liquid State
1. I (CH3OH) : Surface tension decreases as concentration increases.
II (KCl) : Surface tension increases with concentration for ionic
salt.
III [CH3 (CH2 )11 OSO−3 Na + ] : It is an anionic detergent.
There is decrease in surface tension before micelle formation,
and after CMC (Critical Micelle Concentration) is attained, no
change in surface tension.
1/0.875 = 1.14
8
⇒
Empirical formula = C7H8
 w
From gas equation : pV =   RT
M
wRT 2.8 × 0.082 × 400
=
= 91.84 ≈ 92
pV
1× 1
Q Molar mass (M ) is same as empirical formula weight.
Molecular formula = Empirical formula = C7 H8
1
98. For same p and V , n ∝
T
n (gas) T (H2 )
=
⇒
n (H2 ) T (gas)
0.184
n(H2 ) =
= 0.092
2
290
n(gas) =
× 0.092 = 0.0895
⇒
298
Q 0.0895 mole of gas weigh 3.7 g
3.7
∴ 1 mole of gas will weigh
= 41.32 g
0.0895
99. Moles of CO2 can be calculated using ideal gas equation as :
n=
w
RT
M
w
RT = d RT where, ‘d’ is density.
V
pM
5 × 17
d=
=
= 3.42 g L−1.
RT
0.082 × 303
⇒
Solving gives :
pg = 1500 mm
⇒ After 74 min, p (O2 ) : p (g ) = 1271.5 : 1500
Also, in a mixture, partial pressure ∝ number of moles
⇒
n (O2 ) : n (g ) = 1 : 1.18
M =
0.05 mol
0.05 mole MCO3 = 4.21 5 g
4.215
1.0 mole MCO3 =
= 84.3 g (molar mass)
0.05
84.3 = MW of M + 12 + 48
Molecular weight of metal = 24.3
Metal is bivalent, equivalent weight
Molecular weight
=
= 12.15
2
1
pV  700  1336
= 0.05
=
 ×
 
RT  760  1000 0.082 × 300
Surface tension
For O2
KCl (II)
CH3OH (I)
−
CH3(CH2)11OSO3 Na + (III)
Concentration
2. Let us consider, 1.0 L of liquid water is converted into steam .
Volume of H2O (l) = 1L, mass = 1000 g
1000
Volume of 1000 g steam =
⇒
cm 3
0.0006
1000
cm 3 steam = 1000 cm 3
Q Volume of molecules in
0.0006
∴ Volume of molecules in
1000
1000 cm 3 steam =
× 0.0006 × 1000 = 0.60 cm 3
1000
3. Critical temperature is directly proportional to intermolecular
force of attraction. H2O is a polar molecule, has greater
intermolecular force of attraction than O2, hence higher critical
temperature.
4. At liquid-vapour equilibrium at boiling point, molecules in two
phase posses the same kinetic energy.
6
Chemical and Ionic Equilibrium
Topic 1 Chemical Equilibrium
Objective Questions I (Only one correct option)
-
1. Consider the following reaction:
N 2O4 ( g ) q
2NO2 ( g ); ∆H ° = + 58 kJ
For each of the following cases (A , B ), the direction in which the
equilibrium shifts is
(2020 Main, 5 Sep I)
(A) temperature is decreased.
(B) pressure is increased by adding N 2 at constant T.
(a) (A) towards product, (B) towards reactant
(b) (A) towards reactant, (B) no change
(d) (A) towards product, (B) no change
2. The incorrect match in the following is
(a) 10
25
(2019 Main, 12 April II)
(b) ∆G ° = 0, K = 1
(d) ∆G ° < 0, K < 1
c
c
c
c
2SO2 ( g ) + O2 ( g ) → 2SO3 ( g ), ∆H = − 57.2 kJ mol −1 and
K c = 1.7 × 1016 . Which of the following statement is incorrect?
(2019 Main, 10 April II)
(a) The equilibrium constant decreases as the temperature increases
(b) The addition of inert gas at constant volume will not affect the
equilibrium constant
(c) The equilibrium will shift in forward direction as the pressure
increases
(d) The equilibrium constant is large suggestive of reaction going to
completion and so no catalyst is required
5. For the following reactions, equilibrium constants are given :
52
- SO (g ); K = 10
2S( s ) + 3O ( g ) - 2SO ( g ); K = 10
3
2
- 2C + D,
1
(a)
4
(c) 1
(b) 16
(d) 4
- B (g ) + C(g ); K
D ( s ) - C( g ) + E ( g ); K
A (s )
p1
p2
= x atm2
= y atm2
The total pressure when both the solids dissociate
simultaneously is
(2019 Main, 12 Jan I)
4. For the reaction,
2
(d) 10181
K
7. Two solids dissociate as follows:
2CO(g )
(a) 2C(s) + O2 (g )
H2 (g ) + I2 (g )
(b) 2HI(g )
(c) NO2 (g ) + SO2 (g )
NO(g ) + SO3 (g )
(d) 2NO(g )
N2 (g ) + O2 (g )
1
(c) 10
the
initial concentration of B was 1.5 times of the
concentration of A, but the equilibrium concentrations of
A and B were found to be equal. The equilibrium
constant (K ) for the aforesaid chemical reaction is
(2019 Main, 12 April II)
2
(2019 Main, 8 April II)
154
6. In a chemical reaction, A + 2B
3. In which one of the following equilibria, K p ≠ K c ?
S( s ) + O2 ( g )
(b) 10
77
(2019 Main, 12 Jan I)
(c) (A) towards reactant, (B) towards product
(a) ∆G ° < 0, K > 1
(c) ∆G ° > 0, K < 1
The equilibrium constant for the reaction,
2SO2 ( g ) + O2 ( g )
2SO3 ( g ) is
129
(a) x + y atm
(b) x 2 + y2 atm
(c) (x + y) atm
(d) 2( x + y ) atm
8. Consider the reaction,
N 2 ( g ) + 3H 2 ( g )
=2NH (g )
3
The equilibrium constant of the above reaction is K p . If
pure ammonia is left to dissociate, the partial pressure
of ammonia at equilibrium is given by (Assume that
(2019 Main, 11 Jan I)
pNH 3 < < p total at equilibrium)
(a)
(c)
33/ 2 K p1/ 2P 2
4
Kp
(b)
1/ 2 2
16
P
(d)
33/ 2 K p1/ 2P 2
16
Kp
1/ 2 2
P
4
9. 5.1 g NH4 SH is introduced in 3.0 L evacuated flask at
327°C. 30% of the solid NH4 SH decomposed to NH3
and H2 S as gases. The K p of the reaction at 327° C is
Chemical and Ionic Equilibrium 85
(R = 0.082 atm mol −1 K −1 , molar mass of S = 32 g mol −1 ,
−1
molar mass of N = 14 g mol )
(2019 Main, 10 Jan II)
(a) 0. 242 × 10−4 atm 2
(b) 0. 242 atm 2
(c) 4 . 9 × 10−3 atm 2
(d) 1 × 10−4 atm 2
10. The values of
Kp
KC
then the value of x is (assuming ideality)
(a) 1, 24.62 dm 3 atm mol −1 , 606.0 dm 6 atm 2 mol −2
(c) 24.62 dm 3 atm mol −1, 606.0 dm6 atm −2 mol 2,
1.65 × 10−3 dm −6 atm −2 mol 2
(d) 1, 4.1 × 10−2 dm −3atm −1 mol, 606 dm6 atm 2 mol −2
11. Consider the following reversible chemical reactions,
K1
K2
2
2
The relation between K 1 and K 2 is
(a) K 2 = K13
(c) K 2 = K1− 3
…(i)
…(ii)
(2019 Main, 9 Jan II)
(b) K 1K 2 = 3
1
(d) K 1K 2 =
3
12. An aqueous solution contains 0.10 M H2S and 0.20 M HCl. If
the equilibrium constants for the formation of HS− from H2S
is 10
. × 10−7 and that of S2− from HS− ions is 12
. × 10−13 then
the concentration of S2− ions in aqueous solution is :
(a) 5 × 10−8 M
−21
(c) 6 × 10
M
(b) 3 × 10−20 M
−19
(d) 5 × 10
(2018 Main)
M
13. The equilibrium constant at 298 K for a reaction,
A + B q C + D is 100. If the initial concentrations of all
the four species were 1 M each, then equilibrium
concentration of D (in mol L−1 ) will be
(2016 Main)
(a) 0.818
(c) 1.182
(b) 1.818
(d) 0.182
14. The standard Gibbs energy change at 300 K for the reaction,
B + C is 2494. 2 J. At a given time, the composition
1
1
of the reaction mixture is [A]= , [ B ] = 2 and [C ] = . The
2
2
reaction proceeds in the
(R = 8.314JK / mol, e = 2.718)
(2015, Main)
2A a
(a) forward direction because Q > K c
(b) reverse direction because Q > K c
(c) forward direction because Q < K c
(d) reverse direction because Q < K c
(d) CO2 , H2CO3
17. N2 + 3H2 r 2NH3
Which is correct statement if N2 is added at equilibrium
condition?
(2006, 3M)
(b) 1, 24.62 dm 3 atm mol −1, 1.65 × 10−3 dm −6 atm −2 mol 2
-2 AB (g )
6 AB ( g ) - 3 A ( g ) + 3B ( g )
(2006 Main)
(c) HCO−3 , CO23−
3
(2019 Main, 10 Jan I)
A2 ( g ) + B2 ( g )
16. The species present in solution when CO2 is dissolved in
(b) H2CO3 , CO2−
3
2
2
(d) 1
(a) CO2 , H2CO3 , HCO3− , CO32−
=2NO(g )
=2NO (g )
N (g ) + 3H (g ) =2NH (g )
2
(2014 Main)
1
(c)
2
water are
−1
respectively (At 300 K, RT = 24.62 dm atm mol )
N 2 ( g ) + O2 ( g )
N 2O4 ( g )
1
(b) −
2
(a) − 1
for the following reactions at 300 K are,
3
1
SO3 ( g )
2
if K p = K C ( RT )x where, the symbols have usual meaning,
15. For the reaction, SO2 ( g ) + O2 ( g ) q
(a) The equilibrium will shift to forward direction because
according to IInd law of thermodynamics, the entropy must
increases in the direction of spontaneous reaction
(b) The condition for equilibrium is G(N2) + 3G(H2)
= 2G(NH3) where, G is Gibbs free energy per mole of the
gaseous species measured at that partial pressure. The
condition of equilibrium is unaffected by the use of
catalyst, which increases the rate of both the forward and
backward reactions to the same extent
(c) The catalyst will increase the rate of forward reaction by α
and that of backward reaction by β
(d) Catalyst will not alter the rate of either of the reaction
18. Ag + + NH3 s
[Ag(NH3 )]+ ; K 1 = 3. 5 × 10−3
[Ag (NH3 )]+ + NH3 s
[Ag (NH3 )2 ]+ ; K 2 = 1.7 ×10−3
then the formation constant of [Ag(NH3 )2 ]+ is
(a) 6.08 × 10
−6
(b) 6.08 × 10
−9
(d) None of these
(c) 6.08 × 10
(2006, 3M)
6
19. Consider the following equilibrium in a closed container
N2 O4 ( g ) r 2NO2 ( g )
At a fixed temperature, the volume of the reaction container is
halved. For this change, which of the following statements
hold true regarding the equilibrium constant ( K p ) and degree
of dissociation (α ) ?
(2002, 3M)
(a) Neither K p nor α changes
(b) Both K p and α change
(c) K p changes but α does not change
(d) K p does not change but α changes
20. At constant temperature, the equilibrium constant ( K p ) for
the decomposition reaction, N2 O4 r 2NO2 , is expressed
4 x2 p
by K p =
, where, p = pressure, x = extent of
(1 − x2 )
decomposition. Which one of the following statement is true?
(2001, 1M)
86 Chemical and Ionic Equilibrium
(a) K p increases with increase of p
(b) K p increases with increase of x
(c) K p increases with decrease of x
(d) K p remains constant with change in p and x
(c) concentration of NH3 does not change with pressure
(d) concentration of hydrogen is less than that of nitrogen
28. For the reaction, H2 ( g ) + I2 ( g ) r 2HI( g )
21. When two reactants, A and B are mixed to give products, C
and D, the reaction quotient, (Q ) at the initial stages of the
reaction
(2000)
(b) decreases with time
(d) increases with time
22. For the reversible reaction,
N2 ( g ) + 3H2 ( g ) r 2NH3 ( g )
at 500° C , the value of K p is 1.44 × 10–5 when partial
pressure is measured in atmosphere. The corresponding value
of K c with concentration in mol/L is
(2000, S, 1M)
1.44 × 10−5
(0.082 × 500)−2
(b)
1.44 × 10−5
(8.314 × 773)−2
(c)
1.44 × 10−5
(0.082 × 773)2
(d)
1.44 × 10–5
(0.082 × 773)−2
Objective Questions II
(One or more than one correct option)
29. For a reaction, A
P, the plots of [A] and [P] with time at
temperatures T1 and T2 are given below.
10
5
T2
T1
23. For the chemical reaction,
3 X ( g ) + Y ( g ) r X 3Y ( g )
the amount of X 3Y at equilibrium is affected by
(1999, 2M)
24. For the reaction ,
CO( g ) + H2 O( g ) r CO2 ( g ) + H2 ( g ) ,
at a given temperature, the equilibrium amount of CO2 ( g )
can be increased by
(1998)
Time
If T2 > T1 , the correct statement(s) is are
(Assume ∆H s and ∆S s are independent of temperature and
ratio of ln K at T1 to ln K at T2 is greater than T2 / T1 . Here
H , S , G and K are enthalpy, entropy, Gibbs energy and
equilibrium constant, respectively.)
(2018 Adv.)
(a) ∆H s < 0, ∆S s < 0
(b) ∆Gs < 0, ∆H s > 0
(c) ∆Gs < 0, ∆S s < 0
(d) ∆Gs < 0, ∆S s > 0
30. The % yield of ammonia as a function of time in the reaction,
N2( g )+ 3H2( g ) w
2 NH3( g ); ∆H < 0
% yield
25. One mole of N2 O4 ( g ) at 300 K is kept in a closed container
under one atmosphere. It is heated to 600 K when 20% by
mass of N2 O4 ( g ) decomposes to NO2 (g). The resultant
pressure is
(1996, 1M)
(2015 Adv.)
(d) 1.0 atm
If this reaction is conducted at ( p , T1 ), with T2 > T1 the %
yield by of ammonia as a function of time is represented by
(1985, 1M)
(a) Pb(NO3 )2 (aq) + 2NaI (aq) = PbI2 (s) + 2NaNO3 (aq)
(b) AgNO3 (aq) + HCl (aq) = AgCl (s) + HNO3 (aq)
(c) 2Na (s) + 2H2O (l ) = 2NaOH (aq) + H2 (g )
(d) KNO3 (aq) + NaCl (aq) = KCl (aq) + NaNO3 (aq)
(a)
T2
T1
(b)
Time
T1
T2
(d)
% yield
% yield
(c)
T2
Time
27. Pure ammonia is placed in a vessel at a temperature where its
dissociation constant (α ) is appreciable. At equilibrium,
N2 + 3H2 s 2NH3
(1984, 1M)
T1
% yield
26. An example of a reversible reaction is
T1
Time
% yield
(c) 2.0 atm
5
at( p, T1 ) is given below.
(a) adding a suitable catalyst
(b) adding an inert gas
(c) decreasing the volume of the container
(d) increasing the amount of CO(g )
(b) 2.4 atm
T1
T2
Time
(a) temperature and pressure
(b) temperature only
(c) pressure only
(d) temperature, pressure and catalyst
(a) 1.2 atm
-
10
[A]/(mol L–1)
(a)
(1981, 1M)
(a) total pressure
(b) catalyst
(c) the amount of H2 and I2 present
(d) temperature
[P]/(mol L–1)
(a) is zero
(c) is independent of time
the equilibrium constant K p changes with
T2
T1
(a) K p does not change significantly with pressure
(b) α does not change with pressure
Time
Time
Chemical and Ionic Equilibrium 87
31. The initial rate of hydrolysis of methyl acetate (1 M) by a
weak acid (HA, 1M) is 1/100th of that of a strong acid
(2013 Adv.)
(HX, 1M), at 25°C. The K a (HA) is
(a) 1 × 10−4
(b) 1 × 10−5
(c) 1 × 10−6
(d) 1 × 10−3
I
0
32. The equilibrium 2 Cu r Cu + Cu
(c) SCN−
(d) CN−
forward reaction at constant temperature is favoured by
(1991, 1M)
(a) introducing an inert gas at constant volume
(b) introducing chlorine gas at constant volume
(c) introducing an inert gas at constant pressure
(d) increasing the volume of the container
(e) introducing PCl 5 at constant volume
40. The rate of an exothermic reaction increases with increasing
temperature.
(1993, 1M)
(1987, 1M)
42. If equilibrium constant for the reaction, A2 + B2 r 2 AB,
is K, then for the backward reaction
1
1
1
AB r A2 + B2 , the equilibrium constant is .
2
2
K
(1984, 1M)
43. When a liquid and its vapour are at equilibrium and the pressure
is suddenly decreased, cooling occurs.
34. The equilibrium SO2 Cl 2 ( g ) r SO2 ( g ) + Cl 2 ( g ) is
attained at 25° C in a closed container and an inert gas,
helium is introduced. Which of the following statements
are correct?
(1989, 1M)
(a) Concentration of SO2 ,Cl 2 and SO2Cl 2 change
(b) More chlorine is formed
(c) Concentration of SO2 is reduced
(d) None of the above
35. When NaNO3 is heated in a closed vessel, oxygen is liberated
and NaNO2 is left behind. At equilibrium,
(1994, 1M)
True/False
41. Catalyst makes a reaction more exothermic.
33. For the reaction, PCl 5 ( g ) r PCl 3 ( g ) + Cl 2 ( g ) the
(1986, 1M)
(a) addition of NaNO2 favours reverse reaction
(b) addition of NaNO3 favours forward reaction
(c) increasing temperature favours forward reaction
(d) increasing pressure favours reverse reaction
36 For the gas phase reaction,
( ∆H = −32.7 kcal)
C2 H4 + H2 r C2 H6
carried out in a vessel, the equilibrium concentration of
C2 H4 can be increased by
(1984, 1M)
(a) increasing the temperature
(b) decreasing the pressure
(c) removing some H2
(1984, 1M)
Subjective Questions
44. (a) In the following equilibrium N2O4 (g ) r 2NO2 (g )
when 5 moles of each are taken, the temperature is kept at 298
K the total pressure was found to be 20 bar. Given that
∆G °f (N2O4 ) = 100 kJ, ∆G °f (NO2 ) = 50 kJ
(i) Find ∆G of the reaction.
(ii) The direction of the reaction in which the equilibrium shifts.
(b) A graph is plotted for a real gas which follows van der Waals’
equation with pVm taken on Y-axis and p on X-axis. Find the
(2004, 4M)
intercept of the line where Vm is molar volume.
45. When 3.06 g of solid NH4 SH is introduced into a two litre
evacuated flask at 27° C, 30% of the solid decomposes into
gaseous ammonia and hydrogen sulphide.
(i) Calculate K c and K p for the reaction at 27°C.
(ii) What would happen to the equilibrium when more solid
NH4SH is introduced into the flask?
(1999, 7M)
46. (a) The degree of dissociation is 0.4 at 400 K and 1.0 atm for the
gaseous reaction PCl 5 r PCl 3 + Cl 2. Assuming ideal
behaviour of all the gases, calculate the density of equilibrium
mixture at 400 K and 1.0 atm (relative atomic mass of P = 31.0
and Cl = 35.5).
(b) Given, [Ag(NH3 )+2 ] r Ag+ + 2NH3 ,
K c = 6.2 × 10−8 and K sp of AgCl
(d) adding some C2H6
= 1.8 × 10−10 at 298 K.
Fill in the Blanks
37. For a gaseous reaction 2B → A, the equilibrium
constant K p is …… to/than K c .
(1996, 1M)
equilibrium constant K p and K c are related by .....
II
(2011)
(b) Cl −
ten-fold increase in pressure on the reaction,
N2 ( g ) + 3H2 ( g ) r 2NH3 ( g ) at equilibrium, results in
.............. in K p .
39. For a given reversible reaction at a fixed temperature,
in aqueous
medium at 25° C shifts towards the left in the presence of
(a) NO −3
38. A
(1997 C, 1M)
If ammonia is added to a water solution containing excess of
AgCl(s) only. Calculate the concentration of the complex in
1.0 M aqueous ammonia.
(1998, 3M+5M)
88 Chemical and Ionic Equilibrium
53. The equilibrium constant of the reaction
47. The progress of reaction,
A2 ( g ) + B2 ( g ) r 2 AB ( g ) at 100°C is 50. If a one litre
flask containing one mole of A2 is connected to a two litre
flask containing two moles of B2 , how many moles of AB
will be formed at 373 K?
(1985, 4M)
(Concentration/mol L 1 )
A r nB
with time, is represented in fig. use given below.
0.5
54. One mole of N2 and 3 moles of PCl 5 are placed in a 100 L
vessel heated to 227°C. The equilibrium pressure is 2.05
atm. Assuming ideal behaviour, calculate the degree of
dissociation for PCl 5 and K p for the reaction,
0.3
0.1
PCl 5 ( g ) r PCl 3 ( g ) + Cl 2 ( g )
1
5
3
Time/h
7
55. One mole of nitrogen is mixed with three moles of hydrogen
in a four litre container. If 0.25 per cent of nitrogen is
converted to ammonia by the following reaction
Determine :
(i) the value of n
(ii) the equilibrium constant, K and
(iii) the initial rate of conversion of A.
(1984, 6M)
N2 ( g ) + 3H2 ( g ) r 2NH3 ( g ), then
(1994, 3M)
48. 0.15 mole of CO taken in a 2.5 L flask is maintained at 750 K
along with a catalyst so that the following reaction can take place:
CO ( g ) + 2H2 ( g ) r CH3 OH( g )
Hydrogen is introduced until the total pressure of the system
is 8.5 atm at equilibrium and 0.08 mole of methanol is
formed. Calculate (i) K p and K c and (ii) the final pressure if
the same amount of CO and H2 as before are used, but with no
catalyst so that the reaction does not take place. (1993, 5M)
49. For the reaction, CO( g ) + 2H2 ( g ) r CH3 OH( g )
hydrogen gas is introduced into a five litre flask at 327° C,
containing 0.2 mole of CO( g ) and a catalyst, until the
pressure is 4.92 atm. At this point 0.1 mole of CH3 OH( g ) is
formed. Calculate the equilibrium constant, K p and K c .
(1990, 5M)
50. The equilibrium constant K p of the reaction,
2SO2 ( g ) + O2 ( g ) r 2SO3 ( g )
is 900 atm at 800 K. A mixture containing SO3 and O2 having
initial pressure of 1 and 2 atm respectively is heated at
constant volume to equilibrate. Calculate the partial pressure
of each gas at 800 K.
(1989, 3M)
51. N2 O4 is 25% dissociated at 37° C and one atmosphere
pressure. Calculate (i) K p and (ii) the percentage dissociation
at 0.1 atm and 37° C.
(1988, 4M)
52. At a certain temperature, equilibrium constant ( K c ) is 16 for
the reaction;
SO2 ( g ) + NO2 ( g ) r SO3 ( g ) + NO( g )
If we take one mole each of all the four gases in a one litre
container, what would be the equilibrium concentrations of
NO and NO2 ?
(1987, 5M)
calculate the equilibrium constant, K c in concentration units.
What will be the value of K c for the following equilibrium?
1
3
(1981, 4M)
N2 ( g ) + H2 ( g ) r NH3 ( g )
2
2
Passage Based Questions
Thermal decomposition of gaseous X 2 to gaseous X at 298 K takes
place according to the following equation:
X 2 (g ) s
2X (g )
The standard reaction Gibbs energy, ∆ r G°, of this reaction is
positive. At the start of the reaction, there is one mole of X 2 and no
X . As the reaction proceeds, the number of moles of X formed is
given by β. Thus, βequilibrium is the number of moles of X formed at
equilibrium. The reaction is carried out at a constant total pressure
of 2 bar. Consider the gases to behave ideally.
(Given, R = 0.083 L bar K −1 mol −1 )
56. The equilibrium constant K p for this reaction at 298 K, in
terms of β equilibrium is
(a)
(c)
8 β 2equilibrium
2 − β equilibrium
4 β 2equilibrium
2 − β equilibrium
(2016 Adv.)
(b)
8 β 2equilibrium
4 − β 2equilibrium
(d)
4 β 2equilibrium
4 − β 2equilibrium
57. The incorrect statement among the following for this
reaction, is
(2016 Adv.)
(a) Decrease in the total pressure will result in the formation of
more moles of gaseous X
(b) At the start of the reaction, dissociation of gaseous X 2 takes
place spontaneously
(c) β equilibrium = 0.7
(d) KC < 1
Chemical and Ionic Equilibrium 89
Topic 2 Ionic Equilibrium
Objective Questions I (Only one correct option)
1/ 6
1. 100 mL of 0.1M HCl is taken is a beaker and to it 100 mL of
 Ksp 
(a) S = 

 144 
 Ksp 
(b) S = 

 6912
0.1 M NaOH is added in steps of 2 mL and the pH
continuously measured. Which of the following graphs
correctly depicts the change in pH?
(2020 Main, 3 Sep II)
 Ksp 
(c) S = 

 929
1/ 9
 Ksp 
(d) S = 

 216
1/ 7
1/ 7
7. If K sp of Ag 2 CO3 is 8 × 10− 12 , the molar solubility of
(a) pH 7
Ag 2 CO3 in 0.1 M AgNO3 is
(b) pH 7
vol. of NaOH
vol. of NaOH
(2019 Main, 12 Jan II)
(a) 8 × 10− 12 M
(b) 8 × 10− 13 M
(c) 8 × 10− 10 M
(d) 8 × 10− 11 M
8. 20 mL of 0.1 M H2 SO4 solution is added to 30 mL of 0.2 M
NH4 OH solution. The pH of the resultant mixture is [pK b of
(2019 Main, 9 Jan I)
NH4 OH = 4.7]
(c) pH 7
(a) 9.3
(c) 9.0
(d) pH 7
(b) 5.0
(d) 5.2
9. An aqueous solution contains an unknown concentration of
vol. of NaOH
vol. of NaOH
2. The molar solubility of Cd (OH)2 is 184
. × 10−5 m in water.
The expected solubility of Cd(OH)2 in a buffer solution of
(2019 Main, 12 April II)
pH = 12 is
(a) 184
. × 10
−9
M
(c) 6.23 × 10−11 M
2.49
(b)
× 10−9 M
184
.
(d) 2.49 × 10−10 M
3. What is the molar solubility of Al(OH)3 in 0.2 M NaOH
solution? Given that, solubility product of
Al(OH)3 = 2.4 × 10−24
(2019 Main, 12 April II)
(a) 3 × 10−19
(c) 3 × 10−22
(b) 12 × 10−21
(d) 12 × 10−23
4. The pH of a 0.02 M NH4 Cl solution will be [Given
K b (NH4 OH) = 10−5 and log 2 = 0.301]
(a) 4.65
(c) 5.35
(2019 Main, 10 April II)
−9
(a) 5 × 10
−9
M(b) 2 × 10
(c) 11
. × 10−9 M
M
(d) 10
. × 10−10 M
10. Which of the following are Lewis acids?
(a) PH3 and BCl 3
(c) PH3 and SiCl 4
(2018 Main)
(2018 Main)
(b) AlCl 3 and SiCl 4
(d) BCl 3 and AlCl 3
11. Which of the following salts is the most basic in aqueous
solution?
(a) Al(CN)3
(c) FeCl 3
(2018 Main)
(b) CH3COOK
(d) Pb(CH3COO)2
12. pK a of a weak acid (HA) and pK b of a weak base (BOH) are
(2017 Main)
I. The pH of a mixture containing 400 mL of 0.1 M H2SO4 and
400 mL of 0.1 M NaOH will be approximately 1.3.
II. Ionic product of water is temperature dependent.
III. A monobasic acid with K a = 10−5 has a pH = 5. The degree
of dissociation of this acid is 50%.
IV. The Le-Chatelier’s principle is not applicable to
common-ion effect.
(a) I, II and IV
(c) I and II
original concentration of Ba 2+ ?
3.2 and 3.4, respectively. The pH of their salt (AB) solution is
(b) 2.65
(d) 4.35
5. Consider the following statements.
The correct statements are
Ba2 + . When 50 mL of a 1 M solution of Na2SO4 is added,
BaSO4 just begins to precipitate. The final volume is 500 mL.
The solubility product of BaSO4 is 1 × 10−10 . What is the
(2019 Main, 10 April I)
(b) II and III
(d) I, II and III
6. If solubility product of Zr3 (PO4 )4 is denoted by K sp and its
molar solubility is denoted by S , then which of the following
relation between S and K sp is correct? (2019 Main, 8 April I)
(a) 7.2
(c) 7.0
(b) 6.9
(d) 1.0
13. How many litres of water must be added to 1 L of an aqueous
solution of HCl with a pH of 1 to create an aqueous solution
with pH of 2?
(2013 Main)
(a) 0.1 L
(c) 2.0 L
(b) 0.9 L
(d) 9.0 L
14. Solubility product constant ( K sp ) of salts of types MX , MX 2
and M 3 X at temperature ‘T ’ are 4.0 × 10−8 , 3.2 × 10−14 and
2.7 × 10−15 , respectively. Solubilities (mol dm− 3 ) of the salts
at temperature ‘T ’ are in the order
(2008, 3M)
(a) MX > MX 2 > M 3 X
(b) M 3 X > MX 2 > MX
(c) MX 2 > M 3 X > MX
(d) MX > M 3 X > MX 2
90 Chemical and Ionic Equilibrium
15. 2.5 mL of 2/5 M weak monoacidic base (K b = 1 × 10− 12 at
25°C) is titrated with 2/15 M HCl in water at 25°C. The
concentration of H+ at equivalence point is
(K w = 1 × 10− 14 at 25°C)
(2008, 3M)
(a) 3.7 × 10
(c) 3.2 × 10
− 13
−2
(d) 2.7 × 10
M
M
(b) 10−5 M (Ag+ ) and 10−5 M (Cl − )
M
(c) 10−6 M (Ag+ ) and 10−6 M (Cl − )
HCl and the solution is diluted to one litre, resulting hydrogen
ion concentration is
(2005, 1M)
(a) 1.6 × 10− 11
(b) 8 × 10− 11
(c) 5 × 10− 5
(d) 8 × 10− 2
17. HX is a weak acid ( K a = 10−5 ). It forms a salt NaX (0.1M) on
reacting with caustic soda. The degree of hydrolysis
of NaX is
(2004, 1M)
(b) 0.0001%
(c) 0.1%
(d) 0.5%
18. A solution which is 10−3 M each in Mn 2+ , Fe2+ , Zn 2+ and
Hg 2+ is treated with 10−16 M sulphide ion. If K sp of
MnS, FeS, ZnS and HgS are 10−15 , 10−23 , 10−20 and 10−54
respectively, which one will precipitate first?
(2003, 1M)
(a) FeS
(b) MgS
(c) HgS
(d) ZnS
19. Identify the correct order of solubility of Na 2 S, CuS and ZnS
in aqueous medium.
(a) CuS > ZnS > Na 2S
(c) Na 2S > CuS > ZnS
(2002)
(b) ZnS > Na 2S > CuS
(d) Na 2S > ZnS > CuS
20. For a sparingly soluble salt A p Bq , the relationship of its
solubility product ( Ls ) with its solubility (S) is
(a) Ls = S p + q ⋅ pp ⋅ qq
(b) Ls = S p + q ⋅ pq ⋅ qp
(c) Ls = S
(d) Ls = S
pq
⋅ p ⋅q
p
q
pq
⋅ ( p. q)
(2001, 1M)
the order
(1999, 2M)
23. Amongst the following hydroxides, the one which has the
lowest value of K sp at ordinary temperature (about 25° C) is
(1990, 1M)
(d) Be(OH)2
24. Which of the following is the strongest acid?
(d) SO2 (OH) 2
27. The compound that is not a Lewis acid is
(a) BF3
(b) AlCl 3
28. The conjugate acid of
(a) NH3
(c) BeCl 2
NH–2
(1985, 1M)
(d) SnCl 4
is
(1985, 1M)
(c) NH+4
(b) NH2OH
(d) N2H4
29. The best indicator for detection of end point in titration of a
weak acid and a strong base is
(1985, 1M)
(a) methyl orange (3 to 4)
(b) methyl red (5 to 6)
(c) bromothymol blue (6 to 7.5) (d) phenolphthalein (8 to 9.6)
30. A certain weak acid has a dissociation constant of 1.0 × 10−4 .
The equilibrium constant for its reaction with a strong base is
−4
(a) 1.0 × 10
−10
10
(c) 1.0 × 10
(b) 1.0 × 10
(d) 1.0 × 1014
31. A certain buffer solution contains equal concentration of X −
and HX. The K b for X − is 10−10 . The pH of the buffer is
(b) 7
(c) 10
(d) 14
equal volumes of which of the following are mixed?
(a) 100 mL of (M/10) HCl + 100 mL of (M/10) NaOH
(b) 55 mL of (M/10) HCl + 45 mL of (M/10) NaOH
(c) 10 mL of (M/10) HCl + 90 mL of (M/10) NaOH
(d) 75 mL of (M/5) HCl + 25 mL of (M/5) NaOH
(c) SO(OH)2
(a) unionised in the small intestine and in the stomach
(b) completely ionised in the small intestine and in the stomach
(c) ionised in the stomach and almost unionised in the small
intestine
(d) ionised in the small intestine and almost unionised in the
stomach
32. The precipitate of CaF2 , ( K sp = 1.7 × 10−10 ) is obtained, when
(1992, 1M)
(b) ClO2 (OH)
gastric juice in human stomach is about 2-3 and the pH in the
small intestine is about 8. Aspirin will be
(1988, 1M)
(a) 4
22. Which of the following solutions will have pH close to 1.0 ?
(a) ClO3 (OH)
26. The pK a of acetyl salicylic acid (aspirin) is 3.5. The pH of
(1984, 1M)
(a) NaCl < NH4Cl < NaCN < HCl
(b) HCl < NH4Cl < NaCl < NaCN
(c) NaCN < NH4Cl < NaCl < HCl
(d) HCl < NaCl < NaCN < NH4Cl
(c) Ba(OH)2
(d) 10−10 M (Ag+ ) and 10−10 M (Cl − )
(1984, 1M)
(p + q)
21. The pH of 0.1 M solution of the following salts increases in
(a) Mg(OH)2 (b) Ca(OH)2
(a) 10−4 M (Ag+ ) and 10−4 M (Cl − )
−2
16. CH3 NH2 (0.1 mole, K b = 5 × 10− 4 ) is added to 0.08 mole of
(a) 0.01%
precipitation of AgCl ( K sp = 1.8 × 10−10 ) will occur only
with
(1988, 1M)
−7
(b) 3.2 × 10
M
25. When equal volumes of the following solutions are mixed,
(1989, 1M)
(1982, 1M)
(a) 10
−4
M Ca
2+
−4
+ 10
MF
−
(c) 10−5 M Ca 2+ + 10−3 M F−
(b) 10
−2
M Ca
2+
+ 10−3 M F−
(d) 10−3 M Ca 2+ + 10−5 M F−
33. An acidic buffer solution can be prepared by mixing the
solution of
(1981, 1M)
(a) acetate and acetic acid
(b) ammonium chloride and ammonium hydroxide
(c) sulphuric acid and sodium sulphate
(d) sodium chloride and sodium hydroxide
34. Of the given anions, the strongest base is
(a) ClO−
(b) ClO−2
(c) ClO−3
(1981, 1M)
(d) ClO−4
35. At 90°C, pure water has [H3 O+ ] as 10−6 mol L−1 . What is the
value of K w at 90°C ?
(a) 10−6
(b) 10−12
(1981, 1M)
(c) 10−14
(d) 10−8
Chemical and Ionic Equilibrium 91
36. The pH of 10−8 M solution of HCl in water is
(a) 8
(b) −8
(c) between 7 and 8
(d) between 6 and 7
(1981, 1M)
H 2S. What is the minimum molar concentration (M) of H +
required to prevent the precipitation of ZnS?
Use K sp ( ZnS) = 1. 25 × 10−22
Objective Questions II
(One or more than one correct option)
mol/L) of Ag 2 CrO4 in a 0.1 M AgNO3 solution is
(b) 1.1 × 10−10
(c) 1.1 × 10−12
(d) 1.1 × 10−9
(2013 Adv.)
CH3 COONa of identical concentrations are provided. The
pair(s) of solutions which form a buffer upon mixing is(are)
HNO3 and CH3COOH
KOH and CH3COONa
HNO3 and CH3COONa
CH3COOH and CH3COONa
(2010)
(1999, 3M)
(b) sodium acetate and HCl in water
(c) ammonia and ammonium chloride in water
(d) ammonia and sodium hydroxide in water
(1998, 2M)
(c) Autoprotolysis constant of water increases with temperature
(d) When a solution of a weak monoprotic acid is titrated against
 1
a strong base, at half-neutralisation point pH =   pK a
 2
Numerical Answer Type Questions
41. A solution of 0.1 M weak base (B) is titrated with 0.1 M of a
strong acid (HA). The variation of pH of the solution with the
volume of HA added is shown in the figure below. What is the
pK b of the base? The neutralisation reaction is given by
12
10
pH
8
0.06 M Fe2 + ( aq ) and
0. 2 M S2 − ( aq ) solutions are mixed, the equilibrium
concentration of Fe2 + ( aq ) is found by Y × 10− 17 M. The
value of Y is ...........
(2019 Adv.)
When
equal
volumes
of
44. The solubility of a salt of weak acid (AB) at pH 3 is
ionisation constant of HB ( K a ) = 1 × 10−8
Match the Column
(2020 Adv.)
Note Degree of dissociation (α ) of weak acid and weak base is
<< 1; degree of hydrolysis of salt << 1; [H+ ] represents the
concentration of H+ ions
List-I
(10 mL of 0.1 M NaOH +
20 mL of 0.1 M acetic
acid) diluted to 60 mL
Q. (20 mL of 0.1 M NaOH +
20 mL of 0.1 M acetic
acid) diluted to 80 mL
List-II
1.
the value of [H+ ] does not
change on dilution
2.
the value of [H+ ] changes
to half of its initial value
on dilution
R.
(20 mL of 0.1M HCl + 20 3.
mL of 0.1 M ammonia
solution) diluted to 80 mL
the value of [H+ ] changes
to two times of its initial
value on dilution.
S.
4.
10 mL saturated solution
of Ni(OH)2 in equilibrium
with exces solid Ni(OH)2
is diluted to 20 mL (solid
Ni(OH)2 is still present
after dilution).
5.
the value of [H+ ] changes
1
times of its initial
to
2
value on dilution
P.
6
4
(2018 Adv.)
are given in List-I. The effects of dilution of the solution on
[H + ] are given in List-II.
(b) The conjugate base of H2PO4− is HPO2−
4
B + HA → BH+ + A − .
FeS ( s )
-
45. Dilution processes of different aqueous solutions, with water,
40. Which of the following statement(s) is (are) correct?
(a) The pH of 1.0 × 10−8 M solution of HCl is 8
K is 16
. × 1017 .
Y × 10−3 mol L−1 . The value of Y is__ (Given that the value
of solubility product of AB ( K sp ) = 2 × 10−10 and the value of
39. A buffer solution can be prepared from a mixture of
(a) sodium acetate and acetic acid in water
(2020 Adv.)
43. For the following reaction, the equilibrium constant K c at 298
Fe2 + ( aq ) + S2 − ( aq )
38. Aqueous solutions of HNO3 KOH, CH3 COOH and
(a)
(b)
(c)
(d)
and overall dissociation
constant of H2 S, K net = K 1 K 2 = 1 × 10−21 .
37. The K sp of Ag 2 CrO4 is 1.1 × 10−12 at 298 K. The solubility (in
(a) 1.1 × 10−11
42. An acidified solution of 0.05 M Zn 2+ is saturated with 0.1 M
the value of [H+ ] changes
to 2 times of its initial
value on dilution
Match each process given in List-I with one or more effect(s)
in List-II. The correct option is
(2018 Adv.)
2
0
2
4 6 8 10 12
Volume of HA (mL)
(a) P → 4; Q → 2; R → 3; S → 1
(b) P → 4; Q → 3; R → 2; S → 3
(c) P → 1; Q → 4; R → 5; S → 3
(d) P → 1; Q → 5; R → 4; S → 1
92 Chemical and Ionic Equilibrium
Fill in the Blanks
46. In the reaction, I− + I2 → I3− , the Lewis acid is .............
(1997, 1M)
47. Silver chloride is sparingly soluble in water because its lattice
energy is greater than .............. energy.
(1987, 1M)
48. An element which can exist as a positive ion in acidic solution
and also as a negative ion in basic solution is said to
be...................
(1984, 1M)
49. The conjugate base of HSO–4 in aqueous solution is ……
(1982, 1M)
(ii) If 6 g of NaOH is added to the above solution, determine the
final pH (assuming there is no change in volume on mixing,
K a of acetic acid is 1.75 × 10−5 mol/L.
(1984, 1M)
59. The average concentration of SO2 in the atmosphere over a
city on a certain day is 10 ppm, when the average temperature
is 298 K. Given that the solubility of SO2 in water at 298 K is
1.3653 mol/L and pK a of H2 SO3 is 1.92, estimate the pH of
rain on that day.
(2000, 5M)
60. The solubility of Pb(OH)2 in water is 6.7 × 10−6 M. Calculate
the solubility of Pb(OH)2 in a buffer solution of pH = 8.
(1999, 4M)
True/False
50. The following species are in increasing order of their acidic
property : ZnO, Na 2 O2 , P2 O5 , MgO.
(1985, 1/2M)
51. Solubility of sodium hydroxide increases with increase in
temperature.
(1985, 1/2M)
52. Aluminium chloride ( AlCl 3 ) is a Lewis acid because it can
donate electrons.
(1982, 1M)
61. (a) Find the solubility product of a saturated solution of
Ag2CrO4 in water at 298 K if the emf of the cell
Ag|Ag+ (saturated. Ag2CrO4 solution.) || Ag+ (0.1 M) | Ag is
0.164 V at 298 K.
(1998, 6M)
(b) What will be the resultant pH when 200 mL of an aqueous
solution of HCl (pH = 2.0) is mixed with 300 mL of an
aqueous solution of NaOH (pH = 12.0) ?
(1998, 6M)
62. A sample of AgCl was treated with 5.00 mL of 1.5 M
Integer Answer Type Questions
53. The molar conductivity of a solution of a weak acid HX (0.01
M) is 10 times smaller than the molar conductivity of a
solution of a weak acid HY (0.10 M). If λ 0X − ≈ λ 0Y − , the
difference in their pK a values, pK a (HX ) − pK a (HY ), is
(consider degree of ionisation of both acids to be <<1).
(2015 Adv.)
54. In 1 L saturated solution of AgCl [K sp ( AgCl ) = 1.6 × 10−10 ],
0.1 mole of CuCl [K sp ( CuCl )= 1.0 × 10−6 ] is added. The
Na 2 CO3 solution to give Ag 2 CO3 . The remaining solution
contained 0.0026 g of Cl – ions per litre. Calculate the
solubility product of AgCl. [K sp (Ag 2 CO3 ) = 8.2 × 10−12 ]
(1997, 5M)
63. An acid type indicator, HIn differs in colour from its
conjugate base (In − ). The human eye is sensitive to colour
differences only when the ratio [ In – ] / [ HIn ] is greater than
10 or smaller than 0.1. What should be the minimum change
in the pH of the solution to observe a complete colour
change? ( K a = 1.0 × 10−5 )
(1997, 2M)
resultant concentration of Ag + in the solution is 1.6 × 10− x .
The value of ‘x’ is
(2011)
64. The ionisation constant of NH+4 in water is 5.6 × 10−10 at
55. Amongst the following, the total number of compounds
25° C. The rate constant for the reaction of NH+4 and OH− to
whose aqueous solution turns red litmus paper blue is
form NH3 and H2 O at 25° C is 3.4 × 1010 L/mol/s. Calculate
the rate constant per proton transfer from water to NH3 .
KCN
Zn(NO3 )2
LiCN
K 2SO4
FeCl 3
(NH4 )2 C2O4
K 2CO3
NaCl
NH4NO3
(2010)
56. The dissociation constant of a substituted benzoic acid at
25°C is 1.0 × 10− 4 . The pH of 0.01 M solution of its sodium
(2009)
57. 0.1 M of HA is titrated with 0.1 M NaOH, calculate the pH at
end point. Given, K a (HA) = 5 × 10− 6 and α << 1.
65. What is the pH of a 0.50 M aqueous NaCN solution?
(pK b of CN− = 4.70).
Subjective Questions
salt is
(1996, 3M)
(2004)
58. 500 mL of 0.2 M aqueous solution of acetic acid is mixed
with 500 mL of 0.2 M HCl at 250°C.
(i) Calculate the degree of dissociation of acetic acid in the
resulting solution and pH of the solution.
(1996, 2M)
66. Calculate the pH of an aqueous solution of 1.0 M ammonium
formate assuming complete dissociation.
(pK a of formic acid = 3.8 and pK b of ammonia = 4.8)
(1995, 2M)
–
+
−
67. For the reaction, [Ag(CN)2 ] r Ag + 2CN
The equilibrium constant, at 25° C, is 4.0 × 10−19 . Calculate
the silver ion concentration in a solution which was originally
0.10 M in KCN and 0.03 M in AgNO3 .
(1994, 3M)
Chemical and Ionic Equilibrium 93
68. An aqueous solution of a metal bromide MBr2 (0.05 M) is
saturated with H2 S. What is the minimum pH at which MS
will precipitate? K sp for MS = 6.0 × 10−21 , concentration of
−7
saturated H2 S = 0.1 M, K 1 = 10
H2 S.
−13
and K 2 = 1.3 × 10
, for
(1993, 3M)
69. The pH of blood stream is maintained by a proper balance of
H2 CO3 and NaHCO3 concentrations. What volume of 5 M
NaHCO3 solution should be mixed with a 10 mL sample of
blood which is 2 M in H2 CO3 , in order to maintain a pH of
7.4? (K a for H2 CO3 in blood is 7.8 × 10−7 )
(1993, 2M)
70. The solubility product ( K sp ) of Ca(OH)2 at 25° C is
4.42 × 10−5 . A 500 mL of saturated solution of Ca(OH)2 is
mixed with equal volume of 0.4 M NaOH. How much
Ca(OH)2 in milligrams is precipitated?
(1992, 4M)
71. A 40 mL solution of a weak base, BOH is titrated with 0.1N
HCl solution. The pH of the solution is found to be 10.04 and
9.14 after the addition of 5.0 mL and 20.0 mL of the acid
respectively. Find out the dissociation constant of the base.
(1991, 6M)
72. The
solubility
−11
product
of
Ag 2 C2 O4
at
25° C is
3 −3
1.29 × 10 mol L . A solution of K 2 C2 O4 containing
0.1520 mole in 500 mL water is shaken at 25° C with excess of
Ag 2 CO3 till the following equilibrium is reached
Ag 2 CO3 + K 2 C2 O4 r Ag 2 C2 O4 + K 2 CO3
At equilibrium, the solution contains 0.0358 mole of
K 2 CO3 . Assuming the degree of dissociation of K 2 C2 O4 and
K 2 CO3 to be equal, calculate the solubility product of
Ag 2 CO3 .
(1991, 4M)
73. What is the pH of a 1.0 M solution of acetic acid? To what
volume must one litre of this solution be diluted so that the pH
of the resulting solution will be twice the original value?
Given, K a = 1.8 × 10−5
(1990, 4M)
74. Freshly precipitated aluminium and magnesium hydroxides
are stirred vigorously in a buffer solution containing
0.25 mol/L of NH4 Cl and 0.05 M of ammonium hydroxide.
Calculate the concentration of aluminium and magnesium
ions in solution.
K b [NH4 OH] = 1.8 × 10−5
−12
K sp [Mg(OH)2 ] = 8.9 × 10
K sp [Al(OH)3 ] = 6 × 10−32
(1989, 3M)
75. How many gram-mole of HCl will be required to prepare one
litre of buffer solution (containing NaCN and HCl) of pH 8.5
using 0.01 g formula weight of NaCN?
K HCN = 4.1 × 10−10
(1988, 4M)
76. What is the pH of the solution when 0.20 mole of HCl is
added to one litre of a solution containing
(i) 1 M each of acetic acid and acetate ion,
(ii) 0.1 M each of acetic acid and acetate ion?
Assume the total volume is one litre.
K a for acetic acid = 1.8 × 10−5.
(1987, 5M)
77. The solubility of Mg(OH)2 in pure water is 9.57 × 10−3 g / L.
Calculate its solubility (in g/L) in 0.02 M Mg(NO3 )2 solution.
(1986, 5M)
78. The concentration of hydrogen ions in a 0.20 M solution of
formic acid is 6.4 × 10−3 mol/L. To this solution, sodium
formate is added so as to adjust the concentration of sodium
formate to one mole per litre.
What will be the pH of this solution? The dissociation
constant of formic acid is 2.4 × 10−4 and the degree of
dissociation of sodium formate is 0.75.
(1985, 3M)
79. A solution contains a mixture of Ag + (0.10 M) and
Hg 2+ (0.10 M) which are to be separated by selective
precipitation. Calculate the maximum concentration of
iodide ion at which one of them gets precipitated almost
completely. What percentage of that metal ion is
precipitated?
(1984, 4M)
K sp : AgI = 8.5 × 10−17 , HgI2 = 2.5 × 10−26
80. The dissociation constant of a weak acid HA is 4.9 × 10−8 .
After making the necessary approximations, calculate
(i) pH
(ii) OH− concentration in a decimolar solution of the acid.
(Water has a pH of 7).
(1983, 2M)
81. Give reason for the statement that “the pH of an aqueous
solution of sodium acetate is more than seven”.
(1982, 1M)
82. 20 mL of 0.2 M sodium hydroxide is added to 50 mL of 0.2 M
acetic acid solution to give 70 mL of the solution. What is the
pH of this solution?
Calculate the additional volume of 0.2 M NaOH required to
make the pH of the solution 4.74.
(Ionisation constant of CH3 COOH = 1.8 × 10−5 ). (1982, 3M)
83. How many moles of sodium propionate should be added to
1 L of an aqueous solution containing 0.020 mole of
propionic acid to obtain a buffer solution of pH 4.75? What
will be pH if 0.010 moles of HCl are dissolved in the above
buffer solution? Compare the last pH value with the pH of
0.010 M HCl solution. Dissociation constant of propionic
(1981, 4M)
acid, K a at 25°C is1.34 × 10−5 .
Answers
Topic 1
(b)
(a)
(b)
(b)
(b)
(d)
(b)
29. (a,c)
33. (c, d, e)
37. smaller
2.
6.
10.
14.
18.
22.
26.
30.
34.
38.
40. T
47. (1.2)
57. (c)
41. F
53. (1.86)
42. F
54. (0.33)
43. T
56. (b)
2. (d)
6. (b)
10. (d)
3. (c)
7. (c)
11. (b)
4. (c)
8. (a)
12. (b)
1.
5.
9.
13.
17.
21.
25.
(d)
(d)
(b)
(b)
(a)
(d)
(d)
(b)
(d)
no change
3.
7.
11.
15.
19.
23.
27.
31.
35.
39.
(a)
4.
(d)
8.
(c)
12.
(b)
16.
(d)
20.
(a)
24.
(a)
28.
(a)
32.
(c, d)
36.
K p = Kc ( RT ) ∆n
(d)
(b)
(b)
(a)
(d)
(d)
(d)
(b, c, d)
(a, b, c, d)
Topic 2
1. (a)
5. (d)
9. (c)
13.
17.
21.
25.
29.
33.
37.
41.
45.
49.
(d)
(a)
(b)
(a)
(d)
(a)
(b)
(3.00)
(d)
SO 2−
4
14.
18.
22.
26.
30.
34.
38.
42.
46.
50.
53. (3)
54.
58. (4.75)
59.
62. (2 × 10 −8)
65.
69. (80)
71.
10 −11) 73. (27.78 × 10 3)
75. (0.177)
77.
79. (99.83)
81.
(d)
(c)
(d)
(d)
(c)
(a)
(c, d)
(0.20)
I2
F
15.
19.
23.
27.
31.
35.
39.
43.
47.
51.
(d)
(d)
(d)
(c)
(a)
(b)
(a, b, c)
(8.9)
hydration
F
(1.6 × 10 −7)
(4.86)
(11.5)
(1.8 × 10 −5)
56.
60.
66.
72.
(8)
57. (9)
(1.2 × 10 −3 M)
(6.50)
68. (1)
(9.67 ×
(8.7 × 10 −4 gL −1)
(>7)
16.
20.
24.
28.
32.
36.
40.
44.
48.
52.
(b)
(a)
(a)
(a)
(b)
(d)
(b, c)
(4.47)
amphoteric
F
78. (4.20)
Hints & Solutions
Topic 1 Chemical Equilibrium
1. N 2O 4 (s) r 2NO 2 (s); ∆H ° = + 58 kJ
Because in case of an endothermic reaction (∆H = + ve), the
equilibrium constant increases with rise in temperature and
hence, the reaction moves in forward direction.On adding N 2,
pressure is increased at constant T, and volume would also be
constant, so no change is observed.
2. The incorrect match is ∆G ° < 0, K < 1.
For an ideal gas ∆G ° = − RT ln K .
∆G °
and K = e− ∆G ° / RT
ln K = −
∴
RT
The above equation is helpful in predicting the spontaneity of the
reaction. e.g.
(i) If ∆G ° < 0, – ∆G °/ RT = + ve and e− ∆G ° / RT > 1and hence,
K > 1. It means that the reaction occur spontaneously in the
forward direction or products predominate over reactants.
(ii) If ∆G ° > 0; − ∆G °/ RT = − ve and
e− ∆G ° / RT < 1and hence, K < 1. It means that the reaction is
non-spontaneous in forward direction (i.e. product side) but
spontaneous in reverse direction (i.e. reactants predominate
over products or the reaction occurs rarely).
(iii) When K = 1, then ∆G ° = 0. This situation generally occur at
equilibrium.
3.
Key Idea The relationship between K p and K c is
K p = K c (RT )
∆ng
where, ∆ng = nproducts − nreactants
If ∆ng = 0 then K p = K c
If ∆ng = + ve then K p > K c
If ∆ng = − ve then K p < K c
Consider the following equilibria reactions
(a) 2C(s) + O2 (g )
2CO(g )
-
∆ng = nproduct − nreactant = 2 − (1) = 1
∆ng ≠ 0 ⇒ So, K p ≠ K c
(b) 2HI(g )
H2(g ) + I2 (g )
∆ng = nproduct − nreactant = 2 − 2 = 0
∆ng = 0 ⇒ So, K p = K c
(c) NO2 (g ) + SO2 (g )
NO(g ) + SO3 (g )
∆ng = nproduct − nreactant = 2 − 2 = 0
∆ng = 0 ⇒ So, K p = K c
(d) 2NO(g )
N2(g ) + O2 (g )
-
-
-
∆ng = nproduct − nreactant = 2 − 2 = 0
∆ng = 0 ⇒ So, K p = K c
Chemical and Ionic Equilibrium 95
At equilibrium
K p1 = x = pB ⋅ pC = p1( p1 + p2 )
Similarly, D(s)
C (g)+ E(g)
4. The explanation of given statements are as follows:
(a) For the given equilibrium, ∆H is negative, so the equilibrium
constant will decrease with increase in temperature and the
equilibrium will shift in the backward direction.
Thus, statement (a) is correct.
(b) When inert gas is added at constant volume and constant
temperature, an equilibrium remains undisturbed.
Thus, statement (b) is correct.
(c) For the equilibrium,
∆ng = 2 − (2 + 1) = − 1, i.e. (−ve)
So, increase in pressure will shift the equilibrium in the
forward direction.
Thus, statement (c) is correct.
(d) The reaction takes place in the presence of a catalyst which is
V2O5 (s) in contact process or NO(g ) in chamber process.
Thus, statement (d) is incorrect.
5. S + O2
SO2 , K 1
-
∴ SO2
or, 2SO2
2S + 2O2, K 1′′ = (K 1′ )2 =
1
K 12
⇒
2S + 3O2
2SO3 , K 2
Now, [(i) + (ii)] gives
2SO2 + O2
2SO3 , K 3
The value of equilibrium constant,
1
K 3 = K 2 × K 1′′ = K 2 × 2
K1
1
= 10129 ×
= 10129 − 104 = 1025
(1052 )2
-
… (i)
… (ii)
6. For the given chemical reaction,
At, t = 0
a0
t = t eq
a0 − x
K
w
1.5a 0
1.5a 0 − 2x
or
0
x
Now, substituting the values in above equation, we get
(a0 )2 × (0.5a0 )
K=
=4
(0.5a0 ) × (0.5a0 )2
7. The equilibrium reaction for the dissociation of two solids is
given as:
e
B(g)+ C (g)
p1
…(iii)
Now, total pressure is given as
pT = pB + pC + pE
…(iv)
= p1 + ( p1 + p2)+ p2 = 2 ( p1 + p2)
On substituting the value of p1 + p2 from Eq. (iii) to Eq. (iv), we
get
pT = 2 x + y atm
N 2 (g ) +
3H2 (g )
p1 + p 2
2NH3 (g )
=
At equilibrium: pN 2 = P, pH 2 = 3P,
pNH 3 = 2P
~p + p
⇒
p(total ) = pN 2 + pH 2 + pNH 3 −
N2
H2
[Q P(total ) >> pNH 3 ]
= p + 3p = 4 p
2
pNH
p2NH 3
3
Now,
=
Kp =
3
pN 2 × pH2 p × (3 p)3
=
Kp =
pNH 3 =
2
pNH
3
27 × p
4
=
2
pNH
3
 P
27 ×  
 4
2
× 44
pNH
3
2
3 ×3×P
4
[QP = 4 p]
4
2
⇒ pNH
=
3
3 × 31/ 2 × P 2 × K p1/ 2
42
2C + D
0
2x
…(ii)
x + y = p1 + p2
⇒
[x = degree of dissociation]
Given, at equilibrium.
[ A] =[ B]
a0 − x = 15
. a0 − 2x
x = 0.5a0
∴ [A] = a0 − x = a0 − 0.5a0 = 0.5a0
[ B] = 15
. a0 − 2x = 15
. a0 − 2 × 0.5a0 = 0.5a0
[C] = 2x = 2 × 0.5a0 = a0
[D] = x = 0.5a0
[C ]2[ D]
Now,
K=
[ A][ B]2
A(s)
p1 + p2 ⋅ p2
K p 2 = y = pC ⋅ pE =( p1 + p2) p2
On adding Eq. (i) and (ii), we get.
K p1 + K p 2 = x + y = p1( p1 + p2)+ p2( p1 + p2)
= ( p1 + p2 )2
8.
1
S + O2 , K 1′ =
K1
A + 2B
At equilibrium
…(i)
e
=
32 × 3 × P 4 × K p
44
33/ 2 × P 2 × K p1/ 2
9. Molar mass of NH4SH = 18 + 33 = 51g mol
16
−1
Number of moles of NH4SH introduced in the vessel
Weight
5.1
=
=
= 01
. mol
Molar mass 51
NH4 SH( s )
Number of
moles at t = 0
At t = t eq
0.1
c
01
. (1 − 0. 03)
Active mass
(mol L −1)
KC =
⇒
∴
NH3 ( g ) + H2 S( g )
0
0
30% of
01
. = 0. 03
0. 03
= 0. 01
3
30% of 0.1
= 0. 03
0. 03
= 0. 01
3
[ NH3 ][ H2S ] 0.01 × 0.01
=
= 10−4 (mol L
[ NH4HS(s) ]
1
K p = KC (RT )
−1 2
)
∆ng
[where, ∆ng = Σnproduct − Σnreactant ]= 2 − 0 = 2
K p = KC (RT )2
= 10− 4 × [0.082 × (273 + 327)]2 atm 2
= 0.242 atm 2
96 Chemical and Ionic Equilibrium
10. We know that, the relationship between K p and KC of a chemical
14. Given, ∆G ° = 2494.2 J
1
2×
[ B ][C ]
2 =4
=
Q=
2
[ A ]2
 1
 
 2
equilibrium state (reaction) is
K p = KC (RT )
Kp
⇒
= (RT )
KC
∆ng
∆ng = ΣnProducts − ΣnReactants
N2(g ) + O2(g )
2NO(g )
2 − (1 + 1)
0
(RT )
= (RT ) = 1
where,
(i)
⇒
(ii)
⇒
(iii)
∆ng
(RT )
N2O4 (g )
2− 1
c
c
c
∴ We know,
∆G = ∆G °+ RT ln Q
= 2494.2 + 8.314 × 300 ln 4
= 28747.27 J (+ ve value)
Q
Also, we have
∆G = RT ln
K
If ∆G is positive, Q > KC
Therefore, reaction shifts in reverse direction.
2NO2 (g )
= RT = 24.62 dm3 atmmol−1
N2 (g ) + 3H2 (g )
⇒ (RT )2− (3 + 1) = (RT )− 2
=
2NH3 (g )
1
(24.62 dm atm mol −1 )2
15. For the given reaction, ∆ng = nP − nR
3
where, nP = number of moles of products
= 1.649 × 10−3 dm –6 atm − 2 mol 2
11. (i) A2(g )+ B2(g )
⇒
2AB(g );
c
c
K p = K c ( RT )
∆ng = −
HCO−3 r H+ + CO23−
Therefore, in solution, all of the above mentioned species exist.
12. Given [H2S] = 010
. M
[HCl] = 0.20 M So, [H+ ] = 0.20 M
H2S
+
+ HS− , K 1 = 10
. × 10−7
+
+ S2− , K 2 = 12
. × 10−13
-H
HS - H
−
It means for,
H2S
+
- 2H
∆G = 0
G(reactants) = G (products)
G (N2 ) + 3G (H2 ) = 2G (NH3 )
17. At equilibrium,
+ S2−
A catalyst does not affect either equilibrium composition or
equilibrium constant, it just increases rate of both forward and
backward reaction but by the same factor.
K = K 1 × K 2 = 10
. × 10−7 × 12
. × 10−13
Now
= 12
. × 10−20
K × [ H2S]
[according to the final equation]
[ S2− ] =
[ H+ ]2
13.
[Ag(NH3 )+ ] + NH3 r [ Ag(NH3 )+2 ] K 2 = 1.7 × 10− 3
Initially at t = 0
At equilibrium
−2
4 × 10 M
1
1−x
K eq =
or
or
∴
Adding : Ag+ + 2NH3 r [ Ag(NH3 )2+ ]
. × 10−20 × 1 × 10−1 M
12
A
+
K = K 1 × K 2 = 5.95 × 10− 6
= 3 × 10−20 M
C
1
1
+
1+x
1−α
D
1
pi :
1+x
1−α
p
1+ α
Total
2α
2α
p
1+ α
1 +α
Kp =
2
[C ][ D ] (1 + x)(1 + x) (1 + x)
=
=
[ A ][ B ] (1 − x)(1 − x) (1 − x)2
 1 + x
100 = 

 1 − x
N2O4 (g ) r 2NO2 (g )
19.
B q
1−x
K 1 = 3.5 × 10− 3
18. Ag+ + NH3 r [ Ag(NH3 )+ ]
12
. × 10−20 × 01
. M
=
(0.2M)2
=
2
or
∆ng
1
2
16. When CO2 is dissolved in water, following equilibria are established:
H2O + CO2 r H2CO3
H2CO3 r H+ + HCO−3
3 A2(g ) + 3 B2 (g );
3
[ A ] [ B2 ]3
1
1
= 3,
=
K2 = 2
3
[ AB ]6
K1
 [ AB ]2 


 [ A2 ][ B2 ]
K 2 = K 1−3
(ii) 6 AB(g )
n R = number of moles of reactants
[ AB ]2
K1 =
[ A2 ][ B2 ]
10 =
1+ x
1− x
10 − 10x = 1 + x
10 − 1 = x + 10x
9 = 11x
9
x=
= 0.818
11
[ D ] = 1 + x = 1 + 0.818 = 1818
.
4α 2
1 − α2
p
At constant temperature, halving the volume will change both p
and α but K p remains constant.
20. N2O4 r 2NO2, K p =
1− x
2x
4 x2 p
. K p is function of temperature
1 − x2
only, does not change with either p or x.
[C ][ D ]
21. A + B r C + D, Q =
[ A ][ B ]
As time passes, amount of products ‘C’ and ‘D’ increases, hence
Q increases.
Chemical and Ionic Equilibrium 97
∆n = − 2
22. N2 (g ) + 3H2 (g ) r 2NH3 (g )
Ka =
K p = K c ( RT )∆n
Kc =
Kp
(RT )∆n
(Rate)HA = k [ H+ ]HA
1.44 × 10− 5
=
(0.082 × 773)− 2
(Rate) HX = k [ H+ ]HX
(Rate)
23. Both temperature and pressure will change the equilibrium
amount of X 3Y (g ). Temperature changes the value of equilibrium
constant.
order to restore equilibrium. Therefore, addition of CO (g) will
increase the equilibrium amount of CO2.
N2O4 r 2NO2
At 300 K :
1.0 atm
0
At 600 K :
2.0 – 0.40
0.80
(Rate) HX
(Rate) HA
∴
26. In reactions (a), (b) and (c), atleast one of the product is either
27. Kp for a given reversible reaction depends only on temperature.
28. Equilibrium constant of a given reversible reaction depends only
on temperature.
29. For the reaction, A
Given,
…(i)
It shows, On increasing the temperature, K decreases so reaction
is exothermic i.e., ∆H o < 0
Besides, graph shows K >1
So
∆ G º< 0
Now from equation (i)
T1 ln K 1 > T2 ln K 2
− ∆G º1> − ∆G º2
Likewise (− ∆H º+ T1∆S º ) > (− ∆H º+ T2 ∆S º )
or simply
T1∆S º> T2∆S º
So,
(T2 − T1 ) ∆S ° < 0
∴
∆S º< 0
In other words, increase of ∆G with increase in temperature is
possible only when ∆S ° < 0.Hence, options (a) and (c) are correct.
30. Since, the reaction is exothermic, there will be less ammonia at
equilibrium and higher temperature. However, rate of reaction
increases with rise in temperature, NH3 will be formed at faster
rate in the initial stage when the temperature is high.
31.
H+
PLAN RCOOR′ + H 2O → RCOOH + R′ OH
Acid hydrolysis of ester is follows first order kinetics.
For same concentration of ester in each case, rate is dependent on
[ H+ ] from acid.
Rate = k [ RCOOR′ ]
Also for weak acid, HA r H+ + A −
∴
= 100 =
[ H + ] HX
[ H + ] HA
=
1
[ H+ ]HA
1
100
HA r H+ + A −
0
x
0
x
x = 0.01
[ H+ ][ A − ] 0.01 × 0.01
Ka =
=
= 1.01 × 10−4
[ HA ]
0.99
32. Cl − , CN − and SCN − forms precipitate with Cu (I), remove Cu (I)
ion from equilibrium and reaction shifts in backward direction
according to Le-Chatelier's principle.
33. If inert gas is introduced at constant pressure, volume of
container will have to be increased and this will favour the
forward reaction. Also adding PCl 5 (g ) at constant volume will
favour forward reaction because PCl 5 (g ) is a reactant.
=P
T1 < T2
ln K 1 T2
>
ln K 2 T1
= 100(Rate)HA
[ H + ] HA =
1
(1 − x)
Total pressure = 2.40 atm
insoluble precipitate or a gas that drive the reaction continuously
to right and do not allow equilibrium to be established.
Following is the reversible reaction.
KNO3 (aq) + NaCl (aq) r KCl (aq) + NaNO3 (aq)
HX
+
∴ Also in strong acid, [ H ] = [ HX ] = 1M
24. Adding reactant will drive the reaction in forward direction in
25.
[ H+ ][ A − ]
[ HA ]
34. SO2Cl 2 (g ) r SO2 (g ) + Cl 2 (g ),
Adding inert gas at constant volume will not affect partial
pressure of reactant or products, hence will not affect
equilibrium amount of either reactant or products.
1
35. NaNO3 (s) r NaNO2 (s) + O2 (g ) , ∆H > 0
2
NaNO3 and NaNO2 are in solid state, changing their amount has
no effect on equilibrium. Increasing temperature will favour
forward reaction due to endothermic nature of reaction. Also,
increasing pressure will favour backward reaction in which some
O2 (g ) will combine with NaNO2 (s) forming NaNO3.
36. C2H4 + H2 r C2H6, ∆H = − 32.7 kcal
The above reaction is exothermic, increasing temperature will
favour backward reaction, will increase the amount of C2H4.
Decreasing pressure will favour reaction in direction containing
more molecules (reactant side in the present case). Therefore,
decreasing pressure will increase amount of C2H4.
Removing H2 , which is a reactant, will favour reaction in
backward direction, more C2H4 will be formed.
Adding C2H6 will favour backward reaction and some of the
C2H6 will be dehydrogenated to C2H4.
K
37. Smaller : K p = c
RT
38. changing pressure has no effect on equilibrium constant.
39. K p = K c (RT )∆n, where, ∆n = Σn (products) − Σn (reactants)
40. Rate of any reaction increases on rising temperature.
41. Catalyst has no effect on thermodynamics of reaction.
98 Chemical and Ionic Equilibrium
1
.
K
43. Evaporation is an endothermic process.
47. Observing the graph indicates that when 0.20 mole of A is
42. It is
reacted, 0.40 mole of product is formed.
44. (a) N2O4 (g ) r 2NO2 (g )
Also ∆G ° = − RT ln K = 0 , K = 1
Let the reaction shifts in forward direction.
Total
N2O4 (g ) r 2NO2 (g )
pi :
⇒
K =
x − 0.16
0.15 – 0.08
5 + 2x
10 + x
5 + 2x
× 20
10 + x
0.08
0.07
× 8.5
0.34
0.18
H2 =
× 8.5
0.34
0.08
CH3OH =
× 8.5
0.34
2
0.08
 0.34 
Kp =
×
 = 0.056
2
 8.5 
(0.07) (0.18)
(i) Partial pressures : CO =
81x + 405x + 450 = 0
x = − 1.66 and – 3.33
Both values of x indicates that reaction actually proceeds in
backward direction.
a 

(b)  p +
 (Vm − b) = RT

Vm2 


ap2   pV
− b = RT
p+
 

( pV )2   p

0.08
= 0.032 M
2.5
0.18
[H2 ] =
= 0.072 M
2.5
0.07
[CO] =
= 0.028 M
2.5
0.032
Kc =
= 213.33
(0.028) (0.072)2
(ii) Concentrations : [CH3OH] =
⇒[( pV 2 ) p + ap2 ][( pV ) − b ] = p ( pV )2 RT
⇒ p ⋅ [ pV 2 + ap ] ( pV − bp) = p ( pV 2 ) RT
⇒ ( pV )3 = ( pV )2 RT
45. (i) Mole of solid NH4HS taken initially = 3.06 = 0.06
51
At equilibrium NH4HS (s) r NH3 (g ) + H2S (g )
0.018
CO (g ) + 2H2 (g ) r CH3OH (g )
49.
0.018
Mole :
2
0.2 – 0.10
x − 0.20
0.10
⇒ Total moles = x
4.92 × 5
x=
= 0.5
0.082 × 600
 0.018
−5
Kc = 
 = 8.1 × 10
 2 
⇒
0.018 × 0.082 × 300
= 0.22 atm
2
K p = (0.22)2 = 4.84 × 10− 2
⇒ moles of H2 at equilibrium = x − 0.2 = 0.3
0.1
0.3
Partial pressures : CO =
p, H2 =
p,
0.5
0.5
0.1
CH3OH =
p
0.5
p
25
25
5
= 2=
Kp =
= 0.11 atm − 2
2
9 (4.92)2
9p
 p  3 
   p
 5  5 
0.1
0.3
Concentrations : [CO] =
M , [H2] =
M,
5
5
0.1
(0.1/ 5)
[CH3OH] =
= 277.77 M −2.
M ⇒ KC =
5
(0.1/ 5) (0.3 / 5)2
p (NH3 ) =
(ii) Addition of solid NH4HS will have no effect on equilibrium.
46. (a) PCl 5 (g ) r PCl 3 (g ) + Cl 2 (g ) Total moles
1−α
α
α
1+ α
208.5
Average molar mass =
= 148.9
1.4
pM
1 × 148.9
ρ (density) =
=
RT
0.082 × 400
= 4.54 g/L
(b) AgCl (s) + 2NH3 (aq) r [ Ag(NH3 )2+ ] + Cl −
−x
n=2
Total moles at equilibrium = x − 0.01
8.5 × 2.5
x − 0.01 =
= 0.34 ⇒ x = 0.35
0.082 × 750
2
But p = 0
Intercept = RT
⇒
CO (g ) + 2H2 (g ) r CH3OH (g )
48.
(5 + 2x )2 10 + x
×
× 20 = 1
5−x
(10 + x )2
⇒
nB
+ 0. 40
At equilibrium, [A] = 0.30 M, [B] = 0.60 M
[ B ] 2 0.36
Kc =
=
= 1.2
[ A ] 0.30
∆G ° = 2 ∆G f° (NO2 ) − ∆G f° (N2O4 ) = 0
5−x
5−x
× 20
10 + x
r
A
− 0.20
1 − 2x
K =
 x 
= 2.9 × 10− 3 = 

 1 − 2 x
Kc
K sp
x = 0.049 M
x
x
2
50.
2SO2 (g ) + O2 (g ) r SO3 (g )
Initial pi :
0
Equilibrium pi : 2p
Kp = 900 =
2
2+ p
(1 − 2 p)2
(2 + p) (2 p)2
1
1− 2p
[Ignoring p in comparison to 2]
Chemical and Ionic Equilibrium 99
1
atm
87
p=
Kc =
2
atm
87
Partial pressure of SO2 = 2 p =
1 175
atm
=
87 87
Partial pressure of O2 = 2 + p = 2 +
Also for :
 1  85
atm
Partial pressure of SO3 = 1 − 2 p = 1 − 2   =
 87 87
pi :
1−α
1−α
p
1+ α
Kp =
4α 2
4 (0.25)2
p
=
= 0.26 atm
1 − α2
1 − (0.25)2
4α 2 (0.1)
⇒ α = 0.62
1 − α2
52. SO2 (g ) + NO2 (g ) r SO3 (g ) + NO (g )
1− x
1− x
x
x
Qc = 1 < K c , i.e. reaction proceed in forward direction to attain
equilibrium.
 x 
16 = 

 1 − x
t=0
2
∴
⇒ x = 0.80
2
1
0
x

1 − 

2
x
K p = pX2 / pX 2
[NO] = 0.80 M, [NO2 ] = 0.20 M
53. A2 (g ) + B2 (g ) r 2 AB (g )
K =
⇒
50 =
∆n = 0
(2x )2
[ AB ]2
(n )2
= AB =
[ A2 ][ B2 ] nA2 ⋅ nB 2 (1 − x ) (2 − x )
4 x2
⇒ 23x 2 − 75x + 50 = 0
x 2 − 3x + 2
752 − 4 × 23 × 50
= 0.93, 2.32
46
2.32 is not acceptable because x cannot be greater than 1.
⇒
x=
75 ±
Mole of AB = 2x = 2 × 0.93 = 1.86
54. Total moles of gases at equilibrium =
pV
2.05 × 100
=
= 5.0
RT 0.082 × 500
Out of this 5 moles, 1.0 mole is for N2 (g ) and remaining 4 moles
for PCl 5 and its dissociation products.
PCl 5 r PCl 3 + Cl 2
3−x
x
x
3+ x = 4 ⇒ x =1
1
Degree of dissociation = = 0.33
3
N2
55.
+
Initial :
1.0
Equilibrium 1 – 0.25
= 0.75
[N2 ] =
3H2
r
3.0
3 – 0.75
= 2.25
2 X (g )
(where, x = β eq )
x

1 − 

x
2

Total moles = 1 +  and Mole fraction, X 2 (g ) =

x
2

1 + 

2


 x 
X (g ) = 
 and p = 2 bar
 1+ x 

2
x

 1− 
2  . p and p = p ⋅ x
Partial pressure, pX 2 = 
X
x

 1+ x 
1 + 



2
2
At equilibrium
When p = 0.10 atm
0.26 =
X 2 (g )
At
1+ α
2α
2α
p
1+ α
= 0.468 L2 mol − 2
1
3
N2 + H2 r NH3
2
2
K ′c = K c = 0.68
56.
Total
N2O4 r 2NO2
51.
[NH3 ]2
(0.50)2
=
× 16
[N2 ][H2 ]3 (0.75) (2.25)3
=
x 


 px / 1 + 2 

=
(1 − x / 2)
p
x

1 + 

2
8 β 2eq
4 px 2
=
(4 − x 2 ) (4 − β 2eq )
4 px 2
(Q 4 > > > x)
= px 2
(4 − x 2 )
1
∴
x∝
p
If p decreases, x increases. Equilibrium is shifted in the forward
side. Thus, statement (a) is correct.
57. (a)
Kp =
(b) At the start of the reaction, Q = 0 where, Q is the reaction
quotient ∆G = ∆G ° + 2.303RT log Q
Since, ∆G ° > 0, thus ∆G is −ve.
Hence, dissociation takes place spontaneously.
Thus, (b) is correct.
4 × 2(0.7)2
(c) If we use x = 0.7 and p = 2 bar then K p =
[ 4 − (0.7)2 ]
= 1.16 > 1
Thus, (c) is incorrect.
(d) At equilibrium, ∆G = 0
∴
∆G ° = − 2.303RT log K p
Since,
∆G ° = + ve
2NH3
0
0.05
0.75
2.25
0.50
, [H2 ] =
, [NH3 ] =
4
4
4
2
Hence, K p < 1
KC =
Kp
(RT )
Then KC < 1. Thus, (d) is correct.
100 Chemical and Ionic Equilibrium
Ksp = [Cd 2 + ][ OH− ]2
Topic 2 Ionic Equilibrium
Ksp = (S )(2S )2 = 4 S 3 = 4 (184
. × 10− 5 )3
1. Given 100 mL of 0.1 M HCl is taken in beaker and to it 100 mL of
0.1 M NaOH is added.
This is acid (HCl) and base (NaOH) titration. Here,
phenolphthalein act as an indicator and colour change is pink.
The correct graph that depicts the change in pH is as follows.
At first HCl (acid) is take in beaker and base (NaOH) is taken in
burette.
When base is added drop wise then acid-base reaction is occurs
and following changes are observed.
Ksp = 24.9 × 10− 15
[Cd 2 + ] =
[Cd 2 + ] =
The expected solubility of Cd(OH)2 in a buffer solution of
pH = 12 is 2.49 × 10− 10 M.
3.
A
Volume of NaOH
pH → 0 to 7 is acidic.
pH → 7 is neutral.
pH → 7 to 14 is basic.
Initially, the graph increases steeply but when concentration of
acid becomes equal to base then no change is seen in graph.
This point is called neutral point or equivalence point.
Equivalence point, where pH = 7.
After the equivalence point (pH = 7), base (NaOH) is
continuously mixed drop by drop then pH increases by 7 and
graph also increases slowly. At the point B, solution become
basic, again base (NaOH) is mixed drop by drop and the solution
becomes basic at point B. Following inference can be seen in the
graphs.
(a) According to this graph, pH = 7 is equivalence point and use
of NaOH (base) drop by drop leads to basic solution as a
result, value of pH increases slowly. This is correct option.
(b) According to graph, it shows straight line that mean graph
increases instantly but this titration show pH increases slowly
because base (NaOH) is mix drop by drop.
(c) After equivalence point (pH = 7) graph is constant. It is
wrong because graph increases and the solution becomes
pure basic then graph show constant value.
(d) According to graph, equivalence point (pH = 7) graph should
increase but it decreases.
Key Idea The concentration of substance in a saturated
solution is defined as its solubility (S). Its value depends
upon the nature of solvent and temperature.
-
24.9 × 10− 15
= 24.9 × 10− 15 × 10 + 4
(10− 2 )2
[Cd 2 + ] ⇒ 2.49 × 10− 10 M
pH 7
Ax By
[ OH− ]2
= 24.9 × 10− 11 M
⇒
B
2.
Ksp
xA y + + yB x − Ksp = [ A y + ]x [ B x − ] y
Solubility of Cd (OH)2 (S ) = 184
. × 10− 5 M
Given, pH = 12 [for Cd (OH)2 in buffer solution]
So, pOH = 2
(QpH + pOH = pK w )
12 + pOH = 14
pOH = 14 − 12 = 2
∴ [ OH− ] = 10− 2 in buffer solution.
For reaction Cd (OH)2 → Cd 2 + + 2 OH−
S
S2
Key Idea Concentration of substance in a saturated solution is
defined as its solubility (S). Its value depends upon the nature
of solvent and temperature. For reaction,
AB
-A
+
+ B−
+
K sp = [ A ][ B − ]
Al(OH)3
Initially
At equilibrium
1
1− S
-
Al 3+ +
3OH−
0
S
0
3S + 0.2
NaOH → Na + + OH −
0. 2
0. 2
Ksp of Al(OH)3 = 2.4 × 10−24 (Given)
Ksp = [ Al3+ ][ OH− ]3
2.4 × 10−24 = [ S ][ 3S + 0.2 ]3
2.4 × 10
−24
[Q 0.2 >> S ]
= [ S ][ 0.008 ]
[ S ] = 3 × 10−22
4.
Key Idea NH4Cl is a salt of weak base (NH4OH) and strong
acid (HCl). On hydrolysis, NH4Cl will produce an acidic
solution (pH < 7) and the expression of pH of the solution is
1
pH = 7 − (pK b + logC )
2
Given, K b (NH4OH) = 10−5
∴
pK b = − log K b = − log(10−5 ) = 5
C = concentration of salt solution = 0.02 M
= 2 × 10−2 M
1
Now, pH = 7 − (pK b + logC )
2
On substituting the given values in above equation, we get
1
1
= 7 − [ 5 + log(2 × 10−2 )] = 7 − [ 5 + log 2 − 2 ]
2
2
1
= 7 − [ 5 + 0.301 − 2 ]= 7 − 165
. = 5.35
2
5. The explanation of given statements are as follows:
In statement (I), millimoles of H+ = 400 × 01
. × 2 = 80
Millimoles of OH− = 400 × 01
. = 40 (Limiting reagent)
Chemical and Ionic Equilibrium 101
∴ Millimoles of H+ left = 80 − 40 = 40
40
40
1
[ H+ ] =
=
M =
M
400 + 400 800
20
 1
pH = − log[ H+ ] = − log 
⇒
 20
Thus, the relation between molar solubility(S) and solubility
product (K sp ) will be
 K sp 
S =

 6912
= − log1 + log 2 + log10 = −0 + 0.301 + 1
⇒ 1.30
Hence, the option (a) is correct.
In statement (II), ionic product of H2O is temperature dependent.
K w = [ H+ ][ OH− ] ≈ 10−14 (mol / L)2 at 25ºC
With increase in temperature, dissociation of H2O units into H+
and OH− ions will also increase. As a result, the value of ionic
product, [H+ ] × [ OH− ]will be increased. e.g.
Temperature
K w (mol/L2)
5ºC
0186
.
× 10−14
25ºC
1008
.
× 10−14
45ºC
4.074 × 10−14
7. Let the solubility of Ag2CO3 is S. Now, 0.1 M of AgNO3 is added
to this solution after which let the solubility of Ag2CO3 becomes
S′.
∴
[Ag+ ] = S + 0.1 and [CO2−
3 ] = S′
K sp = (S + 0.1)2 (S ′ )
Given,
Ka =
8. The reaction takes place when H2SO4 is added to NH4OH is as
follows :
H2SO4
αC M
−5
αC × αC 10 × α
=
(1 − α )C
1− α
10−5 × α
1− α
⇒
α = 0.5 ⇒ α % = 50
Hence, the option (c) is correct.
In statement (IV), Le-Chatelier’s principle is applicable to
common ion effect. Because, in presence of common ion (given)
10−5 =
⇒
by strong
electrolyte
+ −
(say, Na A), the product of the
concentration terms in RHS increases. For the weaker
electrolyte, HA (say) the equilibrium shifts to the LHS,
HA
H⊕ + A s.
-
As a result dissociation of HA gets suppressed. Hence, the option
(d) is incorrect.
6.
Key Idea The concentration of a substance in a saturated
solution is defined as its solubility(S).
For Ax B y - xA y+ + yB x− ; K sp = [A y+ ] x [B x− ] y
For, Zr3 (PO4 )4,
Zr3 (PO4 )4 (s) -3Zr 4+ (aq) + 4 PO34− (aq)
3S M
K sp = [ Zr
4+ 3
]
4S M
[ PO34− ]4
1
K sp
 K sp  7
= (3S ) (4 S ) = 6912 S or S = 

 6912
3
4
7
S′ = 8 × 10− 12 × 102 = 8 × 10− 10 M
Thus, molar solubility of Ag2CO3 in 0.1 M
AgNO3 is 8 × 10− 10 M.
⇒ pH of the solution is 5, i.e.
[H+ ] = 10−5 M = αC
⇒
K sp = 8 × 10
8 × 10− 12 = 0.01 S ′
or
-
αC M
...(i)
− 12
Q K sp is very small, we neglect S′ against S in Eq. (i)
∴
K sp = (01
. )2 S ′
or
Hence, the option (b) is correct.
In statement (III), for a weak monobasic acid HA
HA
H⊕ + A s
(1 − α) C M
1/ 7
+ 2NH4OH
Strong acid
Weak base
Millimoles at t = 0 20 × 0.1 = 2
Millimoles at t = t
0
30 × 0.2 = 6
2
→
(NH4 )2 SO4 + 2H2O
Salt of strong acid
+ weak base
0
2
So, the resulting solution is a basic buffer
[NH4OH + (NH4 )2 SO4 ].
According to the Henderson’s equation,
[(NH4 )2SO4 ]
pOH = pK b + log
[NH4OH]
2
= 4.7 + log = 4.7
2
⇒
pH = 14 − pOH = 14 − 4.7 = 9.3
9. Its given that the final volume is 500 mL and this final volume
was arrived when 50 mL of 1 M Na 2SO4 was added to unknown
Ba 2+ solution.
So, we can interpret the volume of unknown Ba 2+ solution as
450 mL i.e.
450mL + 50mL → 500mL
Ba 2+
solution
Na 2SO 4
solution
BaSO 4
solution
From this we can calculate the concentration of SO2−
4 ion in the
solution via
M 1V1 = M 2V2
1 × 50 = M 2 × 500
(as 1M Na 2SO4 is taken into consideration)
1
M2 =
= 01
. M
10
Now for just precipitation,
Ionic product = Solubility product (K sp )
i.e.
[Ba 2+ ][SO24− ] = K sp of BaSO4
Given K sp of BaSO4 = 1 × 10−10
So,
[Ba 2+ ][0.1] = 1 × 10−10 or [Ba 2+ ] = 1 × 10−9 M
102 Chemical and Ionic Equilibrium
Remember This is the concentration of Ba 2+ ions in final
solution. Hence, for calculating the [Ba 2+ ] in original solution
we have to use
M 1V1 = M 2V2
as
M 1 × 450 = 10−9 × 500
M 1 = 1.1 × 10−9 M
so,
CH 3COOK is the salt of strong base and weak acid.
Hence, the solution of CH3COOK will be most basic because of
the following reaction.
CH3COOK + H2O
CH3COOH + KOH
-
12. For a salt of weak acid and weak base,
pH = 7 +
10. Key Idea Lewis acids are defined as,
‘‘Electron deficient compounds which have the ability to accept
atleast one lone pair.’’
The compound given are
PH 3-Octet complete although P has vacant 3d-orbital but does
not have the tendency to accept lone pair in it. Hence, it cannot
be considered as Lewis acid.
BCl 3-Incomplete octet with following orbital picture.
1s
2s
2p
∴
3p
3s
pH = 2 ∴ [ H+ ] = 10−2 = 0.01 M
For dilution of HCl, M 1V1 = M 2V2
0.1 × 1 = 0.01 × V2
V2 = 10 L
Volume of water to be added = 10 − 1= 9 L
14. MX :
Vacant p-orbital
3d vacant
1
1
pH = 7 + (3.2) − (3.4) = 7 + 1.6 − 1.7 = 6. 9
2
2
13. pH = 1 ∴ [ H+ ] = 10−1 = 0.1 M
B-
Hence, vacant p-orbital of B can accept one lone pair thus it can
be considered as Lewis acid.
AlCl 3-Similar condition is visible in AlCl3 as well i.e.
Al ( Valence orbital only) =
1
1
pKa − pKb
2
2
Given, pK a (HA ) = 3.2, pK a ( BOH) = 3.4
Vacant
p- orbital
Used in bond
formation with
Cl having one electron
each from B and Cl
(Strong base)
(Weak acid)
K sp = S 2 = 4 × 10− 8 ⇒ S = 2 × 10− 4
MX 2 :
K sp = 4 S 3 = 3.2 × 10− 14
⇒ S = 2 × 10− 5
M3X :
K sp = 27S 4 = 2.7 × 10− 15 ⇒ S = 10− 4
MX > M 3 X > MX 2
2
15. mmol of base = 2.5 × = 1
5
mmol of acid required to reach the end point = 1
Order of solubility is
Volume of acid required to reach the end point =
Used in bond
formation with Cl
15
+ 2.5 = 10 mL
2
1
Molarity of salt at the end point =
= 0.10
10
B+
+ H2O r BOH + H+
Total volume at the end point =
Hence this compound can also be considered as Lewis acid.
SiCl 4 - Although this compound does not have incomplete octet
but it shows the tendency to accept lone pair of electrons in its
vacant d-orbital. This tendency of SiCl 4 is visible in following
reaction.
Cl
Cl
+ H 2O
Si
Cl
Cl
Cl
Cl
C (1 − α )
H
Cl
H
⇒
Lone pair acceptance
in d-orbital
⇒
Cl
Si
⇒
OH + HCl
Cl
Cl
Thus, option (b) and (d) both appear as correct but most suitable
answer is (d) as the condition of a proper Lewis acid is more well
defined in BCl 3 and AlCl 3.
11. Among the given salts
FeCl 3 is acidic in nature i.e., have acidic solution as it is the salt
of weak base and strong acid.
Al(CN) 3 and Pb(CH 3COO) 2 are the salts of weak acid and weak
base.
Cα
Cα
Kw
= 10−2
Kb
Cα 2 0.1 α 2
=
K h = 10− 2 =
1−α 1−α
10α 2 + α − 1 = 0
− 1 + 1 + 40
α=
= 0.27
20
+
[H ] = Cα = 0.1 × 0.27 = 0.027 M
Kh =
O
Si
Cl
15
mL
2
CH3NH2 + HCl → CH3NH3+ + Cl −
16.
Initial :
Final :
0.10
0.02
0.08
0
pOH = pK b + log
0
0.08
0
0.08
[CH3NH+3 ]
[CH3NH2 ]
= − log (5 × 10− 4 ) + log
pH = 14 – pOH = 10.1
[H+ ] = 8 × 10−11
0.08
= 3.9
0.02
Chemical and Ionic Equilibrium 103
17. K h ( X − ) =
K w 10− 14
= − 5 = 10− 9 ⇒α =
Ka
10
10− 9
Kh
=
= 10− 4
0.10
C
29. When a weak acid (HX) is titrated against a strong base NaOH,
basic salt (NaX) is present at the end point which makes end
point slightly basic with pH around 8. Hence,
phenolphthalein, that changes its colour in this pH range,
would be the best choice of indicator to detect the end point.
% hydrolysis = 100 α = 0.01
18. Minimum S2− concentration would be required for precipitation
of least soluble HgS.
30. The reaction of HA with strong base is
2−
HA + OH− r H2O + A −
For HgS, S required for precipitation is
K sp
10−54
= −3 = 10−51 M
[S2− ] =
2+
[Hg ] 10
10− 4
[ A− ]
[H+ ] K
× + = a = − 14 = 1010
−
10
[HA ][OH ] [H ] K w
Kw
31. K a (HX ) =
= 10− 4
Kb
K =
19. Alkali metal salts are usually more soluble than the salts of
transition metals. Also, CuS is less soluble than ZnS because of
3d 9 configuration of Cu 2+ . Therefore, solubility order is
pH = pK a + log
Na 2S > ZnS > CuS
20. Ap Bq r pA + qB
pS
32. For precipitation reaction,
qS
21. NaCN is basic salt, has highest pH while HCl has lowest pH.
NaCl is neutral salt has pH = 7 while NH4Cl is acidic salt, has pH
less than 7.
pH : HCl < NH4Cl < NaCl < NaCN
M
22. 75 mL HCl = 15 mmol HCl
5
M
NaOH = 5 mmol NaOH
25 mL
5
After neutralisation, 10 mmol HCl will be remaining in 100 mL
of solution.
10
Molarity of HCl in the final solution =
= 0.10
100
pH = − log [H+ ] = − log (0.10) = 1
24. HClO4 is the strongest acid among these.
25. For precipitation to occur, K sp < Qsp .
 10− 4   10− 4 
Qsp = 
 = 2.5 × 10− 9 > K sp
 
 2   2 
Hence, precipitate will be formed in this case. In all other case,
Qsp < K sp and no precipitation will occur.
26. In stomach, pH is 2-3, i.e. strongly acidic and aspirin will be
almost unionised here due to common ion effect. However, pH in
small intestine is 8, basic, aspirin will be neutralised here.
33. Acidic buffer is prepared by mixing weak acid with salt of its
conjugate base. Therefore, acetic acid and sodium acetate can be
used to prepare acidic buffer.
34. The order of acidic strength of conjugate acids is
HOCl < HClO2 < HClO3 < HClO4
Reverse is the order of basic strength of their conjugate base, i.e.
ClO− is the strongest base.
35. K w = [H3O+ ][OH− ] = 10− 6 × 10− 6 = 10− 12
36. No matter, what is the concentration of HCl, its pH will always
be less than 7 at 25°C. In the present case, the solution is very
dilute, pH will be between 6 and 7.
37.
PLAN In presence of common ion (in this case Ag + ion) solubility of
sparingly soluble salt is decreased.
Let solubility of Ag2CrO4 in presence of 0.1 M
AgNO 3 = x
Ag2CrO4 a 2 Ag+ + CrO24−
2x
AgNO3 a
Cl
28. NH2− + H2O r
Base
Cl
Cl
NH3
Conjugate
acid
Be
+ OH−
Cl
Cl
Be
0.1
0.1
Total [ Ag ] = (2x + 0.1) M ≈ 0.1 M
as x <<< 0.1 M
[CrO 2−
4 ]= xM
Thus,
[Ag+ ]2 [CrO24− ] = K sp
(0.1)2 (x ) = 1.1 × 10−12
Q
Cl
x
Ag+ + NO3−
+
27. BeCl 2 exist in polymeric forms and has no electron deficiency,
not a Lewis acid.
2
= 1.25 × 10− 9 > K sp , precipitate will be formed.
23. In case of hydroxides of Group II A, solubility increases down
the group. Therefore, Be(OH)2 is least soluble, has lowest value
of K sp .
[Q [ X − ] = [HX ]]
QIP > K sp .
 10− 2   10− 3 
QIP = [Ca 2+ ][F− ]2 = 

 ×
 2   2 
K sp = ( pS )p (qS )q = S (p + q) ⋅ pp ⋅ qq
Be
[X −]
⇒ pK a = 4
[HX ]
x = 1.1 × 10−10 M
38. In HNO3 and CH3COONa combination, if HNO3 is present in
limiting amount, it will be neutralised completely, leaving
behind some excess of CH3COONa.
CH3COONa + HNO3 → CH3COOH + NaNO3
Buffer combination
104 Chemical and Ionic Equilibrium
39. CH3COOH + CH3COONa = Buffer solution
KC =
CH3COONa + HCl → CH3COOH + NaCl
If HCl is taken in limited quantity, final solution will have both
CH3COOH and CH3COONa needed for buffer solution.
Ammonia and ammonium chloride forms basic buffer.
40. pH of 10− 8 M solution will be between 6 and 7 but never 8. The
conjugate base of an acid is formed by removing a proton (H+ )
−
from acid. Therefore, HPO2−
4 is a conjugate base of H 2PO 4 .
H2O r H+ + OH−
∆H > 0
Increasing temperature will increase equilibrium constant of the
above endothermic reaction.
At the mid-point of titration
pH = pK a
41. From the given diagram, 6 mL volume of HA used till equivalene
point. At half of equivalence point, solution will be basic buffer
with B and BH+ .
[ BH+ ]
Q
pOH = pK b + log
[B]
At half equivalence point :[ BH+ ] = [ B ]
(QpH = 11)
pOH = pK b = 14 − 11 = 3
pK b = 3.00
Therefore,
Q
0.05 × [ S2− ] = 125
. × 10−22
X = Y × 10−17 = 8.9 × 10−17
Y = 8.9
44. Key Idea Solubility of salt of weak acid (AB) in presence
of H + ions from buffer solution can be calculated with the
help of following formula.
[H+ ] 
Solubility = K sp 
+ 1

 ka
Given, pH = 3, so [H+ ] = 10−3
K a = 1× 10−8
⇒ Ksp = 2 × 10−10
after putting the values in above formula
 10−3 
Solubility = 2 × 10−10  −8 + 1 ≈ 2 × 10−5 = 4.47 × 10−3M
 10

NaOH + CH 3COOH
−22
. × 10
125
⇒
⇒ 25 × 10−22 M
[ S2− ]=
0.05
For H2S , H2S → 2H+ +
S −2
Initial millimol
Final millimol
[ H+ ]2 [ S2 − ]
[ H2S]
2
1
3
+
+ H 2O
1
For Q, i.e. (20 mL of 0.1 M NaOH + 20 mL of 0.1 M
CH3COOH) is diluted to 80 mL
[ H+ ]2 × 25 × 10−22
[ 01
. ]
1
[ H+ ]2 =
25
1
[ H+ ] = ⇒0.2 M
5
1 × 10−21 =
43. Given, equilibrium constant (KC ) at 298 K = 16
. × 1017
Fe2+ (aq) + S2− (aq) 1
1
−
- CH COO Na
Final volume – 30 mL (20 + 10) in which millimoles of
CH3COOH and CH3COO −Na + are counted.
(25 × 10−22 M)
3
Given,
∴
is diluted to 60 mL
The correct match is 1, i.e. the value of[H+] does not change on
dilution due to the formation of following buffer.
K sp (ZnS ) = [ Zn 2+ ][ S2− ] = 125
. × 10−22
At initial concentration
(Before mixing)
At initial concentration
(After mixing)
At equilibrium
(Pure solid)
1
. × 1017 =
16
X × 0.07
1
X =
= 8.9 × 10−17
16
. × 1017 × 0.07
45. For P, i.e. (10 mL of 0.1 M NaOH + 20 mL of 0.1 M acetic acid)
(0.05 M)
K net = 1 × 10−21 =
[For FeS(s) = 1mol L −1]
Hence, the value of y= 4.47
42. ZnS(s) → Zn2+ (aq) + S2− (aq)
(0. 1 M)
1
X × 0.07
FeS (s)
0.06 M
0.2 M
–
0.03 M
0.1 M
–
0.03- X
0.1 – 0.03 =
0.07
–
[Here, KC >> 10 , thus limiting reagent will be consumed almost
completely, 0.03 − X = 0 ∴ X = 0.03]
From equilibrium constant,
[FeS]
KC =
[Fe2+][S2− ]
The correct match is 5, i.e. the value of [H+] changes to
2 times of its initial value on dilution.
As per the condition given in Q the resultant solution before
dilution contain 2 millimoles of CH3COO −Na+ in 40 mL
solution. Hence, it is the salt of weak acid and strong base. So,
KW K a
[H + ]initial =
C
C
After dilution to 80 mL, the new ‘C ’ becomes , So,
2
K wK a
+
+
or [H ] initial × 2
[H ]new =
C /2
For R, i.e. (20 mL of 0.1 M HCl + 20 mL of 0.1 M NH 3) is
diluted to 80 mL
1
The correct match is 4, i.e. the value of [H+] changes to
times
2
of its initial value of dilution.
Chemical and Ionic Equilibrium 105
As per the condition given in R the resultant solution before
dilution contains 2 millimoles of NH4Cl in 40 mL of solution.
Hence, a salt of strong acid and weak base is formed.
For this,
Kw × C
[H+]initial =
Kb
Now on dilution upto 80 mL new conc. becomes C /2.
C
Kw ×
+
2
So,
[H ]new =
Kb
46. I2 : I
→ I2 =
= [Ag+] × 10–3
⇒
+
[Ag ] = 1.6 × 10–7
55. Basic salts solution will have pH > 7, will change colour of
KCN, K2CO3 and LiCN are the only basic salts among these.
56. The hydrolysis reaction is
- Ni
+ 2OH
−
A − + H2O r AH + OH−
Kh =
Kw
= 10− 10
Ka
[OH− ] = K hC = 10− 6
pH = 8
and
−
57. At the end-point, [ A ] = 0.05
as per the condition given it is a sparingly soluble salt. Hence, on
dilution the concentration of OH− ions remains constant in
saturated solution.
So for this solution,
[H+]new =[H+]initial
−
Now, for AgCl, K sp = 1.6 × 10–10 = [Ag+] [Cl – ]
pOH = 6
2+
K sp (CuCl) = 10–3 M
litmus paper red to blue
1
or
[H+]new = [H + ]initial ×
2
For S , i.e. 10 mL saturated solution of Ni(OH)2 in
equilibrium with excess solidNi(OH)2 is diluted to 20 mL and
solid Ni(OH)2 is still present after dilution.
The correct match is 1.
Ni(OH)2 (s)
[ Cl – ] =
I3−
K b ( A − ) = K w / K a = 2 × 10− 9
[OH− ] = K bC = 2 × 10− 9 × 0.05 = 10− 5
pOH = 5
and
58. (i) CH3COOH r CH3COO
C (1 − α )
−
Cα
pH = 9
+ H+
Cα
47. Hydration energy facilitate solubility.
48. Amphoteric
49. SO2–
4 Conjugate base is formed by removing a proton from acid.
If no HCl is present,
50. P2O5 is strongest acid and MgO is strongest base.
The major contributor of H+ in solution is HCl.
C α (0.1)
Ka =
= 1.75 × 10− 5
C (1− α )
51. NaOH + H2O → NaOH (aq); ∆H < 0
52. Lewis acid accept lone pair of electron.
53. Degree of ionisation (α) =
Let
⇒
Also :
Λm
Λ∞
α = 1.75 × 10− 4
X
Λ m (HY ) = x ⇒ Λ m (HX ) =
10
Λ m (HX ) 1 α (HX )
=
=
[QΛ∞ (HX ) = Λ∞ (HY )]
Λ m ( XY ) 10 α (HY )
K a(HX ) = (0.01) [α( HX )]2
...(i)
K a (HY ) = (010
. ) [α( HY )]
2
= 010
. [10 α (HX )]2 = 10 [α (HX )]2
⇒
1
K a (HX ) 0.01
=
=
K a (HY ) 10 1000
⇒
log K a (HX ) − log K a (HY ) = − 3
... (ii)
⇒ − log K a (HX ) − [ − log K a (HY )] = 3
⇒
0.2
= 0.10 M
2
[CH3COOH] = 0.10 M
[HCl] =
6
× 1000 = 150
40
mmol of HCl = 500 × 0.2 = 100
mmol of CH3COOH = 500 × 0.2 = 100
After neutralisation, mmol of CH3COOH = 50
mmol of CH3COONa = 50
pH = pK a = 4.75
(ii) mmol of NaOH added =
59. Partial pressure of SO2 in air = 10− 5 atm
[SO2 ]aq = 1.3653 × 10− 5 mol L− 1
QpK a = 1.92and concentration of H2SO3 is very low, it is almost
completely ionised as
H2SO3 r H+ + HSO−3
[H+ ] = 1.3653 × 10− 5 M
pK a (HX ) − pK a (HY ) = 3
54. It is a case of simultaneous solubility of salts with a common ion.
Here, solubility product of CuCl is much greater than that of
AgCl, it can be assumed that Cl– in solution comes mainly from
CuCl.
pH = – log (1.3653 × 10− 5 ) = 4.86
60. In water,
K sp = 4 S 3 = 4 (6.7 × 10− 6 )3 = 1.2 × 10− 15
In buffer of pH = 8, pOH = 6, [OH− ] = 10−6
106 Chemical and Ionic Equilibrium
K sp = S [OH− ] 2
S =
64. K a (NH+4 ) = 5.6 × 10− 10
1.2 × 10− 15
= 1.2 × 10− 3 M
10− 12
61. (a) E = 0.164 = – 0.059 log
[Ag ] anode = 1.66 × 10
i.e. NH3 + H2O r NH4+ + OH−
k2
0.10
k
K = 1 = 1.8 × 10− 5
k2
M
[Ag+ ]
= 8.3 × 10− 5 M
2
[CrO24− ] =
k 1 = Kk 2 = 1.8 × 10−5 × 3.4 × 1010 = 6.12 × 105
65. CN− + H2O r HCN + OH−
K sp = [Ag+ ]2 [CrO24− ]
K h = 2 × 10− 5
= (1.66 × 10− 4 )2 (8.3 × 10− 5 ) = 2.3 × 10− 12
[OH− ] = K hC = 2 × 10− 5 × 0.5 = 10− 5
(b) pH of HCl = 2
∴ [ H+ ] = 10−2 M
pOH = 2.5
+
−2
10−2
=
× 200 = 2 × 10−3
1000
Similarly, pH of NaOH = 12
∴ [ H+ ] = 10−12 M
−2
or [OH ] = 10
+
pH = 7 +
[Q[H ][ OH ] = 10
M
−14
m]
Moles of OH ion in 300 mL of 10 M NaOH solution
10−2
=
× 300 = 3 × 10−3
1000
Total volume of solution after mixing = 500 mL
Moles of OH− ion left in 500 mL of solution
= (3 × 10−3 ) − (2 × 10−3 ) = 10−3
1
= 2.5 × 1018
4 × 10− 19
0.03
K = 2.5 × 1018 =
(0.04)2 x
⇒
x = 7.50 × 10− 18 M Ag+
68. For H2S, H2S r 2H+ + S2−
Minimum [S2− ] required to begin precipitation of
6 × 10− 21
= 1.2 × 10− 19
0.05
[H+ ]2 [S2− ]
(1.2 × 10− 19 )
K = 1.3 × 10− 20 =
= [H+ ] 2
[H2S]
0.10
MS =
62. 2AgCl (s) + CO23− r Ag2CO3 (s) + 2Cl −
K =
− 2
+ 2
[H+ ] = 0.10 M
2
[ K sp (AgCl)]
[Cl ]
[Cl ]
[Ag ]
=
×
=
[CO32− ] [CO23− ] [Ag+ ]2 K sp (Ag2CO3 )
0.0026
[Cl ] =
M = 7.3 × 10− 5 M
35.5
The above concentration of Cl − indicates that [CO2−
3 ] remains
almost unchanged.
2
7.3 × 10− 5 [K sp (AgCl)]
=
− 12
1.5
8.2 × 10
−
K sp (AgCl) = 2 × 10− 8
63. pH = pK In + log 10 = pK In + 1
= pK In + log (0.1) = pK In
pH range is pK In
−1
to pK In
−1
+ 1.
[In − ]
= 10
[HIn]
[In − ]
When
= 0.1
[HIn]
When
0
0.03
K =
pH = 14 − 2.699 = 11.301
− 2
0.10
0.10 – 0.06
K = K 1 × K 2 = 1.3 × 10− 20
Molar concentration of OH− ions in the resulting
10−3
solution =
× 1000 = 2 × 10−3 M
500
pOH = − log (2 × 10−3 )
~ 103 = 2.699
= − log 2 + 3 log10 = − 0.3 −
∴
1
1
(pK a − pK b ) = 7 + (3.8 − 4.8) = 6.50
2
2
Initial :
0.03
Equilibrium : x
−2
−
pH = 11.5
Ag+ + 2CN− r Ag(CN)−2
67.
−
and
66. For salts of weak acid and weak base .
Moles of H ions in 200 mL of 10 M HCl solution
−
10− 14
= 1.8 × 10− 5
5.6 × 10− 10
k1
[Ag+ ] anode
−4
+
K b (NH3 ) = K w / K a =
⇒ pH = 1
69. Mixing H2CO3 with NaHCO3 results in buffer solution.
pH = pK a + log
[NaHCO3 ]
n (NaHCO3 )
= pK a + log
[H2CO3 ]
n (H2CO3 )
7.4 = – log (7.8 × 10− 7 ) + log
⇒
⇒
x
20
x = 400 mmol
NaHCO3 = 5 × V
3
⇒ V = 80 mL
−5
70. K sp = 4 S = 4.42 × 10
S = 0.022 M
mmol of Ca(OH)2 in 500 mL saturated solution = 11
mmol of NaOH in 500 mL 0.40 M solution = 200
Total mmol of OH− = 200 + 2 × 11 = 222
[OH− ] = 0.222 M
Chemical and Ionic Equilibrium 107
=
Also, desired pH = 2 × 2.37 = 4.74
K sp
Solubility in presence of NaOH =
[OH− ]2
[H+ ] = 1.8 × 10− 5 = x α
1.8 × 10− 5 α
1−α
α = 0.5 and x = 3.6 × 10− 5 M
4.42 × 10− 5
= 9 × 10− 4 M
(0.222)2
mmol of Ca 2+ remaining in solution = 0.9
mmol of Ca(OH)2 precipitated = 10.1
mg of Ca(OH)2 precipitated = 10.1 × 7.4 = 747.4 mg
K a = 1.8 × 10− 5 =
Volume (final) = 1/3.6 × 10− 5 = 27.78 × 103 L.
74. pOH of buffer solution = pK b + log
71. Let 40 mL of base contain x mmol of BOH.
BOH + HCl → BCl + H2O
x − 0.5
x− 2
Subtracting Eq. (i) from Eq. (ii) gives
 2
x − 0.5
0.90 = log 
×

x − 2
0.5 
⇒
= − log (1.8 × 10− 5 ) + log
0.5 When 5 mL acid is added
2.0 When 20 mL of acid is added
When pH is 10.04, pOH = 3.96 and when pH is 9.14, pOH is
4.86. Therefore,
0.50
3.96 = pK b + log
…(i)
x − 0.5
2.0
…(ii)
3.96 = pK b + log
x− 2
4 (x − 0.5)
⇒ 28 =
x−2
x = 3.5, substituting in equation (i) gives
0.5
3.96 = pK b + log
3
K b = 1.8 × 10− 5
72. Initial concentration of K2C2O4 =
[Al 3+ ] =
[Mg2+ ] =
x
K sp (Ag2CO3 ) = K × K sp (Ag2C2O4 )
= 7.5 × 1.29 × 10− 11 = 9.675 × 10− 11
73. CH3COOH r CH3COO− + H+
When concentration of CH3COOH is 1.0 M, ‘α’ is negligible,
[H+ ] = K aC = 4.24 × 10− 3 M
pH = − log (4.24 × 10− 3 ) = 2.37
Now, let us assume that solution is diluted to a volume where
concentration of CH3COOH (without considering ionisation)
is x.
CH3COOH r CH3COO− + H+
xα
Ka =
1−α
xα
[OH ]
K sp
− 2
[OH ]
=
6 × 10− 32
= 1.28 × 10− 15 M
(3.6 × 10− 6 )3
=
8.9 × 10− 12
= 0.68 M
(3.6 × 10− 6 )2
NaCN + HCl → NaCl + HCN
0.01
mmol of NaCN present initially =
× 1000 = 0.2
49
8.5 = − log (4.1 × 10− 10 ) + log
0.2 − x
x
x = 0.177 mmol
76. (i) 0.20 mole HCl will neutralise 0.20 mole CH3COONa,
producing 0.20 mol CH3COOH. Therefore, in the solution
moles of CH3COOH = 1.20
Moles of CH3COONa = 0.80
[Salt]
pH = pK a + log
[Acid]
(0.80)
= 4.56
= − log (1.8 × 10− 5 ) + log
(1.20)
Given, 0.304 – x = 0.0358 ⇒ x = 0.2682
0.2682
⇒
K =
= 7.5
0.0358
xα
− 3
75. HCN for buffer will be formed by the reaction
[CO23− ] [Ag+ ]2 K sp (Ag2CO3 )
×
=
K =
[C2O24− ] [Ag+ ]2 K sp (Ag2C2O4 )
2
K sp
Let x mmol of HCl is added so that x mmol of NaCN will be
neutralised forming x mmol of HCN.
[NaCN]
pH = pK a + log
[HCN]
Also for the following equilibrium:
Ag2CO3 (s) + K2C2O4 (aq) r Ag2C2O4 (s) + K2CO3
x (1 − α )
0.25
= 5.44
0.05
[OH− ] = 3.6 × 10− 6 M
0.152
= 0.304 M,
0.50
0.304 − x
[NH+4 ]
[NH4OH]
CH3COONa + HCl → CH3COOH + NaCl
(ii)
Initial
Final
0.10
0
0.20
0.10
0
0.10
0
0.10
Now, the solution has 0.2 mole acetic acid and 0.1 mole HCl.
Due to presence of HCl, ionisation of CH3COOH can be
ignored (common ion effect) and H+ in solution is mainly due
to HCl.
[H+ ] = 0.10
pH = – log (0.10) = 1.0
77. In pure water, solubility =
9.57
× 10− 3 M = 1.65 × 10− 4 M
58
K sp = 4 S 3 = 4 (1.65 × 10− 4 )3 = 1.8 × 10− 11
In 0.02 M Mg(NO3 )2;
108 Chemical and Ionic Equilibrium
solubility of Mg(OH)2 =
K sp
2+
[Mg ]
×
pH = 4.15
1
2
[OH− ] =
= 1.5 × 10− 5 mol L− 1
= 1.43 × 10− 10 M
= 1.5 × 10− 5 × 58 g L− 1
= 8.7 × 10− 4 g L− 1
81. Sodium acetate (CH3COONa) is a basic salt (salt of strong base
and weak acid) therefore, its aqueous solution has pH > 7.
78. HCOOH r H+ + HCOO−
82. mmol of NaOH = 20 × 0.2 = 4
HCOONa r Na + + HCOO−
1 − 0.75
mmol of acetic acid = 50 × 0.2 = 10
0.75
[H+ ] (0.75)
[HCOOH]
After neutralisation, buffer solution is formed which contain 6
mmol CH3COOH and 4 mmol CH3COONa.
[CH3COONa]
pH = pK a + log
[CH3COOH]
4
= − log (1.8 × 10− 5 ) + log = 4.56
6
2.4 × 10− 4 × 0.20
= 6.4 × 10− 5
0.75
Now, let x mmol of NaOH is further added so that pH of the
resulting buffer solution is 4.74.
In the above buffer solution, the significant source of formate ion
(HCOO− ) is HCOONa. Hence,
K a = 2.4 × 10− 4
=
[H+ ] =
Now, the buffer solution contains (4 + x ) mmol CH3COONa and
(6 − x ) mmol of CH3COOH.
4+x
4.74 = − log (1.8 × 10− 5 ) + log
6−x
pH = – log (6.4 × 10−5 ) = 4.20
79. K sp (AgI) = 8.5 × 10− 17 = [Ag+ ][I− ]
[I− ] required to start precipitation of AgI
8.5 × 10− 17
= 8.5 × 10− 16 M
0.10
(HgI2 ) = 2.5 × 10− 26 = [Hg2+ ][I− ]2
=
K sp
Kw
10− 14
=
+
[H ] 7 × 10− 5
⇒
⇒
−
[I ] required to start precipitation of HgI2
=
⇒
2.5 × 10− 26
= 5 × 10− 13 M
0.10
−
The above calculation indicates that lower [I ] is required for
precipitation of AgI. When [I− ] reaches to 5 × 10− 13 , AgI gets
precipitated almost completely.
When HgI2 starts precipitating,
8.5 × 10− 17
= 1.70 × 10− 4 M
[Ag+ ] =
5 × 10− 13
1.70 × 10− 4 × 100
= 0.17
0.10
% Ag+ precipitated = 100 – 0.17 = 99.83
% Ag+ remaining =
80. Molarity (C ) = 0.10
[H+ ] = K a ⋅ C = 7 × 10− 5 M (α is negligible)
4+x
=1
6−x
x = 1.0 mmol = 0.2 × V
V = 5.0 mmol NaOH.
83. For acidic buffer, the Henderson’s equation is
pH = pK a + log
(mole of salt)
(mole of acid)
4.75 = – log (1.34 × 10− 5 ) + log
x
0.02
⇒ x = 0.015 mole of sodium propionate.
Addition of 0.01 mole HCl will increase moles of propionic acid
by 0.01 and moles of sodium propionate will decrease by same
amount.
New moles of acid = 0.02 + 0.01 = 0.03
New moles of salt = 0.015 – 0.01 = 0.005
 0.005
pH = – log (1.34 × 10− 5) + log 
 = 4.09
 0.030
pH of 0.01 HCl = 2, just half of the pH of final buffer solution.
7
Thermodynamics and
Thermochemistry
Topic 1 Thermodynamics
Objective Questions I (Only one correct option)
1. An ideal gas is allowed to expand from 1 L to 10 L against a
constant external pressure of 1 bar. The work done in kJ is
(2019 Main, 12 April I)
(a) − 9.0
(b) + 10.0
(c) − 0.9
(d) − 2.0
2. The difference between ∆H and ∆U ( ∆H − ∆U ), when the
combustion of one mole of heptane (l) is carried out at a
temperature T, is equal to
(2019 Main, 10 April II)
(a) − 4 RT (b) 3 RT
(c) 4 RT
(d) − 3 RT
3. A process will be spontaneous at all temperature if
(2019 Main, 10 April I)
(a) ∆H > 0 and ∆S < 0
(c) ∆H < 0 and ∆S < 0
(b) ∆H < 0 and ∆S > 0
(d) ∆H > 0 and ∆S > 0
(a)
(b)
(c)
(d)
8. For
Cyclic process : q = − W
Adiabatic process : ∆U = − W
Isochoric process : ∆U = q
Isothermal process : q = − W
C p ( J K −1 mol −1 ) = 23 + 0.01 T.
silver,
If
the
temperature (T) of 3 moles of silver is raised from 300 K to
1000 K at 1 atm pressure, the value of ∆H will be close to
(2019 Main, 8 April I)
(a) 62 kJ
(c) 21 kJ
(b) 16 kJ
(d) 13 kJ
9. For a diatomic ideal gas in a closed system, which of the
following plots does not correctly describe the relation
between various thermodynamic quantities?
(2019 Main, 12 Jan I)
4. During compression of a spring the work done is 10 kJ and 2
kJ escaped to the surroundings as heat. The change in
internal energy, ∆U (in kJ) is
(2019 Main, 9 April II)
(a) 8
(b) −12
(c) 12
(d) −8
(a)
Cp
(C) W
(D) H − TS
(b) (A), (B) and (C)
(d) (B) and (C)
reversible compression till its temperature becomes 200 K.
If CV = 28 JK −1 mol −1 , calculate ∆U and ∆pV for this
process. ( R = 8.0 JK −1 mol−1 )
∆U
∆U
∆U
∆U
T
(2019 Main, 9 April I)
6. 5 moles of an ideal gas at 100 K are allowed to undergo
(a)
(b)
(c)
(d)
Cv
p
5. Among the following the set of parameters that represents
path functions, is
(A) q + W (B) q
(a) (A) and (D)
(c) (B), (C) and (D)
(b)
(2019 Main, 8 April II)
= 2.8 kJ; ∆ ( pV ) = 0.8 kJ
= 14 J; ∆ ( pV ) = 0.8 J
= 14 kJ; ∆ ( pV ) = 4 kJ
= 14 kJ; ∆ ( pV ) = 18 kJ
7. Which one of the following equations does not correctly
represent the first law of thermodynamics for the given
processes involving an ideal gas ? (Assume non-expansion
work is zero)
(2019 Main, 8 April I)
(c)
(d)
U
Cv
T
V
10. The standard electrode potential E O− and its temperature
−
 dE O

−4
−1
coefficient 
 for a cell are 2V and − 5 × 10 VK at
 dT 
300 K respectively. The cell reaction is
Zn( s ) + Cu2 + ( aq ) → Zn2 + ( aq ) + Cu( s )
−
The standard reaction enthalpy (∆ r H O ) at 300 K in kJ mol −1
is,
[Use, R = 8 JK −1 mol−1 and F = 96,000 C mol−1 ]
(2019 Main, 12 Jan I)
(a) − 412.8
(c) 206.4
(b) − 384.0
(d) 192.0
110 Thermodynamics and Thermochemistry
11. The reaction, MgO( s ) + C( s ) → Mg ( s ) + CO( g ),
for which ∆ r H º = + 491.1 kJ mol
−1
of ice at 273 K to water vapours at 383 K is
and
∆ r S º = 198.0 JK − 1 mol − 1 , is not feasible at 298 K.
Temperature above which reaction will be feasible is
(a) 2040.5 K
(b) 1890.0 K
(c) 2380.5 K
(d) 2480.3 K
12. The standard reaction Gibbs energy for a chemical reaction
at an absolute temperature T is given by, ∆ r G º = A − BT
Where A and B are non-zero constants.
Which of the following is true about this reaction?
(2019 Main, 11 Jan II)
(a)
(b)
(c)
(d)
Endothermic if, A < 0 and B > 0
Exothermic if, B < 0
Exothermic if, A > 0 and B < 0
Endothermic if, A > 0
13. For the chemical reaction, X
18 The entropy change associated with the conversion of 1 kg
(Specific heat of water liquid and water vapour are 4.2 kJ
K −1kg−1 and 2.0 kJK −1 kg−1; heat of liquid fusion and
vapourisation of water are 334 kJ kg−1 and 2491 kJkg−1
respectively).
(log 273 = 2.436,
log 373 = 2.572,
(2019 Main, 9 Jan II)
log 383 = 2.583 )
(a) 9.26 kJ kg −1 K −1
(b) 8.49 kJ kg −1 K −1
(c) 7.90 kJ kg −1 K −1
19 Consider the reversible isothermal expansion of an ideal gas
in a closed system at two different temperatures T1 and T2
(T1 < T2 ). The correct graphical depiction of the dependence
of work done (W) on the final volume (V) is
(2019 Main, 9 Jan I)
- Y , the standard reaction
Gibbs energy depends on temperature T (in K) as
3
∆ rG ° (in kJ mol–1 ) = 120 − T
8
The major component of the reaction mixture at T is
T2
|W|
|W|
(b) X if T = 350 K
(d) Y if T = 300 K
(b)
O
T + T2 
(d) 2 C p ln 1

 2T1T2 
15. The process with negative entropy change is
(2019 Main, 10 Jan II)
(a) synthesis of ammonia from N 2 and H 2
(b) dissociation of CaSO4 ( s ) to CaO( s ) and SO3 ( g )
(c) dissolution of iodine in water
(d) sublimation of dry ice
T2
|W|
−2
17 A process has ∆H = 200 J mol −1 and ∆S = 40 JK −1 mol −1 .
Out of the values given below, choose the minimum
temperature above which the process will be spontaneous
(2019 Main, 10 Jan I)
(b) 4 K
(c) 5 K
T2
T1
T1
(c)
(d)
O
ln V
O
(d) 12 K
ln V
20. The combustion of benzene ( l ) gives CO2 ( g ) and H2 O( l ).
Given that heat of combustion of benzene at constant volume
is −3263.9 kJ mol −1 at 25° C; heat of combustion (in kJ
mol −1 ) of benzene at constant pressure will be
(R = 8.314 JK −1 mol −1 )
(2018 Main)
(b) −452.46
(d) −3267.6
(a) 4152.6
(c) 3260
(a) isochoric work
(c) adiabatic work
to 1 m against a constant external pressure of 4 Nm . Heat
released in this process is used to increase the temperature
of 1 mole of Al. If molar heat capacity of Al is 24 J mol−1 K −1 ,
the temperature of Al increases by
(2019 Main, 10 Jan II)
3
2
(a) K
(b) 1K
(c) 2 K
(d) K
2
3
(a) 20 K
ln V
21. ∆U is equal to
16. An ideal gas undergoes isothermal compression from 5 m 3
3
O
ln V
14. Two blocks of the same metal having same mass and at
 (T + T2 )2 
(c) C p ln 1

 4T1T2 
T1
(a)
|W|
temperature T1 and T2 respectively, are brought in contact
with each other and allowed to attain thermal equilibrium at
constant pressure. The change in entropy, ∆S , for this
process is
(2019 Main, 11 Jan I)
1/ 2 


(
T
T
)
+
T + T2 
2
(b) 2 C p ln 1
(a) 2 C p ln 1


T1T2
 4T1T2 


T2
T1
(2019 Main, 11 Jan I)
(a) Y if T = 280 K
(c) X if T = 315 K
(d) 2.64 kJ kg −1 K −1
(2017 Main)
(b) isobaric work
(d) isothermal work
22. The standard state Gibbs free energies of formation of
C(graphite) and C(diamond) at T = 298 K are
∆ f G ° [C(graphite)] = 0 kJ mol −1
∆ f G ° [C(diamond)] = 2.9 kJ mol −1
The standard state means that the pressure should be 1 bar,
and substance should be pure at a given temperature. The
conversion of graphite [C(graphite)] to diamond
[C(diamond)] reduces its volume by 2 × 10−6 m 3 mol −1 . If
C(graphite) is converted to C(diamond) isothermally at
T = 298K, the pressure at which C(graphite) is in
equilibrium with C(diamond), is
(2017 Adv.)
Thermodynamics and Thermochemistry 111
[Useful information : 1 J = 1kg m 2 s −2 ,
1 Pa = 1kg m
(a) 58001 bar
(c) 14501 bar
−1
−2
∆S ( A → C ) = 50 eu
∆S (C → D ) = 30 eu
∆S (D → B ) = –20 eu where, eu is entropy unit
Given that
5
s ; 1 bar = 10 Pa]
(b) 1450 bar
(d) 29001 bar
23. One mole of an ideal gas at 300 K in thermal contact with
surroundings expands isothermally from 1.0 L to 2.0 L
against a constant pressure of 3.0 atm.
In this process, the change in entropy of surroundings
( ∆S surr ) in JK −1 is (1 L atm = 101.3 J)
(2016 Adv.)
(a) 5.763
(c) − 1.013
(b) 1.013
(d) − 5.763
24. The following reaction is performed at 298K
2NO( g ) + O2 ( g ) r 2NO2 ( g )
The standard free energy of formation of NO(g) is
86.6 kJ/mol at 298 K. What is the standard free energy of
formation of NO2 ( g ) at 298 K? ( K p = 1.6 × 1012 ) (2015 Main)
(a) R( 298 ) ln (1. 6 × 1012 ) – 86600
(b) 86600 + R( 298 ) ln (1. 6 × 1012 )
(c) 86600 –
ln (1. 6 × 1012 )
R ( 298 )
(2006, 3M)
(b) +60 eu
(d) –60 eu
30. A monoatomic ideal gas undergoes a process in which the
ratio of p to V at any instant is constant and equals to 1. What
is the molar heat capacity of the gas ?
(2006, 3M)
4R
3R
5R
(a)
(b)
(c)
(d) 0
2
2
2
31. One mole of monoatomic ideal gas expands adiabatically at
initial temperature T against a constant external pressure of
1 atm from 1 L to 2 L. Find out the final temperature
(R = 0.0821 L atm K − 1 mol − 1 )
(2005, 1M)
T
(a) T
(b)
( 2 )5 / 3 − 1
2
2
(c) T −
(d) T +
3 × 0.082
3 × 0.082
32. 2 moles of an ideal gas expanded isothermally and
(d) 0. 5 [ 2 × 86600 – R ( 298 ) ln (1. 6 × 1012 )]
25. For the process, H2 O ( l ) → H2 O ( g )
at T = 100° C and 1 atmosphere pressure, the correct choice is
(a) ∆S system > 0 and ∆S surrounding > 0
(2014 Adv.)
(b) ∆S system > 0 and ∆S surrounding < 0
(c) ∆S system < 0 and ∆S surrounding > 0
(d) ∆S system < 0 and ∆S surrounding < 0
26. A piston filled with 0.04 mole of an ideal gas expands
reversibly from 50.0 mL to 375 mL at a constant temperature
of 37.0°C. As it does so, it absorbs 208 J of heat. The values
of q and W for the process will be
(2013 Main)
(R = 8.314 J / mol K, ln 7.5 = 2.01)
(a) q = + 208 J, W = − 208 J
(b) q = − 208 J, W = − 208 J
(c) q = − 208 J, W = + 208 J
(d) q = + 208 J, W = + 208 J
27. For the process H2 O( l ) (1 bar, 373 K) → H2 O( g )
(1 bar, 373 K), the correct set of thermodynamic
parameters is
(2007, 3M)
(a) ∆G = 0, ∆S = + ve
(b) ∆G = 0, ∆S = − ve
(c) ∆G = + ve, ∆S = 0
(d) ∆G = − ve, ∆S = + ve
28. The value of log 10 K for a reaction A r B is
(Given : ∆ r H ° 298 K = − 54.07 kJ mol −1 ,
∆ r S ° 298 K = 10 JK −1 mol −1 and R = 8.314 JK−1 mol−1 ;
2. 303 × 8. 314 × 298 = 5705)
(a) 5
(b) 10
(c) 95
(2007, 3M)
(d) 100
29. The direct conversion of A to B is difficult, hence it is carried
out by the following shown path
C → D
↑
↓
A
Then, ∆S ( A → B ) is
(a) + 100 eu
(c) –100 eu
B
reversibly from 1 L to 10 L at 300 K. What is the enthalpy
change?
(2004, 1M)
(a) 4.98 kJ (b) 11.47 kJ (c) –11.47 kJ (d) 0 kJ
33. Spontaneous adsorption of a gas on solid surface is an
exothermic process because
(2004, 1M)
(a) ∆H increases for system (b) ∆S increases for gas
(c) ∆S decreases for gas
(d) ∆G increases for gas
34. One mole of a non-ideal gas undergoes a change of state
(2.0 atm, 3.0 L, 95 K) → (4.0 atm, 5.0 L, 245 K) with a
change in internal energy, ∆E = 30.0 L-atm. The change in
enthalpy ( ∆H ) of the process in L-atm is
(2002, 3M)
(a) 40.0
(b) 42.0
(c) 44.0
(d) not defined, because pressure is not constant
35. Which of the following statements is false?
(2001, 1M)
(a) Work is a state function
(b) Temperature is a state function
(c) Change in the state is completely defined when the
initial and final states are specified
(d) Work appears at the boundary of the system
36. In thermodynamics, a process is called reversible when
(2001, 1M)
(a) surroundings and system change into each other
(b) there is no boundary between system and surroundings
(c) the surroundings are always in equilibrium with the system
(d) the system changes into the surroundings spontaneously
37. For an endothermic reaction, where ∆H represents the
enthalpy of the reaction in kJ/mol, the minimum value for
the energy of activation will be
(1992, 1M)
(a) less than ∆H
(b) zero
(c) more than ∆H
(d) equal to ∆H
112 Thermodynamics and Thermochemistry
38. The difference between heats of reaction at constant pressure
and constant volume for the reaction
2C6 H6 ( l ) + 15O2 → 12CO2 ( g ) + 6H2 O( l )
at 25° C in kJ is
(a) − 7.43
(c) − 3.72
42. An ideal gas is expanded form ( p1 , V1 , T1 ) to ( p2 , V2 , T2 )
under different conditions. The correct statement(s)
among the following is (are)
(2017 Adv.)
(1991, 1M)
(b) + 3.72
(d) + 7.43
Objective Questions II
(One or more than one correct option)
39. In thermodynamics, the p-V work done is given by
w = − ∫ dV pext
For a system undergoing a particular process, the work done is
 RT
a
− 2
w = − ∫ dV 
V − b V 
(a) The work done by the gas is less when it is expanded
reversibly from V1 to V2 under adiabatic conditions as
compared to that when expanded reversibly form V1 to V2
under isothermal conditions.
(b) The change in internal energy of the gas is (i) zero, if it is
expanded reversibly with T1 = T2, and (ii) positive, if it is
expanded reversibly under adiabatic conditions with T1 ≠ T2
(c) If the expansion is carried out freely, it is simultaneously
both isothermal as well as adiabatic
(d) The work done on the gas is maximum when it is
compressed irrversibly from ( p2 , V2 ) to ( p1, V1 ) against
constant pressure p1
43. For a reaction taking place in a container in equilibrium
This equation is applicable to a
(2020 Adv.)
(a) system that satisfies the van der Waals’ equation of state
(b) process that is reversible and isothermal
(c) process that is reversible and adiabatic
(d) process that is irreversible and at constant pressure
40. Choose the reaction(s) from the following options, for which
the standard enthalpy of reaction is equal to the standard
enthalpy of formation.
(2019 Adv.)
(a) 2C( g ) + 3 H2 ( g ) → C2 H6 ( g )
(b) 2H2 ( g ) + O2 ( g ) → 2H2 O( l )
3
(c) O2 ( g ) → O3 ( g )
2
1
(d) S8 ( s ) + O2 ( g ) → SO2 ( g )
8
with its surroundings, the effect of temperature on its
equilibrium constant K in terms of change in entropy is
described by
(2017 Adv.)
(a) With increase in temperature, the value of K for
endothermic reaction increases because unfavourable change
in entropy of the surroundings decreases
(b) With increase in temperature, the value of K for exothermic
reaction decreases because favourable change in entropy of
the surrounding decreases
(c) With increase in temperature, the value of K for
endothermic reaction increases because the entropy change
of the system is negative
(d) With increase in temperature, the value of K for exothermic
reaction decreases because the entropy change of the system
is positive
44. An ideal gas in thermally insulated vessel at internal
41. A reversible cyclic process for an ideal gas is shown below.
Here, p , V and T are pressure, volume and temperature,
respectively. The thermodynamic parameters q , w , H and U
are heat, work, enthalpy and internal energy, respectively.
pressure = p1 , volume = V1 and absolute temperature = T1
expands irreversibly against zero external pressure, as
shown in the diagram. The final internal pressure,
volume and absolute temperature of the gas are p2 , V2
and T2 , respectively. For this expansion
(2018 Adv.)
Volume (V)
pext=0
A(p1, V1, T1)
C ( p 2 , V 1 , T2 )
B(p2, V2, T1)
Irreversible
pext=0
p1,V1,T1
p2,V2,T2
Thermal insulation
Temperature (T)
The correct options is (are)
(a) q AC = ∆U BC and w AB = p2 (V2 − V1 )
(b) wBC = p2 (V2 − V1 ) and qBC = ∆H AC
(c) ∆HCA < ∆UCA and q AC = ∆U BC
(d) qBC = ∆H AC and ∆HCA > ∆UCA
(2014 Adv.)
(a) q = 0
(b) T2 = T1
(c) p2V2 = p1V1
(d) p2V2γ = p1V1γ
45. Benzene and naphthalene form an ideal solution at room
temperature. For this process, the true statement(s) is (are)
(2013 Adv.)
(a) ∆G is positive
(c) ∆S surroundings = 0
(b) ∆S system is positive
(d) ∆H = 0
Thermodynamics and Thermochemistry 113
46. The reversible expansion ob an ideal gas under adiabatic
and isothermal conditions is shown in the figure. Which of
the following statement(s) is (are) correct?
(2012)
What is the ratio of the standard Gibbs energy of the reaction
at 1000 K to that at 2000 K?
(2020 Adv.)
Partial pressure of B (bar)
115
(p1,V1,T1)
Iso
the
rma
l
ia
ba
tic
p
Ad
(p2,V2,T2)
(p3,V3,T3)
V
(a) T1 = T2
(b) T3 > T1
(c) Wisothermal > Wadiabatic
(d) ∆U isothermal > ∆U adiabatic
t′
p(atmosphere)
initial state X to the final state Z. The final state Z can be
reached by either of the two paths shown in the figure.
Y
X
Z
V(litre)
[Take ∆S as change in entropy and W as work done].
Which of the following choice(s) is (are) correct?
(a) ∆S X → Z = ∆S X → Y + ∆S Y → Z
(b) W X → Z = W X → Y + WY → Z
(c) W X → Y → Z = W X → Y
(d) ∆S X → Y → Z = ∆S X → Y
(2012)
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
52. Statement I There is a natural asymmetry between
converting work to heat and converting heat to work.
Statement II No process is possible in which the sole result
is the absorption of heat from a reservoir and its complete
conversion into work.
(2008, 3M)
standard Gibbs energy of reaction is zero.
(2010)
(b) electromotive force
(d) heat capacity
Statement II At constant temperature and pressure,
chemical reactions are spontaneous in the direction of
decreasing Gibbs energy.
(2008, 3M)
54. Statement I The heat absorbed during the isothermal
49. Among the following, the state function(s) is(are)
(a) internal energy
(b) irreversible expansion work
(c) reversible expansion work
(d) molar enthalpy
(2009)
expansion of an ideal gas against vacuum is zero.
Statement II The volume occupied by the molecules of an
ideal gas is zero.
(2000, S, 1M)
Passage Based Questions
50. Identify the intensive quantities from the following.
(1993, 1M)
(b) temperature
(d) refractive index
A fixed mass m of a gas is
subjected to transformation of
states from K to L to M to N and
back to K as shown in the figure.
(2013 Adv.)
Numerical Answer Type Questions
51. Consider the reaction, A
B at 1000 K. At time t’, the
temperature of the system was increased to 2000 K and the
system was allowed to reach equilibrium. Throughout this
experiment the partial pressure of A was maintained at 1 bar.
Given, below is the plot of the partial pressure of B with time.
q
Time
53. Statement I For every chemical reaction at equilibrium,
48. Among the following, extensive property is (properties are)
(a) enthalpy
(c) volume
50
10
47. For an ideal gas, consider only p -V work in going from
(a) molar conductivity
(c) resistance
100
K
L
N
M
Pressure
Volume
55. The pair of isochoric processes among the transformation of
states is
(a) K to L and L to M
(c) L to M and M to N
(b) L to M and N to K
(d) M to N and N to K
114 Thermodynamics and Thermochemistry
56. The succeeding operations that enable this transformation of
states are
(a) heating, cooling, heating, cooling
(b) cooling, heating, cooling, heating
(c) heating, cooling, cooling, heating
(d) cooling, heating, heating, cooling
67. One mole of an ideal gas is taken from a to b along two paths
denoted by the solid and the dashed lines as shown in the
graph below. If the work done along the solid line path is Ws
and that along the dotted line path is Wd , then the integer
(2010)
closest to the ratio Wd / Ws is
4.5
Match the Columns
4.0
57. Match the thermodynamic processes given under Column I
with the expressions given under Column II.
Column I
Column II
q=0
B.
Expansion of 1 mole of an ideal gas into
a vacuum under isolated conditions
q.
W =0
Mixing of equal volumes of two ideal
gases at constant temperature and
pressure in an isolated container
r.
D. Reversible heating of H2 ( g ) at 1 atm
from 300 K to 600 K, followed by
reversible cooling to 300 K at 1 atm
s.
∆S sys < 0
options in Column II.
Column I
V (L)
∆G = 0
Column II
p. Phase transition
r. ∆H is positive
D. P(white, solid) → P(red, solid)
s.
∆S is positive
t.
∆S is negative
Fill in the Blanks
59. Enthalpy is an ............... property.
(1997, 1M)
60. When Fe(s) is dissolved in aqueous hydrochloric acid in a
closed vessel, the work done is ................ .
(1997)
61. The heat content of the products is more than that of the
(1993, 1M)
62. A system is said to be ........ if it can neither exchange matter
(1993, 1M)
(1984, 1M)
64. The total energy of one mole of an ideal monatomic gas at
(1984, 1M)
True/False
65. First law of thermodynamics is not adequate in predicting
(1982, 1M)
66. Heat capacity of a diatomic gas is higher than that of a
monoatomic gas.
0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0
moles of CO and one mole of O2 are taken in a container of
volume 1 L. They completely form two moles of CO2 , the
gases deviate appreciably from ideal behaviour. If the
pressure in the vessel changes from 70 to 40 atm, find the
magnitude (absolute value) of ∆U at 500 K. (1 L-atm = 0.1 kJ)
C. 2H• → H2 ( g )
the direction of a process.
b
68. For the reaction, 2CO + O2 → 2CO2 ; ∆H = − 560 kJ. Two
B. CaCO3 ( s ) → CaO( s ) + CO2 ( g ) q. Allotropic change
27° C is ..............cal.
0.5
∆U = 0
(2011)
63. C p − CV for an ideal gas is .................
1.0
Subjective Questions
58. Match the transformations in Column I with appropriate
nor energy with the surroundings.
2.0
0.0
t.
reactants in an .............. reaction.
p 2.5
(atm)
1.5
p.
A. CO2 ( s ) → CO2 ( g )
a
3.5
3.0
A. Freezing of water at 273 K and 1 atm
C.
Integer Answer Type Question
(1985, 1/2 M)
(2006, 3M)
69. 100 mL of a liquid contained in an isolated container at a
pressure of 1 bar. The pressure is steeply increased to
100 bar. The volume of the liquid is decreased by 1 mL at
this constant pressure. Find the ∆H and ∆U .
(2004, 2M)
3R
3R
70. CV value of He is always
but CV value of H2 is
at low
2
2
5R
at moderate temperature and more than
temperature and
2
5R
at higher temperature. Explain in two or three lines.
2
(2003, 2M)
71. Two moles of a perfect gas undergo the following processes :
(a) a reversible isobaric expansion from (1.0 atm, 20.0 L)
to (1.0 atm, 40.0 L)
(b) a reversible isochoric change of state from (1.0 atm,
40.0 L) to (0.5 atm, 40.0 L)
(c) a reversible isothermal compression from (0.5 atm,
40.0 L) to (1.0 atm, 20.0 L)
(i) Sketch with labels each of the processes on the same
p-V diagram.
(ii) Calculate the total work (W) and the total heat change
(Q) involved in the above processes.
(iii) What will be the values of ∆ U, ∆H and ∆S for the
overall process?
(2002, 5M)
72. When 1-pentyne (A) is treated with 4 N alcoholic KOH at
175° C, it is converted slowly into an equilibrium mixture of
1.3% 1-pentyne (A), 95.2% 2-pentyne (B) and 3.5% of
Thermodynamics and Thermochemistry 115
1, 2-pentadiene (C). The equilibrium was maintained at
175° C. Calculate ∆G° for the following equilibria.
A,
∆G°1 = ?
B r C,
∆G° 2 = ?
B r
Calculate the enthalpy change in this process CV m for argon
is 12.49 JK −1 mol −1 .
(2000, 4M)
75. A gas mixture of 3.67 L of ethylene and methane on
From the calculated value of ∆G°1 and ∆G° 2 indicate the
order of stability of (A), (B) and (C). Write a reasonable
reaction mechanism showing all intermediates leading to
(A), (B) and (C).
(2001, 10M)
1
O2 ( g ) → CO2 ( g ) at
2
300 K, is spontaneous and exothermic, when the standard
entropy change is − 0.094 kJ mol −1 K −1 . The standard
Gibbs’ free energies of formation for CO2 and CO are
–394.4 and –137.2 kJ mol −1 , respectively.
(2000, 3M)
73. Show that the reaction, CO( g ) +
74. A sample of argon gas at 1 atm pressure and 27° C expands
reversibly and adiabatically from 1.25 dm3 to 2.50 dm3 .
complete combustion at 25° C produces 6.11 L of CO2 . Find
out the amount of heat evolved on burning 1 L of the gas
mixture. The heat of combustion of ethylene and methane
are − 1423 and − 891 kJ mol −1 at 25°C.
(1991, 5M)
76. An athlete is given 100 g of glucose (C6 H12 O6 ) of energy
equivalent to 1560 kJ. He utilizes 50 per cent of this gained
energy in the event. In order to avoid storage of energy in the
body, calculate the weight of water he would need to
perspire. The enthalpy of evaporation of water is 44 kJ/mol.
(1989, 2M)
77. Following statement is true only under some specific
conditions. Write the conditions for that in not more than
two sentences
“The heat energy q, absorbed by a gas is ∆H.’’ (1984, 1M)
Topic 2 Thermochemistry
Objective Questions I (Only one correct option)
1. Lattice enthalpy and enthalpy of solution of NaCl are 788 kJ
mol −1 and 4 kJ mol −1 , respectively. The hydration enthalpy
of NaCl is
(2020 Main, 5 Sep II)
(a) −780 kJ mol −1
(b) 780 kJ mol −1
(c) −784 kJ mol −1
(d) 784 kJ mol −1
2. The variation of equilibrium constant with temperature is
given below:
Temperature
(2020 Main, 6 Sep I)
Equilibrium constant
−1
at 200°C. If
specific heat ofI2( s) andI2 (vap.) are 0.055 and 0.031 cal g −1 K −1
respectively, then enthalpy of sublimation of iodine at 250°C
in cal g −1 is
(2019 Main, 12 April I)
3. Enthalpy of sublimation of iodine is 24 cal g
(c) 22.8
(b) x = y − z
(d) x = y + z
5. Given, C(graphite) + O2 ( g ) → CO2 ( g );
∆ r H ° = − 393.5 kJ mol − 1
Based on the above thermochemical equations, the value of
(2017 Main)
∆ r H ° at 298 K for the reaction,
C(graphite) + 2 H2 ( g ) → CH4 ( g ) will be
[use R = 8.314 JK −1 mol −1 ]
(a) 28.4, − 714
and − 5.71
.
(b) 0.64, − 714
and − 5.71
.
(c) 28.4, − 5.71 and − 14. 29
(d) 0.64, − 5.71 and − 14. 29
(b) 5.7
(2019 Main, 12 Jan II)
(a) y = 2 z − x
(c) z = x + y
1
H2 ( g ) + O2 ( g ) → H2 O( l ); ∆ r H ° = − 285.8 kJ mol −1
2
CO2 ( g ) + 2 H2 O( l ) → CH4 ( g ) + 2O2 ( g );
∆ r H ° = + 890.3 kJ mol −1
T1 = 25° C
K 1 = 10
T2 = 100° C
K 2 = 100
The values of ∆H °, ∆G° at T1 and ∆G° at
T2 (in kJ mol −1 ) respectively, are close to
(a) 2.85
1
(iii) CO ( g ) + O2 ( g ) → CO2 ( g ); ∆ r H È = z kJ mol− 1
2
Based on the above thermochemical equations, find out
which one of the following algebraic relationships is correct?
(d) 11.4
4. Given :
(i) C(graphite) + O2 ( g ) → CO2 ( g ); ∆ r H È = x kJ mol− 1
1
(ii) C(graphite) + O2 ( g ) → CO2 ( g );
2
∆ r H È = y kJ mol− 1
(a) + 78 .8 kJ mol − 1
(b) + 144.0 kJ mol − 1
(c) − 74 .8 kJ mol − 1
(d) − 144.0 kJ mol − 1
6. The heats of combustion of carbon and carbon monoxide are
− 393.5 and − 283.5 kJ mol−1 , respectively. The heat of
formation (in kJ) of carbon monoxide per mole is
(2016 Main)
(a) 676.5
(c) −110.5
(b) −676.5
(d) 110.5
7. For the complete combustion of ethanol,
C2 H 5 OH( l ) + 3O2 ( g ) → 2CO2 ( g ) + 3H2 O( l ), the
amount of heat produced as measured in bomb
calorimeter, is 1364.47 kJ mol −1 at 25°C. Assuming
116 Thermodynamics and Thermochemistry
ideality the enthalpy of combustion, ∆C H, for the reaction
will be (R = 8.314 J K–1 mol –1 )
(2014 Main)
(a) − 1366. 95 kJ mol −1
(b) − 1361. 95 kJ mol −1
(c) − 1460. 50 kJ mol −1
(d) − 1350. 50 kJ mol −1
8. The standard enthalpies of formation of CO2 ( g ), H2 O( l ) and
glucose(s) at 25°C are − 400 kJ/mol, − 300 kJ/mol and
− 1300 kJ/mol, respectively. The standard enthalpy of
combustion per gram of glucose at 25°C is
(2013 Adv.)
(a) + 2900 kJ
(b) − 2900 kJ
(c) − 16. 11 kJ
(d) + 16.11kJ
9. Using the data provided, calculate the multiple bond energy
−1
(kJ mol ) of a C≡≡ C bond C2 H2 . That energy is (take the
bond energy of a C  H bond as 350 kJ mol −1 )
(a) 1165
(2012)
∆H = 225 kJ mol −1
2C( s ) + H2 ( g ) → C2 H2 ( g ) ;
2C( s ) → 2C( g ) ;
∆H = 1410 kJ mol −1
H2 ( g ) → 2H( g ) ;
∆H = 330 kJ mol −1
(b) 837
(c) 865
(d) 815
10. The species which by definition has zero standard molar
enthalpy of formation at 298 K is
(a) Br2 ( g ) (b) Cl 2 ( g ) (c) H2 O( g )
(2010)
(d) CH4 ( g )
11. The bond energy (in kcal mol −1 ) of C— C single bond is
approximately
(a) 1
(b) 10
(2010)
(c) 100
(d) 1000
12. ∆H vap = 30 kJ/mol and ∆S vap = 75 Jmol –1 K –1 . Find the
temperature of vapour, at one atmosphere
(2004, 1M)
(a) 400 K
(b) 350 K
(c) 298 K
(d) 250 K
13. Which of the following reactions defines ∆H f° ?
(a) C(diamond) + O2 ( g ) → CO2 ( g )
1
1
(b) H2 ( g ) + F2 ( g ) → HF( g )
2
2
(c) N2 ( g ) + 3H2 ( g ) → 2NH3 ( g )
1
(d) CO ( g ) + O2 ( g ) → CO2 ( g )
2
(2003, 1M)
− 393.5, − 110.5 and −241.8 kJ mol −1 respectively. The
−1
standard enthalpy change (in kJ mol ) for the reaction
CO2 ( g ) + H2 ( g ) → CO( g ) + H2 O( g ) is
(2000, 1M)
(b) + 41.2
(d) − 41.2
Objective Question II
(One or more than one correct option)
15. The following is/are endothermic reaction(s)
(a) Combustion of methane
(b) Decomposition of water
(c) Dehydrogenation of ethane to ethylene
(d) Conversion of graphite to diamond
studied under different conditions.
CaCO3 ( s ) r CaO( s ) + CO2 ( g )
(2013 Adv.)
For this equilibrium, the correct statement(s) is/are
(a) ∆H is dependent on T
(b) K is independent of the initial amount of CaCO3
(c) K is dependent on the pressure of CO2 at a given T
(d) ∆H is independent of the catalyst, if any
Subjective Questions
17. In a constant volume calorimeter, 3.5 g of a gas with
molecular weight 28 was burnt in excess oxygen at 298.0 K.
The temperature of the calorimeter was found to increases
from 298.0 K to 298.45 K due to the combustion process.
Given that the heat capacity of the calorimeter is 2.5 kJ K − 1 ,
the numerical value for the enthalpy of combustion of the
gas in kJ mol − 1 is
(2009)
18. Diborane is a potential rocket fuel which undergoes
combustion according to the reaction
B2 H6 ( g ) + 3O2 ( g ) → B2 O3 ( s ) + 3H2 O ( g )
From the following data, calculate the enthalpy change for
the combustion of diborane.
(2000, 2M)
3
2B ( s ) + O2 ( g ) → B2 O3 ( s ) ; ∆H = − 1273 kJ mol −1
2
1
H2 ( g ) + O2 ( g ) → H2 O ( l ) ; ∆H = − 286 kJ mol –1
2
H2 O ( l ) → H2 O ( g ) ;
∆H = 44 kJ mol −1
2B ( s ) + 3H2 ( g ) → B2 H6 ( g ) ; ∆H = 36 kJ mol −1
19. Estimate the average S–F bond energy in SF6 . The values of
standard enthalpy of formation of SF6 (g), S( g ) and F( g )
are : – 1100, 275 and 80 kJ mol –1 respectively. (1999, 3M)
20. From the following data, calculate the enthalpy change for
the combustion of cyclopropane at 298 K. The enthalpy of
formation of CO2 ( g ), H2 O ( l ) and propane (g) are –393.5,
−285.8 and 20.42 kJ mol −1 respectively. The enthalpy of
isomerisation of cyclopropane to propene is −33.0 kJ mol −1 .
(1998, 5M)
14. The ∆H f° for CO2 ( g ),CO( g ) and H2 O ( g ) are
(a) 524.1
(c) −262.5
16. The thermal dissociation of equilibrium of CaCO3 ( s ) is
21. Compute the heat of formation of liquid methyl alcohol in
kJ mol −1 , using the following data. Heat of vaporisation of
liquid methyl alcohol = 38 kJ/mol. Heat of formation of
gaseous atoms from the elements in their standard states :
H = 218 kJ/mol, C = 715 kJ/mol, O = 249 kJ/mol.
Average bond energies:
(1997, 5M)
C— H = 415 kJ/mol, C— O = 356 kJ/mol,
O— H = 463 kJ/mol
22. The
(1999, 3M)
standard molar enthalpies of formation of
cyclohexane (l) and benzene (l) at 25° C are −156 and
+ 49 kJ mol −1 respectively. The standard enthalpy of
hydrogenation of cyclohexene (l) at 25° C is −119 kJ mol −1 .
Use these data to estimate the magnitude of the resonance
energy of benzene.
(1996, 2M)
Thermodynamics and Thermochemistry 117
23. The polymerisation of ethylene to linear polyethylene is
represented by the reaction,
]n
n [CH2 == CH2 ] → [ CH2 — CH2
where, n has large integral value. Given that the average
enthalpies of bond dissociation for C == C and C  C at
298 K are +590 and +311 kJ/mol respectively, calculate the
enthalpy of polymerization per mole of ethylene at 298 K.
(1994, 2M)
24. In order to get maximum calorific output, a burner should
have an optimum fuel to oxygen ratio which corresponds
to 3 times as much oxygen as is required theoretically
for complete combustion of the fuel. A burner which has been
adjusted for methane as fuel (with x litre/hour of CH4 and
6x litre/hour of O2 ) is to be readjusted for butane, C4 H10 .
29. The standard molar heat of formation of ethane, carbon
dioxide and liquid water are −21.1, −94.1 and − 68.3 kcal
respectively. Calculate the standard molar heat of
combustion of ethane.
(1986, 2M)
30. The bond dissociation energies of gaseous H2 ,Cl 2 and HCl
are 104, 58 and 103 kcal/mol respectively. Calculate the
enthalpy of formation of HCl gas.
(1985, 2M)
31. Given the following standard heats of reactions
(i) heat of formation of water = − 68.3 kcal
(ii) heat of combustion of acetylene = − 310.6 kcal
(iii) heat of combustion of ethylene = − 337.2 kcal
Calculate the heat of reaction for the hydrogenation of
acetylene at constant volume (25° C).
(1984, 4M)
In order to get the same calorific output, what should be the
rate of supply of butane and oxygen? Assume that losses due
to incomplete combustion etc., are the same for both fuels
and that the gases behave ideally. Heats of combustions:
32. The molar heats of combustion of C2 H2 ( g ), C (graphite) and
CH4 = − 809 kJ/mol, C4 H10 = − 2878 kJ/mol
33. The standard heats of formation of CCl 4 ( g ), H2 O( g ),
(1993, 3M)
25. Determine the enthalpy of the reaction,
C3 H8 ( g ) + H2 ( g ) → C2 H6 ( g ) + CH4 ( g ), at 25° C, using
the given heat of combustion values under standard conditions.
Compound :
H2 ( g ) CH4 ( g ) C2 H6 ( g ) C(graphite)
∆H ° (kJ/mol): −285.8 − 890.0 −1560.0
−393.0
The standard heat of formation of C3 H8 ( g ) is −103 kJ/mol.
(1992, 3M)
26. Using the data (all values are in kilocalories per mol at
25° C) given below, calculate the bond energy of C  C and
C  H bonds.
C( s ) → C( g );
∆H = 172
H2 ( g ) → 2H ( g );
∆H = 104
1
H2 ( g ) + O2 ( g ) → H2 O( l );
∆H = − 68.0
2
∆H = − 94.0
C( s ) + O2 ( g ) → CO2 ( g );
Heat of combustion of C2 H6 = −372.0
Heat of combustion of C3 H8 = −530.0
(1990, 5M)
27. The standard enthalpy of combustion at 25° C of hydrogen,
cyclohexene (C6 H10 ) and cyclohexane (C6 H12 ) are − 241,
− 3800 and − 3920 kJ/mol respectively. Calculate the heat
of hydrogenation of cyclohexene.
(1989, 2M)
28. An intimate mixture of ferric oxide, Fe2 O3 , and aluminium,
Al, is used in solid fuel rockets. Calculate the fuel value per
gram and fuel value per cc of the mixture. Heats of
formation and densities are as follows:
H f (Al 2 O3 ) = − 399 kcal/mol
H f (Fe2 O3 ) = − 199 kcal/mol
Density of Fe2 O3 = 5.2 g/cc, Density of Al = 2.7 g/cc
(1989, 2M)
H2 ( g ) are 310.62 kcal, 94.05 kcal and 68.32 kcal
respectively. Calculate the standard heat of formation of
(1983, 2M)
C2 H2 ( g ) .
CO2 ( g ) and HCl( g ) at 298 K are −25.5, −57.8 , −94.1 and
−22.1 kcal/mol respectively. Calculate ∆H ° (298 K) for the
reaction
CCl 4 ( g ) + 2H2 O ( g ) → CO2 ( g ) + 4 HCl ( g ) (1982, 2M)
34. The enthalpy for the following reactions ( ∆H ° ) at 25°C are
given below
1
1
(i) H2 ( g ) + O2 ( g ) → OH( g ) ∆H ° = − 10.06 kcal
2
2
(ii) H2 ( g ) → 2H( g )
∆H ° = 104.18 kcal
(iii) O2 ( g ) → 2O( g )
∆H ° = 118.32 kcal
Calculate the O — H bond energy in the hydroxyl radical.
(1981, 2M)
Passage Based Questions
When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M
NaOH in an insulated beaker at constant pressure, a temperature
increase of 5.7°C was measured for the beaker and its contents
(Expt. 1). Because the enthalpy of neutralisation of a strong acid
with a strong base is a constant ( −57.0 kJ mol−1 ), this experiment
could be used to measure the calorimeter constant. In a second
experiment (Expt. 2), 100 mL of 2.0 M acetic acid
( K a = 2.0 × 10−5 ) was mixed with 100 mL of 1.0 M NaOH (under
identical conditions to Expt. 1) where a temperature rise of 5.6°C
was measured.
35. Enthalpy of dissociation (in kJ mol−1 ) of acetic acid
obtained from the Expt. 2 is
(a) 1.0
(b) 10.0
(c) 24.5
(d) 51.4
36. The pH of the solution after Expt. 2 is
(a) 2.8
(b) 4.7
(c) 5.0
(d) 7.0
Answers
Topic 1 Thermodynamics
59. (extensive)60. (zero)
61. (exothermic reaction)
1. (c)
2. (a)
3. (b)
4. (a)
62. (isolated) 63. (R)
64. (900)
5. (d)
6. (c)
7. (b)
8. (a)
66. (T)
67. (2)
68. (−563 kJ)
73. (−285.4 kJ) 74. (− 116.4 J) 75. (49.82 kJ)
9. (a)
10. (a)
11. (d)
12. (d)
13. (c)
14. (c)
15. (a)
16. (d)
17. (c)
18. (a)
19. (c)
20. (d)
21. (c)
22. (c)
23. (c)
24. (d)
25. (b)
26. (a)
27. (a)
28. (b)
29. (b)
30. (a)
31. (c)
32. (d)
33. (c)
34. (c)
35. (a)
36. (c)
37. (c)
38. (a)
39. (a, b, c)
40. (c, d)
41. (b,c)
42. (a, c, d)
43. (a, b)
44. (a,b,c)
45. (b,c,d)
46. (a,c,d)
47. (a,c)
48. (c,d)
49. (a,c,d)
50. (b, d)
51. (0.25)
52. (b)
53. (d)
54. (b)
55. (b)
56. (c)
57. A → r, t; B → p, q, s; C → p, q, s; D → p,q, s, t
58. A → p, r, s; B → r, s; C → t; D → p, q, t
65. (T)
72. (12.3 kJ)
76. (318.96 g)
Topic 2 Thermochemistry
1. (c)
2. (c)
3. (c)
4. (d)
5. (c)
6. (c)
7. (a)
8. (c)
9. (d)
10. (b)
11. (c)
12. (a)
13. (b)
14. (b)
15. (b,c,d)
16. (a, b, c, d)
17. (9 kJ)
18. (−2035 kJ)
19. (309.16 kJ)
20. (−2091.32 kJ)
21. (−116.4 kJ)
22. (−152 kJ/mol)
23. (−32 kJ/mol)
25. (−55 kJ)
24. (5.46 xL/h)
27. (−121kJ/mol)
29. (−372 kcal/mol)
30. (−22 kcal/mol)
31. (−41.7 kcal)
32. (54.2 kcal)
33. (−41.4 kcal)
34. (121.31 kcal)
35. (1 kJ/mol)
36. (4.7)
Hints & Solutions
Topic 1 Thermodynamics
3. A process will be spontaneous when its free energy (Gibb’s
Key Idea Work done during isothermal expansion of an
ideal gas is given by the equation.
W = − pext (V2 − V1 )
1.
According to the given conditions, the expansion is against
constant external pressure. So, the work done is given by
following formula;
W = − pext (V2 − V1 )
= − 1bar (10L − 1L)= − 9 L bar
2.
(Q1Lbar =100J)
energy) change will be negative, i.e. ∆G < 0.
Spontaneity of a process is decided by the value of ∆G, which
can be predicted from the Gibb’s equation, ∆G = ∆H − T∆S for
positive/negative signs of ∆H and ∆S at any/higher/lower
temperature as:
∆H
∆S
Comment on
temperature
(T)
∆G
Comment on the
process
<0
>0
at any temp.
<0
spontaneous
= −9 × 100 J = −0.9 kJ
>0
<0
at any temp.
>0
non-spontaneous
Key Idea The relation between ∆H and ∆U is
∆H = ∆U + ∆ngRT
where, ∆ng = Σnp − ΣnR
= number of moles of gaseous products − number of
moles of gaseous reactants.
<0
<0
at lower temp.
<0
spontaneous
>0
>0
at higher temp.
<0
spontaneous
4. In the given system, during the compression of a spring the
workdone is 10 kJ and 2 kJ of heat is escaped to the
surroundings. So, q = − 2 kJ and W = 10 kJ
The general combustion reaction of a hydrocarbon is as follows :
y
y

CxH y +  x +  O2 → xCO2 + H2O

4
2
According to the first law of thermodynamics,
For heptane, x = 7, y = 16
⇒ C7H16 (l ) + 11O2(g ) → 7CO2(g ) + 8H2O(l )
The change in internal energy, ∆U (in kJ) is 8 kJ.
∴
∆ng = 7 − 11 = − 4
Now, from the principle of thermochemistry,
⇒
∆H = ∆U + ∆ngRT
∆H − ∆U = ∆ngRT = − 4RT
∆U = q + W = − 2 kJ + 10 kJ
∆U = 8 kJ
5. q (heat) and W (work) represents path functions. These variables
are path dependent and their values depends upon the path
followed by the system in attaining that state. They are inexact
differentials whose integration gives a total quantity depending
upon the path.
Thermodynamics and Thermochemistry 119
Option (a), i.e. q + W and option (d), i.e. H–TS are state
functions. The value of state functions is independent to the way
in which the state is attained. All the state functions are exact
differentials and cyclic integration involving a state functions is
zero.
6. Given,
n = 5 mol, T2 = 200 K, T1 = 100 K
CV = 28 JK −1mol−1
∆U = nCV ∆T = nCV (T2 − T1 )
= 5 mol × 28 JK −1mol−1 × (200 − 100) K
= 14 ,000 J = 14 kJ
∆pV = nR∆T = nR (T2 − T1 )
= 5 mol × 8 JK −1mol−1 × (200 − 100) K
= 4000 J = 4 kJ
7. From the 1st law of thermodynamics,
∆U = q + W
where, ∆U = change in internal energy
q = heat
W = work done
The above equation can be represented for the given processes
involving ideal gas as follows:
(a) Cyclic process For cyclic process, ∆U = 0
∴
q = −W
Thus, option (a) is correct.
(b) Adiabatic process For adiabatic process,
q=0
∴
∆U = W
Thus, option (b) is incorrect.
(c) Isochoric process For isochoric process, ∆V = 0.
Thus,
(QW = p∆V ).
W =0
∴
∆V = q
Thus, option (c) is correct.
(d) Isothermal process For isothermal process,
∴
q = −W
Thus, option (d) is correct.
∆U = 0
9. For diatomic ideal gases,
CV =
where, f = degree of freedom
f = translational degree of freedom + rotational degree of
freedom
= 3 + 2 = 5 [at normal temperature]
The explanation of various plots are as follows.
(a) We know that, C p is heat capacity at constant pressure. Thus,
it does not vary with the variation in pressure. Hence, plot
given in option (a) is incorrect.
(b) In this plot, CV first increases slightly with increase in
temperature and then increases sharply with temperature.
The sharp increase is due to increase in degree of freedom.
Thus, plot given in option (b) is correct.
(c) For ideal gases,
Internal energy (U ) ∝ T
Thus, as temperature increases internal energy also
increases. As temperature increases further degree of
freedom also increases thus, there is slight variation in the
graph. First translational degree of freedom is present
followed by rotational and vibrational degree of freedom.
Hence, plot given in option (c) is also correct.
(d) CV is heat capacity at constant volume. Thus, it does not vary
with variation in volume. Hence, plot given in option (d) is
correct.
10. Given,
 dE °
−4
−1
E° = 2V, 
 = − 5 × 10 VK
 dT 
T = 300K, R = 8JK −1mol −1,
F = 96000Cmol −1
According to Gibbs-Helmholtz equation,
Also,
= − 4 × 96000J mol −1
…(i)
where, ∆H = Change in enthalpy.
C p = Heat capacity at constant pressure.
Given, n = 3 moles, T1 = 300 K, T2 = 1000 K, C p = 23 + 0.01 T
On substituting the given values in Eq. (i), we get
1000
∫ (23 + 0.01 T )dT = 3
300
∫ 23dT + 0.01T dT
300
2  1000

0.01 T
= 3 23T +
2  300

0.01


(10002 − 3002 )
= 3 23 (1000 − 300) +


2
= 3 [16100 + 4550 ] = 61950 J ≈ 62 kJ
…(ii)
[Q n = 2 for the given reaction]
= − 384000J mol −1
T1
1000
…(i)
∆G = − nFE °cell
∆G = − 2 × 96000 C mol−1 × 2 V
T2
∆H = 3
∆ G = ∆ H − T∆ S
On substituting the given values in equation (ii), we get
8. According to Kirchoff’s relation,
∆H = n ∫ C p dT
f
 f

R and C p =  + 1 R
2

2
Now,
 dE °
∆S = nF 

 dT 
or
∆S = 2 × 96000 C mol−1 × (−5 × 10−4 VK −1 )
= − 96 JK −1 mol −1
Thus, on substituting the values of ∆G and ∆S in Eq. (i), we get
− 384000Jmol−1
= ∆H − 300 K × (−96 JK −1mol−1 )
∆H = − 384000 − 28800 Jmol−1
= − 412800J mol −1
= − 412.800 kJ mol −1
120 Thermodynamics and Thermochemistry
11. According to Gibbs-Helmholtz equation,
∆ rG ° = ∆ r H ° − T∆ r S °
For a reaction to be feasible (spontaneous)
∆ r G° < 0
∆ r H °−T∆ r S ° <0
Given, ∆ r H ° = + 491.1kJ mol − 1 ,
∆ r S ° = 198 JK − 1 mol − 1
(a) N2(g ) + 3H2(g ) → 2NH3 (g ),
∆ng = 2 − (1 + 3) = − 2
So, ∆S is also negative (entropy decreases)
3
491.1 × 10
= 2480.3 K
198
∴Above 2480.3 K reaction will become spontaneous.
12. According to Gibb’s Helmholtz equation,
∆ rG ° = ∆ r H ° − T∆ r S °
Given,
∆ rG ° = A − BT
On comparing above two equations, we get,
A = ∆H ° and ∆S ° = B
We know that, if ∆H ° is negative, reaction is exothermic and
when it is positive, reaction is endothermic.
∴ If A > 0, i.e. positive, reaction is endothermic.
13. For a given value of T ,
(i) If ∆ rG ° becomes < 0, the forward direction will be spontaneous
and then the major and minor components will be Y and X
respectively.
(ii) If ∆ rG ° becomes > 0, the forward direction will be
non-spontaneous and then the major and minor components
will be X and Y respectively.
3
(a) ∆ rG ° = 120 − × 280 = 15
8
i.e. ∆ rG ° > O 0, major component = X ;
3
(b) ∆ rG ° = 120 − × 350 = − 1125
.
8
i.e. ∆ rG ° < 0, major component = Y
3
(c) ∆ rG ° = 120 − × 315 = 1875
.
8
i.e. ∆ rG ° > 0, major component = X
3
(d) ∆ rG ° = 120 − × 300 = 7.5
8
i.e. ∆ rG ° > 0, major component = X
14. At the thermal equilibrium,
final temperature T f =
T1 + T2
2
⇒ for the 1st block, ∆S I = C p ln
⇒ for the 2nd block, ∆S II
  T1 + T2  2 
 
 
 (T + T2 )2 
2  
= C p ln  1
= C p ln 


 T1T2
 4T1T2 




15. The explanation of all the options are as follows :
∴ 491.1 × 103 − T × 198 < 0
T>
 T f2 
 Tf Tf 
= C p ln  ×  = C p ln 

 T1 T2 
T1T2 
Tf
T1
Tf
= C p ln
T2
When brought in contact with each other,
Tf
Tf
+ C p ln
∆ S = ∆S I + ∆S II = C p ln
T2
T1
∆
(b) CaSO4 (s) → CaO(s) + SO3 (g ),
∆ng = (1 + 0) − 0 = + 1
So,
∆S = + ve
(c) In dissolution, ∆S = + ve because molecules/ions of the
solid solute (here, iodine) become free to move in
solvated/dissolved state of the solution,
I2 (s) Water
→ I2 (aq)
(KI)
(d) In sublimation process, molecules of solid becomes quite
free when they become gas,
CO2(s) → CO2 (g )
Dry ice
So, ∆S will be positive.
16. It is an irreversible isothermal compression of an ideal gas.
(i) dE = dq + p(V f − Vi )
where, dE = Internal energy change
dq = amount of heat released
⇒ 0 = dq + p (V f − Vi )
[Q dE = 0 for an isothermal process]
⇒ dq = − 4 (1 − 5) = 16 J
(ii)
dq = n × C × ∆T (for Al)
⇒ 16 J = 1mol × 24 J mol − 1 K −1 × ∆T
16
2
K= K
⇒ ∆T =
24
3
17. ∆G = ∆H − T∆S
The process will be spontaneous, when
∆G = − ve, i.e. |T∆S | > | ∆ H |
Given : ∆H = 200 Jmol−1 and ∆S = 40 JK −1mol−1
| ∆ H | 200
T >
=
=5K
⇒
40
| ∆S |
So, the minimum temperature for spontaneity of the process is
5 K.
18. The conversion of 1 kg of ice at 273 K into water vapours at
383 K takes place as follows:
H2O(s)
273K
∆S1
H2O(l)
∆S2
H2O(l)
373K
273K
∆S3
H2O(g)
383K
∆S4
H2O(g)
373K
Thermodynamics and Thermochemistry 121
∆S 1 =
∆H Fusion 334 kJ kg −1
= 122
. kJ kg −1K −1
=
∆TFusion
273 K
∆S 2 = C ln
 373 K 
T2
= 4.2 kJ K −1kg −1 ln 

 273 K 
T1
= 4.2 × 2.303 (log 373 − log 273)kJ K −1kg −1
= 4.2 × 2.303 (2.572 − 2.436) = 131
. kJ K −1kg −1
−
1
∆H vap. 2491 kJ kg
= 6.67 kJ kg −1K −1
∆S 3 =
=
373 K
∆Tvap.
∆S 4 = Cln
 383 K 
T2
= 2 kJ K −1kg −1 ln 

 373 K 
T1
= 2 × 2.303 (log 383 − log 373) kJ K −1kg −1
= 2 × 2.303 (2.583 − 2.572) kJ K −1kg −1 = 0.05 kJ K −1kg −1
∆S Total = ∆S 1 + ∆S 2 + ∆S 3 + ∆S 4
= 122
. + 131
. + 6.67 + 0.05 = 9.26 kJ kg −1K −1
19. For isothermal reversible expansion,
| W | = nRT ln
Vf
Vi
= nRT ln
V
Vi
where, V = final volume, Vi = initial final.
or | W | = nRT ln V − nRT ln Vi
On comparing with equation of straight line, y = mx + c, we get
slope = m = + nRT
intercept = − nRT ln Vi
Thus, plot of |W | with lnV will give straight line in which slope
of 2(T2 ) is greater than slope of 1(T1 ) which is given in all
options.
Now, if Vi < 1then y intercept (− nRT Vi ) becomes positive and
if it is positive for one case then it is positive for other case also.
Thus, it is not possible that one y-intercept goes above and other
y-intercept goes below. Thus, option (b) and (d) are incorrect.
If we extent plot given in option (a) it seems to be merging
which is not possible because if they are merging they give same
+ve y-intercept. But they cannot give same y-intercept because
value of T is different.
Now, if we extent the line of T1 and T2 given in option (c) it
seems to be touching the origin. If they touch the origin then
y-intercept becomes zero which is not possible. Thus, it is not
the exactly correct answer but among the given options it is the
most appropriate one.
20.
Key idea Calculate the heat of combustion with the help of
following formula
∆H p = ∆U + ∆ngRT
where, ∆H p = Heat of combustion at constant pressure
∆U = Heat at constant volume (It is also called ∆E)
∆ng = Change in number of moles (In gaseous state).
R = Gas constant; T = Temperature.
From the equation,
15
C6H6( l ) + O2( g ) → 6CO2( g ) + 3H2O( l )
2
Change in the number of gaseous moles i.e.
15
3
∆ng = 6 −
= − or −15
.
2
2
Now we have ∆ng and other values given in the question are
∆U = − 3263.9 kJ/mol
T = 25° C = 273 + 25 = 298 K
R = 8.314 JK−1mol −1
So, ∆H p = (−3263.9) + (−15
. ) × 8.314 × 10−3 × 298
= − 3267.6 kJ mol −1
21. According to first law of thermodynamics,
∆ U = q + W = q − p∆ V
In isochoric process (∆V = 0), ∆U = q
In isobaric process (∆p = 0), ∆U = q
In adiabatic process (q = 0), ∆U = W
In isothermal process (∆T = 0) and ∆U = 0
∴ ∆U is equal to adiabatic work.
22. G = H − TS = U + pV − TS
⇒ dG = dU + pdV + Vdp − TdS − SdT = Vdp − SdT
[Q dU + pdV = dq = TdS ]
⇒ dG = Vdp if isothermal process (dT = 0)
⇒ ∆G = V∆p
Now taking initial state as standard state
…(i)
Ggr − Ggr ° = Vgr ∆p
…(ii)
G d − G d ° = Vd ∆p
Now (ii)-(i) gives,
° − G°
(Vd − Vgr )∆p = Gd − Ggr + (Ggr
d
At equilibrium, Gd = Ggr
⇒
(Vgr − Vd )∆p = Gd ° − Ggr ° = 2.9 × 103 J
29
29000
2.9 × 103
bar
Pa =
× 108 Pa =
2
2
2 × 10−6
29000
29000
p = p0 +
=1+
= 14501bar
2
2
∆p =
⇒
∆E = Q + W
23. By first law,
For isothermal expansion, ∆E = 0
∴
Q = −W
− Qirrev = W irrev = p∆V = 3(2 − 1) = 3 L atm
Q
(− 3 × 101.3) J
303.9
Also, ∆S surr = irrev =
=−
= − 1.013JK−1
T
300 K
300
24. For the given reaction,
2NO(g ) + O2 (g ) s
2NO2 (g )
∆G ° (NO) = 86.6 kJ/mol
Given,
f
∆G °f (NO2 ) = ?
K p = 1.6 × 1012
Now, we have,
∆G ° = 2∆G °
f
f (NO )
2
− [ 2∆G f°(NO) + ∆G f°(O ) ]
2
= − RT lnK p = 2∆G °f (NO ) − [ 2 × 86,600 + 0 ]
2
∆G °f (NO
2)
∆G °f (NO
2)
1
= [ 2 × 86600 − R × 298 ln (1.6 × 1012 )]
2
= 0.5 [ 2 × 86,600 − R × (298)ln (1.6 × 1012 )]
122 Thermodynamics and Thermochemistry
25.
PLAN This problem is based on assumption that total entropy change
of universe is zero.
32. In case of reversible thermodynamic process,
∆H = nC p ∆T
∴ Process is isothermal, ∆T = 0 ⇒ ∆H = 0
At 100°C and 1 atmosphere pressure,
H2O (l ) r H2O(g ) is at equilibrium.
For equilibrium,
∆S total = 0
and ∆S system + ∆S surrounding = 0
Also; ∆G = ∆H − T∆S
As we know during conversion of liquid to gas entropy of
system increases, in a similar manner entropy of surrounding
decreases.
∴
∆S system > 0 and ∆S surrounding < 0
26. The process is isothermal expansion, hence
q= −W
∆E = 0
V2
V1
335
= − 208J
50
(expansion work)
35. Work is not a state-function, it depends on path followed.
36. In a reversible thermodynamic process, system always remains
with vapour phase, therefore ∆G = 0. As vaporisation occur,
degree of randomness increases, hence ∆S > 0.
3
3
38. ∆H = ∆E + ∆ngRT ⇒ ∆H − ∆E = ∆ngRT = − 3RT
39. Given, w = − ∫ pext dV
= – 57.05 × 10 J
∆G ° = − 2.303 RT log K
− ∆G °
57.05 × 103
log K =
=
= 10
2.303 RT
5705
For 1 mole van der Waals’ gas
 RT
a
−
p=

 V − b V 2
29. Entropy is a state function hence,
For reversible process, pext = pgas
 RT
a
−
w=−∫
 dV
 V − b V 2
∆ S A → B = ∆ S A → C + ∆ SC → D + ∆ S D → B
...(i)
Also from first law : dq = CV dT + pdV
For one mole of an ideal gas : pV = RT
⇒
pdV + Vdp = RdT
From (i)
pdV = Vdp
Substituting in Eq. (ii) gives
R
2pdV = RdT ⇒ pdV = dT
2
R
⇒
dq = CV dT + dT
2
dq
R 3
R
⇒
∫ dT = CV + 2 = 2 R + 2 = 2R
Minimum value of activation energy must be greater than ∆H .
= − 3 × 8.314 × 298 = − 7433 J = – 7.43 kJ
28. ∆G ° = ∆H ° − T∆S ° = – 54.07 × 10 J – 298 × 10 J
= 50 eu + 30 eu + (– 20 eu) = 60 eu
p
30. Given, = 1 ⇒ p = V
V
∆H > 0
E
Reaction coordinates
27. At transition point (373 K, 1.0 bar), liquid remains in equilibrium
⇒
34. ∆H = ∆U + ∆ ( pV ) = 30 + 2(5 − 3) + 5(4 – 2) = 44 L atm.
37.
= − 2.303 × 0.04 × 8.314 × 310 × log
Also,
For adsorption of gas on solid surface, ∆S < 0. Therefore, in
order to be ∆G < 0, ∆H must be negative.
in equilibrium with surroundings.
W = – 2.303 nRT log
q = + 208 J
W = − 208 J
33. For a spontaneous process ∆ G < 0
...(ii)
31. For an irreversible, adiabatic process;
0 = CV (T2 − T1 ) + pe (V2 − V1 )
Substituting the values
CV (T − T2 ) = 1(2 − 1)atm L
1
2
2
T − T2 =
=
⇒
⇒ T2 = T −
CV 3R
3 × 0.082
But, it is not applicable for irreversible process which are carried
out very fast. So, work done is calculated assuming final
pressure remains constant throughout the process.
Thus, statement (a), (b) and (c) correct while statement (d) is
incorrect.
40. The standard enthalpy of formation is defined as standard
enthalpy change for formation of 1 mole of a substance from its
elements, present in their most stable state of aggregation.
3
O 2 (g ) → O 3 (g ) ;
2
1
S8 (s) + O 2 (g ) → SO 2 (g )
8
In the above two reactions standard enthalpy of reaction is equal
to standard enthalpy of formation.
41. In the given curve AC represents isochoric process as volume
at both the points is same i.e., V1
Similarly, AB represents isothermal process (as both the points
are at T1 temperature) and BC represents isobaric process as
both the points are at p2 pressure.
Now (i) for option (a)
qAC = ∆U BC = nCV (T2 − T1 )
Thermodynamics and Thermochemistry 123
where, n= number of moles
Cv = specific heat capacity at constant volume
However, W AB ≠ p2 (V2 − V1 ) instead
V 
W AB = nRT1 ln  2 
 V1 
43. ∆S surr =
For endothermic reaction, if Tsurr increases, ∆Ssurr will increase.
For exothermic reaction, if Tsurr increases, ∆Ssurr will decrease.
44.
So, this option is incorrect.
qBC = ∆H AC = nC p (T2 − T1 )
where, C p = specific heat capacity at constant pressure
Likewise,
W BC = − p2 (V1 − V2 )
Hence, this option is correct.
The change in internal energy of an ideal gas depends only on
temperature and change in internal energy (∆U ) = 0 therefore,
∆T = 0 hence, process is isothermal and
(iii) For option (c)
nC p (T2 − T1 ) < nCV (T2 − T1 )
so
∆HCA < ∆UCA
T2 = T1 and
reversible process.
Hence, only (a), (b) and (c) are correct choices.
(iv) For option (d)
Although qBC = ∆H AC
45.
but ∆HCA >/ ∆UCA
Hence, this option is incorrect.
42.
(a)
p2
p1
–Wirr
V1
p2
–Wrev
p2
V2
Irreversible compression
V1
p2
V2
(b)
A
B
C
V1
PLAN When an ideal solution is formed process is spontaneous thus
According to Raoult’s law, for an ideal solution
∆H = 0, ∆Vmin = 0
From the relation
∆ G = ∆ H − T∆ S
Since,
∆H = 0
∴
∆G = − ve
i.e. less than zero. and ∆S surroundings = 0
Therefore,
∆S sys = + ve
i.e. more than zero.
46. (a) Since, change of state ( p1 , V1 , T1 ) to ( p2 , V2 , T2 ) is isothermal
therefore, T1 = T2.
Reversible compression
Maximum work is done on the system when compression
occur irreversibly and minimum work is done is reversible
compression.
p1
p2V2 = p1V1
(d) p2V2γ = p1V1γ is incorrect, it is valid for adiabatic
and
qAC = ∆U BC
Hence, this option is also correct.
p1
PLAN This problem includes concept of isothermal adiabatic
irreversible expansion.
Process is adiabatic because of the use of thermal insolution
therefore, q = 0
Q
pext = 0
w = pext ⋅ ∆V = 0 × ∆V = 0
Internal energy can be written as
∆U = q + W = 0
(ii) For option (b)
as
− ∆H
Tsurr
V2
AB is isothermal and AC is
adiabatic path. Work done is
area under the curve. Hence,
(b) Since, change of state ( p1 , V1 , T1 ) to ( p3 , V3 , T3 ) is an adiabatic
expansion it brings about cooling of gas, therefore, T3 < T1.
(c) Work done is the area under the curve of p-V diagram. As
obvious from the given diagram, magnitude of area under the
isothermal curve is greater than the same under adiabatic
curve, hence Wisothermal > Wadiabatic
(d) ∆U = nCv ∆T
In isothermal process, ∆U = 0 as ∆T = 0
In adiabatic process, ∆U = nC v (T3 − T1 ) < 0 as T3 < T1 .
⇒
NOTE
∆U isothermal > ∆U adiabatic
Here only magnitudes of work is being considered otherwise
both works have negative sign.
47. (a) Entropy is a state function, change in entropy in a cyclic
process is zero.
(c) It is incorrect. In adiabatic expansion cooling is observed,
hence ∆U = nCv∆T < 0.
Therefore, ∆S X → Y + ∆S Y → Z + ∆S Z →
(d) q = 0 (adiabatic), W = 0 (Free expansion)
Hence, ∆U = 0, ∆T = 0 (Isothermal)
Analysis of options (b) and (c)
⇒
X
=0
− ∆S Z → X = ∆S X → Y + ∆SY → Z = ∆S X → Z
Work is a non-stable function, it does depends on the path
followed. WY → Z = 0 as ∆V = 0.
124 Thermodynamics and Thermochemistry
Therefore, W X → Y → Z = W X → Y. Also, work is the area under the
curve on p-V diagram.
X
Y
X
Z
p
Y
WX→Y
PLAN By Boyle’s law at constant temperature, p ∝
1
V
By Charles’ law at constant pressure, V ∝ T
Process taking place at
Constant temperature — isothermal
Constant pressure — isobaric
Constant volume — isochoric
Constant heat — adiabatic
WX → Y
= WX→Y→Z
p
56.
Z
K → L
At constant p, volume
increases
thus, heating
As shown above W X → Y + WY → Z = W X → Y = W X → Y→ Z but not
equal to W X → Z.
L → M
At constant V, pressure
decreases
thus, cooling
48. Resistance and heat capacity are mass dependent properties,
M → N
At constant p, volume
decreases
thus, cooling
N → K
At constant V, pressure
increases
thus, heating
V
V
hence they are extensive.
49. Internal energy, molar enthalpy are state function. Also,
reversible expansion work is a state function because between
given initial and final states, there can be only one reversible path.
50. Intensive properties are those property which do not depends
on amount of sample. Both temperature and refractive index
are intensive properties while enthalpy and volumes are
extensive properties as they depends on amount of sample.
51. Given : A
q B (p
A
= 1bar)
Using
∆G = ∆G ° + RT ln K p
At equilibrium : ∆G ° = − RT ln K p
∆G °1 = − RT1 ln K p1
∆G2° = − RT2 ln K p 2
… (i)
… (ii)
From Eqs. (i) and (ii),
ln K p1
∆G1°
T
1000
ln(10) 1
= 1×
=
×
= = 0.25
T2 ln K p 2
2000 ln(100) 4
∆G2°
52. Statement I is true, it is statement of first law of thermodynamics.
Statement II is true, it is statement of second law of
thermodynamics. However, Statement II is not the correct
explanation of statement I.
53. Statement I is false. At equilibrium
∆G = 0, G ≠ 0.
Statement II is true, spontaneous direction of reaction is towards
lower Gibbs free energy.
54. Statement I is true.
∴
∴
dq = dE + pext dV = 0
∆T = 0
dE = 0 ; pext = 0
pext dV = 0
Statement II is true. According to kinetic theory of gases,
volume occupied by molecules of ideal gas is zero.
However, Statement II is not the correct explanation of Statement I.
55. L → M At constant V — isochoric,
N→K
57. (A) → r, t ; (B) → p, q, s ; (C) → p, q, s ; (D) → q, s, t
0° C
(A) H2O(l ) s
1 atm
H2O(s)
q< 0, W < 0 (expansion)
∆S sys < 0 (solid state is more ordered than liquid state)
∆U < 0 ; ∆G = 0 (At equilibrium)
(B) q = 0 (isolated), W = 0 (pext = 0)
∆ Ssys > 0 QV2 > V1
∆U = 0 Q q = W = 0
∆G < 0 Q p2 < p1
(C) q = 0 (isothermal mixing of ideal gases at constant p)
W = 0Q ∆U = 0; q = 0, ∆S sys > 0
Q V2 > V1, ∆U = 0
Q∆ T = 0
∆ G < 0Q mixing is spontaneous.
(D) q = 0 (returning to same state and by same path)
W =0
∆S sys = 0 (same initial and final states)
∆U = 0
Q Ti = T f , ∆ G = 0
58. (A) CO2 (s) 
→ CO2 (g )
It is just a phase transition (sublimation) as no chemical
change has occurred. Sublimation is always endothermic.
Product is gas, more disordered, hence ∆S is positive.
(B) CaCO3 (s) 
→ CaO(s) + CO2 (g )
It is a chemical decomposition, not a phase change. Thermal
decomposition occur at the expense of energy, hence
endothermic. Product contain a gaseous species, hence, ∆S > 0.
(C) 2H 
→ H2 (g )
A new H—H covalent bond is being formed, hence, ∆H < 0.
Also, product is less disordered than reactant, ∆S < 0.
Thermodynamics and Thermochemistry 125
(D) Allotropes are considered as different phase, hence
P(white, solid) → P(red, solid) is a phase transition as well as
allotropic change.
Also, red phosphorus is more ordered than white
phosphorus, ∆S < 0.
71. (i)
1
1
2
p
0.5
3
59. Extensive : Enthalpy is an extensive property while molar
enthalpy is an intensive property.
20 L
60. Zero : − W = p∆V = 0Q ∆V = 0
61. Exothermic reaction.
62. Isolated This system neither exchange matter nor energy with
surroundings.
63. R : For an ideal gas, C p − CV = R
3
2
64. 900 cal : E = RT =
3
× 2 × 300 cal
2
65. True First law deals with conservation of energy while second
law deals with direction of spontaneous change.
66. True Diatomic gases have more degree of freedom than a
monatomic gas.
67. Work done along dashed path |− W | = Σp∆V
= 4 × 1.5 + 1 × 1 +
2
× 2.5 = 8.65 L atm
3
Work done along solid path − W = nRT ln
= 2 × 2.3 log
⇒
V2
V
= p1V1 ln 2
V1
V1
5.5
= 2 × 2.3 log 11 = 4.79
0.5
W d 8.65
=
= 1.80 ≈ 2
W s 4.79
68. ∆H = ∆U + ∆ ( pV ) = ∆U + V∆p
⇒
− W 1 = p ∆V = 20 L atm
W 2 = 0 Q ∆V = 0
40
= 20 ln 2
W 3 = nRT ln
20
Total work done = W 1 + W 2 + W 3
= − 20 L atm + 0 + 20 ln 2
= – 6.14 atm
From first law : q = ∆E + (− W ) = − W
(Q ∆E = 0for cyclic process)
⇒
q = 6.14 L atm = 622.53 J
(ii)
(iii) All the states function, ∆U , ∆H and ∆S are zero for cyclic
process.
72. At equilibrium :
⇒
K1 =
⇒
= – 8.314 × 448 × 2.303 log
70. He is monatomic, so it has only three degree of freedom
(translational only) at all temperature hence, CV value is
3
always R.
2
Hydrogen molecule is diatomic, has three translational, two
rotational and one vibrational degree of freedom. The energy
spacing between adjacent levels are in the order of :
translational < rotational < vibrational
At lower temperature only translational degree of freedom
contribute to heat capacity while at higher temperature
rotational and vibrational degree of freedom starts contributing
to heat capacity.
13
= 16 kJ
952
∆G2° = − RT ln K 2
= – 8.314 × 448 × 2.303 log
35
= 12.3 kJ
952
73. ∆ rG ° = ∆ f G ° (products) – ∆ f G ° (reactants)
∆H = ∆U + ∆ ( pV )
∆pV = p2V2 − p1V1
3.5%
∆G1° = − RT ln K 1
For adiabatic process, q = 0, hence ∆U = W
where,
13
952
35
K2 =
952
⇒
69. ∆U = q + W
W = − p (∆V ) = − p (V2 − V1 )
1.3%
B r C
95.2%
∆U = ∆H − V∆p
∆U = − 100 (99 − 100) = 100 bar mL
B r A
95.2%
= – 560 – 1 × 30 × 0.1 = – 563 kJ
⇒
40 L
V
= – 394.4 – (– 137.2) = – 257.2 kJ < 0
The above negative value of ∆G indicates that the process is
spontaneous.
Also, ∆G ° = ∆H ° − T∆S °
⇒
∆ H ° = ∆ G ° + T∆ S °
= – 257.2 + 300 (– 0.094)
= – 285.4 kJ < 0
74. Given : CV = 12.49 ⇒ C p = 20.8
⇒
Cp
CV
= γ = 1.66
In case of reversible adiabatic expansion :
TV γ− 1= constant
126 Thermodynamics and Thermochemistry
γ−1
⇒
T2  V1 
= 
T1  V2 
V 
T2 = T1  1 
 V2 
0.66
⇒
 1
= 300  
 2
0.66
V 
=  1
 V2 
0.66
2. T1 = 298 K, T2 = 373 K, K 1 = 10, K 2 = 100
K 
1
∆H °  1
log 2  =
 − 
 K 1  2.303R  T1 T2 
∴
∴
∆H ° = 28.4 kJ/mol
We know that, ∆G ° = − RT ln K
∴∆G ° at T1 = − 8.314 × 298 × 2.303 × log(10)
⇒ −5.71 kJ/mol
∆G ° at T2 = − 8.314 × 373 × 2.303 × log(100)
⇒ −14.29 kJ/mol
Hence, the correct option is (c).
= 189.86 K
⇒ ∆H = nCp ∆T
1 × 1.25
=
× 20.8 × (189.86 − 300) J
0.082 × 300
= – 116.4 J
75. Let the mixture contain x litre of CH4 and 3.67 − x litre of
ethylene.
3.
CH4 + O2 → CO2
x
C2H4
3.67 − x
x
+ O2 →
2CO2
2 (3.67 − x)
∆H °
1 
 1
 100
−
log


 =
 10  2.303 × 8.314  298 373
Key Idea When q is the amount of heat involved
in a system then at constant pressure
q = qp and C p ∆T = ∆H
Given reaction :
Given : x + 2 (3.67 − x ) = 6.11 L
⇒
x = 1.23 L
Volume of ethylene = 2.44 L
pV
1×1
Total moles of gases in l litre =
=
= 0.04
RT 0.082 × 298
I2 (s) → I2 (g )
Specific heat of I2 (s) = 0.055 cal g −1 K −1.
Also, CH4 and ethylene are in 1 : 2 volume (or mole) ratio, moles
0.04
2 × 0.04
and moles of ethylene =
of CH4 =
3
3
0.04
⇒ Heat evolved due to methane =
× 891= 11.88 kJ
3
2 × 0.04
Heat evolved due to ethylene =
× 1423 = 37.94 kJ
3
⇒ Total heat evolved on combustion of 1.0L gaseous mixture
at 25°C is 11.88 + 37.94 = 49.82 kJ
If q is the amount of heat involved in a system then at constant
pressure q = qp and
∆H = C p ∆T
76. Moles of H2O needs to perspire =
1560
= 17.72
2 × 44
Weight of water needs to perspire = 17.72 × 18 = 318.96 g
77. At constant pressure, q = ∆H .
Topic 2 Thermochemistry
1. The enthalpy of solution of an ionic solid is numericlly equal to
the sum of its hydration and lat energies,
°
°
°
i.e. ∆H sol
= ∆H hydration
+ ∆H lattice
∴
∆H = 4
NaCl (s)
NaCl (aq)
∆H° =788
∆H°
+
–
Na (g) + Cl (g)
∆sol H ° = ∆ latticeH ° + ∆ hydH °
4 = 788 + ∆ hydH °
∆H hydH ° = − 784 kJ/mol
hyd
Specific heat of I2 (vap) = 0.031 cal g –1K –1.
Enthalpy (H 1 ) of sublimation of iodine = 24 cal g –1
H 2 − H 1 = C p (T2 − T1 )
H 2 = H 1 + ∆C p (T2 − T1 )
H 2 = 24 + (0.031 − 0.055) (250 − 200)
H 2 = 24 + (−0.024 ) (50) = 24 − 12
. = 22. 8 cal/g
Thus, the enthalpy of sublimation of iodine at 250°C is
22.8 cal/g.
4. Second equation given in this question is wrong. Hence, No
answer in correct.
If corrected second equation is given,
1
i.e.
C(graphite) + O2 (g ) → CO (g )
2
and if we take the above reaction in consideration then x = y + z
will be the answer as:
1
(ii) C(graphite) + O2 (g ) → CO(g ), ∆ r H ° = y kJ/mol
2
1
(iii) CO(g ) + O2 (g ) → CO2 (g ),
∆ r H ° = z kJ/mol
2
Summing up both the equation you will get equation (i):
C(graphite) + O2 (g ) → CO2 (g ),
∆ r H °= x kJ/mol
Hence, x, y and z are related as:
x =y+ z
5. Based on given ∆ rH °
° = − 393.5 kJ mol − 1
∆ f H ° = H CO
2
…(i)
∆ f H ° = H H° 2 O = − 285.8 kJ mol − 1
…(ii)
Thermodynamics and Thermochemistry 127
∆ f H ° = H °O 2 = 0.00 (elements)
(8.314 J K−1 mol −1) in JK−1 mol −1 must be converted into kJ by
dividing the unit by 1000.
…(iii)
Required thermal reaction is for ∆ f H ° of CH4
9. For calculation of C ≡≡ C bond energy, we must first calculate
Thus, from III
890.3 = [ ∆ f H ° (CH4 ) + 2 ∆ f H ° (O2 )]
dissociation energy of C2H2 as
…(i)
C2H2 (g ) → 2C(g ) + 2H(g )
Using the given bond energies and enthalpies :
…(ii)
C2H2 (g ) → 2C (g ) + 2H (g ) ; ∆H = − 225 kJ
…(iii)
2C( s) → 2C(g );
∆H = 1410 kJ
…(iv)
H2 (g ) → 2H(g ) ;
∆H = 330 kJ
Adding Eqs. (ii), (iii) and (iv) gives Eq. (i).
⇒
C2H2 (g ) → 2C(g ) + 2H(g ) ; ∆H = 1515 kJ
⇒
1515 kJ = 2 × (C  H) BE + (C ≡≡ C) BE
= 2 × 350 + (C ≡≡ C) BE
− [ ∆ f H ° (CO2 ) + 2 ∆ f H ° (H2O)]
= ∆ f H ° (CH4 ) + 0] − [ − 393.5 − 2 × 285.5]
∴
∆ f H ° (CH4 ) = − 74.8 kJ / mol
6. C(s)+ O2 (g ) → CO2 (g ) ; ∆H = − 393.5 kJ mol −1
…(i)
1
CO+ O2 → CO2( g ); ∆H = − 283.5kJ mol −1
2
On subtracting Eq. (ii) from Eq. (i), we get
1
C (s) + O2 (g ) → CO(g );
2
∆H = (− 393.5 + 283.5)kJ mol −1
…(ii)
= − 110 kJ mol−1(approx.)
7. C2H5OH ( l ) + 3O2 (g ) → 2CO2 (g ) + 3H2O(l )
∆U = − 1364.47 kJ/mol
∆H = ∆U + ∆ngRT
∆ng = − 1
− 1 × 8.314 × 298
∆H = − 1364.47 +
1000
[Here, value of R in unit of J must be converted into kJ]
= − 1364.47 − 2. 4776
= − 1366.9476 kJ/mol
or
= − 1366.95 kJ/mol
8.
PLAN ∆ c H ° (Standard heat of combustion) is the standard enthalpy
change when one mole of the substance is completely oxidised.
Also standard heat of formation (∆ f H ° ) can be taken as the
standard of that substance.
H °CO 2 = ∆ f H ° (CO2 ) = − 400 kJ mol −1
H H° 2 O = ∆ f H ° (H2O ) = − 300 kJ mol −1
°
Hglucose
= ∆ f H ° (glucose) = − 1300 kJ mol −1
H O° 2 = ∆ f H ° (O2 ) = 0.00
C6H12O6 (s) + 6 O2 (g ) → 6 CO2 (g ) + 6H2O(l )
∆ cH ° (glucose) = 6[ ∆ f H ° (CO2 ) + ∆ f H °(H2O )]
− [ ∆ f H ° (C6H12O6 ) + 6∆ f H ° (O 2 )]
= 6[ −400 − 300 ] − [ −1300 + 6 × 0 ]
= − 2900 kJ mol −1
Molar mass of C 6H12O6 = 180 g mol −1
Thus, standard heat of combustion of glucose per gram
−2900
=
= − 16.11 kJ g−1
180
To solve such problem, students are advised to keep much
importance in unit conversion. As here, value of R
⇒
(C ≡≡ C) BE = 1515 − 700 = 815 kJ / mol
10. Elements in its standard state have zero enthalpy of formation.
Cl 2 is gas at room temperature, therefore ∆H °f of Cl 2 (g ) is zero.
11. C—C bond energy is approximately 100 kcal.
12. T =
13.
∆H vap
∆S vap
=
30,000
= 400 K
75
1
1
H2 (g ) + F2 (g ) → HF(g )
2
2
Here ∆H ° = Standard molar enthalpy of formation of HF(g).
14. CO2 (g ) + H2 (g ) → CO(g ) + H2O(g )
∆H = Σ∆ f H ° (products) − Σ∆ f H ° (reactants)
= – 110.5 – 241.8 – (– 393.5) = + 41.20 kJ
1
2
It is reverse of combustion of H2 (g ), hence endothermic.
C2H6 → C2H4 + H2;
∆H > 0
15. H2O → H2 + O2, ∆H > 0
Here, more stable (saturated) hydrocarbon is being transformed
to less stable (unsaturated) hydrocarbon, hence endothermic.
C(gr) → C(d) ,
∆H > 0
More stable allotrope is being converted to less stable allotrope.
16. PLAN Heat of reaction is dependent on temperature (Kirchhoff’s
equation) in heterogeneous system, equilibrium constant is
independent on the molar concentration of solid species.
Heat of reaction is not affected by catalyst. It lowers activation
energy.
CaCO3 (s) r CaO(s) + CO2 (g )
By Kirchhoff’s equation,
∆H °2 (at T2 ) = ∆H °1 ( at T1 ) + ∆C p (T2 − T1 )
∆H ° varies with temperature. Thus, (a) is correct.
K = pCO2
K is dependent on pressure of CO2 but independent of molar
concentration of CaCO3. Thus, (b) and (c) are correct. At a given
temperature, addition of catalysis lowers activation energy, ∆H
remaining constant. Thus, (d) is also correct.
128 Thermodynamics and Thermochemistry
22.
+ H2
Ea
CaO + CO2
T
E'a
CaCO3
⇒
∆H°
– 357 kJ = ∆H °f (cyclohexane) − ∆H °f (C6H6 )
⇒ ∆H °f (C6H6 )Theoretical = – 156 + 357 = 201 kJ
Ea′ = Activation energy in presence of catalyst
17. Temperature rise = T2 − T1 = 298.45 – 298 = 0.45 K
⇒ Resonance energy = ∆H °f (exp.) − ∆H °f (Theoretical)
q = heat capacity × ∆T = 2.5 × 0.45 = 1.125 kJ
1.125
⇒ Heat produced per mole =
× 28 = 9 kJ
3.5
18. ∆H r° = ∆H °f (B2O3 ) + 3∆H °f (H2O) – ∆H °f (B2H6 )
∆H °f (H2O)(g ) = ∆H °f (H2O)(l ) + 44 = – 242 kJ
= 49 – 201 = – 152 kJ/mol
23. Per mole of ethylene polymerized, one C == C bond is broken
and two C—C bonds are formed.
∆H ° (Polym.) = 590 − 2 × 311 = – 32 kJ/mol
24. At same temperature and pressure, equal volumes contain equal
moles of gases.
Let 1.0 L of CH4 contain ‘n’ mol
∆H r° = – 1273 – 3 × 242 – 36
= – 2035 kJ
19. SF6 (g ) → S (g ) + 6F (g )
∆H = Σ ∆H °f (products) − Σ ∆H °f (reactants)
= 275 + 6 × 80 + 1100 = 1855 kJ
1855
= 309.16 kJ/mol
⇒ Average S—F bond energy =
6
⇒ x L of CH4 contain nx mol
⇒ Heat evolved in combustion by x L CH4 = 809 nx kJ
1 
Now, 2878 kJ energy is evolved from 1 mole  L C4H10.
n 
⇒ 809 nx kJ energy will be evolved from
20. Given : Cyclopropane →Propene (C3H6 ); ∆H = − 33 kJ
9
Propene (C3H6 ) + O2 → 3CO2 (g ) + 3H2O (l );
2
∆H = – 3 (393.5 + 285.8) – 20.42 = – 2058.32 kJ
Adding :
9
O2 (g ) → 3CO2 (g ) + 3H2 (g ) ;
2
∆H = H 1 + H 2
= −33 + (−2058.32)kJ
∆H = − 2091.32 kJ
Cyclopropane +
21. Given : CH3OH (g ) → CH3OH (l );
;
+ 3H2
∆H = –119 × 3 = – 357 kJ (Theoretical)
E'a <E a
Ea = Activation energy in absence of catalyst
⇒
∆H = –119
;
∆H = − 38 kJ
C (g ) + 4H (g ) + O (g ) → CH3OH (g );
∆H = − (3 × 415 + 356 + 463)
Q
H = H 1 + H 2 = − 2064 kJ
C(g ) → C(g );
∆H = 715 kJ
2H2 (g ) → 4H (g );
∆H = 2 × 2 × 218= 872kJ
1
O2 (g ) → O (g );
∆H = 249 kJ
2
1
Adding : C (gr) + 2H2 (g ) + O2 (g ) → CH3OH (l )
2
∆H = − 266 kJ/mol
1 × 1.25
=
× 20.8 × (189.86 − 300) J = −116.4 J
0.082 × 300
809 nx
L of C4H10
2878 n
=0.28 x L of C4H10
Also, the combustion reaction of butane is
13
C4 H10 +
O2 → 4CO2 + 5H2O
2
13
× 0.28 x × 3 = 5.46 x L/h
⇒ Rate of supply of oxygen =
2
25. First we need to determine heat of combustion of C3H8 .
3C(gr) + 4 H2 (g ) → C3H8 (g )
∆H °f = − 103 kJ
°
⇒ – 103 kJ = – 3 × 393 – 4 × 285.80 – ∆ H comb
(C3H8 )
°
⇒ ∆ H comb
(C3H8 ) = – 2219.20 kJ
°
°
(reactants) − Σ ∆ H comb
(products)
⇒ ∆H r° = Σ ∆ H comb
= – 2219.20 – 285.80 + 1560 + 890
= – 55 kJ
26. Let x kcal be the C—C bond energy and y kcal be the C—H bond
energy per mole.
⇒
2C(gr) + 3H2 (g ) → C2H6 (g ) ;
∆H ° = – 2 × 94 – 3 × 68 + 372
⇒
= – 20 kcal
– 20 kcal = 2 × 172 + 3 × 104 − BE (C2H6 )
⇒ BE (C2H6 ) = 676 kcal
Thermodynamics and Thermochemistry 129
= – 310.6 – 68.3 – (– 337.2)
= – 41.7 kcal
Similarly, 3C(gr) + 4 H2 (g ) → C3H8 (g ) ;
∆H ° =
– 3×
94 – 4 × 68 + 530 = – 24
kcal
⇒
32. The standard state formation reaction of C2H2 (g ) is :
– 24 kcal = 3 × 172 + 4 × 104 − BE (C3H8 )
2C(g ) + H2 (g ) → C2H2 (g ); ∆H °f
⇒ BE (C3H8 ) = 956 kcal
BE (C2H6 ) = 676 kcal = x + 6 y
BE (C3H8 ) = 956 kcal = 2x + 8 y
Also,
…(i)
°
°
∆H r° = Σ ∆ H comb
(reactants) − Σ ∆ H comb
(products)
= – 2 × 94.05 – 68.32 – (– 310.62)
…(ii)
= 54.2 kcal = ∆H °f (C2H2 )
Solving Eqs. (i) and (ii) gives
y = 99 kcal (C—H) BE
33. ∆H r° = Σ ∆ f H ° (products) − Σ ∆ f H ° (reactants)
x =82 kcal (C—C) BE
= – 94.1 + 4 (– 22.1) – (– 25.5 – 2 × 57.8)
= – 41.4 kcal
+ H2
27.
°
°
∆H = Σ ∆ H comb
(reactants) − Σ ∆ H comb
(products)
= – 3800 – 241 – (– 3920) = – 121 kJ/mol
28. Fe2O3 (s) + 2Al (s) → Al 2O3 (s) + 2Fe(s)
∆H r° = ∆H °f (products) − ∆H °f (reactants)
= – 399 – (– 199) = – 200 kcal
Mass of reactants = 56 × 2 + 16 × 3 + 27 × 2 = 214 g
200
⇒ Fuel value/gram =
= 0.93 kcal/g
214
160
54
cc +
Volume of reactants =
cc = 50.77 cc
5.2
2.7
200
Fuel value/cc =
= 3.94 kcal/cc
⇒
50.77
29. ∆H = Σ ∆H °f (products) − Σ ∆H °f (reactants)
= – 2 × 94.1 – 3 × 68.3 – (– 21.1) = – 372 kcal/mol
1
1
30.
H2 (g ) + Cl 2 (g ) → HCl (g ); ∆H °f
2
2
∆H °f = Σ BE (reactants) − Σ BE (products)
=
1
(104 + 58) − 103 = – 22 kcal/mol
2
31. C2H2 + H2 → C2H4
°
°
∆H ° = Σ ∆ H comb
(reactants) − Σ ∆ H comb
(products)
34. ∆H ° = Σ BE (reactants) − Σ BE (products)
1
1
(104.18) + (118.32) − BE (O  H)
2
2
BE (O—H) = 121.31 kcal
⇒
− 10.06 =
35. Let C JK−1 be the heat capacity of calorimeter.
Mass of solution = 200 mL × 1 g mL−1 = 200 g
Heat evolved in Expt.1
= 57 × 1000 × 0. 1(mol ) = 5700 J
⇒ 5700 J = (200 × 4.2 + C ) × 5.7
…(i)
⇒ 1000 = 200 × 4.2 + C
Let x kJ/mol is heat evolved in neutralisation of acetic acid.
⇒ x × 1000 × 010
. = (200 × 4.2 + C ) × 5.6
x × 100
…(ii)
= 200 × 4.2 + C
⇒
5.6
From (i) and (ii) : x = 56 kJ/mol
⇒ Enthalpy of ionisation of acetic acid
= − 56 − (− 57) = 1 kJ/mol
36. CH3COOH + NaOH → CH3COONa + H2O
200 mmol
100 mol
0
100 mmol
0
100 mmol
0
A buffer is now formed.
[H+ ][ CH3COO− ]
Ka =
= [ H+ ]
[CH3COOH]
[Q [ CH3COOH] = [ CH3COO− ]]
⇒
pH = pK a = − log (2 × 10−5 ) = 5 − log 2 = 4.7
8
Solid State
Objective Questions I (Only one correct option)
1. A crystal is made up of metal ions M 1 and M 2 and oxide ions.
Oxide ions form a ccp lattice structure. The cation M 1
occupies 50% of octahedral voids and the cation M 2 occupies
12.5% of tetrahedral voids of oxide lattice. The oxidation
numbers of M 1 and M 2 are, respectively (2020 Main, 6 Sep II)
(a) +2, + 4
(c) +3, + 1
(b) +1, + 3
(d) +4 , + 2
2. A diatomic molecule X 2 has a body-centred cubic (bcc)
structure with a cell edge of 300 pm. The density of the
molecule is 6.17 g cm −3 . The number of molecules present in
200 g of X 2 is (Avogadro constant ( N A ) = 6 × 1023 mol −1 )
(2020 Main, 5 Sep I)
(b) 2N A
(a) 40N A
(c) 8N A
(d) 4N A
3. The ratio of number of atoms present in a simple cubic, body
centered cubic and face centered cubic structure are,
respectively.
(2019 Main, 12 April II)
(a) 8 : 1 : 6
(b) 1 : 2 : 4
(c) 4 : 2 : 1
(d) 4 : 2 : 3
4. An element has a face-centred cubic (fcc) structure with a cell
edge of a. The distance between the centres of two nearest
tetrahedral voids in the lattice is
(2019 Main, 12 April I)
a
3
(a) 2a
(d) a
(b) a
(c)
2
2
5. Consider the bcc unit cells of the solids 1 and 2 with the
position of atoms as shown below. The radius of atom B is
twice that of atom A. The unit cell edge length is 50% more is
solid 2 than in 1. What is the approximate packing efficiency
in solid 2?
(2019 Main, 8 April II)
A
A
A
A
A
A
A
A
B
A
A
A
A
A
A
(b) 90%
A
Solid 2
Solid 1
(a) 65%
A
A
(c) 75%
(d) 45%
6. The statement that is incorrect about the interstitial
compounds is
(a) they are very hard
(b) they have metallic conductivity
(c) they have high melting points
(d) they are chemically reactive
(2019 Main, 8 April II)
7. Element ‘B ’ forms ccp structure and ‘A ’ occupies half of
the octahedral voids, while oxygen atoms occupy all the
tetrahedral voids. The structure of bimetallic oxide is
(2019 Main, 8 April I)
(a) A2BO4
(c) A2B2O
(b) AB2O4
(d) A4B2 O
8. The radius of the largest sphere which fits properly at the
centre of the edge of a body centred cubic unit cell is
(Edge length is represented by ‘a’)
(2019 Main, 11 Jan II)
(a) 0.134 a (b) 0.027 a (c) 0.047 a
(d) 0.067 a
9. A solid having density of 9 × 103 kgm −3 forms face centred
cubic crystals of edge length 200 2 pm. What is the molar
mass of the solid?
[Avogadro constant = 6 × 1023 mol−1 , π = 3]
(2019 Main, 11 Jan I)
(a) 0.03050 kg mol−1
(c) 0.0432 kg mol−1
(b) 0.4320 kg mol−1
(d) 0.0216 kg mol−1
10. A compound of formula A2B3 has the hcp lattice. Which
atom forms the hcp lattice and what fraction of tetrahedral
voids is occupied by the other atoms ? (2019 Main, 10 Jan II)
2
(a) hcp lattice- A, tetrahedral voids-B
3
1
(b) hcp lattice-A, tetrahedral voids-B
3
1
(c) hcp lattice-B, tetrahedral voids-A
3
2
(d) hcp lattice-B, tetrahedral voids-A
3
11. Which primitive unit cell has unequal edge lengths
( a ≠ b ≠ c ) and all axial angles different from 90°?
(2019 Main, 10 Jan I)
(a) Hexagonal
(c) Tetragonal
(b) Monoclinic
(d) Triclinic
Solid State
12. At 100°C, copper (Cu) has FCC unit cell structure with cell
edge length of x Å. What is the approximate density of Cu (in g
cm −3 ) at this temperature?
[Atomic mass of Cu = 63.55 u]
(2019 Main, 9 Jan II)
211
205
(b) 3
(a) 3
x
x
105
422
(c) 3
(d) 3
x
x
13. The one that is extensively used as a piezoelectric material is
(2019 Main, 9 Jan I)
(a) quartz
(c) amorphous silica
(b) tridymite
(d) mica
131
21. A compound M p X q has cubic close packing (ccp)
arrangement of X . Its unit cell structure is shown below. The
empirical formula of the compound, is
(2012)
M
X
(a) MX
(b) MX 2
(c) M 2 X
(d) M 5 X 14
22. The packing efficiency of the two-dimensional square unit
cell shown below is
(2010)
14. Which type of ‘defect’ has the presence of cations in the
interstitial sites?
(a) Schottky defect
(c) Frenkel defect
(2018 Main)
(b) Vacancy defect
(d) Metal deficiency defect
l
15. A metal crystallises in a face centred cubic structure. If the
(a) 39.27% (b) 68.02%
(c) 74.05%
(d) 78.54%
edge length of its unit cell is ‘a’, the closest approach between
two atoms in metallic crystal will be
(2017 Main)
a
(a) 2 a
(b) 2 2 a
(c) 2 a
(d)
2
23. Which of the following fcc structure contains cations in
16. Sodium metal crystallises in a body centred cubic lattice with
24. A substance Ax B y crystallises in a face centred cubic (fcc)
a unit cell edge of 4.29Å. The radius of sodium atom is
approximately
(2015 Main)
(a) 1.86Å
(b) 3.22Å
(c) 5.72Å
(d) 0.93Å
17. CsCl crystallises in body centred cubic lattice. If ‘a’ its edge
length, then which of the following expressions is correct?
(2014 Main)
(a) r
Cs
(c) r
+
Cs +
+ rCl − = 3a
+ rCl − =
3
a
2
(b) r
Cs
(d) r
+
Cs +
+ rCl − =
3a
2
+ rCl − = 3a
18. The arrangement of X − ions around A + ion in solid AX is
given in the figure (not drawn to scale). If the radius of X − is
250 pm, the radius of A + is
(2013 Adv.)
alternate tetrahedral voids?
(a) NaCl
(b) ZnS
(c) Na 2 O
(2005, 1M)
(d) CaF2
lattice in which atoms A occupy each corner of the cube and
atoms B occupy the centres of each face of the cube. Identify
the correct composition of the substance Ax B y
(2002, 1M)
(b) A4 B3
(c) A3 B
(a) AB3
(d) composition cannot be specified
25. In a solid AB having the NaCl structure, A atoms occupy the
corners of the cubic unit cell. If all the face centred atoms
along one of the axes are removed, then the resultant
stoichiometry of the solid is
(2001, S, 1M)
(a) AB2
(b) A2 B
(c) A4 B3
(d) A3 B4
26. The coordination number of a metal crystallising in a
hexagonal close-packed structure is
(a) 12
(b) 4
(c) 8
(1999, 2M)
(d) 6
Objective Questions II
(One or more than one correct option)
–
X
A+
27. The cubic unit cell structure of a compound containing cation
(a) 104 pm
(c) 183 pm
M and anion X is shown below. When compared to the anion,
the cation has smaller ionic radius. Choose the correct
statement(s).
(2020 Adv.)
(b) 125 pm
(d) 57 pm
19. Experimentally it was found that a metal oxide has formula
M 0.98O. Metal M, present as M 2+ and M 3 + in its oxide.
Fraction of the metal which exists as M 3+ would be
(a) 7.01%
(b) 4.08%
(2013 Main)
(c) 6.05%
(d) 5.08%
20. Which of the following exists as covalent crystals in the solid
state?
(a) Iodine
(c) Sulphur
(2013 Main)
(b) Silicon
(d) Phosphorus
M
X
132
Solid State
(a) The empirical formula of the compound is MX.
(b) The cation M and anion X have different coordination
geometries.
(c) The ratio of M — X bond length of the cubic unit cell edge
length is 0.866.
(d) The ratio of the ionic radii of cation M to anion X is 0.414.
28. The correct statement(s) for cubic close packed (ccp) three
dimensional structure is (are)
(2016 Adv.)
(a) The number of the nearest neighbours of an atom present
in the topmost layer is 12
(b) The packing efficiency of atom is 74%
(c) The number of octahedral and tetrahedral voids per atom
are 1 and 2, respectively
(d) The unit cell edge length is 2 2 times the radius of the atom
29. If the unit cell of a mineral has cubic close packed (ccp) array
of oxygen atoms with m fraction of octahedral holes
occupied by aluminium ions and n fraction of tetrahedral
holes occupied by magnesium ions, m and n respectively, are
(2015 Adv.)
1
(a) ,
2
1
(c) ,
2
1
8
1
2
1
(b) 1 ,
4
1 1
(d) ,
4 8
30. The correct statement(s) regarding defects in solids is/are
(a) Frenkel defect is usually favoured by a very small
difference in the sizes of cation and anion
(1999)
(b) Frenkel defect is a dislocation defect
(c) Trapping of an electron in the lattice leads to the formation
of F-centre
(d) Schottky defects have no effect on the physical properties
of solids
31. Which of the following statement(s) is/are correct?
(a) The coordination number of each type of ion in CsCl crystal
is 8
(1998, 2M)
(b) A metal that crystallises in bcc structure has a coordination
number of 12
(c) A unit cell of an ionic crystal shares some of its ions with
other unit cells
(d) The length of the unit cell in NaCl is 552 pm.
(rNa + = 95 pm; rCl − =181pm)
Numerical Answer Type Question
Assertion and Reason
Read the following questions and answer as per the direction
given below :
(a) Statement I is correct Statement II is correct Statement II is
the correct explanation of Statement I
(b) Statement I is correct Statement II is 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
33. Statement I In any ionic soid (MX ) with Schottky defects,
the number of positive and negative ions are same.
Statement II Equal numbers of cation and anion vacancies
are present.
(2001, 1M)
Passage Based Questions
Passage
In hexagonal systems of crystals, a frequently encountered
arrangement of atoms is described as a hexagonal prism. Here, the
top and bottom of the cell are regular hexagons and three atoms are
sandwiched in between them. A space-filling model of this
structure, called hexagonal close-packed (hcp), is constituted of a
sphere on a flat surface surrounded in the same plane by six
identical spheres as closely as possible. Three spheres are then
placed over the first layer so that they touch each other and
represent the second layer. Each one of these three spheres touches
three spheres of the bottom layer.
Finally, the second layer is covered with a third layer that is
identical to the bottom layer in relative position. Assume radius of
every sphere to be ‘r ’.
34. The number of atoms in one of this hcp unit cell is
(2008, 3 × 4M = 12M)
(a) 4
(b) 6
(c) 12
(d) 17
35. The volume of this hcp unit cell is
(a) 24 2 r3 (b) 16 2 r3
(c) 12 2 r3
(d)
3 3
36. The empty space in this hcp unit cell is
(a) 74 %
(b) 47.6 %
(c) 32 %
(d) 26 %
Match the Column
37. Match the crystal system / unit cells mentioned in Column I
with their characteristic features mentioned in Column II.
(2007, 6M)
32. Consider an ionic solid MX with NaCl structure. Construct a
new structure (Z) whose unit cell is constructed from the unit
cell of MX following the sequential instruction given below.
Neglect the charge balance.
(2018 Adv.)
(a) Remove all the anions (X ) except the central one
(b) Replace all the face centered cations (M ) by anions (X )
(c) Remove all the corner cations (M )
(d) Replace the central anion (X ) with cation (M )
 Number of anions 
The value of 
 in Z is ___
 Number of cations 
64 r3
Column I
A. Simple cubic and face
centred cubic
Column II
p. have these cell
parameters a = b = c and
α = β = γ = 90°
B. Cubic and rhombohedral q. are two crystal systems
C. Cubic and tetragonal
r. have only two
crystallographic angles
of 90°
Hexagonal
and
D.
s. belong to same crystal
monoclinic
system
Solid State
133
44. (i) Marbles of diameter 10 mm are to be put in a square area
Integer Answer Type Questions
of side 40 mm so that their centres are within this area.
38. A crystalline solid of a pure substance has a face-centred
cubic structure with a cell edge of 400 pm. If the density of
the substance in the crystal is 8 g cm −3 , then the number of
atoms present in 256 g of the crystal is N × 1024 . The value
of N is
(2017 Adv.)
39. The number of hexagonal faces that are present in a truncated
octahedron is
(2011)
−1
40. Silver (atomic weight = 108 g mol ) has a density of
10.5 g cm−3 . The number of silver atoms on a surface of area
10−12 m2 can be expressed in scientific notation as y × 10 x.
The value of x is
(2010)
Subjective Questions
41. The edge length of unit cell of a metal having molecular
weight 75 g/mol is 5 Å which crystallises in cubic lattice. If
the density is 2 g/cc then find the radius of metal atom.
(N A = 6 × 1023 ). Give the answer in pm.
(2006, 3M)
42. An element crystallises in fcc lattice having edge length
400 pm. Calculate the maximum diameter of atom which can
be placed in interstitial site without distorting the structure.
(2005, 2M)
43. The crystal AB (rock salt structure) has molecular weight
6.023 y u. Where, y is an arbitrary number in u. If the
minimum distance betweeen cation and anion is y1 / 3 nm and
the observed density is 20 kg / m3 . Find the (i) density in
(2004, 2M)
kg / m3 and (ii) type of defect.
(ii) Find the maximum number of marbles per unit area and
deduce an expression for calculating it.
(2003, 4M)
45. The figures given below show the location of atoms in three
crystallographic planes in a fcc lattice. Draw the unit cell for
the corresponding structures and identify these planes in
your diagram.
(2000)
46. A metal crystallises into two cubic phases, face centred cubic
(fcc) and body centred cubic (bcc), whose unit cell lengths
are 3.5 and 3.0 Å, respectively. Calculate the ratio of
densities of fcc and bcc.
(1999, 3M)
47. Chromium metal crystallises with a body centred cubic
lattice. The length of the unit edge is found to be 287 pm.
Calculate the atomic radius . What would be the density of
chromium in g/cm 3 ?
(1997, 3M)
48. A metallic element crystallises into a lattice containing a
sequence of layers of ABABAB…… Any packing of layers
leaves out voids in the lattice. What percentage of this lattice
is empty space?
(1996, 3M)
49. Sodium crystallises in a bcc cubic lattice with the cell edge,
a = 4.29 Å. What is the radius of sodium atom?
(1994, 2M)
Answers
1. (a)
2. (d)
3. (b)
4. (c)
5. (b)
6. (d)
7. (b)
8. (d)
9. (a)
10. (c)
11. (d)
12. (d)
13. (a)
14. (c)
15. (d)
16. (a)
17. (c)
18. (a)
19. (b)
20. (b)
21. (b)
22. (d)
23. (b)
24. (a)
25. (d)
26. (a)
27. (a,c)
28. (b,c,d)
29. (a)
30. (b,c)
31. (a,c,d)
32. (c)
33. (a)
34. (b)
35. (a)
36. (d)
38. (2)
39. (8)
40. (7)
41. (217 pm) 42. (117 pm)
37. A → p, s; B → q; C → q; D → q, r
46. (1.26)
47. (7.3 g/cm 3) 48. (0.74)
49. (1.86 Å)
Hints & Solutions
1.
So, formula is (M 1 )2 (M 2 ) O4.
This must be neutral. Both metals must have + 8charge on total.
From given options :
Oxidaion number of M 1 = + 2
Oxidaion number of M = + 4

2
∴
(2 × + 2) + (1 × + 4 ) = + 8
Hence, correct option is (a).
Number of O2− ions in ccp = 4
Total charge on oxide = − 8
Number of tetrahedral voids = 8
Number of voids = 4
Number of cation M 1 occupies 50% of octahedral voids,
50
×4=2
100
Number of cation M 2 occupies 12.5% of tetrahedral voids of
oxide lattice,
12.5
× 8=1
100
2.
We know that,
d=
Z×M
a3 × N A
Given, density d = 6 . 17 g,
134
Solid State
edge length a = 300 pm = 3 × 10− 8 cm,
23
N A = 6 × 10
[As the atoms A are present at the edges only zA =
−1
mol
2×M
∴6 . 17 =
(3 × 10− 8 )3 × 6 × 1023
⇒ M = 50 g / mol
atom B is present only at the body centre zB = 1]
(For bcc, Z = 2)
∴
Hence, number of molecules present in 200 g of X 2 is
w
200
N =
× N A = 4N A
× NA =
M
50
4. In fcc unit cell, two tetrahedral voids are formed on each of the
four non-parallel body diagonals of the cube at a distance of
3a / 4 from every corner along the body diagonal.
One of the body diagonal
of cubic unit cell
A
√3 a
—
4
1
— AB
4
A′
1
—(AB)
2
B′
B
Position of 2 TVs
on one of the
body diagonal
of the unit cell
1
—(AB)
4
√3 a
—
4
The angle between body diagonal and an edge is cos−1 (1/ 3 ).
So, the projection of the line on an edge is a/ 4. Similarly, other
tetrahedral void also will be a/4 away. So, the distance between
a a a

these two is a −
− = .
4  4 2

5.
Key Idea Packing efficiency
Volume occupied by sphere
=
× 100
Volume of cube
4 3  4 3 
 πrA × 1 +  πrB × 1

 3
3
PE 2 =
a23
4 3
4 3 4
πrA × 9
πrA + π (2rA )3
π
3
3
=
=
= 3
3
8 × 3 3 rA3 2 3
(2 3 rA )
3. The ratio of number of atoms present in simple cubic, body
centred cubic and face centered cubic structure are 1 : 2 : 4
respectively.
1
× 8 = 1,
8
= 90.72% ≈ 90%
6. Interstitial compounds are formed when a neutral atom with a
small radius occupies in an interstitial hole (tetrahedral or
octahedral voids) in a transition metal’s hcp or ccp lattices
(host lattice). Examples of small atoms (guest atom) are H, B, C
and N.
Interstitial compounds are non-stoichiometric (Birtholide) in
composition. They are very hard with very high melting points.
The electrical conductivity of interstitial compounds are
comparable to that of the pure metal. These are chemically
unreactive in nature.
7. The number of element ‘B ’ in the crystal structure = 4 N
Number of tetrahedral voids = 2N
Number of octahedral voids = N
N 4
= =2
∴Number of ‘A’ in the crystal =
2 2
Number of oxygen (O) atoms = 2N = 2 × 4 = 8
∴The structure of bimetallic oxide = A2 B4 O8 = AB2 O4
8. For body centred cubic bcc structure,
3
...(i)
a
4
Where, a = edge length
According to question, the structure of cubic unit cell can be
shown as follows:
radius (R ) =
2r
R
R
Given,
rB = 2rA
a2 = a1 +
50
a1 = 15
. a1
100
For bcc lattice
4 rA = 3 a1
rA =
∴
3 a1
4
a
⇒ a1 =
 4r  3
.  A =
a2 = 15
 3 2
4 rA
3
 4 rA 


 3
a2 = 2 3 rA
4 3
4
πrA × zA + πrB3 × zB
3
3
Packing efficiency =
a23
...(ii)
∴
a = 2(R + r)
On substituting the value of R from Eq. (i) to Eq. (ii), we get
3
a
=
a+ r
2 4
3
a
2a − 3a
r= −
a=
2
4
4
a(2 − 3 )
r=
4
r = 0.067a
Solid State
9. Density of a crystal
14. It is the ‘‘Frenkel defect’’ in which cations leave their original
site and occupy interstitial site as shown below.
M ×Z
d × N A × a3
⇒M =
d=
3
Z
NA × a
3
Given, d = 9 × 10 kg m
+
–
+
–
+
–
+
–
−3
M = Molar mass of the solid
Z = 4 (for fcc crystal)
N A = Avogadro’s constant = 6 × 1023 mol −1
a = Edge length of the unit cell
= 200 2 pm = 200 2 × 10−12 m
(9 × 103 ) kg m −3 × (6 × 1023 ) mol −1 × (200 2 × 10−12 )3 m 3
=
4
= 0.0305 kg mol −1
10. Total effective number of atoms in hcp unit lattice = Number of
octahedral voids in hcp = 6
∴Number of tetrahedral voids (TV) in hcp
= 2 × Number of atoms in hcp lattice
= 2 × 6 = 12
As, formula of the lattice is A2 B3 .
Suppose,
A
B

1
(hcp)
 × TV

3
⇒
So, A =
– + – +
+ – + –
– + – +
– + –
– + – +
+
+ – + –
– + – +
+ – + –
+
–
+
–
+
–
+
–
3
3
a=
× 4.29Å = 1.85Å
4
4
r = 1.85Å ≈ 1.86Å
r=
17. In CsCl, Cl − lies at corners of simple cube and Cs+ at the body
centre. Hence, along the body diagonal, Cs+ and Cl − touch each
other so rCs + + rCl − = 2r
Calculation of r
In ∆EDF,
G
B
H
r
c
A
a
r
3
C
F
a
1
tetrahedral voids, B = hcp lattice
3
E
11. Triclinic primitive unit cell has dimensions as, a ≠ b ≠ c and
b
a
D
Body centred cubic unit cell
α ≠ β ≠ γ ≠ 90° .
Among the seven basic or primitive crystalline systems, the
triclinic system is most unsymmetrical. In other cases, edge
length and axial angles are given as follows :
Hexagonal : a = b ≠ c and α = β = 90°, γ = 120°
Monoclinic : a ≠ b ≠ c and α = γ = 90°, β ≠ 90°
FD = b = a2 + a2 = 2 a
In ∆AFD,
c2 = a2 + b2 = a2 + ( 2 a)2 = a2 + 2 a2
c2 = 3 a2 ⇒ c = 3 a
As ∆ AFD is an equilateral triangle.
∴
3a = 4r
Tetragonal : a ≠ b ≠ c and α = β = γ = 90°
12. For fcc, rank of the unit cell (Z ) = 4
⇒
Mass of one Cu-atom, M = 63.55 u
Avogadro’s number, N A = 6.023 × 1023 atom
electric current when they are placed under mechanical stress.
Crystalline solids can be used as piezoelectric material hence,
quartz is a correct answer.
Cation in
interstitial site
3 a = 4a
16. For bcc unit cell,
1
13. Piezoelectric materials are those materials that produce an
Original
vacant site
of cation
2a
a
=
2
2
∴ Closest distance = 2 r =
6
Edge length, a = x Å = x × 10−8 cm
Z×M
density (d ) =
N A × a3
4 × 63.55
422.048
g cm−3
=
=
6.023 × 1023 × (x × 10−8 )3
x3
–
+
–
+
–
+
–
+
where, r = radius and a = edge length
2r
⇒
1
× 12
3
2
3
2
–
+
–
+
–
+
–
+
15. For fcc arrangement, 4 r = 2a
On substituting all the given values, we get
⇒
135
r=
3a
4
Hence, rCs + + rCl− = 2 r = 2 ×
18.
[QC = 3r + r + r]
3
3
a=
a
4
2
PLAN Given arrangement represents octahedral void and for this
r+ (cation)
= 0.414
r− (anion)
r( A + )
= 0.414
r( X − )
r( A + ) = 0.414 × r( X − ) = 0.414 × 250pm
= 103.5 pm ≈ 104 pm
136
Solid State
Number of B is removed because it is not present on face centres.
⇒ A remaining = 4 − 1 = 3, B remaining = 4,
19. From the valency of M 2+ and M 3+ , it is clear that three
M 2+ ions will be replaced by M 3+ causing a loss of one
M 3+ ion. Total loss of them from one molecule of
MO = 1 − 0.98 = 0.02
Total M 3+ present in one molecule of
MO = 2 × 0.02 = 0.04
That M 2+ and M 3+ = 0.98
0.04 × 100
Thus, % of M 3+ =
= 4.08%
0.98
Formula = A3 B4
26. Three consecutive layers of atoms in hexagonal close packed
lattice is shown below:
A
20. Silicon exists as covalent crystal in solid state. (Network like
structure, as seen in diamond).
B
X
21. Contribution of atom from the edge centre is 1/4. Therefore,
number of
1
M = × 4 (from edge centre) + 1(from body centre) = 2
4
1
1
Number of X = × 8 (from corners) + × 6
8
2
(from face centre) = 4
⇒ Empirical formula = M 2 X 4 = MX 2
1
22. Contribution of circle from corner of square =
4
⇒ Effective number of circle per square
1
= × 4 + 1(at centre) = 2
4
⇒ Area occupied by circle = 2πr2, r = radius.
Also, diagonal of square 4 r = 2 L, where L = side of square.
Area occupied by circles
⇒ Packing fraction =
Area of square
2πr2 2πr2 π
= 2 =
= = 0.785
4
L
8r2
⇒ % packing efficiency = 78.5%.
2−
23. In ZnS, S (sulphide ions) are present at fcc positions giving
four sulphide ions per unit cell. To comply with 1 : 1
stoichiometry, four Zn 2+ ions must be present in four alternate
tetrahedral voids out of eight tetrahedral voids present.
In NaCl, Na + ions are present in octahedral voids while in Na 2O,
Na + ions are present in all its tetrahedral voids giving the desired
2 : 1 stoichiometry. In CaF2 , Ca 2+ ions occupies fcc positions
and all the tetrahedral voids are occupied by fluoride ions.
1
24. In cubic system, a corner contribute th part of atom to one unit
8
1
cell and a face centre contribute part of atom to one unit cell.
2
Therefore,
1
Number of A per unit cell = × 8 = 1
8
1
Number of B per unit cell = × 6 = 3
2
Formula = AB3
⇒
+
25. In NaCl, Na occupies body centre and edge centres while Cl
−
occupies corners and face centres, giving four Na + and four Cl −
per unit cell. In the present case A represent Cl − and B represents
Na + . Two face centres lies on one axis.
1
⇒ Number of A removed = 2 × = 1
2
A
Atom X is in contact of 12 like atoms, 6 from layer B and 3 from
top and bottom layers A each.
27. (a) The empirical formula of the compound:
Contribution of M and X : M 
1
2× 

2
X
1
4× 

4
⇒ MX
(b) Coordination number of both M and X is 8.
(c) Distance between M and X =
a2 a2
+
=
4
2
3
3
a=
a
4
2
⇒ 0.866a
(d) rM : rX = ( 3 − 1) : 1 ⇒ 0.732:1, thus statement (d) is
incorrect.
28. (a) Nearest neighbour in the topmost layer of ccp structure is 9
thus, incorrect.
(b) Packing efficiency is 74% thus, correct.
(c) Tetrahedral voids = 2
Octahedral voids = 1 per atom thus, correct.
A
a
B
(d) Edge length, a =
a
4
r = 2 2r
2
thus, correct
Explanation Edge length = a
Radius = r
AC 2 = AB 2 + BC 2
(4 r)2 = a2 + a2 = 2a2
4 r = 2a
C
Solid State
2
a
a=
4
2 2
⇒
r=
⇒
a=2 2 r
137
32. The unit cell of initial structure of ionic solid MX looks like
Cl– (at face centre)
Cl– (at corner)
In ccp structure, number of spheres is 4.
4

Hence, volume of 4 spheres = 4  πr3
3

Na+ (at face)
Total volume of unit cell = a3 = (2 2r)3
Na+ (at corner edge)
% of packing efficiency
Volume of 4 spheres
=
Volume of unit cell
4

4  πr3
3

=
× 100
[ 2( 2r)]3
~ 74%
= 74.05% −
In NaCl type of solids cations (Na + ) occupy the octahedral
voids while anions (Cl − ) occupy the face centre positions.
However, as per the demand of problem the position of cations
and anions are swapped.
We also know that (for 1 unit cell)
(A) Total number of atoms at FCC = 4
(B) Total number of octahedral voids = 4
(as no. of atoms at FCC = No. of octahedral voids)
Now taking the conditions one by one
29. Oxide ions are at ccp positions, hence 4O2– ions. Also, there
are four octahedral voids and eight tetrahedral voids. Since ‘m’
fraction of octahedral voids contain Al 3+ and ‘n’ fraction of
tetrahedral voids contain Mg2+ ions, to maintain
etectroneutrality 2(2Al 3+ = + 6charge) and
(Mg2+ = + 2charge), will make unit cell neutral
2 1
1
Hence: m = = , n=
4 2
8
(i) If we remove all the anions except the central one than
number of left anions.
= 4 −3 = 1
(ii) If we replace all the face centred cations by anions than
effective number of cations will be = 4 − 3 = 1
Likewise effective number of anions will be = 1+ 3 = 4
30. (a) Wrong statement. A small difference in sizes of cation and
anion favour Schottky defect while Frenkel defect is
favoured by large difference in sizes of cation and anion.
(b) Correct statement. In Frenkel defect the smaller atom or ion
gets dislocated from its normal lattice positions and occupies
the interstitial space.
(c) Correct Statement In F-centre defect, some anions leave
the lattice and the vacant sites hold the electrons trapped in it
maintaining the overall electroneutrality of solid.
(d) Wrong statement : In Schottky defect, some of the atoms or
ions remaining absent from their normal lattice points
without distorting the original unit cell dimension. This
lowers the density of solid.
(iii) If we remove all the corner cations then effective number of
cations will be 1− 1= 0
(iv) If we replace central anion with cation then effective
number of cations will be 0 + 1= 1
Likewise effective number of anions will be 4 − 1= 3
Thus, as the final outcome, total number of cations present in
Z after fulfilling all the four sequential instructions = 1
Likewise, total number of anions = 3
Number of anions 3
Hence, the value of
= =3
Number of cations 1
33. In ionic solid MX (1 : 1 solid) same number of M n+ and X n− ions
are lost in Schottky defect to maintain electroneutrality of solid.
31. (a) The unit cell of CsCl has bcc arrangement of ions in which
each ion has eight oppositely charged ions around it in the
nearest neighbours as shown below :
A
34.
Cs+
B
Q
Cl–
M
C
P
Unit cell of CsCl
(b) In bcc, coordination number of atom is 8.
(c) In an unit cell, a corner is shared in eight unit cells and a face
centre is shared between two adjacent unit cells.
(d) In NaCl unit cell; 2(rNa+ + rCl − ) = a
⇒
a = 2 (95 + 181) = 552 pm
Hence, a, c, d are correct.
R
S
N
A hcp unit cell
Contribution of atoms from corner = 1/ 6
Contribution from face centre = 1 / 2
⇒ Total number of atoms per unit cell = 12 ×
1
1
+ 2 × + 3= 6
6
2
138
Solid State
⇒ a3 (Volume of unit cell) = 6.83 × 10−23 cm 3
35. In close packed arrangement, side of the base = 2r
⇒
RS = r
Also MNR is equilateral triangle, ∠ PRS = 30°
3
RS
In triangle PRS , cos 30° =
=
2
PR
2
2
⇒
PR =
RS =
r
3
3
In right angle triangle PQR : PQ = QR 2 − PR 2 = 2
⇒ a = 4 × 10−8 cm = 4 × 10−10 m
⇒ Surface area of unit cell = a2 = 1.6 × 10−19 m 2
⇒ Number of unit cells on 10−12 m 2 surface
=
2
r
3
2
r
⇒ Height of hexagon = 2PQ = 4
3
⇒ Volume = Area of base × height = 6
3
2
(2r)2 × 4
r
4
3
= 24 2 r3
Volume occupied by atoms
Volume of unit cell
4
1
= 6 × πr3 ×
= 0.74
3
24 2 r3
36. Packing fraction =
⇒ Fraction of empty space = 1 − 0.74 = 0.26 = 26%
37. A. Simple cubic and face centred cubic both have cell
parameters a = b = c and α = β = γ = 90°. Also both of them
belongs to same, cubic, crystal system.
B. The cubic and rhombohedral crystal system belongs to
different crystal system.
C. Cubic and tetragonal are two different types of crystal
systems having different cell parameters.
D. Hexagonal and monoclinic are two different crystal system
and both have two of their crystallographic angles of 90°.
4×M
38. Density (ρ) = 8 =
N A (4 × 10−8 cm )3
⇒
−24
M = 128 × 10
=
NA
256
⇒ No. of atoms =
× NA
M
256
× N A = 2 × 1024
128 × 10−24 N A
39. The truncated octahedron is the 14-faced Archimedean solid,
with 14 total faces : 6 squares and 8 regular hexagons.
The truncated octahedron is formed by removing the six right
square pyramids one from each point of a regular octahedron as :
10−12
= 6.25 × 106
1.6 × 10−19
Q There are two atoms (effectively) on one face of unit cell
Number of atoms on 10−12 m 2 surface = 2 × number of unit cell
[Q y × 10x ]
= 1.25 × 107 .
⇒
x = 7 ⇒ y = 125
.
41. From the given information, the number of atoms per unit cell
and therefore, type of unit cell can be known as
NM
ρ=
NA a3
ρ N A a3 2 × 6 × 1023 × (5 × 10−8 cm )3
=
= 2 ( bcc)
M
75
⇒ In bcc, 4 r = 3a
N =
⇒
3
3
a=
× 5 × 10−10 m
4
4
= 2.17 × 10−10 m = 217 pm
⇒
r=
42. In a cubic crystal system, there are two types of voids known as
octahedral and tetrahedral voids. If r1 is the radius of void and r2
is the radius of atom creating these voids then
 r1 
= 0.414
 
 r2  octa
and
 r1 
= 0.225
 
 r2  tetra
The above radius ratio values indicate that octahedral void has
larger radius, hence for maximum diameter of atom to be present
in interstitial space :
r1 = 0.414 r2
Also in fcc,
4 r2 = 2a
⇒ Diameter required (2r1 ) = (2r2 ) × 0.414
a
=
× 0.414
2
400 × 0.414
=
= 117 pm
2
43. (i) In rock salt like crystal AB, there are four AB units per unit
cell. Therefore, density (d ) is
4 × 6.023 y
d=
6.023 × 1023 × 8 y × 10−27
Truncated octahedron
Truncated octahedron unfolded in two-dimension
40. Ag crystallises in fcc unit cell with 4 atoms per unit cell.
ρ=
4 × 108
= 10.5 g cm −3.
6.023 × 1023 × a3
[Q a = 2 y1/ 3 nm = 2 y1/ 3 × 10−9 m ]
= 5 × 103 g/m 3 = 5 kg/m 3
(ii) Since, observed density is greater than expected, theoretical
density, there must be some excess metal occupying
interstitial spaces. This type of defect is known as metal
excess defect.
Solid State
44. (i) Side of square = 40 mm
Diameter of marble = 10 mm
Number of marble spheres along an edge of square with their
centres within the square = 5 (shown in diagram)
10 mm
46. Density ∝
N
a3
3
3
d1 N 1  a2 
4 3 
=
  =   = 1.26
d2 N 2  a1 
2  3.5
⇒
47. In bcc unit cell, 4 r = 3a
⇒
r (Cr) =
Density of solid =
3a
3
=
× 287 pm = 124.3 pm
4
4
NM
N A ⋅ a3
N = Number of atoms per unit cell,
3
a = Volume of cubic unit cell,
40 mm
Maximum number of marbles per unit area = 5 × 5 = 25
(ii) If x mm is the side of square and d is diameter of marble then
maximum number of marbles on square area with centres
within square area can be known by the following general
formula :
2
x 
N =  + 1
d 
139
M = Molar mass
N A = Avogadro’s number
3


2 × 52 g
1
×
=
 = 7.3 g / cm 3
6.023 × 1023  2.87 × 10−8 cm 
48. The given arrangement : ABABAB...... represents hexagonal
close-packed unit cell in which there are six atoms per unit cell.
Also, volume of unit cell = 24 2r3.
Volume occupied by atoms
Volume of unit cell
4
1
= 6 × πr3 ×
= 0.74
3
24 2 r3
⇒ Packing fraction =
⇒ Percent empty space = 100 (1 − 0.74) = 26%
45.
49. In bcc arrangement of atoms : 4 r = 3a, atoms on body
diagonal remain in contact
3a
3 × 4.29
r=
=
= 1.86 Å
⇒
4
4
9
Solutions and
Colligative Properties
Topic 1 Solution and Vapour Pressure of Liquid Solutions
Objective Questions I (Only one correct option)
1. Henry’s constant (in kbar) for four gasesα, β, γ and δ in water
at 298 K is given below :
5. For the solution of the gases w , x, y and z in water at 298 K,
the Henry’s law constants ( K H ) are 0.5, 2, 35 and 40 K bar,
respectively. The correct plot for the given data is
50
2
2 × 10−5
0.5
3
−3
(density of water = 10 kg m at 298 K) This table implies
that
(2020 Main, 3 Sep I)
(a) α has the highest solubility in water at a given pressure
(b) solubility of γ at 308 K is lower than at 298 K
(c) The pressure of a 55.5 molal solution of γ is 1 bar
(d) The pressure of a 55.5 molal solution of δ is 250 bar
(a)
(2019 Main, 12 April I)
(a) 13.88 × 10−2
(b) 13.88 × 10−1
−3
(d) 13.88 × 10
(c) 13.88
3. What would be the molality of 20% (mass/mass) aqueous
solution of KI? (Molar mass of KI = 166 g mol −1 )
(2019 Main, 9 April I)
(a) 1.48
(b) 1.51
(c) 1.35
(d) 1.08
4. Liquid M and liquid N form an ideal solution. The vapour
pressures of pure liquids M and N are 450 and 700 mmHg,
respectively, at the same temperature. Then correct statement
is
(2019 Main, 9 April I)
x M = mole fraction of M in solution;
x N = mole fraction of N in solution;
y M = mole fraction of M in vapour phase;
y N = mole fraction of N in vapour phase
x
y
x
y
(b) M = M
(a) M > M
xN
yN
xN
yN
xM
yM
(c)
(d) ( x M − y M ) < ( x N − y N )
<
xN
yN
z
y
x w
(0, 0) Mole fraction
of water
z
y
x
w
(0, 0) Mole fraction
of water
z
2. The mole fraction of a solvent in aqueous solution of a solute is
0.8. The molality (in mol kg −1 ) of the aqueous solution is
(b)
Partial pressure
δ
(c)
z
y
(d)
x
w
(0, 0) Mole fraction
of water
Partial pressure
γ
Partial pressure
KH
β
Partial pressure
(2019 Main, 8 April II)
α
y
xw
(0, 0) Mole fraction
of water
6. The vapour pressures of pure liquids A and B are 400 and 600
mmHg, respectively at 298 K. On mixing the two liquids, the
sum of their initial volumes is equal to the volume of the final
mixture. The mole fraction of liquid B is 0.5 in the mixture.
The vapour pressure of the final solution, the mole fractions
of components A and B in vapour phase, respectively are
(2019 Main, 8 April I)
(a) 450 mmHg, 0.4, 0.6
(c) 450 mmHg, 0.5,0.5
(b) 500 mmHg, 0.5, 0.5
(d) 500 mmHg, 0.4,0.6
7. Liquids A and B form an ideal solution in the entire
composition range. At 350 K, the vapour pressures of pure A
and pure B are 7 × 103 Pa and 12 × 103 Pa, respectively. The
composition of the vapour in equilibrium with a solution
containing 40 mole percent of A at this temperature is
(2019 Main, 10 Jan I)
Solutions and Colligative Properties
(a)
(b)
(c)
(d)
xA
xA
xA
xA
= 0.76; xB = 0.24
= 0.28; xB = 0.72
= 0.4; xB = 0.6
= 0.37; xB = 0.63
Objective Questions II
(One or more than one correct option)
15. For a solution formed by mixing liquids L and M , the vapour
8. Which one of the following statements regarding Henry’s law
is not correct?
(2019 Main, 8 Jan I)
(a) Different gases have different K H (Henry’s law
constant) values at the same temperature
(b) Higher the value of K H at a given pressure, higher is the
solubility of the gas in the liquids
(c) The value of K H increases with increase of temperature
and K H is function of the nature of the gas
(d) The partial pressure of the gas in vapour phase is
proportional to the mole fraction of the gas in the solution
9. 18 g of glucose (C6 H12 O6 ) is added to 178.2 g water. The
vapour pressure of water (in torr) for this aqueous solution is
(2016 Main)
(a) 76.0
(c) 759.0
(b) 752.4
(d) 7.6
10. The vapour pressure of acetone at 20°C is 185 torr.
When 1.2 g of a non-volatile substance was dissolved in 100 g
of acetone at 20°C, its vapour pressure was 183 Torr. The
molar mass of the substance is
(2015, 1M)
(a) 32
(b) 64
(a) 128
(b) 488
11. The Henry’s law constant for the solubility of N2 gas in water
at 298 K is 1.0 × 105 atm. The mole fraction of N2 in air is 0.8.
The number of moles of N2 from air dissolved in 10 moles of
water of 298 K and 5 atm pressure is
(2009)
(b) 4.0 × 10− 5
(a) 4.0 × 10− 4
(c) 5.0 × 10− 4
(d) 4.0 × 10− 6
12. A molal solution is one that contains one mole of a solute in
(1986, 1M)
(a) 1000 g of the solvent
(c) 1 L of the solution
141
(b) 1 L of the solvent
(d) 22.4 L of the solution
13. For a dilute solution, Raoult’s law states that
(1985, 1M)
(a) the lowering of vapour pressure is equal to the mole
fraction of solute
(b) the relative lowering of vapour pressure is equal to the
mole fraction of solute
(c) the relative lowering of vapour pressure is proportional
to the amount of solute in solution
(d) the vapour pressure of the solution is equal to the mole
fraction of solvent
14. An azeotropic solution of two liquids has boiling point lower
than either of them when it
(a) shows negative deviation from Raoult’s law
(b) shows no deviation from Raoult’s law
(c) shows positive deviation from Raoult’s law
(d) is saturated
(1981, 1M)
pressure of L plotted against the mole fraction of M in
solution is shown in the following figure. Here xL and x M
represent mole fractions of L and M , respectively, in the
solution. The correct statement(s) applicable to this system is
(are)
(2017 Adv.)
Z
pL
1
XM
0
(a) The point Z represents vapour pressure of pure liquid M
and Raoult’s law is obeyed from xL = 0 to xL = 1
(b) Attractive intermolecular interactions between L - L in
pure liquid L and M - M in pure liquid M are stronger
than those between L - M when mixed in solution
(c) The point Z represents vapour pressure of pure liquid M
and Raoult’s law is obeyed when xL → 0
(d) The point Z represents vapour pressure of pure liquid L
and Raoult’s law is obeyed when xL → 1
16. Mixture(s) showing positive deviation from Raoult’s law at
35°C is (are)
(a) carbon tetrachloride + methanol
(b) carbon disulphide + acetone
(c) benzene + toluene
(d) phenol + aniline
(2016 Adv.)
Numerical Answer Type Questions
17. Liquids A and B form ideal solution for all compositions of A
and B at 25°C. Two such solutions with 0.25 and 0.50 mole
fractions of A have the total vapour pressure of 0.3 and 0.4
bar, respectively. What is the vapour pressure of pure liquid B
in bar?
(2020 Adv.)
18. The mole fraction of urea in an aqueous urea solution
containing 900 g of water is 0.05. If the density of the solution
is 1.2 g cm −3 , then molarity of urea solution is …… (Given
data : Molar masses of urea and water are 60 g mol −1 and 18 g
mol −1 , respectively)
(2019 Adv.)
19. Liquids A and B form ideal solution over the entire range of
composition. At temperature T, equimolar binary solution of
liquids A and B has vapour pressure 45 torr. At the same
temperature, a new solution of A and B having mole fractions
x A and xB , respectively, has vapour pressure of 22.5 torr. The
value of x A / xB in the new solution is ____.
(Given that the vapour pressure of pure liquid A is 20 Torr at
temperature T)
(2018 Adv.)
142
Solutions and Colligative Properties
True/False
20. Following statement is true only under some specific
conditions. Write the condition for it.
“Two volatile and miscible liquids can be separated by
fractional distillation into pure components.”
(1994)
the solution is 600 mm Hg. What is the molecular weight of
the solid substance?
(1990, 3M)
26. The vapour pressure of a dilute aqueous solution of glucose
(C6 H12 O6 ) is 750 mm of mercury at 373 K.
Calculate (i) molality and (ii) mole fraction of the solute.
(1989, 3M)
Subjective Questions
21. The vapour pressure of two miscible liquids A and B are 300
27. The vapour pressure of ethanol and methanol are 44.5 and
and 500 mm of Hg respectively. In a flask 10 moles of A is
mixed with 12 moles of B. However, as soon as B is added, A
starts polymerising into a completely insoluble solid. The
polymerisation follows first-order kinetics. After 100 min,
0.525 mole of a solute is dissolved which arrests the
polymerisation completely. The final vapour pressure of the
solution is 400 mm of Hg. Estimate the rate constant of the
polymerisation reaction. Assume negligible volume change
on mixing and polymerisation and ideal behaviour for the
final solution.
(2001, 4M)
28. An organic compound (Cx H2 y O y ) was burnt with twice the
22. The molar volume of liquid benzene (density = 0.877 g/mL)
increases by a factor of 2750 as it vaporises at 20° C and that
of liquid toluene (density = 0.867 g mL−1 ) increases by a
factor of 7720 at 20° C. A solution of benzene and toluene at
20° C has a vapour pressure of 45.0 torr. Find the mole
fraction of benzene in the vapour above the solution.
(1996, 3M)
23. What
weight of the non-volatile solute urea
(NH2 — CO —NH2 ) needs to be dissolved in 100 g of water,
in order to decrease the vapour pressure of water by 25%?
What will be the molality of the solution?
(1993, 3M)
24. The degree of dissociation of Ca(NO3 )2 in a dilute aqueous
solution, containing 7.0 g of the salt per 100 g of water at
100° C is 70%. If the vapour-pressure of water at 100° C is
760mm, calculate the vapour pressure of the solution.
(1991, 4M)
25. The vapour pressure of pure benzene at a certain temperature
is 640 mm Hg. A non-volatile, non-electrolyte solid weighing
2.175 g is added to 39.0 g of benzene. The vapour pressure of
88.7 mm Hg respectively. An ideal solution is formed at the
same temperature by the mixing 60 g of ethanol with 40 g of
methanol. Calculate the total vapour pressure of the solution
and the mole fraction of methanol in the vapour. (1986, 4M)
amount of oxygen needed for complete combustion to CO2
and H2 O. The hot gases when cooled to 0°C and 1 atm
pressure, measured 2.24 L. The water collected during
cooling weight 0.9 g. The vapour pressure of pure water at
20°C is 17.5 mm Hg and is lowered by 0.104 mm when 50 g
of the organic compound are dissolved in 1000 g of water.
Give the molecular formula of the organic compound.
(1983, 5M)
29. Two liquids A and B form ideal solution. At 300 K, the
vapour pressure of a solution containing 1 mole of A and
3 moles of B is 550 mm of Hg. At the same temperature, if
one more mole of B is added to this solution, the vapour
pressure of the solution increases by 10 mm of Hg.
Determine the vapour pressure of A and B in their pure
states.
(1982, 4M)
30. The vapour pressure of pure benzene is 639.70 mm of Hg and
the vapour pressure of solution of a solute in benzene at the
same temperature is 631.9 mm of Hg. Calculate the molality
of the solution.
(1981, 3M)
31. What is the molarity and molality of a 13% solution
(by weight) of sulphuric acid with a density of 1.02 g/mL ? To
what volume should 100 mL of this solution be diluted in
order to prepare a 1.5 N solution ?
(1978, 2M)
Topic 2 Colligative Properties
Objective Questions I (Only one correct option)
1. A solution is prepared by dissolving 0.6 g of urea (molar mass
= 60 g mol −1 ) and 1.8 g of glucose (molar mass = 180 g
mol −1 ) in 100 mL of water at 27°C. The osmotic pressure of
the solution is ( R = 0.08206 L atm K −1 mol −1 )
(2019 Main, 12 April II)
(a) 8.2 atm
(c) 4.92 atm
(b) 2.46 atm
(d) 1.64 atm
2. 1 g of a non-volatile, non-electrolyte solute is dissolved in
100 g of two different solvents A and B, whose ebullisocopic
constants are in the ratio of 1 : 5. The ratio of the elevation in
∆T ( A )
their boiling points, b
, is
∆Tb ( B )
(2019 Main, 10 April II)
(a) 5 : 1
(b) 10 : 1
(c) 1 : 5
(d) 1 : 0.2
3. At room temperature, a dilute solution of urea is prepared by
dissolving 0.60 g of urea in 360 g of water. If the vapour
pressure of pure water at this temperature is 35 mm Hg,
lowering of vapour pressure will be
(Molar mass of urea = 60 g mol −1 )
(2019 Main, 10 April I)
(a) 0.027 mmHg
(b) 0.031 mmHg
(c) 0.017 mmHg
(d) 0.028 mmHg
Solutions and Colligative Properties
143
4. Molal depression constant for a solvent is 4.0 K kg mol −1 .
12. For 1 molal aqueous solution of the following compounds,
The depression in the freezing point of the solvent for 0.03
mol kg −1 solution of K 2SO4 is
(Assume complete dissociation of the electrolyte)
which one will show the highest freezing point? (2018 Main)
(a) [Co(H2 O)6 ]Cl 3
(b) [Co(H2 O)5 Cl]Cl 2 ⋅ H2 O
(c) [Co(H2 O)4 Cl 2 ]Cl ⋅ 2H2 O (d) [Co(H2 O)3 Cl 3 ] ⋅ 3H2 O
(2019 Main, 9 April II)
13. The freezing point of benzene decreases by 0.45°C when 0.2 g
(c) 0.12 K
(d) 0.24 K
5. The osmotic pressure of a dilute solution of an ionic
compound XY in water is four times that of a solution of
0.01 M BaCl2 in water. Assuming complete dissociation of
the given ionic compounds in water, the concentration of
XY (in mol L−1 ) in solution is
(2019 Main, 9 April I)
−2
−4
(b) 16 × 10
(a) 4 × 10
−4
−2
(c) 4 × 10
(d) 6 × 10
6. Molecules of benzoic acid (C6H5COOH) dimerise in
benzene. ‘w’ g of the acid dissolved in 30 g of benzene
shows a depression in freezing point equal to 2 K. If the
percentage association of the acid to form dimer in the
solution is 80, then w is
of acetic acid is added to 20 g of benzene. If acetic acid
associates to form a dimer in benzene, percentage association
of acetic acid in benzene will be
. K kg mol − 1)
(K f for benzene = 512
(2017 Main)
(a) 64.6 %
(b) 80.4 %
of ethanol to 500 g of water changes the freezing point of the
solution. Use the freezing point depression constant of water
as 2 K kg mol −1 . The figures shown below represent plots of
vapour pressure (V.P.) versus temperature (T ). [Molecular
weight of ethanol is 46 g mol −1 ]
(2017 Adv.)
Among the following, the option representing change in the
freezing point is
(b) 1.0 g
(2019 Main, 12 Jan II)
(c) 2.4 g
(a)
(d) 1.5 g
V.P./bar
(a) 1.8 g
(c) 3A
(d) A
8. K 2HgI4 is 40% ionised in aqueous solution. The value of its
van’t Hoff factor (i) is
(a) 1.6
(b) 1.8
(c) 2.2
1
(d)
Water+Ethanol
270 273
T/K
Water+Ethanol
271 273
T/K
Water
Ice
Ice
1
T/K
Water
Ice
Water+Ethanol
271 273
T/K
15. Consider separate solution of 0.500 M C2 H5 OH (aq),
(d) 2.0
−0.2° C, while it should have been −0.5°C for pure milk.
How much water has been added to pure milk to make the
diluted sample?
(2019 Main, 11 Jan I)
(a) 2 cups of water to 3 cups of pure milk
(b) 1 cup of water to 3 cups of pure milk
(c) 3 cups of water to 2 cups of pure milk
(d) 1 cup of water to 2 cups of pure milk
10. Elevation in the boiling point for 1 molal solution of
glucose is 2 K . The depression in the freezing point for 2
molal solution of glucose in the same solvent is 2 K. The
relation between K b and K f is
(2019 Main, 10 Jan II)
(b) K b = 0.5 K f
(a) K b = 15
. Kf
(c) K b = K f
(d) K b = 2K f
11. A solution contain 62 g of ethylene glycol in 250 g of water
−1
is cooled upto –10º C. If K f for water is 1.86 K kg mol ,
then amount of water (in g) separated as ice is
(2019 Main, 9 Jan II)
(b) 48
Water+Ethanol
Water
1
(2019 Main, 11 Jan II)
9. The freezing point of a diluted milk sample is found to be
(a) 32
(c)
V.P./bar
(b) 2A
Ice
(b)
270 273
freezing point of 12% aqueous solution of Y . If molecular
weight of X is A, then molecular weight of Y is
(a) 4A
Water
1
7. Freezing point of a 4% aqueous solution of X is equal to
(2019 Main, 12 Jan I)
(d) 94.6 %
14. Pure water freezes at 273 K and 1 bar. The addition of 34.5 g
(Given that K f = 5 K kg mol− 1 , molar mass of benzoic
acid = 122 g mol− 1 )
(c) 74.6 %
V.P./bar
(b) 0.36 K
V.P./bar
(a) 0.18 K
(c) 64
(d) 16
0.100 M Mg 3 (PO4 )2 (aq), 0.250 M KBr (aq) and 0.125 M
Na 3 PO4 (aq) at 25°C. Which statement is true about these
solution, assuming all salts to be strong electrolytes?
(a) They all have the same osmotic pressure
(2014 Main)
(b) 0.100 M Mg 3 (PO4 )2 (aq) has the highest osmotic pressure
(c) 0.125 M Na 3 PO4 (aq) has the highest osmotic pressure
(d) 0.500 M C2 H5 OH (aq) has the highest osmotic pressure
16. For a dilute solution containing 2.5 g of a non-volatile
non-electrolyte solute in 100 g of water, the elevation in
boiling point at 1 atm pressure is 2°C. Assuming concentration
of solute is much lower than the concentration of solvent, the
vapour pressure (mm of Hg) of the solution is (take K b = 0.76
K kg mol −1 ).
(2012)
(a) 724
(b) 740
(c) 736
(d) 718
17. The freezing point (in° C) of solution containing 0.1 g of
K 3 [Fe(CN)6 ] (mol. wt. 329) in 100 g of water
( K f = 1.86 K kg mol −1 ) is
(a) − 2.3 × 10−2
(b) −5.7 × 10−2
(c) −5.7 × 10−3
(d) −1.2 × 10−2
(2011)
Solutions and Colligative Properties
(a) 0.5
(b) 1
(c) 2
(d) 3
19. The elevation in boiling point, when 13.44 g of freshly
prepared CuCl 2 are added to one kilogram of water, is. [Some
useful data, K b = 0.52 K kg mol − 1 , molecular weight of
(2005, 1M)
CuCl 2 =134.4 g].
(a) 0.05
(b) 0.1
(c) 0.16
(d) 0.21
20. 0.004 M Na 2 SO4 is isotonic with 0.01 M glucose. Degree of
dissociation of Na 2 SO4 is
(2004, S, 1M)
(a) 75%
(b) 50%
(c) 25%
(d) 85%
21. During depression of freezing point in a solution the
22.
23.
24.
25.
following are in equilibrium
(2003)
(a) liquid solvent, solid solvent
(b) liquid solvent, solid solute
(c) liquid solute, solid solute
(d) liquid solute, solid solvent
The molecular weight of benzoic acid in benzene as
determined by depression in freezing point method
corresponds to
(a) ionisation of benzoic acid
(1996, 1M)
(b) dimerisation of benzoic acid
(c) trimerisation of benzoic acid
(d) solvation of benzoic acid
The freezing point of equimolal aqueous solutions will be
highest for
(1990, 1M)
(a) C6 H5 NH3 Cl (aniline hydrochloride)
(b) Ca(NO3 )2
(c) La(NO3 )3
(d) C6 H12 O6 (glucose)
Which of the following 0.1 M aqueous solution will have the
lowest freezing point?
(1989, 1M)
(a) Potassium sulphate
(b) Sodium chloride
(c) Urea
(d) Glucose
When mercuric iodide is added to the aqueous solution of
potassium iodide
(1987, 2M)
(a) freezing point is raised
(b) freezing point is lowered
(c) freezing point does not change
(d) boiling point does not change
Objective Questions II
(One or more than one correct option)
26. In the depression of freezing point experiment, it is found that
the
(1999, 3M)
(a) vapour pressure of the solution is less than that of pure
solvent
(b) vapour pressure of the solution is more than that of pure
solvent
(c) only solute molecules solidify at the freezing point
(d) only solvent molecules solidify at the freezing point
27. The osmotic pressure of a solution of NaCl is 0.10 atm and
that of a glucose solution is 0.20 atm. The osmotic pressure of
a solution formed by mixing 1 L of the sodium chloride
solution with 2 L of the glucose solution is x × 10−3 atm. x is
……… (nearest integer).
(2020 Main, 4 Sep II)
28. On dissolving 0.5 g of a non-volatile non-ionic solute to 39 g of
benzene, its vapour pressure decreases from 650 mmHg to 640
mmHg. The depression of freezing point of benzene (in K)
upon addition of the solute is ............
(Given data : Molar
mass and the molal freezing point depression constant of
benzene are 78 g mol − 1 and 512
. K kg mol − 1 , respectively).
(2019 Adv.)
29. The plot given below shows p −T curves (where p is the
pressure and T is the temperature) for two solvents X and Y
and isomolal solution of NaCl in these solvents. NaCl
completely dissociates in both the solvents.
1
2
3
4
760
1. Solvent X
2. Solution of NaCl in solvent X
3. Solvent Y
4. Solution of NaCl in solvent Y
367
368
(2007, 3M)
Numerical Answer Type Questions
362
of benzene (K f = 1.72 K kg mol − 1 ), a freezing point
depression of 2 K is observed. The van’t Hoff factor ( i ) is
360
18. When 20 g of naphthoic acid (C11 H8 O2 ) is dissolved in 50 g
Pressure (mmHg)
144
Temperature (K)
On addition of equal number of moles of a non-volatile solute
S in equal amount (in kg) of these solvents, the elevation of
boiling point of solvent X is three times that of solvent Y.
Solute S is known to undergo dimerisation in these solvents.
If the degree of dimerisation is 0.7 in solvent Y , the degree of
dimerisation in solvent X is ____.
(2018 Adv.)
Subjective Questions
30. 75.2 g of C 6 H5 OH (phenol) is dissolved in a solvent of
K f = 14. If the depression in freezing point is 7 K, then find
the percentage of phenol that dimerises.
(2006, 2M)
31. 1.22 g C6 H5 COOH is added into two solvents and data of ∆Tb
and K b are given as :
(i) In 100 g CH3COCH3 ∆Tb = 0.17, K b = 1.7 K kg/mol
(ii) In 100 g benzene, ∆Tb = 0.13 and K b = 2.6 K kg/mol
Find out the molecular weight of C6 H5 COOH in both the
cases and interpret the result.
(2004, 2M)
Solutions and Colligative Properties
32. Consider the three solvents of identical molar masses. Match
their boiling point with their K b values
Solvents
Boiling point
Kb values
X
100°C
0.92
Y
27°C
0.63
Z
283°C
0.53
145
Freezing point depression constant of ethanol
( K ethanol
) = 2.0 K kg mol −1
f
Boiling point elevation constant of water
( K bwater ) = 0.52 K kg mol −1
Boiling point elevation constant of ethanol
( K bethanol ) = 1.2 K kg mol −1
Standard freezing point of water = 273 K
(2003)
33. To 500 cm3 of water, 3.0 × 10–3 kg of acetic acid is added. If
23% of acetic acid is dissociated, what will be the
depression in freezing point? K f and density of water are
1.86 K kg –1 mol –1 and 0.997 g cm−3 , respectively.
(2000, 3M)
34. Nitrobenzene is formed as the major product along with a
minor product in the reaction of benzene with a hot mixture of
nitric acid and sulphuric acid. The minor product consists of
carbon : 42.86%, hydrogen : 2.40%, nitrogen : 16.67% and
oxygen : 38.07%,
(i) Calculate the empirical formula of the minor product.
(ii) When 5.5 g of the minor product is dissolved in 45 g of
benzene, the boiling point of the solution is 1.84°C higher
than that of pure benzene. Calculate the molar mass of the
minor product then determine its molecular and structural
formula. (Molal boiling point elevation constant of benzene is
2.53 K kg mol − 1 ).
(1999)
35. A solution of a non-volatile solute in water freezes at
− 0.30° C.The vapour pressure of pure water at 298 K is 23.51
mm Hg and K f for water is 1.86 K kg mol −1 . Calculate the
vapour pressure of this solution at 298 K.
(1998, 4M)
Standard freezing point of ethanol = 155.7 K
Standard boiling point of water = 373 K
Standard boiling point of ethanol = 351.5 K
Vapour pressure of pure water = 32.8 mm Hg
Vapour pressure of pure ethanol = 40 mm Hg
Molecular weight of water = 18 g mol −1
Molecular weight of ethanol = 46 g mol −1
In answering the following questions, consider the solutions
to be ideal dilute solutions and solutes to be non-volatile and
non-dissociative.
(2008, 3 × 4M = 12M)
37. The freezing point of the solution M is
(a) 268.7 K
(c) 234.2 K
(b) 268.5 K
(d) 150.9 K
38. The vapour pressure of the solution M is
(a) 39.3 mm Hg
(c) 29.5 mm Hg
(b) 36.0 mm Hg
(d) 28.8 mm Hg
39. Water is added to the solution M such that the mole fraction
of water in the solution becomes 0.9. The boiling point of this
solution is
(a) 380.4 K
(b) 376.2 K
(c) 375.5 K
(d) 354.7 K
36. Addition of 0.643 g of a compound to 50 mL of benzene
(density : 0.879 g/mL) lowers the freezing point from 5.51° C
to 5.03° C. If K f for benzene is 5.12, calculate the molecular
weight of the compound.
(1992, 2M)
Passage Based Questions
Passage 1
Properties such as boiling point, freezing point and vapour pressure
of a pure solvent change when solute molecules are added to get
homogeneous solution. These are called colligative properties.
Applications of colligative properties are very useful in day-to-day
life.
One of its examples is the use of ethylene glycol and water mixture
as anti-freezing liquid in the radiator of automobiles.
A solution M is prepared by mixing ethanol and water. The mole
fraction of ethanol in the mixture is 0.9.
Given, freezing point depression constant of water
( K water
) = 1.86 K kg mol −1
f
Fill in the Blank
40. Given that ∆T f is the depression in freezing point of the
solvent in a solution of a non-volatile solute of molality, m,
the quantity lim ( ∆T f / m ) is equal to ......
(1994, 1M)
m→ 0
Integer Answer Type Question
41. If the freezing point of a 0.01 molal aqueous solution of a
cobalt (III) chloride-ammonia complex (which behaves as a
strong electrolyte) is − 0.0558°C, the number of chloride(s)
in the coordination sphere of the complex is
[ K f of water = 186
. K kg mol −1 ]
(2015 Adv.)
42. MX 2 dissociates into M 2+ and X − ions in an aqueous
solution, with a degree of dissociation (α ) of 0.5. The ratio of
the observed depression of freezing point of the aqueous
solution to the value of the depression of freezing point in the
absence of ionic dissociation is
(2014 Adv.)
Answers
Topic 1
Topic 2
1. (d)
2. (c)
3. (b)
4. (a)
5. (a)
6. (d)
7. (b)
8. (b)
9. (b)
10. (b)
11. (a)
12. (b)
13. (b)
14. (c)
15. (b, d)
16. (a, b)
17. (0.2)
18. (2.98 M)
19. (19)
20. (T)
22. (0.72)
23. (18.5)
24. (746.32 mm)
25. (65.25)
26. (0.75)
27. (0.657)
30. (0.158)
31. (180.40 mL)
1.
5.
9.
13.
17.
21.
25.
29.
36.
40.
(c)
(d)
(c)
(d)
(a)
(a)
(a)
(0.05)
(156 g/mol)
(K f )
2.
6.
10.
14.
18.
22.
26.
30.
37.
42.
(c)
(c)
(d)
(b)
(a)
(b)
(a,d)
(75%)
(d)
(2)
3.
7.
11.
15.
19.
23.
27.
32.
38.
(c)
(c)
(c)
(a)
(c)
(d)
(167.00)
(0.23°C)
(a)
4.
8.
12.
16.
20.
24.
28.
35.
39.
(b)
(b)
(d)
(a)
(a)
(a)
(1.02)
(23.44 mm)
(b)
Hints & Solutions
Topic 1 Solution and Vapour Pressure
2.
of Liquid Solutions
1. Henry’s equation for solubility (S ) of a gas (β) in a liquid is
expressed in terms of mole- fraction of the gas (χ B ) at a given
temperature,
p = KH × χ B = KH × S
1
So, solubility of gas, S ∝
at T (K) and given pressure.
KH
Order of solubility of the gases (high the value of K H , lower is
the solubility) :
γ >δ >β >α
So, option (a) is not correct.
Again, K H ∝ temperature, i.e. solubility of the gas will decrease
with increase in temperature also.
But this conclusion cannot be drawn from table.
So, option (b) is not correct.
We know, mole-fraction of a solute ( B ) in a binary aqueous
solution,
18 m
[Qm = molality]
χB =
1000 + 18 m
and also,
X solvent = 0.8 (Given) It means that nsolvent (n1 ) = 0.8 and
nsolute (n2 ) = 0.2
1000
1000
Using formula m = n2 ×
= 0.2 ×
= 13.88 mol kg
n1 × M 1
0.8 × 18
−1
3.
The molality of 20% (mass/mass) aqueous solution of KI
can be calculated by following formula.
w × 1000
m= 2
Mw2 × w1


18 × 55.5
p = K Hγ × m = (2 × 10−5 ) × 

 1000 + 18 × 55.5
20% aqueous solution of KI means that 20 gm of KI is present in
80 gm solvent.
20 1000
m=
×
= 1. 506 ≈ 1. 51mol/kg
166
80
= 1. 81 × 10−5 K bar = 1. 8 × 10−2 bar
So, option (c) is not correct.
For δ at m = 55.5 molal
So, option (d) is correct.
Key Idea Molality is defined as number of moles of solute
per kg of solvent.
w2
1000
m =
×
Mw2
w1
w2 = mass of solute, Mw2 = molecular mass of solute
w1 = mass of solvent.
For γ at m = 55.5 molal


18 × 55.5
p = K Hδ = (0.5) × 

 1000 + 18 × 55.5
~ 250bar
= 0.2498k bar −
Mass of solute (w2 ) × 1000
Molar mass of solute (M 2 ) ×
mass of solvent (w1 )
w2 1000
m=
×
M2
w1
1000
m = n2 ×
n1 × M1
Key Idea Molality ( m) =
4.
Key Idea For a solution of volatile liquids the partial vapour
pressure of each component of the solution is directly
proportional to its mole fraction present in solution. This is
known as Raoult’s law.
Liquid M and N form an ideal solution. Vapour pressures of pure
liquids M and N are 450 and 700 mm Hg respectively.
Solutions and Colligative Properties
∴
pºN > pºM
So, by using Raoult’s law
yN > xN
and
xM > yM
Multiplying (i) and (ii) we get
yN xM > yM xN
xM yM
∴
>
xN
yN
147
p°A = 7 × 103 Pa, p°B = 12 × 103 Pa
…(i)
…(ii)
On substituting the given values in Eq. (i),
we get
p = 0.4 × 7 × 103 + 0.6 × 12 × 103
= 10 × 103 Pa = 1 × 104 Pa
In vapour phase,
0.4 × 7 × 103
p
x′ p°
= 0.28
xA = A = A A =
p
p
1 × 104
Thus, correct relation is (a).
∴
5. According to Henry’s law (at constant temperature)
pgas = K H × χ gas (solute) = K H × [1 − χ H 2 O (solvent) ]
pgas = K H − K H χ H 2 O
pgas = partial pressure of the gas above its solution with a liquid
(solvent) say water.
χ gas = mole fraction of the gas (solute) in the solution.
χ H 2 O = mole fraction of water (solvent).
xB = 1 − 0.28 = 0.72
[Q xA + xB = 1]
8. At constant temperature, solubility of a gas (S ) varies inversely
with Henry’s law constant (K H )
Pressure
P
KH =
=
Solubility of a gas in a liquid S
Thus, higher the value of K H at a given pressure, the lower is the
solubility of the gas in the liquid.
9. Key Idea Vapour pressure of water ( p° ) = 760 torr
pgas
KH
Mass (g)
Molecular mass (g mol −1 )
18 g
=
= 0.1 mol
180 gmol −1
Number of moles of glucose =
pgas
KH
Molar mass of water = 18 g/mol
χH2O=0
χgas=1
χH2O=0
χgas=0
[ i.e. pgas = K H ] Higher the value of K H, higher
will be the partial pressure of the gas ( pgas ), at a given
temperature. The plot of pgas vs χ H2O gives a (−ve) slope.
pgas = K H − K H × χ H2O
Comparing the above equation with the equation of straight line
y = mx + c
Slope = − K H , intercept = K H
So, (i) Higher the value of K H, more (−ve) will be the slope and it
is for z (K H = 40 K bar )
(ii) Higher the value of K H, higher with the value of intercept, i.e.
partial pressure and it is also for z.
6. (d) According to Dalton’s law of partial pressure
ptotal = pA + pB = pA° χ A + pB° χ B
Given,
pAº
= 400 mm Hg,
pBº
= 600 mm Hg
χ B = 0.5, χ A + χ B = 1 ∴ χ A = 0.5
On substituting the given values in Eq. (i). We get,
ptotal = 400 × 0.5 + 600 × 0.5 = 500 mm Hg
Mole fraction of A in vapour phase,
pAº χ A 0.5 × 400
p
YA = A =
=
= 0.4
ptotal
ptotal
500
Mole of B in vapour phase,
YA + YB = 1
YB = 1 − 0. 4 = 0.6
7. For ideal solution,
Q
p = x′ A p°A + x′ B p°B
x′ A = 0.4, x′ B = 0.6
…(i)
Mass of water (given) = 178.2g
Number of moles of water
Mass of water
178. 2g
=
=
= 9.9 mol
Molar mass of water 18 g / mol
Total number of moles = (0.1 + 9.9) moles = 10 moles
Now, mole fraction of glucose in solution = Change in pressure
with respect to initial pressure
∆p 01
.
or ∆p = 0.01p° = 0.01 × 760 = 7.6 torr
i.e.
=
p° 10
∴ Vapour pressure of solution = (760 − 7.6) torr = 752.4 torr
10. Given, p° = 185 Torr at 20°C
ps = 183 Torr at 20°C
Mass of non-volatile substance, m = 1.2 g
Mass of acetone taken = 100g
M =?
p° − ps
n
As, we have
=
ps
N
Putting the values, we get,
1.2
185 − 183 M
2
1.2 × 58
=
⇒
=
100
183
183 100 × M
58
183 × 1. 2 × 58
M =
∴
2 × 100
M = 63.684 = 64 g/mol
11. Give, K H = 1 × 105 atm, χ N 2 = 0.8
nH 2 O = 10 moles, ptotal = 5 atm
pN 2 = ptotal × χ N 2 = 5 × 0.8 = 4 atm
148
Solutions and Colligative Properties
According to Henry’s law,
pN 2 = K H × χ N 2
4 = 105 × χ N 2
χ N 2 = 4 × 10
nN 2
nN 2 + nH 2 O
nN 2
nN 2 + 10
17. Using Raoult’s law equation for a mixture of volatile liquids.
pT = p°A χ A + pB° χ B
−5
= 4 × 10−5
nN 2 = 4 × 10−4
12. Molality = moles of solute present in 1.0 kg of solvent.
13. The relative lowering of vapour pressure :
(mole fraction of solute)
14. In case of positive deviation from Raoult's law, the observed
vapour pressure is greater than the ideal vapour pressure and
boiling point of azeotrope becomes lower than either of pure
liquid.
Z
Real
= 88157
. mL
Now, molarity
= Number of moles of solute ×
al
Ide
=
xM
… (ii)
nurea = 2.6315 moles
wurea = nurea × (M ⋅ wt )urea = (2. 6315 × 60)g
2.6315 × 60 + 900 
Mass of solution 
V=
Q Density =

12
.
Volume of solution 
because the observed vapour pressure of L is greater than the
ideal pressure
1
0.4 = 0.5 χ p°A + 0.5 χpB°
Thus, the vapour pressure of pure liquid B in bar is 0.2.
Number of moles of solute × 1000
18. Key Idea Molarity (M ) =
Volume of solution (in mL)
Mass
Also, volume =
Density
Given, mole fraction of urea (χ urea ) = 0.05
Mass of water = 900g
Density = 12
. g/cm 3
nurea
900
[Q Moles of water =
χ urea =
= 50]
nurea + 50
18
nurea
0.05 =
⇒ 19nurea = 50
nurea + 50
15. The graph shown indicates that there is positive deviation
pL
… (i)
By solving equation (i) and (ii)
p°A = 0.6 bar and pB° = 0.2 bar
= 4 × 10−5
− ∆p
= χ2
p°
0.3 = 0.25 χ pA° + 0.75 χpB°
0
Since, deviation is positive, the intermolecular force between L
and M is smaller than the same in pure L and pure M.
Also as xL → 1, xM → 0, the real curve approaching ideal curve
where Raoult’s law will be obeyed.
16. When intermolecular attraction between two components A and
B in the mixture is same as between A and A or B and B, hence it
is a case of ideal solution.
When intermolecular attraction between A and B in a mixture is
smaller than that between A and A or B and B, then mixture is
more vaporised, bp is lowered. It is a case of positive deviation
from Raoult’s law.
When intermolecular attraction between A and B is higher than
that between A and A or B and B, then mixture is less vaporised,
bp is increased. It is a case of negative deviation.
(a) Methanol molecules (CH3OH) are hydrogen bonded. In a
mixture of CCl 4 and CH3OH, extent of H-bonding is
decreased. Mixture is more vaporised thus, positive deviation
from Raoult’s law.
(b) Acetone molecules have higher intermolecular attraction due
to dipole-dipole interaction. With CS2, this interaction is
decreased thus, positive deviation.
(c) Mixture of benzene and toluene forms ideal solution.
(d) Phenol and aniline have higher interaction due to
intermolecular H-bonding. Hence, negative deviation.
1000
Volume of solution (mL)
2.6315 × 1000
= 2.98 M
88157
.
19. Key Idea Use the formula
pTotal = pA° × χ A + pB° × χ B
1
and for equimolar solutions χ A = χ B =
2
Given, pTotal = 45 torr for equimolar solution
pA° = 20 torr
1
1 1
So,
45 = pA° × + pB° × = ( pA° + pB° )
2
2 2
or
pA° + pB° = 90 torr
…(i)
But we know pA° = 20 torr
so,
pB° = 90 − 20 = 70torr
(From Eq. (i))
Now, for the new solution from the same formula
Given,
pTotal = 22 . 5 torr
(As χ A + χ B =1)
So,
22 . 5 = 20χ A + 70 (1− χ A )
or
22.5 = 70 − 50χ A
70 − 22 . 5
So,
χA =
= 0.95
50
Thus
(as χ A + χ B =1)
χ B = 1 − 0.95 = 0.05
Hence, the ratio
χ A 0.95
=
= 19
χ B 0.05
Solutions and Colligative Properties
20. It will be true only if boiling points of two liquids are
significantly different.
n1
n1 + in2
 100


 18 
= 0.982
=
7
 100
+
2.4
×


 18 
164
⇒ Mole fraction of solvent =
21. Let after 100 min, x moles of A are remaining unpolymerised
moles of B = 12
Moles of non-volatile solute = 0.525
Mole fraction of A =
⇒
Mole fraction of B =
⇒
χ
χ + 12 + 0.525
⇒
χ = 9.9
⇒ Moles of A polymerised in 100 min = 10 – 9.9 = 0.10
1
10
1
10
k = ln
=
ln
min −1
⇒
t
9.9 100
9.9
= 1.005 × 10−4 min −1
78
mL = 88.94 mL
22. Volume of 1.0 mole liquid benzene =
0.877
⇒ Molar volume of benzene vapour at 20°C
88.94 × 2750
L = 244.58 L
=
1000
0.082 × 293
× 760 mm
⇒ VP of pure benzene at 20°C =
244.58
= 74.65 mm
Similarly; molar volume of toluene vapour
92
7720
=
L = 819.2L
×
0.867 1000
0.082 × 293
⇒ VP of pure toluene =
× 760 mm = 22.3mm
819.2
Now, let mole fraction of benzene in the liquid phase = χ
⇒
4.65 χ + 22.3 (1 – χ ) = 45
⇒
χ = 0.43
⇒ Mole fraction of benzene in vapour phase
Partial vapour pressure of benzene
=
Total vapour pressure
74.65 × 0.43
=
= 0.72
45
23. Vapour pressure of solution = 0.75 × VP of water
⇒
75 = 100 χ 1 : χ 1 = mole fraction of solute
1
3
χ 1 = and χ 2 = 1 − χ 1 =
4
4
100
n
χ 2 n2 1
= 1.85
=
= ⇒ n2 = 1 =
3 18 × 3
χ 1 n1 3
⇒
⇒
⇒ Weight of urea = 1.85 × 60 = 111 g
n
1000 1 1000
Molality = 2 ×
= 18.5
= ×
n1
M1
3
18
24. Ca(NO3 )2 r Ca 2+ + 2NO3−
1−α
α
⇒
2α
i = 1 + 2α where, α = 0.7
⇒ i = 1 + 2 × 0.7= 2.4
p = p0 χ1 = 760 × 0.982
(VP of H2O at 100°C = 760 mm of Hg)
= 746.32 mm
12
χ + 12 + 0.525




12
χ
400 = 
 × 300 + 
 × 500
 χ + 12.525
 χ + 12.525
149
25. According to Raoult’s law :
p = p0 χ1
 n1 
600 = 640 

 n1 + n2 
⇒
n2 64
1
=
−1=
n1 60
15
39 1
×
= 0.033
n2 =
78 15
2.175
= 0.033
M
M = 65.25
⇒
⇒
⇒
⇒
26. At 373 K (bp) of H2O, Vapour pressure = 760 mm
VP of solution at 373 K = 750 mm
⇒
p = p0χ 1 or 750 =760 χ 1
75
⇒
χ1 =
= mole fraction of H2O
76
75 1
χ2 = 1 −
=
= mole fraction of solute
⇒
76 76
n2
1
Now
=
n1 + n2 76
n1
= 75
n2
⇒
⇒
Molality =
n2
1000
× 1000 =
= 0.74 molal
n1M 1
75 × 18
60
= 1.3
46
40
Moles of methanol =
= 1.25
32
27. Moles of ethanol =
1.3
= 0.51
1.3 + 1.25
⇒ Vapour pressure of solution = pethanol + pmethanol
= 0.51 × 44.5 + 0.49 × 88.7
= 66.16 mm
Mole fraction of methanol in vapour phase
pmethanol
43.463
=
=
= 0.657
Total vapour pressure
66.16
⇒ Mole fraction of ethanol =
28. From lowering of vapour pressure information :
⇒
0.104
n2
= χ2 =
17.5
n1 + n2
n1
+ 1 = 168.27
n2
150
Solutions and Colligative Properties
⇒
n1
= 167.27
n2
⇒
1000 M
×
= 167.27
18
50
For the relation, π = CRT =
Given, mass of urea = 0.6 g
⇒
M = 150 g/mol
Also, the combustion reaction is :
CxH2 yO y + xO2 → xCO2 + yH2O
Q 18 y g of H2O is produced from 1.0 mole of compound.
0.9
1
=
mol
∴0.9 g of H2O will be produced from
18 y 20 y
x
⇒ At the end, moles of O2 left =
20 y
x
moles of CO2 formed =
20 y
2x
2.24
=
⇒ Total moles of gases at STP =
20 y 22.4
Molar mass of urea = 60 g mol− 1
Mass of glucose = 1.8 g
Molar mass of glucose = 180 g mol− 1
[ n (urea ) + n2 (glucose)]
π= 2
RT
V
. 
 0.6 18
+


 60 180
=
× 1000 × 0.0821 × 300
100
= (0.01 + 0.01) × 10 × 0.0821 × 300
π = 4 . 92 atm
2. The expression of elevation of boiling point,
⇒
x= y
⇒ Molar mass; 150 = 12x + 2x + 16x = 30x
150
⇒
x=
=5
30
Formula = C5H10O5
⇒
∆Tb = K b × m × i = k b ×
When 1.0 mole of A is mixed with 4 moles of B.
560 = 0.20 pA° + 0.80 pB°
…(i)
…(ii)
Now, solving (i) and (ii) pA° = 400 mm
pB° = 600mm.
30. According to Raoult’s law :
⇒
⇒
p = p0χ 1 ⇒631.9 = 639.7 χ 1
χ 1 = 0.9878 ⇒ χ 2 = 0.0122
0.0122
Molality =
× 1000 = 0.158
0.9878 × 78
31. Let us consider 1.0 L of solution.
Weight of solution = 1000 × 1.02 = 1020 g
13
= 132.60 g
Weight of H2SO4 = 1020 ×
100
Weight of H2O = 1020 – 132.60 = 887.40 g
132.60
Molarity =
⇒
= 1.353 M
98
132.60
1000
Molality =
×
= 1.525 m
98
887.40
Normality = 2 × M = 2.706
2.706 × 100 = 1.5 V ⇒ V = 180.40 mL
⇒
Topic 2 Colligative Properties
1.
Key Idea Osmotic pressure is proportional to the molarity
(C) of the solution at a given temperature (T).
Thus, π ∝ C, π = CRT (for dilute solution)
n
π =
RT
V
w2 × 1000
×i
M 2 × w1
where, m = molality
i = van’t Hoff factor = 1(for
non-electrolyte/non-associable)
w2 = mass of solute in g = 1g (present in both of the solutions)
M 2 = molar mass of solute in g mol −1 (same solute in both
of the solutions)
w1 = mass of solvent in g = 100 g (for both of the
solvents A and B)
K b = ebullioscopic constant
So, the expression becomes,
∆Tb ∝ K b

∆Tb ( A ) K b ( A ) 1
K b ( A) 1 
=
=
⇒
Given K ( B ) = 5 
∆Tb ( B ) K b ( B ) 5


b
29. When 1.0 mole of A is mixed with 3 moles of B.
550 = 0.25 pA° + 0.75 pB°
n
RT
V
3.
Key Idea For dilute solution, lowering of vapour pressure
∆p
p0
which is a colligative property of solutions.
∆p
= χ B × i ⇒ ∆p = χ B × i × p0
p0
(∆p) = p0 − p and relative lowering of vapour pressure =
where, p0 = vapour pressure of pure solvent
i = van’t Hoff factor
χ B = mole fraction of solute
Given,
p° = vapour pressure of pure water of 25º C = 35 mm Hg
χ B = mole fraction of solute (urea)
0.60
nB
0.01
0.01
60
=
=
=
=
= 0.0005
360
0
.
60
nA + nB
20 + 0.01 20.01
+
18
60
i = van’t Hoff factor = 1 (for urea)
Now, according to Raoult’s law
∆p = χ B × i × pº
On substituting the above given values, we get
∆p = 0.0005 × 1 × 35 = 0.0175 mm Hg
Solutions and Colligative Properties
4.
Key Idea Depression in freezing point (∆ Tf ) is given by
∆Tf = iK f m
i = vant Hoff factor
K f = molal depression constant
m = molality
K f = 4.0 K kg mol −1
m = 0.03 mol kg
(Given)
−1
(Given)
∆T f = ?
For K 2SO4, i = 3
It can be verified by the following equation :
K 2SO4
2K + + SO2−
4
-
Using formula
∆T f = iK f × m
∆T f = 3 × 4 × 0.03 = 0.36 K
5. Key Idea Osmotic pressure is proportional to the molarity (C)
of the solution at a given temperature, π = CRT
Concentration of BaCl2 = 0.01M
π XY = 4 π BaCl 2
i × CRT = 4 × i × CRT
For the calculation of i,
XY → X + + Y −
BaCl2 → Ba 2+ + 2Cl−
On substituting all the given values in Eq. (i), we get
0.6 × 5 × w × 1000
, w = 2.44 g
2=
122 × 30
Thus, weight of acid (w) is 2.4 g.
7. Given, Freezing point of 4% aqueous solution of X .
= Freezing point of 12% aqueous solution of Y
[Q∆T f = T f° − T f ]
or
(∆T f )X = (∆T f )Y
K f × mX = K f mY
where, mX and mY are molality of X and Y, respectively.
or
mX = mY
Number of moles of solute (n)
Now,
molality =
Mass of solvent (in kg)
Weight
n=
Molecular mass
wX
wY
=
M X × (wsolvent )1 MY × (wsolvent )2
Given,
(Given)
(Given)
…(i)
(Here, i = 2)
(Here, i = 3)
∴
Thus,
Putting the values of i in (i)
2 × [ XY ] = 4 × 3 × [ BaCl2 ]
2 × [ XY ] = 12 × 0.01
12 × 0.01
[ XY ] =
2
So, the concentration of XY = 0.06 mol L−1
= 6 × 10−2 mol L−1
6. Molecules of benzoic acid dimerise in benzene as:
-
2(C6H5COOH)
(C6H5COOH)2
Now, we know that depression in freezing point (∆T f ) is given
by following equation:
i × K f × wsolute × 1000
...(i)
∆T f = i × K f × m =
Mwsolute × wsolvent
Given, wsolute (benzoic acid) = w g
wsolvent (benzene) = 30g
Mw Solute (benzoic acid) = 122 g mol− 1, ∆T f = 2 K
K f = 5 Kkg mol− 1, %α = 80 or α = 0.8
2(C6H5COOH)
Initial
1
Final
1−α
= 1 − 0. 8 = 0. 2
-
(C6H5COOH)2
0
α/2
0. 8 / 2 = 0. 4
Total number of moles at equilibrium = 0.2 + 0.4 = 0.6
Number of moles at equilibrium
i=
Number of moles present initially
0.6
i=
= 0.6
1
151
wX = 4 and w(solvent )1 =96
wY = 12and w(solvent )2 = 88
MX = A
4 × 1000 12 × 1000
=
M X × 96 MY × 88
12 × 1000 × M X × 96
MY =
4 × 1000 × 88
96 × 12
=
× A = 3.27A ≈ 3A
4 × 88
8. The ionisation of K2HgI4 in aqueous solution is as follows:
K 2 [ HgI4 ]
-
2K + + [ HgI4 ]2 −
van’t Hoff factor (i) for ionisation reaction is given as,
i = 1 + α ( n − 1)
where,
n = number of ions,
α = degree of ionisation or dissociation
From above equation, it is clear that n = 3
i = 1 + 0.4 (3 − 1)
[Given, % α = 40% or α = 0.4]
= 1.8
9. We know that,
Depression in freezing points (∆T f )
T ° f − Tf = K f × m × i
where,
K f = molal depression constant
wsolute × 1000
m = molality =
M solute × wsolvent (in g)
i = van’t Hoff factor
For diluted milk
∆T f1 = K f × m1 × i
wmilk × 1000
⇒ 0 − (0.2) ⇒ 0.2 = K f ×
×1
M milk × w1 (H2O)
For pure milk
∆T f2 = K f × m2 × i
152
Solutions and Colligative Properties
⇒ 0 − (−0.5) = 0.5 = K f ×
wmilk × 1000
×1
M milk × w2 (H2O)
[Co(H2O)5 Cl]Cl 2 ⋅ H2O
2
wmilk × 1000
M
× w2 (H2O) w2 (H 2O)
0.2 K f
So,
× milk
=
×
=
M milk × w1 (H2O)
wmilk × 1000
0.5 K f
w1 (H2O)
⇒
1
Moles at equilibrium
‘i’ is 3
−
2
‘i’ is 2
(CH3 COOH)2
0
1−α
α
2
α
α
α
or i = 1 −
=1−
2
2
2
Now, depression in freezing point (∆T f ) is given as
∆T f = i K f m
where, K f = molal depression constant or
cryoscopic constant.
m = molality
K b = 2K f
…(i)
K(i)
number of moles of solute 0.2 1000
=
×
20
weight of solvent (in kg) 60
Molality =
= 0°C, T f = − 10°C
Putting the values in Eq. (i)
wB = mass of ethylene glycol = 62 g
M B = molar mass of ethylene glycol
∴







wA = mass of water in g as liquid solvent,
i = van’t-Hoff factor = 1 (for ethylene glycol in water)
K f = 1.86 K kg mol −1
On substituting in Eq. (i), we get
62 × 1000
0 − (− 10) = 186
. ×
×1
62 × wA
186
. × 62 × 1000
⇒
wA =
= 186 g
10 × 62
⇒
∴
 α
 0.2 1000 
0.45 = 1 −
(512
. )
×

 60
2 
20 
1−
0.45 × 60 × 20
α
=
2 512
. × 0.2 × 1000
1−
α
= 0.527
2
⇒
α
= 1 − 0.527
2
α = 0.946
Thus, percentage of association = 94.6%
34.5
14. −∆T f = ik f m 2 = 1 × 2 ×
× 1000 = 3
46 × 500
Vapour pressure curves shown in (b) is in agreement with the
calculated value of −∆T f . (a) is wrong, vapour pressure
decreases on cooling.
So, amount of water separated as ice (solid solvent)
= 250 − wA = (250 − 186)g = 64 g
15.
12. Key idea ‘‘Addition of solute particles to a pure solvent results
to depression in its freezing point.’’
All the compounds given in question are ionic in nature so,
consider their van’t Hoff factor (i ) to reach at final conclusion.
The solution with maximum freezing point must have minimum
number of solute particles. This generalisation can be done with
the help of van’t Hoff factor (i ) i.e.
Number of solute particles ∝ van’t Hoff factor (i )
Thus, we can say directly
Solution with maximum freezing point will be the one in which
solute with minimum van’t Hoff factor is present
3+
Now, for Co(H2 O)6 ]Cl 3
[Co(H2 O)6 ]
van’t Hoff factor ( i ) is 4. Similarly for,
-
4
+
∴ Total moles = 1 − α +
∆T f = K f × m × i
wB × 1000
T f° − T f = K f ×
×i
M B × wA (in g )
= 62 g mol
2
2CH3 COOH q
a solution,

 CH2 — CH2


 

OH
 OH

−1
2
Initial moles
11. Considering the expression of the depression in freezing point of
Here, T f°
−
benzene is α, then
Depression is freezing point (∆T f ) = K f × m × i
where, m = molality
For the glucose solution (van’t Hoff factor, i = 1),
∆Tb1m = ∆T f2m = 2K
Kb × 1 × 1 = K f × 2 × 1 ⇒
2
5
13. Let the degree of association of acetic acid (CH3COOH) in
10. Elevation in boiling point (∆Tb ) = K b × m × i
So,
4
2
and for [Co(H2O)3 Cl 3 ]⋅ 3H2O, ‘i’ is 1 as it does not show
ionisation. Hence, [Co(H2O)3 Cl 3 ]⋅ 3H2O have minimum number
of particles in the solution.
So, freezing point of its solution will be maximum.
2
w2 (H2O) (in pure milk)
=
w1 (H2O) (in diluted milk) 5
i.e. 3 cups of water has to be added to 2 cups of pure milk.
2+
-[Co(H O) Cl] + 2Cl
[Co(H O) Cl ]Cl ⋅ 2H O -[Co(H O) Cl ] + Cl
+ 3Cl
−
PLAN This problem includes concept of colligative properties (osmotic
pressure here) and van’t Hoff factor. Calculate the effective
molarity of each solution.
i.e. effective molarity = van’t Hoff factor × molarity
0.5 M C2H5OH (aq)
i=1
Effective molarity = 0 .5
0.25 M KBr (aq)
i=2
Effective molarity = 0 .5 M
i=5
0.1 M Mg3 (PO4 )2(aq)
Effective molarity = 0 . 5 M
i=4
0.125 M Na 3PO4 (aq)
Effective molarity = 0. 5 M
Molarity is same hence, all colligative properties are also same.
NOTE This question is solved by assuming that the examiner has
taken Mg 3(PO 4 )2 to be completely soluble. However, in real it is
insoluble (sparingly soluble).
Solutions and Colligative Properties
16. The elevation in boiling point is
n
∆Tb = K b ⋅ m : m =molality = 2 × 1000
w1
[n2 = Number of moles of solute, w1 = Weight of
solvent in gram]
5
n2
2 = 0.76 ×
× 1000 ⇒ n2 =
⇒
19
100
Also, from Raoult’s law of lowering of vapour pressure :
− ∆p
n2
n
= x2 =
≈ 2
[Q n1 >> n2 ]
p°
n1 + n2 n1
5
18
− ∆p = 760 ×
×
= 36 mm of Hg
⇒
19 100
⇒
p = 760 − 36 = 724 mm of Hg
3–
17. van’t Hoff factor (i) = 4 {3K++ [Fe(CN)6 ]
Molality =
⇒
0.1 1000
1
×
=
329 100 329
– ∆T f = iK f . m
1
= 2.3 × 10–2
329
T f = –2 .3 × 10–2 ° C
= 4 × 1.86 ×
⇒
(As % freezing point of water is 0ºC)
18. Molality = 
20  1000
= 2.325 m
 ×
 172
50
⇒
− ∆T f = 2 = iK f ⋅ m
2
= 0.5
i=
1.72 × 2.325
⇒
19. Molality =
13.44
= 0.1
134.1
i=3
∆Tb = iK b ⋅ m = 3 × 0.52 × 0.1
= 0.156
⇒
20. For isotonic solutions, they must have same concentrations of
ions, Therefore,
0.004 i (Na 2SO4) = 0.01
0.01
= 2.5
⇒
i=
0.004
Also Na 2SO4 r 2Na + + SO24−
2α
1− α
⇒
α
i = 1 + 2α = 2.5
α = 0.75 = 75%
21. During freezing, liquid solvent solidify and solid solvent remains
in equilibrium with liquid solvent.
22. In benzene, benzoic acid dimerises as :
C6H5COOH r
23. C6 H5 NH3 Cl : i = 2;
Ca(NO3 )2 :
i=3
La(NO3 )3 :
C6H12O6 :
i = 4;
i=1
1
(C H COOH)2
2 6 5
153
Lower the value of i, smaller will be the depression in freezing
point, higher will be the freezing temperature, if molalities are
equal. Hence, glucose solution will have highest freezing
temperature.
i=3
NaCl : i = 2
Urea : i = 1
Glucose : i = 1
Greater the value of i, greater the lowering in freezing point,
lower will be the freezing temperature, if molarity in all cases are
same. Therefore, K2SO4 solution has the lowest freezing point.
24. K2SO4 :
25. Addition of HgI2 to KI solution establishes the following
equilibrium :
HgI2 + 2KI r K2[HgI4 ]
The above equilibrium decreases the number of ions (4 ions on
left side of reactions becomes three ions on right side), hence
rises the freezing point.
26. In depression of freezing point experiment, vapour pressure of
solution is less than that of pure solvent as well as only solvent
molecules solidify at freezing point.
27. Osmotic pressure,
π = i × C × RT
Here, i = van’t Hoff factor, T = temperature
C = concentration and R = gas constant.
For NaCl, i = 2
So, π NaCl = i × C NaCl × RT
01
. = 2 × C NaCl × RT
0.05
C NaCl =
RT
For glucose, i = 1because it cannot ionise
So,
π glucose = i × Cglucose × RT
0.2 = 1 × Cglucose × RT
0.2
Cglucose =
RT
(Q nNaCl = numbers of moles NaCl)
nNaCl in 1 L = C NaCl × Vlitre
0.05
=
(nglucose = number of moles of glucose)
RT
0.4
nglucose in 2 L = Cglucose × Vlitre =
RT
Vtotal = 1 + 2 = 3L
0.05
So, final conc. NaCl =
3RT
0.4
Final conc. glucose =
3RT
π total = π NaCl + π glucose = [ i × C NaCl + Cglucose ] RT
0.4 
0.5
 2 × 0.05
atm
=
+
 × RT =
 3RT
3RT 
3
atm = 166.6 × 10−3 atm = 167.00 × 10−3 atm
= 01666
.
So, x = 167.00
154
28.
Solutions and Colligative Properties
Key Idea First calculate, molar mass of solute using the
p° − ps
nsolute
formula,
and then calculate ∆T f
=
p°
nsolute + nsolvent
by applying the formula; ∆T f = K f × m.
When 0.5 g of non-volatile solute dissolve into 39 gm of benzene
then relative lowering of vapour pressure occurs. Hence, vapour
pressure decreases from 650 mmHg to 640 mmHg.
Given, vapour pressure of solvent ( p° ) = 650 mmHg
Vapour pressure of solution ( ps ) = 640 mmHg
Weight of non-volatile solute = 0.5 g
Weight of solvent (benzene) = 39 g
From relative lowering of vapour pressure,
p° − ps
nsolute
= xSolute =
p°
nsolute + nsolvent
0.5
650 − 640
molar
mass
=
0.5
39
650
+
molar mass 78
0.5
10
molar
mass
=
0.5
650
+ 0.5
molar mass
0.5 + 0.5 × molar mass = 65 × 0.5
∴ Molar mass of solute = 64 g
From molal depression of freezing point,
K f × wsolute
∆T f = K f × molality =
(MW )solute × wsolvent
0.5 × 1000
∆T f = 512
. ×
⇒ ∆T f = 102
. K
64 × 39
29. From the graph we can note
∆Tb for solution X i.e.,
∆Tb (X) = 362 − 360 = 2
Likewise, ∆Tb for solution Y i.e., ∆Tb (Y) = 368 − 367 = 1
Now by using the formula
∆Tb = i × molality of solution× K b
For solution X
…(i)
2 = i × mNaCl × K b (X)
Similarly for solution y
……(ii)
1= i × mNaCl × K b (Y)
from Eq. (i) and (ii) above
K b (X) 2
= or 2 or K b (X) = 2K b (Y)
K b (Y) 1
For solute S
2 S → S 2
Initial
α
Final
(1 − α)
So, here
(given due to dimerisation)
0
α
2
 α
i =  1− 
 2
 α 
∆Tb [ X ](s) = 1 − 1  K b (X)

2
 α 
∆Tb [Y] (s) = 1− 2  K b (Y)

2
Given,
∆Tb (X)(s) = 3∆Tb (Y)(s)
 α 
 α1 
1−  K b (X) = 3 × 1− 2  × K b (Y)


2
2
or
 α 
 α 
2 1− 1  = 3 1− 2 


2
2
[Q K b (X) = 2K b (Y) ]
or
 0.7
 α 
2  1− 1  = 3  1− 


2
2
(as given, α 2 = 0.7)
4 − 2α 1 = 6 − 2.1 or 2α 1 = 01
.
α 1 = 0.05
1000 × K f × W B
30. Molar mass of solute (M B ) =
W A × ∆T f
or
so,
MB =
⇒
1000 × 14 × 75.2
1000 × 7
M B = 150.4 g per mol
Actual molar mass of phenol = 94 g/mol
Calculated molar mass
Now, van’t Hoff factor, i =
Observed molar mass
94
i=
= 0.625
∴
150.4
Dimerisation of phenol can be shown as :
2C6H5OH r (C6 H5OH)2
Initial
1
At equilibrium
0
α
2
1–α
Total number of moles at equilibrium, i = 1 − α +
i =1−
α
2
α
2
α
2
α /2 = 1 − 0.625
α = 0.75
Thus, the percentage of phenol that dimerises is 75%.
But i = 0.625, thus,
0.625 = 1 −
31. (i) ∆Tb = K b ⋅ m2
1.22 1000
×
⇒ M = 122
M
100
1.22 1000
(ii) 0.13 = 2.6 ×
×
⇒ M = 244
M
100
0.17 = 1.7 ×
⇒
The above molar masses suggests thapt benzoic acid is
monomeric in acetone while dimeric in benzene.
32. Higher the value of K b of a solvent suggest that there is larger
polarity of solvent molecules, which in turn implies higher
boiling point due to dipole-dipole interaction.
Therefore, the correct order of K b values of the three given
solvents is
Solvents
Boiling point
Kb
X
Y
Z
100°C
27°C
283°C
0.63
0.53
0.92
Solutions and Colligative Properties
33. Mass of water = 500 × 0.997 g = 498.5 g
Also CH3COOH r CH3COO− +
α
1−α
⇒
⇒
H+
⇒
α
3
1000
⇒ − ∆T f = iK f ⋅ m = 1.23 × 1.86 ×
×
= 0.23°C
60 498.5
H
2.40
2.40
2
M = 156 g/mol
Therefore, molality of H2O =
N
16.67
1.19
1
O
38.07
2.38
2
⇒
Empirical formula = C3H2NO2
5.5 1000
(ii) ∆Tb = 1.84 = 2.53 ×
×
M
45
⇒
M = 168
Q Empirical formula weight (84) is half of molar mass,
molecular formula is C6H4N2O4 a dinitrobenzene :
NO2
0.1
× 1000 = 2.4
0.9 × 46
− ∆T f = K fethanol × 2.4 = 2 × 2.4 = 4.8
⇒
34. (i) Empirical formula determination
C
42.86
3.57
3
0.643
1000
×
M
50 × 0.879
37. In the given solution ‘M’, H2O is solute.
i = 1 + α = 1.23
Elements
Weight %
Moles
Simplest ratio
0.48 = 5.12 ×
155
⇒
T f = 155.7 – 4.8 = 150.9 K
38. Vapour pressure = p (H2O) + p(ethanol )
= 32.8 × 0.1 + 40 × 0.9
= 3.28 + 36 = 39.28 mm
39. Now ethanol is solute.
Molality of solute =
⇒
⇒
40.
0.1
× 1000 = 6.17
0.9 × 18
∆Tb = 6.17 × 0.52 = 3.20
Tb = 373 + 3.2 = 376.2 K
 ∆T f 
lim 
 = K f (Cryoscopic constant)
m→ 0 m 
41. 1+ ∆T f = iK f m
(C6H4N2O4)
NO2
35. − ∆T f = K f ⋅ m2
⇒
Also,
0.3
= 0.1613
1.86
n
1000
= 0.1613
m2 = 2 ×
n1
M1
m2 =
⇒
n2 0.1613 × 18
= 2.9 × 10−3
=
n1
1000
⇒
n2
n + n1
+ 1= 2
= 2.9 × 10−3 + 1
n1
n1
⇒
⇒
n1
1
= 0.997
= χ1 =
n1 + n2
1 + 2.9 × 10−3
p = p0χ 1 = 23.51 × 0.997 = 23.44 mm
36. − ∆T f = 5.51 − 5.03 = 0.48
⇒
− ∆T f = 0.48 = K f ⋅ m
∆T f = 0 – (–0 .0558° C) = 0.0558° C
0.0558
⇒ i (vant Hoff’s factor) =
=3
186
. × 0.01
This indicates that complex upon ionisation produces three
ions as:
[Co(NH3 )5 Cl]Cl 2 → [Co(NH3 )5 Cl]2+ (aq) + 2Cl − (aq)
Thus, only one Cl is inside the coordination sphere.
42. MX 2 → M 2+ + 2 X −
van’t Hoff factor for any salt can be calculated by using equation
i = 1 + α (n − 1)
where,
n = number of constituent ions
∴
i (MX 2 ) = 1 + α (3 − 1) = 1 + 2α
(∆T f )observed
= i = 1 + 2α
(∆T f )theoretical
∴
i = 1 + 2 × 0.5 ⇒ i = 2
10
Electrochemistry
Topic 1 Electrochemical Cells
Objective Questions I (Only one correct option)
1. Given,
.
Co 3 + + e− → Co 2 + ; E ° = + 181V
4+
−
2+
Pb + 2e → Pb ; E ° = + 167
. V
Ce4 + + e− → Ce3 + ; E ° = + 161
. V
Bi 3+ + 3e− → Bi; E ° = + 0.20 V
Oxidising power of the species will increase in the order
(2019 Main, 12 April I)
(a)
(b)
(c)
(d)
4+
4+
3+
3+
Ce < Pb < Bi < Co
Bi 3 + < Ce4 + < Pb 4 + < Co 3 +
Co 3 + < Ce4 + < Bi 3 + < Pb 4 +
Co 3 + < Pb 4 + < Ce4 + < Bi 3 +
electrodes using 0.1 Faraday electricity. How many mole of
(2019 Main, 9 April II)
Ni will be deposited at the cathode?
(a) 0.20
(b) 0.10
(c) 0.15
(d) 0.05
3. Calculate the standard cell potential (in V) of the cell in which
following reaction takes place
E ° Ag + / Ag = x V
E ° Fe3 + / Fe = z V
(2019 Main, 8 April II)
(b) x − y
(d) x − z
4. Given, that EOs2 / H 2O = +1.23V;
2 8
4
2
/ Br s
s
= +1.09V, EAu
3+ / Au = +1.4V
The strongest oxidising agent is
(a) Au 3+
(b) O2
(c) S2 O2−
8
(2019 Main, 8 April I)
5. Consider the following reduction processes:
Zn 2+ + 2e− 
→ Zn (s); E ° = − 0.76 V
Ca
2+
6. The anodic half-cell of lead-acid battery is recharged using
electricity of 0.05 Faraday. The amount of PbSO4
electrolysed in g during the process is (Molar mass of
PbSO4 = 303g mol −1 )
(2019 Main, 9 Jan I)
(a) 11.4
(b) 7.6
(c) 15.2
(d) 22.8
7. How long (approximate) should water be electrolysed by
passing through 100 amperes current so that the oxygen
released can completely burn 27.66 g of diborane?
(Atomic weight of B = 10.8 µ)
(2018 Main)
(a) 6.4 hours (b) 0.8 hours (c) 3.2 hours (d) 1.6 hours
2 /Cl
−
° 3+
= 1.36 V, ECr
= − 0.74 V
/Cr
° 2 − 3 + = 1.33 V, E ° −
= 1.51 V
ECr
O /Cr
MnO / Mn 2 +
2 7
4
Among the following, the strongest reducing agent is
(2017 Main)
(a) Cr
(b) Mn
2+
(c) Cr
3+
(d) Cl −
Pt ( s ) | H2 ( g , 1bar ) | H+ ( aq , 1 M)
| | M 4 + ( aq ), M 2 + ( aq ) | Pt ( s )
[ M ( aq )]
Ecell = 0.092 V when
= 10x
[ M 4 + ( aq )]
RT
Given : E °M 4+ / M 2+ = 0.151 V; 2.303
= 0.059 V
F
The value of x is
(2016 Adv.)
(a) − 2
(b) − 1
(c) 1
(d) 2
2+
E ° Fe 2+ / Fe = y V
s
E sS O 2− / SO 2− = 2.05V; EBr
(b) Ni < Zn < Mg < Ca
(d) Ca < Mg < Zn < Ni
9. For the following electrochemical cell at 298 K,
Fe2+ ( aq ) + Ag+ ( aq ) → Fe3 + ( aq ) + Ag( s )
(a) x + 2 y − 3 z
(c) x + y − z
(2019 Main, 10 Jan I)
(a) Zn < Mg < Ni < Ca
(c) Ca < Zn < Mg < Ni
°
8. Given, ECl
2. A solution of Ni(NO3 )2 is electrolysed between platinum
Given that,
The reducing power of the metals increases in the order
−
+ 2e 
→ Ca (s); E ° = − 2.87 V
Mg2+ + 2e− 
→ Mg(s); E ° = − 2.36 V
Ni2+ + 2e− 
→ Ni(s); E ° = − 0.25 V
(d) Br2
10. Two Faraday of electricity is passed through a solution of
CuSO4 . The mass of copper deposited at the cathode is
(at. mass of Cu = 63.5 u)
(2015 Main)
(a) 0 g
(b) 63.5 g
(c) 2 g
(d) 127 g
° 3+ = − 0.74 V; E ° −
11. Given, ECr
= 1.51 V
MnO /Mn 2+
/Cr
4
° 2− 3+ = 1.33 V; E ° − = 1.36 V
ECr
Cl /Cl
O /Cr
2 7
Based on the data given above strongest oxidising agent will be
(2013 Main)
(a) Cl
(b) Cr 3+
(c) Mn 2+
(d) MnO −4
Electrochemistry 157
12. Electrolysis of dilute aqueous NaCl solution was carried out
by passing 10 mA current. The time required to liberate
0.01 mole of H2 gas at the cathode is (1 F = 96500 C mol −1 )
(a) 9.65 × 104 s
(b) 19.3 × 104 s
(2008, 3M)
4
4
(c) 28.95 × 10 s
(d) 38.6 × 10 s
13. In the electrolytic cell, flow of electrons is from
(2003, 1M)
(a) cathode to anode in solution
(b) cathode to anode through external supply
(c) cathode to anode through internal supply
(d) anode to cathode through internal supply
14. Standard electrode potential data are useful for understanding
the suitability of an oxidant in a redox titration. Some
half-cell reactions and their standard potentials are given
below :
MnO−4 ( aq) + 8H+ ( aq ) + 5e− → Mn 2+ ( aq ) + 4H2 O( l ),
E° = 1.51 V
Cr2 O27 − ( aq ) + 14 H+ ( aq ) + 6e− → 2Cr 3+ ( aq ) + 7H2 O( l ) ,
E° = 1.38 V
E° = 0.77 V
Fe3+ ( aq ) + e− → Fe2+ ( aq )
Cl 2 ( g ) + 2e
−
−
→ 2Cl ( aq )
E° = 1.40 V
Identify the incorrect statement regarding the quantitative
estimation of aqueous Fe(NO3 )2
(2002, 3M)
(a) MnO−4 can be used in aqueous HCl
(b) Cr2 O2−
7 can be used in aqueous HCl
−
(c) MnO4 can be used in aqueous H2 SO4
(d) Cr2 O2−
7 can be used in aqueous H2 SO4
15. Saturated solution of KNO3 is used to make ‘salt-bridge’
because
(a) velocity of K + is greater than that of NO–3
(b) velocity of NO−3 is greater than that of K +
(2001, 1M)
(c) velocities of both K + and NO−3 are nearly the same
(d) KNO3 is highly soluble in water
16. The gas X at 1 atm is bubbled through a solution containing a
mixture of 1 M Y − and 1 M Z − at 25° C . If the order of
reduction potential is Z > Y > X , then
(1999, 2M)
(a) Y will oxidise X and not Z
(b) Y will oxidise Z and not X
(c) Y will oxidise both X and Z
(d) Y will reduce both X and Z
17. The standard reduction potential values of three metallic
cations, X, Y, Z are 0.52, − 3.03 and − 1.18 V respectively.
The order of reducing power of the corresponding metals is
(a) Y > Z > X
(b) X > Y > Z
(1998, 2M)
(c) Z > Y > X
(d) Z > X > Y
18. The standard reduction potentials E°, for the half reactions
are as
Zn = Zn 2+ + 2e− , E° = + 0.76 V
Fe = Fe2+ + 2e− , E° = 0.41 V
The emf for the cell reaction,
Fe2+ + Zn → Zn 2+ + Fe is
(a) – 0.35 V
(b) + 0.35 V
(c) + 1.17 V
(d) − 1.17 V
19. When a lead storage battery is discharged
(1989, 1M)
(1986, 1M)
(a) SO2 is evolved
(b) lead is formed
(c) lead sulphate is consumed
(d) sulphuric acid is consumed
20. The reaction,
1
H2 ( g ) + AgCl( s ) r H+ ( aq ) + Cl − ( aq ) + Ag ( s )
2
occurs in the galvanic cell
(1985, 1M)
(a) Ag |AgCl ( s )|KCl(soln) | AgNO3 |Ag
(b) Pt|H2 ( g )|HCl(soln) | AgNO3 (soln)|Ag
(c) Pt |H2 ( g )HCl (soln) | AgCl ( s )|Ag
(d) Pt | H2 ( g ) | KCl (soln) | AgCl( s )|Ag
21. The electric charge for electrode deposition of one gram
equivalent of a substance is
(1984, 1M)
(a) one ampere per second
(b) 96,500 coulombs per second
(c) one ampere for one hour
(d) charge on one mole of electrons
22. A solution containing one mole per litre of each Cu (NO3 )2 ,
AgNO3 ,Hg 2 (NO3 )2 and Mg(NO3 )2 is being electrolysed by
using inert electrodes. The values of standard electrode
potentials in volts (reduction potential) are
Ag + / Ag = + 0.80, Hg 2+
2 / 2Hg = + 0.79
Cu 2+ / Cu = + 0.34, Mg 2+ / Mg = − 2.37
With increasing voltage, the sequence of deposition of metals
on the cathode will be
(1984, 1M)
(a) Ag, Hg, Cu, Mg
(b) Mg, Cu, Hg, Ag
(c) Ag, Hg, Cu
(d) Cu, Hg, Ag
23. Faraday’s laws of electrolysis are related to the (1983, 1M)
(a) atomic number of the cation
(b) atomic number of the anion
(c) equivalent weight of the electrolyte
(d) speed of the cation
24. The standard reduction potentials at 298K for the following
half cells are given :
Zn 2 + ( aq ) + 2e− r Zn ( s ) ; E° = −0.762 V
Cr 3 + ( aq ) + 3e− r Cr ( s ) ; E° = −0.740 V
2H+ ( aq ) + 2e− r H2 ( g ); E° = 0.000 V
Fe3 + ( aq ) + e− r Fe2 + ( aq ); E° = 0.770 V
Which is the strongest reducing agent?
(1981, 1M)
(a) Zn( s )
(b) Cr( s )
(c) H2 ( g )
(d) Fe2+ ( aq )
158 Electrochemistry
Objective Questions II
(One or more than one correct option)
Passage Based Questions
25. In a galvanic cell, the salt-bridge
(a)
(b)
(c)
(d)
(2014 Adv.)
does not participate chemically in the cell reaction
stops the diffusion of ions from one electrode to another
is necessary for the occurrence of the cell reaction
ensures mixing of the two electrolytic solutions
26. For the reduction of NO–3 ion in an aqueous solution E° is
+ 0.96 V. Values of E° for some metal ions are given below
E° = − 1.19 V
V2+ ( aq ) + 2e– → V;
Fe3+ ( aq ) + 3e– → Fe;
E° = − 0.04V
E° = + 1.40 V
Au 3+ ( aq ) + 3e– → Au;
Hg 2+ ( aq ) + 2e– → Hg ;
E° = + 0.86V
The pair(s) of metals that is/are oxidised by NO–3 in aqueous
solution is (are)
(2009)
(a) V and Hg (b) Hg and Fe (c) Fe and Au (d) Fe and V
27. Consider a 70% efficient hydrogen-oxygen fuel cell working
under standard conditions at 1 bar and 298 K. Its cell reaction is
1
H 2( g )+ O 2( g ) → H 2O( l )
2
The work derived from the cell on the consumption of
. × 10−3 mole of H2 ( g ) is used to compress 1.00 mole of a
10
monoatomic ideal gas in a thermally insulated container.
What is the change in the temperature (in K) of the ideal gas?
The standard reduction potentials for the two half-cells are
given below:
O 2 (g) + 4H+ (aq) + 4 e− → 2H 2O(),
l
E° = 1.23 V,
2H+ ( aq ) + 2e− → H 2 ( g ),
Use, F = 96500 C mol , R = 8.314 J mol
E° = 0.00 V
−1
−1
K . (2020 Adv.)
28. For the electrochemical cell,
Mg( s )|Mg 2 + ( aq, 1 M) || Cu 2 + ( aq , 1 M) | Cu( s )
The standard emf of the cell is 2.70 V at 300 K. When the
concentration of Mg 2+ is changed to x M, the cell potential
changes to 2.67 V at 300 K. The value of x is _____.
F
(Given, = 11500 K V−1 , where F is the Faraday constant
R
and R is the gas contant, In (10) = 2.30)
(2018 Adv.)
29. Consider
A( s ) | A
n+
(2007, 3 × 4M = 12M)
30. The total number of moles of chlorine gas evolved is
(a) 0.5
(c) 2.0
Numerical Answer Type Questions
−1
Chemical reactions involve interaction of atoms and
molecules. A large number of atoms/molecules
(approximately 6.023 × 1023 ) are present in a few grams of
any chemical compound varying with their atomic/molecular
masses. To handle such large numbers conveniently, the
mole concept was introduced. This concept has implications
in diverse areas such as analytical chemistry, biochemistry,
electrochemistry and radiochemistry. The following example
illustrates a typical case, involving chemical/electrochemical
reaction, which requires a clear understanding of the mole
concept. A 4.0 M aqueous solution of NaCl is prepared and
500 mL of this solution is electrolysed. This leads to the
evolution of chlorine gas at one of the electrodes
(atomic mass : Na = 23, Hg = 200 , 1F = 96500 C).
an
electrochemical
cell
:
( aq , 2M )|| B 2 n + ( aq ,1M)| B( s ). The value of ∆H s
for the cell reaction is twice of ∆Gs at 300 K. If the emf of the
cell is zero, the ∆S s (in J K −1 mol −1 ) of the cell reaction per
mole of B formed at 300 K is …… .
(Given : ln( 2 ) = 0.7, R (universal gas constant) = 8.3 J K −1
mol −1 . H , S and G are enthalpy, entropy and Gibbs energy,
respectively.)
(2018 Adv.)
(b) 1.0
(d) 3.0
31. If the cathode is a Hg electrode, the maximum weight
(in gram) of amalgam formed from this solution is
(a) 200
(b) 225
(c) 400
(d) 446
32. The total charge (coulombs) required for complete
electrolysis is
(a) 24125
(c) 96500
(b) 48250
(d) 193000
Subjective Questions
33. The following electrochemical cell has been set-up :
Pt (1) | Fe3+ , Fe2+ ( a = 1) | Ce4+ , Ce3+ ( a = 1) | Pt (2)
E° (Fe3+ , Fe2+ ) = 0.77 V
and
E° (Ce4+ , Ce3+ ) = 1.61 V
If an ammeter is connected between the two platinum
electrodes, predict the direction of flow of current, will the
current increases or decreases with time?
(2000, 2M)
34. Copper sulphate solution (250 mL) was electrolysed using a
platinum anode and a copper cathode. A constant current of
2 mA was passed for 16 min. It was found that after
electrolysis the absorbance of the solution was reduced to
50% of its original value. Calculate the concentration of
copper sulphate in the solution to begin with.
(2000, 3M)
35. A cell, Ag | Ag + || Cu 2+ | Cu , initially contains 1 M Ag + and 1
M Cu 2+ ions. Calculate the change in the cell potential after
the passage of 9.65 A of current for 1 h.
(1999, 6M)
36. How many grams of silver could be plated out on a serving
tray by electrolysis of a solution containing silver in +1
oxidation state for a period of 8.0 h at a current of 8.46 A?
What is the area of the tray, if the thickness of the silver
plating is 0.00254 cm? Density of silver is 10.5 g/cm3 .
(1997, 3M)
Electrochemistry 159
44. An acidic solution of Cu 2+ salt containing 0.4 g of Cu 2+ is
37. The Edison storage cell is represented as:
Fe ( s ) / FeO( s) / KOH( aq ) /Ni 2 O3 ( s ) / Ni( s )
The half-cell reactions are :
Ni 2 O3 ( s ) + H2 O ( l ) + 2e− r 2NiO( s ) + 2OH− ,
E° = + 0.40 V
FeO ( s ) + H2 O( l ) + 2e− r Fe( s ) + 2OH− ,
E° = −0.87 V
(i) What is the cell reaction?
(ii) What is the cell emf ? How does it depend on the
concentration of KOH?
(iii) What is the maximum amount of electrical energy that
can be obtained from one mole of Ni 2O3? (1994, 4M)
38. The standard reduction-potential for the half-cell
NO−3 ( aq ) + 2H+ + e− → NO2 ( g ) + H2 O is 0.78 V
electrolysed until all the copper is deposited. The electrolysis is
continued for seven more minutes with the volume of solution
kept at 100 mL and the current at 1.2 A. Calculate the volume
of gases evolved at NTP during the entire electrolysis.
(1989, 5M)
45. In a fuel cell hydrogen and oxygen react to produces
electricity. In the process hydrogen gas is oxidised at the anode
and oxygen at the cathode. If 67.2 L of H2 at STP react in 15
min, what is the average current produced? If the entire current
is used for electro deposition of copper from copper (II)
solution, how many grams of copper will be deposited?
Anode reaction :
H2 + 2OH− → 2H2 O + 2e−
Cathode reaction : O2 + 2H2 O + 2e− → 4OH−
(1988, 4M)
46. A cell contains two hydrogen electrodes. The negative electrode
(i) Calculate the reduction-potential in 8M H+ .
(ii) What will be the reduction-potential of the half-cell in a
neutral solution? Assume all the other species to be at unit
(1993, 2M)
concentration.
is in contact with a solution of 10−6 M hydrogen ions. The emf
of the cell is 0.118 V at 25° C . Calculate the concentration of
hydrogen ions at the positive electrode.
(1988, 2M)
39. Chromium metal can be plated out from an acidic solution
47. A 100 watt, 110 V incandecent lamp is connected in series with
containing Cr O3 according to the following equation.
Cr O3 ( aq ) + 6H+ ( aq ) + 6e− → Cr ( s ) + 3H2 O
Calculate (i) How many grams of chromium will be plated
out by 24,000 C and (ii) How long will it take to plate out
1.5 g of chromium by using 12.5 A current? (1993, 2M)
40. An aqueous solution of NaCl on electrolysis gives
H2 ( g ), Cl 2 ( g ) and NaOH according to the reaction.
2Cl – ( aq ) + 2H2 O r 2OH− ( aq ) + H2 ( g ) + Cl 2 ( g )
A direct current of 25 A with a current efficiency of 62% is
passed through 20 L of NaCl solution (20% by weight).
Write down the reactions taking place at the anode and
cathode. How long will it take to produce 1kg of Cl 2 ? What
will be the molarity of the solution with respect to hydroxide
ion? (Assume no loss due to evaporation)
(1992, 3M)
an electrolyte cell containing cadmium sulphate solution.
What weight of cadmium will be deposited by the current
flowing for 10 h?
(1987, 5M)
48. During the discharge of a lead storage battery, the density of
sulphuric acid fell from 1.294 to 1.139 g/mL. Sulphuric acid of
density 1.294 g/mL is 39% H2 SO4 by weight and that of
density 1.139 g/mL is 20% H2 SO4 by weight. The battery
holds 3.5 L of the acid and the volume remained practically
constant during the discharge.
Calculate the number of ampere-hours for which the battery
must have been used. The charging and discharging reactions
are
Pb + SO24 − = PbSO4 + 2e− (charging)
−
PbO2 + 4H+ + SO2−
4 + 2e
= PbSO4 + 2H2 O (discharging)
(1986, 5M)
49. How long a current of 3 A has to be passed through a solution
41. For the galvanic cell,
Ag | AgCl( s ) , KCl (0.2 M) || KBr (0.001 M), AgBr (s) | Ag
Calculate the emf generated and assign correct polarity to
each electrode for a spontaneous process after taking into
account the cell reaction at 25° C.
[ K sp (AgCl) = 2.8 × 10−10 , K sp (AgBr) = 3.3 × 10−13] (1992, 4M)
42. A current of 1.70 A is passed through 300.0 mL of 0.160M
solution of a ZnSO4 for 230 s with a current efficiency of
90%. Find out the molarity of Zn 2+ after the deposition Zn.
Assume the volume of the solution to remain constant
during the electrolysis.
(1991, 4M)
43. Calculate the quantity of electricity that would be required
to reduce 12.3 g of nitrobenzene to aniline, if the current
efficiency for the process is 50%. If the potential drop across
the cell is 3.0 V, how much energy will be consumed?
(1990, 3M)
of silver nitrate to coat a metal surface of 80 cm2 with a
0.005 mm thick layer?
Density of silver is 10.5 g/cm3 .
(1985, 3M)
50. In an electrolysis experiment current was passed for 5 h
through two cells connected in series. The first cell contains a
solution of gold and the second contains copper sulphate
solution. 9.85 g of gold was deposited in the first cell. If the
oxidation number of gold is +3, find the amount of copper
deposited on the cathode of the second cell. Also calculate the
magnitude of the current in ampere.
(Atomic weight of Au = 197 and atomic weight of
Cu = 63.5)
(1983, 3M)
51. A current of 3.7 A is passed for 6 h between nickel electrodes
in 0.5 L of a 2.0 M solution of Ni(NO3 )2 . What will be the
molarity of solution at the end of electrolysis?
(1978, 2M)
Topic 2 Conductivity of Electrolytic Solutions and their
Measurement and Nernst Equation
Objective Questions I (Only one correct option)
1. Molar conductivity ( Λ m ) of aqueous solution of sodium
stearate, which behaves as a strong electrolyte, is recorded at
varying concentrations (C ) of sodium stearate. Which one of
the following plots provides the correct representation of
micelle formation in the solution?
(critical micelle concentration (CMC) is marked with an
arrow in the figures)
(2019 Adv.)
Λm
Λm
(a)
CMC
(b)
4. Consider the statements S1 and S 2 :
S 1 : Conductivity always increases with decrease in the
concentration of electrolyte.
S 2 : Molar conductivity always increases with decrease in the
concentration of electrolyte.
The correct option among the following is
(2019 Main, 10 April I)
(a)
(b)
(c)
(d)
S1 is correct and S2 is wrong
S1 is wrong and S2 is correct
Both S1 and S2 are wrong
Both S1 and S2 are correct
5. The standard Gibbs energy for the given cell reaction in kJ
CMC
√C
√C
mol−1 at 298 K is
Zn( s ) + Cu2 + ( aq ) → Zn2 + ( aq ) + Cu( s ),
E° = 2Vat 298 K
(Faraday’s constant, F = 96000 C mol−1 )
(2019 Main, 9 April I)
CMC
Λm
(c)
Λm
CMC
(d)
(a) 384
(b) 192
(c) −384
(d) −192
6. Λ°m for NaCl, HCl and NaA are 126.4, 425.9 and
100.5 S cm 2 mol− 1 , respectively. If the conductivity of 0.001
M HA is 5 × 10− 5 S cm − 1 , degree of dissociation of HA is
√C
√C
(2019 Main, 12 Jan II)
(a) 0.25
2. The decreasing order of electrical conductivity of the
following aqueous solution is
0.1 M formic acid (A),
0.1 M acetic acid (B),
0.1 M benzoic acid (C).
(2019 Main, 12 April II)
(a) A > C > B
(b) C > B > A
(c) A > B > C
(d) C > A > B
3. Which one of the following graphs between molar
conductivity (Λ m ) versus C is correct?
(2019 Main, 10 April II)
(b) 0.50
(d) 0.125
7. Given the equilibrium constant ( K C ) of the reaction :
Cu( s ) + 2Ag+ ( aq ) → Cu2 + ( aq ) + 2Ag( s )
°
is 10 × 1015 , calculate the Ecell
of this reaction at 298 K.
RT


2.303 F at 298 K = 0.059 V
(2019 Main, 11 Jan II)
(a) 0.4736 V
(b) 0.04736 mV
(c) 0.4736 mV
(d) 0.04736 V
8. For the cell, Zn( s)|Zn2+ ( aq )||M x+ ( aq )|M ( s), different half
cells and their standard electrode potentials are given below.
Na
Cl
KC
l
(b) Λ m
Cl
Na
l
KC
(a) Λ m
(c) 0.75
√C
(d) Λ m
Na
l
KC
Cl
Na
√C
√C
Au 3+ (aq)/
Au(s)
Ag + (aq)/
Ag(s)
Fe 3+ (aq)/
Fe 2+ (aq)
Fe 2+ (aq)/
Fe(s)
E ° M x + / M /V
1.40
0.80
0.77
− 0.44
If E° Zn 2+ / Zn = − 0.76 V, which cathode will give a maximum
value of E° cell per electron transferred? (2019 Main, 11 Jan I)
Ag+
Fe2+
Au3+
Fe3 +
(b)
(c)
(d)
(a)
Ag
Fe
Au
Fe2+
√C
l
KC
(c) Λ m
M x+ (aq)/M(s)
Cl
9. In the cell,
Pt(s) H 2 ( g , 1 bar) HCl( aq )|AgCl( s ) Ag( s ) Pt( s ) the cell
potential is 0.92 V when a 10−6 molal HCl solution is used.
Electrochemistry 161
The standard electrode potential of ( AgCl/ Ag,Cl− )
2.303RT


electrode is Given,
= 0.06 V at 298 K
F


(a) 0.40 V (b) 0.20 V
(c) 0.94 V
(d) 0.76 V
Match E° of the rebox pair in Column I with the values given
in Column II and select the correct answer using the code
given below the lists.
(2013 Adv.)
Column I
10. If the standard electrode potential for a cell is 2V at 300 K, the
equilibrium constant (K) for the reaction,
Zn( s ) + Cu2+ ( aq )
- Zn
2+
( aq ) + Cu( s )
3+
Column II
P.
E° (Fe
/ Fe)
Q.
E° ( 4H2 O r 4H+ + 4OH− )
2+
+
+ Cu → 2 Cu )
at 300 K is approximately
R.
E° ( Cu
( R = 8 JK −1 mol−1 , F = 96000 C mol−1 )
S.
E° (Cr 3 + , Cr 2 + )
(2019 Main, 9 Jan II)
(a) e−160
(c) e−80
(b) e160
Codes
P
(a) 4
(c) 1
(d) e320
11. For the following cell,
Zn ( s )| ZnSO4 ( aq )||CuSO4 ( aq )|Cu( s )
when the concentration of Zn2+ is 10 times the concentration
of Cu2+ , the expression for ∆G (in J mol −1 ) is
[F is Faraday constant; R is gas constant;
T is temperature; E° (cell) = 11
. V]
(a) 2.303 RT +11
(b) 1.1 F
. F
(c) 2.303 RT − 2.2 F
(d) −2.2 F
(b) Cu
(c) Zn
R
2
3
S
3
4
P
2
3
(b)
(d)
(2014 Main)
Mn 2 + + 2e− → Mn ; E° = − 1.18 eV
P.
(C2 H5 )3 N + CH3 COOH
Q.
X
Y
KI(0.1 M) + AgNO3 (0.01 M) 2.
X
Y
R.
CH3 COOH + KOH
X
S.
1.
NaOH + HI
(c) 5 × 103
(d) 5 × 102
16. The standard reduction potential data at 25°C is given below.
E° (Fe3 + / Fe2 + ) = + 0.77 V; E° (Fe2 / Fe) = −0.44 V;
E° (Cu 2 + / Cu) = + 0.34 V; E° (Cu + / Cu) = + 0.52 V;
−
E ° ( O2 ( g ) + 4H + 4 e ) → 2H2 O) = + 1.23 V;
Q
4
3
E° (Cr 3 + / Cr) = −0.74 V; E° (Cr 2 + / Cr) = +0.91 V
S
1
2
4.
Conductivity does not
change much and then
increases
S
1
1
(b)
(d)
P
4
1
Q
3
4
R
2
3
S
1
2
18. Consider the following cell reaction,
2Fe( s ) + O2 ( g ) + 4H+ ( aq ) → 2Fe2+ ( aq ) + 2H2 O ( l ),
E ° = 1.67 V
At [Fe2+ ] = 10−3 M, P(O 2 ) = 0.1 atm and pH = 3, the cell
potential at 25° C is
(2011)
(a) 1.47 V
(b) 1.77 V
(c) 1.87 V
(d) 1.57 V
19. AgNO3 (aqueous) was added to an aqueous KCl solution
gradually and the conductivity of the solution was measured.
The plot of conductance (Λ) versus the volume of AgNO3 is
Λ
×
×
× ×
×
−
E ° ( O2 ( g ) + 2H2 O + 4 e ) → 4OH) = + 0.40 V
R
2
4
Volume
(P)
(a) (P)
×Λ
×
××
Volume
(Q)
(b) (Q)
Λ
Λ
× ××
×
+
R
4
1
Conductivity increases
and then does not
change much
×
×
×
×
(b) 5 × 10−3
Codes
P
(a) 3
(c) 2
××
(a) 5 × 10−4
– 0.83 V
3.
Y
×
specific conductance of the solution of 0.5 M solution of same
electrolyte is 1.4 S m−1 and resistance of same solution of the
same electrolyte is 280 Ω . The molar conductivity of 0.5 M
solution of the electrolyte in Sm 2mol−1 is
(2014 Main)
4.
Conductivity decreases
and then does not
change much
Y
X
15. Resistance of 0.2 M solution of an electrolyte is 50 Ω . The
– 0.04 V
Conductivity decreases
and then increases
14. The equivalent conductance of NaCl at concentration C and
at infinite dilution are λ C and λ ∞ , respectively. The correct
relationship between λ C and λ ∞ is given as (where, the
constant B is positive)
(2014 Main)
(a) λ C = λ ∞ + ( B ) C
(b) λ C = λ ∞ − ( B ) C
(d) λ C = λ ∞ + ( B ) C
(c) λ C = λ ∞ − ( B ) C
3.
Column II
2 (Mn 3+ + e− → Mn 2+ ) ; E° = + 1.51eV
The E ° for 3Mn 2 + → Mn + 2Mn 3+ will be
(a) − 2. 69 V; the reaction will not occur
(b) − 2. 69 V; the reaction will occur
(c) − 0. 33 V; the reaction will not occur
(d) − 0. 33 V; the reaction will occur
– 0.4 V
solution of Y as shown in Column I. The variation in
conductivity of these reactions is given in Column II. Match
Column I with Column II and select the correct answer using
the codes given below the Columns.
(2013 Adv.)
(d) Pb
13. Given below are the half-cell reactions
2.
Q
3
4
Column I
(2016 Main)
− 0.18 V
17. An aqueous solution of X is added slowly to an aqueous
(2017 Adv.)
12. Galvanisation is applying a coating of
(a) Cr
Q
1
2
1.
Volume
(R)
(c) (R)
××× ×
×
×
×
Volume
(S)
(d) (S)
(2011)
162 Electrochemistry
At 1250 K : 2Cu( s)+ 1 / 2 O 2 ( g ) → Cu 2O( s);
20. The half cell reactions for rusting of iron are :
1
2H + 2e + O2 → H2 O( l );
2
Fe2 + + 2e− → Fe( s );
+
−
E° = − 0.44 V
∆G° (in kJ) for the reaction is
(a) – 76
(b) – 322
(2005, 1M)
(c) – 122
∆Gs = − 78,000 J mol −1
E° = + 1.23V
(d) – 176
21. Zn | Zn 2+ ( a = 0.1M ) || Fe2+ ( a = 0.01M) | Fe.
The emf of the above cell is 0.2905 V. Equilibrium constant
for the cell reaction is
(2004, 1M)
(a) 100.32/ 0.059
(b) 100.32/ 0.0295
(c) 100.26/ 0.0295
(d) 100.32/ 0.295
1
H 2( g )+ O2 ( g ) → H 2O ( g );
2
∆Gs = − 1,78,000 J mol −1 ; G is the Gibbs energy (2018 Adv.)
Passage Based Questions
Passage I
The electrochemical cell shown below is a concentration cell.
M | M 2+ (saturated solution of a sparingly soluble salt,
22. The correct order of equivalent conductance at infinite
MX 2 )||M 2+ (0.001 mol dm −3 )|M . The emf of the cell depends
dilution of LiCl, NaCl and KCl is
(2001, 1M)
(a) LiCl > NaCl > KCl
(b) KCl > NaCl > LiCl
(c) NaCl > KCl > LiCl
(d) LiCl > KCl > NaCl
on the difference in concetration of M 2+ ions at the two
electrodes. The emf of the cell at 298K is 0.059 V.
+
23. For
−
the electrochemical cell, ( M |M )||( X | X ),
E° ( M + / M ) = 0.44 Vand E° ( X / X − ) = 0.33 V.
From this data one can deduce that
(2000, 1M)
(a) M + X → M + + X − is the spontaneous reaction
(b) M + + X – → M + X is the spontaneous reaction
(c) Ecell = 0.77 V
(d) Ecell = − 0.77 V
(2012)
28. The solubility product (K sp : mol 3 dm −9 ) of MX 2 at 298 based
on the information available the given concentration cell is
(take 2.303 × R × 298/ F = 0.059 V)
(a) 1 × 10−15 (b) 4 × 10−15 (c) 1 × 10−12 (d) 4 × 10−12
29. The value of ∆G (kJ mol −1 ) for the given cell is
(take 1 F = 96500 C mol −1 )
(a) − 5.7
(b) 5.7
(c) 11.4
24. The standard reduction potentials of Cu 2+ /Cu and Cu 2+ / Cu +
are 0.337 V and 0.153 V respectively. The standard electrode
potential of Cu + /Cu half-cell is
(1997, 1M)
(a) 0.184 V (b) 0.827 V (c) 0. 521 V (d) 0.490 V
Numerical Answer Type Questions
25. For
the
disproportionation
reaction
2Cu+ ( aq )
Cu ( s ) + Cu 2 + ( aq ) at 298 K, In K (where K
is the equilibrium constant) is ………× 10−1 .
c
° 2+
Given : ( ECu
. V,
= 016
/ Cu +
RT
= 0.025)
F
(2020 Main, 2 Sep II)
26. An oxidation-reduction reaction in which 3 electrons are
transferred has a ∆G° of 17.37 kJ mol −1 at 25° C. The value of
°
(in V) is …… ×10−2 .
Ecell
° +
ECu
= 0.52 V and
/ Cu
(1 F = 96,500 C mol −1 )
(2020 Main, 5 Sep I)
27. The surface of copper gets tarnished by the formation of
copper oxide. N 2 gas was passed to prevent the oxide
formation during heating of copper at 1250 K. However, the
N 2 gas contains 1 mole % of water vapour as impurity. The
water vapour oxidises copper as per the reaction given below
2Cu( g )+ H 2O( g ) → Cu 2O( s)+ H 2( g )
pH 2 is the minimum partial pressure of H 2 (in bar) needed to
prevent the oxidation at 1250 K. The value of ln( pH 2 ) is …… .
(Given : total pressure = 1 bar, R (universal gas constant)
= 8 J K −1 mol −1 , ln(10 ) = 2.30 Cu ( s ) and Cu 2O ( s ) are
mutually immiscible.)
(d) –11.4
Passage II
The concentration of potassium ions inside a biological cell is
at least twenty times higher than the outside. The resulting
potential difference across the cell is important in several
processes such as transmission of nerve impulses and
maintaining the ion balance. A simple model for such a
concentration cell involving a metal M is :
M ( s ) | M + ( aq; 0.05 molar) || M + ( aq; 1 molar) | M ( s )
For the above electrolytic cell the magnitude of the cell
potential | Ecell | = 70 mV.
(2010)
30. For the above cell
(a) Ecell < 0; ∆G > 0
(c) Ecell < 0; ∆G° > 0
(b) Ecell > 0; ∆G < 0
(d) Ecell > 0; ∆G° < 0
31. If the 0.05 molar solution of M + is replaced by a 0.0025
molar M + solution, then the magnitude of the cell potential
would be
(a) 35 mV
(b) 70 mV
(c) 140 mV
(d) 700 mV
Passage III
Redox reaction play a pivotal role in chemistry and biology.
The values of standard redox potential ( E° ) of two half-cell
reactions decide which way the reaction is expected to
proceed. A simple example is a Daniell cell in which zinc
goes into solution and copper gets deposited. Given below are
a set of half-cell reactions (acidic medium) along with their
E ° (V with respect to normal hydrogen electrode) values.
Using this data obtain the correct explanations to Questions
17–19.
(2007, 4 × 3M = 12M)
Electrochemistry 163
I2 + 2e− → 2I−
E° = 0.54
Cl 2 + 2e → 2Cl
−
E° = 1.36
Mn 3+ + e− → Mn 2+
E° = 1.50
−
3+
Fe
−
2+
+ e → Fe
O2 + 4H+ + 4e− → 2H2O
Subjective Questions
38. We have taken a saturated solution of AgBr, K sp is
12 × 10− 14 . If 10− 7 M of AgNO3 are added to 1 L of this
solution, find conductivity (specific conductance) of this
solution in terms of 10 − 7 Sm− 1 units.
(2006, 6M)
°
−3
2
−1
Given,
λ (Ag + ) = 6 × 10 Sm mol ,
E° = 0.77
E° = 1.23
32. Among the following, identify the correct statement.
(a) Chloride ion is oxidised by O2
(b) Fe2+ is oxidised by iodine
(c) Iodide ion is oxidised by chlorine
(d) Mn 2+ is oxidised by chlorine
λ °(Br − ) = 8 × 10− 3 Sm2 mol −1 ,
λ °(NO – ) = 7 × 10− 3 Sm2 mol −1 .
3
39. Calculate
33. While Fe3+ is stable, Mn 3+ is not stable in acid solution
(a) Ag (aq) + Cl − (aq) → AgCl (s)
Given
37. Ammonia is always added in this reaction. Which of the
following must be incorrect?
(a) NH3 combines with Ag + to form a complex
(b) Ag(NH3 ) 2+ is a stronger oxidising reagent than Ag +
(c) In the absence of NH3 silver salt of gluconic acid is
formed
(d) NH3 has affected the standard reduction potential of
glucose/gluconic acid electrode
∆G °f (Cl )−
−129kJ /mol
∆G ° (Ag+ )
77kJ /mol
(b) 6.539 × 10−2 g of metallic Zn(u = 65.39) was added to
100 mL of saturated solution of AgCl. Calculate
[ Zn 2+ ]
. Given that
log10
[ Ag+ ]2
Ag+ + e− → Ag;
E° = 0.80V
2+
E° = −76V
Zn
−
+ 2e → Zn;
Also find how many moles of Ag will be formed?
40. Find the equilibrium constant for the reaction
Cu 2+ + In 2 + r Cu + + In 3+
Given that E°Cu 2+ /Cu + = 0.15V,
E° In 2+ / In + = − 0.4V,
°
Ered
Ag(NH 3 )+2 + e− → Ag ( s) + 2NH 3;
= 0.337 V
RT
F
[Use 2.303 ×
= 0.0592 and
= 38.92 at 298 K]
F
RT
Find ln K of this reaction.
(a) 66.13
(b) 58.38
(c) 28.30
(d) 46.29
36. When ammonia is added to the solution, pH is raised to 11.
Which half-cell reaction is affected by pH and by how much?
°
(a) Eoxi will increase by a factor of 0.65 from Eoxi
°
(b) Eoxi will decrease by a factor of 0.65 from Eoxi
°
(c) E red will increase by a factor of 0.65 from E red
(d) E red will decrease by a factor of 0.65 from E °red
−109 kJ /mol
Represent the above reaction in form of a cell. Calculate E°
of the cell. Find log10 K sp of AgCl.
(2005, 6M)
C 6 H12 O6 + H2 O → C 6 H12 O7 + 2H+ + 2e− ;
°
Gluconic acid
= − 0.05 V
Eoxi
(2006, 3 × 4M = 12M)
∆G °f (AgCl )
f
Passage IV
35. 2Ag + + C 6 H12 O6 + H2 O → 2Ag ( s) + C 6 H12 O7 + 2H+
of the following reaction:
+
because
(a) O2 oxidises Mn 2+ to Mn 3+
(b) O2 oxidises both Mn 2+ to Mn 3+ and Fe2+ to Fe3+
(c) Fe3+ oxidises H2 O to O2
(d) Mn 3+ oxidises H2 O to O2
34. Sodium fusion extract, obtained from aniline, on treatment
with iron (II) sulphate and H2 SO4 in the presence of air gives
a Prussian blue precipitate. The blue colour is due to the
formation of
(a) Fe4 [Fe (CN)6 ] 3
(b) Fe3 [Fe(CN)6 ] 2
(c) Fe4 [Fe(CN) 6 ] 2
(d) Fe3 [Fe(CN)6 ] 3
Tollen’s reagent is used for the detection of aldehydes. When
a solution of AgNO3 is added to glucose with NH4 OH, then
gluconic acid is formed.
°
= 0.80 V
Ag + + e− → Ag; Ered
∆Gr°
E°
In 3+ / In +
41.
= −0.42 V
(2004, 4M)
(a) Will pH value of water be same at temperature 25°C and
4°C? Justify in not more than 2 or 3 sentences.
(b) Two students use same stock solution of ZnSO4 and a
solution of CuSO4. The emf of one cell is 0.03 V higher
than the other. The concentration of CuSO4 in the cell with
higher emf value is 0.5 M. Find out the concentration of
CuSO4 in the other cell. Given : 2.303 RT /F = 0.06V .
(2003, 2M)
42. The standard potential of the following cell is 0.23 V at 15° C
and 0.21 V at 35° C.
Pt | H2 ( g ) | HCl( aq ) | AgCl ( s ) | Ag ( s )
(i) Write the cell reaction.
(ii) Calculate ∆H ° and ∆S ° for the cell reaction by assuming
that these quantities remain unchanged in the range 15° C to
35° C.
(iii) Calculate the solubility of AgCl in water at 25° C.
Given, the standard reduction potential of the
(Ag+ (aq)/Ag (s) is 0.80 V at 25° C.
(2001, 10M)
164 Electrochemistry
43. Find the solubility product of a saturated solution of
Ag 2 CrO4 in water at 298 K, if the emf of the cell Ag | Ag +
(Saturated Ag 2 CrO4 solution. ) ||Ag + (0.1 M) | Ag is 0.164 V
at 298 K.
2H2 O + 2e− r H2 + 2OH− , is −0.8277 V. Calculate the
equilibrium constant for the reaction,
2H2 O r H3 O+ + OH− at 25° C.
(1998, 6M)
44. Calculate the equilibrium constant for the reaction,
2Fe3+ + 3I− r 2Fe2+ + I3− . The standard reduction
potentials in acidic conditions are 0.77 V and 0.54 V
respectively for Fe3+ / Fe2+ and I−3 / I− couples.
(1998, 3M)
45. Calculate the equilibrium constant for the reaction
Fe2+ + Ce4+ r Fe3+ + Ce3+
4+
Given, E° ( Ce /Ce
3+
3+
) = 1.44 V, E° ( Fe /Fe
2+
) =0.68 V
46. The standard reduction potential for Cu 2+ /Cu is +0.34 V.
Calculate the reduction potential at pH = 14 for the above
couple. K sp of Cu (OH)2 is 1.0 × 10−19 .
(1996, 3M)
47. An excess of liquid mercury is added to an acidified solution
of 1.0 × 10−3 M Fe3+ . It is found that 5 % of Fe3+ remains at
equilibrium at 25° C. Calculate E° (Hg 2+ / Hg ) assuming that
the only reaction that occurs is
2+
2Hg + 2Fe3+ → Hg 2+
2 + 2Fe
E° (Fe3+ / Fe2+ ) = 0.77 V
(1989, 3M)
52. The emf of a cell corresponding to the reaction.
Zn( s ) + 2H+ ( aq ) → Zn 2+ (0.1M) + H2 , ( g ,1atm)
is 0.28 V at 25° C.
Write the half-cell reactions and calculate the pH of the
solution at the hydrogen electrode.
E° ( Zn 2+ / Zn ) = −0.76 V E ° +
=0
H /H2
(1997, 2M)
Given,
51. The standard reduction potential at 25° C of the reaction,
(1995, 4M)
(1986, 4M)
53. Give reasons in one or two sentences.
“Anhydrous HCl is a bad conductor of electricity but aqueous
HCl is a good conductor.”
(1985, 1M)
54. Consider the cell,
Zn | Zn 2 + ( aq ) (1.0 M ) || Cu 2 + ( aq ) (1.0 M ) | Cu
The standard reduction potentials are 0.350 V for
Cu 2 + ( aq ) + 2e− → Cu
and −0.763 V for Zn 2 + ( aq ) + 2e− → Zn
(i) Write down the cell reaction.
(ii) Calculate the emf of the cell.
(iii) Is the cell reaction spontaneous or not?
(1982, 2M)
+
48. The standard reduction potential of the Ag /Ag electrode at
298 K is 0.799 V. Given that for AgI, K sp = 8.7 × 10−17 ,
evaluate the potential of the Ag + / Ag electrode in a saturated
solution of AgI. Also calculate the standard reduction
potential of the I− / AgI/Ag electrode.
(1994, 3M)
49. Zinc granules are added in excess to a 500 mL of 1.0 M nickel
nitrate solution at 25° C until the equilibrium is reached. If the
standard reduction potential of Zn 2+ / Zn and Ni 2+ / Ni are
– 0.75 V and – 0.24 V respectively. Find out the
concentration of Ni 2+ in solution at equilibrium. (1991, 2M)
50. The standard reduction potential of Cu
2+
+
/ Cu and Ag / Ag
electrodes are 0.337 and 0.799 V respectively. Construct a
galvanic cell using these electrodes so that its standard emf is
positive. For what concentration of Ag + will the emf of the
cell, at 25° C, be zero if the concentration of Cu 2+ is 0.01 M?
(1990, 3M)
Integer Answer Type Questions
55. The conductance of a 0.0015 M aqueous solution of a weak
monobasic acid was determined by using a conductivity cell
consisting of platinised Pt electrodes. The distance between
the electrodes is 120 cm with an area of cross section of 1
cm 2 . The conductance of this solution was found to be
5 × 10−7 S. The pH of the solution is 4. The value of limiting
molar conductivity ( Λ°m ) of this weak monobasic acid in
aqueous solution is Z × 102 S cm −1 mol −1 . The value of Z is
(2017 Adv.)
56. All
the
energy
released
from
the
reaction
X → Y , ∆ r G ° = − 193 kJmol −1 is used for oxidising M + as
M + → M 3 + + 2e− , E ° = − 0.25 V.
Under standard conditions, the number of moles of M +
oxidised when one mole of X is converted to Y is
[F = 96500 C mol − ]
(2015 Adv.)
Answers
Topic 1
1.
5.
9.
13.
17.
21.
(b)
(b)
(d)
(c)
(a)
(d)
25. (a, b)
2.
6.
10.
14.
18.
22.
(d)
(b)
(b)
(a)
(b)
(c)
3.
7.
11.
15.
19.
23.
(a)
(c)
(d)
(c)
(d)
(c)
4.
8.
12.
16.
20.
24.
(c)
(a)
(b)
(a)
(c)
(a)
26. (a,b,d)
29. (–11.62 JK mol −1)
32. (d)
27. (13.32)
28. (10)
30. (b)
31. (d)
34. (8 × 10 −5 M) 35. (0.01 V)
36. (300 cm 2)
37. (−245.11 kJ) 40. (1.4085 M) 41. (−0.037 V) 42. (0.154 M)
43. (347.40 kJ)
45. (190.50 g)
46. (10 −4 M)
47. (19.1 g)
48. (265 Ah)
49. (125 s)
50. (0.80 A)
51. (1.172 M)
Electrochemistry 165
29. (d)
Topic 2
1. (b)
2. (a)
3. (c)
30. (b)
31. (c)
32. (c)
4. (b)
33. (d)
34. (a)
35. (b)
36. (c)
37. (d)
38. (55)
40. (1010)
41. (0.05 M)
5. (c)
6. (d)
7. (a)
8. (a)
9. (b)
10. (b)
11. (c)
12. (c)
43. (2.45 × 10 −12)
13. (a)
14. (c)
15. (a)
16. (d)
45. (6.88 × 1012) 46. (− 0.222 V) 47. (0.7926 V) 49. (1.7 × 1017)
17. (a)
18. (d)
19. (d)
20. (b)
50. (1.57 × 10 −9) 51. (1.04 × 10 −14 )
55. (6 × 10 S cm
21. (b)
22. (b)
23. (b)
24. (c)
25. (144)
26. (–6)
27. (–14.16)
28. (b)
2
44. (5.89 × 10 7)
−1
−1
mol )
52. (8.6)
56. (4 mol)
Hints & Solutions
Topic 1 Electrochemical Cells
⇒ Eº
Fe3+ / Fe2 +
Key Idea Negative E ° means that redox couple is weaker
oxidising agent than H+ /H2 couple. Positive E° means that
1.
redox couple is a stronger oxidising agent than H+ / H2
couple
So, from equation (i)
º
Ecell
= xV − (3z − 2 y) V = (x − 3z + 2 y) V
4. Higher the standard reduction potential (E ºMn + / M ), better is
oxidising agent. Among the given, E °S2O 82− / SO 24− is highest, hence
Given, Co3+ + e− → Co2+ ; E ° = + 1.81 V
Pb
4+
−
S2O82− is the strongest oxidising agent.
The decreasing order of oxidising agent among the given option
is as follows:
S2O82− > Au 3+ > O2 > Br2
2+
+ 2e → Pb ; E ° = + 1.67 V
Ce4+ + e− → Ce3+ ; E ° = + 1.61 V
Bi3+ + 3e− → Bi; E ° = + 0.20 V
Oxidising power of the species increases in the
order of Bi3 + < Ce4+ < Pb4+ < Co3+ .
Higher the emf value, stronger the oxidising power. The
maximum value of emf is possessed by Co3+ . Hence, it has
maximum oxidising power. Whereas Bi3+ possess the lowest
emf value. Hence, it has minimum oxidising power.
5. Reducing power of an element
∝
Here, E °
M2 + / M
Metals
electrodes using 0.1 Faraday electricity. It means that 0.1
equivalent of Ni2+ will be discharged.
+
Ag /Ag
− Eº
3+
Fe
2+
/ Fe
= xV − E º
3+
Fe
2+
/ Fe
… (i)
Now, for two half-cells
º
(i) Fe2+ + 2e− → Fe; EFe
= E1º = yV ∆G2º = − 2FE1º
2+
/ Fe
º
(ii) Fe3 + + 3e− → Fe ; EFe
= E2º = zV ∆G2º = − 3FE2º
3+
/ Fe
So, Fe3 + + e– → Fe2+ ; E º
Again,
⇒
⇒
Fe3+ / Fe2 +
º
; ∆G3 = − 1 × FE3º
º
º
∆G3 = ∆G2 − ∆G1º
− FE3º = − 3FE2º − (− 2FE1º )
− E3º = 2E1º − 3E2º ⇒ E3º = 3E2º
= E3º = ?
− 2E1º
Zn
Mg
Ca
Thus, the correct order of increasing reducing power of the given
metal is,
Ni < Zn < Mg < Ca.
Ni2+ + 2e− → Ni (Atomic mass of Ni = 58.7)
3. Fe2+ (aq) + Ag + (aq) → Fe3 + (aq) + Ag(s)
values of the given metals are as,
Reducing power
Electrolysis of Ni(NO3 )2 gives
Number of equivalents = Number of moles × number of electrons.
01
. = Number of moles × 2
01
.
∴Number of moles of Ni =
= 0.05
2
Ni
1
Standard reduction potential
E°( V ) − 0.25 − 0.76 − 2.36 − 2.87
→
2. A solution of Ni(NO3 )2 is electrolysed between platinum
º
= Eº
∴ Ecell
= (3z − 2 y) V
6.
Key Idea This question is based upon Faraday’s first law
which states that “Mass of any substance deposited or
liberated at any electrode is directly proportional to the
quantity of electricity passed.”
During charging:
Pb + SO24 − → PbSO4 + 2e−
⇒
1 F ≡ 1g-equiv. of PbSO4
1
303
g PbSO4
= mol of PbSO 4 ⇒
2
2
303
∴ 0.05 F ≡
× 0.05 g of PbSO4 = 7.575 g of PbSO4
2
7. Given that, i = 100 amp. also, 27.66 g of diborane (B2H6 )
Molecular mass of B2 H6 = 10.8 × 2 + 6 = 27.6
Given mass 27.66
Number of moles of B2 H6 in 27.66 g =
=
≈1
Molar mass
27.6
Now consider the equation
B2H6 + 3O2 → B2O3 + 3H2O
166 Electrochemistry
From the equation we can interpret that 3 moles of oxygen is
required to burn 1 mole (i.e. 27.6 g) B2 H6 completely.
Also consider the electrolysis reaction of water i.e.
H2O s 2H+ + O−−
12. 0.01 mol of H2 = 0.02 g equivalent
⇒ Coulombs required = 0.02 × 96500 = 1930 C
⇒
Q = It = 1930 C
1930
⇒
t=
= 19.3 × 104 s
10 × 10− 3
+2e −
2H+ → 2H → H2 ↑
Cathode
2 such
O−− →
O →
O2 ↑
−
atoms
Anode
13. In electrolytic cell electrolysis occur at the cost of electricity :
At cathode : M n+ + ne → M
−2e
From the above equation it can be easily interpreted that in
electrolysis of water for the production of 1 mole of oxygen from
1 mole of H2O at anode 4 moles electrons are required.
Likewise for the production of 3 moles of O2 12(3 × 4 ) moles of
electrons will be needed.
So, the total amount of charge required to produce 3 moles of
oxygen will be 12 × F or 12 × 96500
We know
Q = it
So,
12 × 96500 = 100 × tin seconds
12 × 96500
or
= tin hours = 3.2 hours
100 × 3600
8. The substances which have lower reduction potentials are
At anode :
Net: M 4+ (aq) + H2 (g ) → M 2+ (aq) + 2H+ (aq);
∴
∴
[ M 2+ ] [ H+ ] 2
[ M 2+ ]
° = 0.151 V )=
(Ecell
K=
4+
[ M ] pH 2
[ M 4+ ]
0.059
° −
Ecell = Ecell
log K
2
0.059
[ M 2+ ]
0.092 = 0151
. −
log 4+
2
[M ]
0.059
x
0.059 =
log10
2
x
log10 = 2
x=2
10. Given, Q = 2F
Atomic mass of Cu = 63.5u
Valency of the metal Z = 2
We have,
CuSO4 → Cu 2+ + SO24−
because the following reaction is spontaneous :
MnO−4 + Cl − → Mn 2+ + Cl 2;
2mol
2F
1mol =63.5g
Alternatively.
E
2 × 63.5
= 63.5
⋅ 2F = 2E =
F
2
11. Higher the standard reduction potential, better is oxidising agent.
°
Among the given EMnO
is highest, hence MnO −4 is the
−
/ Mn 2+
W = ZQ =
4
strongest oxidising agent.
E° = 1.51 − 1.40 = 0.11 V
In all other cases, the redox process between oxidising agent and
medium (HCl or H2SO4) are non-spontaneous, would not
interfere oxidation of Fe2+ .
15. One of the requirement for electrolyte used in salt-bridge is, both
cation and anion must have comparable size so that they migrate
towards electrodes of opposite polarity at comparable speeds.
16. Higher the value of reduction potential, stronger the oxidising
agent.
E° : Z > Y > X
Q
⇒ Y will oxidise X but not Z.
17. Lower the value of E° , stronger the reducing agent.
Reducing power:
Y (E ° = − 3.03 V) > Z (E ° = − 1.18 V) > X (E ° = 0.52 V).
18.
Fe2+ + 2e− → Fe ;
⇒ Fe
2+
Zn → Zn
2+
+ Zn → Zn
2+
E° = − 0.41V
−
+ 2e ; E° = + 0.76V
E° = + 0.35V
+ Fe ;
19. In a lead storage battery, sulphuric acid is consumed as :
Pb + PbO2 + 2H2SO4 → 2PbSO4 + 2H2O
20. In a galvanic cell, oxidation occur in the left hand electrode
chamber and reduction in right hand electrode chamber. In the
following cell.
Pt | H2 (g ) | HCl (l )||AgCl (s) | Ag(s)
The cell reactions are :
1
H (g ) → H+
2 2
+ e−
−
AgCl(s) + e → Ag + Cl
Cu 2+ + 2 e− → Cu
1mol
→ X + ne−
14. MnO−4 cannot be used for oxidation of Fe2+ in HCl medium
9. Oxidation at anode
° = 0.00 V
H2 (g ) → 2H+ (aq)+ 2e−; ESHE
Reduction at cathode
M 4+ (aq)+ 2e− → M 2+ (aq); EM° 4+ / M 2+ = 0.151 V
(electron gone in solution)
(electron supplied to anode)
Therefore, electron is moving from cathode to anode via internal
circuit.
° 3 + / Cr = − 0.74 V) is
stronger reducing agents. Therefore, Cr (ECr
the strongest reducing agent among all the other given options.
X
n−
Net :
−
At anode
At cathode
1
H2 (g ) + AgCl (s) → H+ + Ag(s) + Cl −
2
21. One gram equivalent of an electrolyte required 1.0 mole of
electronic charge for discharging.
22. In aqueous solution, only those ions who are less electropositive
than hydrogen (E° > 0) would be deposited.
Therefore, in the present case, only Ag, Hg and Cu would be
deposited on passing electricity through aqueous solution of
these ions, Mg will not be deposited.
Electrochemistry 167
Also, higher the value of E°, easier will be their reduction,
therefore, the sequence in which ions will be deposited on
increasing voltage across the electrodes is :
Ag, Hg, Cu.
23. Faraday’s law of electrolysis is related to equivalent weight of
electrolytes as “the number of Faraday’s passed is equal to the
number of gram equivalent of electrolytes discharged.”
24. Lower the value of E°, stronger the reducing agent.
25.
PLAN This problem is based on characteristics of salt-bridge.
Functions of salt-bridge are
(i) It connects the two half-cells and completes the cell circuit.
(ii) It keeps the solutions of two half-cells and complete the cell
circuit but does not participate chemically in the cell reaction.
(iii) It maintains the diffusion of ions from one electrode to
another electrode.
(iv) A cell reaction may also occur in the absence of salt-bridge.
Sometimes, both the electrodes dip in the same electrolyte
solution and in such cases we do not require a salt-bridge.”
So, option (c) is incorrect.
(v) This prevent mixing of two electrolytic solutions hence,
option (d) is incorrect choice.
Hence, correct choices are (a), (b).
26. Metals with E° value less than 0.96 V will be able to reduce NO−3
in aqueous solution.Therefore, metals V (E° = − 1.19 V),
Fe (E° = – 0.04 V), Hg (E° = 0.86 V) will all reduce NO−3 but Au
(E° = 1.40 V) cannot reduce NO−3 in aqueous solution.
27. Vessel is insulated, thus q = 0
For the given reaction :
H2 (g ) + 1/ 2 O2 (g ) → H2O(l ); E° = 123
. − 0.00 = 123
. V
∆G ° = − nFE ° = − 2 × 96500 × 123
. J / mol
Therefore, work derived from this fuel cell using 70% efficiency
and on consumption of 10
. × 10− 3 mol of H2 (g )
= 2 × 96500 × 123
. × 0.7 × 1 × 10− 3
= 16617
. J
This work done = change in internal energy (for monoatomic
gas, CV , m = 3R / 2 ) ,
16617
. ×2
166.17 = nCV , m∆T ⇒ ∆T =
⇒ 13.32 K
1 × 3 × 8.314
28. Equation of cell reaction according to the cell notation given, is
Reduction
Mg(s) + Cu2+(aq)
Mg2+(aq) + Cu(s)
Oxidation
° = 2.70 V, T = 300 K
Given, Ecell
with
[Mg 2+ (aq)]=1 M and [Cu 2+ (aq)]=1 M
and n= 2
Further,
with
Ecell = 2 . 67 V
[Cu 2+(aq) ]=1M
[Mg 2+ (aq)]= xM
and
F
= 11500 KV −1
R
where F = Faraday constant, R = gas constant
From the formula,
2+
RT [Mg (aq)]
° −
Ecell = Ecell
ln
nF [Cu 2+ (aq)]
and
After putting the given values
RT x
2.67 = 2.70 −
ln
2F 1
R × 300
or
2.67 = 2.70 −
× ln x
2F
− R × 300
−0.03 =
× ln x
2F
0.03 × 2 × 11500
0.03 × 2 F
or
ln x =
= 2.30
× =
R
300
300
So,
ln x = 2.30
or x =10 (as given ln (10) = 2.30)
29. Given,
A (s)| A n+ (aq, 2 M ) || B 2n+ (aq, 1 M ) | B (s)
So, reactions at respective electrode will be
Anode
A (s)→ A n+ + ne− × 2
Cathode
B 2n+ + 2 ne− → B (s)
Overall reaction
2 A (s) + B 2n+ (aq) → 2 A n+ (aq) + B (s)
Further,
∆H ° = 2∆G ° and Ecell = 0 is also given
Now by using the Nernst equation
RT [Product]
Ecell = E°cell −
ln
nF [Reactant]
After putting the values
RT [ A n+ ]2
ln
0 = Ecell
° −
2nF [ B 2n+ ]
or
E° =
RT
[ 2 ]2 RT
ln
=
ln 4
2 nF
[1] 2nF
…(i)
Further from the formula,
∆G ° = − nFE ° ⇒ ∆G ° = − 2nFE °
Now putting the value of E° from eq. (i)
RT
…(ii)
∆G ° = − 2nF ×
ln 4
2nF
∆G ° = − RT ln 4
Finally, using the formula
∆ G ° = ∆ H ° − T∆ S °
(as ∆H ° = 2∆G °, given)
∆G ° = 2∆G ° − T∆S °
∆ G ° = T∆ S °
∆G ° − RT ln 4
or
∆S ° =
=
T
T
(from eq. (ii), ∆G ° = − RT ln 4)
= − R ln 4 = − 8.3 × 2 × 0.7
(as all values given)
= − 1162
. J/K-mol
168 Electrochemistry
30. Moles of NaCl electrolysed = 4 ×
500
= 2.0
1000
⇒ moles of Cl 2 produced = 1.0
2Cl − → Cl 2 + 2e−
Hg
31. At cathode Na + + e− → Na(Hg)
Amalgam
Two moles of Na formed during electrolysis would produce two
moles of Na(Hg) amalgam.
⇒ mass of amalgam = 2 × (23 + 200) = 446 g
32. Two Faraday of electric charge would be required for
electrolysis of 2.0 moles of NaCl.
⇒
total coulombs = 2 × 96500 = 193000 C
° . Also, left
33. Since, activities of all the ions are unity, Ecell = Ecell
hand electrode is at lower reduction potential, it act as anode and
E ° = E ° (Ce4+ , Ce3+ ) − E ° (Fe3+ , Fe2+ ) = 0.84
i.e. electrons will flow from left to right hand electrode and
current from right hand electrode [Pt (2)] to left hand electrode
[Pt(1)].
[Fe3+ ][Ce3+ ]
Also,
E = E ° − 0.0592 log
[Fe2+ ][Ce4+ ]
As electrolysis proceeds, E will decrease and therefore, current.
2 × 10− 3 × 16 × 60
96500
=1.99 × 10− 5
34. The number of Faraday’s passed =
⇒ number of gram equivalent of Cu 2+ deposited = 1.99 × 10− 5
⇒ number of moles of Cu 2+ deposited =
1.99
× 10− 5 ≈ 10− 5
2
Absorbance is directly proportional to [Cu 2+ ]. Therefore,
if ‘C’ be the initial molarity, 0.5 C will be the final molarity.
0.5 C × 0.25 = 10− 5 ⇒ C = 8 × 10− 5 M
⇒
35. The number of Faraday’s passed =
9.65 × 60 × 60
= 0.36 F
96500
After electrolysis : [Ag+ ] = 1.36 M
[Cu 2+ ] = 1 –
0.36
= 0.82 M
2
E1 (before electrolysis) = E °
E2 (after electrolysis) = E ° −
⇒ E1 − E2 =
0.0592
[Ag+ ]2
log
2
[Cu 2+ ]
0.0592
(1.36)2
= 0.01 V (decreased)
log
2
0.82
36. Coulombs passed = 8.46 × 8 × 60 × 60 = 243648 C
243648
= 2.52
96500
63.5
Weight of Cu plated = 2.52 ×
g = 80.01 g
2
80.01
Volume of Cu plated =
= 7.62 cm 3
10.5
7.62
= 3000 cm 2
⇒ Area plated out =
0.00254
Number of Faraday’s passed =
37. Given, FeO (s)/Fe (s)
and
Ni 2O3 /NiO (s)
E° = – 0.87 V
E° = + 0.40 V
Electrode at lower reduction potential act as anode and that at
higher reduction potential act as cathode.
(i) Electrodes reaction :
Fe(s) + 2OH− → FeO (s) + H2O (l )
E° = + 0.87 V
Ni 2O3 (s) + H2O (l ) + 2e− → 2NiO (s) + 2OH− E° = 0.40 V
Net : Fe (s) + Ni 2O3 (s) → 2NiO (s) + FeO (s)
E° = 1.27 V
(ii) Emf is independent of concentration of KOH.
(iii) Maximum amount of energy that can be obtained = ∆G °
⇒ ∆G ° = − nE ° F = – 2 × 1.27 × 96500 J = – 245.11 kJ
i.e. 245.11 kJ is the maximum amount of obtainable energy.
1
38. (i) E = 0.78 – 0.0592 log 2 = 0.887 V
8
1
= – 0.0488 V
(ii) E = 0.78 – 0.0592 log
(10− 7 )2
39. Molar mass of Cr = 52 g
52
g
6
(i) Mass of Cr deposited on passing 24000 Coulombs
24000 52
=
×
g = 2.15 g
96500 6
1.5
9
(ii) Number of gram equivalent of Cr =
×6=
52
52
9
× 96500 = It
⇒ Coulombs required for 1.5 g Cr =
52
9 × 96500
s = 22.27 min
⇒
t=
52 × 12.5
Equivalent mass of Cr =
40. At anode
2Cl − → Cl 2 + 2e−
At cathode 2H2O + 2e− → H2 + 2OH−
1000
1 kg Cl 2 =
equivalent of Cl 2 = 28.17 equivalent
35.5
⇒ Theoretical electricity requirement = 28.17 F
Q Efficiency is only 62%
∴ Electricity requirement (experimental)
28.17 × 100
=
F = 45.44 F
62
45.44 × 96500 = 25 t (in second)
⇒
t = 48.72 h
⇒
Also, gram equivalent of HO− produced = 28.17
28.17
⇒ Molarity of HO− =
= 1.4085 M
20
41. [Ag+ ] in left hand electrode chamber =
= 1.4 × 10− 9 M
2.8 × 10− 10
0.2
[Ag+ ] in right hand electrode chamber =
= 3.3 × 10− 10 M
3.3 × 10− 13
0.001
Electrochemistry 169
emf = 0 – 0.0592 log
[Ag+ ] anode
[Ag+ ] cathode
1.4 × 10− 9
= – 0.0592 log
= – 0.037 V
3.3 × 10− 10
Therefore, the cell as written is non-spontaneous and its reverse
will be spontaneous with emf = 0.037 V.
1.7 × 230
42. Faraday’s passed =
= 4.052 × 10− 3 F
96500
Faradays used for reduction of Zn 2+ = 4.052 × 10− 3 × 0.9
= 3.65 × 10− 3
⇒ Meq. of Zn 2+ reduced = 3.65
Initial meq. of Zn 2+ = 300 × 0.16 × 2 = 96
⇒ Meq. of Zn 2+ remaining = 96 − 3.65 = 92.35
⇒ Molarity of Zn 2+ =
43.
NO2
92.35
1
×
= 0.154 M
2
300
NH2
46. Q Emf = 0.118 V > 0, it is galvanic cell and anode is negative
electrode :
At anode :
H2 (g ) → 2H+ (10− 6 M ) + 2e−
At cathode : 2H+ (x ) + 2e− → H2
Cell reaction :
H+ (x ) → H+ (10− 6 M )
Emf = 0.118 V = 0 – 0.0592 log
10− 6
x
⇒ x = 10− 4 M
47. 100 W lamp will produce 100 Js − 1.
10
Coulombs
11
Therefore, total Coulomb passed in 10 h
10
=
× 10 × 60 × 60 = 32727.27 C
11
⇒
100 J = 110 × C
⇒ C=
Number of gram equivalent of Cd 2+ deposited
32727.27
= 0.34
=
96500
112.4
Weight of Cd deposited = 0.34 ×
g = 19.1 g
2
48. For 1.0 L H2SO4 :
39
= 504.66 g
100
20
Final mass of H2SO4 = 1139 ×
= 227.80 g
100
Initial mass of H2SO4 = 1294 ×
Change in oxidation number at nitrogen = 4 − (− 2) = 6
123
Equivalent weight of nitrobenzene =
g
6
12.3 × 6
= 0.60
⇒ gram equivalent of nitrobenzene =
123
⇒ Theoretical requirement = 0.60 × 96500 C = 57900 C
⇒ Actual requirement of electricity = 2 × 57900 = 115800 C
Q
V ×C = J
⇒ Energy consumed = 115800 × 3 J = 347.40 kJ
44. If the salt is CuSO4
During deposition of Cu at cathode, O2 (g ) will evolve at anode
0.4 × 2
gram-equivalent of Cu deposited =
= 0.0126
63.5
Volume of O2 liberated at NTP at anode
= 0.0126 × 5600 mL = 70.56 mL
In the next 7 min, H2 at cathode and O2 at anode would be
produced.
1.2 × 7 × 60
Faraday’s passed =
= 5.22 × 10− 3
96500
⇒ Volume of H2 (at NTP) = 5.22 × 10− 3 × 11200 mL
= 58.46 mL
Volume of O2 (at NTP) = 5.22 × 10− 3 × 5600 mL = 29.23 mL
Therefore, O2 (g) at NTP = 70.56 + 29.23 = 99.79 mL
H2 (g ) at NTP = 58.46 mL
67.2
45. Total number of gram equivalent of H2 used =
=6
11.2
⇒
6 × 96500 = 15 × 60 × I ⇒ I = 643.33A
63.5
g = 190.50 g
Mass of Cu deposited = 6 ×
2
⇒ H2SO4 consumed/litre = 504.66 – 227.80 = 276.86 g
⇒ Total H2SO4 used up = 276.86 × 3.5 = 969.01 g
969.01
mol = 9.888 mol
=
98
Q 1 mole of H2SO4 is associated with transfer of 1.0 mole of
electrons, total of 9.888 moles of electron transfer has occurred.
Coulomb produced = 9.888 × 96500
9.888 × 96500
Ampere-hour =
= 265 Ah
3600
0.005
49. Volume of Ag coating = 80 cm 2 ×
cm = 0.04 cm 3
10
⇒ mass of Ag coating = 0.04 × 10.5 g = 0.42 g
0.42
= number of Faraday’s
⇒ gram equivalent of Ag =
108
0.42
× 96500 C = 3 × t ⇒ t = 125 s
⇒
108
9.85
50. Moles of Au deposited =
= 0.05
197
⇒ gram equivalent of Au deposited = 0.05 × 3 = 0.15
Now, according to Faraday’s law of electrolysis, if same
quantity of electricity is passed through different cells
connected in series, same number of gram equivalents of
electrolytes are discharged at respective electrodes.
⇒ gram equivalent of Cu deposited = 0.15
63.5
⇒ amount of Cu deposited = 0.15 ×
= 4.7625 g
2
Also, Coulombs passed = 0.15 × 96500 = I × 5 × 60 × 60
0.15 × 96500
⇒
I=
= 0.80 A
5 × 3600
170 Electrochemistry
51. During electrolysis, Ni 2+ will be reduced at cathode and H2O
will be oxidised at anode.
Number of Faraday’s passed =
3.7 × 6 × 60 × 60
= 0.828
96500
⇒ 0.828 g equivalent of Ni 2+ will be deposited at cathode.
Initial moles of Ni 2+ ion = 2 × 0.5 = 1.0
Moles of Ni 2+ ion remaining after electrolysis = 1.0 –
⇒ Molarity of Ni
2+
0.828
2
= 0.586
0.586
in final solution =
= 1.172 M
0.50
3. NaCl and KCl are strong electrolytes. So, the study of their molar
conductances (λ m ) can be experimentally verified by
Debye-Huckel Onsagar equation,
Λcm = Λ0m − B C
Λcm = molar conductance at concentration.
Λ0m = molar conductance at infinite dilution. i.e. C → 0
B = Debye-Huckel Onsagar constant.
For (both NaCl and KCl) a strong binary electrolyte like AB, the
nature of the plot of Λ m vs C will be
Λm
Topic 2 Conductivity of Electrolytic
Solutions and their Measurement
and Nernst Equation
1.
Key Idea The aqueous solution of ionic surfactant, i.e.
–
+
sodium stearate (C17H35COONa) acts as a strong univalent
type of electrolyte in the concentration range below the CMC
and the linear function of dependence of Λ m on C has a
small negative slope.
√C
⊕
Size of Na is being smaller than K ⊕ and Na ⊕ will remain in
more hydrated state, i.e. larger sized in aqueous solution. As a
result, ionic mobility as well as ionic conductance of Na ⊕
(or NaCl as ClÈ is common to NaCl and KCl) will be lower than
K ⊕ (or KCl). Thus, the plot of Λ m vs C for NaCl and KCl is as
follows :
CMC of Lm on ÖC has a
small negative slope.
Lm
Λm
¾
√C
At normal or low concentration, sodium stearate
[CH3(CH2)16COO −Na+ ] behaves as strong electrolyte and for
strong electrolyte, molar conductance (Λm ) decreases with
increase in concentration.
Above particular concentration, sodium stearate forms aggregates
known as micelles. The concentration is called as CMC. Since,
number of ions decreases and hence Λm also decreases.
Hence, option (b) is correct.
2. Electrical conductivity of the given aqueous solutions depends
on the degree of ionisation. Degree of ionisation is directly
proportional to the acidic strength. Electron withdrawing groups
(EWGs) increases the stability of the carboxylate ion by
dispersing the negative charge through resonance effect on the
conjugate while electron donating groups (EDGs) decreases the
stability of the carboxylate ion by intensifying the negative
charge.
O
C s
O
EWG
O
EDG
C s
O
Acidity of carboxylic
acids decreases due
to the presence of
electron donating groups
Acidity of carboxylic
acids increases due
to the presence of electron
withdrawing groups (EWGs)
The correct order of acidic strength and electrical conductivity is
as follows:
HCOOH > PhCOOH > CH3COOH
A
C
B
Na
KC
l
Cl
√C
4. The explanation of statements (S 1 and S 2) are as follows :
In conductivity cell, conductivity (κ ) is equal to the sum of ionic
conductances (c), of an electrolytic solution present is unit
volume of the solution enclosed by two electrodes of unit area
(a ≠ 1) separated by a unit length (l = 1).
l
κ = c×
a
⇒
κ = cwhen l = 1, a = 1
So, with decrease in the concentration of electrolyte, number of
ions in the given unit volume also decreases, i.e. κ [conductivity]
also decreases.
Thus, statement S 1 is wrong. S 2 : Molar conductivity (λ m ) is
defined as the conducting power of all the ions present in a
solution containing 1 mole of an electrolyte.
1000
λ m = κ × VmL = κ ×
M
where, VmL = volume in mL containing 1 mole of electrolyte
m = molar concentration (mol/L)
So, in a conductivity cell
1
λm ∝
M
i.e. molar conductivity increases with decrease in the
concentration (M) of electrolyte.
Thus, statement S 2 is correct.
Electrochemistry 171
5.
Key Idea Gibbs energy of the reaction is related to E°cell by
the following formula
∆Gº = − nFE °cell
∆G º = Gibbs energy of cell
nF = amount of charge passed
E = EMF of a cell
Given reaction is,
Zn + Cu 2+ → Zn 2+ + Cu
Eºcell = 2.0 V
F = 96000 C
n=2
To find the value of ∆G º (kJ mol), we use the formula
∆G º = − nFE ºcell
∆G º = −2 × 96000 × 2 = −384000 J/mol
−384000
In terms of kJ/mol, ∆G º =
= −384 kJ/mol
1000
6. According to Kohlrausch’s law, the molar conductivity of HA at
infinite dilution is given as,
Λ°m (HA ) = [ Λ°m (H+ ) + Λ°m (Cl− ) ] + [ Λ°m (Na + ) + Λ°m ( A − ) ]
− [ Λ°m (Na + ) + Λ°m (Cl− ) ]
= 425.9 + 100.5 − 126.4 = 400 S cm 2 mol − 1
Also, molar conductivity at given concentration is given as,
8.
1000 × κ
M
Given, κ = conductivity ⇒ 5 × 10− 5 S cm − 1
Λm =
M = Molarity ⇒ 0.001 M
1000 × 5 × 10− 5 S cm− 1
= 50 S cm 2 mol − 1
Λm =
10− 3 M
∴
Therefore, degree of dissociation (α), of HA is,
Λ
50 S cm2mol− 1
= 0.125
α= m =
°
Λ m 400 S cm2mol− 1
7. According to Nernst equation,
Ecell = E °cell −
Given,
2.303 RT
log Q
nF
2.303 RT
= 0.059 V
F
Ecell = E °cell −
∴
0.059
log Q
n
At equilibrium, Ecell = 0
0.059
log KC
E °cell =
n
For the given reaction, n = 2
[given]
Also,
KC = 10 × 1015
0.059
E°cell =
log (10 × 1015 ) = 0.472V ≈ 0.473V
∴
2
Cell
Anode (A) Cathode (C)
E °cell (SRP) = E °C − E ° A
Ag
Zn
1. [Zn + 2Ag+ → Zn2+ + 2Ag]
0.80 − (−0.76) = + 156
. V for 2e−
+
156
.
= + 0.78 V
2
Fe
Zn
2. [Zn + Fe 2+ → Zn2+ + Fe]
− 0.44 − (− 0.76) = + 0.32 V for 2e−
+
0.32
= + 016
. V
2
Au
Zn
3. [ 3Zn + 2Au 3+ → 3Zn 2+ + 2Au]
1.40 − (−0.76) = + 2.4 V for 6e−
+
216
.
= + 0.36 V
6
Fe
Zn
4. [ 3Zn + 2Fe3+ → Zn 2+ + 2Fe2+ ]
0.77 − (−0.76) = + 153
. V for 2e−
+
153
.
= + 0.765 V
2
9. It is an electrochemical cell. The overall cell reaction can be
written, as
H 2 ( g ) + 2AgCl( s ) → 2HCl( aq ) + 2Ag(s)
(1 bar)
(10−6 M)
(i) According to Nernst equation,
2.303 × RT
[HCl]2 [ Ag]2
°
°
Ecell = ( Ecathode
− Eanode
)−
log
n×F
pH [ AgCl]2
2
°
°
Here, (i) Ec° = EAgCl/
= Ecathode
Ag, Cl −
°
°
(ii)Eanode
= E2H
= 0.00 V
+
/H
2
(Standard hydrogen electrode)
⇒
E°cell free e transfer
0.92 = ( Ec° − 0 ) − 0.06 × log
(10−6 )2 × 12
1 × 12
= Ec° + 0.06 × 6 × 2
⇒
Ec°
= 0.92 − 0.72 = 0.20 V
Note 10 −6 molal HCl is a very dilute solution.
So, 10 −6 m ~
− 10 −6 M
10. The relationship between standard electrode potential (E° ) and
equilibrium constant (K ) of the cell reaction,
Zn(s)+ Cu 2+(aq)
Zn 2+(aq)+ Cu(s)
c
can be expressed as,
RT
E° =
ln K ⇒ K = enFE ° / RT
nF
172 Electrochemistry
Given, n = 2, F = 96000 C mol −1
−1
E° = 2 V, R = 8 JK mol
Now, molar conductivity
κ
λm=
1000 × m
1/ 4
1
=
=
1000 × 0 .5 2000
−1
T = 300 K
∴
K =e
2 × 96000 × 2
8 × 300
= e160
11. The redox reaction is : Zn(s) + Cu 2+ → Zn 2+ + Cu
2.303 RT
log10
2F
2.303RT
= 11
. −
2F
2.303 RT 

Also, ∆G = − nEF = −2F 11
. −


2F 
The Nernst equation is E = E ° −
= − 2.2F + 2.303RT
= 2.303RT − 2. 2F
12. Zinc metal is the most stable metal to cover iron surfaces. The
process of coating the iron surface by zinc is called
galvanisation.
13. Standard electrode potential of reaction [ E° ]can be calculated as
o =E − E
Ecell
R
P
where, ER = SRP of reactant , EP = SRP of product
o
If Ecell
= +ve, then reaction is spontaneous otherwise
non-spontaneous.
E ° = 1.51 V
1
→ Mn 2+
Mn 3+  
Mn
∴ For Mn
2+
E 2° = − 1.18 V
 
→ Mn
2+
disproportionation,
E° = − 1. 51 V − 1.18 V = − 2. 69 V < 0
Thus, all reaction will not occur.
= 5 × 10−4 S m 2 mol −1
16.
PLAN When different number of electrons are involved in a redox
reaction
° = ∆G °1 + ∆G °2
∆Gnet
− n3FE °3 = − n1FE °1 − n2FE °2
n E ° + n2E °2
E °3 = 1 1
n3
∴
(P) E3° Fe3+ / Fe
Net reaction Fe3+ → Fe
is obtained from
Fe
Fe
3+
2+
where, λ C = limiting equivalent conductivity at concentration C
λ ∞ = limiting equivalent conductivity at infinite dilution
C = concentration
15. In order to solve the problem, calculate the value of cell constant
of the first solution and then use this value of cell constant to
calculate the value of k of second solution. Afterwards, finally
calculate molar conductivity using value of k and m.
For first solution,
k =1.4 Sm −1, R = 50 Ω, M = 0.2
1
l
Specific conductance (κ ) = ×
R A
1
l
1.4 Sm −1 =
×
50 A
l
= 50 × 1.4 m −1
⇒
A
l
For second solution, R = 280, = 50 × 1.4 m −1
A
1
1
κ=
× 1.4 × 50 =
280
4
+ e → Fe
n
2+
E°
d
°
E1 = 0.77 V
n1 = 1
−
+ 2e → Fe
n2 = 2
E2° = − 0.44 V
3+
n3 = 3
E3° = ?
Q Fe
E °3 =
−
+ 3e → Fe
n1E °1 + n2E °2 0.77 + 2(−0.44) −0.11
=
=
= −0.04 V
n3
3
3
Thus, P — (3)
Net reaction
4 H2O r 4 H+ + 4 OH−
is obtained from
n
2H2O → O 2 + 4 H+ + 4 e−
n1 = 4
2H2O + O 2 + 4 e− → 4OH−
+
−
4H2O → 4H + 4 e
14. According to Debye Huckel Onsager equation,
λC = λ ∞ − B C
−
E °3 =
?
n1E °1 + n2E °2
= E °1 + E °2
n3
Cu → Cu + + e−
Cu
E3 ° =
2+
−
+ e → Cu
+
n
2
E°
0.34 V
1
− 0.52 V
E°3
?
n1E1 ° + n2E2 ° 2 × 0.34 + 1 × (−0.52)
=
= 0.16 V
n3
1
Also,
Cu → Cu + + e−
Cu 2+ + e− → Cu +
Cu
+ 0.40 V
n3 = 4
= − 1.23 + 0.40 = −0.83 V
Thus, Q — (4)
(R) Cu 2+ + Cu → 2Cu +
For thus E° of Cu 2+ → Cu +
is also required.
Cu 2+ + 2e− → Cu
2+
n2 = 4
E°
−1.23 V
+ Cu → 2Cu
n
n1 = 1,
n2 = 1
+
E° = − 0.52 + 0.16 = − 0.36 V
Thus, (R) — (1)
E°
− 0.52 V
0.10 V
Electrochemistry 173
(S) Cr 3+ → Cr 2+
is obtained from
Cr 3+ + 3e− → Cr
n
3
E°
− 0.74 V
2
+ 0.91V
1
?
Cr → Cr 2+ + 2e−
Cr 3+ + e− → Cr 2+
IV.
In burette
acid
Weak
[(C 2H 5) 3N]
V.
− 0.74 × 3 + 2 × 0.91
= −0.4 V
1
Thus, S = (2)
P — (3), Q — (4), R — (1), S — (2)
E3 ° =
In flask base
Weak
Conductivity increases due
(CH 3COOH) to formation of ions and
then remains constant due
to addition of weak base.
KX
AgNO3
17. The variation is conductivities in general can be seen as :
In flask base
Curve
Strong
(NaOH)
Conductance first
decreases due to formation
of H2O and then increases
due to addition of strong
electrolyte.
Volume of acid added
II.
Strong
(CH 3COOH)
Weak
(KOH)
Insoluble salt AgX is
formed, hence
conductance remains
constant. It increases due
to addition of KX.
18. The half reactions are Fe( s ) → Fe2+ ( aq ) + 2e– × 2
O2 (g ) + 4H+ + 4 e– → 2H2O
2Fe(s) + O2 (g ) + 4H+ → 2Fe2+ (aq) + 2H2O(l ) ;
E = E º–
0.059
(10–3 )2
= 1.57V
log
4
(10–3 )4 (0.1)
19. As AgNO3 is added to solution, KCl will be displaced according
to following reaction.
AgNO3 (aq) + KCl(aq) 
→ AgCl(s) + KNO3 (aq)
Conductance
I.
In burette
acid
Strong
(HI)
Curve
Conductance increases
slightly as NH+4 (salt) is
hydrolysed forming HCl.
After neutral point, it acid
increases rapidly due to
addition of strong
For every mole of KCl displaced from solution, one mole of
KNO3 comes in solution resulting in almost constant
conductivity. As the end point is reached, added AgNO3 remain
in solution increasing ionic concentration, hence conductivity
increases.
20. The net reaction is
1
O2 + Fe → H2O + Fe2+ ; E° = 1.67 V
2
2 × 1.67 × 96500
kJ = – 322.31 kJ
∆G ° = − nE ° F = −
1000
2H+ +
21. The cell reaction is :
Conductance
Zn + Fe2+ r Zn 2+ + Fe ; Ecell = 0.2905 V
⇒
⇒
Volume of acid added
Weak
(CH 3COOH)
Strong
(KOH)
Conductivity decreases
due to neutralisation of
conducting strong base and
then remains constant due
to addition of weak acid.
Also
⇒
⇒
22. In LiCl, NaCl and KCl, anions are same.
Cations have same charge but different size. Smaller cations are
more heavily hydrated in aqueous solution giving larger
hydrated radius and thus smaller ionic speeds and equivalent
conductance.
Conductance
III.
0.059
[Zn 2+ ]
log
2
[Fe2+ ]
0.059
0.1
E° = 0.2905 +
= 0.32 V
log
2
0.01
0.059
E° =
log K
n
2E °
0.32
log K =
=
0.059 0.0295
K = (10)0.32/ 0.0295
E = E° −
⇒
Weak acid added
to strong base
Equivalent conductance : KCl > NaCl > LiCl
23. The spontaneous cell reaction is
X − + M + → M + X ; E° = 0.11V
174 Electrochemistry
24. E° is an intensive property :
(i) Cu 2+ + 2e−
2+
E°
∆G ° = − nE ° F
0.337 V
→ Cu
−
– 0.674 F
+
(ii) Cu + e → Cu
0.153 V
– 0.153 F
Subtracting (ii) from (i) gives :
Cu + + e− → Cu ∆G ° = − 0.521 F = − nE ° F
⇒
E° = 0.521 V
Q
n=1
25. 2Cu+ (aq)
For the cell reaction,
°
°
°
ECell
= ECu
− ECu
= (0.52 − 016
. ) V = 0.36 V
+
2+
/ Cu
/ Cu+
⇒
⇒
°
∆G ° = − nF ECell
°
− RT ln K = − nF ECell
E
1
°
ln K = n ×
× ECell
=1×
× 0.36
RT
0.025
= 14.4 = 144 × 10−1
º
26. ∴ ∆G º = − nFEcell
Here, ∆G º = 17.37 kJ mol − 1
n = number of electrons
F = Faraday constant = 96500 C /mol
= 17.37 × 1000 J mol − 1
or
ln pH 2 ≥ −10 + ln pH 2O
or ln pH 2 ≥ −10 + 2.3 log (0.01) (as pH 2O = 1%)
≥ −10 − 4.6
so
ln pH 2 ≥ −14.6
M → M 2 + at left hand electrode.
M 2+ → M at right hand electrode
⇒ M 2+ (RHS electrode) → M 2+ (LHS electrode)
E° = 0
Applying Nernst equation
0.059
[ M 2 + ] at LHS electrode
Ecell = 0.059 = 0 −
log
2
0.001
[ M 2 + ] at LHS electrode
=−2
⇒ log
0.001
⇒ [ M 2+ ] at LHS electrode = 10−2 × 0.001 = 10−5 M
The solubility equilibrium for MX 2 is
MX 2 (s) r M 2+ (aq) + 2 X − (aq)
Solubility product, K sp = ][ M 2+ ][ X − ]2
º
17.37 × 1000 = − 3 × 96500 × Ecell
17. 37 × 1000
º
Ecell
=
3 × 96500
º
Ecell = − 0.06 = − 6 × 10− 2
= 10−5 × (2 × 10−5 )2 = 4 × 10−15
[Q In saturated solution of MX 2 , [ X − ] = 2 [ M 2+ ]]
2 × 0.059 × 96500
kJ = − 11.4 kJ
1000
30. M (s) + M + (aq, 1 M ) → M + (aq, 0.05 M ) + M (s)
2.303 RT
0.05
Ecell = 0 −
log
>0
F
1
Hence, | Ecell | = Ecell = 0.70 V and ∆G < 0 for spontaneity of
reaction.
0.0538
31. Ecell = E ° −
log 0.0025 = 0.139 V ≈ 140 mV
1
29. ∆G = − nEF = −
27. Given
1
(i) 2Cu(s)+ O 2 (g ) → Cu 2O(s);
2
or
28. For the given concentration cell, the cell reaction are
Cu(s)+ Cu 2+(aq)
c
⇒
 pH 
105 + 104 ln  2  ≥ 0
 pH 2O 
4
10 (ln pH 2 − ln pH 2O) ≥ −105
or
∆G º= −78000 J mol −1
=−78 kJ mol −1
1
(ii) H2 (g ) + O 2 (g ) → H2O(g ); ∆G º= −178000 J mol −1
2
= −178 kJ mol −1
So, net reaction is (By (i)-(ii))
2Cu(s)+H2O(g) → Cu 2O(s)+H2(g);
∆G =100000 J/ mol or 105 J/mol = 100 kJ mol −1
Now , for the above reaction
 pH 
∆G = ∆G º+ RT ln  2 
 pH 2O 
and to prevent above reaction,
∆G ≥ 0
So,
 pH 
∆G º + RT ln  2  ≥ 0
 pH 2O 
After putting the values,
 pH 
105 + 8 × 1250 ln  2  ≥ 0
 pH 2O 
° >0
32. For spontaneous redox reaction : Ecell
2I− + Cl 2 → 2Cl − + I2
E° = 1.36 – 0.54 = 0.82 V > 0
i.e. Cl 2 will spontaneously oxidise I− .
For
° < 0, they are non-spontaneous.
In other cases Ecell
33. For the reaction :
(i) 4Fe3+ + 2H2O → 4Fe2+ + 4H+ + O2; E° = − 0.46 V
(ii) 4Mn 3+ + 2H2O → 4Mn 2+ + 4H+ + O2;E° = + 0.27 V
As evidenced above, reaction (i) is non-spontaneous, therefore,
Fe3+ is stable in acid solution. However, reaction (ii) is
spontaneous Mn 3+ oxidises H2O to O2 and itself reduced to
Mn 2+ in acidic medium.
34. Sodium fusion extract from aniline produces NaCN which reacts
with Fe2+ to form [Fe(CN)6 ] 4− . The complex ion then reacts with
Fe3+ to give blue precipitate of prussian blue.
Fe3+ + [Fe(CN)6 ]4− r Fe4[Fe(CN)6 ]3
Prussian blue
Electrochemistry 175
35. E° for 2Ag+ + C6H12O6 + H2O r 2Ag(s)
40. Given,
+ C6H12O7 + 2H+ is 0.75 V
0.0592
2E °
Also E ° =
= 25.33
log K ⇒ log K =
2
0.0592
⇒ ln K = 2.303 log K = 58.35
36. On increasing concentration of NH3 , the concentration of H+ ion
decreases, therefore,
° − 0.0592 log [ H+ ]2 = 0 − 0.0592 × 2 log 10− 11
Ered = Ered
2
2
= 0.65 V
i.e. Ered increases by 0.65 V.
In 2+ + e− → In +
38. The solubility of AgBr in10 M AgNO3 solution is determined as
AgBr r
Ag+ + Br −
S
S + 10− 7
In 3+ + 2e− → In +
−14
K sp = 14 × 10
In 3+ + e → In 2+ ;
⇒
Now, for :
= S (S + 10 )
K w = [H+ ][OH− ]
Q K w is a function of temperature, [H+ ] will change with
temperature.
(b) Let the emf of first cell be X volt.
⇒ emf of 2nd cell = (X + 0.03) volt
[Cu 2+ ] in 2nd cell = 0.50 M
[NO−3 ] = 10− 7 M
⇒ κ (sp. conductance) = κ Br − + κ Ag + + κ NO −
= [8 × 10
−7
× 3 × 10
−3
+ 6 × 10
= 0.15 – (– 0.44) = 0.59 V
E° = 0.0590 log K
E°
log K =
= 10 ⇒ K = 1010
0.059
Also
In pure water, [H+ ] depends on value of K w which is
[Ag+ ] = 4 × 10−7 M,
−3
3
[Cu 2+ ] in 1st cell = ?
−7
× 4 × 10
−3
+ 7 × 10
E1 = E1° −
−7
× 10 ] 1000
= 24 × 10−7 + 24 × 10−7 + 7 × 10−7
−7
= 55 × 10 S m
−1
−7
= 55 (in terms of 10
Sm )
= – 109 – (– 129 + 77) kJ = – 57 kJ
Ag | AgCl, Cl − || Ag+ | Ag
For K sp ; reaction is AgCl (s) r Ag+ + Cl −
∆G ° = + 57 kJ ⇒ ∆G ° = − RT ln K sp
∆G °
57 × 1000
log K sp = −
=−
= − 10
2.3 RT
2.3 × 8.314 × 298
⇒
E° of Ag+ + Cl − r AgCl
Now,
∆G ° 57000
=
= 0.59 V
nF
96500
(b) The cell reaction is :
Zn + 2Ag+ r Zn 2+ + 2Ag; E ° = 1.56 V
E° = −
0 = E° −
⇒
⇒
log
0.059
[Zn 2+ ]
log
2
[Ag+ ] 2
2 × 1.56
2E °
[Zn 2+ ]
=
= 52.88
=
0.059
[Ag+ ] 2 0.059
Moles of Zn added =
6.539 × 10− 2
= 10− 3
65.39
⇒ Moles of Ag formed = 2 × 10− 3.
2.303 RT
[Zn 2+ ]
log
2F
[Cu 2+ ]
2.303 RT
[Zn 2+ ]
log
E2 = E1° −
2F
[Cu 2+ ]2
−1
39. (a) ∆G ° = Σ ∆G °f (products) − Σ ∆G °f (reactants)
Cell :
Cu 2+ + In 2+
∆G ° = 0.44 F = – E ° F
E° = − 0.44 V
→ Cu + + In 3+
E ° = E ° (Cu 2+ / Cu + ) − E ° (In 3+ / In 2+ )
−7
[Br − ] = 3 × 10−7 M,
⇒
…(ii)
41. (a) pH = – log [H+ ]
S = 3 × 10−7 M
Solving for S gives :
E° = – 0.42
⇒ ∆G ° = 0.84 F
Subtracting (i) from (ii)
⇒
Ag+ + NO−3
S + 10− 7
10−7
AgNO3 →
…(i)
⇒ ∆G ° = 0.40 F
37. NH3 has no effect on the E° of glucose/gluconic acid electrode.
−7
E° = – 0.40
⇒
⇒
⇒
42. At anode
2.303 RT 
[Cu 2+ ]2 
log


2F
[Cu 2+ ]1 

0.50
0.03 = 0.03 log
[Cu 2+ ]1
E2 − E1 =
0.50
= 10
[Cu 2+ ]1
⇒ [Cu 2+ ]1 = 0.05 M
1
H2 → H+ + e− ; E° = 0
2
AgCl (s) + e− → Ag + Cl − ; E° = ?
1
(i) Cell reaction : H2 + AgCl (s) → Ag + H+ + Cl −
2
(ii) ∆G ° = − nE ° F = ∆H ° − T ∆S °
At 15°C : – 0.23 × 96500 = ∆H ° − 288 ∆S °
…(i)
At 35°C : – 0.21 × 96500 = ∆H ° − 308 ∆S °
…(ii)
⇒ 96500 (0.23 − 0.21) = − 20 ∆S °
96500 × 0.02
⇒
∆S ° = −
= − 96.5 J
20
Substituting value of ∆S ° in (i)
∆H ° = 288 × (− 96.5) − 0.23 × 96500 = – 49.987 kJ
At cathode
176 Electrochemistry
(iii) At 25°C
− E ° × 96500 = − 49987 − 298 (− 96.5)
E° = 0.22 V
−
AgCl (s) + e → Ag + Cl − ; E° = 0.22 V
⇒
⇒
Ag → Ag+ + e− ;
Adding :
⇒
⇒
43. E = 0 −
⇒
2+
2Hg + 2Fe3+ r Hg2+
2 + 2Fe
AgCl (s) → Ag+ + Cl − ;
K =
E° = – 0.58 V
=
Q E° =
[Fe2+ ]2 [Hg2+
2 ]
[Fe3+ ]2
(9.5 × 10− 4 )2 (4.75 × 10− 4 )
= 0.17
(5 × 10− 5 )2
0.0592
log K
2
= E ° (Fe3+ / Fe2+ ) − E ° (Hg22+ / Hg)
[Ag+ ]anode
0.10
⇒ [Ag+ ]anode = 1.7 × 10− 4 M
In saturated Ag2CrO4 solution present in anode chamber :
Ag2CrO4 (s) r
2Ag+
+
CrO24−
−4
1.7 × 10 M 1.7 × 10− 4 M
2
K sp = [Ag+ ]2 [CrO24− ]
⇒ E° (Hg2+
2 / Hg) = 0.77 + 0.0226
= 0.7926 V
48. In a saturated AgI solution;
[Ag+ ] = 8.7 × 10− 17 M
= 9.32 × 10− 9 M
EAg +/ Ag = E ° − 0.0592 log
⇒
 1.7

= (1.7 × 10− 4 )2 
× 10− 4 = 2.45 × 10− 12
 2

E ° = E ° (Fe3+ / Fe2+ ) − E ° (I−3 / I− )
= 0.77 – 0.54 = 0.23 V
0.0592
E° =
log K
(n = 2)
2
2E °
2 × 0.23
= 7.77
log K =
=
0.0592 0.0592
K = 5.89 × 107
Q
⇒
4+
E° = E° (Ce
Q
⇒
⇒
3+
3+
/ Ce ) – E° (Fe
2+
/ Fe )
= 1.44 – 0.68 = 0.76 V
E° = 0.0592 log K
E°
0.76
log K =
=
= 12.83
0.0592 0.0592
12
K = 6.88 × 10
46. pH = 14
⇒
pOH = 0 ⇒ [OH− ] = 1.0 M
K sp = 10− 19 = [Cu 2+ ][OH− ]2
⇒
[Cu 2+ ] =
10− 19
= 10− 19
[OH− ]2
For reaction : Cu 2+ + 2e− → Cu; E° = 0.34 V
E = E° −
= 0.34 –
0.0592
1
log
2
[Cu 2+ ]
0.0592
log 1019 = – 0.222 V
2
1
9.32 × 10− 9
= 0.324 V
Also, for
AgI r Ag+ + I− ;
E° = 0.0592 log K sp
= – 0.95 V
Ag → Ag+ + e− ; E° = – 0.799 V
AgI + e− → Ag + I−
Adding :
⇒
Fe2+ + Ce4+ r Fe3+ + Ce3+
45.
1
[Ag+ ]
= 0.799 – 0.0592 log
2 Fe3+ + 3I− r 2Fe2+ + I−3
44. For
0
0
4.75 × 10 − 4 9.5 × 10 − 4
= − 0.0226 V
0.0592
[Ag + ]anode
log
1
[Ag + ]cathode
0.164 = − 0.0592 log
10 − 3 M
5 × 10 − 5
Initial :
Equilibrium :
E° = – 0.80 V
E° = 0.0592 log K sp
− 0.58
log K sp =
= − 9.79
0.0592
K sp = 1.6 × 10−10
⇒
47. For reaction,
E° = x
AgI → Ag+ + I− ; E° = – 0.95 V
= x – 0.799
x = − 0.151 V
49. The redox reaction is
Zn + Ni 2+ r Zn 2+ + Ni E° = + 0.51 V
0.0592
⇒
E° =
log K
2
0.51 × 2
log K =
= 17.23
⇒
0.0592
17
⇒
K = 1.7 × 10
Such a high value of equilibrium constant indicates that the
reaction is almost complete. Therefore, concentration of
Zn 2+ in solution will be equal to initial concentration of Ni 2+
ion, i.e. 1.0 M.
50. The galvanic cell is : Cu | Cu 2+ || Ag + | Ag
Cell reaction : Cu + 2Ag+ r Cu 2+ + 2Ag; E° = 0.462 V
0.0592
(0.01)
log
E = 0 = 0.462 −
2
[Ag+ ]2
⇒
[Ag+ ] = 1.57 × 10− 9 M
Electrochemistry 177
H2 O + e− r
51.
1
H2 + HO− ; E° = – 0.8277 V
2
1
H2 + H2O r H3O+ + e− ;
2
2H2O r H3O+ + HO−
E° = – 0.8277 V
⇒
52. At anode
Zn → Zn 2+ + 2e− E° = 0.76 V
At cathode 2H+ + 2e− → H2 (g )
⇒ For
+
Zn + 2H → Zn
2+
⇒
⇒
Therefore, the cell reaction is spontaneous.
55. pH = Cα = 10−4
10−4
0.0015
 A
Also, conductance (G ) = κ  
 l
⇒
α=
⇒
120
 l
κ = G   = 5 × 10−7 ×
 A
1
E° = 0.00 V
+ H2 (g ) E° = 0.76 V
0.0592
[Zn 2+ ]
log
2
[H+ ]2
2 (E − E ° )
1
= − log [Zn 2+ ] − 2 log +
0.0592
[H ]
E = E° −
⇒
Ecell = 1.113 V > 0
∆G = − nEF < 0
E° = 0 V
E° = 0.0592 log K
− 0.8277
log K =
= − 13.98
0.0592
− 14
K = 1.04 × 10 .
⇒
(iii) Q
− 16.2 = – log (0.1) – 2 pH
1 + 16.2
pH =
= 8.6
2
⇒
⇒
56. Energy obtained as one mole X is converted into Y is 193 kJ.
53. For conductivity, the charge carriers are required. In
anhydrous state, HCl is not ionised and no charge carrier
ions are available, hence bad conductor. However, in aqueous
solution, HCl is fully ionised producing H+ and Cl − and
conducts electricity.
54. (i) The cell reaction is
Zn + Cu 2+ → Zn 2+ + Cu
° = Ecathode
°
° = 0.350 – (– 0.763) = 1.113 V
(ii) Ecell
− Eanode
Q Both Zn 2+ and Cu 2+ are at unit concentrations,
E = E ° = 1.113 V
= 6 × 10−5
κ × 1000
Λc =
C
6 × 10−5 × 1000
=
0.0015
Λc 6 × 10−5 × 1000 0.0015
∞
=
×
Λ =
α
0.0015
10−4
2
−1
−1
= 600 = 6 × 10 S cm mol
H(1s1)
Ground state
3s
3p
3d
Second excited
state of H-atom
All are degenerate
degeneracy = 9
Energy consumed in converting one mole of M + to M 3+
96500
J
= − nE ° F = 2 × 96500 × 0.25 J =
2
 96500
⇒ 193 × 103 = n 
 ⇒ n = 4 mol
 2 
11
Chemical Kinetics
Objective Questions I (Only one correct option)
1. Consider the following reactions
A → P1; B → P2; C → P3; D → P4,
The order of the above reactions are a , b , c and d,
respectively. The following graph is obtained when log[rate]
vs log[conc.] are plotted:
[D]
4. For the reaction of H 2 with I2 , the rate constant is
. dm 3 mol− 1 s− 1
2.5 × 10− 4 dm3 mol− 1 s− 1 at 327ºC and 10
at 527ºC. The activation energy for the reaction, in
kJ mol− 1 is (R = 8.314 JK − 1 mol− 1 ) (2019 Main, 10 April II)
(a) 59
(c) 150
(b) 72
(d) 166
5. A bacterial infection in an internal wound grows as
[B]
log [rate]
[A]
[C]
N ′ ( t ) = N0 exp ( t ), where the time t is in hours. A dose of
antibiotic, taken orally, needs 1 hour to reach the wound.
Once it reaches there, the bacterial population goes down as
dN
N
= − 5N 2 . What will be the plot of 0 vs t after 1 hour ?
dt
N
(2019 Main, 10 April I)
log [conc.]
(a)
N0
N
Among the following, the correct sequence for the order of
the reactions is
(2020 Main, 6 Sep I)
(a) D > A > B > C
(c) C > A > B > D
(b) A > B > C > D
(d) D > B > A > C
(c)
decomposition of N2 O5 in CCl 4 as per the equation,
The initial concentration of N2 O5 is 3.00 mol L−1 and it is
2.75 mol L
is
−1
after 30 minutes. The rate of formation of NO2
(a) 4167
.
× 10−3 mol L −1 min −1
(b) 1667
.
× 10
−2
mol L
−1
min
t(h)
t(h)
6. Consider the given plot of enthalpy of the following reaction
between A and B. A + B → C+D
Identify the incorrect statement.
Enthalpy 15
(kJ mol–1)10
(d) 2.083 × 10−3 mol L −1 min −1
5
3. In the following reaction; xA → yB
 d [ B ]
 d [ A ]
+ 0.3010
= log 10 
log 10 −

 dt 
 dt 
n-butane and iso-butane
C2H2 and C6 H6
C2H4 and C4 H8
N2O4 and NO2
N0
N
(d)
(2019 Main, 9 April II)
20
(c) 8.333 × 10−3 mol L −1 min −1
(a)
(b)
(c)
(d)
N0
N
(2019 Main, 12 April II)
−1
A and B respectively can be
t(h)
t(h)
2. NO2 required for a reaction is produced by the
2N2 O5 ( g ) → 4NO2 ( g ) + O2 ( g )
N0
N
(b)
(2019 Main, 12 April I)
D
A +B
C
Reaction
coordinate
(a)
(b)
(c)
(d)
D is kinetically stable product.
Formation of A and B from C has highest enthalpy of activation.
C is the thermodynamically stable product.
Activation enthalpy to form C is 5 kJ mol −1 less than that to
form D.
Chemical Kinetics 179
7. The given plots represent the variation of the concentration of
a reaction R with time for two different reactions (i) and (ii).
The respective orders of the reactions are
13. If a reaction follows the Arrhenius equation, the plot lnk vs
1/(RT) gives straight line with a gradient (− y) unit.
The energy required to activate the reactant is
(2019 Main, 11 Jan I)
(2019 Main, 9 April I)
(i)
(ii)
In [R]
[R]
y
unit
R
(a)
(b) − y unit
(c) yR unit
(d) y unit
14. For an elementary chemical reaction,
k1
time
(a) 1, 1
(c) 0, 1
A2
time
k −1
(2019 Main, 10 Shift II)
(b) 0, 2
(d) 1, 0
k1
= 2A , the expression for d[dtA ] is
k2
8. For a reaction scheme, A → B → C , if the rate of
formation of B is set to be zero then the concentration of B is
given by
(2019 Main, 8 April II)
(a) k1k2[ A ]
k 
(b)  1 [ A ]
 k2 
(c) (k1 − k2 )[ A ]
(d) (k1 + k2 )[ A ]
(a) 2k1[ A2 ] − k−1[ A ]2
(b) k1[ A2 ] − k−1 [ A ]2
(c) 2k1[ A2 ] − 2k−1 [ A ]2
(d) k1[ A2 ] + k−1[ A ]2
15. Consider the given plots for a reaction obeying Arrhenius
equation (0°C < T < 300°C) : ( k and E a are rate constant and
activation energy, respectively)
(2019 Main, 10 Jan I)
k
k
9. For the reaction, 2 A + B → C , the values of initial rate at
different reactant concentrations are given in the table below.
The rate law for the reaction is
(2019 Main, 8 April I)
[A](mol L−1 )
[B](mol L−1 )
0.05
0.05
0.045
0.10
0.05
0.090
0.20
Initial rate
(mol L−1s−1 )
0.10
0.72
(a) rate = k [A ][B ]
(b) rate = k [A ] [B ]
(c) rate = k [A ][B ]
(d) rate = k [A ]2[B ]
2
2
2
10. For a reaction, consider the plot of ln k versus1/ T given in the
figure. If the rate constant of this reaction at 400 K is
10− 5 s − 1 , then the rate constant at 500 K is
(2019 Main, 12 Jan II)
Ea
I
T(°C)
II
Choose the correct option.
(a) Both I and II are wrong
(b) Both I and II are correct
(c) I is wrong but II is right
(d) I is right but II is wrong
16. For the reaction, 2A + B → products
When concentration of both (A and B) becomes double, then
rate of reaction increases from 0.3 mol L −1 s −1 to
2.4 mol L −1 s −1 .
When concentration of only A is doubled, the rate of reaction
increases from 0.3 mol L −1 s −1 to 0.6 mol L −1 s −1 .
Which of the following is true?
(2019 Main, 9 Jan II)
(a) The whole reaction is of 4th order
(b) The order of reaction w.r.t. B is one
(c) The order of reaction w.r.t. B is 2
(d) The order of reaction w.r.t. A is 2
Slope = –4606
ln k
17. The following results were obtained during kinetic studies of
the reaction;
1/ T
(a) 4 × 10− 4 s− 1
(c) 10
−4
−1
(d) 2 × 10
s
2A + B → Products
(b) 10− 6 s− 1
−4
−1
s
11. Decomposition of X exhibits a rate constant of 0.05 µg/year.
(b) 25
(c) 40
[B]
(in mol L−1)
I.
0.10
0.20
6.93 × 10−3
II.
0.10
0.25
6.93 × 10−3
III.
0.20
0.30
1386
.
× 10−2
(d) 50
12. The reaction, 2 X → B is a zeroth order reaction. If the initial
concentration of X is 0.2 M, the half-life is 6 h. When the
initial concentration of X is 0.5 M, the time required to reach
its final concentration of 0.2 M will be (2019 Main, 11 Jan II)
(a) 7.2 h
(b) 18.0 h
(c) 12.0 h
(d) 9.0 h
Initial rate of
reaction
(in mol L−1 min −1)
[A]
(in mol L−1 )
Experiment
How many years are required for the decomposition of 5 µg
of X into 2.5 µg?
(2019 Main, 12 Jan I)
(a) 20
(2019 Main, 9 Jan I)
The time (in minutes) required to consume half of A is
(a) 5
(c) 100
(b) 10
(d) 1
180 Chemical Kinetics
18. Which of the following lines correctly show the temperature
dependence of equilibrium constant, K , for an exothermic
reaction?
(2018 Main)
In K
A
B
1
T(K)
(0, 0)
dC
= k[ A ][ B ]
dt
dC
(c)
= k[ A ][ B ] 2
dt
dC
= k[ A ] 2[ B ]
dt
dC
(d)
= k[ A ]
dt
(a)
(b)
25. In the reaction, P + Q → R + S , the time taken for
75% reaction of P is twice the time taken for 50% reaction
of P. The concentration of Q varies with reaction time as
shown in the figure. The overall order of the reaction
(2013 Adv.)
is
××
××
××
C
××
××
××
D
(a) A and B
(c) C and D
The rate law for the formation of C is
(b) B and C
(d) A and D
[Q]0
19. At 518°C, the rate of decomposition of a sample of gaseous
acetaldehyde, initially at a pressure of 363 Torr, was 1.00 Torr
s −1 when 5% had reacted and 0.5 Torr s −1 when 33% had
reacted. The order of the reaction is :
(2018 Main)
(a) 2
(c) 1
factors. Activation energy of R1 exceeds that of R2 by 10 kJ
mol− 1 . If k1 and k2 are rate constants for reactions R1 and R2 ,
k 
respectively at 300 K, then ln  2  is equal to
 k1 
(2017 Main)
( R = 8.314 J mol − 1K − 1 )
(c) 6
(d) 4
21. Decomposition of H2 O2 follows a first order reaction. In 50
min, the concentration of H2 O2 decreases from 0.5 to 0.125 M
in one such decomposition. When the concentration of H2 O2
reaches 0.05 M, the rate of formation of O2 will be (2016 Main)
(a) 6.93 × 10−4 mol min −1
(c) 1.34 × 10−2 mol min −1
disappearance of M increases by a factor of 8 upon doubling the
concentration of M . The order of the reaction with respect to M
is
(2014 Adv.)
(b) 3
(d) 1
24. For the non-stoichiometric reaction, 2 A + B → C + D, the
following kinetic data were obtained in three separate
experiments, all at 298 K.
(2014 Main)
(i)
0.1 M
(d) 1
(b) 48.6 kJ mol −1 (2013 Main)
(d) 60.5 kJ mol −1
27. Plots showing the variation of the rate constant ( k ) with
temperature (T ) are given below. The plot that follows
Arrhenius equation is
(2010)
(a)
k
(b)
k
T
(2015 Main)
23. For the elementary reaction, M → N , the rate of
Initial
concentration
[A]
(a) 53.6 kJ mol −1
(c) 58.5 kJ mol −1
T
(a) low probability of simultaneous collision of all the reacting
species
(b) increase in entropy and activation energy as more molecules are
involved
(c) shifting of equilibrium towards reactants due to elastic collisions
(d) loss of active species on collision
(a) 4
(c) 2
(c) 0
changes from 300 K to 310 K. Activation energy of such a
reaction will be ( R = 8.314 JK −1 mol −1 and log 2 = 0.301)
(b) 2.66 L min −1 at STP
(d) 6.93 × 10−2 mol min −1
22. Higher order (>3) reactions are rare due to
(b) 3
26. The rate of a reaction doubles when its temperature
20. Two reactions R1 and R2 have identical pre- exponential
(b) 12
Time
(a) 2
(b) 3
(d) 0
(a) 8
[Q]
Initial
concentration
[B]
Initial rate of
formation of C
(mol L −1s −1)
0.1 M
1. 2 × 10−3
−3
(ii)
0.1 M
0.2 M
1. 2 × 10
(iii)
0.2 M
0.1 M
2. 4 × 10−3
(c)
k
(d)
T
k
T
28. For a first order reaction, A → P, the temperature (T )
dependent rate constant ( k ) was found to follow the
equation :
2000
log k =
+ 6.0
T
the pre-exponential factor A and the activation energy Ea ,
respectively, are
(2009)
(a) 1.0 × 106 s− 1 and 9.2 kJ mol − 1
(b) 6.0 s− 1 and 16.6 kJ mol − 1
(c) 1.0 × 106 s− 1 and 16.6 kJ mol − 1
(d) 1.0 × 106 s− 1 and 38.3 kJ mol − 1
29. Under the same reaction conditions, initial concentration
of 1.386 mol dm −3 of a substance becomes half in 40 s and
20 s through first order and zero order kinetics
Chemical Kinetics 181
( k1 ) and zero order ( k0 ) of the reaction is
−1
(a) 0.5 mol dm
(c) 1.5 mol dm −3
−3
3
(2008, 3M)
(b) 1.0 mol dm
(d) 2.0 mol −1 dm 3
aG + bH → products. When
concentration of both the reactants G and H is doubled, the
rate increases by eight times. However, when concentration
of G is doubled keeping the concentration of H fixed, the rate
is doubled. The overall order of the reaction is (2007, 3M)
30. Consider
a
reaction,
(a) 0
(c) 2
(b) 1
(d) 3
31. Which one of the following statement(s) is incorrect about
order of reaction?
(2005, 1M)
(a) Order of reaction is determined experimentally
(b) Order of reaction is equal to sum of the power of concentration
terms in differential rate law
(c) It is not affected with stoichiometric coefficient of the reactants
(d) Order cannot be fractional
32. (A) follows first order reaction, ( A ) → product. Concentration
of A, changes from 0.1 M to 0.025 M in 40 min. Find the rate
of reaction of A when concentration of A is 0.01 M.
(2004, 1M)
(a) 3.47 × 10–4 M min –1
(b) 3.47 × 10−5 M min –1
(c) 1.73 × 10−4 M min –1
(d) 1.73 × 10−5 M min –1
33. In a first order reaction the concentration of reactant
decreases from 800 mol/dm3 to 50 mol/dm3 in 2 × 104 s. The
rate constant of reaction ins −1 is
(a) 2 × 104
(b) 3.45 × 10−5
(c) 1.386 × 10−4
(d) 2 × 10−4
(2003, 1M)
34. Consider the chemical reaction,
N2 ( g ) + 3H2 ( g ) → 2NH3 ( g )
The rate of this reaction can be expressed in terms of time
derivatives of concentration of N2 ( g ), H2 ( g ) or NH3 ( g ).
Identify the correct relationship amongst the rate expressions
d [N2 ]
1 d [H2 ] 1 d[NH3 ]
=−
=
dt
2
dt
3 dt
d [N2 ]
d [H2 ]
d [NH3 ]
(b) Rate = −
= −3
=2
dt
dt
dt
d [N2 ] 1 d [H2 ] 1 d [NH3 ]
(c) Rate =
=
=
dt
2
dt
3 dt
d [N2 ]
d [H2 ] d [NH3 ]
(d) Rate = −
=−
=
dt
dt
dt
(a) Rate = −
(2002, 3M)
35. If I is the intensity of absorbed light and C is the concentration
of AB for the photochemical process.
AB + hν → AB* , the rate of formation of AB* is directly
proportional to
(2001, 1M)
(a) C
(c) I 2
(b) I
(d) C ⋅ I
36. The rate constant for the reaction, 2N2 O5 → 4NO2 + O2 is
3.0 × 10−5 s −1 . If the rate is 2.40 × 10−5 mol L − 1 s − 1 , then the
concentration of N2 O5 (in mol L − 1 ) is
(a) 1.4
(c) 0.04
(2000, 1M)
(b) 1.2
(d) 0.8
37. The half-life period of a radioactive element is 140 days.
After 650 days, one gram of the element will reduce to
(a)
1
g
2
(b)
1
g
4
(c)
1
g
8
(d)
38. A catalyst is a substance which
1
g
16
(1986)
(1983, 1M)
(a) increases the equilibrium concentration of the product
(b) changes the equilibrium constant of the reaction
(c) shortens the time to reach equilibrium
(d) supplies energy to the reaction
39. The specific rate constant of a first order reaction depends on
the
(1983, 1M)
(a) concentration of the reactant
(b) concentration of the product
(c) time
(d) temperature
40. The rate constant of a reaction depends on
(a)
(b)
(c)
(d)
(1981, 1M)
temperature
initial concentration of the reactants
time of reaction
extent of reaction
Objective Questions II
(One or more than one correct option)
41. Which of the following plots is(are) correct for the given
reaction?
(2020 Adv.)
([ P ]0 is the initial concentration of P)
CH3
CH3
H3C
OH + NaBr
H3C
Br + NaOH
CH3
CH3
(P)
( Q)
(a) t
1/2
(b)
Initial rate
k 
respectively. Ratio  1  of the rate constants for first order
 k0 
[P]0
(c) [Q]
[P]0
[P]0
(d) In [P]
[P]0
Time
Time
42. For a first order reaction, A( g ) → 2B( g )+ C ( g ) at
constant volume and 300 K, the total pressure at the beginning
( t = 0 ) and at time t are p0 and pt , respectively.
182 Chemical Kinetics
(b)
t1/3
(a)
In(3p0–pt)
Initially, only A is present with concentration [ A ]0 , and t1 / 3 is
the time required for the partial pressure of A to reach 1/3 rd of
its initial value. The correct option(s) is (are) (Assume that all
these gases behave as ideal gases)
(2018 Adv.)
(d)
Time
(1984, 1M)
increases the average kinetic energy of reacting molecules
decreases the activation energy
alters the reaction mechanism
increases the frequency of collisions of reacting species
Numerical Answer Type Questions
[A ] 0
49 If 75% of a first order reaction was completed in 90 minutes,
43. In a bimolecular reaction, the steric factor P was
experimentally determined to be 4.5. the correct option(s)
among the following is(are)
(2017 Adv.)
(a) The activation energy of the reaction is unaffected by the value
of the steric factor
(b) Experimentally determined value of frequency factor is higher
than that predicted by Arrhenius equation
(c) The value of frequency factor predicted by Arrhenius equation
is higher than that determined experimentally
(d) Since P = 4.5, the reaction will not proceed unless an effective
catalyst is used
44. According to the Arrhenius equation,
(2016 Adv.)
(a) a high activation energy usually implies a fast reaction
(b) rate constant increases with increase in temperature. This is due
to a greater number of collisions whose energy exceeds the
activation energy
(c) higher the magnitude of activation energy, stronger is the
temperature dependence of the rate constant
(d) the pre-exponential factor is a measure of the rate at which
collisions occur, irrespective of their energy
2N2 O5 ( g ) → 4NO2 ( g ) + O2 ( g )
(Take : log 2 = 0.30; log 2.5 = 0.40)
(2020 Main, 4 Sep I)
50. Consider the kinetic data given in the following table for the
reaction A + B + C → Product
Experiment
[A]
[B]
[C]
No.
(mol dm − 3 ) (mol dm − 3 ) (mol dm − 3 )
Rate of
reaction
(mol dm − 3s− 1 )
1
0.2
0.1
0.1
6.0 × 10− 5
2
0.2
0.2
0.1
6.0 × 10− 5
3
0.2
0.1
0.2
12
. × 10− 4
4
0.3
0.1
0.1
9.0 × 10− 5
The rate of the reaction for [ A ] = 015
. mol dm − 3 ,
. mol dm − 3 is found to
[ B ] = 0. 25 mol dm − 3 and [C ] = 015
be Y × 10− 5 mol dm − 3s − 1 . The value of Y is ............
(2019 Adv.)
∆
(2011)
(a) the concentration of the reactant decreases exponentially with time
(b) the half-life of the reaction decreases with increasing temperature
(c) the half-life of the reaction depends on the initial concentration
of the reactant
(d) the reaction proceeds of 99.6% completion in eight half-life
duration
46. The following statement (s) is are correct
60% of the same reaction would be completed in
approximately (in minutes) ........... .
51. The decomposition reaction
45. For the first order reaction,
(1999, 3M)
1
is linear
T
(b) A plot of log [X] vs time is linear for a first order reaction,
x→ p
1
(c) A plot of log p vs is linear at constant volume
T
1
(d) A plot of p vs is linear at constant temperature
V
(a) A plot of log K p vs
(b) a plot of reciprocal concentration of the reactant vs time gives a
straight line
(c) the time taken for the completion of 75% reaction is thrice the
1
of the reaction
2
(d) the pre-exponential factor in the Arrhenius equation has the
dimension of time, T −1
(a)
(b)
(c)
(d)
Rate constant
In(p0–pt)
(c)
(1998, 2M)
(a) the degree of dissociation is equal to (1 − e− kt )
48. A catalyst
[A ] 0
Time
47. For the first order reaction,
2N 2O 5 ( g ) → 2N 2O 4 ( g ) + O 2 ( g ) is started in a closed
cylinder under isothermal isochoric condition at an initial
pressure of 1 atm. After Y × 103 s, the pressure inside the
cylinder is found to be 1.45 atm. If the rate constant of the
reaction is 5 × 10−4 s −1 , assuming ideal gas behaviour, the
value of Y is ………
(2019 Adv.)
52. Consider the following reversible reaction,
A( g )+ B( g )
-
AB( g )
The activation energy of the backward reaction exceeds that
of the forward reaction by 2RT (in J mol −1 ). If the
pre-exponential factor of the forward reaction is 4 times that
of the reverse reaction, the absolute value of ∆Gs (in J mol −1 )
for the reaction at 300 K is ……… .
(Given ; ln( 2 ) = 0.7 RT = 2500 J mol −1 at 300 K and G is the
Gibbs energy)
(2018 Adv.)
Chemical Kinetics 183
57. For the reaction : N2 ( g ) + 3H2 ( g ) → 2NH3 ( g )
Passage Based Questions
Carbon-14 is used to determine the age of organic material. The
procedure is based on the formation of 14 Cby neutron capture in the
upper atmosphere.
14
7 N
+ 0 n1 →
14
6 C
+
1
1p
14
C is absorbed by living organisms during photosynthesis. The 14 C
content is constant in living organism once the plant or animal dies,
the uptake of carbon dioxide by it ceases and the level of 14 C in the
dead being, falls due to the decay which C-14 underoges
14
14
−
6 C → 7 N + β
The half-life period of 14 C is 5770 yr.
The decay constant ( λ ) can be calculated by using the following
0.693
formula λ =
.
t1 / 2
The comparison of the β − activity of the dead matter with that of the
carbon still in circulation enables measurement of the period of the
isolation of the material from the living cycle. The method however,
ceases to be accurate over periods longer than 30,000 yr. The
proportion of 14 C to 12 C in living matter is 1 : 1012.
(2006, 3 × 4M = 12M)
53. Which of the following option is correct?
(a) In living organisms, circulation of 14 C from atmosphere is high
so the carbon content is constant in organism
(b) Carbon dating can be used to find out the age of earth crust and
rocks
(c) Radioactive absorption due to cosmic radiation is equal to the
rate of radioactive decay, hence the carbons content remains
constant in living organisms
(d) Carbon dating cannot be used to determine concentration of
14
C in dead beings
54. What should be the age of fossil for meaningful
determination of its age?
(a)
(b)
(c)
(d)
concentration of C14 in nearby areas. C14 concentration is C1
in nearby areas and C 2 in areas far away. If the age of the fossil
is determined to be T1 and T2 at the places respectively then
(a) the age of fossil will increase at the place where explosion has
C
1
taken place and T1 − T2 = ln 1
λ
C2
(b) the age of fossil will decrease at the place where explosion has
C
1
taken place and T1 − T2 = ln 1
λ
C2
(c) the age of fossil will be determined to be the same
T
C
(d) 1 = 1
T2 C 2
Fill in the Blanks
as the rate constant at .........
order reaction.
(1986, 1M)
59. The rate of chemical change is directly proportional to
.............
(1985, 1M)
True/False
60. For a first order reaction, the rate of the reaction doubles as
the concentration of the reaction (s) doubles.
. A may be termed
(1997, 1M)
(1986, 1M)
Integer Answer Type Questions
61. An organic compound undergoes first order decomposition.
The time taken for its decomposition to 1/8 and 1/10 of its
initial concentration are t1 / 8 and t1 / 10 respectively. What is the
[t ]
value of 1 / 8 × 10 ? (log 10 2 = 0.3)
[ t1 / 10 ]
(2012)
62. The concentration of R in the reaction R → P was
measured as a function of time and the following data is
obtained :
[ R ] (molar)
t (min)
1.0
0.0
0.75
0.05
0.40
0.12
0.10
0.18
The order of the reaction is
(2010)
Subjective Questions
63. 2 X ( g ) → 3Y ( g ) + 2 Z ( g )
Partial pressure of
X (in mm of Hg)
55. A nuclear explosion has taken place leading to increase in
(− E a / RT )
58. The hydrolysis of ethyl acetate in ........... medium is a ..........
Time
(in min)
6 yr
6000 yr
60,000 yr
It can be used to calculate any age
56. In Arrhenius equation, k = A exp
Under certain conditions of temperature and partial pressure of
the reactants, the rate of formation of NH3 is 0.001 kg/h−1. The
rate of conversion of H2 under the same condition is ....
(1994, 1M)
kg /h −1 .
0
100
200
800
400
200
Assuming ideal gas condition. Calculate
(a) order of reaction
(b) rate constant
(c) time taken for 75% completion of reaction
(d) total pressure when px = 700 mm
(2005, 4M)
64. For the given reaction, A + B → Products
Following data are given
Initial conc.
(m/L)
Initial conc.
(m/L)
[ A ]0
[ B ]0
0.1
0.1
0.2
0.1
0.1
0.1
0.2
0.05
(a) Write the rate equation.
(b) Calculate the rate constant.
Initial rate
[mL–1s –1 ]
0.05
(2004, 2M)
184 Chemical Kinetics
Cu (half-life = 12.8 h) decays by β emission (38%), β +
emission (19%) and electron capture (43%). Write the decay
products and calculate partial half-lives for each of the decay
processes.
(2002)
(i) the order of the reaction with respect to A and with respect
to B.
(ii) the rate constant at 300 K.
(iii) the pre-exponential factor.
(1994, 5M)
66. The rate of first order reaction is 0.04 mol L–1s –1 at10 min and
74. The gas phase decomposition of dimethyl ether follows first
0.03 mol L–1s –1 at 20 min after initiation. Find the half-life of the
reaction.
(2001, 5M)
67. A hydrogenation reaction is carried out at 500 K .If the same
reaction is carried out in the presence of a catalyst at the same
rate, the temperature required is 400 K. Calculate the
activation energy of the reaction if the catalyst lowers the
activation barrier by 20 kJ mol −1 .
(2000, 3M)
order kinetics
CH3 — O— CH3 ( g ) → CH4 ( g ) + H2 ( g ) + CO ( g )
The reaction is carried out in a constant volume container at
500° C and has a half-life of 14.5 min. Initially only dimethyl
ether is present at a pressure of 0.40 atm. What is the total
pressure of the system after 12 min? Assume ideal gas
behaviour.
(1993, 4M)
75. A first order reaction, A → B, requires activation energy of
70 kJ mol −1 . When a 20% solution of A was kept at 25° C for
20 min, 25% decomposition took place. What will be the
percentage decomposition in the same time in a 30% solution
maintained at 40°C ? Assume that activation energy remains
constant in this range of temperature.
(1993, 4M)
65.
64
68. The rate constant for an isomerisation reaction, A → B is
4.5 × 10−3 min. If the initial concentration of A is 1 M,
calculate the rate of the reaction after 1 h.
(1999, 4M)
69. (i) The rate constant of a reaction is 1.5 × 107 s−1 at 50° C and
4.5 × 107 s−1 at 100° C. Evaluate the Arrhenius parameters
A and Ea .
(1998, 5M)
1
(ii) For the reaction, N2O5 (g ) → 2NO2 (g ) + O2 (g ),
2
calculate the mole fraction N2O5 (g ) decomposed at a
constant volume and temperature, if the initial pressure is
600 mm Hg and the pressure at any time is 960 mm Hg.
Assume ideal gas behaviour.
70. The rate constant for the first order decomposition of a certain
76. Two reactions (i) A → products (ii) B → products, follow
first order kinetics.The rate of the reaction (i) is doubled when
the temperature is raised from 300 K to 310 K. The half-life
for this reaction at 310 K is 30 min. At the same temperature B
decomposes twice as fast as A. If the energy of activation for
the reaction (ii) is half that of reaction (i), calculate the rate
constant of the reaction (ii) at 300 K.
(1992, 3M)
77. The nucleidic ratio,
− 18
reaction is described by the equation
1.25 × 104 K
log k (s −1 ) = 14.34 −
T
(i) What is the energy of activation for the reaction?
(ii) At what temperature will its half-life period be 256 min?
(1997, 5M)
3
1H
to
1
1H
in a sample of water is
: 1. Tritium undergoes decay with a half-life
8.0 × 10
period of 12.3 yr. How many tritium atoms would 10.0 g of
such a sample contain 40 yr after the original sample is
collected.
(1992, 4M)
78. The decomposition of N2 O5 according to the equation,
Sr 90 and its subsequent incorporation in bones. This nucleide
has a half-life of 28.1 yr. Suppose one microgram was
absorbed by a new-born child, how much Sr 90 will remain in
his bones after 20 yr.
(1995, 2M)
2N2 O5 ( g ) → 4NO2 ( g ) + O2 ( g )
is a first order reaction. After 30 min from the start of the
decomposition in a closed vessel, the total pressure developed
is found to be 284.5 mm of Hg. On complete decomposition,
the total pressure is 584.5 mm of Hg. Calculate the rate
constant of the reaction.
(1991, 6M)
72. At 380° C, the half-life period for the first order
79. In Arrhenius equation for a certain reaction, the value of A and
71. One of the hazards of nuclear explosion is the generation of
decomposition of H2 O2 is 360 min. The energy of activation
of the reaction is 200 kJ mol −1 . Calculate the time required for
75% decomposition at 450° C.
(1995, 4M)
73. From the following data for the reaction between A and B
[A], (mol/L)
2. 5 × 10−4
5.0 × 10−4
1. 0 × 10−3
Calculate
[B], (mol/L)
Initial rate (mol L–1s–1) at
300 K
320 K
3. 0 × 10−5
5. 0 × 10−4
2.0 × 10−3
6.0 × 10−5
4.0 × 10−3
—
6. 0 × 10−5
1. 6 × 10−2
—
Ea (activation energy) are 4 × 1013 s −1 and 98.6 kJ mol –1
respectively. If the reaction is of first order, at what
temperature will its half-life period be 10 min?
(1990, 3M)
80. An experiment requires minimum beta activity produced at
the rate of 346 beta particles per minute. The half-life period
of 42 Mo 99 , which is a beta emitter, is 66.6 h. Find the
minimum amount of 42 Mo 99 required to carry out the
experiment in 6.909 h.
(1989, 5M)
−6
81. A first order gas reaction has k = 1.5 × 10 per second at
200° C. If the reaction is allowed to run for 10 h, what
percentage of the initial concentration would have change in
the product? What is the half-life of this reaction? (1987, 5M)
Chemical Kinetics 185
82. While studying the decomposition of gaseous N2 O5 , it is
observed that a plot of logarithm of its partial pressure versus
time is linear. What kinetic parameters can be obtained from
this observation?
(1985, 2M)
85. Rate of reaction, A + B → products is given below as a
function of different initial concentrations of A and B
[A] mol/L
[B] (mol/L)
Initial rate
(mol L–1 min –1 )
0.01
0.01
0.005
0.02
0.01
0.010
0.01
0.02
0.005
83. Radioactive decay is a first order process. Radioactive carbon
in wood sample decays with a half-life of 5770 yr. What is the
rate constant (in yr − 1 ) for the decay? What fraction would
remain after 11540 yr?
(1984, 3M)
84. A first order reaction is 20% complete in 10 min. Calculate (i)
the specific rate constant of the reaction, and
(ii) the time taken for the reaction to go to 75% completion.
Determine the order of the reaction with respect to A and B.
What is the half-life of A in the reaction ?
(1982, 4M)
(1983, 2M)
Answers
1. (d)
2. (b)
3. (c)
4. (d)
49. (60)
50. (6.75)
51. (2.3)
52. (+8500J/ mol)
55. (a)
56. T = ∞
61. (9)
62. (0)
5. (a)
6. (d)
7. (d)
8. (b)
53. (c)
54. (b)
9. (a)
10. (c)
11. (d)
12. (b)
57. 0.0015
60. T
13. (d)
14. (c)
15. (b)
16. (c)
63. (950 mm Hg) 66. (25 min)
17. (b)
18. (a)
19. (a)
20. (d)
21. (a)
22. (a)
23. (b)
24. (d)
68. (3.26 × 10 −3 mol L −1 min −1)
(20.74 min)
25. (d)
26. (a)
27. (a)
28. (d)
74. (0.75 atm)
29. (a)
30. (d)
31. (d)
32. (a)
77. (5.6 × 10 )
33. (c)
34. (a)
35. (d)
36. (d)
37. (d)
38. (c)
39. (d)
40. (a)
79. (311.34 K)
(128.33 h)
41. (a)
42. (a,d)
43. (a,c)
44. (b,c,d)
45. (a,b,d)
46. (a,b,d)
47. (a,d)
48. (b,c)
5
82. (128.33 h)
71. (6.1 × 10 −7g)
−3
min
−1
−1
)
)
80. (3.56 × 10 −16 g)
83. (0.25)
72.
76. (3.26 × 10 −2 min
75. (67 %)
78. (5.2 × 10
67. (100 kJ mol −1)
81.
85. (1.386 min)
Hints & Solutions
Given, [N2O5 ]initial = 3.00 mol L− 1
1. We know that,
Rate = k (concentration)
n
[n = order of reaction and k = rate constant]
Taking log both side,
log (Rate) = log k + n log [concentration]
Slope of graph is the order of reaction greater the slope, greater is
the order of reaction.
∴Correct sequence for the order of reaction is D > B > A > C.
Hence, the correct option is (d).
2.
Key Idea The rate of a chemical reaction means the speed
with which the reaction takes place.
For
R → P
Rate of disappearance of R
Decrease in conc.of R
∆[ R]
=
=−
Time taken
∆t
Rate of appearance of P
Increase in conc. of P
∆[ P]
=
=+
Time taken
∆t
After 30 min, [N2O5 ] = 2 .75 mol L− 1
2N2O5 (g ) → 4NO2 (g ) + O2 (g )
t= 0
3.0 M
t = 30
2.75 M
From the equation, it can be concluded that
1 − ∆[N2O5 ] 1 ∆[NO2 ]
×
= ×
2
4
∆t
∆t
0.25
− ∆[N2O5 ] − (2.75 − 3.00) mol L− 1
⇒
=
30
∆t
30
∆[NO2 ]
0.25
∆[NO2 ]
∆ (N2O5 )
and
⇒
=− 2 ×
=−2
∆t
30
∆t
∆t
= − 1667
.
× 10− 2 mol L− 1 min − 1
=
3. In the given reaction; x A → y B
 − d[ A ]
 d[ B ]
= log10
+ 0.3010
 dt 
 dt 
Value of log 2 = 0.3010
log10
186 Chemical Kinetics
Substituting 0.3010 by log2
 d[ A ]
 d[ B ]
log10 −
= log10
+ log 2
 dt 
 dt 
Using logarithm rules,
1  d [ A ]  d[ B ]
 − d[ A ]
 d[ B ]
= 2×
⇒−
=
2  dt   dt 
 dt 
 dt 
…(i)
On comparing the above equation with equation of straight line,
y = mx + c
1
We get m = 5N 0, c = − 5N 0
e
N0
vs t is shown aside.
∴ Plot of
N
Using the rate equation (i) to determine the reaction involved is
2A → B
Option that fits correct in the above reaction is (c).
2C2H4 → C4H8 .
4.
N0
—
N
t(h)
Key Idea The Arrhenius equation for rate constants at two
different temperatures is
k
Ea T2 − T1 
[where, T2 > T1]
log 2 =
k 1 2.303R  T1T2 
where, k 1 and k 2 are rate constants at temperatures T1 and T2,
respectively.
R = Gas constant, Ea = Activation energy
For the reaction, H2 + I2 → 2HI
Given k 1 = 2.5 × 10−4 dm3mol−1s −1
T1 = (273 + 327) K = 600 K
k 2 = 1 dm 3mol −1 s −1 at T2 = (273 + 527) K = 800 K
Now, log
⇒ log
k2
Ea  T2 − T1 
=


k 1 2.303R  T1T2 
 800 − 600
Ea
1
=


2.5 × 10−4 2.303 × 8.314 × 10−3  600 × 800
E
(10 × 103 )
200
= a ×
2.5
0.019 48 × 104
~ E × 0.022
⇒ log 4 + 3 log10 −
⇒ log
a
⇒
2 × log 2 + 3
3.6 ~
Ea =
=
– 163.6kJ mol −1
0.022
0.022
5. The expression for bacterial growth is
∫eN
0
t
N −2dN = −5 ∫ dt
1
‘‘Activation enthalpy to form C is 15 kg mol −1 more than 5 kg
mol −1 that is required to form D.’’ It can be easily explained by
following graph.
20
15
Enthalpy
(kJ mol–1) 10
5
Activation
enthalpy
D
A+B
C
Reaction
coordinate
Activation enthalpy (or energy) is the extra energy required by
the reactant molecules that result into effective collision between
them to form the products.
7. In first order reaction, the rate expression depends on the
concentration of one species only having power equal to unity.
nR → products
− d[ R ]
= k [R ]
dt
On integration, − ln[ R ] = kt − ln[ R0 ]
or
N = N 0et
N0
= e− t
N
From 0 to 1 hour N ′ (t ) = N 0et
dN
From 1 hour onwards,
= −5N 2
dt
On differentiating the above equation from N ′ to N we get.
N
6. Only statement (d) is incorrect. Corrected statement is
ln(R ) = ln (R0 ) − kt
y = c + mx
m = slope = −k (negative)
c = intercept = ln (r0 )
The graph for first order reactions is
ln (R)
[QAt 1 hour, N ′ = eN 0]
1
1 
 N − eN  = 5(t − 1)

0
Multiply both sides by N 0, we get
N0 1
N
1
− = 5N 0 (t − 1) or, 0 = 5N 0 (t − 1) +
N e
N
e
N0
1

= 5N 0t + − 5N 0
 e

N
t
In zero order reaction,
[ R ] → product
− d[ R ]t
= k or − d[ R ]t = kdt
∴
dt
On integrating, −[ R ]t = kt + c
If
t = 0,[ R ]t = [ R ]0
Chemical Kinetics 187
∴
−[ R ]t = kt − [ R ]0
10. The temperature dependence of a chemical reaction is expressed
[ R ]t = [ R ]0 − kt
Thus, the graph plotted between [ r ]t and t gives a straight line
with negative slope (−k ) and intercept equal to [ R ]0.
The graph for zero order reaction is
[R]
t
8.
K1
K2
A → B → C
Rate of formation of B is
d[ B ]
= k 1[ A ] − k 2[ B ]
dt
⇒
⇒
d[ B ]


Q Given,
=0


dt
0 = k 1[ A ] − k 2[ B ]
k 2[ B ] = k 1[ A ]
⇒ Concentration of B, [ B ] =
k1
[A]
k2
9. Let the rate equation be k [ A ]x [ B ]y
From Ist values,
0.045 = k [ 0.05 ]x [ 0.05 ]y
…(i)
From 2nd values,
0.090 = k [ 010
. ]x [ 0.05 ]y
…(ii)
From 3rd values,
0.72 = k [ 0.20 ] [ 010
. ]
x
y
On dividing equations (i) by (ii), we get
0.045  0.05 
=
0.09  010
. 
x
1
x
 0.05 
 0.05 
=
 010
. 
. 
 010
∴
x =1
Similarly on dividing Eq. (ii) by (iii) we get
x
0.09  01
.   0.05 
=
0.72  0.2   010
. 
0.01 01
.  0.05 
=
0.08 0.2  01
. 
0.25 =
 0.05 
 010
. 
y
y
0.25 = [ 0.5 ]y
[ 0.5 ]2 = [ 0.5 ]y
∴
y=2
Hence, the rate law for the reaction
Rate = k [ A ][ B ]2
y
…(iii)
by Arrhenius equation,
k = Ae− E a$ / RT
…(i)
Taking natural logarithm on both sides, the Arrhenius equation
becomes,
E
ln k = ln A − a
RT
E
where, − a is the slope of the plot and ln A gives the intercept.
R
Eq. (i) at two different temperatures for a reaction becomes,
k
E 1
1
…(ii)
ln 2 = a  − 
k1
R  T1 T2 
⇒ In the given problem,
T1 = 400K, T2 = 500 K
k 1 = 10− 5 s − 1, k 2 = ?
E
− a (Slope) = − 4606
R
On substituting all the given values in Eq. (ii), we get
k
1 
 1
ln −2 5 = 4606 
−

 400 500
10
k
ln −2 5 = 2.303
10
k2
= 10 ⇒ k 2 = 10− 4 s − 1
10− 5
Therefore, rate constant for the reaction at
500 K is 10− 4s − 1.
11. Given, rate constant (k) = 0.05 µg/year
Thus, from the unit of k, it is clear that the reaction is zero order.
Now, we know that
a
half-life (t1/ 2 ) for zero order reaction = o
2k
where, ao = initial concentration,
k = rate constant
5 µg
t1/ 2 =
= 50 years
2 × 0.05 µg/year
Thus, 50 years are required for the decomposition of 5 µg of X
into 2.5 µg.
12. For zero order reaction,
[ A0 ] − [ At ] = kt
where, [ A0 ] = initial concentration
[ At ] = final concentration at time ‘t’
k = rate constant
[A ]
Also, for zero order reaction, t1/ 2 = 0
2k
Given, t1/ 2 = 6 h and [ A0 ] = 0.2 M
0.2
6=
∴
2k
0.2
1
or,
k=
=
2 × 6 60
Now, from Eq. (i)
[ A0 ] − [ At ] = kt
...(i)
188 Chemical Kinetics
Given, [ A0 ] = 0.5 M, [ At ] = 0.2 M
1
∴
0.5 − 0.2 =
×t
60
1
0.3 =
× t ⇒ t = 0.3 × 60 = 18h
60
a
1

Qk =
60 

13. The temperature dependence of rate of a chemical reaction is
expressed by Arrhenius equation as, k = Ae− E α / RT
…(i)
where, A = Arrhenius factor or frequency factor or
pre-exponential factor
R = Gas constant, Ea = Activation energy
Taking log on both sides of the
Eq. (i), the equation becomes
E
ln/k
ln k = ln A − a
RT
On comparing with equation of
straight line
1/RT
( y = mx + c), the nature of the
1
plot of lnk vs
will be:
RT
(i) Intercept = C = ln A
(ii) Slope/gradient = m = − Ea = − y ⇒ Ea = y
So, the energy required to activate the reactant, (activation
energy of the reaction, Ea is = y )
14. The elementary reaction, A2
k1
2A
c
k −1
follows opposing or reversible kinetics,
(i) Rate of the reaction,
r = rforward − rbackward = k 1[ A2 ] − k − 1[ A ]2
… (i)
(ii) Again, rate of the reaction can be expressed as,
r= −
1 d[ A ]
d [ A2 ]
=+
2 dt
dt
Expt 2
⇒
∴
⇒
⇒
⇒
r2  2 A   B 
=   
r1  A   B 
b
0.6
= 2a × 1 ⇒ 21 = 2 a ⇒ a = 1
0.3
From Eq. (i), 1 + b = 3 ⇒ b = 2
Order of the reaction (n) = a + b = 1 + 2 = 3
Order of the reaction wrt. A = 1
Order of the reaction wrt. B = 2
17. Let, the rate expression is r ∝ [ A ]a [ B ] b .
From experiment I,
r2  0.1
6.93 × 10− 3
 5
 0.25
=1 ×  
=
 ⇒
 ×
 4
 0.20
r1  0.1
6.93 × 10− 3
a
⇒
 5
1=  
 4
b
b
0
 5
 5
⇒   = 
 4
 4
b
⇒ b =0
a
From experiment II,
⇒
⇒
⇒ So,
r3  0.2
 0.30
=

 ×
 0.20
r1  0.1
b
1.386 × 10− 2
= (2)a × (1.5)0
0.693 × 10− 2
2 = 2a × 1 ⇒ 21 = 2a ⇒ a =1
r ∝ [ A ]1[ B ]0 ⇒ r ∝ [ A ]
Order of the reaction (n) = 1
⇒ Now, let for the 1st experiment,
r1 = k ⋅ [ A ]
r1 6.93 × 10− 3
⇒
k=
= 6.93 × 10− 2 s− 1
=
[A]
0.1
0.693
0.693
t50 =
=
= 10 s
⇒
k
6.93 × 10− 2
−∆H ° ∆S °
…(i)
+
RT
R
Mathematically, the equation of straight line is
…(ii)
y = c + mx
After comparing Eq. (ii) with (i) we get,
− ∆H °
∆S °
slope =
and intercept =
R
R
Now, we know for exothermic reaction ∆H is negative (−)ve.
But here,
−∆H °
is positive
Slope =
R
So, lines A and B in the graph represent temperature dependence
of equilibrium constant K for an exothermic reaction as shown
below
ln k =
15. The Arrhenius equation is,
k = A. e− E a / RT
where, k = rate constant,
A = Arrhenius constant, Ea = activation energy,
and
T = temperature in K
From the equation, it is clear that k decreases exponentially with
Ea . So, the plot-I is correct.
In the plot-II, k is plotted with temperature (in °C but not in K).
So, at 0°C, k ≠ 0 and k will increase exponentially with
temperature upto 300°C. Therefore, the plot-II is also correct.
16. For the reaction, 2A + B → products.
Let, the rate expression is
r ∝ [ A ]a [ B ]b
a
⇒
ln K
r2  2 A   2 B 
=   
r1  A   B 
2.4
= 2a × 2b ⇒ 23 = 2a + `b
0.3
b
⇒ 3=a+ b
b
18. From thermodynamics,
So, the rate of appearance of A, i.e.
d[ A ]
= 2r = 2k 1[ A2 ] − 2 k − 1 [ A ]2 [from Eq. (i)]
dt
Expt 1
…(ii)
A
B
(0, 0)
… (i)
1
T(K)
Chemical Kinetics 189
Let order of reaction with respect to CH3CHO is m.
Its given, r1 = 1 torr/sec. when CH3CHO is 5% reacted i.e. 95%
unreacted. Similarly, r2 = 0.5 torr/sec when CH3CHO is 33%
reacted i.e., 67% unreacted.
Use the formula, r ∝ (a − x )m
Alternative Method
In fifty minutes, the concentration of H2O2 decreases from 0.5 to
0.125 M or in one half-life, concentration of H2O2 decreases
from 0.5 to 0.25 M. In two half-lives, concentration of H2O2
decreases from 0.5 to 0.125 M or 2 t1/ 2 = 50 min
t1/ 2 = 25 min
 0.693
−1
k =
∴
 min
 25 
where (a − x ) = amount unreacted
or
19. For the reaction,
Decomposes
CH3CHO( g ) → CH4 + CO
r1  a − x1 
=
r2  a − x2 
r1 (a − x1 )
or
=
r2 (a − x2 )m
m
so,
m
d [O2 ]
1 d [H2O2 ] k [H2O2 ]
=−
=
= 6.93 × 10−4 mol min −1
2
dt
2
dt
22. The main conditions for the occurrence of a reaction is proper
orientation and effective collision of the reactants.
Now putting the given values
Since, the chances of simultaneous collision with proper
orientation between two species in high order reactions are very
rare, so reaction with order greater than 3 are rare.
m
1  0.95
=
. )m or m = 2
 ⇒ 2 = (141
0.5  0.67
23. For the elementary reaction, M → N
20. According to Arrhenius equation
Rate law can be written as
Rate ∝ [ M ] n
k = Ae− E a / RT
where, A = collision number or pre-exponential factor.
Rate = k [ M ] n
R = gas constant, T = absolute temperature
Ea = energy of activation
− E a1 / RT
For reaction R1, k 1 = Ae
When we double the concentration of [ M ],
rate becomes 8 times, hence new rate law can be written as
…(ii)
8 × Rate = k [ 2M ] n
…(i)
− E a2 / RT
For reaction R2, k 2 = Ae
…(ii)
Rate
k [M ]n
⇒
=
8 × Rate k [ 2M ] n
On dividing Eq. (ii) by Eq. (i), we get
−
k2
=e
k1
(E a
2
− Ea )
1
RT
⇒
…(iii)
[Q Pre-exponential factor ‘A’ is same for both reactions]
∴
10,000 J mol − 1
k2
=4
=
k 1 8.314 J mol − 1K− 1 × 300 K
21. For first order reaction, k =
2.303
a
log
t
a−x
M
t = 50 min, a = 0.5 M, a − x = 0125
.
2.303
0.5
−1
∴
k=
= 0.0277 min
log
50
0125
.
Now, as per reaction
As shown above, rate of reaction remains constant as the
concentration of reactant (B) changes from 0.1 M to 0.2 M and
becomes double when concentration of A change from 0.1 to 0.2,
(i.e. doubled).
Given,
2H2O2 → 2H2O + O2
1 d [H2O2 ] 1 d [H2O] d [O2 ]
−
=
=
2
dt
2 dt
dt
d [H2O2 ]
Rate of reaction, −
= k [H2O2 ]
dt
d [O2 ]
1 d [H2O2 ] 1
…(i)
∴
=−
= k [H2O2 ]
dt
dt
2
2
When the concentration of H2O2 reaches 0.05 M,
d [O2 ] 1
[from Eq. (i)]
= × 0.0277 × 0.05
dt
2
d [O2 ]
or
= 6.93 × 10−4 mol min −1
dt
[2]n = 8 = [2]3 ⇒ n = 3
w.r.t. each reactant and then writing rate law equation of the
given equation accordingly as
dC
R=
= k [ A ]x [ B ] y
dt
where, x = order of reaction w.r.t A
y = order of reaction w.r.t B
1. 2 × 10−3 = k (0 .1)x (0 .1) y
1. 2 × 10−3 = k (01
. )x (0.2) y
−3
2.4 × 10 = k (0.2)x (01
. )y
1
0
R = k [ A] [ B]
E a 1 = E a 2 + 10 kJ mol − 1 = Ea 2 + 10,000 J mol − 1
ln
1
1
=
8 [2]n
24. This problem can be solved by determining the order of reaction
Taking ln on both the sides of Eq. (iii), we get
 k  E a1− E a 2
ln  2  =
 k1
RT
Given,
…(i)
25.
PLAN Time of 75% reaction is twice the time taken for 50% reaction if it
is first order reaction w.r.t. P. From graph, [Q ] decreases linearly
with time, thus it is zeroth order reaction w.r.t. Q
dx
= bk[ P ]a[Q ]b
dt
Order w.r.t
P = a=1
Order w.r.t
Q =b=0
Thus, overall order of the reaction = 1 + 0 = 1
26. From Arrhenius equation, log
Given,
k2
− Ea  1
1
=
 − 
k1 2.303 R  T2 T1 
k2
= 2 T2 = 310 K
k1
T1 = 300 K
190 Chemical Kinetics
⇒ For
On putting values,
⇒
1 
− Ea
 1
log 2 =
−


2.303 × 8.314  310 300
⇒
Ea = 53.603 kJ/mol
35. Rate will be directly proportional to both concentration and
27. According to Arrehnius equation, rate constant increases
exponentially with temperature :
k = Ae− E a / RT
Ea
2.303 RT
2000
Given :
log k = 6 −
T
Comparing the above two equations :
log A = 6 ⇒ A = 106
Ea
and
= 2000
2.303 R
log k = log A −
Ea = 2000 × 2.303 × 8.314 J= 38.3 kJ mol
38. A catalyst increases the rate of reaction but by the same factor to
both forward and backward directions. Hence, a catalyst shorten
the time required to reach the equilibrium.
−1
…(i)
…(ii)
30. Rate ∝ [G ] m [ H ] n
Q Rate is double on doubling the concentration of G and
maintaining H constant, m = 1, i.e. R ∝ [G ].
Also, when both concentration of G and H are doubled, rate
increases by a factor of 8. Here rate is increasing by a factor of 2
due to G (first order in G), therefore, factor due to H is 4.
⇒ R ∝ [ H ]2 ⇒ Overall order = m + n = 1 + 2 = 3
31. Order of a reaction can take any real value, i.e. negative, integer,
fraction etc.
32. For first order reaction,
2.303
a
2.303
0.1
log
=
= 3.46 × 10−2
log
t
a−x
40
0.025
Rate = [ k ] A = 3.46 × 10−2 × 0.01 = 3.46 × 10−4
⇒
4
n
ln 2
= 40 s
k1
[ A ]0
For zero order reaction t1/ 2 =
= 20 s
2k 0
1 [ A ]0
k
= =
× 1
⇒ Eq. (ii)/(i)
2 2k 0
ln 2
k 1 ln 2 0.693
⇒
=
= 0.5
=
k 0 [ A ]0 1.386
33. For a first order reaction, kt = ln
36. The unit of rate constant (t − 1) indicating that the decomposition
1
 1
 1
g
=   × initial amount =   × 1.0 g =
 2
 2
16
29. For first order reaction t1/ 2 =
k=
intensity, i.e. rate of formation of AB * ∝ C ⋅ I .
reaction following first order kinetics.
Rate = k[ N2O5 ]
⇒
Rate 2.40 × 10−5
= 0. 8 M
=
[ N2O5 ] =
k
3 × 10−5
560
37. 560 days =
= 4 half-lives.
140
Amount of reactant remaining after n-half-lives
28. The logarithmic form of Arrhenius equation is
⇒
N2 + 3H2 → 2NH3
d [ N2 ]
1 d [ H2 ] 1 d [ NH3 ]
Rate = −
=−
=
dt
3 dt
2
dt
[ A ]0
[A]
1 [ A ]0
1
800 4 ln 2 − 1
ln
s
k = ln
=
=
4
t
[ A ] 2 × 10
50 2 × 104
= 1.386 × 10− 4 s− 1
34. For any general reaction,
aA + bB → cC + dD
1 d [A]
1 d [ B ] 1 d [C ] 1 d [ D ]
Rate = −
=−
=
=
a dt
b dt
c dt
d dt
39. Specific rate constant of reaction depends on temperature.
40. The rate constant (k ) of all chemical reactions depends on
temperature.
k = Ae− E a / RT
where, A = pre-exponential factor, Ea = activation energy.
SN 1
41. (CH3 )3 C  Br + NaOH →
( First order reaction)
(CH3 )3C OH+ NaBr
0.693
This is first order reaction and for first order reaction t1/ 2 =
k
So, half-life is independent of initial concentration.
Therefore, the plot (a) correct.
For first order reaction, (r) = k [ (CH3 )3 C  Br ];
 P
P 
ln  0  = k x t or ln   = − k x t
 P
 P0 
Hence, plot (b) and (d) are incorrect.
For first order reaction,
[Q ]
= (1 − ekt )
Q = [ P0 ](1 − ekt ) or
[ P0 ]
Hence, plot (c) is incorrect.
42. Given for the reaction (at T= 300 K and constant volume = V)
at t = 0
at t = t
at t = t1/ 3
A ( g ) → 2B ( g ) + C ( g )
p0
p0 − x
 p − 2 p0  = p0
 0
3  3
–
2x
4 p0
3
–
x
2 p0
3
We can calculate,
pt = p0 − x + 2x + x = p0 + 2x
p − p0
or
2x = pt − p0 or x = t
2
Now for first order reaction,
1
p0
t = ln
k
( p0 − x )
Chemical Kinetics 191
Putting the value of x in the equation,
1
p0
1
2 p0
t = ln
= ln
k
2 p 0 − pt + p 0
 p − p0  k
p0 −  t


2 
or kt = ln
2 p0
or kt = ln 2 p0 − ln ( 3 p0 − pt )
( 3 p 0 − pt )
ln ( 3 p0 − pt ) = − kt + ln 2 p0
or
It indicates graph between ln (3p0 − pt ) vs ‘t’ will be a straight
line with negative slope , so option (a) is correct
t1/3 =
1
p0
1
= ln 3
ln
k
p0 / 3 k
43. If steric factor is considered, the corrected Arrhenius equation
will be
k=
where A = frequency factor by Arrhenius.
Q p > 1, pA > A hence, (a) is correct.
Activation energy is not related to steric factor.
44. Rate constant,
k = Ae− E a / RT
where, Ea = activation energy and A = pre-exponential factor
(a) If Ea is high, it means lower value of k hence, slow reaction.
Thus, incorrect.
(b ) On increasing temperature, molecules are raised to higher
energy (greater than Ea ), hence number of collisions
increases. Thus, correct.
E
d (log k )
E
(c) log k = log A − a ⇒
= a2
RT
dT
RT
Thus, when Ea is high, stronger is the temperature
dependence of the rate constant. Thus, correct.
(d) Pre-exponential factor (A) is a measure of rate at which
collisions occur. Thus, correct.
concentration of reactant remaining
after time t is given by [A]
[ A ] = [ A ] 0 e– kt
Therefore, concentration of reactant
t
decreases exponentially with time.
(b) Rise in temperature increases rate constant (k) and therefore
decreases half-life (t1/ 2) as
ln 2
t 1/ 2 =
k
(c) Half-life of first order reaction is independent of initial
concentration.
(d) For a first order reaction, if 100 moles of reactant is taken
initially, after n half-lives, reactant remaining is given by
⇒
n
i.e. log [ X ] vs ‘t’ will give a straight line.
Also at constant temperature, pV = constant
1
⇒ Plot of p vs will give a straight line.
V
kt = ln
8
 1
= 100   = 0.3906
 2
A reacted = 100 – 0.3906 = 99.6%
1
1−α
where, α = degree of dissociation.
1 − α = e− kt ⇒ α = 1 − e− kt
⇒
Also
ekt
1
, i.e. plot of reciprocal of concentration of
=
[ A ] [ A ]0
reactant vs time will be exponential.
1
100
2 ln 2
Time for 75% = ln
=
= 2 (t1/ 2 )
k
k
100 − 75
The Arrhenius equation is :
Ea
RT
The dimensions of k and A must be same. For first order reaction,
dimensions of k is t − 1.
ln k = ln A −
48. A catalyst lowers the activation energy by enabling the reaction
to continue through an alternative path, i.e. catalyst changes the
reaction mechanism. However, catalyst does not affect either
average kinetic energies of reactants or the collision frequency.
49. Given, first order reaction,
t75 = 90 minute
t60 = ?
Consider the following reaction,
A
→ B
(Reactant)
Time,
t15% = 90 min
45. (a) For a first order reaction, the
 1
% A = 100  
 2
∆H
2.3 RT
⇒ Plot of log K p vs 1/ T will be a straight line.
For the first order reaction X → P
[ X ]0
kt
kt
,
log
=
⇒ log [ X ] = log [ X 0 ] −
2.3
[ X ] 2.3
log K p = constant −
47. For a first order reaction :
It indicates t1/ 3 is independent of initial concentration so, option
(b) is incorrect.
Likewise, rate constant also does not show its dependence over
initial concentration. Thus, graph between rate constant and [ A ]0
will be a straight line parallel to X-axis.
−Ea
pAe RT
46. Equilibrium constant is related to temperature
t 60% = ?
(Product)
t = 0 100
(100 − 75) = 25
(100 − 60) = 40
For first order reaction,
2 . 303  a0  a0 = Initial concentration 
t=
log 
 a   a = Concentration
k

For 75%,
t75% =
2 . 303  100
Initial conc.= 100
log
 …(i) Conc. = 100 − 75 = 25
 25 


k
For 60%,
t60% =
2 . 303  100
Initial conc.= 100
log
 …(ii) Conc. = 100 − 60 = 40
 40 


k
On equating Eqs. (i) and (ii),
2 . 303  100
log

 40 
t60%
k
=
t75% 2 . 303  100
log

 25 
k
192 Chemical Kinetics
 0.3979
t60% = t75% × 

 0.602 
t60%
t60°
52. For the reaction,
0.3979
= 90 ×
0.602
= 59.48 minute ≈ 60 min
50. Rate = k [ A ]x[ B ]y[C ]z
- AB(g )
Af
A f = 4 Ab or
Ea b − Ea f = 2RT
or
Ab
=4
− E a f / RT
k f = Af e
⇒ y=0
6 × 10−5
(Rate)1 [ 0.2 ]x [ 01
. ]y [ 01
. ]z
=
=
z
x
y
(Rate)3 [ 0.2 ] [ 01
. × 10−4
. ] [ 0.2 ] 12
⇒ z =1
(Rate)1 [ 0.2 ]x [ 01
. ]y [ 01
. ]z 6 × 10−5
=
=
(Rate)4 [ 0.3 ]x [ 01
. ]y [ 01
. ]2 9 × 10−5
Likewise, rate constant for backward reaction,
− E ab / RT
k b = Ab e
At equilibrium,
Rate of forward reaction = Rate of backward reaction
kf
i.e.,
k f = k b or
= k eq
kb
− E a f / RT
⇒ x =1
So, rate = k [ A ]1[C ]1
k eq =
so
Af e
− E ab / RT
Ab e
=
Af
Ab
− (E a f − E ab ) / RT
e
After putting the given values
From exp-Ist,
Rate = 6.0 × 10−5 mol dm −3 s −1
k eq = 4 e2 (as Ea b − Ea f = 2RT and
6.0 × 10−5 = k [ 0.2 ]1[ 01
. ]1
Now,
k = 3 × 10−3
Af
Ab
= 4)
∆G ° = − RT ln K eq = − 2500 ln(4 e2 )
= − 2500 (ln 4 + ln e2 ) = − 2500 (1.4 + 2)
[ A ] = 015
. mol dm −3
= − 2500 × 3.4 = − 8500J/mol
Absolute value = 8500 J/mol
[ B ] = 0.25 mol dm −3
[C ] = 015
. mol dm −3
53. Living plants maintain an equilibrium between the absorption of
∴Rate = (3 × 10−3 ) × [ 015
. ]1[ 0.25 ]0[ 015
. ]1
= 3 × 10−3 × 015
. × 015
.
−5
Rate = 6.75 × 10 mol dm −3 s −1
Thus, Y = 6.75
51. At constant V , T
∆
2N 2O 5 (g ) → 2N 2O 4 (g ) + O 2 (g )
At initial t = 0 1
t = Y × 103 sec 1 − 2 p
A (g ) + B (g )
Ea b = Ea f + 2RT
Now, rate constant for forward reaction,
(Rate)1 [ 0.2 ]x [ 01
. ]y [ 01
. ]z 6 × 10−5
=
=
x
y
(Rate)2 [ 0.2 ] [ 0.2 ] [ 01
. ]z 6 × 10−5
Given,
Given
Further
0
0
2p
p
pTotal = 1 − 2 p + 2 p + p
1. 4 = 1 + p
p = 0.45 atm
According to first order reaction,
2.303
pi
log
k=
t
pi − 2 p
pi = 1atm (given)
2 p = 2 × 0.45 = 0.9 atm
On substituting the values in above equation,
1
2k ⋅ t = 2.303 log
1 − 0.9
1
−4
3
2 × 5 × 10 × y × 10 = 2.303 log
01
.
y = 2.303 = 2.3
Note Unit of rate constant (k ), i.e. s−1 represents that it is a first
order reaction.
C14 (produced due to cosmic radiation) and the rate of decay of
C14 present inside the plant. This gives a constant amount of C14
per gram of carbon in a living plant.
54. Fossil whose age is closest to half-life of C-14 (5770 yr) will
yield the most accurate age by C-14 dating.
N
55. λT = ln 0
N
where N 0 = Number of C14 in the living matter and
N = Number of C14 in fossil. Due to nuclear explosion,
amount of C14 in the near by area increases. This will increase
N 0 because living plants are still taking C-14 from atmosphere,
during photosynthesis, but N will not change because fossil will
not be doing photosynthesis.
⇒ T (age) determined in the area where nuclear explosion has
occurred will be greater than the same determined in normal
area.
C
C
C
1
Also, λT1 = ln 1 ⇒ λT2 = ln 2 ⇒ T1 − T2 = = ln 1
C
C
C2
λ
C = Concentration of C-14 in fossil.
56. k = A e− Ea / RT : At T = ∞ , k = A
57. −
1 d [ H2 ] 1 d [ NH3 ]
=
3 dt
2
dt
⇒
− d [ H2 ] 3 d [ NH3 ] 3
= × 0.001 = 0.0015 kg h − 1 .
=
dt
2
dt
2
58. acidic, first or basic, second.
59. Rate is directly proportional to concentration of reactants.
Chemical Kinetics 193
60. R ∝ [Reactant]
100T 100
=
× 12.8 = 33.68 h
38
38
T2 = 2T1 = 67.36 h
38T1 38 × 33.68
T3 =
=
= 29.76 h
43
43
⇒
T1 =
On doubling the concentration of reactant, rate would be double.
[ A ]0
61. For a first order process kt = ln
[A]
where, [ A ]0 = initial concentration.
[ A ] = concentration of reactant remaining at time “t”.
[ A ]0
…(i)
kt1/ 8 = ln
= ln 8
⇒
[ A ]0 / 8
[ A ]0
and
…(ii)
kt1/ 10 = ln
= ln 10
[ A ]0 /10
t1/ 8
ln 8
Therefore,
=
= log 8 = 3 log 2 = 3 × 0.3 = 0.9
t1/ 10 ln 10
t1/ 8
× 10 = 0.9 × 10 = 9
⇒
t1/ 10
66.
R = k [ A ] ⇒ R1 = k [ A ]1 and R2 = k [ A ]2
R1 4 [ A ]1
⇒
= =
R2 3 [ A ]2
4
ln 2
[ A ]1
4
Also k (t2 − t1 ) = ln
× 10 = ln
= ln ⇒
t1/ 2
3
[ A ]2
3
10 log 3
3
⇒
t1/ 2 =
=
= 25 min
log 4 − log 3 0.6 − 0.48
67. k 500 = A e− E1 / RT1
62. Rate of reaction is constant with time.
Q
63. (a) Partial pressure becomes half of initial in every 100 min,
⇒
therefore, order = 1.
800
(b) k × 100 = ln
= ln 2 ⇒ k = 6.93 × 10−3 min − 1
400
(c) For 75% reaction; time required = 2 × half-life = 200 min
(d) 2 X (g ) → 3Y (g ) + 2 Z (g )
800 − x
(3 / 2 ) x
a
(Rate)1 0.05 1  1
=
= =   ⇒ a = 1; order w.r.t A.
(Rate)2 0.10 2  2
64
38%
k1
19%
64
→
29 Cu
k2
29 Cu
64
+
0
+
0
+ 1β +
0
− 1e
Rate = k [ A ] = 4.5 × 10− 3 × 0.76
⇒
= 3.42 × 10−3 mol L−1 min −1
k 2 Ea  T2 − T1 
=


k1
R  T1T2 
Ea
RT
22 × 1000
7
At 50°C : ln A = ln (1.5 × 10 ) −
= 8.33
8.314 × 323
ln k = ln A −
Also
− 1β
→
⇒
 4.5 × 107 

Ea 
50
⇒ ln 
 =
 ⇒ Ea = 22 kJ

 1.5 × 107  8.314  323 × 373
(a) Rate = k [ A ]
Rate 0.05
(b) k =
=
= 0.5 s− 1
[ A ] 0.10
29 Cu
⇒
43%
→
k3
30Zn
E2 T2 400 4
=
=
=
E1 T1 500 5
E1 = E2 + 20000 J
E1 − 20,000 4
=
⇒ E1 = 100,000 J = 100 kJ mol −1
E1
5
[ A ]0
kt = ln
[A]
1
4.5 × 10− 3 × 60 = ln
⇒ [ A ] = 0.76 M
[A]
69. (i) ln
Order w.r.t B = 0
65.
Also
68.
3
x
2
Also 800 − x = 700 ⇒ x = 100
3
⇒ Total pressure = 800 + × 100 = 950 mm Hg
2
64.
k 500 = k 400
E1
E
= 2 ⇒
RT1 RT2
x
Total pressure = 800 +
k 400 = A e− E 2 / RT2
64
⇒
64
28 Ni
A = 4.15 × 103 s− 1
(ii) N2O5 (g ) → 2NO2 (g ) +
64
28 Ni
600 − p
64
Above are the parallel reactions occurring from Cu .
k 1 38
T
k 1 38 T3
=
= 2 = 2 and
=
=
k 2 19
T1
k 3 43 T1
T1 , T2 and T3 are the corresponding partial half-lives.
Also
k = k1 + k2 + k3
ln 2 ln 2 ln 2 ln 3
⇒
=
+
+
T
T1
T2
T3
1
1
43
1
1
1
1
=
+
+
=
+
+
⇒
T T1 T2 T3 T1 2T1 38T1
100
1 
1 43 1  38 + 19 + 43
=

 =
1 + +  =
 38T1
38
T1 
2 38 T1 
2p
1
O2 (g )
2
p/ 2
3
Total pressure = 960 = 600 + p ⇒ p = 240 mm
2
⇒ Partial pressure of N2O5 (g ) remaining = 600 – 240
= 360 mm
360
= 0.375
⇒ Mole fraction =
960
70. (i) The Arrhenius equation is
log k = log A −
Ea
2.303 RT
Comparing with the given equation :
Ea
1.25 × 104 =
⇒ Ea = 239.33 kJ mol −1
2.303 R
194 Chemical Kinetics
(ii) When half-life = 256 min,
ln 2
0.693 − 1
k=
=
s = 4.5 × 10− 5 s− 1
t1/ 2 256 × 60
74. CH3  O  CH3 (g ) → CH4 (g ) + H2 (g ) + CO (g )
At 12 min : 0.40 − p
1.25 × 104
= 14.34 – log 4.5 × 10− 5 = 16.68
T
1.25 × 104
⇒
T =
= 669 K
16.68
⇒
71. k t = ln
k∝
k
ln 
k

⇒ ln 
 t1/ 2
⇒
(450° C) 
 = ln
(380° C) 
1
t1/ 2
 t1/ 2 (380° C)  Ea  450 − 380


=

 t1/ 2 (450° C)  R  727 × 653
 200 × 103
360
70
×
= 3.54
=
(450° C) 
8.314
727 × 653
⇒
t1/ 2 (450°C) =10.37 min
⇒ Time for 75% reaction at 450°C
= 2 × t1/ 2 = 2 × 10.37 = 20.74 min
k
kB
(ii) B → Product
⇒
m = 2, order w.r.t. A
Now comparing the data of experiment number 1 and 2 :
2
n
8 = (2)2 (2)n ⇒ n = 1, order w.r.t. B.
⇒
(i) Order with respect to A = 2, order with respect to B = 1.
(ii) At 300 K, R = k [ A ]2 [ B ]
k=
⇒
15  70 × 1000
15
 k (40° C)  Ea 
×
= 1.35

=
=
8.314
298 × 313
 k (25° C)  R  298 × 313
k (40° C)
⇒
= 3.87
k (25° C)
1
100 1
4
Also k (25°C) =
=
ln
ln
20
75 20
3
⇒
k (40° C) = 3.87 × k (25°C)
1
4
= 3.87 ×
ln = 55.66 × 10− 3 min − 1
20
3
100
Now
k (40°C) × 20 = ln
100 − x
100
55.66 × 10− 3 × 20 = ln
⇒
⇒ x = 67%
100 − x
75. ln 
m
R2 4 × 10−3  5 × 10− 4   6.0 × 10− 5 
=
=
 

R1 5 × 10− 4  2.5 × 10− 4   3.0 × 10− 5 
R
5.0 × 10− 4
=
2
[ A ] [ B ] (2.5 × 10− 4 )2 (3.0 × 10− 5 )
= 2.66 × 108 s− 1 L2 mol −2
(iii) From first experiment :
Rate (320 K) = k (320 K) (2.5 × 10− 4 )2 (3.0 × 10− 5 )
2 × 10−3
⇒ k (320 K) =
(2.5 × 10− 4 )2 (3.0 × 10− 5 )
= 9300 R ln 2 = 53.6 kJ
E (i )
= a
= 26.8 kJ
⇒ Ea (ii)
2
At 310 K
t1/ 2 (i) = 30 min
Q Rate of (ii)
= 2 rate of (i)
⇒
t1/ 2 (ii) = 15 min
Now for reaction (ii) :
 t1/ 2 (300)  Ea (ii ) 
 k (310) 

10
ln  B


=
 = ln 
R  300 × 310
 t1/ 2 (310) 
 k B (300) 
⇒ Ea (i)
⇒
⇒
⇒
⇒
 1.066 × 109 

Ea 
20
ln 
 =


 2.66 × 108  8.314  300 × 320
⇒
At 300 K : ln (2.66 × 108 ) = ln A −
Solving :
55.42 × 103
8.314 × 300
ln A = 41.62 ⇒ A = 1.2 × 1018
10
20
× 2 × 6 × 1023 =
× 1023
8
3
N (1 H1 ) 20 × 1023
1+
=
N (1 H3 ) 3N (1 H3 )
N (1 H3 + 1H1 ) =
 k (320 K)  Ea  T2 − T1 
ln 


=
 k (300 K)  R  T1T2 
Now
 t (300)  ln 2
⇒ t1/ 2 (300) = 21.2 min
ln  1/ 2
=
2
 15 
ln 2 0.693
k B (300) =
=
= 3.26 × 10− 2 min − 1
t1/ 2
21.2
77. Initially :
⇒
Ea = 55.42 kJ mol −1
E
ln k = ln A − a
RT

10
Ea 

 = ln 2
R  300 × 310
For (i)
= 1.066 × 109 s− 1 L2 mol − 2.
⇒
p
A
Product
76. (i) A →
73. Comparing the data of experiment number 2 and 3 :
R3 1.6 × 10− 2  1.0 × 10− 3 
=
=

R2
4 × 10− 3
 5 × 10− 4 
p
⇒ Total pressure = 0.4 + 2p = 0.4 + 2 × 0.175 = 0.75 atm
w0
ln 2
10− 6 g
⇒
⇒ w = 6.1 × 10− 7 g
× 20 = ln
w
28.1
w
72. For 1st order reaction :
p
Total pressure = 0.4 + 2p
0.40
ln 2
Also k × 12 = ln
=
× 12 = 1.77 ⇒ p = 0.175
0.40 − p 14.5
1+
1
20 × 1023
=
≈ 1.25 × 1017
− 18
8 × 10
3N (1 H3 )
⇒
N (1 H 3 ) =
⇒
kt = ln
⇒
20 × 1023
= 5.33 × 106
3 × 1.25 × 1017
N0
5.33 × 106
ln 2
× 40 = ln
⇒
12.3
N
N
N = 5.6 × 105
Chemical Kinetics 195
78. For the reaction :
2N2O5 → 4NO2 + O2
If p0 is the initial pressure, the total pressure after completion of
5
reaction would be p0.
2
5
584.5 = p0 ⇒ p0 = 233.8 mm
⇒
2
Let the pressure of N2O5 decreases by ‘p’ amount after 30 min.
Therefore,
2N2O5 → 4NO2 + O2
p0 − p
At 30 min :
Total pressure = p0 +
2p
3
p = 284.5
2
p
2
2
(284.5 − 233.8) = 33.8
3
p0
kt = ln
p0 − p
1
233.8
k=
ln
min − 1 = 5.2 × 10− 3 min − 1
30
233.8 − 33.8
⇒
p=
Now,
⇒
79. Arrhenius equation is :
Ea
2.303 RT
ln 2
0.693
when t1/ 2 = 10 min, k =
=
= 1.115 × 10− 3 s− 1
t1/ 2 10 × 60
Ea
A
4 × 1013
= log A − log k = log = log
⇒
2.303 RT
k
1.115 × 10− 3
log k = log A −
= 16.54
Ea
98.6 × 1000
=
= 311.34 K
⇒T =
2.303 R × 16.54 2.303 × 16.54 × 8.314
80. The minimum rate of decay required after 6.909 h is
346 particles min − 1.
Rate = kN
⇒
Rate 346 × 66.6 × 60
⇒ N =
=
= 1.995 × 106 atoms
k
0.693
N
ln 2
N
kt = ln 0 ⇒
⇒
× 6.909 = ln 0 = 0.0715
N
66.6
N
N0
= 1.074
⇒
N
⇒
N 0 = 1.074 × N = 1.074 × 1.995 × 106
= 2.14 × 106atoms of Mo
⇒ Mass of Mo required =
2.14 × 106
× 99 = 3.56 × 10− 16 g
6.023 × 1023
81. k = 1.5 × 10− 6 s− 1
kt = ln
⇒
ln
100
100 − x
100
= 1.5 × 10− 6 s− 1 × 10 × 60 × 60 s = 0.0054
100 − x
⇒
100
= 1.055
100 − x
⇒ x = 5.25% reactant is converted into product.
0.693
ln 2
Half-life =
=
= 462000 s = 128.33 h
k
1.5 × 10− 6
[ A ]0
82. For a first order process : ln
= kt ⇒ ln [ A ] = ln [ A ]0 − kt
[A]
If the reactant is in gaseous state
[ A ]0
and kt1/ 10 − ln
= ln 10
[ A ]0 / 10
t
ln 8
Therefore, 1/ 8 =
= log 8 = 3 log 2 = 3 × 0.3 = 0.9
t1/ 10 ln 10
t1/ 8
⇒
× 10 = 0.9 × 10 = 9
t1/ 10
…(ii)
…(i)
ln p = ln p0 − kt
where p is the partial pressure of reactant remaining unreacted at
instant ‘t’ and p0 is its initial partial pressure.
Also, from equation (i), ln p vs t would give a straight line.
Therefore, decomposition of N2O5 following first order kinetics.
ln 2 0.693 − 1
83. k =
=
yr = 1.2 × 10− 4 yr − 1
t1/ 2 5770
1
ln 2
1
Also kt = ln =
× 11540 = ln 4 ⇒ f = = 0.25
f 5770
4
84. For a first order reaction,
kt = ln
[ A ]0
[A]
where [ A ]0 = Initial concentration of reactant
[ A ] = Concentration of reactant remaining
unreacted at time t.
1 [ A ]0
1
100
1
5
(i) ⇒ k = ln
=
=
ln
ln
t
[ A ] 10 100 − 20 10
4
2.303 (log 5 − 2 log 2)
=
min − 1 = 0.023 min –1
10
1 100 2 ln 2 2 × 0.693
(ii)
t = ln
=
=
= 60 min
k
k
25
0.023
85. Looking at the rate data of experiment number 1 and 2 indicates
that rate is doubled on doubling concentration of A while
concentration of B is constant. Therefore, order with respect to A
is 1. Similarly, comparing data of experiment number 1 and 3,
doubling concentration of B, while concentration of A is
constant, has no effect on rate.
Therefore, order with respect to B is zero.
Rate = k [ A ]
⇒
0.005
0.693
k=
= 0.5 min − 1 =
⇒
t1/ 2
0.010
0.693
⇒
t1/ 2 =
= 1.386 min
0.5
12
Nuclear Chemistry
Objective Questions I (Only one correct option)
1. Bombardment of aluminium by α-particle leads to its artificial
disintegration in two ways, (i) and (ii) as shown. Products X , Y
(2011)
and Z respectively, are
27
13Al
30
15P
(ii)
Objective Questions II
(One or more than one correct option)
9. In the decay sequence.
238
92 U
30
14Si
+X
2. A positron is emitted from
(b) x2 is β −
23
11 Na.
(c) x1 will deflect towards negatively charged plate
(d) x3 is γ-ray
10. A plot of the number of neutrons (n) against the number of
protons (p) of stable nuclei exhibits upward deviation from
linearity for atomic number, Z > 20. For an unstable nucleus
having n/p ratio less than 1, the possible mode(s) of decay is
(2016 Adv.)
(are)
(b) 22/11
(d) 23/12
23
Na is the more stable isotope of Na. Find out the process by
which 24
11 Na can undergo radioactive decay.
(2003, 1M)
−
(a) β -emission
(b) α-emission
(c) β + -emission
(d) K-electron capture
4. The number of neutrons accompanying the formation of
139
54 Xe
235
92 U,
and 94
38 Sr from the absorption of a slow neutron by
followed by nuclear fission is
(1999, 2M)
27
13 Al
(b) 2
is a stable isotope.
(c) 1
29
13 Al
(a) α-emission
(c) positron emission
(d) 3
(1996, 1M)
(1984, 1M)
(a) Ge77
32
(b) As77
33
(c) Se77
34
(d) Se78
34
8. If uranium (mass number 238 and atomic number 92) emits an
α-particle, the product has mass number and atomic number
(1981, 1 M)
(b) 234 and 90
(d) 236 and 90
(d) β + -decay (positron emission)
9
4 Be
+ X → 84 Be + Y
(2013 Adv.)
(b) ( p, D)
(d) (γ , p)
(c) (n, D)
12. Decrease in atomic number is observed during
(a) alpha emission
(c) positron emission
(1998, 2M)
(b) beta emission
(d) electron capture
13. The nuclear reactions accompanied with emission of neutron(s)
are
(b) definitely beta rays
(d) either alpha rays or beta rays
7. An isotope of Ge76
32 is
(c) neutron emission
X and Y are
as seen after deflection by a magnet in one direction,(1984,
are 1M)
(a) definitely alpha rays
(c) both alpha and beta rays
(b) orbital or K-electron capture
(a) (γ , n)
is expected to decay by
(b) β-emission
(d) proton emission
(a) β − - decay (β - emission)
11. In the nuclear transmutation,
6. The radiation from a naturally occurring radioactive substance,
(a) 236 and 92
(c) 238 and 90
230
90 Th
(a) Z is an isotope of uranium
+Z
(b) neutron, positron, proton
(d) positron, proton, neutron
(a) 22/10
(c) 23/10
5.
− x3
234
→
91 Pa  
− x4
234
Z 
→
The ratio of the atomic mass
and atomic number of the resulting nuclide is
(2007, 3M)
(a) 0
− x2

→
x1 , x2 , x3 and x4 are particles/radiation emitted by the
respective isotopes. The correct option(s) is(are) (2019 Adv.)
(a) proton, neutron, positron
(c) proton, positron, neutron
3.
234
90 Th
+Y
(i)
30
14Si
− x1

→
(1988, 1 M)
(a)
27
13 Al
(c)
30
15 P
+
4
2He
→
→
30
14Si
+
30
15P
0
1e
(b)
12
6 C
(d)
241
96 Cm
+
1
1H
+
→
4
2He
13
7N
→
244
97Bk
+ 01e
Numerical Answer Type Questions
14.
238
92 U
is known to undergo radioactive decay to form 206
82 Pb
by emitting alpha and beta particles. A rock initially
contained 68 × 10−6 g of 238
92 U. If the number of alpha
particles that it would emit during its radioactive decay of
18
238
206
92 U to 82 Pb in three half-lives is Z × 10 , then what is the
value of Z?
(2020 Adv.)
Nuclear Chemistry 197
15. During the nuclear explosion, one of the products is 90Sr with
half-life of 6.93 years. If 1 µg of 90Sr was absorbed in the bones
of a newly born baby in place of Ca, how much time, in years, is
required to reduce it by 90% if it is not lost metabolically
……… .
(2020 Main, 7 Jan I)
21. Elements of the same mass number but different atomic number
are known as ………
(1983, 1M)
22. An element Z M A undergoes an α-emission followed by two
successive β -emissions. The element formed is ……
(1982, 1M)
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
Integer Answer Type Questions
23. The periodic table consists of 18 groups. An isotope of copper,
on bombardment with protons, undergoes a nuclear reaction
yielding element X as shown below. To which group, element
(2012)
X belongs in the periodic table?
63
29 Cu
controlled nuclear fission to
reaction
26.
→ 2
92 X
234
is
(2010)
92 U
238
→
82Pb
214
is ..... .
(2009)
− 7α
→ Y. Find out atomic number, mass number of Y
−6β
and identify it.
27.
(2004)
238
is radioactive and it emits α and β particles to form
92 U
206
. Calculate the number of α and β particles emitted in this
conversion.
82 Pb
1
18. (a) 235
92 U + 0n →
(b)
undergoes
25. The total number of α and β particles emitted in the nuclear
Fill in the Blanks
82
34 Se
90
142
54 Xe and 38 Sr
Subjective Questions
30
13 Al
is less stable than 40
20 Ca.
Statement II Nuclides having odd number of protons and
neutrons are generally unstable.
(1998)
17. Statement I Nuclide
235
92 U
24. The number of neutrons emitted when
16. Statement I The plot of atomic number ( y-axis) versus number
of neutrons (x-axis) for stable nuclei shows a curvature towards
x-axis from the line of 45° slope as the atomic number is
increased.
Statement II Proton-proton electrostatic repulsions begin to
overcome attractive forces involving protons and neutrons and
neutrons in heavier nuclides.
(2008)
+ 11 H → 610 n + 42α + 211 H + X
0
− 1e
137
52 A
+
97
40 B
+ ......
+ ......
An ore of
238
is found to contain
92 U
238
and
weight ratio of 1 : 0.1. The half-life period of 92 U
yr. Calculate the age of the ore.
(2005, 1M × 2 = 2M)
19. A radioactive nucleus decays by emitting one alpha and two
82 Pb
238
20. The number of neutrons in the parent nucleus which gives N
14
(1985, 1M)
is 4.5 × 109
(2000)
α-particle.
29.
disintegrates to give 82 Pb206 as the final product. How
many alpha and beta particles are emitted during this process?
90 Th
234
(1986, 2M)
Answers
2. (c)
3. (a)
4. (b)
5. (b)
8. (b)
6. (c)
7. (a)
10. (b, d)
11. (a, b)
12. (a, c, d)
13. (a, d)
14. (1.2)
15. (23.03)
16. (a)
17. (b)
18. 2 0n1,
9. (a, b, c)
36 Kr
82
in the
14
(1989, 1M)
1. (a)
206
28. Write a balanced equation for the reaction of N with
beta particles, the daughter nucleus is.... of the parent.
on beta emission is ...........
92 U
19. isotope
20. eight
21. isobars
24. 3
25. (8)
27. (7.12 × 10 yr)
8
12.
26.
ZM
A−4
84 Po
29. (13)
206
23. (8)
Hints & Solutions
1. (i)
27
13 Al
+ 42He (α) →
30
14Si
(a) By β − - decay, 10 n → 11 p + −01e neutron changes to
proton. Thus, (n/ p) ratio further decreases below 1.
Thus, this decay is not allowed.
+ 11X
X is proton 11H.
(ii)
27
13 Al
+ 42He (α ) →
Y is neutron,
Z is positron ,
1
0 n.
30
15 P
0
+1e.
30
15P
→
+ 10Y
(b) By orbital or K- electron capture, 11 p +
30
14Si
+ 10Z
23
→
0
+1e
+
10Na
proton changes to neutron.
Hence, n/ p ratio increases. Thus correct.
23
3. In stable isotope of Na, there are 11 protons and 12 neutrons. In
the given radioactive isotope of sodium (Na 24 ), there are 13
neutrons, one neutron more than that required for stability. A
neutron rich isotope always decay by β-emission as
→
−1β
0
120
+ 1H1
4. The balanced nuclear reaction is
20 n1 +
5.
92U
235
→
139
54Xe
+
38 Sr
94
29
is neutron rich isotope, will decay by β-emission
converting some of its neutron into proton as
1
0
1
0 n → −1β + 1H
13 Al
Neutron-rich
nuclei
100
Number of neutrons
1
0n
60
1.2
40
1.1 ratio
Neutron poor nuclei
20
0
20
40
60
Number of protons
7. Isotopes have same atomic number (Z ) but different mass
76
32 Ge
and
77
32 Ge
are isotopes.
9.
238
→ 2He4 (α ) +
90Th
234
Key Idea The lose of one α-particle will decrease the mass
number by 4 and atomic number by 2. On the other hand, loss
of β-particle will increase the atomic number by 1.
11.
PLAN 94 Be + ba X → 84 Be + dc Y
Atomic number same
4 + a=4 + c
9+ b=8+ d
In decay sequence,
92U
238
234
90Th
+ 2He4 (or a)
If X = 00 γ
a=0
b=0
Y = 10 n
c=0
d =1
If X = 11 p
a=1
b=1
Y = 12 D
c=1
d=2
X1 particle
234
+
92U
234
Z
is isotope
of uranium
(b– or –1e0)
91Pa
234
+ (b– or –1e0)
X2 particle
X3 particle
230
90Th
+ 2He4 (or a)
X4 particle
9
4 Be
9
4 Be
X 1 particle will deflect towards negatively charged plate due to
presence of positive charge on α- particles.
Hence, options (a, b, c) are correct.
10. For the elements with atomic number (Z ) larger than 20,
Neutrons (n) > Protons (p); Thus, n/ p > 1
Thus, there is upward deviation from linearity.
If n < p, Thus n/ p < 1, then
80
Plot of the number of neutrons against the number of protons in
stable nuclei (shown by dots).
8. The nuclear reaction is
92 U
1.5
1.4
80
6. Both α-rays and β - rays are deflected by magnetic field.
number ( A ). Therefore,
→ 10n
proton changes to neutron, hence, (n/ p) ratio increases.
Thus stability increases. Thus correct.
(c) Neutron emission further decreases n/p ratio.
(d) By β + -emission, 11 p → 10n + +01e
2. The required nuclear reaction is
11 Na
0
−1e
+ 00 γ → 84 Be + 10n
+ 11 p → 84 Be + 21D
12. In the following nuclear reactions, there occur decrease in
atomic number (Z )
A
→ 2He4 +
ZX
ZX
A
+
ZX
0
−1e
A
0
+1e
A
Z − 1Y ,
→
→
+
A− 4
, α - emission
Z − 2Y
A
Z − 1Y , positron emission
electron capture
Nuclear Chemistry 199
n
17
(13 Al 30 ) =
= 1.3 > 1, unstable, β-emitters.
p
13
n
20
( Ca 40 ) =
= 1, stable.
p 20
20
In beta emission, increase in atomic number is observed.
ZX
A
→
0
−1e
+
Z + 1Y
, β -emission
A
13. If sum of mass number of product nuclides is less than the sum
of parent nuclides, then neutron emission will occur. In both (a)
and (d), sum of mass number of product nuclides is one unit less
than the sum of parent nuclides, neutron emission will balance
the mass number.
14.
238
92 U
→
206
82 Pb
238
92 U present
−6
68 × 10
=
238
After three half-lifes, moles of
=
= 23.03 (yr)
+ 842 He + 6 0− 1β
Number of moles of
68 × 10
238
−6
Also, nuclei with both neutrons and protons odd are usually
unstable but it does not explain the assertion appropriately.
18. (a)
initially
(b)
1
+ 0n
82
34 Se
→
52 A
→ 2 −1e0 +
137
36Kr
+
40 B
97
+ 2 0n1
82
19. Isotope : Z X A → 2He4 + 2 −1e0 + ZY A − 4
238
92 U decayed
−6
1  68 × 10

× 1 − 3  =
 2 
238
20. 8 : 6 C14 → 7 N14 +
×
7
8
22.
ZM
A−4
: Z M A → 2He4 + 2 −1e0 + ZM A − 4
23. Balancing the given nuclear reaction in terms of atomic number
(charge) and mass number:
63
29 Cu
Thus, the correct answer is 1.2.
+ 1 H1 → 60 n1 + 2 He4 (α ) + 21 H1 + 26 X 52
The atomic number 26 corresponds to transition metal Fe which
belongs to 8th group of modern periodic table.
15. Radioactive decay follows first order kinetics.
∴Time taken for decay from N 0 to N t is
(N = number of nuclei)
1 N
t = ln 0
λ Nt
N
1
t = × 2.303 log 0
Nt
λ
0.693
Also, we know λ = (decay constant) =
t1/ 2
0
−1e
21. Isobars have same mass number but different atomic number.
Therefore, number of α-particles emitted
68 × 10− 6 7
=
× × 8 × 6.023 × 1023
238
8
= 1204
.
× 1018 ≈ 12
. × 1018
24.
92 U
235
→
142
54Xe
+
25.
92 U
238
→
214
82Pb
+ 6 2He4 + 2
38 Sr
90
+ 30 n1
0
−1e
⇒ Number of (α + β ) = 6 + 2 = 8
26.
…(i)
27.
Also, we know 90% nuclei are decayed
N 0 100
∴
=
= 10
Nt
10
N
Put the values of λ and 0 in Eq. (i), we get
Nt
6.93
∴
t=
× 2.303 × log10
0.693
16. After atomic number 20, proton-proton repulsion increases
immensely, more neutrons are required to shield this
electrostatic repulsion, curve of stability incline towards
neutron axis.
17. Upto atomic number of 20, stable nuclei possess neutron to
92 X
234
→ 7 2He4 + 6 −1e0 +
Y is
84 Po
92 U
238
84Y
206
206
.
→
206
82Pb
+ 8 2He4 + 6−1 e0
Present : N 0 − N
N
w (U)
1
Given,
=
= 10
w (Pb) 0.1
N (U)
10
206 N 0 − N
=
×
=
⇒
N (Pb) 238
N
1
N
N0
238 2298
238
⇒
=1+
=
=
⇒
N 0 − N 2060
N0 − N
2060 2060
where, t1/ 2 = 6.93 yr (given)
proton ratio (n/ p) = 1.
92 U
235
Now, applying first order rate law
 ln 2
 N0 
 N0 
(t )

 t = ln 
 ⇒ t = 1/ 2 log 

log 2
 t1/ 2 
N0 − N 
N0 − N 
=
14
28.
7N
29.
90 Th
4.5 × 109
2298
log
= 7.12 × 108 yr
0.3
2060
+ 2He4 →
234
→
82Pb
9F
206
18
+ 7 2He4 + 6 −1e0
13
Surface Chemistry
Objective Questions I (Only one correct option)
1. Simplified absorption spectra of three complexes [(i), (ii) and
(iii)] of M n+ ion are provided below; their λ max values are
marked as A, B and C respectively. The correct match
between the complexes and their λ max values is
(2020 Main, 2 Sep II)
(b) brownian motion in colloidal solution is faster if the viscosity
of the solution is very high.
(c) addition of alum to water makes it unfit for drinking.
(d) colloidal particles in lyophobic sols can be precipitated by
electrophoresis.
5. A gas undergoes physical adsorption on a surface and follows
the given Freundlich adsorption isotherm equation
Adsorption of the gas increases with
(a)
(b)
(c)
(d)
Absorption
C
A
B
x
= Kp 0. 5
m
(2019 Main, 10 April I)
increase in p and increase in T
increase in p and decrease in T
decrease in p and decrease in T
decrease in p and increase in T
6. Match the catalysts Column I with products Column II.
(2019 Main, 9 April I)
λmax
Column I (Catalyst)
λmax
λmax
Wavelength (nm)
(i) [ M (NCS)6 ](− 6 + n)
(ii) [ M F6 ](− 6 + n)
(iii) [ M (NH3 )6 ]n +
(a) A-(iii), B-(i), C-(ii)
(c) A-(ii), B-(iii), C-(i)
(b) A-(ii), B-(i), C-(iii)
(d) A-(i), B-(ii), C-(iii)
2. Among the following, the incorrect statement about colloids
is
(2019 Main, 12 April II)
(a) They can scatter light
(b) They are larger than small molecules and have high molar mass
(c) The osmotic pressure of a colloidal solution is of higher order
than the true solution at the same concentration
(d) The range of diameters of colloidal particles is between 1 and
1000 nm
3. Peptisation is a
(a)
(b)
(c)
(d)
(2019 Main, 12 April I)
process of bringing colloidal molecule into solution
process of converting precipitate into colloidal solution
process of converting a colloidal solution into precipitate
process of converting soluble particles to form colloidal solution
4. The correct option among the following is
(2019 Main, 10 April II)
(a) colloidal medicines are more effective, because they have small
surface area.
Column II (Product)
(A)
V2O5
(i)
Polyethlyene
(B)
TiCl 4 / Al(Me)3
(ii)
Ethanal
(C)
PbCl 2
(iii)
H2SO4
(D)
Iron oxide
(iv)
NH3
(a)
(b)
(c)
(d)
(A)-(ii), (B)-(iii), (C)-(i), (D)-(iv)
(A)-(iv), (B)-(iii), (C)-(ii), (D)-(i)
(A)-(iii), (B)-(i), (C)-(ii), (D)-(iv)
(A)-(iii), (B)-(iv), (C)-(i), (D)-(ii)
7. The number of water molecule(s) not coordinated to copper
ion directly in CuSO4 ⋅ 5H 2O, is
(a) 2
(c) 1
(2019 Main, 9 April I)
(b) 3
(d) 4
8. The aerosol is a kind of colloid in which
(2019 Main, 9 April I)
(a) gas is dispersed in liquid
(b) gas is dispersed in solid
(c) liquid is dispersed in water
(d) solid is dispersed in gas
9. Adsorption of a gas follows Freundlich adsorption isotherm.
x is the mass of the gas adsorbed on mass m of the adsorbent.
x
x
The plot of log versus log p is shown in the given graph.
m
m
is proportional to
(2019 Main, 8 April I)
Surface Chemistry 201
17. Which of the salt-solution is most effective for coagulation of
arsenious sulphide?
x
log —
m
(a) BaCl 2
(c) Na 3PO4
2
3
18. Adsorption of a gas follows Freundlich adsorption isotherm.
log p
(a) p2 / 3
(b) p3/ 2
(2019 Main, 9 Jan II)
(b) AlCl 3
(d) NaCl
(c) p3
(d) p2
In the given plot,x is the mass of the gas adsorbed on mass m
x
of the adsorbent at pressure p ⋅ is proportional to
m
(2019 Main, 9 Jan I)
10. Among the following, the false statement is
(2019 Main, 12 Jan II)
(a) Tyndall effect can be used to distinguish between a colloidal
solution and a true solution
(b) It is possible to cause artificial rain by throwing electrified sand
carrying charge opposite to the one on clouds from an aeroplane
(c) Lyophilic sol can be coagulated by adding an electrolyte
(d) Latex is a colloidal solution of rubber particles which are
positively charged
11. Given, Gas :
H2 , CH4 , CO2 , SO2
Critical temperature/K 33 190
304 630
On the basis of data given above, predict which of the
following gases shows least adsorption on a definite amount
of charcoal?
(2019 Main, 12 Jan I)
(a) CH4
(b) SO2
(c) CO2
(d) H2
12. Among the colloids cheese (C), milk (M) and smoke (S), the
correct combination of the dispersed phase and dispersion
medium, respectively is
(2019 Main, 11 Jan II)
(a)
(b)
(c)
(d)
C : liquid in solid; M : liquid in liquid; S : solid in gas
C : solid in liquid; M : liquid in liquid; S : gas in solid
C : liquid in solid; M : liquid in solid; S : solid in gas
C : solid in liquid; M : solid in liquid; S : solid in gas
13. An example of solid sol is
(a) gem stones
(c) butter
(2019 Main, 11 Jan I)
(b) hair cream
(d) paint
14. Haemoglobin and gold sol are examples of
(2019 Main, 10 Jan II)
(a)
(b)
(c)
(d)
negatively and positively charged sols, respectively
negatively charged sols
positively charged sols
positively and negatively charged sols, respectively
15. Which of the following is not an example of heterogeneous
catalytic reaction?
(a)
(b)
(c)
(d)
(2019 Main, 10 Jan I)
Haber’s process
Combustion of coal
Hydrogenation of vegetable oils
Ostwald’s process
16. The correct match between item-I and Item-II is
(2019 Main, 9 Jan II)
A. Benzaldehyde
B. Alumina
C. Acetonitrile
P. Dynamic phase
Q. Adsorbent
R. Adsorbate
(a) (A) → (R) ; (B) → (Q); (C) → (P)
(b) (A) → (P); (B) → (R); (C) → (Q)
(c) (A) → (Q); (B) → (P); (C) → (R)
(d) (A) → (Q); (B) → (R); (C) → (P)
x
log m
2 unit
4 unit
log p
(a) p2
(c) p
1/ 2
(b) p1/ 4
(d) p
19. The Tyndall effect is observed only when following
conditions are satisfied
(2017 Main)
1. The diameter of the dispersed particles is much smaller
than the wavelength of the light used.
2. The diameter of the dispersed particle is not much smaller
than the wavelength of the light used.
3. The refractive indices of the dispersed phase and
dispersion medium are almost similar in magnitude.
4. The refractive indices of the dispersed phase and
dispersion medium differ greatly in magnitude.
(a) 1 and 4
(c) 1 and 3
(b) 2 and 4
(d) 2 and 3
20. For a linear plot of log ( x / m ) versus log p in a Freundlich
adsorption isotherm, which of the following statements is
correct? (k and n are constants)
(2016 Main)
(a) 1/n appears as the intercept
(b) Only 1/n appears as the slope
1
(c) log   appears as the intercept
 n
(d) Both k and 1/n appear in the slope term
21. Methylene blue, from its aqueous solution, is adsorbed on
activated charcoal at 25°C. For this process, the correct
statement is
(2013 Adv.)
(a)
(b)
(c)
(d)
the adsorption requires activation at 25°C
the adsorption is accompanied by a decreases in enthalpy
the adsorption increases with increase of temperature
the adsorption is irreversible
22. The coagulating power of electrolytes having ions Na + , Al 3+
and Ba 2+ for arsenic sulphide sol increases in the order
(2013 Main)
(a)
(b)
(c)
(d)
Al 3+ < Ba 2+ < Na +
Na + < Ba 2+ < Al 3+
Ba 2+ < Na 2+ < Al 3+
Al 3+ < Na + < Ba 2+
202 Surface Chemistry
the electrolytes Na 2 SO4 , CaCl 2 , Al 2 (SO4 )3 and
NH4 Cl, the most effective coagulating agent for Sb 2 S3 sol is
31. The given graph/data I, II, III and IV represent general
trends observed for different physisorption and
chemisorption processes under mild conditions of
temperature and pressure. Which of the following
choice(s) about I, II, III and IV is (are) correct?
(2012)
(2009, 1M)
(b) CaCl 2
(d) NH4Cl
24. Among the following, the surfactant that will form micelles in
(a) CH3 (CH2 )15 N+ (CH3 )3 Br −
(I)
(b) CH3 (CH2 )11 OSO3− Na +
(c) CH3 (CH2 )6 COO− Na +
(d) CH3 (CH2 )11 N+ (CH3 )3 Br −
(II)
(III)
T
200 K
250 K
(IV)
0
p
26. Spontaneous adsorption of a gas on solid surface is an
exothermic process, because
Amount of gas
absorbed
(2005, 1M)
(a) irreversible sols
(b) prepared from inorganic compounds
(c) coagulated by adding electrolytes
(d) self-stabilising
(2004, 1M)
(a) ∆H increases for system
(b) ∆S increases for gas
(c) ∆S decreases for gas
(d) ∆G increases for gas
(a)
(b)
(c)
(d)
Eads
Distance of molecule
from the surface
Dhads = 150 kJ mol–1
I is physisorption and II is chemisorption
I is physisorption and III is chemisorption
IV is chemisorption and II is chemisorption
IV is chemisorption and III is chemisorption
32. Choose the correct reason(s) for the stability of the
27. Rate of physisorption increases with
(2003, 1M)
(b) increase in temperature
(d) decrease in surface area
28. When the temperature is increased, surface tension of water
(2002, 1M)
(a) increases
(c) remains constant
p constant
T
25. Lyophilic sols are
(a) decrease in temperature
(c) decrease in pressure
Amount of gas
absorbed
p constant
aqueous solution at the lowest molar concentration at ambient
conditions, is
(2008, 3M)
Potential energy
(a) Na 2SO4
(c) Al 2 (SO4 )3
Amount of gas
absorbed
23. Among
(b) decreases
(d) shows irregular behaviour
lyophobic colloidal particles.
(2012)
(a) Preferential adsorption of ions on their surface from the
solution
(b) Preferential adsorption of solvent on their surface from the
solution
(c) Attraction between different particles having opposite
charges on their surface
(d) Potential difference between the fixed layer and the
diffused layer of opposite charges around the colloidal
particles
Objective Questions II
(One or more than one correct option)
29. The correct statement(s) about surface properties is(are)
(2017 Adv.)
(a) The critical temperatures of ethane and nitrogen are 563 K and
126 K, respectively. The adsorption of ethane will be more than
that of nitrogen of same amount of activated charcoal at a given
temperature
33. The correct statement(s) pertaining to the adsorption of a
gas on a solid surface is (are)
(2011)
(a) Adsorption is always exothermic
(b) Physisorption may transform into chemisorption at high
temperature
(b) Cloud is an emulsion type of colloid in which liquid is dispersed
phase and gas is dispersion medium
(c) Physisorption increases with increasing temperature but
chemisorption decreases with increasing temperature
(c) Adsorption is accompanied by decrease in enthalpy and decrease
in entropy of the system
(d) Chemisorption is more exothermic than physisorption,
however it is very slow due to higher energy of activation
(d) Brownian motion of colloidal particles does not depend on the
size of the particles but depends on viscosity of the solution
30. When O2 is adsorbed on a metallic surface, electron transfer
occurs from the metal to O 2 . The true statement(s) regarding
this adsorption is (are)
(2015 Adv.)
(a) O2 is physisorbed
(b) heat is released
(c) occupancy of π ∗ 2 p of O2 is increased
(d) bond length of O2 is increased
Numerical Answer Type Questions
34 The mass of gas adsorbed, x per unit mass of adsorbate, m
was measured at various pressures, p. A graph between
x
log and log p gives a straight line with slope equal to 2
m
x
and the intercept equal to 0.4771. The value of
at a
m
pressure of 4 atm is (Given, log 3 = 0.4771)
(2020 Main, 2 Sep I)
Surface Chemistry 203
(c) Statement I is true; Statement II is false.
(d) Statement I is false; Statement II is true.
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 a
correct explanation of Statement I.
(b) Statement I is true; Statement II is true; Statement II is
not the correct explanation of Statement I.
35. Statement I Micelles are formed by surfactant molecules
above the critical micelle concentration (CMC).
Statement II The conductivity of a solution having surfactant
molecules decreases sharply at the CMC.
(2007)
Answers
1.
5.
9.
13.
17.
(a)
(b)
(a)
(a)
(b)
2.
6.
10.
14.
18.
(c)
(c)
(d)
(d)
(c)
3.
7.
11.
15.
19.
(b)
(c)
(d)
(b)
(b)
4.
8.
12.
16.
20.
(d)
(d)
(a)
(a)
(b)
21.
25.
29.
33.
(b)
(d)
(a, c)
(a, b, d)
22.
26.
30.
34.
(b)
(c)
(b, c, d)
(48)
23.
27.
31.
35.
(c)
(a)
(a, c)
(b)
24. (a)
28. (b)
32. (a, d)
Hints & Solutions
1. Here, same metal ion, M n+ form three homoleptic octahedral
complexes (i), (ii) and (iii) on separate combination with
È
È
three mono-dentate ligands–N CS,F and NH3 respectively.
So, we have to compare their CFSE (∆ 0 ) as well as
wavelength (λ) values, where
hc
1
or, ∆ 0 ∝
∆E(CFSE) =
λ
λ
Again, ∆ 0 value will depend on power of ligand as placed in
spectrochemical series.
1
Power of ligand ∝ ∆ 0 ∝ ,
λ
−
−
F − < NCS− < NH3, F >NC S>NH3
→ Power of Ligand ← λ
→ ∆ 0
So, for the given complexes, it is evidented from the plot
(Absorption vs λ)
ii )
i)
iii )
> λ (max
> λ (max
λ (max
(C )
[F − ]
(B )
[NCS− ]
(A )
[NH 3 ]
2. Statement (c) is incorrect about colloids. Colligative
properties such as relative lowering of vapour pressure,
elevation in boiling point, depression in freezing point and
osmotic pressure of a colloidal solution is of low order than
the true solution at the same concentration.
3. Peptisation is a process of converting precipitate into
colloidal solution. This process involves the shaking of
precipitate with the dispersion medium in the presence of
small amount of electrolyte. The electrolyte added is called
peptising agent.
During peptisation, the precipitate adsorbs one of the ions of
the electrolyte on its surface. This causes the development of
positive or negative charge on precipitates, which ultimately
breakup into smaller particles of the size of a colloid.
4. The explanation of the given statements are as follows :
(a) Colloidal medicines are more effective because they (dispersed
phase) have larger surface area.
Thus, option (a) is incorrect.
(b) Brownian motion of dispersed phase particles in colloidal
solution is faster if the viscosity of the solution is very low.
Thus, option (b) is incorrect.
(c) Addition of alum(K 2SO4 ⋅ Al2(SO4 )3 ⋅ 24H2O), an electrolyte
to water makes it fit for drinking purposes because alum
coagulates mud particles from water.
Thus, option (c) is incorrect.
(d) Precipitation of lyophobic solution particles by electrophoresis
is called cottrell precipitation.
Thus, option (d) is correct.
5. For physisorption or physical adsorption,
Adsorption isotherm (Temperature, T = constant) is shown below:
x
—
m
x
Moderate pressure zone, — ∝ p1/n
m
p
where, x = amount of adsorbate, m = amount of adsorbent,
x
= degree of adsorption
m
1
1
= order of the reaction, where, 0 < < 1 and so,
1< n < ∞
n
n
x
Here,
= Kp1/ 2,
m
x
i.e.
∝ p1/ 2
m
204 Surface Chemistry
Adsorption isobar (Pressure, p = constant)
The logarithm equation of Freundlich adsorption isotherm is
x
1
log = log K + log p
m
n
On comparing the above equation with straight line equation,
( y = mx + c)
x
—
m
we get
m =slope =
T
From the given plot,
So, the rate of physical adsorption of the gas, increases with p
(when,T is constant) and decreases withT (when pis constant).
m=
6. Correct match is
(A) → (iii); (B) → (i); (C) → (ii); (D) → (iv)
(i) TiCl4 + AlCl3 (Ziegler- Natta catalyst) is used to prepare
polyethylene from ethene.
CH2
nCH2
Zieglar-Natta catalyst
CH2
CH2
∴
V 2O 5
It is the key step in the manufacture of H2SO4.
(iii) Fe (Iron) is used as catalyst in Haber’s process for the
manufacture of ammonia.
The closest synthetic latex that can be associated with the
properties of natural latex is SBR, i.e. Styro Butane Rubber.
Rest of all the statements are correct.
11. Same adsorbant (charcoal in this case) at same temperature will
adsorb different gases to different extent. The extent to which
gases are adsorbed is proportional to the critical temperature of
gas.
8a
Q
Tc =
27Rb
Fe( s )
N2 (g ) + 3H2 (g ) → 2NH3 (g )
(iv) Pd (Palladium) is used to prepare ethanal. Reaction involved is
PdCl 2/CuCl 2
H 2C == CH 2 + O2 → CH 3CHO
H 2O
where, a is the magnitude of intermolecular forces between
gaseous molecules.
This reaction is also known as Wacker’s process.
7. In CuSO4 ⋅ 5H2O, one molecule of water is indirectly connected
Thus, higher the cirtical temperature more is the gas adsorbed.
Among the given gases, H2 has the minimum critical
temperature, i.e. 33K thus, it shows least adsorption on a definite
amount of charcoal.
to Cu. In this molecule, four water molecules form coordinate
bond with Cu 2+ ion while one water molecule is associated with
H-bond with SO2−
4 .
Structure of CuSO4 ⋅ 5H2 O
H
H
12.
H
δ–
Cu2+
δ+
O
H
δ–
δ+
O
O
H
H
H
H
O–
H
O–
O
S
O
O
[Cu(H 2O)4 ] SO4 ⋅ H 2O
8. The aerosol is a kind of colloid in which solid is dispersed in gas.
e.g. smoke, dust.
9.
y2 − y1 1 2
= =
x2 − x1 n 3
x
= Kp2 / 3
m
These microparticles belong to rubber and are negatively
charged in nature. Natural latex contains some amount of sugar,
resin, protein and ash as well.
2SO2 ( g ) + O2 ( g ) → 2SO3 ( g )
O
c = log K
dispersion, i.e. emulsion of polymer microparticles in an
aqueous medium.
n
(ii) V2O5 (Vanadium pentoxide) is used as catalyst to prepare
H2SO4 from contact process. Reaction involved is
H
and
10. Statement given as statement (d) is incorrect. Latex is a stable
Polyethylene
Ethene
1
n
Key Idea According to Freundlich,
x
= Kp1/n [ n > 1]
m
where, m = mass of adsorbent, x = mass of the gas
x
adsorbed, = amount of gas adsorbed per unit mass of
m
solid adsorbent, p = pressure, K and n = constants.
Dispersed
phase
Liquid
Dispersion
medium
Solid
Type of
colloid
Gel
Liquid
Solid
Liquid
Gas
Emulsion
Aerosol
Examples
Cheese (C), butter,
jellies
Milk (M), hair cream
Smoke (S), dust
Thus, C : liquid in solid, M : liquid in liquid and S : solid in gas.
13. Solid sol consists of solid as both dispersed phase and dispersion
medium. In gemstones, metal crystals (salt and oxides of metals)
are dispersed in solid (stone) medium. Hair cream is an emulsion
(liquid in liquid). Butter is a colloidal solution of liquid in solid.
Paint is also sol (solid in liquid).
14. Haemoglobin and gold sol both are colloids and always carry an
electric charge. Haemoglobin is a positively charged sol,
because in haemoglobin, Fe2+ ion is the central metal ion of the
octahedral complex.
All metal sols like, Au-sol, Ag-sol etc; are negatively charged
sols.
Surface Chemistry 205
15. In heterogeneous catalytic reactions, physical state of reactants
and that of catalyst(s) used are different.
Haber’s process, hydrogenation of vegetable oils and Ostwald’s
process all are heterogeneous process. Combustion of coal is not
a heterogeneous catalytic reaction.
l
In Haber’s process
(i) The diameter of the colloids should not be much smaller than
the wavelength of light used.
(ii) The refractive indices of the dispersed phase and dispersion
medium should differ greatly in magnitude.
20. According to Freundlich adsorption isotherm,
x
= kp1/ n
m
Fe( s), Mo( s)
N 2 ( g ) + 3H 2( g ) → 2NH 3 ( g )
Hydrogenation of vegetable oils,
[(Ph P) Rh] Cl
3 3
→ Vanaspati (s)
Vegetable oil (l ) 
or Ni ( s)
(Unsaturated)
l
log p
Pt( s)
V 2 O5 (s)
No catalyst is used in combustion of coal. The reaction is
highly spontaneous in nature.
C + O2 → CO2
(Coal)
16. Using the principle of adsorption chromatography, qualitative
and quantitative analysis of benzaldehyde can be done from its
mixture with acetonitrile. Here, a mobile phase moves over a
stationary phase (adsorbent). Adsorbents used are alumina
(Al 2O 3 ) and silica gel. The sample solution of benzaldehyde and
acetonitrile when comes in contact with the adsorbent,
benzaldehyde gets adsorbed on the surface of the adsorbent. So,
benzaldehyde acts as absorbate whereas acetonitrile starts
moving as mobile phase over the stationary phase of the
adsorbate. Hence, act as dynamic phase.
2−
17. Arsenious sulphide sol is a negative colloid, As2S3.(S ). So, it
will be coagulated by the cation of an electrolyte.
According to the Hardy-Schulze rule, the higher the charge of the
ion, the more effective it is in bringing about coagulation. Here,
the cations available are Al 3+ (from AlCl 3), Ba 2+ (from BaCl 2)
and Na + (from Na 3PO 4 and NaCl). So, their power to coagulate
As2S3. (S2− ) will follow the order as
Al 3+ > Ba 2+ > Na+
18. According to Freundlich adsorption isotherm,
x
x
∝ p1/ n ⇒
= Kp1/ n
m
m
On taking log on both sides, we get
1
 x
log   = log K + log p
 m
n
On comparing with equation of straight line, y = mx + c, plot of
x
log vs log p gives,
m
Slope
∴
Slope = 1
n
log k
Ostwald’s process,
4NH3 (g ) + 5O 2 (g ) → 4NO(g ) + 6H2O(g )
l
θ
log x/m
l
( y2 − y1 ) 1
2 1
= ⇒ =
( x2 − x1 ) n
4 2
x
1/ 2
∝p
m
19. Colloidal solutions show Tyndall effect due to scattering of light
by colloidal particles in all directions in space. It is observed
only under the following conditions.
On taking logarithm of both sides, we get
x
log = log k + log p1 / n
m
1
x
or
log = log k + log p
m
n
y = c + mx
x
y = log ,
m
c = intercept = log k
1
m = slope = and x = log p
n
21. Physical adsorption takes place with decrease in enthalpy thus
exothermic change. It is physical adsorption and does not require
activation. Thus, (a) is incorrect.
Being physical adsorption ∆H < 0 thus, (b) is correct. Exothermic
reaction is favoured at low temperature thus (c) is incorrect.
Physical adsorption is always reversible, thus (d) is incorrect.
22. According to Hardy Schulze rule, greater the charge on
oppositely charged ion, greater is its coagulating power. Since
arsenic sulphide is a negatively charged sol, thus, the order of
coagulating power is Na + < Ba 2+ < Al 3+ .
23. Sb2S3 is a negative (anionic) sol. According to Hardy Schulze
rule, greater the valency of cationic coagulating agent, higher its
coagulating power. Therefore, Al 2 (SO4 )3 will be the most
effective coagulating agent in the present case.
24. Larger the hydrophobic fragment of surfactant, easier will be the
micellisation, smaller the crticial micelle concentration.
Therefore, CH3 (CH2 )15 N+ (CH3 )3 Br − will have the lowest
crticial micelle concentration.
25. Lyophilic sols are reversible, not easily coagulated because it is
self-stabilising.
26. ∆G = ∆H − T∆S
As gas is adsorbed on surface of solid, entropy decreases, i.e.
∆S < 0. Therefore, for ∆G < 0, ∆H must be negative.
27. It is an exothermic process, according to Le-Chatelier’s principle,
lowering temperature drive the process in forward direction.
28. As temperature increases surface tension of liquid decreases.
29. (a) Higher the critical temperature, greater the extent of
adsorption.
(c) P (s) + Q (g ) → PQ (s)
Adsorbent
Adsorbate
206 Surface Chemistry
As gaseous adsorbate is adsorbed on solid surface, entropy
decreases, ∆S < 0. Also formation of bond between P and Q
results in release of energy, hence ∆H < 0.
30. Since, adsorption involves electron transfer from metal to O2, it
is chemical adsorption not physical adsorption, hence (a) is
incorrect. Adsorption is spontaneous which involves some
bonding between adsorbent and adsorbate, hence exothermic.
The last occupied molecular orbital in O2 is π * 2p. Hence,
electron transfer from metal to oxygen will increase occupancy
of π * 2p molecular orbitals. Also increase in occupancy of π * 2p
orbitals will decrease bond order and hence increase bond length
of O2.
31. Graph-I represents physisorption as in physisorption, absorbents
are bonded to adsorbate through weak van der Waals’ force.
Increasing temperature increases kinetic energy of adsorbed
particles increasing the rate of desorption, hence amount of
adsorption decreases.
Graph-II represents chemisorption as it is simple activation
energy diagram of a chemical reaction.
Graph-III also represents physical adsorption as extent of
adsorption increasing with pressure.
Graph-IV represents chemisorption as it represents the potential
energy diagram for the formation of a typical covalent bond.
32. Lyophobic sol, which is otherwise unstable, gets stabilised by
preferential adsorption of ions on their surface, thus developing a
potential difference between the fixed layer and the diffused
layer. Thus, option (a) and (d) are correct.
33. (a) In the process of adsorption, a bond is formed between
adsorbate and adsorbent, hence always exothermic.
(b) Physisorption require very low activation energy while
chemisorption require high activation energy. Therefore a
physisorption may transform into chemisorption but only at
high temperature.
(c) It is wrong statement as at higher temperature, physically
adsorbed substance starts desorbing.
(d) In physical adsorption, van der Waals' force hold the
adsorbate and adsorbent together which is a weak
electrostatic attraction. In chemisorption, strong chemical
bond binds the adsorbate to the adsorbent. Therefore,
chemisorption is more exothermic than physical adsorption.
34. From Freundlich adsorption isotherm equation,
log
x
1
= log K + log p
m
n
x
log m
θ
log p
When we plot log x / m vs log p, we get a straight line of
1
1
(i) slope = = 2 ⇒ n =
n
2
(ii) intercept = log K = 0.4771
⇒ log K = log 3 ⇒ K = 3
So,
x
= Kp1/ n = 3 × 4 2 = 48.00
m
(Q p = 4 atm)
35. Both statements are independently correct but Statement II does
not explain Statement I. Critical micelle concentration is the
minimum concentration of surfactant at which micelle formation
commences first. At critical micelle concentration, several
molecules of surfactant coalesce together to form one single
micelle molecule. This decreases the apparent number of
molecule suddenly lowering conductivity sharply.
14
s-Block Elements
4. HF and CH4 are called as molecular hydrides.
Topic 1 Group I Elements
Objective Questions I (Only one correct option)
1. Which of the following liberates O 2 upon hydrolysis?
(2020 Adv.)
(a) Pb3O4
(c) Na 2O2
(b) KO2
(d) Li 2O2
(2020 Main, 6 Sep II)
the reaction of Zn with dilute HCl
the electrolysis of acidified water using Pt electrodes
the electrolysis of brine solution
the electrolysis of warm Ba(OH)2 solution using Ni electrodes.
3. In the following reactions, products (A ) and ( B ),
respectively, are
(2020 Main, 7 Jan II)
NaOH + Cl2 → ( A ) + side products (hot and conc.)
Ca(OH)2 + Cl2 → ( B ) + side products (dry)
(a) NaClO3 and Ca(OCl)2
(c) NaOCl and Ca(OCl)2
(b) NaClO3 and Ca(ClO3 )2
(d) NaOCl and Ca(ClO3 )2
4. The temporary hardness of a water sample is due to
compound X . Boiling this sample converts X to compound
(2019 Main, 12 April II)
Y . X and Y , respectively, are
(a) Mg(HCO3 )2 and Mg(OH)2
(b) Ca(HCO3 )2 and Ca(OH)2
(c) Mg(HCO3 )2 and MgCO3
(d) Ca(HCO3 )2 and CaO
5. The incorrect statement is
(b) 34%
(2019 Main, 8 April II)
(c) 13.6%
(d) 3.4%
9. The correct order of hydration enthalpies of alkali metal ions
is
(2019 Main, 8 April I)
(a) Li + > Na + > K + > Cs+ > Rb+
(b) Na + > Li + > K + > Rb+ > Cs+
(c) Na + > Li + > K + > Cs+ > Rb+
(d) Li + > Na + > K + > Rb+ > Cs+
10. The correct statement(s) among I to III with respect to
potassium ions that are abundant within the cell fluids is/are
(2019 Main, 12 Jan II)
I. They activate many enzymes.
II. They participate in the oxidation of glucose to produce
ATP.
III. Along with sodium ions, they are responsible for the
transmission of nerve signals.
(a) I, and III only
(c) I and II only
(b) I, II and III
(d) III only
hydrolysis with water yields H2 O2 and O2 along with another
product. The metal is
(2019 Main, 12 Jan I)
(2019 Main, 12 April II)
6. The metal that gives hydrogen gas upon treatment with both
acid as well as base is
molar mass of
H = 1g mol −1 and O = 16 g mol −1 ]
11. A metal on combustion in excess air forms X . X upon
(a) lithium is the strongest reducing agent among the alkali metals.
(b) lithium is least reactive with water among the alkali metals.
(c) LiNO3 decomposes on heating to give LiNO2 and O2.
(d) LiCl crystallise from aqueous solution as LiCl ⋅ 2H2O.
(a) magnesium
(c) zinc
(b) (1), (2) and (3) only
(d) (1), (3) and (4) only
8. The strength of 11.2 volume solution of H 2O2 is [Given that
(a) 1.7%
2. Dihydrogen of high purity (> 99.95%) is obtained through
(a)
(b)
(c)
(d)
(a) (1), (2), (3) and (4)
(c) (3) and (4) only
(2019 Main, 12 April I)
(a) Li
(b) Mg
(c) Rb
(d) Na
12. The hardness of a water sample (in terms of equivalents of
CaCO3 ) containing is
(Molar mass of CaSO4 = 136 g mol −1 )
(a) 100 ppm
(b) 10 ppm
(2019 Main, 12 Jan I)
(c) 50 ppm
(d) 90 ppm
13. The hydride that is not electron deficient is
(2019 Main, 11 Jan II)
(b) mercury
(d) iron
(a) AlH3
7. The correct statements among (a) to (d) are:
(2019 Main, 8 April II)
1. Saline hydrides produce H2 gas when reacted with H2O.
2. Reaction of LiAlH4 with BF3 leads to B2H6.
3. PH3 and CH4 are electron rich and electron precise
hydrides, respectively.
(b) B2H6
(d) GaH3
(c) SiH4
14. The correct statements among (a) to (d) regarding H2 as a
fuel are :
(2019 Main, 11 Jan I)
I. It produces less pollutants than petrol.
II. A cylinder of compressed dihydrogen weights ~ 30times
more than a petrol tank producing the same amount of
energy.
208 s-Block Elements
III. Dihydrogen is stored in tanks of metal alloys like NaNi 5 .
IV. On combustion, values of energy released per gram of
liquid dihydrogen and LPG are 50 and 142 kJ,
respectively.
(a) I, II and III only
(c) II and IV only
blue solution due to the formation of
(2019 Main, 10 Jan II)
(a) sodium ammonia complex
(b) sodium ion-ammonia complex
(c) sodamide
(d) ammoniated electrons
17. The total number of isotopes of hydrogen and number of
radioactive isotopes among them, respectively, are
(2019 Main, 10 Jan I)
(b) 3 and 2
(d) 3 and 1
(2019 Main, 10 Jan I)
(a) oxidising and reducing agent in both acidic and basic medium
(b) oxidising and reducing agent in acidic medium, but not in
basic medium
(c) reducing agent in basic medium, but not in acidic medium
(d) oxidising agent in acidic medium, but not in basic medium
19. The metal that forms nitride by reacting directly with N 2 of
air, is
(2019 Main, 9 Jan II)
(b) K
(d) Li
(2019 Main, 9 Jan II)
(a)
(b)
(c)
(d)
(c) region 4
(d) region 1
25. Which one of the following statements about water is false?
(2016 Main)
(a) Water can act both as an acid and as a base
(b) There is extensive intramolecular hydrogen bonding in the
condensed phase
(c) Ice formed by heavy water sinks in normal water
(d) Water is oxidised to oxygen during photosynthesis
26. The main oxides formed on combustion of Li, Na and K in
excess of air respectively are
(2016 Main)
(b) Li 2O2, Na 2O2 and KO2
(d) Li 2O , Na 2O and KO2
27. Which of the following atoms has the highest first ionisation
energy?
(a) Na
(2016 Main)
(b) K
(c) Sc
(d) Rb
28. Hydrogen peroxide in its reaction with KIO4 and NH2 OH
respectively, is acting as a
(a)
(b)
(c)
(d)
(2014 Adv.)
reducing agent, oxidising agent
reducing agent, reducing agent
oxidising agent, oxidising agent
oxidising agent, reducing agent
agent?
(b) CaCl 2
(d) Ca(HCO 3) 2
21. The isotopes of hydrogen are
(b) region 3
29. In which of the following reactions H2 O2 acts as a reducing
20. What is reason of temporary hardness of water?
(a) Na 2SO 4
(c) NaCl
(a) region 2
(a) LiO2, Na 2O2 and K 2O
(c) Li 2O, Na 2O2 and KO2
18. The chemical nature of hydrogen peroxide is
(a) Rb
(c) Cs
(2016 Main)
Region 4
Region 3
Region 2
Region 1
(2019 Main, 11 Jan I)
(b) electron-rich hydride
(d) molecular hydride
16. Sodium metal on dissolution in liquid ammonia gives a deep
(a) 2 and 1
(c) 2 and 0
below is
(b) II, III and IV only
(d) I and III only
15. NaH is an example of
(a) metallic hydride
(c) saline hydride
24. The hottest region of Bunsen flame shown in the figure given
(2019 Main, 9 Jan I)
deuterium and tritium only
protium and deuterium only
protium, deuterium and tritium
tritium and protium only
22. Hydrogen peroxide oxidises [Fe(CN)6 ]4− to [Fe(CN)6 ]3− in
acidic medium but reduces [Fe(CN)6 ]3− to [Fe(CN)6 ]4− in
alkaline medium. The other products formed are,
respectively.
(2018 Main)
(a) (H2O + O2 ) and H2O
(b) (H2O + O2 ) and (H2O + OH− )
(c) H2O and (H2O + O2 )
(d) H2O and (H2O + OH− )
23. Both lithium and magnesium display several similar
properties due to the diagonal relationship; however, the one
which is incorrect is
(2017 Main)
(a) Both form basic carbonates
(b) Both form soluble bicarbonates
(c) Both form nitrides
(d) nitrates of both Li and Mg yield NO2 and O2 on heating
(2014 Main)
I. H2O2 + 2H+ + 2e− → 2H2O
II. H2O2 − 2e− → O2 + 2H+
III. H2O2 + 2e− → 2OH−
IV. H2O2 + 2OH− − 2e− → O2 + 2H2O
(a) I and II
(b) III and IV
(c) I and III
(d) II and IV
30. A sodium salt of an unknown anion when treated with MgCl 2
gives white precipitate only on boiling. The anion is
(a) SO2−
4
(b) HCO−3
(2004, 1M)
(c) CO2−
3
(d) NO−3
31. A dilute aqueous solution of Na 2 SO4 is electrolysed using
platinum electrodes. The products at the anode and cathode
are respectively
(1996, 1M)
(a) O2 , H2
(b) S2O2–
8 , Na
(c) O2 , Na
(d) S2O82– , H2
32. Hydrolysis of one mole of peroxodisulphuric acid produces
(a) two moles of sulphuric acid
(1996, 1M)
(b) two moles of peroxomono sulphuric acid
(c) one mole of sulphuric acid and one mole of peroxomono
sulphuric acid
(d) one mole of sulphuric acid, one mole of peroxomono sulphuric
acid and one mole of hydrogen peroxide
33. The species that do not contain peroxide ions, is
(a) PbO2
(b) H2O2
(c) SrO2
(1992, 1M)
(d) BaO2
s-Block Elements 209
34. The metallic lustre exhibited by sodium metal is explained
by
(1987, 1M)
lattice energy
35. A solution of sodium sulphate in water is electrolysed using
inert electrodes. The products at cathode and anode are
respectively
(1987, 1M)
(b) O2 , H2
(c) O2 , Na
(d) O2 , SO2
36. Nitrogen dioxide cannot be obtained by heating
(a) KNO3
(b) Pb(NO3 )2
(c) Cu(NO3 )2
(1985, 1M)
(d)AgNO3
37. The oxide that gives H2 O2 on treatment with a dilute acid is
(a) PbO2
(c) MnO2
(b) Na 2O2
(d) TiO2
(1985, 1M)
38. Molecular formula of Glauber’s salt is
(1985, 1M)
39. Heavy water is
(1983, 1M)
(a) MgSO4 ⋅ 7H2O
(c) FeSO4 ⋅ 7H2O
(b) CuSO4 ⋅ 5H2O
(d) Na 2SO4 ⋅ 10H2O
(a) H2O18
(b) water obtained by repeated distillation
(c) D2O
(d) water at 4°C
(1981, 1M)
(b) sodium hydride
(d) solvated electrons
41. The temporary hardness of water is due to calcium
bicarbonate can be removed by adding
(b) Ca(OH)2
(c) CaCl 2
(b) Na 2O
(b) K and excess of O2
(d) O2 and 2-ethylanthraquinol
(2007, 2M)
(c) NaO2
(d) NaOH
44. Sodium nitrate decomposes above ≈ 800°C to give
(a) N2
(c) NO2
(b) O2
(d) Na 2 O
(1998, 2M)
(1998, 2M)
46. When zeolite, which is hydrated sodium aluminium silicate,
is treated with hard water, the sodium ions are exchanged
with
(1990, 1M)
(a) H+ ions
(b) SO2–
4 ions
(c) Mg2+ ions
50. Statement I The alkali metals can form ionic hydrides
which contain the hydride ion, H− .
Statement II The alkali metals have low electronegativity,
their hydrides conduct electricity when fused and liberate
hydrogen gas at the anode.
(1994, 2M)
Fill in the Blanks
(1987, 1M)
52. Sodium dissolved in liquid ammonia conducts electricity
because of ……
(1985, 1M)
53. The adsorption of hydrogen by palladium is commonly
known as ……
(1983, 1M)
54. Iodine reacts with hot NaOH solution. The products are NaI
and ……
(1980, 1M)
True/False
55. Sodium when burnt in excess of oxygen gives sodium oxide.
(1987, 1M)
45. Highly pure dilute solution of sodium in liquid ammonia
(a) shows blue colour
(b) exhibits electrical conductivity
(c) produces sodium amide
(d) produces hydrogen gas
(2007, 3M)
concentrated solution of ……… .
43. The compound(s) formed upon combustion of sodium metal
(a) Na 2O2
blue solution.
Statement II Alkali metals in liquid ammonia give solvated
species of the type [ M ( NH 3 )n ]+ (M = alkali metals).
51. Hydrogen gas is liberated by the action of aluminium with
42. The pair(s) of reagents that yield paramagnetic species is/are
in excess air is (are)
48. Statement I Alkali metals dissolve in liquid ammonia to give
(1979, 1M)
(d) HCl
Objective Questions II
(One or more than one correct option)
(a) Na and excess of NH3
(c) Cu and dilute HNO3
Read the following questions and answer as per the direction given
below :
(a) Statement I is correct; Statement II is correct; Statement
II is the correct explanation of Statement I
(b) Statement I is correct; Statement II is 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
Statement II Electronegativity difference between Li and
Cl is too small.
(1998, 2M)
reducing due to the presence of
(a) CaCO3
Assertion and Reason
49. Statement I LiCl is predominantly a covalent compound.
40. A solution of sodium metal in liquid ammonia is strongly
(a) sodium atoms
(c) sodium amide
hydration energy
(c) the lattice energy has no role to play in solubility
(d) the hydration energy of sodium sulphate is less than its
(a) diffusion of sodium ions
(b) oscillation of loose electron
(c) excitation of free protons
(d) existence of body centred cubic lattice
(a) H2 , O2
(b) the lattice energy of barium sulphate is more than its
(d) OH− ions
47. Sodium sulphate is soluble in water, whereas barium
sulphate is sparingly soluble because
(1989, 1M)
(a) the hydration energy of sodium sulphate is more than its
lattice energy
Subjective Questions
56. A white solid is either Na 2 O or Na 2 O2 . A piece of red litmus
paper turns white when it is dipped into a freshly made
aqueous solution of the white solid.
(i) Identify the substance and explain with balanced
equation.
(ii) Explain what would happen to the red litmus if the white
solid were the other compound.
(1999, 4M)
57. Hydrogen peroxide acts both as an oxidising and as a
reducing agent in alkaline solution towards certain first row
transition metal ions. Illustrate both these properties of H2 O2
using chemical equations.
(1998, 4M)
210 s-Block Elements
58. Element A burns in nitrogen to give an ionic compound B.
Compound B reacts with water to give C and D. A solution of
C becomes ‘milky’ on bubbling carbon dioxide gas. Identify
A, B, C and D.
(1997, 3M)
59. Complete and balance the following chemical reaction.
Anhydrous potassium nitrate is heated with excess of
metallic potassium
(1992, 1M)
KNO3 ( s ) + K ( s ) → K + K
60. Give reasons in one or two sentences for the following:
“H2 O2 is a better oxidising agent than H2 O.”
(1986, 1M)
61. Sodium carbonate is prepared by Solvay process but the
same process is not extended to the manufacture of
potassium carbonate, explain.
(1981, 1M)
62. Water is a liquid, while H2 S is a gas at ordinary temperature.
Explain .
(1978, 1M)
Topic 2 Group II Elements
Objective Questions I (Only one correct option)
1. Match the following compounds (Column -I) with their uses
(Columns -II).
(2020 Main, 6 Sep II)
Column -I
Column - II
(I)
Ca(OH)2
(A) Casts of statues
(II)
NaCl
(B) White wash
(III)
1
CaSO 4 ⋅ H2O
2
(C) Antacid
(IV)
CaCO 3
(D) Washing soda preparation
(a) (I)-(D), (II)-(A), (III)-(C), (IV)-(B)
(b) (I)-(B), (II)-(D), (III)-(A), (IV)-(C)
(c) (I)-(B), (II)-(C), (III)-(D), (IV)-(A)
(d) (I)-(C), (II)-(D), (III)-(B), (IV)-(A)
2. In comparison to boron, berylium has
(a)
(b)
(c)
(d)
(2019 Main, 12 April II)
lesser nuclear charge and lesser first ionisation enthalpy
greater nuclear charge and lesser first ionisation enthalpy
greater nuclear charge and greater first ionisation enthalpy
lesser nuclear charge and greater first ionisation enthalpy
3. The correct sequence of thermal stability of the following
carbonates is
(a)
(b)
(c)
(d)
(2019 Main, 12 April I)
BaCO3 < CaCO3 < SrCO3 < MgCO3
MgCO3 < CaCO3 < SrCO3 < BaCO3
MgCO3 < SrCO3 < CaCO3 < BaCO3
BaCO3 < SrCO3 < CaCO3 < MgCO3
(b) CaX 2
(c) MgX 2
(d) BeX 2
9. Match the following items in Column I with the corresponding
items in Column II.
(2019 Main, 11 Jan II)
Column I
Column II
(i)
Na 2CO3 ⋅ 10H2O
A.
Portland cement
ingredient
(ii)
Mg(HCO3 )2
B.
Castner-Kellner process
(iii) NaOH
C.
Solvay process
(iv) Ca 3Al 2O6
D.
Temporary hardness
(a)
(b)
(c)
(d)
(i) - (D); (ii) - (A); (iii) - (B); (iv) - (C)
(i) - (B); (ii) - (C); (iii) - (A); (iv) - (D)
(i) - (C); (ii) - (B); (iii) - (D); (iv) - (A)
(i) - (C); (ii) - (D); (iii) - (B); (iv) - (A)
10. The amphoteric hydroxide is
(a) Be(OH)2
(b) Ca(OH)2
(b) Be
(2019 Main, 11 Jan I)
(c) Sr(OH)2
(d) Mg(OH)2
(c) Mg
(d) Ca
12. The alkaline earth metal nitrate that does not crystallise with
water molecules, is
(a) Ca(NO3 )2
(c) Ba(NO3 )2
(2019 Main, 9 Jan I)
(b) Sr(NO3 )2
(d) Mg(NO3 )2
13. Which one of the following alkaline earth metal sulphates
has its hydration enthalpy greater than its lattice enthalpy?
(a) CaSO 4
(c) BaSO 4
5. The alloy used in the construction of aircrafts is
(2019 Main, 10 April I)
(b) Mg-Mn
(d) Mg-Al
6. The structures of beryllium chloride in the solid state and
vapour phase, respectively are
(2019 Main, 8 April II)
(a) SrX 2
(2019 Main, 10 Jan I)
baking soda and soda ash
washing soda and soda ash
baking soda and dead burnt plaster
washing soda and dead burnt plaster
(a) dimeric and dimeric
(c) dimeric and chain
(b) Mg (NO3 )2 and Mg 3N2
(d) MgO and Mg (NO3 )2
8. The covalent alkaline earth metal halide ( X = Cl, Br, I) is
(a) Na
on heating initially gives a
monohydrated compound Y .
Y upon heating above 373 K leads to an anhydrous white
powder Z. X and Z, respectively, are (2019 Main, 10 April II)
(a) Mg-Zn
(c) Mg-Sn
(a) MgO and Mg 3N2
(c) MgO only
11. The metal used for making X-ray tube window is
4. A hydrated solid X
(a)
(b)
(c)
(d)
7. Magnesium powder burns in air to give (2019 Main, 9 April I)
(2019 Main, 9 April II)
(b) chain and chain
(d) chain and dimeric
(b) BeSO 4
(d) SrSO 4
(2015 Main)
14. The following compounds have been arranged in order of their
increasing thermal stabilities. Identify the correct order.
K 2 CO3 (I), MgCO3 (II), CaCO3 (III), BeCO3 (IV)
(a) I < II < III < IV
(c) IV < II < I < III
(b) IV < II < III < I
(d) II < IV < III < I
(1996, 1M)
15. The oxidation state of the most electronegative element in
the products of the reaction, BaO2 with dil. H2 SO4 are
(a) 0 and –1
(c) –2 and 0
(b) –1 and –2
(d) –2 and –1
(1991, 1M)
s-Block Elements 211
16. Calcium is obtained by
(1980, 1M)
(a) electrolysis of molten CaCl 2
(b) electrolysis of solution of CaCl 2 in water
(c) reduction of CaCl 2 with carbon
(d) roasting of limestone
21. MgCl 2 ⋅ 6H2 O on heating gives anhydrous MgCl 2 .
22. Identify (X) in the following synthetic scheme and write their
structures.
*
17. The reagent(s) used for softening the temporary hardness of
water is(are)
(2010)
(b) Ca(OH)2
(d) NaOCl
white crystalline precipitate. What is its formula?
(2006, 3M)
(b) Mg3 (PO4 )3
(d) MgSO4
(b) Si
(c) Sn
(1993, 1M)
(d) Ti
Fill in the Blank
20. Anhydrous MgCl 2 is obtained by heating the hydrated salt
with ……
(2001, 1M)
23. Give reasons for the following in one or two sentences only :
(1999, 2M)
24. The crystalline salts of alkaline earth metals contain more
water of crystallisation than the corresponding alkali metal
salts. Why?
(1997, 2M)
25. Calcium burns in nitrogen to produce a white powder which
19. The material used in solar cells contains
(a) Cs
Ba CO3 + H2 SO4 → X (gas) (C* denotes C14 )
“BeCl 2 can be easily hydrolysed.”
18. MgSO 4 on reaction with NH4 OH and Na 2 HPO4 forms a
(a) Mg(NH4 )PO4
(c) MgCl 2 ⋅ MgSO4
(1982, 1M)
Subjective Questions
Objective Questions II
(One or more than one correct option)
(a) Ca 3 (PO4 )2
(c) Na 2CO3
True/False
dissolves in sufficient water to produce a gas A and an
alkaline solution. The solution on exposure to air produces a
thin solid layer of B on the surface. Identify the compounds A
and B.
(1996, 3M)
26. Arrange the following in increasing order of basic strength :
MgO, SrO, K 2 O, NiO, Cs2 O
(1980 , 1M)
(1991, 1M)
Answers
Topic 1
45. (a, b)
46. (a, d)
47. (a, b)
1. (b)
2. (d)
3. (a)
4. (a)
49. (c)
50. (a)
51. NaOH
5. (c)
6. (c)
7. (a)
8. (d)
52. (solvated electrons)
9. (d)
10. (b)
11. (c)
12. (a)
13. (c)
14. (a)
15. (c)
16. (d)
17. (d)
18. (a)
19. (d)
20. (d)
21. (c)
22. (c)
23. (a)
24. (a)
25. (b)
26. (c)
27. (c)
28. (a)
29. (d)
30. (b)
31. (a)
32. (c)
33. (a)
34. (b)
35. (a)
36. (a)
37. (b)
38. (d)
39. (c)
40. (d)
41. (b)
42. (a, b, c)
43. (a, b)
44. (a, b, d)
48. (b)
53. (occlusion) 54. (NaIO 3)
55. (F)
Topic 2
1. (b)
2. (d)
3. (b)
4. (b)
5. (d)
6. (c)
7. (a)
8. (d)
9. (d)
10. (a)
11. (b)
12. (c)
13. (b)
14. (b)
15. (d)
16. (a)
17. (b, c, d)
18. (a)
19. (b)
20. dry HCl
21. F
Hints & Solutions
Topic 1 Group I Elements
1. (a) Pb3O4 + H2O → No reaction
Pb3O4 is insoluble in water or do not react with water.
(b) KO2 + 2H2O → KOH + H2O2 + 1/2 O2
Potassium superoxide is a strong oxidant, able to convert
oxides into peroxides or molecular oxygen. Hydrolysis gives
oxygen gas, hydrogen peroxide and potassium hydroxide.
(c) Na 2O2 + 2H2O → 2NaOH + H2O2
When sodium peroxide dissolves in water, it is hydrolysed
and forms sodium hydroxide and hydrogen peroxide. The
reaction is highly exothermic.
(d) Li 2O2 + 2H2O → 2LiOH + H2O2
The reactivity of Li 2O2 toward water differs from LiO2, in
Li 2O2 results in H2O2 as a product.
Hence, the correct option is (b).
2. Electrolysis of warm aqueous Ba(OH)2 solution between nickel
electrodes is a commercial method to obtain highly pure
(> 99.95%) dihydrogen.
In the presence of Ba(OH)2, water dissociates into ions easily
and quickly and H+ ions are produced, which go on cathode,
gets discharged there and liberate hydrogen gas.
212 s-Block Elements
∴34 g of H2O2 is present in 1000 g of solution
34
∴% w/w =
× 100 = 3.4%
1000
3. 6 NaOH + 3Cl2 → NaClO3 + 5NaCl +3H2O
(Hot and conc.)
( A)
2Ca(OH)2 + 2Cl2 → Ca(OCl)2 +CaCl2 +2H2O
( B)
(Dry)
9.
Thus, A : NaClO3
B : Ca (OCl)2
4. The temporary hardness of a water sample is due to compound
X [i.e. Mg(HCO3 )2]. Boiling of this sample converts X [i.e.
Mg(HCO3 )2] to compound Y[i.e. Mg(OH)2 ]. Generally,
temporary hardness is due to presence of magnesium and
calcium hydrogen carbonates. It can be removed by boiling.
During boiling, the soluble Mg(HCO3 )2 is converted into
insoluble Mg(OH)2 and Ca(HCO3 )2 changed to insoluble
CaCO3 . These precipitates can be removed by filteration.
Heating
Mg(HCO3 )2  
→ Mg(OH)2 ↓ + 2CO2 ↑
Heating
Ca(HCO3 )2 → CaCO3 ↓ + H2O + CO2 ↑
5. Statement (c) is incorrect. LiNO3(Lithium nitrate) on heating
gives a mixture of Li 2O, NO2 and O2.
∆
4LiNO3 → 2Li 2O + 4NO2 ↑ + O2 ↑
Among the alkali metals, lithium is the strongest reducing agent.
6. Metal that gives hydrogen gas upon treatment with both acid as
well as base is zinc. Hence, it is amphoteric in nature.
Reactions involved are as follows:
Zn + Dil. NaOH → Na 2 ZnO2 + H2 ↑
Key Idea The amount of energy released when one mole of
gaseous ions combine with water to form hydrated ions is
called hydration enthalpy.
The correct order of hydration enthalpies of alkali metal ions is
Li + > Na + > K+ > Rb+ > Cs+
Li+ possesses the maximum degree of hydration due to its small
size. As a consequence of hydration enthalpy, their mobility
also get affected. Cs+ has highest and Li + has lowest mobility in
aqueous solution.
10. All the statements are correct. K+ being metallic unipositive
ions work as enzyme activators. These also participate in many
reactions of glycolysis and Kreb’s cycle to produce ATP from
glucose.
Being unipositive these are also equally responsible for nerve
signal transmission along with Na + . (Na + ion-pump theory)
11. Metal (A) is rubidium (Rb). In excess of air, it forms RbO2( X ).
X is a superoxide that have O−2 ion. It is due to the stabilisation
of large anion by large cations through lattice energy effects.
RbO2 (X ) gets easily hydrolysed by water to form the
hydroxide, H2O2 and O2.
The reaction involved are as follows:
Rb + O2 → RbO2 (superoxide)
Zn + 2HCl(dil.) → ZnCl 2 + H2 ↑
7. The explanation of given statements are as follows :
1. Saline or ionic hydrides produce H2 with H2O.
⊕ È
M H + H2O → H2 ↑ + MOH
Thus, statement (1) is correct.
Ether
2. 3LiAlH4 + 4BF3 → 2B2H6 + 3LiF + 3AlF3
(Diborane)
Thus, statement (2) is correct.
3. PH3 and CH4 are covalent hydrides and in both of the
hydrides, octet of P and C have been satisfied. But P in PH3
has one lone pair of electrons and C in CH4 does not have so
PH3 (group 15) and CH4 (group 14) are electron rich and
electron precise hydrides, respectively.
Thus, statement (3) is correct.
4. HF and CH4 are called as molecular hydrides because of
their discrete and sterically symmetrical structure.
Thus, statement (4) is also correct.
8. 11.2 volume of H2O2 means that 1 mL of this H2O2 will give
11.2 mL of oxygen at STP.
2H2O2 (l ) →
2 × 34 g
O2 (g )
22.4 L at STP
+ 2H2O(l )
22.4 L of O2 at STP is produced from H2O2 = 68 g
∴ 11.2 L of O2 at STP is produced from
68
H2O2 =
× 112
. = 34 g
22.4
(X)
2RbO2 + 2H2O → 2RbOH + H2O2 + O2
(X)
12. Hardness of water sample can be calculated in terms of ppm
concentration of CaCO3.
Given, molarity = 10−3M
i.e. 1000 mL of solution contains 10−3 mole of CaCO3.
∴Hardness of water = ppm of CaCO3
10−3 × 1000
=
× 106 = 100ppm
1000
13. GaH3, AlH3 and B2H6 are the hydrides of group-13 (ns2np1 ),
whereas SiH4 is an hydride of group 14.
H
H
sp2
Ga
Al
sp2
H
H
H
H
6e – around
6e – around
Al 3+ <8e – (octet)
Ga 3+ <8e – (octet)
(AlH3)
(GaH3)
3c–2e – bridge bond
H
sp3 H
H
B
B
H
H
H
B is sp3, but cannot satisfy octet due
to the attachment of two 3c-2e –
bridge bonds (B–H–B)
So, B2H6, AlH3 and GaH3 are electron deficient hydrides.
s-Block Elements 213
But, SiH4 is an electron precise hydride of group-14 (ns2np2 ), i.e.
these hydrides can have the required number of electrons to write
their conventional Lewis structures.
H
8e– around Si (Octet gets satisfied)
Si
H
Only tritium (T) is radioactive, because of its very high
n
value,
p
n

 = 2 .
p

18. H2O2 can act as both oxidising and reducing agents in both
acidic and basic medium.
H H
H2 O2 as oxidising agent
14. (I) H2 is a 100% pollution free fuel. So, statement (I) is correct.
(II) Molecular weight of H2(2u).
1
=
× molecular weight of butane,
29
C 4H10 (LPG) [58u].
In acidic medium: H2 O2 + 2H+ + 2e− → 2H2 O
l
In basic medium : H2 O2 + 2O H + 2e− → 2H2 O + 2O2 −
−
H2 O2 as reducing agent
So, compressed H2 weighs ~30 times more than a petrol tank
and statement (II) is correct.
(III) NaNi 5, Ti - TiH2 etc. are used for storage of H2 in small
quantities. Thus, statement (III) is correct.
(IV) On combustion values of energy released per gram of liquid
dihydrogen (H2 ) : 142 kJ g −1, and for LPG : 50 kJ g −1. So,
staement (IV) is incorrect.
r
l
l
In acidic medium : H2 O2 → O2 + 2H+ + 2e−
In basic medium :
H2O2 + 2OH− → O2 + 2H2O + 2e−
19. Among the group-1 metals, only Li is able to form its nitride,
Li 3N. [All alkaline earth metals of group-2 form their nitride,
M 3N2]
6Li + N 2 → 2Li 3N (Ruby red solid)
(Air)
s
15. Na H is an example of ionic or saline hydride. These hydrides are
formed when hydrogen combines with metals having less
electronegativity and more electropositive character with
respect to hydrogen.
Except Be and Mg, all s-block metals form saline hydrides.
Hydrides of p-block elements are covalent in nature, viz, electron
deficient hydrides (by group-13 elements), electron-precise
hydrides (by group-14 elements), and electron-rich hydrides (by
group 15-17 elements). Hydrides of d , f -block metals are called
interstitial or metallic hydrides.
16. Sodium metal on dissolution in liquid ammonia gives a deep
blue solution due to the formation of ammoniated electrons. The
reaction is represented as follows:
Na (s) + (x + y) NH3 (l ) → [Na(NH)x ]+ + [ e(NH 3 ) y ]
[Ammoniated Na +
or expanded Na]
[Ammoniated
electrons]
Ammoniated (solvated) electrons show electronic transition in
visible region and the solution becomes deep blue coloured.
This deep blue solution also shows the following properties due
to the presence of ammoniated electrons.
(i) It is strongly reducing in nature.
(ii) It is paramagnetic.
(iii) It is a good conductor of electricity.
17. Hydrogen has three isotopes:
1
1H
2
1H
3
1 H
Protium (P)
Deuterium (D)
Tritium (T)
p
1
1
1
n
0
1
2
n
p
0
1
2
I
II
[ 3M + N 2 → M 3N 2 ]
Li ⊕ is the smallest metal ion of group-1. Smaller size of Li ⊕ and
larger size of nitride ion, N 3− , enable Li ⊕ to polarise the
spherical electron cloud of N 3− and it gives higher stability to
Li 3N.
20. Temporary hardness of water is due to presence of soluble
Ca(HCO3 )2 or Mg(HCO3 )2.
Permanent hardness of water is due to the presence of CaCl 2 or
CaSO4 or MgCl 2 or MgSO4.
Temporary hardness of water is also called carbonate hardness
which can be easily removed by boiling or by treatment with
Ca(OH)2 (Clark’s method).
21. There are three known isotopes of hydrogen, each possessing an
atomic number 1 and atomic masses 1, 2 and 3 respectively.
These are named as protium (1 H), deuterium
(2 H or D) and tritium (3 H or T)
The most common isotope is the ordinary hydrogen usually
called protium. It consists of one proton in the nucleus and an
electron revolving around it.
The second isotope of hydrogen is called heavy hydrogen or
deuterium. It consists of one proton and one neutron in the
nucleus and an electron revolving around it. The third isotope of
hydrogen is called tritium. It consists of one proton and two
neutrons in the nucleus and an electron revolving around it.
22. Both reactions in their complete format are written below
(i) In acidic medium,
−2
−1
[Fe2+ (CN)6 ]4− + H2 O2 + 2H+ → [Fe3+ (CN)6 ]3− + 2H2 O
(ii) In alkaline medium,
−1
[Fe3+ (CN)6 ]3− + H2 O2 + 2OH− →
[Fe2+ (CN)6 ]4− + O2 + 2H2O
Hence, H2O (for reaction (i)) and O2 + H2O
(for reaction (ii)) are produced as by product.
214 s-Block Elements
O
O


HO  S − O  O − S  OH → 2H2SO4 + H2O2
 HO  H H  OH 
O
O
23. Mg can form basic carbonate while Li cannot.
5 Mg
2+
+6
CO2–
3
+ 7H2O → 4MgCO3 ⋅ Mg(OH)2
⋅5H2O + 2 HCO−3
24. Region 1 (Pre-heating zone)
On partial hydrolysis, it gives one mole of H2SO4 and one mole
of peroxomonosulphuric acid as
O
O


HO  S  O  O  S  OH → H2SO4
 H  OH 
O
O
O

+ HO  S  O  O  H

O
Region 2 (Primary combustion zone, hottest zone)
Region 3 (Internal zone)
Region 4 (Secondary reaction zone)
25. There is extensive intermolecular H-bonding in the condensed
phase.
1
26. 2Li + O2 (g ) → Li 2O
2
(Excess)
2Na + O2 (g ) → Na 2O2; K + O2 (g ) → KO2
( Excess)
(Excess)
27. Order of first ionisation energy is Sc > Na > K > Rb.
Due to poor shielding effect, removal of one electron from 4s
orbital is difficult as compared to 3s-orbital.
28.
PLAN This problem can be solved by using concept of oxidant and
reductant.
Peroxomonosulphuric acid
33. In PbO2 , Pb is in +4 oxidation state and oxygen is in –2 oxidation
state. In all other case, peroxide ion (O2−
2 ) is present.
34. Metallic lustre of any metal is due to oscillation of free electrons
present in the metal.
Oxidant Oxidant increases the oxidation number of the
species with which it is reacted.
35. H2O is reduced as well as oxidised giving H2 (g ) at cathode and
Reductant Reductant decreases the oxidation number of the
species with which it is reacted.
36. KNO3 and other nitrates of alkli metals (except LiNO3) are
H2O2 reacts with KIO4 in the following manner:
37. Sodium peroxide on treatment with dilute acid gives H2O2
+7
O2 (g ) at anode.
thermally stable.
Na 2O2 + H2SO4 → Na 2SO4 + H2O2
+5
KIO4 + H2O2 → KIO3 + H2O + O2
On reaction of KIO4 with H2O2, oxidation state of I varies from
+7 to +5, i.e. decreases. Thus, KIO4 gets reduced hence, H2O2 is
a reducing agent here.
With NH2OH, it given following reaction:
+3
−1
N H2OH + H2O2 → N 2 O3 + H2O
In the above reaction, oxidation state of N varies from –1 to +3.
Here, oxidation number increases, hence H2O2 is acting as an
oxidising agent here.
Hence, (a) is the correct choice.
29. Release of electron is known as reduction. So, H2O2 acts as
reducing agent when it releases electrons.
Here, in reaction (II) and (IV), H2O2 releases two electron,
hence reaction (II) and (IV) is known as reduction.
In reaction (I) and (III), two electrons are being added, so (I) and
(III) represents oxidation.
30. Mg(HCO3 )2 on boiling decomposes to give white precipitate of
38. Glauber’s salt is Na 2SO4 ⋅ 10H2O.
39. D2O is commonly known as heavy water.
40. Presence of solvated electrons makes solution of alkali metal in
liquid ammonia makes them strongly reducing agent.
41. Lime treatment remove bicarbonate hardness by forming
insoluble CaCO3 as
Ca(HCO3 )2 + Ca(OH)2 → 2CaCO3 ↓ + 2H2O
42.
PLAN Paramagnetic character of species can be easily explained on
the basis of presence of unpaired electrons, i.e. compounds
containing unpaired electron(s) is/are paramagnetic.
Reaction of alkali metals with ammonia depends upon the
physical state of ammonia whether it is in gaseous state or liquid
state. If ammonia is considered as a gas then reaction will be
1
(a) Na + NH3 → NaNH2 + H2
2
(Excess)
(NaNH2 + 1/2 H2 are diamagnetic)
If ammonia is considered as a liquid then reaction will be
M + (x + y)NH3 → [ M (NH3 )x ]+ + [ e(NH 3 ) y ]−
• Ammoniated electron
• Blue colour
• Paramagnetic
• Very strong reducing agent
MgCO3 as:
Heat
Mg(HCO3 )2 (aq) → MgCO3 ↓ + H2O + CO2 ↑
31. Electrolysis of aqueous Na 2SO4 gives H2 (g ) at cathode and
O2 (g ) at anode.
32. Peroxodisulphuric acid (H2S2O8 ) on complete hydrolysis gives
two moles of H2SO4 and one mole of H2O2 as
(b) K + O2 → KO2 (K+ , O−2 )
(Excess)
Potassium superoxide
paramagnetic
(c) 3Cu + 8HNO3 → 3Cu(NO3 )2 +
Paramagnetic
2NO
Paramagnetic
+ 4H2O
s-Block Elements 215
OH
H2O2 → H2O +
O
Et
Et
O2
(d)
+ H 2O 2
OH
(ii) If the compound is Na 2O, it will hydrolyse to form NaOH.
Na 2O + H2O → 2NaOH
NaOH solution formed above will change colour of red
litmus paper into blue.
O
2-ethylanthraquinol
2-ethylanthraquinone
Hence, option (a), (b) and (c) are correct choices.
43. When sodium metal is burnt in excess of air, mainly sodium
57. KMnO4 + H2O2 → MnO2 + KOH + O2
OA
RA
44. NaNO3 when heated, it decomposes in two stages as:
1
O2
2
T > 800° C
→ Na 2O + N2 + O2
A
B
C
H+ and Mg2+ ions present in hard water.
47. Solubility of a salt is influenced by two major factors, lattice
energy and hydration energy. For greater solubility, there
should be smaller lattice energy and greater hydration energy.
48. Both statements are correct but blue colour is due to presence of
solvated electron NH3 (e− ).
D
Milkyness
M can be either Ca or Ba but essentially not Mg because
Mg(OH)2 is very sparingly soluble in water.
Na + NH3 → Na + + NH3 (e− )
46. Zeolite acts as ion exchange resin and its Na + is exchanged with
C
M (OH)2 + CO2 → MCO3
are present whose emission spectrum gives blue colouration to
solution.
Also, presence of solvated electrons and solvated Na + ion
makes solution highly conducting.
B
M 3N2 + 6H2O → 3M (OH)2 + 2NH3
45. In dilute solution of Na in liquid ammonia, solvated electrons
Solvated electron
OA
3M + N2 → M 3 N2
58.
NaNO3 → NaNO2 +
NaNO2
RA
FeSO4 + H2O2 → Fe3+ + H2O
peroxide (Na 2O2 ) with little sodium oxide (Na 2O) are formed.
T < 500° C
[O]
Bleaches colour
of red litmus
59. 2KNO3 (s) + 10K(s) → 6K2O (s) + N2 (g )
60. In H2O2 , the peroxide ion (O2−
2 ) is unstable, has tendency to pass
into stable oxide state (O2− ). Hence, H2O2 is a good oxidising
agent while H2O is stable.
61. In Solvay process, NaHCO3 is extracted from the solution by
fractional crystallisation, which is then heated to convert it into
Na 2CO3 ⋅ KHCO3 , being more soluble than NaHCO3 , cannot be
extracted by fractional crystallisation. Hence, Solvay process
fails in production of K2CO3.
62. Water forms stronger intermolecular H-bonds, therefore it is
liquid at room temperature while H2S cannot form such strong
intermolecular bonds, gas at room temperature.
49. Statement I is correct. Small size of Li + makes it highly
polarising, introduces predominant covalency in LiCl.
Statement II is incorrect, there is very large difference in
electronegativity of Li and Cl.
50. Alkali metal forms MH in which hydrogen is in –1 oxidation
state. Both statements are correct and statement –2 is correct
explanation of statement I.
51. Al + conc. NaOH → NaAlO2 + H2 ↑
52. Na in liquid ammonia contain NH3 (e− ) which possesses charge
and conduct electricity.
53. Occlusion is a phenomena in which particles are physically
trapped in voids.
Topic 2 Group II Elements
1.
Correct match is I → (B), II → (D), III → (A), IV → (C)
(I) Ca(OH)2 used in white wash.
(II) NaCl is used in the preparation of washing soda.
2NH3 + H2O + CO2 → (NH4 )2 CO3
(NH4 )2CO 3 + H2O + CO 2 → 2NH4HCO 3
NH4HCO 3 + NaCl → NH4Cl + NaHCO 3 (s)
∆
2NaHCO 3 
→ Na 2CO 3 + CO 2 + H2O
(III) CaSO4 ⋅
54. I2 disproportionate in alkali giving NaI and NaIO3.
55. Sodium when burnt in excess of oxygen, gives sodium peroxide
as major product.
∆
Na + O2 →
Na 2O2 + Na 2O
Major
Minor
56. The substance is Na 2O2. When Na 2O2 is dissolved in water, it
forms NaOH and H2O2. In this case, NaOH is a strong base
while H2O2 is a weak acid.
(i) Na 2O2 + 2H2O → 2NaOH + H2O2
H2O2 decolourises red litmus paper due to its bleaching
action which is due to its oxidising character.
1
H2O (plaster of Paris) is used for making casts of
2
statues.
(IV) CaCO 3 is used as an antacid.
2. In comparison to boron, beryllium has lesser nuclear charge and
greater first ionisation enthalpy.
Electronic configuration of Be(4 ) = 1s2 , 2s2.
It possess completely filled s-orbitals. Hence, high amount of
energy is required to pull the electron from the gaseous atom.
Beryllium (4) lies left to the boron (5) and on moving from left
to right an electron is added due to which nuclear charge
increases from Be to B.
216 s-Block Elements
3. The correct sequence of thermal stability of carbonates is
MgCO3 < CaCO3 < SrCO3 < BaCO3
On moving down the group, i.e. from Mg to Ba, atomic radius
generally increases. It is due to the addition of shell. As a result,
the atomic size increases. CO2−
3 is a large anion. Hence, more
stabilised by Ba 2+ (large cation) and less stabilised by Mg2+ .
Therefore, BaCO3 has highest thermal stability followed by
SrCO3, CaCO3 and MgCO3.
4. Baking soda (NaHCO3 ) is not a hydrated solid. Thus, ( X ) is not
baking soda. Thus, option (a) and (c) are incorrect. Dead burnt
plaster (CaSO4 ) is obtained from gypsum via the formation of
plaster of Paris.
1
> 393K
CaSO4 ⋅2H2O → CaSO4 ⋅ H2O →
CaSO4
1
−1/ 2 H 2 O
2
− H 2 O Dead burnt plaster
Gypsum
380-393 K
Plaster of Paris
2
(anhydrous)
Therefore, the reaction takes place as follows :
< 373K
> 373K
Na 2CO3 ⋅ 10H2O → Na 2CO3 ⋅ H2O →
Washing soda
(X )
−9H 2 O
Monohydrate
(Y )
−H 2 O
Na 2CO3
Anhydrous white powder
(soda ash) (Z )
5. Names of magnesium alloys are given by two letters followed by
two numbers. The common alloying elements are A (Aluminium),
Z (zinc), T (tin), M (manganese) etc. Numbers indicate
respective nominal compositions of main alloying elements,
e.g. ‘AZ 91’ implies the composition of the alloy as : Al = 9%,
Zn = 1% and Mg = 100 – (9 + 1) = 90%
Among the alloys given, Mg – Al (Magnalium ; Mg = 5%,
Al = 95%) is being light, tough and strong, hence it is used in
aircrafts.
6. The structures of beryllium chloride in the solid state and vapour
phase, respectively are dimeric and chain. In vapour phase at
above 900°C, BeCl2 is monomeric having a linear structure Cl
BeCl. The bonding in BeCl2 is covalent and Be atom
accommodates 2 + 2 = 4 electrons in the two sp-hybrid orbitals.
Below 900°C, beryllium chloride in vapour phase exists as a
mixture of monomer BeCl2 and dimer Be2Cl4.
7. Magnesium powder burns in air to give MgO and Mg 3N2. MgO
does not combine with excess oxygen to give any superoxide.
Mg reacts with nitrogen to form magnesium nitride (Mg 3N2 ).
Mg + O2 → MgO
3Mg + N2 → Mg3N2
8.
Key Idea According to Fajan’s rule, degree of covalency
(ionic potential), φ ∝-polarisation power of the
cation ∝ charge on the cation
1
.
∝
size of the cation
Alkaline earth metals contains bipositive (H2+ ) ions in their
compounds.
So, here
(i) Charge on cation, i.e. + 2 is constant.
(ii) Halide present (X − ) is also constant.
So, the covalent character depends on the size of alkaline earth
metal. As we move down the group, size of metal ion increases.
Be2+ < Mg2+ < Ca 2+ < Sr 2+ < Ba 2+
So, Be2+ readily forms covalent compounds like BeX 2, because
of very high positive charge density over its small size, so that it
readily polarises anionic spherical electron cloud.
9. (i) Washing soda (Na 2CO3 ⋅ 10H2O) is manufactured in Solvay
process. In this method, CO2 gas is passed through a
conc. solution of NaCl saturated with NH3. It gives ammonium
carbonate followed by ammonium hydrogen carbonate.
The obtained NH4HCO3 is treated with solution of NaCl which
result in the formation of NaHCO3. The crystal obtained are
heated to obtain Na 2CO3.
NaCl + NH3 + CO 2 + H2O → NaHCO3 + NH4Cl
∆
Crystallisation
2NaHCO3 →Na 2CO3 → Na 2CO3 ⋅ 10H2O
− H2O, CO2
(ii) Mg(HCO3 )2 and Ca(HCO3 )2 cause temporary hardness to
water that can be easily removed by boiling.
(iii) NaOH is manufactured by Castner-Kellner process.
In this reaction, Na amalgam flows out and treated with
water to give NaOH and H2 gas. During electrolysis,
hydrogen is evolved at cathode and chlorine is evolved at
anode, which are the by product of this process.
–
Cl2
Carbon anode 2Cl –2e
Electrolysis
2NaCl(aq)
+
Hg cathode
2Na +2e
2Na (Na/Hg)
H2O
2NaOH(aq) +H2
(iv) Portland cement constitutes, tricalcium aluminosilicate,
3CaO ⋅Al 2O3. SiO2, i.e. Ca 3Al 2O6 ⋅ SiO2.
10. For group-2 metal hydroxides, basicity increases down the
group, as:
Be(OH)2 < Mg(OH)2 < Ca(OH)2 < Sr(OH)2 < Ba(OH)2
This is because as the size of metal atom increases, M—OH
bond length increases or M—OH bond become weaker thus
readily breaks to release OH − ions which are responsible for the
basicity of these solutions.
But Be(OH)2 shows amphoteric (basic as well as acidic)
character as it reacts with acid and alkali both which is shown in
the following reactions. Be(OH)2 as a base :
Be(OH)2 + 2HCl → BeCl 2 + 2H2O
Be(OH)2 as an acid :
Be(OH)2 + 2NaOH → Na 2[Be(OH)4 ]
11. Among the four elements given, Na, Be, Mg and Ca, Be has
highest IE value because of its smallest size and 2s2 valence shell
configuration.
So, X-ray cannot cause ionisation from the material used, i.e. Be
in the tube window, which may cause interference in the study.
I
12. A saturated aqueous solution of M (NO3 )2 on crystallisation
II
will produce hydrated crystal, M (NO3 )2 ⋅ nH2O only when
hydration enthalpy (∆H °hyd ) of M 2 + ion will be appreciably
more negative.
s-Block Elements 217
Hydration of an ion depends on its size. Smaller the size of an
ion, higher will be its charge density and as a result it will remain
more solvated (hydrated) through ion dipole interaction.
Size of group-2 metal ions increases on going down the group.
So, their ability to form hydrated crystals follows the order:
Be2 + >> Mg 2 + >> Ca 2 + >> Sr 2 + >> Ba 2 +
Thus, Ba(NO3 )2 is slightly or almost insoluble in water.
13. As we move down the group, size of metal increases. Be has
SO2−
4
has bigger size, that’s why BeSO 4 breaks
lower size while
easily and lattice energy becomes smaller but due to lower size
of Be, water molecules are gathered around and hence hydration
energy increases. On the other hand, rest of the metals, i.e. Ca,
Ba, Sr have bigger size and that’s why lattice energy is greater
than hydration energy.
Time saving technique In the question of finding hydration
energy only check the size of atom. Smaller sized atom has more
hydration energy. Thus, in this question Be is placed upper most
in the group has lesser size and not comparable with the size of
sulphates. Hence, BeSO 4 is the right response.
14. Thermal stability of salts with common anion depends on
polarising power of cation. Greater the polarising power, lower
be their thermal stability. Hence,
BeCO3 (IV) < MgCO3 (II) < CaCO3 (III) < K2CO3 (I)
BaO2 + H2SO4 → BaSO4 + H2O2
The most electronegative atom, oxygen, in BaSO4 and H2O2 has
−2 and −1oxidation state respectively.
16. Electrolysis of molten CaCl 2 gives calcium at cathode
−
Ca + 2e → Ca (at cathode)
In case of electrolysis in aqueous medium, less electropositive
H+ is reduced at cathode rather than Ca 2+.
17. Ca(OH)2 + Ca(HCO3 )2 → 2CaCO3 ↓ +
HO− + HCO3− → CO23− + H2O
Ca(HCO3 )2 + Na 2CO3 → CaCO3 ↓ + 2NaHCO3
18. Magnesium ammonium phosphate is precipitated out.
MgSO4 + NH4OH + Na 2HPO4 → Mg(NH4 )PO4 ↓ + Na 2SO4
19. Si is used in solar cells, because of its semi-conductor properties.
20. Anhydrous MgCl 2 is obtained by heating hydrated salt in stream
of dry HCl.
21. Heating MgCl 2 ⋅ 6H2O brings about partial dehydration as
∆
MgCl 2 ⋅ 6H2O →
Mg(OHCl) + HCl + 5H2O
*
*
22. BaCO
3 + H 2SO 4 → BaSO 4 + H 2O + CO 2
*
(C = C14)
23. Be in BeCl 2 is electron deficient, short of two lone pair of
electrons from stable octet. H2O has lone pair of electrons, reacts
with BeCl 2.
24. Alkaline earth metal salts have M 2+ ions which has very high
polarising power compared to polarising power of monovalent
metal ion (M + ) of alkali metal. Due to high polarising power of
M 2+ , it associate more water than M + .
25. A = NH3 , B = CaCO3.
15. The reaction involved is
2+
NaOCl + H2 → NaOH + HOCl
2H2O
(Clark’s method)
Reactions involved are :
Heat
3Ca + N2 → Ca 3N2
Ca 3N2 + 6H2O → 3Ca(OH)2 + 2NH3
A
Ca(OH)2 + CO2
(From air)
→ CaCO3 + H2O
B
26. Basic strength (i) decreases from left to right in period and
(ii) increases from top to bottom in group. Therefore,
NiO < MgO < SrO < K2O < Cs2O Basic strength
15
p-Block Elements-I
Topic 1 Group 13 Elements
Objective Questions I (Only one correct option)
1. The reaction of H3N3 B3Cl3
(A) with LiBH 4 in
tetrahydrofuran gives inorganic benzene (B). Further, the
reaction of (A) with (C) leads to H3N3 B3 (Me)3 . Compounds
(B) and (C) respectively, are
(2020 Main, 9 Jan II)
(a)
(b)
(c)
(d)
diborane and MeMgBr
boron nitride and MeBr
borazine and MeBr
borazine and MeMgBr
(2019 Adv.)
(b) CrB
(d) Cr2 (B4O7 )3
3. The correct statements among I to III regarding group 13
element oxides are:
(2019 Main, 9 April II)
I. Boron trioxide is acidic.
II. Oxides of aluminium and gallium are amphoteric.
III. Oxides of indium and thallium are basic.
(b) I and III only
(d) II and III only
4. Diborane (B2 H6 ) reacts independently with O2 and H2 O to
produce, respectively.
(2019 Main, 8 April I)
(a) B2O3 and H3BO3
(b) B2O3 and [BH4 ]−
(c) H3BO3 and B2O3
(d) HBO2 and H3BO3
5. The relative stability of + 1 oxidation state of group 13
elements follows the order
(a)
(b)
(c)
(d)
(2019 Main, 11 Jan II)
Al < Ga < Tl < In
Al < Ga < In < Tl
Tl < In < Ga < Al
Ga < Al < In < Tl
9. The increasing order of atomic radii of the following Group
10. B(OH) 3 + NaOH w
(2016 Adv.)
(b) Ga < Al < In < Tl
(d) Al < Ga < Tl < In
NaBO2 + Na[B(OH) 4 ] + H2 O
How can this reaction is made to proceed in forward
direction?
(2006, 3M)
(a)
(b)
(c)
(d)
Addition of cis 1, 2-diol
Addition of borax
Addition of trans 1, 2-diol
Addition of Na 2HPO4
11. H3 BO3 is
(a)
(b)
(c)
(d)
(2003, 1M)
monobasic acid and weak Lewis acid
monobasic and weak Bronsted acid
monobasic and strong Lewis acid
tribasic and weak Bronsted acid
12. In compounds of type ECl 3 , where E = B , P, As or Bi, the
angles Cl  E  Cl for different E are in the order
(1999, 2M)
(a) B > P = As = Bi
(c) B < P = As = Bi
(a) silica
(c) diamond
bonds in B2 H6 , respectively, are
(2019 Main, 10 Jan II)
(b) 2 and 4
(d) 2 and 1
(b) B > P > As > Bi
(d) B < P < As < Bi
(2019 Main, 10 Jan I)
(b) carbon
(d) boron
(1982, 1M)
(b) graphite
(d) None of these
Objective Questions II
(One or more than one correct option)
14. Among the following, the correct statement(s) is(are)
7. The electronegativity of aluminium is similar to
(a) lithium
(c) beryllium
(b) lanthanoid contraction
(d) diagonal relationship
13. Moderate electrical conductivity is shown by
6. The number of 2-centre-2-electron and 3-centre-2-electron
(a) 4 and 2
(c) 2 and 2
(2019 Main, 9 Jan I)
(a) lattice effect
(c) inert pair effect
13 elements is
chromium (III) salt is due to
(a) I, II and III
(c) I and II only
thallium exists in +1and +3 oxidation states. This is due to
(a) Al < Ga < In < Tl
(c) Al < In < Ga < Tl
2. The green colour produced in the borax bead test of a
(a) Cr 2O 3
(c) Cr(BO2 )3
8. Aluminium is usually found in +3 oxidation state. In contrast,
(2017 Adv.)
(a) Al (CH3 )3 has the three-centre two-electron bonds in its dimeric
structure
(b) The Lewis acidity of BCl 3 is greater than that of AlCl 3
p-Block Elements-I 219
(c) AlCl 3 has the three-centre two-electron bonds in its dimeric
structure
(d) BH3 has the three-centre two-electron bonds in its dimeric
structure
15. The crystalline form of borax has
(2016 Adv.)
(a) tetranuclear [ B4O5 (OH)4 ]2 − unit
(b) all boron atoms in the same plane
(c) equal number of sp2 and sp3 hybridised boron atoms
(d) one terminal hydroxide per boron atom
16. The correct statement(s) for orthoboric acid is/are
(2014 Adv.)
(a) It behaves as a weak acid in water due to self ionisation
(b) Acidity of its aqueous solution increases upon addition of
ethylene glycol
(c) It has a three-dimensional structure due to hydrogen
bonding
(d) It is a weak electrolyte in water
+
17. In the reaction, 2 X + B2 H6 → [BH2 ( X )2 ] [BH4 ]
the amine(s) X is/are
−
(2009)
(a) NH3
(b) CH3NH2
(c) (CH3 )2 NH
(d) (CH3 )3 N
Match the Column
22. Match the following.
(2006, 6M)
Column I
Column II
A.
Bi 3 + → (BiO)+
p.
Heat
B.
[AlO2 ]− → Al(OH)3
q.
Hydrolysis
C.
SiO44 − →
r.
Acidification
D.
(B4O72 − ) → [B(OH)3 ]
s.
Dilution by water
Si 2O67 −
Fill in the Blank
23. The two types of bonds present in B2 H6 are covalent and
………
(1994, 1M)
True/False
24. The basic nature of hydroxide of group 13 (group IIIA)
decreases progressively down the group.
(1993, 1M)
25. All the Al  Cl bonds in Al 2 Cl 6 are equivalent.
(1989, 1M)
Integer Answer Type Questions
26. Three moles of B2 H6 are completely reacted with methanol. The
number of moles of boron containing product formed is
(2015 Adv.)
Numerical Answer Type Question
27. The value of n in the molecular formula Ben Al2 Si6 O18 is
18. Among B 2H 6, B 3N 3H 6, N 2O, N 2O 4, H 2S2O 3 and H 2S2O 8,
the total number of molecules containing covalent bond
between two atoms of the same kind is .............(2019 Adv.)
Assertion and Reason
Read the following questions and answer as per the
direction given below :
(a) Statement I is correct; Statement II is correct Statement II is
the correct explanation of Statement I
(b) Statement I is correct; Statement II is 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
19. Statement I Boron always forms covalent bond.
Statement II The small size of B3+ favours formation of
covalent bond.
(2007, 3M)
20. Statement I In water, orthoboric acid behaves as a weak
monobasic acid.
Statement II In water, orthoboric acid acts as a proton
donor.
(2007, 3M)
21. Statement I Al(OH)3 is amphoteric in nature.
Statement II Al  O and O  H bonds can be broken
(1998 , 2M)
with equal ease in Al (OH)3 .
(2010)
Subjective Questions
28. AlF3 is insoluble in anhydrous HF but when little KF is added to
the compound it becomes soluble. On addition of BF3, AlF3 is
precipitated. Write the balanced chemical equations. (2004, 2M)
29. (i) How is boron obtained from borax? Give chemical equations
with reaction conditions.
(ii) Write the structure of B2H6 and its reaction with HCl.
(2002)
30. Compound X on reduction with LiAlH4 gives a hydride Y
containing 21.72% hydrogen alongwith other products. The
compound Y reacts with air explosively resulting in boron
trioxide. Identify X and Y. Give balanced reactions involved in
the formation of Y and its reaction with air.
Draw the structure of Y.
(2001, 5M)
31. Aluminium sulphide gives a foul odour when it becomes damp.
Write a balanced chemical equation for the reaction.
(1997, 2M)
32. Anhydrous AlCl 3 is covalent. From the data given below,
predict whether it would remain covalent or become ionic in
aqueous solution.
(Ionisation energy for Al = 5137 kJ mol –1
∆H hydration for Al 3+ = – 4665 kJ mol –1
∆H hydration for Cl − = − 381 kJ mol −1
(1997, 2M)
220 p-Block Elements-I
Topic 2 Group 14 Elements
Objective Questions I (Only one correct option)
1. The C C bond length is maximum in
(a) graphite
(c) C60
(2019 Main, 12 April II)
(b) C70
(d) diamond
2. The basic structural unit of feldspar, zeolites, mica and asbestos
is
(2019 Main, 12 April I)
(a) (SiO3 )2−
(b) SiO2
R

(d) ( Si  O )
(R = Me)
n

R
(c) (SiO4 )4−
(a) CO
(2019 Main, 10 April I)
(b) Si > Sn > C > Ge
(d) Ge > Sn > Si > C
(2019 Main, 9 April I)
(d) Pb
7. The element that shows greater ability to form pπ- pπ multiple
bonds, is
(2019 Main, 12 Jan II)
(b) Si
(c) Sn
(d) C
8. The chloride that cannot get hydrolysed is (2019 Main, 11 Jan I)
(a) SnCl 4
(c) PbCl 4
(b) CCl 4
(d) SiCl 4
9. Correct statements among (I) to (IV) regarding silicones are:
(2019 Main, 9 Jan I)
IV. Usually, they are resistant to oxidation and used as greases.
(b) I, II, III only
(d) I, II and IV only
are shared is
(a) pyrosilicate
(c) linear chain silicate
(a) SnCl 2 ⋅ 2H2O is a reducing agent.
(b) SnO2 reacts with KOH to form K 2[Sn(OH)6 ].
(c) A solution of PbCl 2 in HCl contains Pb2+ and Cl − ions.
(d) The reaction of Pb3O4 with hot dilute nitric acid to give PbO2
is a redox reaction.
(2005, 1M)
(b) sheet silicate
(d) three-dimensional silicate
11. Me2 SiCl 2 on hydrolysis will produce
(a) (Me)2 Si(OH)2
(b) (Me)2 Si ==O
(c) [  O  (Me)2 Si  O ]n (d) Me2SiCl(OH)
(2012)
(a) Graphite is harder than diamond
(b) Graphite has higher electrical conductivity than diamond.
(c) Graphite has higher thermal conductivity than diamond.
(d) Graphite has higher C C bond order than diamond
Assertion and Reason
(a) Statement I is correc;t Statement II is correct Statement II is
the correct explanation of Statement I
(b) Statement I is correct; Statement II is 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
Statement II The higher oxidation states for the group
14 elements are more stable for the heavier members of the
group due to ‘inert pair effect’.
(2008, 3M)
18. Statement I Between SiCl 4 and CCl 4 , only SiCl 4 reacts
10. Name the structure of silicates in which three oxygen atoms of
[ SiO4 ]
Objective Question II
(One or more than one correct option)
agents than Sn 2+ compounds.
III. In general, they have high thermal stability and low
dielectric strength.
4−
(d) PbI4
17. Statement I Pb 4+ compounds are stronger oxidising
I. They are polymers with hydrophobic character.
II. They are biocompatible.
(a) I and II only
(c) I, II, III and IV
(1996, 1M)
(c) SnI4
statement(s) given below is/are correct?
(2019 Main, 12 Jan II)
(c) Si
(b) GeI4
16. With respect to graphite and diamond, which of the
6. The element that does not show catenation is
(a) Ge
(1996, 1M)
(d) SiO2
(2020 Adv.)
(2019 Main, 9 April II)
(b) kieselguhr
(d) quartz
(b) Sn
(c) ZnO
15. Choose the correct statement(s) among the following:
16 hexagons and 16 pentagons
20 hexagons and 12 pentagons
12 hexagons and 20 pentagons
18 hexagons and 14 pentagons
(a) Ge
(b) SnO2
doubtful existence?
5. C60 an allotrope of carbon contains
(a)
(b)
(c)
(d)
(2002, 3M)
13. Which one of the following oxides is neutral ?
(a) CCl 4
4. The amorphous form of silica is
(a) tridymite
(c) cristobalite
CO2 , CuO, CaO, H2 O.
(a) CaO < CuO < H2O < CO2
(b) H2O < CuO < CaO < CO2
(c) CaO < H2O < CuO < CO2
(d) H2O < CO2 < CaO < CuO
14. Which of the following halides is least stable and has
3. The correct order of catenation is
(a) C > Sn > Si ≈ Ge
(c) C > Si > Ge ≈ Sn
12. Identify the correct order of acidic strength of
(2003, 1M)
with water.
Statement II SiCl 4 is ionic and CCl 4 is covalent.
(2001, S, 1M)
Fill in the Blanks
19. A liquid which is permanently supercooled is frequently
called ……… .
(1997, 1M)
p-Block Elements-I 221
20. The recently discovered allotrope of carbon (e.g. C60 ) is
commonly known as ……… .
(1994, 1M)
21. The hydrolysis of trialkyl chlorosilane R3 SiCl, yields ………
.
(1994, 1M)
22. The hydrolysis of alkyl substituted chlorosilanes gives
……… .
(1991, 1M)
28. Draw the structure of a cyclic silicate, (Si 3 O9 )6– with proper
labelling.
(1998, 4M)
29. Write the balanced equation for the preparation of crystalline
silicon from SiCl 4 .
columns Y and Z. Match the appropriate entries.
X
True/False
Z
Fermentation
Ethanol
(1993, 1M)
Mica
Graphite
Abrasive
(1993, 1M)
Superphosphate
Crystalline cubic
Insulator
Carbon fibres
Layer structure
Fertiliser
Rock salt
Diamond structure
Reinforced
plastics
Carborundum
Bone ash
Preservative
25. Graphite is a better lubricant on the moon than on the earth.
(1987, 1M)
26. Carbon tetrachloride burns in air when lighted to give
phosgene gas.
Y
Yeast
23. The tendency for catenation is much higher for C than for Si.
24. Diamond is harder than graphite.
(1990, 1M)
30. Each entry in column X is in some way related to the entries in
(1983, 1M)
Subjective Questions
(1989, 3M)
27. Starting from SiCl 4 , prepare the following in steps not
exceeding the number given in parenthesis (give reactions
only).
(i) Silicon
(ii) Linear silicon containing methyl group only
(2001, 5M)
(iii) Na 2SiO3
31. Give reasons for the following in one or two sentences :
“Graphite is used as a solid lubricant.”
(1985, 1M)
32. Give reason for the following in one or two sentences :
“Solid carbon dioxide is known as dry ice.”
(1983, 1M)
33. Carbon acts as an abrasive and also as a lubricant, explain.
(1981, 1M)
Answers
Topic 2
Topic 1
1. (d)
2. (c)
3. (a)
4. (a)
1. (d)
2. (c)
3. (c)
4. (b)
5. (b)
6. (a)
7. (c)
8. (c)
5. (b)
6. (d)
7. (d)
8. (b)
9. (b)
10. (a)
11. (a)
12. (b)
13. (b)
14. (a, b, c)
15. (a, c, d)
16. (b, d)
17. (a, b, c)
18. (4.00)
19. (a)
20. (a)
21. (a)
22. (A → q; B → r; C → p ; D → q, r)
23. (Three centre two electron bond or banana bond)
24. (F)
25. (F)
27. (3)
9. (d)
10. (a)
11. (c)
12. (a)
13. (a)
14. (d)
15. (a,c)
16. (b, d)
17. (a)
18. (c)
19. (glass)
20. (Buckminster)
21. ( R3SiO) 2
22. (silicones)
23. (T)
25. (T)
26. (F)
24. (T)
Hints & Solutions
Topic 1 Group 13 Elements
inflammable nature, it catches fire spontaneously when exposed
to air and burns in oxygen releasing an enormous amount of
energy as:
B2H6 + 3O2 → B2O3 + 3H2O + 1976 kJ/mol
It gets hydrolysed readily to give boric acid.
B2H6 + 6H2O → 2H3BO3 + 6H2 ↑
1. Road map of the given reactions is as follows:
+LiBH4
(THF)
H3N3B3Cl3
(A)
+ (C)
(B),
Inorganic
benzene
Borane
H3N3B3(Me)3
Cl H H
MeMgBr (C) H
H
H
H
N
N
B
N
B
N
B
N
H
B
B
Cl
N
H
H
H
H
+LiBH4
(THF)
The presence of two oxidation states in p-block elements is due
to the inert pair effect.
Because of the presence of poor shielding d and f -orbitals, as we
move from Ga to Tl, effective nuclear charge of these elements
increases so as to hold the valence ns2 electrons tightly. It causes
difficulty to the ionisation of ns2-electrons and it remains inert,
only np1-electron ionises to give + 1oxidation state.
H
H
+
B–
–
+
N
N
B
(B3N3H6)
(Borazine)
H
elements will be:
B3 + > Al 3 + > Ga 3 + > In 3 + >>Tl 3+
(order of + 3oxidation state)
B+ <<Al + < Ga + < In + < Tl +
(order of + 1 oxidation state)
H
Cl H H
H
B
Dihydrogen
5. The stability order of + 3 and + 1 oxidation states of group 13
Complete reactions are as follows :
B3N3H3(Me)3
Orthoboric acid
B
N+
+
H
H
6. The structure of B2H6 can be shown as :
B
Sodium
metaborate
acidic nature of oxides decreases and the basic nature of oxides
increases on moving from B to Tl. This is because as we move
down the group, the atomic size of elements goes on increasing,
whereas the ionisation energy decreases, due to which the
strength of metal oxide (MO) bond goes on decreasing. Thus,
boron trioxide or boron oxide is acidic and reacts with basic
oxides to give metal borates. Aluminium and gallium oxides are
amphoteric while oxides of indium and thallium are basic in
nature.
4. Diborane (B2H6 ) reacts independently with O2 and H2O to
produce B2O3 and H3BO3 respectively. Diborane is a colourless,
highly toxic gas, having boiling point 180 K. Because of its
H
B sp
H
H
elements,
Group-1 Group-2 Group-13 Group-19
Period 2 ⇒ Li
Boric anhydride is non-volalite. When it react with Cr(III) salt
then deep green complex is formed.
2Cr 3+ + 3B 2O 3 → 2Cr(BO 2)3
3. All the given statements are correct. For group 13 elements, the
H
B
e H
–2
2c 3
7. Let, us consider the electronegativity values of the given
glassy bead
Hence, option (c) is correct.
H
3c–3e
In B2H6, four 2-centre-2-electron (2c − 2e) bonds are present in
the same plane and two 3-centre-2-electron (3c − 2e) bonds are
present in another plane.
14442444
3
Deep green
Þ
H
∆
Boric
anhydride
B
H
Na2B 4O 7 ⋅10H2O → Na2B 4O 7 +10H2O ↑
∆
H
H
(sodium pyroborate), Na2B 4O 7 ⋅10H2O on heating gets fused and
lose water of crystallisation. It swells up into fluffy white porous
mass which melts into a colourless liquid which later form a clear
transparent glassy bead consisting of boric anhydride and
sodium metaborate.
Na2B 4O 7 → B 2O 3 + 2NaBO 2
H
H
2. Borax bead test is performed only for coloured salt. Borax
(1 . 0)
Period 3 ⇒
Be
B
(2 . 0)
(1 . 5)
C
(2 . 5)
Al
(1. 5)
Be and Al show diagonal relationship which is based on their
Z∗
same
value (Z* is effective nuclear charge, r = atomic
r
radius). So, they have similar electronegativity.
8. Due to inert pair effect, group-13 elements (ns2np1 ) show + 3and
+ 1 oxidation states in their compounds. Stability order of these
oxidation states will be as,
+ 3 oxidation states
l
B3+ > Al3+ > Ga 3+ > In3+ > Tl3+
B3+ does not exist in free states. All B(III) compounds are
covalent.
l
+ 1 oxidation states
B+ < Al+ < Ga + < In + < Tl+
B+ does not exist in ionic as well covalent compounds.
p-Block Elements-I 223
9. Due to poor shielding of d-orbital in Ga, atomic radius of Ga is
smaller than that of Al. Thus, Ga < Al < In < Tl.
10. Orthoboric acid is a very weak acid, direct neutralisation does
15. Na 2B4O7 ⋅ 10H2O (borax) is actually made of two tetrahedral
and two triangular units, and is actually written as
Na 2[B4O5 (OH)4 ]⋅ 5H2O.
not complete. However, addition of cis-diol allow the reaction
to go to completion by forming a stable complex with
[B(OH)4 ]− as:
HO
HO
B
OH
OH
OH
sp3
s
sp2
O—B O
HO—B
B—OH
O
O—B O
2
s
sp
sp3 OH
CH2  OH
2
→
CH2  OH
+
H2C  O

H2C  O
B
O  CH2
 + 2H2O
O  CH2
11. Orthoboric acid is a weak, monobasic, Lewis acid.
OH
has deficiency of a lone-pair
(Lewis acid)
HO—B
OH
pπ - pπ backbonding between ‘B’ and ‘O’ decreases acid
strength greatly :
pπ–p
π
HO—B—OH
OH
12. In BCl 3, bond angle = 120°.
In PCl 3 , AsCl 3 and BiCl 3 , central atom is sp3 hybridised. Since
P, As and Bi are from the same group, bond angle decreases
down the group. Hence, overall order of bond angle is :
B > P > As > Bi
(a) Thus, correct.
(b) Boron atoms are in different planes thus, incorrect.
(c) Two sp2 and two sp3-hybridised B atoms thus, correct.
(d) Each boron has one OH group thus, correct.
16. (a) It does not undergo self ionisation in water but accepts an
electron pair from water, so it behaves as weak monobasic
acid.
H3BO3 + H2O r B(OH)4− + H+
Hence, (a) is incorrect.
(b) When treated with 1, 2-dihydroxy or polyhydroxy
compounds, they form chelate (ring complex) which
effectively remove [ B(OH)4 ]− species from solution and
thereby produce maximum number of H3O+ or H+ ions, i.e.
results in increased acidity.
(c) Boric acid crystallises in a layer structure in which planar
triangular BO3−
3 ` ions are bonded together through hydrogen
bonds.
One trigonal
planar B(OH)3 unit
H
H
H
13. Graphite has layered structure and conducted electricity
moderately. Silica and diamond have 3-dimensional network
structures and non-conducting.
14. (a)
CH3
B
Al
H
H
Al
C
CH3
(b) BCl 3 is stronger Lewis acid than AlCl 3 due to greater extent
of pπ − pπ back bonding in AlCl 3.
(c)
Three centre four
electron bond
Cl
Cl
Al
Cl
(d)
B
H
B
H
H
B
H
H
H
H
H
(d) In water the pK a value of H3BO3 is 9.25.
H3BO3 + H2O r B(OH)−4 + H+ ; pKa = 9.25
So, it is a weak electrolyte in water.
primary and secondary amine while tertiary amine brings about
symmetrical cleavage of B2H6 as :
B
H
H
Cl
H
H
H
17. Diborane (B2H6 ) undergoes unsymmetric cleavage with NH3 ,
Three centre two
electron bond
H
H
Cl
Al
Cl
B
H
H
H H H
B
H H
H
H
H H
H
CH3
C
CH3
H
H
B
H
Three centre two
electron bond
H H H
H
H
H
H
H
H
H
+
B
B
H
H
Unsymmetric
cleavage
NH3 or 1° amine
or 2° amine
X
+
–
[BH2(X)2] [BH4]
224 p-Block Elements-I
H
H
H
H
23. Three centred two electron bonds.
+ 2R3N
B
B
2H3B(R3N)
H
H
H
Symmetrical cleavage
H
18. N 2O, N 2O 4 , H2S2O 3 and H2S2O 8 molecules are containing
covalent bond between two atoms.
O
S
,
O
O
(N2O4)
O
O
S
O¾O
HO
O
(H2S2O8)
OH
H
O
B
H
(B2H6)
H
Cl
Therefore, from 3 moles of B2H6 , 6 moles of B(OCH3 )3 will be
28. 3KF + AlF3 → B2H5Cl + H2
K3AlF6 + 3BF3 → AlF3 ↓ + 3KBF4
29. (i) Na 2B4O7 + HCl → NaCl + H3BO3
B
B
N
H
(B3N3H6)
B 2H6 and B 3N 6H6 have polar bond, but do not have same kind of
atom.
19. Small size and high charge on B3+ makes it highly polarising.
Therefore, in most of its compounds, boron forms covalent bonds.
Hence, both statement I and statement II are correct and
statement II is a correct explanation of statement I.
20. Orthoboric acid is a weak, monobasic, Lewis acid and the poor
acidic character is due to pπ − pπ backbondings as:
pπ–pπ
Backbonding decreases electron
deficiency at boron, decreases
HO—B—OH
its Lewis acid strength.
OH
21. Due to small size and high charge on Al in Al(OH)3 the fission
ability of Al—O and O—H bonds become comparable and
compound can give both H+ and HO− under appropriate
reaction conditions as:
Al(OH)3 + 3HCl → AlCl 3 + 3H2O
Base
Al(OH)3 + NaOH → Na[Al(OH)4 ]
Acid
Therefore, both statements are correct and statement II is a
correct explanation of statement I.
22. (A) Bi
Cl
ppt
H
3+
Cl
The bridged Al—Cl bonds
are different from terminal
Al—Cl bonds.
Al
26. B2H6 + 6 CH3OH → 2[B(OCH3 )3 ] + 6H2
H
N
N
Cl
Cl
27. BenAl 2Si6O18 , 2n + 6 + 24 − 36 = 0 ⇒ n = 3
H
|
B
H
H
H
B
,
25. In Al 2Cl6 , Al–Cl bonds are not equivalent :
Al
H
S
bottom due to increase in electropositive character.
Cl
(H2S2O3)
O
H
H
24. The basic nature of hydroxide of group-13 increases from top to
O
N¾N
OH
HO
Covalent bonds
B
Three centred two electron
bridged B—H—B bonds
S
N ºº N ¾® O ,
(N2O)
H
H
B
∆
H3BO3 + HCl →
BCl 3 + H2O
∆
BCl 3 + Al →
B + AlCl 3
H
H
H
(ii) B2H6 :
B
B
H
H
H
It has 4 terminal B—H bonds. There are two B—H—B, three
centred two electron bridged bonds.
B2H6 + HCl → B2H5Cl + H2
LiAlH4
30. Compound X →
Y a hydride + other compound. Hydride
Y contains 21.72% hydrogen.
∆
Y + O2 → B2O3 + H2O
Therefore, Y is a hydride of boron and it is obtained by reduction
of X with LiAlH4. So, X is either BCl 3 or BF3.
4BCl 3 + LiAlH4 → B2H6 + 3AlCl 3 + 3LiCl
1442443
X
Y
Y
Structure of Y (B2H6 )
+
hydrolysis to (BiO)  q.
(B) [AlO2 ]− exist in basic medium, on acidification gives
Al(OH)3  r.
(C) Orthosilicate (SiO4−
4 ) on heating changes into pyrosilicate
Si 2O6−

p.
7
(D) Tetraborate ion [B4O72− ] on treatment with dil. acid
hydrolysis gradually to orthoboric acid  q, r.
Other products
Molar mass of B2H6 = 2 × 11 + 6 = 28
6
% of H in B2H6 =
× 100 = 21.5 ≈ 21.72
28
B2H6 + 3O2 → B2O3 + 3H2O + Heat
3
1.3
H
B
Å
H
97°
1.1
B
9ÅH
122°
H
H
H
1.77 Å
p-Block Elements-I 225
(a) There are 4 terminal B—H bonds.
(b) There are two 3-centre-2-electron B—H—B bridged bonds.
(c) Terminal H—B—H planes are perpendicular to bridged
B—H—B bonds.
31. Al 2S3 + 6H2O → 2Al(OH)3 ↓ + 3H2S(g ) ↑
Foul odour
Foul odour on damping of Al 2S3 is due to the formation of H2S
gas as shown above.
32. The total hydration energy of AlCl 3
= Hydration energy of Al 3+ + 3 × Hydration energy of Cl −
= − 4665 + 3 (− 381) kJ/mol
= − 5808 kJ/mol
The above hydration energy is more than the energy required for
ionisation of AlCl 3 into Al 3+ and 3Cl − .
Due to this reason, AlCl 3 becomes ionic in aqueous solution. In
aqueous solution, it is ionised completely as
AlCl 3 + 6H2O → [Al(H2O)6 ]3+ + 3Cl −
Topic 2 Groups 14 Elements
1. The C C bond length is maximum in diamond having value
154 pm. Here, each carbon atom undergoes sp3 hybridisation and
linked to four other carbon atoms by using hybridised orbitals in
tetrahedral fashion. It has a rigid three-dimensional network of
carbon atoms.
C C bond length within the layers of graphite is 141.5 pm. In
C60, C C distances between single and double bonds are 143.5
pm and 138.3 pm respectively.
2. The basic structural unit of feldspar,
zeolites, mica and asbestos is (SiO4 )4− . These all are silicates.
All silicates involve two types of Si  O bonds.
(i) Terminal Si O bonds in which oxygen is bonded to a
silicon and not other atom.
(ii) Bridging Si  O Si bonds in which oxygen is bonded to
two silicon atoms.
s
O–
O–
Si
O–
O–
s
s
In SiO4−
4 ion, each Si atom is bonded to four oxygen atoms
tetrahedrally.
3. Catenation property is an unique property of group 14 elements.
Down the group 14, catenation power decreases as:
C > Si > Ge ≈ Sn
Pb does not show catenation.
4. Silica occurs in nature in several amorphous and crystalline
forms. Kieselguhr is the amorphous form of silica. Quartz,
tridymite and cristobalite are crystalline forms of silica.
5. C60 is aromatic allotrope of carbon containing 12 pentagons and
20 hexagons. It is a fullerene having a shape like soccer ball and
called Buckminster fullerene.
6. The property of self-linking of atoms of an element through
covalent bonds to form straight or branched chains and rings of
different sizes is called catenation. Down the group, catenation
tendency decreases due to decrease in element bond strength.
Carbon (C), silicon (Si), germanium (Ge), tin (Sn), lead (Pb) are
group-14 elements.
Catenation tendency is highest in carbon while silicon has second
highest tendency of catenation among all elements of family due
to higher bond energy. The decreasing tendency of catenation
among group 14 elements is as follows:
C >> Si > Ge ≈ Sn
However, Pb does not show catenation.
7. Carbon (C) has greatest ability to form stable pπ-pπ multiple
bonds. 2p-orbitals of this element participate in the process. The
stability of multiple bonds of C is attributed to their closeness
with C-nucleus. Thus, the smaller size of C plays a significant
role in the process.
8. The compounds given are the tetrahalides
(MCl 4) of group 14 elements. For the hydrolysis, (nucleophilic
substitution) of MCl 4 the nature of the M—Cl bond should be as:
δ+
δ–
M——Cl
It must expand its covalency beyond 4 by the use
of its vacant d-orbital which will accommodate the
lone pair of electrons of H2O (the–nucleophile).
sp3
Cl3 M——Cl
d + d – OH2
d0
sp3d
Cl3 M——Cl
H
O
H
Transition state
M(OH)4
3H2O
–3HCl
–HCl
sp3
Cl3 M—(OH)
Here, M can be Si, Sn and Pb because they have vacant
nd-orbital. But, carbon is a member of second period (n = 2,
l = 0, 1),
it does not have d-orbital (l = 2). So, CCl 4 will not be hydrolysed
and correct option is (b).
9. Silicones are polysiloxanes with general chemical formula,
[R2SiO]n, where R is an organic group such as:
CH3, C2H5, C6H5 etc.
Silicones have many useful properties:
(i) They repel water and form watertight seals.
(ii) They are heat resistant because of constancy of properties
over a wide range of temperature (− 100° to 250° C).
(iii) Silicones are non-toxic.
(iv) Silicones are biocompatible because these do not support
microbiological growth and these have high gas permeability
at room temperature.
(v) They are resistant to O2, O3 and UV-radiation.
(vi) Silicones are formulated to be electrically insulative.
226 p-Block Elements-I
(vii) Silicone grease is typically used as a lubricant for brake
components in automobiles, since it is stable at high
temperature, is not water soluble and is a odourless viscous
liquid.
10. In sheet silicates, three out of four oxygen of SiO4−
4 unit are
shared as shown below :
–
II
II
PbCl 2 + 2HCl → H2[ PbCl 4 ]
Hence, statement (c) is incorrect.
+2
+4
(d) Pb3O4 is a mixture of (2 PbO + PbO2 )
–
+4
+2
Pb3O4 + 4 HNO3 → 2Pb(NO3 )2 + PbO2 + 2H2O
–
It is not a redox reaction. Thus, the statement (d) is incorrect.
–
–
–
(c) In conc. HCl, PbCl 2 exists as chloroplumbous acid,
H2[PbCl 4 ]
–
–
–
–
–
Three oxygens of every
tetrahedra are
shared with others
–
–
–
–
–
–
–
–
–
–
In pyrosilicates, there is only one shared oxygen, in linear chain
silicates, two oxygen per tetrahedra are shared while in
three-dimensional silicates, all four oxygens are shared.
11. Me2SiCl 2 on hydrolysis yields a linear chain silicone as :
CH3
CH3


Cl  Si  Cl + 2H2O → HO  Si  OH + 2HCl


CH3
CH3
16. Diamond has a three-dimensional network structure, a hard
substance where graphite is soft due to layered structure.
In graphite, only three valence electrons are involved in bonding
and one electron remain free giving electrical conductivity. In
diamond, all the four valence electrons are covalently bonded
hence, insulator.
Diamond is better thermal conductor than graphite. Electrical
conductivity is due to availability of free electrons, thermal
conduction is due to transfer of thermal vibrational energy from
one atom to another atom. A compact and precisely aligned
crystals like diamond thus facilitate better movement of heat.
In graphite C  C bond acquire some double bond character,
hence, higher bond order than in diamond.
17. In group 13, 14, 15 as we descend down in group, the higher
oxidation state becomes less tenable due to inert pair effect.
Therefore, lead show +2 as stable oxidation state. Hence, Pb4+
act as a strong oxidising agent, itself reduced to Pb2+ very easily.
Both statement I and statement II are correct and statement II is a
correct explanation of statement I.
18. SiCl 4 reacts with water due to vacant d-orbitals available with
Si as:
CH3

Polymerisation
nHO  Si  OH →

CH3
CH3


[ O  Si  O ]n

CH3
12. CO2 is acidic oxide, H2O is neutral, CaO is strongly basic and
CuO is weakly basic. Therefore, order of acid strength is :
CaO < CuO < H2O < CO2
13. Carbon monoxide is a neutral oxide, all others are amphoteric :
H2O
SiCl4
No such vacant d-orbitals are available with carbon, hence CCl 4
does not react wtih water. Otherwise, both SiCl 4 and CCl 4 are
covalent.
Statement I is correct but statement II is incorrect.
SnO2 + 4HCl → SnCl 4 + 2H2O
19. Glass is commonly known as supercooled liquid.
ZnO + 2HCl → ZnCl 2 + H2O
20. Buckminster fullerene is the name of recently discovered
SiO2 + 2NaOH → Na 2SiO3 + H2O
SnO2 and ZnO also react with NaOH. SiO2 is also attacked by
H3PO4.
14. PbI4 is least stable, has doubtful existence. It is due to inert pair
effect, the stable oxidation state of lead is + 2.
15. (a) Sn 2+ of stannous chloride dihydrate (SnCl 2 ⋅ 2H2O) tends to
convert into Sn 4+ .
Hence, statement (a) is correct.
(b) SnO2 reacts with KOH and gives K2SnO3 ⋅ 3H2O or
K2[ Sn(OH)6 ]because it is amphoteric in nature.
SnO2 + KOH → K2SnO3 + H2O or K2[ Sn(OH)6 ]
Hence, statement (b) is correct.
allotrope of carbon.
21. After dimerisation, no reactive function group remains.
R

R2SiCl + H2O → R  Si  OH
− HCl

R
R
R


→ R  Si  O  Si  R


R
R
Dimeric silicone
p-Block Elements-I 227
22. Silicones are organosilicon polymers, obtained by hydrolysis of
heat
Si(OH)4 →
SiO2 + 2H2O
alkyl substituted chlorosilanes.
∆
SiO2 + Na 2CO3 →
Na 2SiO3 + CO2
23. Due to smaller size of carbon than silicon, C—C bond is stronger
than Si—Si bond, hence former is more likely to extend than
later.
24. Graphite has a layered structure of hexagonal carbon rings
O–
–
stacked one over other which makes it slippery.
O
On the other hand, in diamond, each carbon is tetrahedrally bond
to other four carbons extended in three dimensional space, giving
a giant, network structure. Due to this reason, diamond is harder
than graphite.
25. Graphite is better lubricant on moon than on earth because of
absence of gravitational pull on the moon.
26. Phosgene gas is obtained by treatment of CCl 4 with superheated
steam :
CCl 4 + H2O (vapour) → COCl 2 + 2HCl
27.
∆ 3Si + 4AlCl ;
(i) 3SiCl 4 + 4Al →
3
Mg or Zn can also be used.
(ii) SiCl 4 + 2CH3MgCl → (CH3 )2 SiCl 2 + 2MgCl 2
OH

− HCl
(CH3 )2 SiCl 2 + H2O → CH3  Si  CH3

OH
CH3
CH3
CH3



→  O  Si  O  Si  O  Si  O



CH3
CH3
CH3
(iii) SiCl 4 + 4H2O → Si(OH)4 + 4HCl
Unstable
O = Oxygen
= Silicon
28.
6–
O
Si3O9
O
O
29. 3SiCl 4 + 4Al
Vapour
30.
Molten
X
∆
→
O
4AlCl 3 +
Volatilizes
3Si
Crystalline
Y
Z
Yeast
Fermentation
Ethanol
Mica
Layered structure
Insulator
Superphosphate
Bone ash
Fertiliser
Carbon fibres
Graphite
Reinforced plastics
Rock salt
Crystalline cubic
Preservative
Carborundum
Diamond structure
Abrasive
31. Graphite has layered structure and the adjacent layers are weakly
associated giving slippery nature, used as solid lubricant.
32. Carbon dioxide solidifies at very low temperature, hence solid
CO2 is very cold, commonly known as dry ice. Also solid carbon
dioxide sublime, without passing through liquid state.
33. The two common allotropes of carbon are diamond and graphite.
Diamond is the hardest, natural, substance, used as an abrasive
while graphite is soft, used as a lubricant.
16
p-Block Elements-II
Topic 1 Elements and Compounds of Group 15 and 16
Objective Questions I (Only one correct option)
1. The correct statement among the following is
(2019 Main, 12 April I)
(a) (SiH3 )3 N is planar and less basic than (CH3 )3 N.
(b) (SiH3 )3 N is pyramidal and more basic than (CH3 )3N.
(c) (SiH3 )3 N is pyramidal and less basic than (CH3 )3 N.
(d) (SiH3 )3 N is planar and more basic than (CH3 )3 N.
2. The number of pentagons in C60 and trigons (triangles) in
white phosphorus, respectively, are
(a) 20 and 3
(c) 20 and 4
(2019 Main, 10 April II)
(b) 12 and 4
(d) 12 and 3
(a) H2S2O3
(c) H2S2O7
(2019 Main 10 April I)
(b) H2S2O4
(d) H2S4O6
4. The correct order of the oxidation states of nitrogen in NO,
NO2 , NO2 and N2 O3 is
(2019 Main, 9 April I)
(a) NO2 < NO < N2O3 < N2O
(b) N2O < NO < N2O3 < NO2
(c) O2 < N2O3 < NO < N2O
(d) N2O < N2O3 < NO < NO2
5. The pair that contains two PH bonds in each of the
oxoacids is
(a) H4P2O5 and H4P2O6
(c) H4P2O5 and H3PO3
(2019 Main, 10 Jan II)
(b) H3PO3 and H3PO2
(d) H3PO2 and H4P2O5
6. When the first electron gain enthalpy (∆ e g H ) of oxygen is
− 141 kJ/ mol, its second electron gain enthalpy is
(2019 Main, 9 Jan II)
(a) a positive value
(b) a more negative value than the first
(c) almost the same as that of the first
(d) negative, but less negative than the first
7. Good reducing nature of H3 PO2 is attributed to the presence
of
(a) two P H bonds
(c) two P OH bonds
thermal decomposition is
(a) Ba(N3 )2
(c) NH4NO2
(2018 Main)
(b) (NH4 )2 Cr2O7
(d) (NH4 )2 SO4
9. The order of the oxidation state of the phosphorus atom in
H3 PO2 , H3 PO4 , H3 PO3 and H4 P2 O6 is
(2017 Adv.)
(a) H3PO4 > H3PO2 > H3PO3 > H4P2O6
(b) H3PO4 > H4P2O6 > H3PO3 > H3PO2
(c) H3PO2 > H3PO3 > H4P2O6 > H3PO4
3. The oxoacid of sulphur that does not contain bond between
sulphur atoms is
8. The compound that does not produce nitrogen gas by the
(2019 Main, 9 Jan II)
(b) one P H bond
(d) one P  OH bond
(d) H3PO3 > H3PO2 > H3PO4 > H4P2O6
10. The species in which the N-atom is in a state of
sp hybridisation is
(2016 Main)
(a) NO−2
(b) NO−3
(c) NO2
(d) NO+2
11. The pair in which phosphorus atoms have a formal oxidation
state of +3 is
(2016 Main)
(a) pyrophosphorous and hypophosphoric acids
(b) orthophosphorous and hypophosphoric acids
(c) pyrophosphorous and pyrophosphoric acids
(d) orthophosphorous and pyrophosphorous acids
12. The product formed in the reaction of SOCl 2 with white
phosphorus is
(a) PCl 3
(c) SCl 2
(2014 Adv.)
(b) SO2Cl 2
(d) POCl 3
13. Which of the following properties is not shown by NO?
(a) It is paramagnetic in liquid state
(2014 Main)
(b) It is a neutral oxide
(c) It combines with oxygen to form nitrogen dioxide
(d) Its bond order is 2.5
14. Concentrated nitric acid upon long standing, turns
yellow-brown due to the formation of
(a) NO
(c) N 2O
(b) NO2
(d) N 2O4
(2013 Main)
p-Block Elements-II
15. Which of the following is the wrong statement? (2013 Main)
(a)
(b)
(c)
(d)
ONCl and ONO − are not isoelectronic
O3 molecule is bent
Ozone is violet-black in solid state
Ozone is diamagnetic gas
27. Polyphosphates are used as water softening agents because
they
16. The reaction of white phosphorus with aqueous NaOH gives
229
(2002, 3M)
(a) form soluble complexes with anionic species
(b) precipitate anionic species
(c) form soluble complexes with cationic species
(d) precipitate cationic species
phosphine alongwith another phosphorus containing
compound. The reaction type, the oxidation states of
phosphorus in phosphine and the other product respectively
are
(2012)
28. The number of S  S bonds in sulphur trioxide trimer,
(a)
(b)
(c)
(d)
29. Ammonia can be dried by
redox reaction, − 3 and − 5
redox reaction, 3 and + 5
disproportionation reaction, − 3 and + 5
disproportionation reaction, − 3 and + 3
decreasing order of the oxidation state of nitrogen?
(2012)
(b) HNO3 , NO, N2 , NH4Cl
(d) NO, HNO3 , NH4Cl, N2
18. Extra pure N 2 can be obtained by heating
(a) NH3 with CuO
(c) (NH4 )2 Cr2O7
(2011)
(b) NH4NO3
(d) Ba(N3 )2
(a) dry O2
(b) a mixture of O2 and N2
(c) moist O2
(2009)
(d) 75
(2005, 1M)
(b) FeSO4
(d) K 2MnO4
concentrated HNO3?
(2005)
(b) O2
(d) N2O
23. A pale blue liquid obtained by equimolar mixture of two
gases at – 30°C is
(2005, 1M)
32. The number of P  O  P bonds in cyclic metaphosphoric
(2000, 1M)
thermodynamically most stable?
25. Which of the following has —O—O— linkage?
(2005, 1M)
(d) four
(1999, 2M)
(1996, 1M)
(a) reducing Na 2SO4 solution with H2S
(b) boiling Na 2SO3 solution with S in alkaline medium
(c) neutralising H2S2O3 solution with NaOH
(d) boiling Na 2SO3 solution with S in acidic medium
35. There is no S  S bond in
(a) S 2O42–
(b) S 2O52–
(1991, 1M)
(c) S2O32–
(d) S2O72–
(b) NH3
(c) PH3
(d) SbH3
37. Amongst the trihalides of nitrogen, which one is least basic?
(b) NCl 3
(d) NI3
(a) NF3
(c) NBr3
(1987, 1M)
38. Which of the following oxides of nitrogen is a coloured gas ?
(2004, 3M)
(b) H2S2O8
(d) H2S4O6
(a) H3PO3 is dibasic and reducing
(b) H3PO3 is dibasic and non-reducing
(c) H3PO4 is tribasic and reducing
(d) H3PO3 is tribasic and non-reducing
(c) three
(a) one mole of phosphine
(b) two moles of phosphoric acid
(c) two moles of phosphine
(d) one mole of phosphorus pentaoxide
(a) AsH3
(b) White
(d) Yellow
26. For H3PO3 and H3PO4, the correct choice is
(b) two
36. Which one of the following is the strongest base? (1989, 2M)
(b) N2O3
(d) N2O5
24. Which of the following isomers of phosphorus is
(a) H2S2O6
(c) H2S2O3
(2000, 1M)
(b) CO2 > N2O5 > SO3
(d) K 2O > CaO > MgO
34. Sodium thiosulphate is prepared by
22. Which gas is evolved when PbO2 is treated with
(a) Red
(c) Black
(a) Cl 2O7 > SO2 > P4O10
(c) Na 2O > MgO > Al 2O3
water gives
(2007, 3M)
(c) 50
21. Which of the following is not oxidised by O3 ?
(a) N2O
(c) N2O4
(2000, 1M)
33. One mole of calcium phosphide on reaction with excess
P—P bonds in P4 is
(a) NO2
(c) N2
highest boiling point is
(a) H2O because of hydrogen bonding
(b) H2Te because of higher molecular weight
(c) H2S because of hydrogen bonding
(d) H2Se because of lower molecular weight
acid is
20. The percentage of p-character in the orbitals forming
(a) KI
(c) KMnO4
30. Amongst H2 O, H2 S, H2 Se and H2 Te, the one with the
(a) zero
(d) O2 in the presence of aqueous NaOH
(b) 33
(2000, 1M)
(b) P4O10
(d) anhydrous CaCl 2
31. The correct order of acidic strength is
19. The reaction of P4 with X leads selectively to P4 O6 . The X, is
(a) 25
(2001, 1M)
(b) two
(d) zero
(a) conc. H2SO4
(c) CaO
17. Which ordering of compounds is according to the
(a) HNO3 , NO, NH4Cl, N2
(c) HNO3 , NH4Cl, NO, N2
(S3 O9 ) is
(a) three
(c) one
(b) NO
(d) NO2
(a) N2O
(c) N2O4
39. The bonds present in N2 O5 are
(2003, 1M)
(a) only ionic
(c) only covalent
(b) O2
(1986, 1M)
(b) covalent and coordinate
(d) covalent and ionic
40. A gas that cannot be collected over water is
(a) N2
(1987,1M)
(c) SO2
(1985, 1M)
(d) PH3
230 p-Block Elements-II
41. Ammonia gas can be dried by
(a) conc H2SO4
(c) CaCl 2
(1978, 1M)
(b) P2O5
(d) quicklime
42. Which of the following is incorrect statement?
51. The total number of compounds having at least one bridging
(1978, 1M)
(a) NO is heavier than O2
(b) The formula of heavy water is D2O
(c) N2 diffuses faster than oxygen through an orifice
(d) NH3 can be used as a refrigerant
(2018 Adv.)
Assertion and Reason
Read the following questions and answer as per the direction
given below:
43. The compound(s) which generate (s) N2 gas upon thermal
decomposition below 300°C is (are)
(2018 Adv.)
(b) (NH4 )2 Cr2O7
(d) Mg3N2
44. Based on the compounds of group 15 elements, the correct
statement(s) is (are)
(2018 Adv.)
(a) Bi 2O5 is more basic than N2O5
(b) NF3 is more covalent than BiF3
(c) PH3 boils at lower temperature than NH3
(d) The N—N single bond is stronger than the P—P single bond
45. The nitrogen containing compound produced in the reaction
of HNO 3 with P4 O10
(2016 Adv.)
(a)
(b)
(c)
(d)
(d) Ν 2Ο 5
53. Statement I The electronic structure of O3 is
⊕
Os
O
(2013 Adv.)
O
O
(1998, 2M)
54. Statement I HNO3 is a stronger acid than HNO2 .
(2009)
(c) N2 O4
in the atmosphere but these do not react to form oxides of
nitrogen.
Statement II The reaction between nitrogen and oxygen
requires high temperature.
(1998, 2M)
Statement II The following structure is not allowed
because octet around O cannot be expanded.
47. The nitrogen oxide(s) that contain(s) N—N bond(s) is/are
(b) N2O3
52. Statement I Nitrogen and oxygen are the main components
O
O — O bond lengths are equal
thermal decomposition of O3 is endothermic
O3 is diamagnetic in nature
O3 has a bent structure
(a) N2O
(a) Statement I is correct, Statement II is correct, Statement II
is the correct explanation of Statement I
(b) Statement I is correct, Statement II is 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
O
(a) can also be prepared by reaction of P4 and HNO 3
(b) is diamagnetic
(c) contains one NN bond
(d) reacts with Na metal producing a brown gas
46. The correct statement(s) about O 3 is/are
oxo group among the molecules given below is ……… .
N 2O 3, N 2O 5, P4O 6, P4O7 , H 4 P2O 5 , H 5P3O10, H 2S 2O 3,
H 2S 2O 5
Objective Questions II
(One or more than one correct option)
(a) NH4NO3
(c) Ba(N3 )2
Integer Answer Type Question
Statement II In HNO3 , there are two nitrogen to oxygen
bonds whereas in HNO2 there is only one.
(1998, 2M)
55. Statement I Although PF5 , PCl 5 and PBr5 are known, the
pentahalides of nitrogen have not been observed.
48. Ammonia, on reaction with hypochlorite anion, can form
Statement II Phosphorus has lower electronegativity than
nitrogen.
(1994, 2M)
(1999, 3M)
(a) NO
(b) NH4Cl
(c) N2H4
(d) HNO2
49. White phosphorus (P4 ) has
Passage Based Questions
Passage 1
(1998, 2M)
(a) six P  P single bonds
(b) four P  P single bonds
(c) four lone pairs of electrons
(d) PPP angle of 60°
50. Nitrogen (I) oxide is produced by
(a) thermal decomposition of NH4NO3
(b) disproportionation of N2O4
(c) thermal decomposition of NH4NO2
(d) interaction of hydroxylamine and nitrous acid
Upon heating KClO 3 in presence of catalytic amount of
MnO 2 , a gas W is formed. Excess amount of W reacts with
white phosphorus to give X . The reaction of X with pure
HNO 3 gives Y and Z.
(2017 Adv.)
56. Y and Z are, respectively
(1989, 1M)
(a) N2O4 and HPO3
(c) N2O3 and H3PO4
(b) N2O4 and H3PO3
(d) N2O5 and HPO3
57. W and X are, respectively
(a) O2 and P4O10
(c) O3 and P4O6
(b) O2 and P4O6
(d) O3 and P4O10
p-Block Elements-II
Codes
Passage 2
There are some deposits of nitrates and phosphates in earth’s
crust. Nitrates are more soluble in water. Nitrates are
difficult to reduce under the laboratory conditions but
microbes do it easily. Ammonia forms large number of
complexes with transition metal ions. Hybridisation easily
explains the ease of sigma donation capability of NH3 and
PH3 . Phosphine is a flammable gas and is prepared from
white phosphorus.
(2008, 3 × 4M = 12M)
(a) Phosphates have no biological significance in humans
(b) Between nitrates and phosphates, phosphates are less abundant
in earth’s crust
(c) Between nitrates and phosphates, nitrates are less abundant in
earth’s crust
(d) Oxidation of nitrates is possible in soil
59. Among the following, the correct statement is
(a) Between NH3 and PH3, NH3 is a better electron donor because
the lone pair of electrons occupies spherical ‘s’ orbital and is
less directional
(b) Between NH3 and PH3, PH3 is a better electron donor because
the lone pair of electrons occupies sp3-orbital and is more
directional
(c) Between NH3 and PH3, NH3 is a better electron donor because
the lone pair of electrons occupies sp3-orbital and is more
directional
(d) Between NH3 and PH3, PH3 is a better electron donor because
the lone pair of electrons occupies spherical ‘s’ orbital and is
less directional
60. White phosphorus on reaction with NaOH gives PH3 as one
of the products. This is a
(b) disproportionation reaction
(d) precipitation reaction
61. The unbalanced chemical reactions given in Column I show
missing reagent or condition (?) which are provided in
Column II. Match Column I with Column II and select the
correct answer using the codes given below the Columns.
(2013 Adv.)
Column I
Column II
?
PbO2 + H 2SO4 → PbSO4 + O2 +
1.
NO
2.
I2
other product
Q.
P Q R S
(b) 3 2 1 4
(d) 3 4 2 1
Fill in the Blanks
62. The lead chamber process involves oxidation of SO2 by
atomic oxygen under the influence of ………as catalyst.
(1992, 1M)
phosphorus atom is ……… .
(1992, 1M)
64. The basicity of phosphorus acid (H3 PO3 ) is ………
(1990, 1M)
65. ……… phosphorus is reactive because of its highly strained
tetrahedral structure.
(1987, 1M)
True/False
66. Nitric oxide, though an odd electron molecule, is
diamagnetic in liquid state.
(1991, 1M)
67. The H  N  H bond angle in NH3 is greater than the
H  As  H bond angle in AsH3 .
(1984, 1M)
68. In aqueous solution, chlorine is a stronger oxidising agent
than fluorine.
(1984, 1M)
2+
69. Dilute HCl oxidises metallic Fe to Fe .
(1983, 1M)
Numerical Answer Type Questions
70. The amount of water produced (in g) in the oxidation of
1 mole of rhombic sulphur by conc. HNO 3 to a compound
with the highest oxidation state of sulphur is …… (Given
data : Molar mass of water = 18 g mol −1 )
(2019 Adv.)
71. The total number of lone pair of electrons in N2O3 is
(2015 Adv.)
Match the Column
P.
P Q R S
(a) 4 2 3 1
(c) 1 4 2 3
63. In P4 O10 , the number of oxygen atoms bonded to each
58. Among the following, the correct statement is
(a) dimerisation reaction
(c) condensation reaction
231
? NaHSO +
Na2S2O3 + H 2O →
4
72. Among the following, the number of compounds that can
react with PCl 5 to give POCl 3 is O2 , CO2 , SO2 , H2 O,
(2011)
H2 SO4 , P4 O10 .
73. The total number of diprotic acids among the following is
H3PO4
H2CO3
H3PO2
H2SO4
H2S2O7
H2CrO4
H3PO3
H3BO3
H2SO3
(2010)
Subjective Questions
74. Draw the structure of P4O10.
other product
R.
? N + other product
N 2H 4 →
2
3.
Warm
S.
? Xe + other product
XeF2 →
4.
Cl 2
(2005)
75. Arrange the following oxides in the increasing order of
Bronsted basicity.
Cl2O7, BaO, SO3, CO2, B2O3
(2004)
76. Identify the compounds A, B, C, D
SO
Na 2CO3
Elemental S
I2
2
Na 2 CO3 →
A → B → C → D
∆
and give oxidation state of sulphur in each compounds.
(2003, 4M)
232 p-Block Elements-II
77. Write the balanced equations for the reactions of the
following compounds with water:
(i) Al 4C3 (ii) CaNCN (iii) BF3 (iv) NCl 3 (v) XeF4
(iii) Manufacture of phosphoric acid from phosphorus.
(iv) Reaction of aluminium with aqueous sodium hydroxide.
(2002, 5M)
(1997, 1M × 4 = 4M)
78. Give reason(s), why elemental nitrogen exists as a diatomic
85. Draw the structure of P4 O10 and identify the number of
molecule whereas elemental phosphorus is a tetra atomic
molecule?
(2000, 2M)
79. The Haber’s process can be represented by the following
scheme.
+ CO2
NH3.H2O
H2O
B
NaHCO3
sentences only.
(ii) Mg 3 N2 when reacted with water gives of NH3 but HCl is
not obtained from MgCl 2 on reaction with water at room
temperature.
(1995, 2M × 3 = 6M)
NaCl
87. Complete and balance the following reactions.
(1994, 1M)
Heat K + 5CaSO ⋅ 2H O + K
Ca 5 (PO4 )3 F + H2SO4 + H2O →
4
2
C + H2O
NH3 + H2O + E
Identify A, B, C, D and E .
88. In the following reaction, identify the compounds A and B
PCl 5 + SO2 → A + B
(1999, 5M)
(1994, 1M)
89. Complete and balance the following reaction.
80 (a) In the following equation
A + 2B + H2 O → C + 2D
(A = HNO2 , B = H2SO3 , C = NH2OH).
Identify D. Draw the structures of A, B, C and D.
(b) In the contact process for industrial manufacture of
sulphuric acid, some amount of sulphuric acid is used as a
starting material. Explain briefly. What is the catalyst used
in the oxidation of SO2?
(1999, 10M)
81. Complete and balance the following chemical equations.
(i) P4O10 + PCl 5 →
(ii) SnCl 4 + C2H5Cl + Na →
86. Account for the following. Write the answers in four or five
(iii) (SiH3 )3 N is a weaker base than (CH3 )3 N.
+D
A
(1996, 3M)
(i) The experimentally determined N  F bond lengths in NF3
is greater than the sum of the single bond covalent radii of N
and F.
CaCO3
CaO
single and double P  O bonds.
(1998, 1 M × 2 = 2M)
82. (a) Thionyl chloride can be synthesised by chlorinating SO2
using PCl5. Thionyl chloride is used to prepare
anhydrous ferric chloride starting from its hexahydrated
salt. Alternatively, the anhydrous ferric chloride can also
be prepared from its hexahydrated salt by treating with
2,2-dimethoxypropane. Discuss all this using balanced
chemical equations.
(b) Reaction of phosphoric acid with Ca3(PO4)2 yields a
fertiliser “triple superphosphate” represent the same
through balanced chemical equation.
(1998, 5M)
Red phosphorus is reacted with iodine in the presence of
water.
(1992, 1M)
P + I2 + H2 O → K + K
90. Give reasons in two or three sentences only.Sulphur dioxide
is a more powerful reducing agent in the alkaline medium
than in acidic medium.
(1992, 2M)
91. Draw the two resonance structures of ozone which satisfy
the octet rule.
(1991, 1M)
92. Give reasons in one or two sentences.
Ammonium chloride is acidic in liquid ammonia solvent.
(1991, 1M)
93. Write the balanced chemical equations for the following.
(i) Sodium nitrite is produced by absorbing the oxides of
nitrogen in aqueous solution of washing soda.
(ii) Nitrogen is obtained in the reaction of aqueous ammonia
with potassium permanganate.
(iii) Elemental phosphorus reacts with concentrated HNO3 to
give phosphoric acid.
(iv) Sulphur is precipitated in the reaction of hydrogen
sulphide with sodium bisulphite solution.
(i) Phosphorus is treated with concentrated nitric acid.
(v) Carbon dioxide is passed through a suspension of
limestone in water.
(1991, 1 × 5 = 5M)
94. Write the balanced chemical equation for the following
reactions.
(i) Aqueous solution of sodium nitrate is heated with zinc
dust and caustic soda solution.
(ii) Sodium iodate is added to a solution of sodium bisulphite
(ii) Oxidation of hydrogen peroxide with potassium
permanganate in acidic medium.
95. Write the two resonance structures of N2 O that satisfy the
83. A soluble compound of a poisonous element M, when
heated with Zn / H2 SO4 , gives a colourless and extremely
poisonous gaseous compound N, which on passing through
a heated tube gives a silvery mirror of element M. Identify
M and N.
(1997, 2M)
84. Write balanced equations for the following.
(1990, 2M)
octet rule.
(1990, 2M)
p-Block Elements-II
233
105. Write down the balanced equation for the reactions when
96. Draw balanced equations for
(i) the preparation of phosphine from CaO and white
phosphorus.
(ii) the preparation of ammonium sulphate from gypsum,
ammonia and carbon dioxide.
(1990, 2M)
97. Explain the following
(1989, 2M)
(i) H3 PO3 is a dibasic acid.
(ii) Phosphine has lower boiling point than ammonia.
98. Write the balanced chemical equations for the following.
(i) Hypophosphorous acid is heated.
(ii) Sodium chlorate reacts with sulphur dioxide in dilute
sulphuric acid medium.
(i) calcium phosphate is heated with a mixture of sand and
carbon.
(ii) ammonium sulphate is heated with a mixture of nitric
oxide and nitrogen dioxide.
(1985, 2M)
106. Draw the resonance structures of nitrous oxide.
(1985, 90, 2M)
107. Show with balanced chemical reaction what happens when
following are mixed?
Aqueous solution of ferric sulphate and potassium iodide.
(1984, 1M)
108. Write the matched set (of three) for each entry in Column A
99. Arrange the following as indicated. CO2 , N2 O5 , SiO2 , SO3
in the order of increasing acidic character.
100. Give balanced equations for the following.
(i) Phosphorus reacts with nitric acid to give equimolar ratio
of nitric oxide and nitrogen dioxide.
(ii) Carbon dioxide is passed through a concentrated aqueous
solution of sodium chloride saturated with ammonia.
(1988, 3M)
101. Give reason for “valency of oxygen is generally two, whereas
sulphur shows valency of two, four and six.”
(1988, 1M)
102. Explain the following in one or two sentences.
(i) Magnesium oxide is used for the lining of steel making
furnace.
(ii) The mixture of hydrazine and hydrogen peroxide with a
copper (II) catalyst is used as a rocket fuel.
(iii) Orthophosphorous acid is not tribasic acid.
(iv) The molecule of magnesium chloride is linear, whereas
that of stannous chloride is angular.
(1987, 4M)
103. Write balanced equations for the following.
(1987, 2M)
(i) Phosphorus is reacted with boiling aqueous solution of
sodium hydroxide in an inert atmosphere.
(ii) Dilute nitric acid is slowly reacted with metallic tin.
104. Complete and balance the following reactions.
A
B
Asbestos
Paramagnetic
Air pollutant
C
Lithium metal
Silicates of Ca and Mg
Electron donor
Nitric oxide
Reducing agent
(1984, 2M)
109. Complete and balance the following reactions.
(i)
HNO3 + HCl → NO + Cl 2
4+
(ii) Ce3+ +S2O82– → SO2–
4 + Ce
(iii) Cl 2 + OH– → Cl – + ClO–
(1983, 3M)
110. Explain, “orthophosphoric acid, H3 PO4 is tribasic but
phosphorous acid, H3 PO3 is dibasic”.
(1982, 1M)
111. Give structural formula for the following.
(i) Phosphorous acid, H3 PO3
(ii) Pyrophosphoric acid, H4 P2 O7
(1981, 2M)
112. Sulphur melts to a clear mobile liquid at 119°C, but on
further heating above 160° C, it becomes viscous, explain.
(1981, 1M)
113. Explain the following in not more than two sentences .
(i) Conc. HNO3 turns yellow in sunlight.
(i) S + OH– → S2– + S2 O2–
3 + ......
(ii) ClO–3 + I– + H2 SO4 → Cl – + HSO–4 + ...... + ......
(ii) Bleaching powder loses its bleaching properties when it
is kept in an open bottle for a long time.
(1980, 2M)
(1986, 2M)
Topic 2 Elements and Compounds of Group 17 and 18
Objective Questions I (Only one correct option)
1. The electron gain enthalpy (in kJ/mol) of fluorine, chlorine,
bromine and iodine, respectively, are
(a) −333, −325, −349 and −296
(b) −296, −325, −333 and −349
(c) −333, −349, −325 and −296
(d) −349, −333, −325 and −296
(2020 Main, 7 Jan I)
2. The number of bonds between sulphur and oxygen atoms in
and the number of bonds between sulphur and
S2O2−
8
sulphur atoms in rhombic sulphur, respectively, are
(2020 Main, 8 Jan I)
(a) 4 and 6
(b) 8 and 8
(c) 4 and 8
(d) 8 and 6
3. The noble gas that does not occur in the atmosphere is
(2019 Main, 10 April II)
(a) Ra
(b) Kr
(c) He
(d) Ne
234 p-Block Elements-II
4. Chlorine on reaction with hot and concentrated sodium
hydroxide gives
(2019 Main, 12 Jan II)
(a) Cl − and ClO−
(b) Cl − and ClO3−
(c) ClO3− and ClO−2
(d) Cl − and ClO−2
other products. The oxidation state of iodine in Y , is
(2019 Main, 12 Jan I)
(b) 3
(c) 7
(a) H2 + Cl 2 → 2HCl
(c) H2 + F2 → 2HF
(2019 Main, 10 Jan II)
(b) H2 + I2 → 2HI
(d) H2 + Br2 → 2HBr
7. The type of hybridisation and number of lone pair(s) of
electrons of Xe in XeOF 4 , respectively, are
(2019 Main, 10 Jan I)
(a) sp3d 2 and 1
(b) sp3d and 2
(c) sp3d and 1
(d) sp3d 2 and 2
8. Which of the following reactions is an example of a redox
reaction?
(2017 Main)
(a) XeF4 + O2F2 → XeF6 + O2
(d) XeF6 + 2H2O → XeO2F2 + 4HF
(a) ClO− and ClO−3
(b) ClO−2 and ClO−3
(c) Cl − and ClO−
(d) Cl − and ClO−2
(2017 Main)
10. Which among the following is the most reactive?
(2015 Main)
(c) I2
(d) ICl
11. Which one has highest boiling point?
(a) He
(b) Ne
(c) Kr
(2015 Main)
(d) Xe
12. Under ambient conditions, the total number of gases released
as products in the final step of the reaction scheme shown
below is
(2014 Adv.)
XeF6
Complete
hydrolysis
of acidic strength is
HOCl > HClO2 > HClO3 > HClO4
HClO4 > HOCl > HClO2 > HClO3
HClO4 > HClO3 > HClO2 > HOCl
HClO2 > HClO4 > HClO3 > HOCl
(2001, 1M)
(a) HClO < HClO2 < HClO3 < HClO4
(b) HClO4 < HClO3 < HClO2 < HClO
(c) HClO < HClO4 < HClO3 < HClO2
(d) HClO4 < HClO2 < HClO3 < HClO
18. Which one of the following species is not a pseudo halide?
(a) CNO–
(b) RCOO−
(c) OCN−
(1997, 1M)
(d) NNN−
(b) II > I > III
(d) I > III > II
20. KF combines with HF to form KHF2 . The compound
contains the species
(1996, 1M)
(a) K + , F − and H+
(b) K + , F – and HF
(c) K + and [HF2 ] –
(d) [KHF]+ and F –
21. Bromine can be liberated from potassium bromide solution
by the action of
(a) iodine solution
(c) sodium chloride
(1987, 1M)
(b) chlorine water
(d) potassium iodide
22. Chlorine acts as a bleaching agent only in the presence of
(1983, 1M)
(a) dry air
(c) sunlight
(b) moisture
(d) pure oxygen
(1981, 1M)
(b) KMnO4
(d) None of these
Objective Questions II
(One or more than one correct option)
24. With respect to hypochlorite, chlorate and perchlorate ions,
(d) 3
choose the correct statement(s).
13. Among the following oxoacids, the correct decreasing order
(a)
(b)
(c)
(d)
17. The set with correct order of acidic strength is
and HF can reduce
Products
(c) 2
(d) IO−
(a) H2SO4
(c) K 2Cr2O7
Slow disproportionation
in HO–/H2O
(b) 1
(c) IO−4
23. HBr and HI reduce sulphuric acid, HCl can reduce KMnO4
P + Other product
HO–/H2O
Q
(a) 0
(2004, 1M)
(b) I2
(a) I > II > III
(c) III > II > I
9. The products obtained when chlorine gas reacts with cold and
(b) Br2
converts into
(a) IO−3
decreasing acidic strength. Identify the correct order.
(1996, 1M)
ClOH (I), BrOH (II), IOH (III)
(c) XeF6 + H2O → XeOF4 + 2HF
(a) Cl 2
(2008, 3M)
(b) NaHSO4
(d) NaOH
(a) Na 2S4O6
(c) NaCl
19. The following acids have been arranged in the order of
(b) XeF2 + PF5 → [XeF] + PF6−
dilute aqueous NaOH are
(2012)
(b) square planar
(d) see-saw
16. When I− is oxidised by KMnO4 in alkaline medium, I−
(d) 5
6. Among the following reactions of hydrogen with halogens,
the one that requires a catalyst is
(a) trigonal bipyramidal
(c) tetrahedral
15. Aqueous solution of Na 2 S2 O3 on reaction with Cl 2 gives
5. Iodine reacts with concentrated HNO3 to yield Y along with
(a) 1
14. The shape of XeO2 F2 molecule is
(2014 Main)
(2020 Adv.)
(a) The hypochlorite ion is the strongest conjugate base.
(b) The molecular shape of only chlorate ion is influenced by the
lone pair of electrons of Cl.
(c) The hypochlorite and chlorate ions disproportionate to give
rise to identical set of ions.
(d) The hypochlorite ion oxidises the sulphite ion.
p-Block Elements-II
25. The correct statement(s) about the oxoacids, HClO4 and
HClO, is (are)
(2017 Adv.)
(a) The central atom in both HCl O4 and HClO is sp3-hybridised
(b) HCl O4 is formed in the reaction between Cl 2 and H2O
(c) The conjugate base of HCl O4 is weaker base than H2O
(d) HCl O4 is more acidic than HClO because of the resonance
stabilisation of its anion
26. The colour of the X 2 molecules of group 17 elements
changes gradually from yellow to violet down the group.
This is due to
(2017 Adv.)
(a) decrease in π * − σ * gap down the group
(b) decrease in ionisation energy down the group
(c) the physical state of X 2 at room temperature changes from gas
to solid down the group
(d) decreases in HOMO-LUMO gap down the group
27. The compound(s) with two lone pairs of electrons on the
central atom is (are)
(a) BrF5
(2016 Adv.)
(b) ClF3
(c) XeF4
(d) SF4
28. The correct statement(s) regarding,
(i) HClO, (ii) HClO2 , (iii) HClO3 and (iv) HClO4 is (are)
(a) the number of Cl == O bonds in (ii) and (iii) together is two
(b) the number of lone pair of electrons on Cl in (ii) and (iii)
together is three
(c) the hybridisation of Cl in (iv) is sp3
(d) amongst (i) to (iv), the strongest acid is (i)
Passage 1
The reactions of Cl 2 gas with cold-dilute and hot-concentrated
NaOH in water give sodium salts of two (different) oxoacids
of chlorine, P and Q, respectively. The Cl 2 gas reacts with
SO2 gas in the presence of charcoal, to give a product R. R
reacts with white phosphorus to give a compound S. On
hydrolysis, S gives an oxoacid of phosphorus T. (2013 Adv.)
29. P and Q respectively, are the sodium salts of
(a)
(b)
(c)
(d)
32. Bleaching powder contains a salt of an oxoacid as one of its
components. The anhydride of that oxoacid is
hypochlorous and chloric acids
hypochlorous and chlorous acids
chloric and perchloric acids
chloric and hypochlorous acids
Passage 3
The noble gases have closed-shell electronic configuration
and are monoatomic gases under normal conditions. The
low boiling points of the lighter noble gases are due to weak
dispersion forces between the atoms and the absence of
other interatomic interactions.
The direct reaction of xenon with fluorine leads to a series of
compounds with oxidation numbers + 2, + 4 and + 6. XeF4
reacts violently with water to give XeO3 . The compounds of
xenon exhibit rich stereochemistry and their geometries can
be deduced considering the total number of electron pairs in
the valence shell.
(2007, 3 × 4M = 12 M)
33. Argon is used in arc welding because of its
(a) low reactivity with metal
(b) ability to lower the melting point of metal
(c) flammability
(d) high calorific value
34. The structure of XeO3 is
(a) linear
(c) pyramidal
(b) SO2Cl 2 , PCl 3 and H3PO3
(d) SOCl 2 , PCl 5 and H3PO4
Bleaching powder and bleach solution are produced on a
large scale and used in several household products. The
effectiveness of bleach solution is often measured by
iodometry.
(2012)
31. 25 mL of household bleach solution was mixed with 30 mL
of 0.50 M KI and 10 mL of 4 N acetic acid. In the titration of
the liberated iodine, 48 mL of 0.25 N Na 2 S2 O3 was used to
reach the end point. The molarity of the household bleach
solution is
(b) 0.96 M
(c) 0.24 M
(b) reducing
(d) strongly basic
Match the Column
36. All the compounds listed in Column I react with water.
Match the result of the respective reactions with the
appropriate options listed in Column II.
(2010)
Column I
Column II
A.
(CH3 )2 SiCl 2
p.
Hydrogen halide formation
B.
XeF4
q.
Redox reaction
C.
Cl 2
r.
Reacts with glass
D.
VCl5
s.
Polymerisation
t.
O2 formation
Fill in the Blank
37. The increase in solubility of iodine in aqueous solution of KI
Passage 2
(a) 0.48 M
(b) planar
(d) T-shaped
35. XeF4 and XeF6 are expected to be
30. R, S and T, respectively, are
(a) SO2Cl 2 , PCl 5 and H3PO4
(c) SOCl 2 , PCl 3 and H3PO2
(d) Cl 2O6
(c) ClO2
(b) Cl 2O7
(a) Cl 2O
(a) oxidising
(c) unreactive
Passage Based Questions
235
(d) 0.024 M
is due to the formation of …..
(1982, 94, 1M)
True/False
38. HBr is a stronger acid than HI because of hydrogen
bonding.
(1993, 1M)
Numerical Answer Type Questions
39. At 143 K, the reaction of XeF4 with O 2F2 produces a xenon
compound Y . The total number of lone pair(s) of electrons
present on the whole molecule of Y is ............ (2019 Adv.)
236 p-Block Elements-II
40. Reaction of Br2 with Na 2 CO3 in aqueous solution gives
sodium bromide and sodium bromate with evolution of CO2
gas. The number of sodium bromide molecules involved in
the balanced chemical equation is
(2011)
41. Write the balanced equation for the reaction of the following
compound with water.
XeF4
(2002, 5M)
42. Draw molecular structures of XeF2 , XeF4 and XeO2 F2 ,
indicating the locations of lone pair(s) of electrons.
(2000, 3M)
43. Give an example of oxidation of one halide by another
45.
46.
47.
“Chlorine gas is bubbled through a solution of ferrous
bromide.”
(1986, 2M)
51. Complete and balance the following reaction:
ClO–3 + I– + H2SO4 → Cl – + HSO–4 + ......+ ......
Subjective Questions
44.
50. Mention the products formed in the following
halogen. Explain the feasibility of the reaction. (2000, 2M)
Work out the following using chemical equations
“Chlorination of calcium hydroxide produces bleaching
powder.”
(1998, 2M)
Complete the following chemical equations:
(i) KI + Cl 2 →
(ii) KClO3 + I2 → (1996, 2M)
Give reasons in two or three sentences only for
(i) Bond dissociation energy of F2 is less than that of Cl 2 .
(ii) Sulphur dioxide is a more powerful reducing agent in the
alkaline medium than in acidic medium.
(1992, 2M)
Write the balanced chemical equation for the following:
Sodium bromate reacts with fluorine in the presence of alkali.
48. Arrange the following as indicated. HOCl, HOClO2 ,
HOClO3 , HOClO in increasing order of thermal stability
(1988, 2M)
49. Give balanced equation for the following:
Iodate ion reacts with bisulphite ion to liberate iodine. (1988, 3M)
(1986, 2M)
52. Arrange the following in the order of
(i) increasing bond strength HCl, HBr, HF, HI
(ii) increasing oxidation number of iodine
I2 , HI, HIO4 , ICl
(1986, 2M)
53. Give reason in one or two sentences.
Fluorine cannot be prepared from fluorides by chemical
reduction method.
(1985, 1M)
54. Complete and balance the following reaction.
Cl 2 + OH– → Cl – + ClO–
(1983, 3M)
55. Explain the following in not more than two sentences.
Bleaching powder loses its bleaching properties when it is
kept in an open bottle for a long time.
(1980, 2M)
56. Give reasons for the following in one or two sentences.
(i) Hydrogen bromide cannot be prepared by the action of
conc. sulphuric acid on sodium bromide.
(ii) When a blue litmus paper is dipped into a solution of
hypochlorous acid, it first turns red and then later gets
decolourised.
(1979, 2M)
57. Write the balanced equations involved in the preparation of
(i)
(ii)
(iii)
(iv)
bleaching powder from slaked lime
(1979, 10M)
nitric oxide from nitric acid
chlorine from sodium chloride
anhydrous aluminium chloride from alumina
Answers
Topic 1
1. (a)
2. (b)
3. (c)
4. (b)
65. (white)
66. (T)
67. (T)
68. (F)
5. (d)
6. (a)
7. (a)
8. (d)
69. (T)
70. (288)
71. (8)
72. (4)
9. (b)
10. (d)
11. (d)
12. (a)
73. (6)
13. (a)
14. (b)
15. (c)
16. (c)
17. (b)
18. (d)
19. (b)
20. (d)
21. (c)
22. (b)
23. (b)
24. (c)
2. (b)
3. (a)
4. (b)
25. (b)
26. (a)
27. (c)
28. (d)
29. (c)
30. (a)
31. (a)
32. (c)
33. (c)
34. (b)
35. (d)
36. (b)
37. (a)
38. (d)
39. (b)
40. (c)
41. (d)
42. (a)
43. (b,c)
44. (a,b,c)
45. (b, d)
46. (a, c, d)
47. (a, b, c)
48. (c)
49. (a, c, d)
50. (a, d)
51. (6)
52. (a)
53. (a)
54. (a)
55. (b)
56. (a)
57. (b)
58. (c)
59. (c)
60. (b)
61. (d)
62. (NO2 )
63. (Four)
64. (Two)
Topic 2
1. (c)
5. (d)
6. (b)
7. (a)
8. (a)
9. (c)
10. (d)
11. (d)
12. (c)
13. (c)
14. (a)
15. (a)
16. (a)
17. (a)
18. (b)
19. (a)
20. (c)
21. (b)
22. (b)
23. (d)
24. (a,b,d)
25. (a, c, d)
26. (b, c)
27. (b, c)
28. (b, c)
29. (a)
30. (a)
31. (c)
32. (a)
33. (a)
34. (c)
35. (a)
36. (A → p, s
37. (KI3 )
B → p, q, r, t
38. (F)
C → p, q, t
39. (19)
D → p)
40. (5)
Hints & Solutions
Topic 1 Elements and Compounds of
Group 15 and 16
H2 S 2 O3 (thiosulphuric acid), H2 S 2 O4 (hyposulphurous or
dithionous acid) and H2 S 4 O6 (tetrathionic acid) contains SS
bonds.
1. The correct statement is that (SiH3 )3 N is planar and less basic
O
than (CH3 )3 N. The compounds trimethylamine (CH3 )3 N and
trisilylamine (SiH3 )3 N have similar formulae, but have totally
different structures. In trimethylamine the arrangement of
electrons is as follows :
Electronic structure of
nitrogen atom
(ground state)
1s
2s
S
H2S2O4 ⇒
2p
N
N
O
nitrogen oxides is
+1
+2
+4
+3
N 2 O< N O < N 2 O3 < N O2
l
l
2. In C60 (Buckminster fullerene) twenty hexagons and twelve
pentagons are present which are interlocked resulting a shape
of soccer ball. Every ring in this structure is aromatic.
O
O
4. The correct increasing order of oxidation state of nitrogen for
SiH3
N(SiH3)3
molecule
CH3
HO
O
CH3
N(CH3)3
molecule
OH
S——S
H2S4O6 ⇒ HO—S—S—S—S—OH
SiH3
H3Si
O
O
Three unpaired electrons
form bonds with CH3 groups
tetrahedral arrangements of
three bond pairs and one lone pair
In trisilylamine, three sp2 orbitals are used for
σ -bonding, giving a plane triangular structure.
H 3C
H2S2O3 ⇒ HO—S—OH
l
l
Oxidation state of N in N2O is
2(x ) − 2 = 0
2
x = + = +1
2
Oxidation state of N in NO is
x−2= 0
x = +2
Oxidation state of N in N2O3 is
2x + 3(−2) = 0
6
x= =3
2
Oxidation state of N in NO2 is
x + 2(−2) = 0
x−4 = 0
x = +4
5. Let us consider the structure of the phosphorus oxyacids,
Phosphorus has large atomic size and less electronegativity, so
it forms single bond instead of pπ-pπ multiple bond. So, it
consists of discrete tetrahedral P4 molecule as shown below :
H
H
OH
H3PO2
Hypophosphorus
acid
(P—H bonds=2)
P
P
P
P
HO
P
H
OH
H3PO3
Orthophosphorus
acid
(P—H bond=1)
P
∴ Number of trigons (triangles) = 4
H
3. S  S bond is not present in H2 S 2 O7 (pyrosulphuric acid or
O
H2S2O7 ⇒ HO—S—O—S—OH
O
O
While the other given oxoacids of sulphur, i.e.
O
HO
P
H
OH
H 4P 2O 5
Pyrophosphorus
acid
(P—H bonds=2)
oleum).
O
P
O
P
P
OH
OH
OH
H4P2O6
Hypophousphoric
acid
(P—H bond=0)
6. As given, the first electron gain enthalpy of oxygen can be shown
as,
O(g )+ e− → O − (g ), ∆egH 1 = − 141kJ/mol
238 p-Block Elements-II
The expression of second electron gain enthalpy of oxygen will be,
O − (g )+ e− → O 2− (g,) ∆egH 2 = + ve
∆egH 2 of oxygen is positive, i.e. endothermic, because a strong
electrostatic repulsion will be observed between highy negative O −
and the incoming electron (e− ). A very high amount of energy will
be consumed (endothermic) by the system to overcome the
electrostatic repulsion.
7. The structure of H3PO2 (hypophosphorous) acid is
O
White phosphorus on reaction with thionyl chloride (SOCl 2 )
produces phosphorus trichloride.
P4 (s) + 8SOCl 2 (l ) → 4PCl 3 (l ) + 4SO2 (g ) + 2S2Cl 2 (g )
But if amount of thionyl chloride (SOCl 2 ) is in excess then it
produces phosphorus pentachloride.
P4 + 10SOCl 2 (l ) → 4PCl 5 + 10SO2
it has one unpaired electron.
Total number of electrons present = 7 + 8 = 15 e −
H
H
Due to the presence of two P  H bonds, H3 PO 2 acts a strong
reducing agent. e.g.
+1
PLAN This problem is based on chemical properties of
phosphorus.
13. NO is paramagnetic in gaseous state because in gaseous state,
P
HO
12.
+1
0
+5
4 Ag NO 3 + H3PO 2 + 2H2O → 4 Ag ↓ + H3 PO 4 + 4 HNO 3
8. The thermal decomposition of given compounds is shown below
∆
(NH4 )2 Cr2O7 → N2 + 4H2O + Cr2O3
∆
NH4NO2 → N2 + 2H2O
Hence, there must be the presence of unpaired electron in
gaseous state while in liquid state, it dimerises due to
unpaired electron.
14. NO 2 is a brown coloured gas and imparts this colour to
concentrated HNO 3 during long standing.
4 HNO 3 → 2H2O + 2NO 2 + 3O 2
−
15. (a) ONCl = 8 + 7 + 17 = 32 e
ONO− = 8 + 7 + 8 + 1 = 24 e− (correct)
∆
(NH4 )2 SO4 → 2NH3 +H2SO4
Ba(N3 )2 → Ba + 3N2
Thus, only (NH4 )2 SO4 does not gives N2 on heating (It give NH3 ).
While rest of the given compounds gives N2 on their thermal
decomposition.
+5
+4
+3
+1
9. H3 P O 4 > H4 P2 O6 > H3 P O 3 > H3 P O 2
10.
Species
Hybridisation
O
(b)
O
O
Central O-atom is sp 2 -hybridised with 1 lone pair, so
bent shape (correct).
(c) In solid state, ozone is violet-black. Ozone does not
exist in solid state, thus incorrect.
(d) O 3 has no unpaired electrons, so diamagnetic (correct).
Hence, (c) is the correct.
–
O
–
sp2
N
O
O
N
O
O
16. The reaction of white phosphorus with aqueous alkali is
P4 + 3NaOH + 3H2O → PH3 + NaH2PO2
sp2
In the above reaction, phosphorus is simultaneously
+1
oxidised [P4 (0) → NaH2 P O2 ] as well as reduced
−3
N
sp2
O
O
+
O N O
sp
O

11. Orthophosphorous acid, H3PO3 : HO  P  OH

H
x
H3 PO3 = 3 + x + 3( −2 ) = 0 or x = + 3
Pyrophosphorous acid, H 4 P2 O 5 :
O
O


HO  P  O  P  OH


H
H
x
H4 P2 O5 = 4 + 2x + 5 ( − 2) = 0
4 + 2x − 10 = 0, x = + 3
[P4 (0) → P H3 ]. Therefore, this is an example of
disproportionation reaction. Oxidation number of
phosphorus in PH3 is − 3 and in NaH2PO2 is + 1. However,
+ 1 oxidation number is not given in any option, one might
think that NaH2PO2 has gone to further decomposition on
heating.
+5
∆
2NaH2PO2 → Na 2H P O4 + PH3
17. Let oxidation number of N be x.
In HNO3, + 1 + x + 3 (− 2) = 0 ⇒ x = + 5
In NO,
x−2=0 ⇒ x=+ 2
In N2,
x=0
In NH4Cl,
x + 4 −1= 0 ⇒ x = − 3
18. Ba(N3 )2 Heat
→ Ba(s) + 3N2 (g )
Azide salt of barium can be obtained in purest form as well as
the decomposition product contain solid Ba as by product
p-Block Elements-II
alongwith gaseous nitrogen, hence no additional step of
separation is required.
Other reactions are
28. The structure of S3O9 is
S
Heat
O
Heat
2NH3 + 3CuO → 3Cu + 3H2O + N2
O
O
O
S
S
O
O
O
Heat
(NH4 )2 Cr2O7 → Cr2O3 + 4H2O + N2
lower oxide P4O6 while excess of oxygen gives P4O10. A mixture
of O2 and N2 is used for controlled oxidation of phosphorus into
P4O6.
O
O
NH4NO3 → N2O + 2H2O
19. In limited supply of oxygen, phosphorus is oxidised to its
It has no S—S linkage.
29. CaO, a basic oxide, is most suitable for drying of basic ammonia.
30. H2O, due to its ability to form intermolecular H-bonds.
31. Corresponding acids are HClO4, H2SO3 and H3PO4. Hence, the
order of acidic strength is
Cl 2 O7 > SO2 > P4 O10
3
20. In P4, all phosphorus are sp -hybridised and has 75%
p-character.
32. The structure of cyclic metaphosphate is
P
–
sp3
P
P
–
21. In KMnO4, Mn is already in its highest oxidation state (+7),
O
O
O
O
P
P
O
P
P
O
O
O
O
–
There is three P—O—P bonds.
cannot be oxidised by any oxidising agent.
22. PbO2 + HNO3 → Pb(NO3 )2 + H2O + O2
33. Ca 3P2 + 6H2O → 3Ca(OH)2 + 2PH3
23. Equimolar amounts of NO and NO2 at –30°C gives N2O3 (l )
OH
34. Na 2SO3 + S →
Na 2S2O3
which is a blue liquid.
( Blue)
24. Black phosphorus is thermodynamically most stable allotrope
of phosphorus.
It is due to three dimensional, network structure of polymeric
black phosphorus.
25. H2S2O8 is a peroxy acid, has—O—O—linkage
O
O


HO— S — O— O— S — OH


O
O
Peroxodisulphuric acid
26. H3PO3 is a dibasic, reducing acid. H3PO4 is tribasic,
Dibasic, reducing
−
∆
35.
−30 ° C
NO( g ) + NO2 ( g ) → N2 O3( l )
non-reducing acid.
O

H— P— OH

OH
239
O

HO— P— OH

OH
Tribasic, non -reducing
27. Polyphosphates are used as water softening agents because
they form soluble complexes with cationic species of hard
water.
Na 2[Na 4 (PO3 )6 ]+ CaSO4 → Na 2[(Ca 2 (PO3 )6 ] + Na 2SO4
Soluble complex
S2O27 −
has no S—S linkage.
O
O



−
O— S — O— S — O−


O
O
All others have atleast one S—S linkage.
36. Amongst XH3 where ‘X ’ is group-15 elements, basic strength
decreases from top to bottom. Hence, NH3 is strongest base.
37. The electron withdrawing inductive effect of halogen decreases
electron density on nitrogen, lowers basic strength. Since,
fluorine is most electronegative, NF3 is least basic.
38. NO2 (g ) is deep brown coloured.
39. In N2O5, there are σ (sigma) covalent bonds, π (pi) bonds and
coordinate covalent bonds as
O
N
O
O
O
N
O
40. SO2 cannot be collected over water because it reacts with water
forming H2SO3.
SO2 + H2 O → H2 SO3
41. Quicklime (CaO) is used for drying NH3 gas because both are
basic, do not react. On the other hand, H2SO4 and P2O5 are
acidic, reacts with ammonia forming salts. CaCl 2 forms
complex with ammonia.
240 p-Block Elements-II
(b) N2O5 has no unpaired electron and is thus, diamagnetic
thus, (b) is correct.
(c)
O
O
N
O N
O
O
42. NO is lighter than O2.
D2O is commonly known as heavy water.
N2 is lighter than O2, effuse at faster rate under identical
experimental conditions. NH3 liquefies at very low temperature.
Therefore, liquid NH3 is used as a refrigerant.
43. Among the given compounds, those which generate N2 on thermal
There is no N—N bond, thus, (c) is incorrect.
(d) N2O5 + Na → NaNO3 + NO2
N2O5 vapours are of brownish colour. Thus, (d) is
correct.
decomposition below 300°C are ammonium dichromate i.e.,
(NH4 )2Cr2 O7 and barium azide or nitride i.e., Ba(N3 )2 . Reactions
of their thermal decomposition are given below
It is an exothemic reaction with
∆H = − 429.1 ± 3 kcal/ mol.
∆
(ii) Ba(N3 )2 → Ba + 3N2 ↑
46.
Plan Due to resonance, bond lengths between two atoms are
equal. Species is said to be diamagnetic if all electrons are
paired.
Process is endothermic if it takes place with absorption of
heat.
Around 160 ° and above
below 300 °C
However, on rapid heating or explosion
(i.e. above 300°C) it gives off nitrogen as
Rapid heating
2NH4NO3  → 2N2 + O2 + 4H2O
or explosion
44. Statement wise explanation is
s
O
O
Exothermic
Thus, (b) is incorrect. (a, c, d) are correct.
47. The structures of these oxides are
O
N
N
O
→ O
N
N
(b) O
(a)
O
O
O
N
N
O
O
N
O
O
N
O
→
P4O10 + 6H2O → 4H3PO4
(a) P4 + 20HNO3 → 4H3PO4 + 20NO2 + 4H2O
Thus, (a) is incorrect.
→→
O
r

2HNO3 → N2O5 + H2O
O
bent molecule all electrons paired thus, diamagnetic
2 O 3 → 3 O 2 ∆H ° = − 142 kJ mol −1
→
45. P4O10 is a dehydrating agent and converts HNO3 into N2O5
r
O
O

(i) Statement (a) Bi 2O5 is a metallic oxide while N2O5 is a
non-metallic oxide.
Metallic oxides being ionic are basic in nature while non
metallic oxides being covalent are acidic in nature. This
confirms more basic nature of Bi 2O5 in comparison to N2O5.
Hence, this is a correct statement.
(ii) Statement (b) The electronegativity difference between N(3) and
F(4) is less as compared to the electronegativity difference
between Bi (1.7) and F(4). More electronegativity difference
leads to ionic compounds. Thus, NF3 must be more covalent in
nature as compared to BiF3. Hence, this statement is also correct.
(iii) Statement (c) In NH3 intermolecular hydrogen bonding is
present, which is altogether absent in PH3. Thus, PH3 boils at
lower temperature than NH3.
Hence, this is also a correct statement.
(iv) Statement (d) Due to smaller size of N the lone pair-lone pair
repulsion is more in N—N single bond as compared to O—P
single bond. This results to weaker N—N single bond as
compared to P—P single bond. Hence, this statement is incorrect.
O
O
s
→
Magnesium nitride (Mg3N2 ) does not decompose at lower
temperatures being comparatively more stable. Its thermal
decomposition requires a minimum temperature of 700°C and
proceeds as
− 1500° C
Mg3N2 700

→ 3Mg + N2 ↑
116.80°
pm

∆
NH4NO3 → N2O + 2H2O
O 218
→
Ammonium nitrate (NH4NO 3) on heating below 300°C gives
N 2O as

Below 300 °C
→
∆
(i) (NH4 )2 Cr2O7 → N2 ↑ + Cr2O3 + 4H2O
O
(d)
(c)
(a), (b), (c) have N—N bonds.
48. 2NH3 + OCl − → H2N— NH2 + H2O + Cl −
49. The structure of P4 is
P
P
P
P
It has six P—P single bonds.
There are four lone pairs on four phosphorus. P—P—P
bond angles are of 60°.
Heat
50. NH4NO3 → N2O + 2H2O
NH2OH ⋅ HCl + NaNO2 → NaCl + 2H2O + N2O
However, NH4 NO2 on heating gives N2 .
p-Block Elements-II
51. The structures of various molecules given in problem are
8. H2S 2 O5
discussed below—
1. N 2 O3 It is the tautomeric mixture of following two
structures—
O
N
O
N
N
O
O
N
O
O
N
Bridging oxo
group
P
O
O
O
52. Both Statement I and Statement II are true and Statement II is
correct explanation of Statement I.
explains the Statement I appropriately. Nitrate ion (NO−3 ) is
more stable than nitrite ion :
−
O
O


−
O— N→ O ←→ O == N→ O (Resonance structure)
••
O
P
P
O
O
P
O
O
Phosphorus has vacant 3d-orbitals, it can expand its valence
shell beyond eight electrons, its both trihalides and
pentahalides exist.
O
Passage 1
MnO 2
KClO3 → KCl + O2
Conclusion 6 bridging oxo groups are present in the
compound.
5. H4 P2 O5
O
P
P
O
H
OH
O−
but reason is not the correct explanation of Statement I.
Nitrogen does not has any vacant d-orbitals, it cannot expand
its valence shell beyond eight electrons, i.e. it cannot violate
octet. Therefore, nitrogen forms only trihalides
(NX 3 with eight electrons in valence shell of N).
P
O
(Resonance structure)
N
O
55. Both Statement I and Statement II are independently correct
Conclusion 6 bridging oxo groups are present in the
compound.
4. P4 O7
O
O
O
••
←→
N
−
O
P
Conclusion This compound also does not contain any
bridging oxo group.
P
O
O
OH
54. Both Statement I and Statement II are true and Statement II
Conclusion 1 bridging oxo group is present in the compound.
3. P4 O6
P
S
correct explanation of Statement I.
O
O
S
53. Both Statement I and Statement II are true and Statement II is
O
N
O
HO
O
Conclusion 1 bridging oxo group is present in the compound.
2. N 2 O5 It has following structure.
O
O
O
Bridging
oxo group
241
∆
W
HNO 3
∆
O2 + P4 → P4O10 → N2O5 + HPO3
X
56. (a)
H
OH
Conclusion 1 bridging oxo group is present in the compound.
Y
Z
57. (b)
Passage 2
58. Due to greater solubility in water and prone to microbial
attack, nitrates are less abundant in earth’s crust.
6. H 5P3O 10
O
O
O
59. NH3 is stronger Lewis base than PH3. In a group of hydrides,
P
P
P
60. White phosphorus undergo disproportionation in alkaline
O
HO
OH
O
OH
basic strength decreases down the group.
OH
OH
Conclusion 2 bridging oxo groups are present in the
compound.
7. H2S 2 O3
P4 + NaOH → PH3 + NaH2 PO2
Warm (3)
61. (P) 2PbO2 + 2H2SO4 → 2PbSO4 + O2 + 2H2O
Cl 2 ( 4 )
(Q) Na 2S2O3 + H2O → NaHSO4 + HCl
S
S
HO
medium.
OH
O
Conclusion This compound does not contain any bridging
oxo group.
I2 (2 )
(R) N2H4 → N2 + Hl
NO(1)
(S) XeF2 → Xe + NOF
Thus, P—(3), Q—(4), R—(2), S—(1)
242 p-Block Elements-II
Oxides of N2
62. NO2 : 2SO2 (g ) + O2 (g ) →
2SO3 (g )
72. PCl5 produces POCl3 with the following reagents
PCl 5 + SO2 → POCl 3 + SOCl 2
PCl 5 + H2O → POCl 3 + 2HCl
2PCl 5 + H2SO4 → SO2Cl 2 + 2POCl 3 + 2HCl
6PCl 5 + P4O10 → 10POCl 3
(NO2 )
63.
O
O
O
O
HO—S—OH
P
O
— —
— —
HO—Cr—OH HO—S—OH
O
O
O
O
HO—S—O—S—OH
64. H3PO3 [O == PH(OH)2 ] is a dibasic acid.
OH
O
— —
Here four oxygen atoms are bonded to each phosphorus
atom.
H
HO—C—OH
—
O
O
O
Others are
65. White phosphorus has highly strained, tetrahedral
O
—
—
O
structure, therefore highly reactive.
P
66. In liquid state, nitric oxide (NO) dimerises into (NO)2
and odd electrons disappear giving diamagnetic property.
→ O == N— N== O ( l )
2NO
Paramagnetic
P——OH
O
O
O
O
—
P
—
P
73. Diprotic acids = 6
O
O
— —
O
O
P
H H
OH
HO OH OH
Triprotic
Monobasic
Lewis acid
Diamagnetic
67. Both ‘N’ and ‘As’ in corresponding hydrides are
P
HO—B—OH
74.
OH
Monoprotic
O
sp3-hybridised. If central atoms are from same group,
bond angle decreases from top to bottom if all other
things are similar. Hence, H—N—H bond angle in NH3
is greater than H—As—H bond angle in AsH3.
P
O
O
O
P
O
O
P
O
68. Halogens are all good oxidising agent and their oxidising
power decreases from top to bottom (F2 to I2) in group.
Any halogen above in group oxidises halides down in
group from their aqueous solution. Hence, Cl 2 can
oxidise Br − to Br2, I− to I2 but cannot oxidise F− to F2
rather F2 can oxidise Cl − to Cl 2.
69. Fe is more electropositive than hydrogen, displaces H+
ions from acid solution as :
Fe + 2HCl → FeCl 2 + H2 ↑
70.
When rhombic sulphur (S8 ) is oxidised by conc. HNO 3
then H2SO 4 is obtained and NO 2 gas is released.
S8 + 48HNO 3 → 8H2SO 4 + 48NO 2 + 16 H2O
1 mole of rhombic sulphur produces = 16 moles of H2O
∴ Mass of water = 16 × 18 (molar mass of H2O) = 288 g
••
••
••
••
••
••
••
••
••
O
••
(P4O10)
75. Cl 2O7 < SO3 < CO2 < B2O3 < BaO
+4
+4
+2
+2.5
76. A = NaH SO3 ; B = Na 2 SO3 ; C = Na 2 S2O3; D = Na 2 S4 O6
(ii) CaNCN + 5H2O
(iii) 4BF3 + 3H2O
(iv) NCl 3 + 3H2O
(v) 2XeF4 + 3H2O
→
→
→
→
CaCO3 + 2NH4OH
H3BO3 + 3HBF4
NH3 + 3HOCl
Xe + XeO3 + F2 + 6HF
78. Nitrogen in N2 are bonded by one sigma and two pi bonds. Phosphorus
and other elements of this period, due to larger size, are very less likely
to form pi bonds, hence P4 is formed in which there is no pi bonds.
79. In given scheme : A = Ca (OH)2
B = NH4HCO3, C = Na 2CO3
71. N2O3 has two proposed structures.
O == N  O  N == O and
O
O
77. (i) Al 4C3 + 12H2O → 4Al(OH)3 + 3CH4
Key Idea Rhombic sulphur (S8 ) gets oxidised into
sulphuric acid and water, NO2 gas is released on
reaction with conc. HNO3.
••
P
O
••
O•
• ••
N N
D = NH4Cl and E = CaCl 2
80. (a) HNO2 + 2H2SO3 + H2O → NH2OH + 2H2SO4
A
••
O
••
In both cases, number of lone pair of electrons are eight.
B
C
D
(b) In SO3 + H2O → H2SO4, sulphuric acid is obtained in misty
form and the reaction is explosive. By adding H2SO4, above
reaction is prevented :
p-Block Elements-II
H2SO4 + SO3 → H2S2O7 (oleum)
H2S2O7 + H2O → 2H2SO4
In the contact process, V2O5 is used as catalyst.
P4O10 + 6PCl 5 → 10 POCl 3
81. (i)
(ii) SnCl 4 + 2C2H5Cl + 2Na → Na 2SnCl 4 + C4H10
Heat
87. Ca 5 (PO4 )3 F + 5H2SO4 + 10H2O → 3H3PO4
+ 5CaSO4 ⋅ 2H2 O + HF
88. PCl 5 + SO2 → POCl 3 + SOCl 2
A
(b) Ca 3 (PO4 )2 + 4H3PO4 →
3Ca(H2PO4 )2
triple superphosphate
form H3PO3 and HI as–
2P + 3I2 + 6H2 O → 2H3PO3 + 6HI
90. SO2 acts as reducing agent on account of following reaction :
SO2 + 2OH − → SO42 − + 2H + + 2e−
83. The poisonous element M may be As. On the basis of given
information
Hence, the above reaction proceeds in forward direction on
increasing concentration of HO− ion. H+ is on product side,
adding H+ retards the reaction by sending it in backward
direction.
Zn /HCl
AsCl 3 + 6H → AsH3 + 3HCl
N
∆
2AsH3 → 2As + 3H2
M
84. (i) P4 + 20HNO3 → 4H3PO4 + 20NO2 + 4H2O
(ii) 3KMnO4 + 5H2O2 + 3H2SO4 → K2SO4
+ 2MnSO4 + 5O2 + 8H2O
(iii) P4 + 20HNO3 → 4H3PO4 + 20NO2 + 4H2O
(iv) 2Al + 2NaOH + 2H2O → 2NaAlO2 + 3H2
85.
O
O
O
P
4 P==O bonds
O
O
P
P
O
O
B
89. Red phosphorus reacts with iodine in the presence of water to
PCl 5 + SO2 → POCl 3 + SOCl 2
82. (a)
243
+
91.
+
O
O
O
O
–
–
O
O
92. Ammonia, in liquid state undergo self-ionisation as :
2NH3 q NH4+ + NH2−
Thus, addition of NH4 Cl to liquid ammonia increases
concentration of NH4+ in solution and NH4 Cl act as acid.
93. (i) Na 2CO3 + NO + NO2 → 2NaNO2 + CO2
(ii) 2KMnO4 + 2NH3 → 2MnO2 + 2KOH + 2H2 O + N2
(iii) P4 + 20HNO3 → 4H3PO4 + 20NO2 + 4H2 O
(iv) 2H2 S + NaHSO3 + H+ → 3S + 3H2 O + Na +
O
P
O
O
86. (i) The size of both nitrogen and fluorine are very small as well
as they have very high electron density. Thus in NF3, N and
F repel each other stretching the N—F bond. Hence, in NF3,
N—F bond lengths are greater than the sum of their single
bond covalent radii.
(ii) Mg 3N2 + 6H2 O → 3Mg(OH)2 + 2NH3
MgCl 2 is a salt of strong acid HCl and strong base
Mg(OH)2 and therefore, not hydrolysed in aqueous
solution.
(iii) In (SiH3 )3 N, the lone pair of nitrogen is involved in pπ - dπ
bonding, less available on nitrogen for donation to a Lewis
acid, a weaker Lewis base
pπ-dπ
H3Si—N——SiH3
SiH3
Carbon does not have any vacant d-orbitals, no such pπ -dπ
bonding occur in trimethyl amine, lone pair of nitrogen is
available for donation to Lewis acid, hence a stronger Lewis
base.
(v) CaCO3 + CO2 + H2 O → Ca(HCO3 )2
94. (i) NaNO2 + Zn + NaOH → 3Na 2ZnO2 + NH3 + H2O
(ii) 2NaIO3 + 5NaHSO3 → 3NaHSO4 + 2Na 2 SO4
+ I2 + H 2 O
+
−
−
+
95. N ≡≡ N— O ←→ N == N == O
96. (i)
∆
15CaO + 4P4 → 5Ca 3P2 + 3P2O5 ↑
[Ca 3P2 + 6H2 O → 3Ca(OH)2 + 2PH3 ↑ ] × 5
15CaO + 4P4 + 30H2 O → 15Ca(OH)2
+ 3P2 O5 ↑ + 10PH3 ↑
(ii) 2NH3 + CO2 + H2 O → (NH4 )2 CO3
CaSO4 + (NH4 )2 CO3 → CaCO3 ↓ + (NH4 )2 SO4
gypsum
CaSO4 + 2NH3 + CO2 + H2O → CaCO3 ↓ + (NH4 )2 SO4
97. (i) In H3PO3, there is only two replaceable H, hence dibasic
O

H— P — OH H—of OH are acidic, dibasic.

OH
(ii) NH3 molecules are associated by intermolecular
H—bonds.
244 p-Block Elements-II
98. (i)
2H3PO2
∆
→ PH3 + H3PO4 (Disproportionation)
109.
(ii) NaClO3 + SO2  → NaCl + S + 5H2O
110. Orthophosphoric acid (H3PO4 ) has three replaceable (acidic)
99. SiO2 < CO2 < N2O5 < SO3
hydrogen while orthophosphorus acid (H3PO3 ) has only two
replaceable hydrogen.
O
O


O— P — OH
HO— P — OH


OH
OH
100. (i) 4P + 10HNO3 + H2O → 5NO + 5NO2 + 4H3PO4
(ii) NaCl + NH4OH + CO2 → NH4Cl + NaHCO3
101. Oxygen lacks empty d-orbitals in its valence shell, cannot
violate octet rule, hence in most of its compound it show only
divalency. On the other hand, sulphur has vacant 3d-orbitals in
its valence shell, can violate octet rule, show di, tetra and hexa
valency.
(i) MgO is used for the lining of steel making furnace because
it forms slag with impurities, and thus helps in removing
them from iron.
(ii) The mixture of N2 H4 and H2 O2 (in presence of Cu(II)
catalyst) is used as a rocket propellant because the reaction
is highly exothermic and large volumes of gases is evolved.
N2H4 (l ) + 2H2O2 (l ) → N2 (g ) + 4H2O (g )
(iii) In orthophosphorus acid (H3PO3 ) only two of the three H
are replaceable as
O

H— P — OH

OH
(Only H of —OH are acidic)
(iv) In MgCl 2, Mg is sp-hybridised while in SnCl 2, Sn is
sp2-hybridised with a lone pair at Sn. Hence, MgCl 2 is
linear while SnCl 2 is angular.
Inert atm.
103. (i) P4 + 3NaOH + 3H2O → 3NaH2PO2 + PH3
(phosphine)
(ii) 4Sn + 10HNO3 → 4Sn(NO3 )2 + NH4 NO3 + 3H2 O
dil
104. (i) 4S + 6OH− → 2S2 − + S2O32− + 3H2O
(ii) ClO3– + 6I– + 6H2SO4 → Cl – + 6HSO–4 +3I2 + 3H2O
∆
105. (i) 2Ca 3 (PO4 )2 + 6SiO2 + 10C →
6CaSiO3
+ 10CO +
P4
(white)
(ii) (NH4 )2 SO4 + NO + NO2 → 2N2 + 3H2 O + H2 SO4
+
−
−
+
106. N ≡≡ N— O ←→ N== N ==O
107. Fe2 (SO4 )3 + 2KI → 2FeSO4 + K2SO4 + I2
In the above reaction, strong reducing agent, iodide, reducing
ferric salt into ferrous salt.
108.
A
Asbestos
Lithium metal
Nitric oxide
B
Silicates of Ca and Mg
Reducing agent
Paramagnetic
Cl 2 + 2OH− → Cl − + ClO− + H2 O
(iii)
10 H +
102.
(i) 2HNO3 + 6HCl → 2NO + 3Cl 2 + 4H2O
(ii) 2Ce3 + + S2 O28 − → 2SO24 − + 2Ce4 +
hypophosphorus
acid
C
Donar
Electron donor
Air pollutant
orthophosphoric acid
(it has three acidic H)
orthophosphorus acid
(only two acidic H, H directly
bonded to P is not acidic)
O
111.
O
O
(i)
(ii)
H
P
HO
HO
OH
OH
phosphorus acid
(P is sp3-hybridised)
P
P
O
OH
OH
(pyrosphosphoric acid)
112. Rhombic sulphur has a eight membered puckered ring structure.
On heating ring tends to break and linear chain sulphur is
formed. When sulphur melts, the S8 rings slip and roll over
one another very easily. It gives rise to a clear mobile liquid.
When liquid sulphur is further heated to higher temperature,
rings are broken giving long chain sulphur molecules. This
long chain molecules of sulphur gets entangled into one
another increasing viscosity of melt.
113. (i) In the presence of sunlight, concentrated nitric acid
decomposes partially as
hν
Conc. HNO3 → NO2 + H+ + O2
It is the NO2 which impart yellow colouration to nitric acid.
(ii) The bleaching action of bleaching powder is due to
presence of available chlorine, but in contact of moisture,
it releases chlorine decreasing the amount of available
chlorine. Hence, bleaching property decreases gradually as
bleaching powder is kept in open container for long time.
Topic 2 Elements and Compounds of
Group 17 and 18
1. Electron gain enthalpy (∆ eg H ) is the enthalpy change for
converting 1 mol of isolated atoms to anions by adding
electrons. All halogens have negative ∆ eg H (exothermic)
values. Generally, ∆ eg H becomes less negative when
comparing elements of the same group from top to bottom.
But among fluorine and chlorine there is an anomaly because
inter-electron repulsion is stronger in fluorine due to its extra
small size.
∴ ∆ e g H is less exothermic than expected for F-atom.
Thus, the correct values of electron gain enthalpies
F < Cl > Br >
I
kJ mol −1
( −333)
( −349)
( −325)
( −296 )
p-Block Elements-II
2. S 2O2−
8 is
245
8. The reaction in which oxidation and reduction occur
simultaneously are termed as redox reaction.
+ 1
+4
O—S—O—O—S—O
Total number of S  O or S==O bond = 8
S
Rhombic sulphur is
S
S
S
S
S
S
9. Cl 2, Br2 and I2 form a mixture of halide and hypohalites when
react with cold dilute alkalies while a mixture of halides and
haloate when react with concentrated cold alkalies.
S
Cl 2 + 2NaOH → NaCl + NaClO + H2O
Cold and dilute
∴ Cl − and ClO− are obtained as products when chlorine gas
reacts with cold and dilute aqueous NaOH.
3. Radium (Ra) is a radioactive element. Ra belongs to group 2
(alkaline earth metals), it is not a noble gas.
Note In question noble gas which does not exist in the
atmosphere is asked and answer is Ra. But Ra (radium) is an
alkaline earth metal and not noble gas. It can be Rn (radon) and
is misprint in JEE Main Paper.
4. Halogens form halates and halides with hot and concentrated
solution of NaOH as :
3 X 2 + 6NaOH → 5NaX + NaXO3 + 3H2 O
So, Cl2 will also give Cl− (as NaCl) and ClO3−
(as NaClO3 ) in the above reaction.
Thus, option (b) is correct.
Note
When halogens react with cold and dilute solution of
NaOH, hypohalites and halides are produced as:
X 2 + 2NaOH → NaX + NaXO + H 2O
5. Iodine reacts with concentrated HNO 3to yield HIO 3 along
with NO 2 and H2 O. The reaction involved in as follows :
I2 + 10HNO3 → 2HIO3 + 10NO2 + 4H2O
The oxidation state of ‘I’ in HIO 3 is + 5 as calculated below :
1 + x + 3(−2) = 0
x − 5 = 0, x = + 5
10. Interhalogen compounds are generally more reactive than
halogens (except fluorine).
11. Xe has highest boiling point.
12.
PLAN This problem can be solved by using concept involved in
chemical properties of xenon oxide and xenon fluoride.
XeF6 on complete hydrolysis produces XeO3.
XeO3 on reaction with OH− produces HXeO−4 which on further
treatment with OH− undergo slow disproportionation reaction
and produces XeO64− along with Xe(g ), H2O(l ) and O2 (g ) as a
by-product.
Oxidation half-cell in basic aqueous solution
HXeO− + 5OH− → XeO4− + 3H O + 2e–
4
6
4
2
Balanced overall disproportionation reaction is
4HXeO−4 + 8OH− → 3XeO64− + Xe + 6H2O
14
4244
3
2 products
Complete sequence of reaction can be shown as
XeF6+3H2O
The chemical reactivity follows the order.
XeO3+ 3H2F2
–
OH
F2 > Cl2 > Br2 > I2.
The highest reactivity of fluorine is attributed to two factors:
(i) The low dissociation energy of F  F bond (which results in
low attraction energy for the reaction).
(ii) Very strong bonds which are formed. Both properties arise
from, small size of fluorine. I2 is being the least reactive
halogen, it requires a catalyst for the reaction.
H2 + I2 → 2HI
7. In XeOF4 , Xe is sp3d 2-hybridised. Geometry of the molecule is
octahedral, but shape of the molecule is square pyramidal.
According to VSEPR, theory it has one π bond. Remaining six
electron pairs form an octahedron with one position occupied by a
lone pair.
HXeO–4
OH–/H2O (disproportionation)
XeO64-(s) + Xe(g) + H2O(l ) + O2(g)
Thus, (c) is the correct answer.
13. Decreasing order of strength of oxoacids
HClO4 > HClO3 > HClO2 > HOCl
Reason Consider the structures of conjugate bases of each
oxyacids of chlorine.
O
O
Number of bp = 5
Xe
F
Cl
F
Number of lp = 1
F
Here, Xe contains one lone pair of electrons.
2
Reduction half-cell in basic aqueous solution
HXeO− + 3H O + 6e− → Xe + 7OH−
6. Chemical reactivity of halogens decreases down the group.
F
0
Since, Xe undergoes oxidation while O undergoes reduction.
So, it is an example of redox reaction.
Total number of S S bond = 8
Thus, the correct answer is (b).
∴
+ 6
XeF4 + O2 ( F2 ) → X eF6 + O 2
O
O
O
– O
Cl
O
O
–
Cl
O
O–
Cl—O –
Negative charge is more delocalised on ClO4− due to
resonance, hence, ClO−4 is more stable (and less basic).
246 p-Block Elements-II
Hence, we can say as the number of oxygen atom(s) around
Cl-atom increases as oxidation number of Cl-atom increases and
thus, the ability of loose the H+ increases.
Weak acid have strong conjugate base thus hypochlorite ion
has strongest conjugate base. Therefore, statement (a) is
correct.
(b) Hypochlorite ion is linear and perchlorate ion is tetrahedral
and there is no effect of lone pair on hypochlorite ion. Thus
statement (b) is correct.
14. In XeO2F2, the bonding arrangement around the central atom Xe is
Xe
O
F
4σ bonds + 1.0 l p = 5
Hybridisation of Xe = sp3d
O
F
Cl—O–
sp3d-hybridisation corresponds to trigonal bipyramidal geometry.
Remove
lone-pair
F
O
See-saw shape
15. Sodium thiosulphate, Na 2S2O3 gets oxidised by chlorine water
as Na 2S2O3 + 4Cl 2 + 5H2O → 2NaHSO4 + 8HCl
FeCl 3 oxidises Na 2S2O3 to Na 2S4O6.
Thus, statement (d) is correct.
25. (a) ClO−4 is more stable than ClO− .
(b) Incorrect : Cl 2 + H2O → HCl + HOCl
(c)
O
17. Amongst oxyacids of a given halogen, higher the oxidation
number of halogen, stronger the acid. Hence,
HO
HOCl < HClO2 < HClO3 < HClO4.
18. Pseudo halides must contain atleast one nitrogen atom.
19. Among oxyacids of halogens, if there are same number of
oxygens bonded to central atom, higher the electronegativity of
halogen, stronger the acid. Hence,
IOH < BrOH < ClOH
20. All others has at least one S-S linkage.
KF + HF → K+ + HF2−
21. Among halogens, oxidising power decreases from top to bottom.
Hence, the upper halogen oxidises lower halides from aqueous
solution. Chlorine will oxidise bromide into bromine.
22. Moist chlorine gives nascent oxygen, act as oxidising agent :
Cl 2 + H2 O → HCl + HOCl
HOCl → HCl +
Cl
O
23. Fluorine, being the most electronegative, its size is very small.
24. (a) Order of acid strength different oxyacids of chlorine are :
HClO
(Hypochlorous acid)
< HClO3 <
(Chloric acid)
(+7)
HClO4
(Perchloric acid)
Cl
HO
sp3
(d) HClO4 is stronger acid than H2 O.
26. Colour of halogen arises due to transition from HOMO to LUMO
in the visible region. On moving down a group, the difference in
energy between HOMO and LUMO decreases electronic
transition occur more easily and colour intensity increases.
27.
Compounds
Hydridisation
BrF5
sp3d 2
Lone pair on
central atom
Structures
F
F
Br
F
F
1
F
F
ClF3
F
sp3d
Therefore, it does not have a tendency to loose electrons. Hence,
HF does not act as a reducing agent.
(+5)
O
sp3
[O]
nascent oxygen
(bleaching action)
(+1)
Perchlorate
ion
Thus, statement (c) is incorrect.
(d) The hypochlorite ion oxidises the sulphite ion to
sulphate ion, because HOCl is the strongest oxidising
Cl oxyacids,
ClO − + SO 23− → SO 24− + Cl −
NOTE According to Bent’s rule, the more electronegative atoms must
be present on axial position. Hence, F are kept on axial
positions.
2KMnO4 + KI + H2O → 2KOH + 2MnO2 + KIO3
Chlorate
ion
O—Cl==O
While in hypochlorite ion, chlorite ion Cl(+ 1) is oxidised to
chlorate, Cl(+ 5) and reduced to chloride, Cl(− 1) ion.
3ClO− → ClO3− + 2 Cl −
O
16. I– is oxidised by MnO–4 in alkaline medium to form IO–3
–
(c) In the disproportionation reaction, chlorate ion Cl(+ 5 ) is
oxidised to perchlorate, Cl(+ 7) and reduced to chloride,
Cl (−1).
4ClO −3 → 3ClO 4− + Cl −
Xe
F
O—Cl==O
Hypochlorite
ion
Also, in trigonal bipyramidal geometry, lone pairs remain
present on equatorial positions in order to give less electronic
repulsion.
F
F
O
O
Xe
–
2
Cl
F
XeF4
sp3d 2
F
F
SF4
sp3d
F
2
Xe
F
1
p-Block Elements-II
H  O  Cl == O
H  O  Cl == O
⇒ m mol of OCl − =
the same manner.
32. Bleaching powder is Ca(OCl)Cl. Therefore, the oxoacid whose
salt is present in bleaching powder is HOCl. Anhydride of HOCl
is Cl 2O as
O
(iv)
(iii)
2 HOCl → Cl 2O + H2O
−
(a) Number of Cl == O bonds in (ii) and
(iii) together is three. Hence, wrong.
(b) Number of Lone Pair on Cl in (ii) and (iii) together is three.
Hence, correct.
(c) In (iv), Cl is sp3-hybridised. Hence, correct.
(d) Amongst (i) to (iv), the strongest acid is (iv). Hence, wrong.
Passage 1 Q. Nos. (29-30)
Q
33. Ar, being inert, provide inert atmosphere in arc welding, and
prevent from undesired oxidation.
••
34. O == Xe == O
35. Both XeF4 and XeF6 are strong oxidising agent.
NaOH
→ NaOCl
36. (CH3 )2 SiCl 2 + H2O → (CH3 )2 Si(OH)2 + 2HCl
P
CH3
NaOH
HClO3 →
NaClO3
chloric acid
Passage 3 Q.Nos. (33 to 35)
Xe is sp3-hybridised with one lone pair. Hence, molecule of
XeO3 has pyramidal shape.
P
hot
6NaOH + 3Cl 2 → 5NaCl + NaClO3 + 3H2O
hypochlorous
acid
NOTE The oxidation number of element in anhydride and oxoacid
remains the same.

O
Cold
2NaOH + Cl 2 → NaCl + NaOCl + H2O
HOCl
12
= 6 m mol. Remaining part is solved in
2
==
H  O  Cl == O
(ii)
O
==
H  O  Cl
(i)
O
==
28.
Polymerisation
Q
CH3
R
10 SO2Cl 2 + P4 → 4PCl 5 + 10 SO2
Passage 2 Q.Nos. (31-32)
31. The involved redox reactions are :
I2 +
−
→ 2I +
S4O62−
Also the n-factor of S2O32− is one as
2S2O32− → S4O62− + 2e−
[one ‘e’ is produced per unit of S2O32− ]
CH3
…(i)
37. KI + I2 → KI3
…(ii)
38. Among HX, acidic strength increases from HF to HI.
39. XeF4 reacts with O 2F2 to form XeF6 ⋅ O 2F2 is fluoronating
reagent.
143 K
XeF4 + O 2F2 → XeF6 + O 2
⇒ Molarity of Na 2S2O3 = 0.25 N × 1 = 0.25 M
⇒ m mol of Na 2S2O3 used up = 0.25 × 48 = 12
Now from stoichiometry of reaction (ii)
12 m mol of S2O2−
3 would have reduced 6 m mol of I2.
From stoichiometry of reaction (i)
m mol of OCl − reduced = m mol in I2 produced = 6
⇒ Molarity of household bleach solution =
CH3
silicone
3
O2
2
1
Cl 2 + H2O → HCl + HOCl → HCl + O2
2
VCl 5 + H2O → VOCl 3 + 2HCl
T
2H+ + OCl − + 2I− → Cl − + I2 + H2O
Cl
3XeF4 + 6H2O → XeO3 + 2Xe + 12HF +
S
PCl 5 + 4H2O → H3PO4 + 5 HCl
2S2O2−
3
CH3
—Si—O—Si—O—Si—O
Cl 2 + SO2 → SO2Cl 2
R
247
6
= 0.24 M
25
Shortcut Method
Milliequivalent of Na 2S2O3 =milliequivalent of OCl −
= 0.25 × 48 =12
Also n-factor of OCl − = 2 [Cl + → Cl − , gain of 2e− ]
(Y)
The structore of XeF6 is
F
F
F
Xe
F
F
F
Y compound (XeF6 ) has 3 lone pair in each fluorine and one lone
pair in xenon.
Hence, total number of lone pairs electrons is 19.
40. Br2 is disproportionated in basic medium as
3Br2 + 3Na 2CO3 → 5NaBr + NaBrO3 + 3CO2
41. 2XeF4 + 3H2O → Xe + XeO3 + F2 + 6HF
248 p-Block Elements-II
42.
F
F
F
Xe
Xe
linear
O
Xe
F
F
F
F
square planar
F
49. 2IO−3 + 5HSO−3 → I2 + H2O + 3HSO−4 + 2SO24−
50. Cl 2 + FeBr2 → FeCl 3 + Br2
51. ClO−3 + 6I− + 6H2SO4 → Cl − + 6HSO−4 + 3I2 + 3H2O
O
see-saw shaped
43. Halogen above in the group oxidises halide below to it from
their aqueous solution, e.g.
Cl 2 + 2I− (aq) → 2Cl − + I2
40 ° C
44. Ca(OH)2 + Cl 2 →
CaOCl 2 + H2O
45. (i) 2KI + Cl 2 → 2KCl + I2
(ii) 2KClO3 + I2 → 2KIO3 + Cl 2
46. (i) Due to small size and high electron density of fluorine atom,
there exist a significant repulsions between fluorine atoms
in F2 , they have greater tendency to get apart. Hence, bond
energy of F2 is less than that of Cl 2 . This is against to
bond-length bond-energy relationship,.
(ii) Sulphur dioxide is a more powerful reducing agent in
alkaline medium because nascent hydrogen is produced in
the presence of moisture
i.e. SO2 + 2H2O → H2SO4 + 2H
And alkaline solution neutralises the acid i.e. H2SO4 and
shift the equilibrium in the forward direction producing
more nascent hydrogen. But in acidic medium the
equilibrium will suppressed resulting in a lesser amount of
nascent hydrogen.
52. (i) Bond strength is inversely related to bond length. Hence,
bond energy : HI < HBr < HCl < HF
(ii) HI(– 1) < I2 (0) < ICl(+1) < HIO4 (+7)
53. F2 itself, is the strongest oxidising agent. Therefore, chemical
reagent cannot oxidise fluoride to fluorine.
54. Complete and balance the following reactions
Cl 2 + 2OH− → Cl − + ClO− + H2O
55. The bleaching action of bleaching powder is due to presence of
available chlorine, but in contact of moisture, it releases chlorine
decreasing the amount of available chlorine. Hence, bleaching
property decreases gradually as bleaching powder is kept in
open container for long time.
56. (i) HBr is a stronger reducing agent, reduces cencentrated
H2SO4 to SO2. Hence, HBr cannot be prepared by heating
bromide salts with concentrated H2SO4.
(ii) Hypochlorous acid is acidic in nature, therefore it turns blue
litmus paper into red. However, HOCl is also an oxidising
acid (bleaching), it bleaches red colour to finally colourless.
40 ° C
57. (i) Ca(OH)2 + Cl 2 → CaOCl 2 + H2 O
(ii) 3Cu + 8HNO3 (dil) → 3Cu(NO3 )2 + 4H2 O + 2NO
(iii) 2NaCl + 2H2 SO4 + MnO2 → Na 2 SO4 + MnSO4
47. NaBrO3 + 3F2 → 3F2O + NaBr
48. HOCl < HOClO < HOClO2 < HOClO3
+ 2H2O + Cl 2
∆
(iv) Al 2 O3 + 3C + 3Cl 2 → 2AlCl 3 + 3CO
17
Transition and
Inner-Transition Elements
Objective Questions I (Only one correct option)
−
1. The shape/structure of [ XeF5 ] and XeO 3F2, respectively, are
(2020 Main, 2 Sep II)
(a) pentagonal planar and trigonal bipyramidal
(b) octahedral and square pyramidal
(c) trigonal bipyramidal and pentagonal planar
(d) trigonal bipyramidal and trigonal bipyramidal
8. The maximum number of possible oxidation states of
actinoides are shown by
(2019 Main, 9 April II)
(a) berkelium, (Bk) and californium (Cf)
(b) nobelium (No) and lawrencium (Lr)
(c) actinium (Ac) and thorium (Th)
(d) neptunium (Np) and plutonium (Pu)
9. The lanthanide ion that would show colour is
2. The incorrect statement(s) among (1)-(3) is (are)
(2020 Main, 4 Sep II)
1. W(VI) is more stable than Cr(VI).
2. in the presence of HCl, permanganate titrations provide
satisfactory results.
3. some lanthanoid oxides can be used as phosphors.
(a) 2 and 3 only
(c) 1 only
(b) 2 only
(d) 1 and 2 only
3. Thermal decomposition of a Mn compound (X ) at 513 K
results in compound (Y ), MnO2 and a gaseous product. MnO2
reacts with NaCl and concentrated H2 SO4 to give a pungent
gas Z. X , Y and Z, respectively, are
(2019 Main, 12 April II)
(a) K 3MnO4 , K 2MnO4 and Cl 2 (b) K 2MnO4 , KMnO4 and SO2
(c) KMnO4 , K 2MnO4 and Cl 2 (d) K 2MnO4 , KMnO4 and Cl 2
4. The pair that has similar atomic radii is
(a) Mn and Re
(c) Sc and Ni
(b) Ti and Hf
(d) Mo and W
5. The correct order of the first ionisation enthalpies is
(2019 Main, 10 April II)
(a) Mn < Ti < Zn < Ni
(c) Zn < Ni < Mn < Ti
(b) Ti < Mn < Zn < Ni
(d) Ti < Mn < Ni < Zn
6. The highest possible oxidation states of uranium and
plutonium, respectively, are
(a) 7 and 6
(c) 6 and 4
(2019 Main, 10 April II)
correct order of their spin-only magnetic moment is
(2019 Main, 10 April I)
3+
2+
2+
(a) Sc < Ti < Ti < V
(c) Ti 3+ < Ti 2+ < Sc3+ < V2+
3+
(b) Sm 3+
(c) La 3+
10. The correct order of atomic radii is
(a) Ho > N > Eu > Ce
(c) Eu > Ce > Ho > N
(b) Sc < Ti 3+ < V2+ < Ti 2+
(d) V2+ < Ti 2+ < Ti 3+ < Sc3+
(d) Lu3+
(2019 Main, 12 Jan II)
(b) N > Ce > Eu > Ho
(d) Ce > Eu > Ho > N
4 KOH, O
2
11. A →
2B + 2H 2O
(Green)
4 HCl
3B → 2C
+ MnO2 + 2H 2O
(Purple)
H 2O, KI
2C → 2 A + 2KOH + D
In the above sequence of reactions, A and D, respectively, are
(2019 Main, 11 Jan II)
(a) KI and KMnO4
(c) KI and K2MnO4
(b) MnO2 and KIO3
(d) KIO3 and MnO2
12. The element that usually does not show variable oxidation
states is
(a) Sc
(2019 Main, 11 Jan I)
(b) Cu
(c) Ti
(d) V
st
13. The 71 electron of an element X with an atomic number of 71
enters into the orbital
(a) 4 f
(b) 6p
(2019 Main, 10 Jan II)
(c) 5d
(d) 6s
14. The effect of lanthanoid contraction in the lanthanoid series
of elements by and large means
(b) 6 and 7
(d) 4 and 6
7. Consider the hydrated ions of Ti 2 + , V2 + , Ti 3 + and Sc3+ . The
3+
(2019 Main, 8 April I)
(a) Gd3+
(a)
(b)
(c)
(d)
(2019 Main, 10 Jan I)
increase in atomic radii and decrease in ionic radii
decrease in both atomic and ionic radii
increase in both atomic and ionic radii
decrease in atomic radii and increase in ionic radii
15. The transition element having least enthalpy of atomisation is
(2019 Main, 9 Jan II)
(a) Zn
(b) V
(c) Fe
(d) Cu
250 Transition and Inner-Transition Elements
16. In the following reactions, ZnO is respectively acting as a/an
25. Consider the following reaction,
(2013 Main)
(i) ZnO + Na2O → Na2 ZnO2
(ii) ZnO + CO2 → ZnCO3
z
H2 O
2
The values of x, y and z in the reaction are, respectively
(a) base and acid
(c) acid and acid
(a) 5, 2 and 16
(c) 2, 5 and 16
(2017 Main)
(b) base and base
(d) acid and base
17. Sodium salt of an organic acid ‘X ’ produces effervescence
with conc. H2SO4 . ‘X ’ reacts with the acidified aqueous
CaCl2 solution to give a white precipitate which decolourises
acidic solution of KMnO4 . ‘X ’ is
(2017 Main)
18. Which of the following combination will produce H 2 gas?
(2017 Adv.)
(a) Fe metal and conc. HNO3
(b) Cu metal and conc. HNO3
(c) Au metal and NaCN (aq) in the presence of air
(d) Zn metal and NaOH (aq)
(b) VO2
(d) TiO2
respectively, produce
(2016 Main)
(b) NO and N2O
(d) N2O and NO2
Zn 2+ , respectively, are
(2016 Main)
(a) octahedral, square planar and tetrahedral
(b) square planar, octahedral and tetrahedral
(c) tetrahedral, square planar and octahedral
(d) octahedral, tetrahedral and square planar
23. Which series of reactions correctly represents chemical
relations related to iron and its compound?
O2 , Heat
(b) Fe →
FeO
Cl , Heat
H SO ,O
Dil. H 2 SO4
→
Heat, air
2
(c) Fe →
FeCl 3 → FeCl 2
CO, 600°C
2
(d) Fe →
Fe3O4 →
FeO
O , Heat
(2014 Main)
Heat
Fe2 (SO4 ) 3
FeSO4
Heat
→
Zn
→
Fe
(2005)
(d) K 2MnO4
colour in solution?
(a) VOCl 2 ; FeCl 2
(c) MnCl 2 ; FeCl 2
(2005, 1M)
(b) CuCl 2 ; VOCl 2
(d) FeCl 2 ; CuCl 2
I − converts into
(a) IO−3
(b) I2
(2004)
(c) IO−4
(d) IO−
oxidation state is
(2004, 1M)
(a) MnO2 , FeCl 3
(b) [MnO4 ]− , CrO2 Cl 2
(c) [Fe(CN)6]3− , [Co(CN)3]
(d) [NiCl 4 ]2 − , [CoCl 4 ]−
→
Fe
listed below with atomic numbers. Which one of them is
expected to have the highest E °M 3+ / M 2+ value? (2013 Main)
(b) Mn (Z = 25)
(d) Co (Z = 27)
(b) heating NH4NO3
(d) Na(comp.) + H2O2
33. When MnO2 is fused with KOH, a coloured compound is
(a) K 2MnO4, purple green
(c) Mn 2O3, brown
(2003, 1M)
(b) KMnO4, purple
(d) Mn 3O4, black
34. Amongst the following, identify the species with an atom in +
6 oxidation state
(a) MnO−4
(b) Cr(CN)63−
(a) oxygen
(c) nitrous oxide
Fe
CO, 700° C
→
(a) heating NH4NO2
(c) Mg3N2 + H2O
(2000, 1M)
(c) NiF62 −
(d) CrO2Cl 2
35. On heating ammonium dichromate, the gas evolved is
Fe
24. Four successive members of the first row transition elements
(a) Cr (Z = 24)
(c) Fe (Z = 26)
(c) KMnO4
formed, the product and its colour is
(b) K 3 [ Co (NO2 )6 ]
(d) BaCrO 4
2
4
2
→
(b) FeSO4
(2004, 1M)
(2015 Main)
FeSO4
(2012)
(d) violet
32. (NH4 ) 2 Cr2 O7 on heating gives a gas which is also given by
22. Which of the following compounds is not yellow coloured?
Dil. H SO
CuSO4 is
(a) orange-red (b) blue-green (c) yellow
31. The pair of compounds having metals in their highest
21. The geometries of the ammonia complexes of Ni 2 + , Pt 2 + and
2
4
→
27. The colour of light absorbed by an aqueous solution of
30. When I − is oxidised by MnO−4 in alkaline medium,
20. The reaction of zinc with dilute and concentrated nitric acid,
(a) Fe
(2013 Main)
29. Which of the following pair is expected to exhibit same
(2016 Main)
(a) Zn 2 [ Fe (CN)6 ]
(c) (NH4 )3 [ As (Mo3O10 )4 ]
correct order of the property stated against it?
(a) V2+ < Cr 2+ < Mn 2+ < Fe2+ : paramagnetic behaviour
(b) Ni 2+ < Co2+ < Fe2+ < Mn 2+ : ionic size
(a) KI
ferromagnetic?
(a) NO2 and NO
(c) NO2 and N2O
26. Which of the following arrangements does not represent the
28. Which of the following will not be oxidised by O3 ?
19. Which of the following compounds is metallic and
(a) CrO2
(c) MnO2
(b) 2, 5 and 8
(d) 5, 2 and 8
(c) Co3+ < Fe3+ < Cr 3+ < Sc3+ : stability in aqueous solution
(d) Sc < Ti < Cr < Mn : number of oxidation states
(b) HCOONa
(d) Na 2C2O4
(a) C6H5COONa
(c) CH3COONa
xMnO−4 + y C2 O42 − + zH+ → xMn 2+ + 2 y CO2 +
(b) ammonia
(d) nitrogen
36. In the dichromate dianion
(a)
(b)
(c)
(d)
(1999, 2M)
(1999, 2M)
4 Cr—O bonds are equivalent
6 Cr—O bonds are equivalent
all Cr—O bonds are equivalent
all Cr—O bonds are non-equivalent
37. Which of the following compounds is expected to be
coloured?
(a) Ag 2 SO4
(1997, 1M)
(b) CuF2
(c) MgF2
(d) CuCl
Transition and Inner-Transition Elements 251
38. Ammonium dichromate is used in some fireworks. The green
coloured powder blown in the air is
(a) CrO3
(b) Cr2 O3
(1997, 1M)
(c) Cr
(d) CO
39. The reaction which proceed in the forward direction is
(a) Fe2 O3 + 6HCl → 2FeCl 3 + 3H 2 O
(1991, 1M)
(b) NH 3 + H 2 O + NaCl → NH 4 Cl + NaOH
46. Consider the following reactions (unbalanced).
(c) SnCl 4 + Hg 2 Cl 2 → SnCl 2 + 2HgCl 2
(d) 2CuI + I2 + 4H + → 2Cu 2 + + 4KI
40. Zinc-copper couple that can be used as a reducing agent is
obtained by
(a)
(b)
(c)
(d)
(1984, 1M)
mixing of zinc dust and copper gauge
zinc coated with copper
copper coated with zinc
zinc and copper wires welded together
(b) 2
(d) 8
(1981, 1M)
(b) Cu
(c) Mg
(1980, 1M)
(d) Al
43. Which of the following dissolves in concentrated NaOH
solution?
(1980, 1M)
(a) Fe
(b) Zn
(c) Cu
(d) Ag
Objective Questions II
(One or more than one correct option)
(a) Aqua-regia is prepared by mixing conc. HCl and conc. HNO3
in 3 : 1 (v / v) ratio
(b) The yellow colour of aqua-regia is due to the presence of
NOCl and Cl 2
(c) Reaction of gold with aqua-regia produces an anion having Au
in +3 oxidation state
(d) Reaction of gold with aqua regia produces NO2 in the absence
of air
48. The correct statement(s) about Cr 2+ and Mn 3+ is/are [atomic
44. In an experiment, mgrams of a compound X (gas/liquid/solid)
taken in a container is loaded in a balance as shown in figure I
below.
(I)
Balanced;
Magnetic field absent
47. With reference to aqua-regia, choose the correct option(s).
(2019 Adv.)
42. One of the constituent of German silver is
(a) Ag
Zn + Hot conc. H 2SO 4 → G + R + X
Zn + conc. NaOH → T + Q
G + H 2S + NH 4OH → Z (a precipitate) + X + Y
Choose the correct option(s).
(2019 Adv.)
(a) The oxidation state of Zn in T is +1
(b) R is a V-shaped molecule
(c) Bond order of Q is 1 in its ground state
(d) Z is dirty white in colour
41. How many unpaired electrons are present in Ni 2 + ?
(a) 0
(c) 4
(a) Both Y and Z are coloured and have tetrahedral shape
(b) Y is diamagnetic in nature while Z is paramagnetic
(c) In both Y and Z, π-bonding occurs between p-orbitals of
oxygen and d-orbitals of manganese
(d) In aqueous acidic solution, Y undergoes disproportionation
reaction to give Z and MnO2
(II)
Upward deflection;
Magnetic field present
(III)
Downward deflection;
Magnetic field present
number of Cr = 24 and Mn = 25]
(2015 Adv.)
(a) Cr 2+ is a reducing agent
(b) Mn 3+ is an oxidising agent
(c) Both Cr 2+ and Mn 3+ exhibit d 4 electronic configuration
(d) when Cr 2+ is used as a reducing agent, the chromium ion
attains d 5 electronic configuration
49. Fe3+ is reduced to Fe2+ by using
(2015 Adv.)
(a) H2O2 in presence of NaOH (b) Na 2O2 in water
(c) H2O2 in presence of H2SO4 (d) Na 2O2 in presence of H2SO4
m
50. Which of the following halides react(s) with AgNO3 ( aq ) to
X
N
S
N
S
give a precipitate that dissolves in Na 2 S2 O3 ( aq ) ?
(a) HCl
(b) HF
(c) HBr
(d) HI
51. Reduction of the metal centre in aqueous permanganate ion
Magnet
In the presence of a magnetic field, the pan with X is either
deflected upwards (figure II), or deflected downwards (figure
III), depending on the compound X. Identify the correct
statement(s).
(2020 Adv.)
(a) If X is H2O(l ) , deflection of the pan is upwards.
(b) If X is K 4 [Fe(CN)6 ](s), deflection of the pan is upwards.
(c) If X is O 2 (g ), deflection of the pan is downwards.
(d) If X is C 6H6 (l ), deflection of the pan is downwards.
45. Fusion of MnO 2 with KOH in presence of O 2 produces a salt
W . Alkaline solution of W upon electrolytic oxidation yields
another salt X . The manganese containing ions present in W
and X , respectively, areY and Z. Correct statement(s) is (are)
(2019 Adv.)
involves
(a)
(b)
(c)
(d)
(2011)
three electrons in neutral medium
five electrons in neutral medium
three electrons in alkaline medium
five electrons in acidic medium
52. Which of the following statement (s) is/are correct?
(1998)
(a) The electronic configuration of Cr is [Ar] 3d 5 4s1 (Atomic
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 of
the opposite type (Atomic number of Ag = 47)
(d) The oxidation state of nitrogen in HN 3 is – 3
252 Transition and Inner-Transition Elements
53. Which of the following statement(s) is/are correct when a
mixture of NaCl and K 2 Cr2 O7 is gently warmed with
(1998, 2M)
conc. H2 SO4 ?
(a) A deep red vapours is formed
(b) Vapours when passed into NaOH solution gives a yellow
solution of Na 2 CrO4
(c) Chlorine gas is evolved
(d) Chromyl chloride is formed
54. Which of the following alloys contains Cu and Zn?
(a) Bronze
(c) Gun metal
(b) Brass
(d) Type metal
(1993, 1M)
63. Mn 2 + can be oxidised to MnO−4 by ………
(SnO2 , PbO2 , BaO2 )
True/False
64. Dipositive zinc exhibit paramagnetism due to loss of two
electrons from 3d-orbitals of neutral atom.
65. Copper metal reduces Fe
(d) CrCl 3
56. Potassium manganate (K 2 MnO4 ) is formed when
67. An acidified solution of potassium chromate was layered
with an equal volume of amyl alcohol. When it was shaken
after the addition of 1 mL of 3% H2 O2 , a blue alcohol layer
was obtained. The blue colour is due to the formation of a
chromium (VI) compound ‘X’. What is the number of
oxygen atoms bonded to chromium through only single
bond in a molecule of X ?
(2020 Adv.)
(1988, 2M)
(a) chlorine is passed into aqueous KMnO4 solution
(b) manganese dioxide is fused with KOH in air
(c) formaldehyde reacts with potassium permanganate in the
presence of strong alkali
(d) potassium permanganate reacts with conc. H 2 SO4
68. In neutral or faintly alkaline solution, 8 moles of
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.
Statement II
form Zn 2 + .
2+
is diamagnetic.
The electrons are lost from 4s orbital to
permanganate anion quantitative oxidise thiosulphate
anions to produce X moles of a sulphur containing product.
The magnitude of X is ... .
(2016 Adv.)
69. In
59. The compound Y Ba 2 Cu 3 O7 which show super conductivity
has copper in oxidation state …… assuming that the rare earth
element Yttrium in its usual +3 oxidation state. (1994, 1M)
60. The outermost electronic configuration of Cr is ……
(1994, 1M)
61. Fehling’s solution A consists of an aqueous solution of
copper sulphate while Fehling’s solution B consists of an
alkaline solution of ……
(1990, 1M)
62. The salts ……… and …… are isostructural. (FeSO4 ⋅7H2 O,
(1990, 1M)
the
complex
MnO–4 .
+
For this
H2 SO4
change of [MnO−4 ] is
(2015 Adv.)
70. Consider the following list of reagents, acidified K 2 Cr2 O7 ,
alkaline KMnO4 , CuSO4 , H2 O2 , Cl 2 , O3 , FeCl 3 , HNO3 and
Na 2 S2 O3 . The total number of reagents that can oxidise
aqueous iodide to iodine is
(2014 Adv.)
Subjective Questions
71.
(B )
W hite fumes
with pungent smell
Fill in the Blanks
aqueous
reaction, the ratio of the rate of change of [H ] to the rate of
(1998, 2M)
strong acid is added, it changes its colour from yellow to
orange.
Statement II The colour change is due to the change in
oxidation state of potassium chromate.
(1988, 2M)
dilute
diaquadioxalatoferrate (II) is oxidised by
58. Statement I To a solution of potassium chromate if a
CuSO4 ⋅ 5H2 O,MnSO4 ⋅ 4H2 O,ZnSO4 ⋅ 7H2 O
in an acidic medium. (1982, 1M)
of KMnO4 and KI in weakly basic solution, what is the
number of moles of I2 released for 4 moles of KMnO4
consumed?
(2020 Adv.)
(1990, 1M)
57. Statement I Zn
(1987, 1M)
66. In the chemical reaction between stoichiometric quantities
in case of
(c) Co(NO3 )2
2+
Numerical Answer Type Questions
55. The aqueous solution of the following salts will be coloured
(a) Zn(NO3 )2 (b) LiNO3
(e) potash alum
(1981, 1M)
Moist air
←
MCl 4
( M = Transition
element colourless)
Zn
→
(A)
(purple colour)
Identify the metal M and hence MCl 4 . Explain the difference
in colours of MCl 4 and A.
(2005)
72. Give reasons : CrO3 is an acid anhydride.
(1999, 2M )
73. A compound of vanadium has a magnetic moment of
1.73 BM. Work out the electronic configuration of the
vanadium ion of the compound.
(1997)
74. Write balanced equations for the following
(i) Oxidation of hydrogen peroxide with potassium
permanganate in acidic medium.
(1997, 2M)
(ii) Reaction of zinc with dilute nitric acid.
Transition and Inner-Transition Elements 253
75. Complete and balance the following reactions
(i) [MnO4 ]
2−
+H
+
79. Give reason in one or two sentences
−
“Most transition metal compounds are coloured.”(1986, 1M)
→ ..... + [MnO4 ] + H2 O
(ii) SO2 ( aq) + Cr2 O72 − + 2H+ → … + … +
(1994, 2M)
80. Show with balanced equations for the reactions when
(1993, 1M)
(i) potassium permanganate interacts with manganese
dioxide in the presence of potassium hydroxide.
(ii) potassium ferricyanide is heated with concentrated
(1985, 2M)
sulphuric acid.
77. Write the balanced chemical equations for the following
81. State the conditions under which the following
76. Complete and balance the following reaction.
(NH4 ) 2 S2 O8 + H2 O + MnSO4 → …… + …… + ……
reactions.
preparations are carried out. Give necessary equations
which need not be balanced.
“Potassium permanganate from manganese dioxide”
(i) A mixture of potassium dichromate and sodium chloride is
heated with concentrated H2 SO4 .
(ii) Potassium permanganate is added to a hot solution of
manganous sulphate.
(1990, 2M)
(1983, 1M)
82. Complete and balance the following reactions
78. Complete and balance the following reactions.
(1983, 2M)
(i) Zn + NO−3 → Zn 2+ + NH+4
(i) Mn 2 + + PbO2 → MnO4− + H2 O
(ii) Cr2 O27 − + C2 H4 O → C2 H4 O2 + Cr 3+
(ii) Ag + + AsH3 → H3AsO3 + H+
(1987, 2M
Answers
1. (a)
2. (b)
3. (c)
4. (d)
37. (b)
38. (b)
39. (a)
40. (b)
41. (b)
42. (b)
43. (b)
44. (a,b,c)
46.
50.
54.
58.
62.
5. (d)
6. (b)
7. (a)
8. (d)
9. (b)
10. (c)
11. (b)
12. (a)
13. (c)
14. (b)
15. (a)
16. (d)
17. (d)
18. (d)
19. (a)
20. (d)
21. (a)
22. (a)
23. (d)
24. (d)
25. (c)
26. (a)
27. (a)
28. (c)
45.
49.
53.
57.
61.
29. (b)
30. (a)
31. (b)
32. (a)
63. PbO 2
64. F
65. F
66. (6)
33. (a)
34. (d)
35. (d)
36. (b)
67. (4)
68. (6)
69. (8)
70. (7)
(a,c,d)
(a, b)
(a, b, d)
(b)
Rochelle salt
(b,c,d)
47. (a,b,c)
48. (a, b, c)
(a, c, d)
51. (a, c, d)
52. (a, b, c, d)
(b, c)
55. (c, d)
56. (b, c)
(c)
59. x = +7 / 3 60. 3d 5 4s 1
FeSO 4 ⋅ 7 H 2O and ZnSO 4 ⋅ 7H 2O
Hints & Solutions
3. Thermal decomposition of Mn compound (X), i.e. KMnO4 at
1. [XeF5 ]
F
F
s
Xe
F
F
F
Xe : sp3d3-hybridised
Geometry = Pentagonal
bipyramidal
Shape/Structure = Pentagonal
planar
F
Xe
O
2.
F
K
2KMnO4 513

→ K2MnO4 + MnO2 + O2 (g )
( X)
∆
(Y )
MnO2 + 4NaCl + 4 H2SO4 →
MnCl 2 + 4NaHSO4 + 2H2O + Cl 2 (g )
XeO3F2
O
513 K results in compound Y(i.e. K 2MnO4), MnO2 and a
gaseous product. MnO2 reacts with NaCl and concentrated
H2SO4 to give a pungent gas Z(i.e. Cl2). The reactions
involved are as follows :
Xe: sp3d-hybridised
Geometry = Shape/ structure
O
= Trigonal bipyramidal
[because, Xe in XeO3F2 does not
have pair of electrons]
Only statement 2 is incorrect, whereas 1 and 3 are correct.
KMnO4 will not give satisfactory result when it is titrated by HCl,
because same amount of KMnO4 is consumed to oxidise HCl into Cl 2.
(Z )
Pungent gas.
4. The pair that has similar atomic radii is Mo and W. It is due to
lanthanoid contraction. The factor responsible for lanthanoid
contraction is the imperfect shielding of one electron by
another in the same set of orbitals. Shielding of 4 f is very less
due to its diffused shape. As a result, nuclear charge increases.
Hence, Mo and W have similar atomic radii.
254 Transition and Inner-Transition Elements
5. The 3d-transition series is
Atomic number
Sc
Ti
V
Cr Mn Fe Co Ni
21
22
23
24
25
26
27
Cu Zn
28
29
30
⇒
⇒
⇒
⇒
Outermost
Electronic
Configuration
10. The correct order of atomic radii is
3d24s2
3d54s2
3d84s2
3d104s2
In 1st ionisation, one electron will be removed from 4 s2
subshell/orbital.
With increase in atomic number (Z ), i.e. with increase in number
of protons in the nucleus, effective nuclear charge (Z* ) also
increases from Sc to Zn.
IE ∝ Z*
So, IE order of the given elements will be,
Ti < Mn < Ni < Zn
6. Actinoids show a variety of oxidation states due to comparable
energies of 5 f , 6d and 7s energy levels.
In the actinoids family (5 f -block), uranium (U) neptunium
(Np), plutonium (Pu) and americium (Am) have highest possible
oxidation states of +6, + 7, + 7 and +6 respectively.
7. The spin only magnetic moment (µ ) of each ion can be calculated
Europium (Eu) > Cerium (Ce) > Holmium (Ho) > Nitrogen (N)
199 pm
183 pm
Note
(i) N being the member of p-block and second period, have the
smallest radii.
(ii) Rest of all the 3 members are lanthanides with Eu having
stable half-filled configuration thus with bigger size than rest
two.
(iii) Among Ce and Ho, Ce has larger size which can be explained
on the basis of “Lanthanoid contraction”.
11. When MnO2(A) is fused with alkali in presence of air then
potassium manganate (B) is formed. Potassium manganate (B) is
of green colour which disproportionate in a neutral or acidic
solution to produce potassium permanganate (C). Potassium
permanganate (C) in presence of acidic medium oxidises iodide
to iodate.
The reaction can be shown as:
0
+ 4
∆
(A )
+6
[Q n = No. of unpaired electron(s)] ⇒ µ ∝ n, i.e. higher the
number of unpaired electron, higher will be the value of µ.
Metal ion
Z
n (for metal ion) M (BM)
2
22
2 (3d )
8
Paramagnetic
V2+
23
3 (3d 3 )
15
Paramagnetic
Ti 3+
22
1 (3d 1 )
3
Paramagnetic
Ti
3+
Sc
21
0
0 (3d )
0
Diamagnetic
Thus, the correct order of spin only magnetic moments of given
hydrated ions will be
Sc3+ < Ti 3+ < Ti 2+ < V2+
8. The maximum number of possible oxidation states of actinoids
are shown by neptunium (Np) and plutonium (Pu). These
actinoids exhibit oxidation states of +3, +4, +5 and +6.
3+
9. The lanthanide ion that would show colour is Sm . Colour of a
compound depends on the number of electrons in 4 f -orbitals.
Electronic configuration of given lanthanides are as follows:
Gd 3+ = 4 f 7
Sm 3+ = 4 f 5
La 3+ = 4 f 0
Lu 3+ = 4 f 14
Gd 3+ have half-filled 4 f -orbitals.
La 3+ have no electron in 4 f -orbitals.
Lu 3+ have fully-filled 4 f -orbitals.
Only Sm 3+ contain 4 f 5. The electrons can easily undergoes
excitation. That result in a formation of colour.
2K 2 Mn O4
+ 2H2O
(B )
Potassium manganate
(Green)
+ 7
4 HCl
+7
(ii) 3K2 Mn O4 → 2K Mn O4 + Mn O2 + 2H2O + 4KCl
(B )
(C )
Potassium
permanganate
(purple)
Nature
2+
−2
+6 −2
4 KOH, O2
(i) 2 Mn O2 →
as :
µ = n (n + 2) BM
65 pm
176 pm
+ 7
−1
K I , H2O
+ 4
+5
(iii) 2K MnO4  → 2 MnO2 + 2KOH + K IO3
(C )
(A )
(D)
Thus, A and D are MnO2 and KIO3 respectively.
12. The most stable oxidation states in the compounds of the given
transition metals of 3d-series are,
Sc : + 3; Ti : + 3, + 4; V : +2, + 3, + 4 , + 5; Cu : + 1, + 2
The electronic configuration of Sc (Z = 21) is [Ar] 3d 1 , 4 s2.
Due to the presence of only one 3d-electron
(no pairing energy) and two 4s-electrons, they easily ionise to
achieve most stable +3 oxidation state.
13. In the lanthanoid series, atomic number of fourteen 4 f -block
elements ranges from 58 (Ce) to 71 (Lu).
Ytterbium, Yb(Z = 70) has electronic configuration :
[ Xe ] 4 f 14 6s2. So, the 71nth electron of lutetium, Lu (Z = 71)
should enter into 5d orbital and its (here, Lu is ‘X ’) electronic
configuration will be : [ Xe ]4 f 14 5d 1 6s2. It happens so, because
f -block elements have general electronic configuration,
(n − 2) f 1 − 14 (n − 1)d 1− 10ns2. Therefore, option (c) is correct.
14. Lanthanoid contraction in the lanthanoid series takes place due
to the presence of electron(s) in the 4 f -orbitals. f -orbitals have
poor shielding effect. As a result, the effective nucleur charge
will be more experienced by the 5d and 6s- electrons and it will
cause contraction or decrease in both atomic and ionic radii.
15. For transition metals,
°
∆H Atomisation
∝ Strength of metallic bonding
∝ Number of unpaired electrons in the metal atom
Transition and Inner-Transition Elements 255
For the given 3d-transition metals,
V
Fe
Cu
Zn
3d 34 s2 3d64 s2 3d 104 s1 3d 104 s0
n=3 n=4
n=0
n=0
[Q n = no. of unpaired electrons]
The correct reactions are as follows:
(a) Fe + dil. H2SO4 → FeSO4 + H2
1
H2SO4 + 2FeSO4 + O2 → Fe2 (SO4 )3 + H2O
2
∆° H Atomisation (kJ mol −1) = 515 418
The given reaction is incorrect in question
∆
Fe2 (SO4 )3 → Fe2O3 (s) + 3SO3 ↑
339 130
So, absence of unpaired d-electrons and larger size of Zn atoms,
make the crystal lattice of Zn less closely packed.
24. SRP value normally increases from left to right in the period of
d-block elements. Some SRP value are exceptionally higher due
to stability of product ion. e.g.
= + 1.97 V
E°
= + 1.57 V; E°
16. Zinc oxide (ZnO) when react with Na 2O it act as acid while with
Mn 3+ / Mn 2+
CO2 it act as base. Therefore, it is an amphoteric oxide.
Thus, EM° 3+ / M2+ is highest for Co.
ZnO + Na 2 O → Na 2 ZnO2
Acid
Base
Salt
25. The half equations of the reaction are
ZnO + CO2 → ZnCO3
Base
Acid
MnO −4 → Mn 2+
Salt
C2O42− → CO2
17. The reaction takes place as follows
Na 2C2O4 + H2SO4 → Na2 SO4 + H2O
(X)
(Conc.)
The balanced half equations are
MnO−4 + 8H+ + 5e− → Mn 2+ + 4H2O
+ CO ↑ + CO2 ↑
Effervescence
C2O24− → 2CO2 + 2e−
Na 2C2O4 + CaCl 2 → CaC2O4 + 2NaCl
White ppt.
(X)
On equating number of electrons, we get
2MnO−4 + 16H+ + 10e− → 2Mn 2+ + 8H2O
5CaC2O4 + 2KMnO4 + 8H2SO4 → K2SO4 + 5CaSO4
Purple
5C2O42− → 10CO2 + 10e−
+ 2MnSO4 + 10CO2 + 8H2O
On adding both the equations, we get
Colourless
Hence, X is Na 2C2O4.
18.
Zn
Amphoteric
2MnO−4 + 5C2O−4 + 16H+ → 2Mn 2+ + 2 × 5CO2 +
+ 2NaOH → Na 2ZnO2 + H2
19. Only three elements iron (Fe), cobalt (Co) and nickel (Ni) show
26. (a) V2+ = 3 unpaired electrons
Cr 2+ = 4 unpaired electrons
Mn 2+ = 5 unpaired electrons
Fe2+ = 4 unpaired electrons
Hence, the order of paramagnetic behaviour should be
V2+ < Cr 2 + < Fe2 + < Mn 2 +
(b) Ionic size decreases from left to right in the same period.
(c) (As per data from NCERT)
20. Zn + 4HNO3 → Zn (NO3 )2 + 2H2O + 2NO2
(Conc.)
4Zn +10HNO3 → 4Zn (NO3 )2 + N2O + 5H2O
(Dil.)
21. [Ni(NH3 )6 ] 2+ sp3d 2 octahedral
Co3+ / Co2+ = 1.97;
[Pt(NH3 )4 ] 2+ dsp2 square planar
Fe3+ /Fe2+ = 0.77;
[Zn(NH3 )4 ] 2+ sp3 tetrahedral
Cr 3+ / Cr 2+ = − 0.41
22. Zn 2[ Fe(CN)6 ], K3[ Co(NO2 )6 ] and (NH4 )3As [ Mo3O10 ] 4 show
colour due to d-d transition while BaCrO4 is coloured due to
charge transfer phenomenon.
Further according to spectrochemical series the strong ligand
possessing complex has higher energy and hence lower
wavelength. Therefore, complexes containing NO2 , NH+4 , O2−
etc., ligands show yellow colour while CN− forces the complex
to impart white colour.
Spectrochemical series
I− < Br − < S2− < SCN− < Cl − < NO3− < N3− < F− < OH−
< C2O42− ≈ H2O < NCS− < CH3CN < py < NH3 < en
< bipy < Phen < NO−2 < PPh 3 < CN− ≈ CO
PLAN Analyse each reaction given in the question and choose the
correct answer on the basis of oxidation state and stability of
iron compounds. Use the concept of Ellingham diagram to
solve this problem.
16
H2O
2
Thus x , y and z are 2, 5 and 16 respectively.
ferromagnetism at room temperature. CrO2 is also a metallic and
ferromagnetic compound which is used to make magnetic tapes
for cassette recorders.
23.
Co3+ / Co2 +
3+
Sc
is highly stable (It does not show + 2).
(d) The oxidation states increases as we go from group 3 to
group 7 in the same period.
27. The aqueous solution of CuSO4 consist of the complex
[Cu(H2O)4 ]2+ ion which absorbed in orange-red region and
impart deep blue colouration to solution.
28. KMnO4 is itself a very strong oxidising agent, O3 cannot oxidise it.
29. In CuCl 2, Cu 2 + has d 9 configuration, exhibit d-d transition and
show colour. Similarly in VOCl 2, V 4 + has d 1 configuration, can
exhibit d-d transition and show colour.
30.
MnO−4 + I− + OH− → MnO42− + IO−3
256 Transition and Inner-Transition Elements
is in highest oxidation state possible for Cr.
32. Ammonium dichromate on heating produces N2 (g). NH4NO2
also gives N2 on heating :
∆
(NH4 )2 Cr2O7 → N2 + Cr2O3 + 4H2O
∆
NH4NO2 → N2 + 2H2O
33. K2MnO4 (purple green) is formed which is the first step of
(c) X = O2 (g)
[Here, O2 (g ) is paramagnetic due to two-unpaired electrons
present in π * (antibonding orbitals). Hence, statement (c) is
correct.
(d) X = C6H6(l)
(Here, C6H6 is diamagnetic due to presence of 0 unpaired
electrons). Hence, statement (d) is incorrect.
1
2
∆
45. MnO 2 + 2KOH+ O 2 →
K 2MnO 4 + H2O
preparation of KMnO4.
2MnO2 + 4KOH + O2 →
K2MnO4 + 2H2O
K2MnO4 (aq)
Purple green
(W)
O
é
–
oxygen is in (–2) oxidation states.
O
Cr
O
O
O
–
Cr
O
–
O
Green coloured
complex
O
é
MnO 2−
4 ion has one unpaired electrons, therefore it gives
d-d transition to form green colour. Y complex has paramagnetic
nature due to presence of one unpaired electron.
In aqueous solution,
Electrolytic oxidation
(W)
(X)
O
KMnO4(aq)
9
37. Cu (3d ) undergo d-d transition, exhibit colour.
D
+
K + MnO4
(Z)
-
é
producing green powder of Cr2O3 and N2 (g ) is evolved.
39. Fe2O3 is a basic oxide, neutralised by HCl spontaneously
forming FeCl 3 and water.
sp3, tetrahedral
(purple coloured
complex ion)
O
Mn
O
38. Ammonium dichromate [(NH4 )2 Cr2O7 ] on heating decomposes
é
O
MnO −4
ions gives charge transfer spectrum in which a fraction of
electronic charge is transferred between the molecular entities.
Electrolytic
40. Zinc coated with copper is used as a reducing agent.
41. The valence shell electronic configuration of Ni 2+ is :
[Ar]
4s0
; two unpaired
electrons
42. German silver is an alloy of copper (56%), Zn (24%) and
Ni(20%).
43. Zn being amphoteric, dissolves in both acid and base :
Zn + 2NaOH → Na 2ZnO2 + H2
44. Paramagnetism is a form of magnetism whereby some materials
are attracted by an externally applied magnetic field, and form
internal, induced magnetic fields in the direction of the applied
magnetic field. So, magnetic balance shows downward
deflection. While diamagnetic substance shows repulsion in
magnetic field and magnetic balance shows upward deflection.
(a) X = H2O (l)
(Water has no unpaired electrons and is thus diamagnetic).
Hence, statement (a) is correct.
(b) X = K4[Fe(CN)6 ](s)
(CN is a strong field ligand which forces the d-orbital
electrons to pair up (t2g6eg0 ) and making it diamagnetic).
Hence, statement (b) is correct
(aq)
sp- hybridisation,
tetrahedral (manganate ion)
O
Exhibit resonance phenomena. Except the bridged Cr—O—Cr,
all Cr—O bonds are equivalent.
3d 8
2–
K 2MnO 4 + H2O → H2 + KOH+ KMnO 4
Cr2O72–
2+
4
3
Mn
O
Heat
35. (NH4 )2 Cr2O7 → N2 + Cr2O3 + 4H2O
36. The structure of dichromate ion is :
+
92K (aq) + MnO
(Y)
34. In CrO2Cl 2 , Cr is in + 6 oxidation state because Cl is in (–1) and,
O
(W ) potassium
manganate
é
CrO2Cl 2, Cr
6+
é
31. In MnO−4 , Mn 7+ is in highest oxidation state possible for Mn. In
MnO 24− → MnO 4− + e−
Q
oxidation
(Y)
( Z)
In acidic medium, Y undergoes disproportionation reaction.
3MnO 24− (aq) + 4H+ → 2MnO 4− + MnO 2 + 2H2O
(Y )
(Z )
−
MnO 2−
4 and MnO 4 both ions form π-bonding between p-orbitals
(Y)
(Z)
of oxygen and d-orbitals of manganese.
Thus, options (a, c, d) are correct.
46. When Zn react with hot conc. H2SO 4 then SO 2 is released and
ZnSO 4 is obtained.
Zn + 2H2SO 4 → ZnSO 4 + SO 2 ↑ + 2H2O
(Hot + Conc.)
(G)
(R)
(X)
R(SO 2 ) molecule is V-Shaped
S
O
O
Thus, option (b) is correct.
When Zn is react with conc. NaOH then H2 gas is evolved and
Na 2ZnO 2 is obtained.
Zn + 2 NaOH (conc.) → Na 2ZnO 2 + H2 ↑
(T )
(Q)
Transition and Inner-Transition Elements 257
In ground state, H—H (Q) (bond order = 1)
Thus, option (c) is correct.
The oxidation state of Zn in T (Na2ZnO 2 ) is +2
Thus, option (a) is incorrect.
ZnSO 4 + H2S + NH4OH → ZnS↓ + 2H2O + (NH4 )2SO 4
(Z)
(G)
(X)
(Y)
ZnS (Z) compound is dirty white coloured.
Thus, option (d) is correct.
47. The explanation of given statements are as follows:
(a) Aqua-regia is prepared by mixing conc. HCl and conc. HNO 3
in 3:1 (v/v) ratio and is used in oxidation of gold and platinum.
Hence, option (a) is correct.
(b) Yellow colour of aqua-regia is due to its decomposition into
NOCl (orange yellow) and Cl 2 (greenish yellow).
Hence, option (b) is correct.
(c) When gold reacts with aqua-regia then it produces AuCl −4
anion complex in which Au has +3 oxidation state.
+3
0
Na 2S2O3 solution dissolve all three, AgCl, AgBr, AgI by forming
complex [Ag(S2O3 )2 ]3− as S2O2−
3 is a stronger complexing agent
than ammonia.
51. In neutral medium
MnO–4 → MnO2 (Mn 7 + + 3e– → Mn 4+ )
In alkaline medium
MnO–4 → MnO2 (Mn 7 + + 3e– → Mn 4+ )
In acidic medium
MnO–4 → Mn2+ (Mn7 + + 5e– → Mn2+ )
52. Cr : [Ar]3d 5 4 s1
Magnetic quantum number : – l……0……+ l.
Ag(4 d 10 5s1 ) All paired electrons have opposite spin. The last
one has unpaired spin.
53. 4NaCl + K2Cr2O7 + 6H2SO4 → 2CrO2Cl 2
Chromyl chloride
(red vapour)
+ 4NaHSO4 + 2KHSO4 + 3H2O
CrO2 Cl 2 + 4NaOH → Na 2 CrO4 + 2NaCl + 2H2 O
Au + HNO3 +4HCl → Au Cl –4 + H3O+ + NO + H2O
yellow solution
Oxidation
Hence, option (c) is correct.
(d) Reaction of gold with aqua-regia produces NO gas in
absence of air.
Hence, option (d) is incorrect.
2+
4
48. In aqueous solution Cr (3d )acts as a reducing agent, oxidising
3+
3
itself to Cr (3d ) that gives a completely half-field t2g level in
octahedral ligand field of H2O.
(b) Mn 3+ (3d 4 ) is an oxidising agent as it is reduced to
Mn 2+ (3d 5 ) , a completely half-filled stable configuration.
(c) Both Cr 2+ and Mn 3+ have d 4 configuration.
A
(d) 3d 4 Cr 2+ (aq) R.

→ Cr 3+ (aq)+ e–
54. Brass = Cu and Zn
Bronze = Cu and Sn
Gun metal = Cu, Sn, Zn
Type metal = Pb, Sn, Sb
55. Co2+ (3d 7 ) and Cr 3+ (3d 3 ) have allowed d-d transition, therefore
produces coloured aqueous solution.
∆
56. 2KOH + MnO2 + O2 → K2MnO4 + H2O
HCHO + KMnO4 + 2KOH → K2MnO4 + H2O + HCOOH
57. Both Statement I and Statement II are independently true but
Statement II is not the correct explanation of Statement I.
Diamagnetism is due to lack of unpaired electron in Zn 2+ (3d 10 ).
58. Statement I is true but Statement II is false :
K2CrO4 + H2SO4 → K2Cr2 O7 + K2SO4 + H2O
Hence (d) is wrong statement.
Yellow
49. H2O2 is alkaline medium acts as reducing agent, reduces Fe3+ to
2+
2+
Fe . In acidic medium the same H2O2 oxidises Fe
3+
to Fe .
50. Solubilities of silver halides in water decreases from fluoride
(AgF) to iodide (AgI). Silver fluoride is readialy soluble in
water, hence when AgNO3 solution is added to HF solution (HF
being weak acid, its solution maintain very low concentration of
F − ) no precipitate of AgF is formed.
HCl, HBr and HI being all strong acid, forms precipitates of
AgCl, AgBr and AgI when AgNO3 solution is added to their
aqueous solution.
HCl(aq) + AgNO3(aq) → AgCl(s)+ HNO3(aq)
Curdy white
HBr (aq) + AgNO3(aq) → AgBr (s) + HNO3(aq)
Pale yellow
Hl (aq) + AgNO 3(aq) → AgI (s) + HNO 3(aq)
Yellow
The solubilities decreases from AgCl to AgI, AgCl dissolves in
aqueous ammonia, AgBr dissolves only slightly in concentrated
ammonia while AgI does not dissolve in ammonia solution.
Orange
In both K2CrO4 and K2Cr2O7 , chromium is in +6 oxidation state.
59. Y = +3, 2Ba = 2 × 2 = 4
7 ‘O’ = 7 × (−2) = − 14
3 + 4 + (−14 ) + 3x = 0 ⇒ x = +
7
3
60. 3d 5 4 s1
61. Rochelle salt.
62. FeSO4 ⋅ 7H2O and ZnSO4 ⋅ 7H2O
63. PbO2, a strong oxidising agent, oxidises Mn 2+ to MnO−4 .
64. Zn 2+ (3d 10 ) has no unpaired electron–diamagnetic.
65. Cu cannot reduce Fe2+
66. In alkaline medium : Iodide is oxidised to iodate
2MnO −4 + H2O + I− → 2MnO 2 + 2OH− + IO 3−
But, in weakly basic solution :
+7
−1
+4
0
K Mn O4 + KI → Mn O2 + I2
Eq. of KMnO4 = Eq. of I2
4 ×3=n×2 ⇒ n=6
258 Transition and Inner-Transition Elements
67. When a solution of K2CrO4 is treated with amyl alcohol and
71. A = [Ti(H2O)6 ]3+ and M = Ti, B = TiO2, Ti(IV) has no electron
acidified H2O2, the layer of amyl alcohol turns blue because
acidified H2O2 converts K2CrO4 to CrO5 to given the blue
colouration,
+ 3H2O
CrO24− + 2H + 2H2O2 →
CrO5
in 3d-orbital, no d-d transition is possible, therefore MCl 4 is
colourless. In A, there is one electron in 3d-orbital and its d-d
transition is responsible for colour.
(Blue coloured
compound)
CrO3 + H2O →
O
O
O
CrO5 ⇒
1.73 = n (n + 2) ⇒ n = 1; V 4+ = 3d 1
O
Number of oxygen atom bonded with chromium with single
bond is (4).
68. In neutral or faintly alkaline solution,
MnO−4
is reduced to MnO2
and S2O32− is oxidised to SO2−
4 .
–
MnO4
or
8MnO−4
2–
+
SO 4
MnO2
+4
+6
+2
Change in ON = 3 units
Thus, 4MnO4− +
+
Thus, moles of
SO2−
4
→
6SO24−
(ii) 3SO2 (aq) + Cr2O72− + 2H+ → 3SO24− + 2Cr 3+ + H2O
76. (NH4 )2 S2O8 + 2H2O + MnSO4 → MnO2
+ 4NaHSO4 + 3H2 O + 2KHSO4
MnO−4
=6
69. The balanced redox reaction is
MnO−4 + [Fe(H2O)2 (C2O4 )2 ]2 − + 8H+ → Mn 2+ + Fe3+
⇒
(ii) 4Zn + 10HNO3 → 4Zn(NO3 )2 + N2 O + 5H2 O
77. (i) K2Cr2O7 + 4NaCl + 6H2SO4 → 2CrO2Cl 2
+ 8MnO2
formed by 8 moles of
8
r [ H+ ]
= =8
r [ MnO 4– ] 1
+ 2MnSO4 + 5O2 + 8H2O
+ 2H2SO4 + (NH4 )2 SO4
3
S2O23− → 3SO24− + 4 MnO2
2
3S2O32−
74. (i) 2KMnO4 + 5H2O2 + 3H2SO4 → K2SO4
75. (i) 3MnO24− + 4H+ → MnO2 + 2MnO4− + 2H2O
Change in ON = 4 units
2–
+ 1/2 S2 O3
+7
H2CrO4
Chromic acid
73. µ = n (n + 2) BM where ‘n’ is number of unpaired electrons.
Cr
O
72. CrO3 is anhydride of chromic acid :
+ 4CO2 + 6H2O
70. Acidified K2Cr2O7 , CuSO4 , H2O2, Cl 2, O3, FeCl 3 and HNO3
oxidise aq. iodide to iodine. Alkaline KMnO4 oxidise aq. iodide
to IO−3 .
Na 2S2O3 is a strong reducing agent which on reaction with I 2
produces I − .
Na 2S2O3 + I2 → 2I− + Na 2S4O6
Therefore, no reaction takes place between Na 2S2O3 and iodide
ion.
Hence, correct integer is (7).
(ii) 2KMnO4 + 3MnSO4 + 2H2 O → 5MnO2
+ K 2 SO4 + 2H2 SO4
78. (i) 2Mn 2+ + 5PbO2 + 4H+ → 2MnO−4 + 2H2O + 5Pb2+
(ii) 6Ag + + AsH3 + 3H2 O → 6Ag + H3AsO3 + 6H+
79. Most transition metals have partially filled d-orbitals which
absorb in visible region and undergo d-d transition, which is
responsible for colour.
80. (i) 2KMnO4 + 4KOH + MnO2 → 3K2MnO4 + 2H2O
(ii) K 4 Fe(CN)6 + 6H2 SO4 + 6H2 O →
2K 2 SO4 + FeSO4 + 3(NH4 )2 SO4 + 6CO
81. Potassium permanganate can be prepared from MnO2 under the
following conditions :
Heat
MnO2 + KOH + O2 → K 2 MnO4 + H2 O
K 2 MnO4 + Cl 2 → KMnO4 + KCl
82. (i) 4Zn + NO3− + 10H+ → 4Zn 2+ + NH+4 + 3H2O
(ii) Cr2O27 − + 3C2H4O + 8H+ → 3C2H4O2 + 2Cr 3+ + 4H2O
18
Coordination Compounds
Topic 1 Nomenclature and Isomerism of Coordination Compounds
Objective Questions I (Only one correct option)
1. If AB4 molecule is a polar molecule, a possible geometry of
AB4 is
(a) square pyramidal
(c) rectangular planar
(2020 Main, 2 Sep I)
(2019 Main, 12 April II)
(c) 6 and 6
(d) 5 and 6
3. The species that can have a trans-isomer is (en = ethane -1,
2-diamine, ox = oxalate)
(2019 Main, 10 April I)
(a) [Pt(en)Cl 2 ]
(b) [Cr(en)2 (ox)]
(c) [Pt(en)2 Cl 2 ]2+
(d) [Zn(en)Cl 2 ]
+
4. The maximum possible denticities of a ligand given below
towards a common transition and inner-transition metal ion,
respectively, are
(2019 Main, 9 April II)
sOOC
COOs
N
N
sOOC
(a) 8 and 8
(b) 8 and 6
N
COOs
COOs
(c) 6 and 6
(d) 6 and 8
5. The one that will show optical activity is (en = ethane-1,
2-diamine)
(2019 Main, 9 April I)
A
A
(a)
B
(b)
M
A
B
B
[Co(NH3 )4 Br2 ]+ + Br − → [Co(NH3 )3 Br3 ] + NH3
I. Two isomers are produces if the reactant complex ion is
a cis-isomer.
II. Two isomers are produced if the reactant complex ion is
a trans-isomer.
III. Only one isomer is produced if the reactant complex ion
is a trans-isomer.
IV. Only one isomer is produced if the reactant complex ion
is a cis-isomer.
The correct statements are
(2018 Main)
(a) (I) and (II)
(c) (III) and (IV)
B
en
(d) en
(2016 Main)
M
en
(b) 3
6. The following ligand is
NEt2
(2015 Main)
(d) 6
(2013 Main)
(a) [Co(en)3 ]3+
(b) [Co(en)2 Cl 2 ]+
(c) [Co (NH3 )3Cl 3 ]
(d) [Co(en)(NH3 )Cl 2 ]+
13. As per IUPAC nomenclature, the name of the complex
[Co (H2 O)4 (NH3 )2 ]Cl 3 is
N
–
(c) 4
12. Which of the following complex species is not expected to
exhibit optical isomerism?
A
O–
(b) trans [Co(en)2 Cl 2 ]Cl
(d) [Co(NH3 )3 Cl 3 ]
[Pt(Cl)(py)(NH3 )(NH2 OH)]+ is (py = pyridine).
B
B
(b) (I) and (III)
(d) (II) and (IV)
11. The number of geometric isomers that can exist for square planar
A
(a) 2
M
(2018 Main)
(b) +3, +2 and +4
(d) +3, 0 and +4
9. Consider the following reaction and statements :
isomerism?
A
(c)
(d) 8
K 2 [Cr(CN)2 (O)2 (O2 )(NH3 )] respectively are
(a) +3, +4 and +6
(c) +3, 0 and +6
(a) cis [Co(en)2 Cl 2 ]Cl
(c) [Co(NH3 )4 Cl 2 ]Cl
A
A
(2019 Main, 10 Jan I)
(c) 4
8. The oxidation states of Cr, in[Cr(H2 O)6 ]Cl 3 ,[Cr(C6 H6 )2 ], and
B
M
B
(b) 16
10. Which one of the following complexes shows optical
A
B
7. The total number of isomers for a square planar complex
(a) 12
K 3 [Al(C2 O4 )3 ], respectively, are (en = ethane-1, 2-diamine)
(b) 3 and 3
(b) tetradentate
(d) tridentate
[M(F)(Cl)(SCN)(NO 2 )] is
(b) square planar
(d) tetrahedral
2. The coordination numbers of Co and Al in [CoCl(en)2 ]Cl and
(a) 5 and 3
(a) hexadentate
(c) bidentate
O
(2019 Main, 8 April I)
(a)
(b)
(c)
(d)
tetraaquadiaminecobalt (III) chloride
tetraaquadiamminecobalt (III) chloride
diaminetetraaquacobalt (III) chloride
diamminetetraaquacobalt (III) chloride
(2012)
260
Coordination Compounds
14. Geometrical shapes of the complexes formed by the reaction
of Ni 2+ with Cl − , CN− and H2 O, respectively, are
(2011)
21. Statement I The geometrical isomers of the complex
(a) octahedral, tetrahedral and square planar
(b) tetrahedral, square planar and octahedral
(c) square planar, tetrahedral and octahedral
(d) octahedral, square planar and octahedral
[ M (NH3 )4 Cl 2 ] are optically inactive.
Statement II Both geometrical isomers of the complex
(2008, 3M)
[ M (NH3 )4 Cl 2 ] possess axis of symmetry.
15. The correct structure of ethylenediaminetetraacetic acid
(EDTA) is
(a)
(b)
(c)
(d)
N — CH == CH — N
HOOCCH2
N — CH — CH — N
HOOCCH2
HOOCCH2
CH2COOH
HOOC—H2C
H
H 2C
NiCl 2 + KCN (excess) → A (cyano complex)
COOH
COOH
NiCl 2 + conc. HCl (excess) → B (chloro complex)
22. The IUPAC name of A and B are
CH2COOH
CH2
CH2COOH
H
CH2—COOH
16. The ionisation isomer of [Cr(H2 O)4 Cl(NO2 )]Cl is
(a) [Cr(H2O)4 (O2N)]Cl 2
(c) [Cr(H2O)4 Cl(ONO)]Cl
(2010)
(b) [Cr(H2O)4 Cl 2 ](NO2 )
(d) [Cr(H2O)4 Cl 2 (NO2 )] ⋅ H2O
17. The IUPAC name of [Ni(NH3 )4 ][NiCl 4 ] is
[CoL2 Cl 2 ]− ( L= H2 NCH2 CH2 O− ) is (are) ...
(2016 Adv.)
25. Among the complex ions,
[Co(NH2 CH2 CH2  NH2 )2 Cl 2 ]+ , [CrCl 2 (C2 O4 )2 ]3− ,
[Fe(H2 O)4 (OH)2 ]+ , [Fe(NH3 )2 (CN)4 ]− ,
(b) Optical and ionisation
(d) Geometrical only
[Co(NH2  CH2  CH2  NH2 )2 (NH3 ) Cl]2+ and
[Co(NH3 )4 (H2 O)Cl]2+ the number of complex ion(s)
that show(s) cis-trans isomerism is
(2015 Adv.)
Objective Questions II
(One or more than one correct option)
19. The pair(s) of coordination complexes/ions exhibiting the
(2013 Adv.)
[Cr(NH3 )5 Cl]Cl 2 and [Cr(NH3 )4 Cl 2 ]Cl
[Co(NH3 )4 Cl 2 ]+ and [Pt(NH3 )2 (H2O)Cl]+
[CoBr2Cl 2 ]2− and [PtBr2Cl 2 ]2−
[Pt(NH3 )3 (NO3 )]Cl and [Pt(NH3 )3 Cl]Br
26. The volume (in mL) of 0.1 M AgNO3 required for complete
precipitation of chloride ions present in 30 mL of 0.01 M
solution of [Cr(H2 O)5 Cl]Cl 2 , as silver chloride is close to
(2011)
27. Total number of geometrical isomers for the complex
[RhCl(CO)(PPh 3 )(NH3 )] is
20. The compound(s) that exhibit(s) geometrical isomerism
is/are
(1994, 1M)
24. The possible number of geometrical isomers for the complex
(2005, 1M)
(a) [Pt(en)Cl2]
(c) [Pt(en)2Cl2]Cl2
23. The type of magnetism exhibited by [Mn(H2 O)6 ]2+ ion is …
Integer Answer Type Questions
18. Which kind of isomerism is shown by Co(NH3 )4 Br2 Cl?
same kind of isomerism is/are
Fill in the Blank
(2008, 3M)
(a) Tetrachloronickel (II)-tetraamminenickel (II)
(b) Tetraamminenickel (II)-tetrachloronickel (II)
(c) Tetraamminenickel (II)-tetrachloronickelate (II)
(d) Tetrachloronickel (II)-tetraamminenickelate (0)
(a) Geometrical and ionisation
(c) Geometrical and optical
(2006, 3 × 4M =12M)
(a) potassium tetracyanonickelate (II), potassium
tetrachloronickelate (II)
(b) tetracyanopotassiumnickelate (II),
tetrachloropotassiumnickelate (II)
(c) tetracyanonickel (II), tetrachloronickel (II)
(d) potassium tetracyanonickel (II), potassium
tetrachloronickel (II)
COOH
N—CH—CH—N
Passage
The coordination number of Ni 2 + is 4.
CH2COOH
N — CH2 — CH2 — N
HOOC
(a)
(b)
(c)
(d)
Passage Based Question
(2010)
HOOCCH2
HOOC
HOOC
(c) Statement I is true; Statement II is false.
(d) Statement I is false; Statement II is true.
(2009)
(b) [Pt(en)2]Cl2
(d) [Pt(NH3)2]Cl2
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.
(2010)
Subjective Questions
28. Write the formulae of the following complexes :
(i) Pentamminechlorocobalt (III) ion
(ii) Lithium tetrahydridoaluminate (III)
29. Write the IUPAC name for [Cr(NH3 )5 CO3 ]Cl.
(1997, 2M)
(1996, 1M)
30. Write the IUPAC name of the following compounds :
(i) [Co(NH3 )5 ONO] Cl 2
(ii) K 3[Cr(CN)6 ]
(1995, 2M)
Topic 2 Bonding and Important Property of
Coordination Compounds
Objective Questions I (Only one correct option)
6
1. Consider that a d metal ion ( M
7. The incorrect statement is
2+
) forms a complex with
aqua ligands and the spin only magnetic moment of the
complex is 4.90 BM. The geometry and the crystal field
stabilisation energy of the complex is
(2020 Main, 2 Sep I)
(a)
(b)
(c)
(d)
tetrahedral and −1.6∆ t + 1 P
octahedral and −2.4 ∆ 0 + 2 P
octahedral and −1.6 ∆ 0
tetrahedral and −0.6 ∆ t
(d) the spin only magnetic moment of [Ni(NH3 )4 (H2O)2 ]2 + is 2.83
BM
d-electron configuration
[Fe(H 2O)6 ]Cl2 , respectively are
of [ Ru(en)3 ]Cl2
and
(2020 Main, 3 Sep II)
(a) t62g eg0 and t62geg0
(b) t62g eg0 and t24geg2
(c) t24g eg2 and t62geg0
(d) t24g eg2 and t24geg2
3. The compound used in the treatment of lead poisoning is
(2019 Main, 12 April II)
(b) desferrioxime-B
(d) EDTA
4. Complete removal of both the axial ligands (along the z-axis)
from an octahedral complex leads to which of the following
splitting patterns? (relative orbital energies not on scale).
(2019 Main, 12 April I)
dxy
(b) E
[CoCl(NH3 )5 ]2+
3+
[Co(NH3 )6 ]
(I),
[Co(NH3 )5 H2 O]3+
(II)
and
(III)
absorb light in the visible region. The correct order of the
wavelength of light absorbed by them is (2019 Main, 10 April I)
(a) II > I > III
(c) III > I > II
(b) I > II > III
(d) III > II > I
9. The degenerate orbitals of [Cr(H2 O)6 ]3+ are
(2019 Main, 9 April I)
(a) d 2 and dxz
z
(c) d 2 2 and dxy
x − y
(b) dxz and d yz
(d) d yz and d 2
z
10. The calculated spin only magnetic moments (BM) of the
anionic and cationic species of [Fe(H2 O)6 ]2 and [Fe(CN)6 ],
respectively, are
(2019 Main, 8 April II)
dx2 – y2
(a) 0 and 4.9
(c) 0 and 5.92
dz2
dxz, dyz
dxz, dyz
dxy
dx2 – y2
dx2 – y2
dz2
dz2
(c) E
8. Three complexes,
dz2
dx2 – y2
(a) E
light
(c) the spin only magnetic moments of Fe(H2O)6 ]2 + and
[Cr(H2O)6 ]2 + are nearly similar
2. The
(a) D-penicillamine
(c) cis-platin
(2019 Main, 10 April II)
(a) the gemstone, ruby, has Cr 3 + ions occupying the octahedral
sites of beryl
(b) the color of [CoCl(NH3 )5 ]2 + is violet as it absorbs the yellow
(d) E
(b) 2.84 and 5.92
(d) 4.9 and 0
11. The compound that inhibits the growth of tumors is
(2019 Main, 8 April II)
(a) trans-[Pt(Cl)2 (NH3 )2 ]
(b) cis-[Pd(Cl)2 (NH3 )2 ]
(c) cis-[Pt(Cl)2 (NH3 )2 ]
(d) trans-[Pd(Cl)2 (NH3 )2 ]
12. The correct order of the spin only magnetic moment of metal
dxy
dyz, dxz
ions in the following low spin complexes, [V(CN)6 ]4− ,
dxz, dyz
dxy
[Fe(CN)6 ]4− , [Ru(NH3 )6 ]3+ , and [Cr(NH3 )6 ]2+ , is
(2019 Main, 8 April I)
5. The complex ion that will lose its crystal field stabilisation
energy upon oxidation of its metal to +3 state is
(a) Cr 2+ > Ru3+ > Fe2+ > V2+ (b) V2+ > Cr 2+ > Ru3+ > Fe2+
(c) V2+ > Ru3+ > Cr 2+ > Fe2+ (d) Cr 2+ > V2+ > Ru3+ > Fe2+
13. The magnetic moment of an octahedral homoleptic Mn(II)
complex is 5.9 BM. The suitable ligand for this complex is
(a) CN−
(c) NCS−
(Phen =
N
Ignore pairing energy
(a) [Co(phen)3 ]
2+
(c) [Zn(phen)3 ]2+
N
(2019 Main, 12 April I)
(b) [Ni(phen)3 ]
2+
(d) [Fe(phen)3 ]2+
6. The crystal field stabilisation energy (CFSE) of
[Fe(H2 O)6 ]Cl 2 and K 2 [NiCl 4 ], respectively, are
(2019 Main, 10 April II)
(a) − 0.4 ∆ o and − 12
. ∆t
(c) − 2.4 ∆ o and − 12
. ∆t
(b) − 0.4 ∆ o and − 0.8 ∆ t
(d) − 0.6 ∆ o and − 0.8 ∆ t
(b) ethylenediamine
(d) CO
14. The pair of metal ions that can given a spin-only magnetic
moment of 3.9 BM for the complex [M (H2 O)6 ]Cl 2 , is
(2019 Main, 12 Jan I)
(a) Co2+ and Fe2+
(c) V2+ and Co2+
(b) Cr 2+ and Mn 2+
(d) V2+ and Fe2+
15. The metal d-orbitals that are directly facing the ligands in
K 3 [Co(CN)6 ] are
(a) dxz , d yz and d 2
z
(c) dxy , dxz and d yz
(2019 Main, 12 Jan I)
(b) d 2 2 and d 2
x − y
z
(d) dxz and d 2 2
x − y
262
Coordination Compounds
16. Mn 2 (CO)10 is an organometallic compound due to the
presence of
(2019 Main, 12 Jan I)
(a) Mn  C bond
(c) C O bond
(b) Mn  O bond
(d) Mn  Mn
17 The number of bridging CO ligand(s) and Co Co bond(s) in
Co2 (CO )8 , respectively are
(a) 2 and 0
(b) 0 and 2
(2019 Main, 11 Jan II)
(c) 4 and 0
(d) 2 and 1
18. The coordination number of Th in K 4 [Th(C2 O4 )4 (OH2 )2 ] is (
C2O24 − = Oxalato)
(a) 14
(2019 Main, 11 Jan II)
(b) 10
(c) 8
(d) 6
19. Match the metals (Column I) with the coordination
compound(s)/enzyme(s) (Column II).
Column I
(2019 Main, 11 Jan I)
Column II
26. Two complexes [Cr(H 2O)6 ]Cl3 (A) and [Cr(NH3 )6 ]Cl3 (B)
are violet and yellow coloured, respectively. The incorrect
statement regarding them is
(2019 Main, 9 Jan I)
(a) ∆ o value for (A) is less than that of (B)
(b) both absorb energies corresponding to their complementary
colours
(c) ∆ o values of (A) and (B) are calculated from the energies of
violet and yellow light, respectively
(d) both are paramagnetic with three unpaired electrons
27. The recommended concentration of fluoride ion in drinking
water is up to 1 ppm as fluoride ion is required to make teeth
enamel harder by converting [3Ca 3 (PO4 )2 ⋅ Ca(OH)2 ] to :
(2018 Main)
(a) [CaF2 ]
(c) [3Ca 3 (PO4 )2 ⋅ CaF2 ]
(b) [3(CaF2 ) ⋅ Ca(OH)2 ]
(d) [3{Ca 3 (PO4 )2} ⋅ CaF2 ]
(A)
Co
(i)
Wilkinson catalyst
(B)
Zn
(ii
Chlorophyll
(C)
Rh
(iii)
Vitamin B 12
CoCl3 .6H2O with excess of AgNO3 ; 1.2 × 1022 ions are
precipitated. The complex is
(2017 Main)
(D)
Mg
(iv)
Carbonic anhydrase
(a) [Co(H2O)4 Cl 2 ] Cl ⋅ 2H2O
(b) [Co(H2O)3 Cl 3 ] ⋅ 3H2O
(c) [Co(H2O)6 ]Cl 3
(d) [Co(H2O)5 Cl] Cl 2 ⋅ H2O
A
(a) (i)
(b) (iv)
(c) (iii)
(d) (ii)
B
(ii)
(iii)
(iv)
(i)
C
(iii)
(i)
(i)
(iv)
D
(iv)
(ii)
(ii)
(iii)
29. The pair having the same magnetic moment is
[at. no. Cr = 24, Mn = 25, Fe = 26 and Co = 27]
(a) [Cr(H2O)6 ]2+ and [Fe(H2O)6 ]2+
20. The difference in the number of unpaired electrons of a metal
ion in its high-spin and low-spin octahedral complexes is two.
The metal ion is
(2019 Main, 10 Jan II)
(a) Mn
2+
2+
(b) Fe
(c) Ni
2+
(d) Co
2+
21. A reaction of cobalt (III) chloride and ethylene diamine in a
1 : 2 mole ratio generates two isomeric products A (violet
coloured) and B (green coloured). A can show optical
activity, but B is optically inactive. What type of isomers does
A and B represent ?
(2019 Main, 10 Jan II)
(a) Ionisation isomers
(c) Geometrical isomers
(b) Coordination isomers
(d) Linkage isomers
22. Wilkinson catalyst is
(a) [(Et 3P)3 RhCl]
(c) [(Ph 3P)3 RhCl]
28. On treatment of 100 mL of 0.1 M solution of
(2019 Main, 10 Jan I)
(b) [(Et 3P)3 IrCl](Et = C2H5)
(d) [(Ph 3P)3 IrCl]
23. Homoleptic octahedral complexes of a metal ion ‘M 3 + ’ with
(2016 Main)
(b) [Mn(H2O)6 ]2+ and [Cr(H2O)6 ]2+
(c) [CoCl 4 ]2− and [Fe(H2O)6 ]2+
(d) [Cr(H2O)6 ]2+ and [CoCl 4 ]2−
[Ni(CO)4 ], [ NiCl 4 ]2 − ,
[ Co(NH3 )4 Cl 2 ] Cl,
Na 3 [ CoF6 ], Na 2 O2 and CsO2 , the total number of
paramagnetic compounds is
(2016 Adv.)
30. Among
(a) 2
(c) 4
(b) 3
(d) 5
31. The colour of KMnO4 is due to
(2015 Main)
(a) M → L charge transfer transition
(b) d → d transition
(c) L → M charge transfer transition
(d) σ → σ∗ transition
32. The equation which is balanced and represents the correct
product(s) is
three monodentate ligands L1 , L2 and L3 absorb wavelengths
in the region of green, blue and red respectively. The
increasing order of the ligand strength is (2019 Main, 9 Jan II)
(a) Li 2O + 2KCl → 2LiCl + K 2O
(a) L1 < L2 < L3
(c) L3 < L1 < L2
(c) [Mg(H2 O)6 ]2+ + (EDTA)4 − →
(b) L2 < L1 < L3
(d) L3 < L2 < L1
( ∆ ), is
(2019 Main, 9 Jan II)
(b) [Co(NH3 )5 (H2O)]Cl 3
(d) K 2[CoCl 4 ]
25. The highest value of the calculated spin only magnetic
moment (in BM) among all the transition metal complexes is
(2019 Main, 9 Jan I)
(a) 5.92
(c) 6.93
(b) 3.87
(d) 4.90
(b) [CoCl(NH3 )5 ]+ + 5H+ → Co2+ + 5NH4+ + Cl −
Excess NaOH
24. The complex that has highest crystal field splitting energy
(a) [Co(NH3 )5 Cl] Cl 2
(c) K 3[Co(CN)6 ]
(2014 Main)
[Mg(EDTA)]2+ + 6H2 O
(d) CuSO4 + 4KCN → K 2 [Cu(CN)4 ] + K 2SO4
33. The octahedral complex of a metal ion M 3+ with four
monodentate ligands L1 , L2 , L3 and L4 absorb wavelengths in
the region of red, green, yellow and blue, respectively. The
increasing order of ligand strength of the four ligands is
(2014 Main)
(a) L4 < L3 , L2 < L1
(c) L3 < L2 < L4 < L1
(b) L1 < L3 < L2 < L4
(d) L1 < L2 < L4 < L 3
Coordination Compounds
34. Consider the following complex ions, P, Q and R.
3−
2+
2+
P = [ FeF6 ] , Q = [ V(H 2O )6 ] and R = [ Fe(H 2O )6 ]
The correct order of the complex ions, according to their
spin-only magnetic moment values (in BM) is
(2013 Adv.)
(a) R < Q < P
(c) R < P < Q
(b) Q < R < P
(d) Q < P < R
tetrahedral and tetrahedral
square planar and square planar
tetrahedral and square planar
square planar and tetrahedral
36. Among the following complexes (K-P),
(2011)
K 3 [Fe(CN)6 ] (K), [Co(NH3 )6 ]Cl 3 (L),
2.82 BM is
(2010)
(d) [Ni(CN)4 ]2−
units) of Cr(CO)6 is
(b) 2.84
(2009)
(c) 4.90
(d) 5.92
39. Among the following, the coloured compound is
(a) CuCl
(c) CuF2
(2008, 3M)
(b) K 3[Cu(CN)4 ]
(d) [Cu(CH3CN)4 ]BF4
hybridisations of nickel in these complexes respectively, are
2
(c) dsp , sp
(b) sp3 , dsp2
3
(2008, 3M)
(d) dsp2 , dsp2
41. Among the following metal carbonyls, the C—O bond order
is lowest in
(2007, 3M)
(a) [Mn(CO)6 ]+
(d) [V(CO)6 ]
−
then what is the value of CO bond length in Fe(CO)5? (2006)
(b) 1.128 Å
(c) 1.72 Å (d) 1.118 Å
is
(2004, 1M)
(b) 15
(c) 24
(d) 8
44. The compound having tetrahedral geometry is
(a) [Ni(CN)4 ]
(c) [PdCl 4 ]
2–
2–
(b) [Pd(CN)4 ]
(d) [NiCl 4 ]2–
47. The geometry of Ni(CO)4 and Ni(PPh 3 )2 Cl 2 are
(1999, 2M)
(a) both square planar
(b) tetrahedral and square planar, respectively
(c) both tetrahedral
(d) square planar and tetrahedral, respectively
48. Which of the following is formed when excess of KCN is
added to aqueous solution of copper sulphate?
(1996, 1M)
(b) K 2[Cu(CN)4 ]
(d) K 3[Cu(CN)4 ]
(1993, 1M)
(a) [Cr(H2O)6 ]3+
(b) [Fe(H2O)6 ]2+
(c) [Cu(H2O)6 ]2+
(d) [Zn(H2O)6 ]2+
50. Amongst Ni(CO)4 , [Ni(CN)4 ]2– and NiCl 2–
4
(1991, 1M)
2–
(a) Ni(CO)4 and NiCl 2–
is
4 are diamagnetic and [Ni(CN)4 ]
paramagnetic
(c) Ni(CO)4 and [Ni(CN)4 ]2– are diamagnetic and [NiCl 4 ]2− is
paramagnetic
(d) Ni(CO)4 is diamagnetic and [NiCl 4 ]2− and [Ni(CN)4 ]2– are
paramagnetic
per mole of the compound at 298 K will be shown by
(1988, 2M)
(a) MnSO4 ⋅ 4H2O
(c) FeSO4 ⋅ 6H2O
(b) CuSO4 ⋅ 5H2O
(d) NiSO4 ⋅ 6H2O
Objective Question II
(One or more than one correct option)
2–
(2020 Adv.)
(b) [Co(en)(NH3 )2 Cl 2 ]+ has 2 geometrical isomers.
(c) [FeCl 4 ]− has higher spin-only magnetic moment than
[Co(en)(NH3 )2 Cl 2 ]+ .
43. Spin only magnetic moment of the compound Hg[Co(SCN)4 ]
(a) 3
(d) [Cr(H2O)6 ]3+
(a) [FeCl 4 ]− has tetrahedral geometry.
42. If the bond length of CO bond in carbon monoxide is 1.128 Å,
(a) 1.15 Å
(c) [Fe(CN)6 ]
52. Choose the correct statement(s) among the following:
(b) [Fe(CO)5 ]
(c) [Cr(CO)6 ]
(b) [Co(NH3 )6 ]
3–
51. Amongst the following, the lowest degree of paramagnetism
40. Both [Ni(CO)4 ] and [Ni(CN)4 ]2− are diamagnetic. The
(a) sp3 , sp3
(2001, 1M)
3+
paramagnetic
(b) [NiCl 4 ]2− and [Ni(CN)4 ]2– are diamagnetic and Ni(CO)4 is
38. The spin only magnetic moment value (in Bohr magneton
(a) 0
(a) [MnO4 ]
–
paramagnetism?
37. The complex showing a spin only magnetic moment of
(c) Ni(PPh 3 )4
(2003)
(d) 0.02, 0.02
49. Among the following ions, which one has the highest
(b) K, M, O, P
(d) L, M, N, O
(b) [NiCl 4 ]2−
(c) 0.01, 0.02
metal atom is
(a) Cu (CN)2
(c) K[Cu(CN)2 ]
Na 3 [Co (ox)3 ] (M), [Ni(H2 O)6 ] Cl 2 (N),
K 2 [Pt(CN)4 ](O), [Zn(H2 O)6 ](NO3 )2 (P)
the diamagnetic complexes are
(a) Ni(CO)4
[Co(NH3)5Br]SO4 was prepared in 2 L of solution.
1 L of mixture X + excess AgNO3 → Y
1 L of mixture X + excess BaCl2 → Z
Number of moles of Y and Z are
46. The complex ion which has no ‘d’-electrons in the central
magnetic behaviour (paramagnetic/diamagnetic) the
coordination geometries of Ni 2+ in the paramagnetic and
diamagnetic states respectively, are
(2012)
(a) K, L, M, N
(c) L, M, O, P
45. Mixture X = 0.02 mole of [Co(NH3)5SO4]Br and 0.02 mole of
(a) 0.01, 0.01 (b) 0.02, 0.01
35. NiCl 2 {P(C2 H5 )2 (C6 H5 )}2 exhibits temperature dependent
(a)
(b)
(c)
(d)
263
(d) The cobalt ion in [Co(en)(NH3 )2 Cl 2 ]+ has sp3d 2 hybridisation.
53. A tin chloride Q undergoes the following reactions (not
balanced)
(2004, 1M)
Q + Cl − → X
Q + Me3N → Y
Q + CuCl 2 → Z + CuCl
264
Coordination Compounds
X is a monoanion having pyramidal geometry. Both Y and Z
are neutral compounds.
(2019 Adv.)
Choose the correct option(s).
(a) There is a coordinate bond in Y
(b) The central atom in Z has one lone pair of electrons
(c) The oxidation state of the central atom in Z is + 2
(d) The central atom in X is sp3 hybridised
54. The correct statement (s) regarding the binary transition
metal carbonyl compounds is (are) (Atomic numbers :
(2018 Adv.)
Fe = 26, Ni = 28)
(a) Total number of valence shell electrons at metal centre in
Fe(CO)5 or Ni(CO) 4 is 16
(b) These are predominantly low spin in nature
(c) Metal-carbon bond strengthens when the oxidation state of the
metal is lowered
(d) The carbonyl C—O bond weakens when the oxidation state of
the metal is increased
55. The correct option(s) regarding the complex
[Co(en)(NH3 )3 (H2 O)]3+ (en = H2 NCH2 CH2 NH2 ) is (are)
(a) It has two geometrical isomers
(2018 Adv.)
(b) It will have three geometrical isomers, if bidentate ‘en’ is
replaced by two cyanide ligands
(c) It is paramagnetic
(d) It absorbs light at longer wavelength as compared to
[Co(en)(NH3 )4 ]3+
56. Addition of excess aqueous ammonia to a pink coloured
aqueous solution of MCl 2 ⋅ 6H2 O( X ) and NH4 Cl gives an
octahedral complex Y in the presence of air. In aqueous
solution, complex Y behaves as 1 : 3 electrolyte. The reaction
of X with excess HCl at room temperature results in the
formation of a blue colured complex Z. The calculated spin
only magnetic moment of X and Z is 3.87 B.M., whersas it is
zero for complex Y .
Among the following options, which statement(s) is (are)
correct?
(2017 Adv.)
(a) The hybridisation of the central metal ion in Y is d 2sp3
(b) Addition of silver nitrate to Y given only two equivalents of
silver chloride
(c) When X and Y are in equilibrium at 0°C, the colour of the
solution is pink
(d) Z is a tetrahedral complex
Numerical Answer Type Question
57. Total number of cis N  Mn Cl bond angles (that is
Mn N and Mn Cl bonds in cis positions) present in a
molecule of cis [Mn(en)2 Cl 2 ] complex is ………
(2019 Adv.)
(en = NH 2CH 2CH 2NH 2 )
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.
58. Statement I [Fe(H2 O)5 NO]SO4 is paramagnetic.
Statement II The Fe in [Fe(H2 O)5 NO]SO4 has three
unpaired electrons.
(2008, 3M)
Passage Based Questions
The coordination number of Ni 2 + is 4.
NiCl 2 + KCN (excess) → A (cyano complex)
NiCl 2 + conc. HCl (excess) → B (chloro complex)
59. Predict the magnetic nature of A and B.
(a) Both are diamagnetic
(b) A is diamagnetic and B is paramagnetic with one unpaired
electron
(c) A is diamagnetic and B is paramagnetic with two unpaired
electrons
(d) Both are paramagnetic
60. The hybridisation of A and B are
(a) dsp2 , sp3
(b) sp3 , sp3
(c) dsp2 , dsp2
(d) sp3d 2 , d 2sp3
Match the Columns
61. Match each set of hybrid orbitals from List–I with complexes
given in List-II.
List–I
P.
dsp
Q.
sp3
2
List–II
1.
[FeF6 ]
4−
2.
[Ti(H2O)3 Cl 3 ]
R.
3 2
sp d
3.
[Cr(NH3 )6 ]3+
S.
d 2sp3
4.
[FeCl 4 ]2−
5.
[Ni(CO)4 ]
6.
[Ni(CN)4 ]2−
The correct option is
(2018 Adv.)
(a) P → 5; Q → 4, 6; R → 2, 3; S → 1
(b) P → 5,6; Q → 4; R → 3; S → 1,2
(c) P → 6; Q → 4, 5; R → 1; S → 2, 3
(d) P → 4,6; Q → 5, 6; R → 1,2; S → 3
62. Match each coordination compound in Column I with an
appropriate pair of characteristics from Column II and
select the correct answer using the codes given below the
Columns (en = H2 NCH2 CH2 NH2 ; atomic numbers : Ti = 22;
Cr = 24; Co = 27; Pt = 78)
(2014 Adv.)
Column I
Column II
(A)
[Cr(NH3 )4 Cl 2 ]Cl
1.
Paramagnetic and exhibits
ionisation isomerism
(B)
[Ti(H2O)5 Cl](NO3 )2
2.
Diamagnetic and exhibits
cis-trans isomerism
(C)
[Pt(en)(NH3 )Cl]NO3
3.
Paramagnetic and exhibits
cis-trans isomerism
(D)
[Co(NH3 )4 (NO3 )2 ]NO3
4.
Diamagnetic and exhibits
ionisation isomerism
Coordination Compounds
Codes
A B C D
(a) 4 2 3 1
(c) 2 1 3 4
(a) Draw its structure and show H-bonding
(b) Give oxidation state of Ni and its hybridisation
(c) Predict whether it is paramagnetic or diamagnetic
A B C D
(b) 3 1 4 2
(d) 1 3 4 2
in Column II.
(2007, 6M)
(2003, 4M)
(A)
[Co(NH3 )4 (H2O)2 ]Cl 2
p.
Geometrical isomers
(B)
[Pt(NH3 )2 Cl 2 ]
q.
Paramagnetic
(C)
[Co(H2O)5 Cl]Cl
r.
Diamagnetic
(D)
[Ni(H2O)6 ]Cl 2
s.
Metal ion with +2
oxidation state
64. The IUPAC name of [Co(NH3 )6 ] Cl 3 is ……
(1994, 1M)
True/False
65. Both potassium ferrocyanide and potassium ferricyanide are
diamagnetic.
(1989, 1M)
y2
orbital is zero.
(1986, 1M)
Integer Answer Type Questions
in SCN−
−
(thiocyanato-S) and in CN ligand environments, the
difference between the spin only magnetic moments in Bohr
magnetons (when approximated to the nearest integer) is
[atomic number of Fe = 26 ]
(2015 Adv.)
67. For the octahedral complexes of Fe3+
68. In the complex acetylbromidodicarbonylbis (triethylphosphine)
iron (II), the number of Fe  C bond (s) is
2–
72. Deduce the structures of [NiCl 4 ]
and [Ni(CN)4 ]2–
considering the hybridisation of the metal ion. Calculate the
magnetic moment (spin only) of the species.
(2002, 5M)
Fill in the Blank
69. EDTA
the
IUPAC
name
of
the
compound
K 2 [Cr(NO)(CN)4 (NH3 )] . Spin magnetic moment of the
complex µ = 1.73 BM. Give the structure of anion.
Column II
66. The electron density in the xy plane in 3dx 2 −
(2004, 4M)
71. Write
63. Match the complexes in Column I with their properties listed
Column I
265
(2015 Adv.)
4−
is ethylenediaminetetraacetate ion. The total
number of N  Co  O bond angles in [Co(EDTA)] −
complex ion is
(2013 Adv.)
73. A metal complex having composition Cr(NH3 )4 Cl 2 Br has
been isolated in two forms A and B. The form A reacts with
AgNO3 to give a white precipitate readily soluble in dilute
aqueous ammonia, whereas B gives a pale yellow precipitate
soluble in concentrated ammonia.
Write the formula of A and B and state the hybridisation of
chromium in each. Calculate their magnetic moments
(spin-only value).
(2001, 5M)
74. Draw the structures of [Co (NH3 )6 ]3+ , [Ni(CN)4 ]2– and
[Ni(CO)4 ] . Write the hybridisation of atomic orbitals of the
transition metal in each case.
(2000, 4M)
75. A, B and C are three complexes of chromium (III) with the
empirical formula H12 O6 Cl 3 Cr. All the three complexes have
water and chloride ion as ligands.
Complex A does not react with concentrated H2 SO4 , whereas
complexes B and C lose 6.75% and 13.5% of their original
mass, respectively, on treatment with concentrated H2 SO4 .
Identify A, B and C.
(1999, 2M)
76. Identify the complexes which are expected to be coloured.
Explain
(1994, 2M)
(i) [Ti(NO3 )4 ]
(iii) [Cr(NH3 )6 ] Cl3
(ii) [Cu(NCCH3 )]+ BF4
(iv) K3 [VF6 ]
77. Give reasons in two or three sentences only for the
Subjective Questions
70. NiCl 2 in the presence of dimethyl glyoxime (DMG) gives a
complex which precipitates in the presence of NH4 OH, giving
a bright red colour.
following :
“The species[CuCl 4 ]2– exists, while[CuI4 ]2– does not exist.”
(1992, 1M)
Answers
Topic 1
Topic 2
1. (a)
2. (d)
3. (c)
4. (d)
1. (d)
2. (b)
3. (d)
4. (a)
5. (c)
6. (b)
7. (a)
8. (c)
5. (d)
6. (b)
7. (a)
8. (b)
9. (b)
10. (a)
11. (b)
12. (c)
9. (b)
10. (a)
11. (c)
12. (b)
13. (d)
14. (b)
15. (c)
16. (b)
13. (c)
14. (c)
15. (b)
16. (a)
17. (c)
18. (a)
19. (b,d)
20. (c,d)
17. (d)
18. (b)
19. (c)
20. (d)
21. (b)
22. (a)
23. (paramagnetism)
21. (c)
22. (c)
23. (c)
24. (c)
24. (5)
25. (6)
26. (6)
25. (a)
26. (c)
27. (c)
28. (d)
29. (a)
30. (b)
31. (c)
32. (b)
27. (3)
266
Coordination Compounds
33. (b)
34. (b)
35. (c)
36. (c)
57. (6)
58. (a)
37. (b)
38. (a)
39. (c)
41. (b)
42. (a)
43. (b)
45. (a)
46. (a)
49. (b)
53. (a, d)
59. (c)
40. (b)
61. (c)
62. (b)
44. (d)
63. (A → p, q, s B → p, r, s C → q, s D → q, s)
47. (c)
48. (d)
64. (hexaammine cobalt (III) chloride)
65. (F)
50. (c)
51. (b)
52. (a,c)
66. (F)
69. (8)
54. (b, c)
55. (a, b, d)
56. (a, b, d)
67. (4)
60. (a)
68. (3)
Hints & Solutions
Topic 1 Nomenclature and Isomerism of
Coordination Compounds
In first complex, ‘en’ is a didentate ligand and ‘Cl’ is a unidentate
ligand.
[Co(Cl)(en)2]Cl, coordination number = 1 + 2 × 2 = 1 + 4 ⇒ 5
1. If AB4 molecule is a polar moelcule, a possible geometry of AB4
So, the coordination number is 5.
is square pyramidal.
All possible structure of AB4 molecule are as follows:
(i) AB4 ⇒ L = 0 ⇒ A is sp3 (tetrahedral) or
For K 3 [Al(C2O4 )3 ], ‘C2O24 − ’ is a didentate ligand.
Coordination number = 3 × 2 = 6.
Hence, coordination number is 6.
Ais dsp2 (square planar)
⇒ Dipole moment, µ = 0(Non-polar)
(ii) AB4L2 ⇒ L = 2 ⇒ A is sp3d 2 (octahedral)
3.
⇒µ = 0(Non-polar), because
B
B
⇒ µnet = 0 (Non-polar)
A
B
Key Idea Square planar complexes of general formulae :
[ M(a − a)b2 ] and [ M(a − a) (b − b) ] do not show geometrical
isomerism. Whereas, an octahedral complex of general
formula [ M (a − a)2 b2 ] can show geometrical (cis-trans)
isomerism.
[ Pt(en)2 (Cl 2 )]2+ with formula [ M (a − a)2 b2 ] will show
geometrical isomerism as follows:
2+
Cl
2+
B
en
(Assuming, B is more electronegative than A)
(iii) AB4L ⇒ L = 1 ⇒ A is sp3d (square pyramidal or trigonal
bipyramidal)
⇒µ ≠ 0 (polar), because
B
B
B
µplane = 0
µaxis ≠ 0
µnet ≠ 0 (polar)
[Square pyramidal]
en
Cl
en
Cl
cis(optically active)
trans(optically inactive)
(I)
common transition and inner transition metal ion, are 6 and 8
respectively.
B
B
µplane ≠ 0
µaxis ≠ 0
µnet ≠ 0 (polar)
[Trigonal pyramidal]
So, it can be seen that when AB4 molecule is a polar molecule
then possible geometry of AB4 is square pyramidal.
2.
Pt
en
Pt
4. The maximum possible denticities of given ligand towards a
A
A
B
B
B
Cl
Key Idea The total number of ligands to which the metal is
directly attached is called coordination number.
The coordination numbers of Co and Al in [Co(Cl)(en)2]Cl and
K 3[ Al(C2O4 )3 ]are 5 and 6 respectively.
–OOC
COO–
N
N
N
–OOC
COO–
COO–
The given ligand act as hexadentate ligand in transition metal ion
because the common oxidation state shown by them is +3.
Whereas in case of inner transition metal ion, its denticity is 8
because their common oxidation state is +4.
5. Optical activity is the ability of a chiral molecule to rotate the
plane of polarised light, measured by a polarimeter. A chiral
molecule does not have any plane of symmetry. If a molecule
possess any plane of symmetry, then it is an achiral molecule.
Given options (a), (b) and (d) possess plane of symmetry.
Coordination Compounds
A
A
B
B
en
M
M
en
or
2 + x − 2 − 4 − 2 = 0 or x − 6 = 0
hence x = + 6
Thus, +3, 0 and +6 is the answer.
9. If the reactant is cis isomer than following reaction takes place.
NH3
B
Br
Br
Br–
A
(b)
NH3
B
NH3 NH3
Br
Br NH3
Cis-isomer
NH3
NH3
Br
Facial
Meridonial
i.e. two isomers are produced. If the reactant is trans isomer than
following reaction takes place.
M
A
B
Br
NH3
+
NH3
B
A
Br
NH3
NH3
NH3
Br
NH3
Br–
A
(a)
NH3
Only molecule (c) does not possess any plane of symmetry.
Hence, it is a chiral molecule and shows optical activity.
6.
Br
NH3
Br
B
A
(d)
267
Key Idea Denticity of ligand is defined as donor sites or
number of ligating groups.
The given ligand is tetradentate. It contains four donor atoms. It
can bind through two nitrogen and two oxygen atom to the
central metal ion.
Ligand bound to the central atom or ion through coordinate bond
in the coordination entity. It act as a Lewis base. The attacking
site of the given ligand is given in bold.
Br
Br
Br
Trans
Meridonial
i.e. only 1 isomer is produced. Thus, statement (I) and (III) are
correct resulting to option (b) as the correct answer.
Cl
10.
Cl
Cl
en
en
Co
en
cis-[Co(en)2Cl2]Cl
(optically active)
NEt2
en
Co
Cl
trans-[Co(en)2Cl2]Cl
(optically inactive due
to plane of symmetry)
[Co(NH3 )4 Cl 2 ]Cl can exist in both cis and trans forms that are
given below:
N
O–
NH3
NH3
+
Cl
–O
H3N
NH3
H 3N
Co
7. A square planar complex of general formula, M abcd gives three
geometrical isomers only.
Let, a = F− , b = Cl − , c = S CN− , d = NO−2
SCN− and NO−2 are ambidentate ligands and they also show
linkage isomerism (structural). Considering both linkage and
geometrical isomerism.
Total number of possible isomers given by the complex,
= 3 × (2 + 2) = 12
H3N
Cl
Co
NH3
Cl
H 3N
NH3
Cl
trans-[Co(NH3)4Cl2]Cl
(optically inactive)
cis-[Co(NH3)4Cl2]Cl
(optically inactive)
[Co(NH3 )3 Cl 3 ]exists in fac and mer-isomeric forms and both are
optically inactive.
NH3
Cl
8. Let the oxidation state of Cr in all cases is ‘ x’
(i) Oxidation state of Cr in [Cr(H2O)6 ]Cl 3
x + (0 × 6) + (−1 × 3 ) = 0
or
x + 0 − 3 = 0 or x = + 3
(ii) Oxidation state of Cr in [Cr(C6H6 )2 ]
x + (2 × 0) = 0 or x = 0
(iii) Oxidation state of Cr in
K2[Cr(CN)2 (O)2 (O2 )(NH3 )]
1 × 2 + x + (−1 × 2 ) + (−2 × 2 ) + (−2 ) + 0 = 0
+
NH3
NH3
NH3
Cl
Co
Cl
NH3
Cl
fac-isomer
(optically inactive)
NH3
Co
Cl
Cl
NH3
mer-isomer
(optically inactive)
11. [Pt(Cl)(py)(NH3 )(NH2OH)]+ is square planar complex. The
structures are formed by fixing a group and then arranging all the
groups.
268
Coordination Compounds
Py
Py
NH3
NH3
Pt
Pt
Cl
NH2OH HOH2N
Py
16. Ionisation isomers are the complexes that produces different
ions in solution, i.e. they have ions interchanged inside and
outside the coordination sphere.
[Cr(H2O)4 Cl(NO2 )]Cl and [Cr(H2O)4 Cl 2 ](NO2 ) have different
ions inside and outside the coordinate sphere and they are
isomers.
Cl
Cl
Therefore, they are ionisation isomers.
17. [Ni(NH3 )4 ]2 + = tetraamminenickel (II)
Pt
HOH2N
[NiCl 4 ]2− = tetrachloronickelate (II)
Cationic part is named first, hence :
tetraamminenickel (II)-tetrachloronickelate(II)
NH3
Hence, this complex shows three geometrical isomers.
12. Optical isomerism is exhibited by only those complexes which
lacks elements of symmetry. [Co(NH3 )3 Cl 3] shows facial as well
as meridional isomerism. But both the forms contain plane of
symmetry. Thus, this complex does not exhibit optical
isomerism.
18. [Co(NH3 )4 Br2 ]Cl and [Co(NH3 )4 BrCl]Br are ionisation isomers.
H 3N
Br
NH3
+
Br
Br +
H 3N
Co
Co
13. First of all, the compound has complex positive part
‘‘[Co(H2O)4 (NH3 )2 ]3+ therefore, according to IUPAC
conventions, positive part will be named first. Secondly, in
writing name of complex, ligands are named first in alphabetical
order, irrespective of its charge, hence “ammine” will be written
prior to “aqua”. Therefore, name of the complex is
[Co(H2O)4 (NH3 )2 ]Cl 3.Diamminetetraaqua cobalt (III) chloride.
NOTE In alphabetical order, original name of ligands are
considered not the initials of prefixes. Also, special precaution
should be taken in spelling name of NH3 ligand as it is ammine.
14. Ni 2+ + 4 CN– → [ Ni(CN)4 ]2–
Here, Ni 2+ has d 8 -configuration with CN– as strong ligand.
4s
4p
3d
H 3N
Br
H3N
NH3
Geometrical isomers
19.
PLAN Depending on the structure of the complex,different types of
isomerism are shown.
Complex
A.
B.
Isomerism
Neither of structural nor
stereoisomerism
[Cr(NH3 )5 Cl]Cl 2
[Cr(NH3 )4 Cl 2 ] Cl
[Co (NH3 )4 Cl 2 ]+
H3N
:
:
:
:
H3N
Cl
H3N
H3N
[Pt(H2O) ⋅ (NH3 )2 Cl]
Cl
H2O
NH3
Pt
:
:
:
:
:
:
:
:
4p
cis
:
:
:
:
:
:
Therefore, [Ni(H 2O)6] 2+ has octahedral geometry.
N—CH2CH2—N
CH2COOH
CH2COOH
NH3
H2O
NH3
Pt
H3N
C.
4d
sp3d2
Cl
NH3
Cl
sp
d 8 -configuration in weak ligand field gives sp 3-hybridisation,
hence tetrahedral geometry. Ni 2+ with H 2O forms
[Ni(H 2O)6] 2+ complex and H 2O is a weak ligand.
3d
NH3
cis w.r.t. Cl
+
2
HOOCH2C
NH3
Co
Here, Ni 2+ has d 8 -configuration with Cl – as weak ligand.
3d
15.
NH3
trans w.r.t. Cl
d 8 -configuration in strong ligand field gives dsp2-hybridisation,
hence square planar geometry.
Ni 2+ + 4Cl – 
→ [NiCl 4 ]2–
HOOCH2C
Cl
Co
dsp2
4s
NH3
NH3
D.
Cl
trans
[Co Br2Cl 2 ]2−
sp3 tetrahedral
[PtBr2Cl 2 ]2−
dsp2 square planar
[Pt (NH3 )3 (NO3 )]Cl
[ Pt (NH3 )3 (NO3 )Cl ]
[Pt(NH3 )3 Cl]Br
[ Pt (NH3 )3 Cl ] NO3 ]
ionisation
[ Pt (NH3 )3 Cl ] Br
[ Pt (NH3 )3 Br ]
ionisation
Coordination Compounds
20. Both [Pt(en)2 Cl 2 ]Cl 2 and [Pt(NH3 )2 Cl 2 ] are capable of showing
25. All six complex will show cis-trans isomerism
geometrical isomerism.
en
Cl
Cl
+
Pt
en
en
Cl
Pt
H 3N
cis
O
Pt
Cl
Cl
trans
O
Cl
NH3
O
Cl
NH3
H3N
Cl
M
H 3N
M
NH3
H3N
Cl
NH3
Cl
Trans
2+
CN
H 3N
23. Paramagnetism : In [Mn(H2O)6 ] , Mn(II) has 3d configuration.
Co3+
Cl
NH2
O
Cis
O
Cl
3+
O
and en
en
Co3+
Cl
en
NH3
trans
2+
2+
Cl
H3N
Cl
Co3+
and
OH2
NH3
NH3
H 3N
OH2
trans
⇒ mmol of Cl – ion produced from its 0.3 mmol = 0.6
Hence, 0.6 mmol of Ag+ would be required for precipitation.
NH2
NH2
Co
2+
Cl
[Cr(H2O)5 Cl]Cl 2 → [Cr(H2O)5 Cl]2+ + 2Cl –
Geometrical isomers are
Co
and
NH3
Also, 1 mole of complex [Cr(H2O)5Cl]Cl2 gives only two moles
of chloride ion when dissolved in solution
CH2 — NH2

CH2 O
Cl
Fe3+
26. mmol of complex = 30 ×0.01 = 0.3
(weak ligand field)
3+
NH3
NH3
cis
Mn(II) :
NH2
2+
Co3+
–
NH3
CN
cis
cis
H3N
Since, H2O is a weak ligand, all five d-electrons are unpaired :
4p
4d
4s
3d
OH2
OH
trans
NH3
5
O
NC
en
CN
A : Potassium tetracyanonickelate (II)
Cl
–
NH3
trans
B = K2[NiCl 4 ]
OH2 NC
H2O
OH
NC
O
CN
Fe3+
OH2
cis
NC
+
H2O
and
Fe
Cis
Cl
OH
OH
Fe3+
NH3
B : Potassium tetrachloronickelate (II)
24. Ligand is
+
3+
Both have several planes of symmetry
22. A = K2[Ni(CN)4 ] ;
H 2O
O
trans
OH2
H2O
O
O
O
NH3
O
Cl
3–
O
Cr3+
and
O
21. Both statements are true. However, axis of symmetry is not a
O
O
Cr3+
criteria of optical isomerism. Optical inactivity of the two
geometrical isomers of [ M (NH3 )4 Cl 2 ] is due to the presence of
plane of symmetry.
Cl
3–
O
Square planar complex
H 3N
Cl
trans
O
H3N
Cl
en
Co3+
cis
cis
H3N
+
Cl
en
en
Cl
and en
Co3+
Pt
trans
Cl
+
Cl
+
Cl
en
269
O
NH2
Trans
NH2
O
⇒ 0.60 mmol of Ag+ = 0.1M × V(in mL) ⇒ V = 6 mL
O
Co3+
Cl
Cl
Cis
27.
PPh3
Cl
CO
Cl
Rh
Rh
H 3N
CO
PPh3
H 3N
Cl
PPh3
Rh
OC
28. (i) [Co(NH3 )5 Cl]2+
H 3N
(ii) Li[AlH4 ]
270
Coordination Compounds
3. The compound used in the treatment of lead poisoning is EDTA.
29. [Cr(NH3 )5 CO3 ]Cl
Medication occurs through chelation therapy. Calcium disodium
ethylenediamine tetraacetic acid chelates divalent metal ion such
as Pb2 + from plasma and interstitial body fluids.
: pentaamminecarbonatochromium (III) chloride.
30.
(i) [Co(NH3 )5 ONO]Cl 2
: pentaamminenitritocobalt (III) chloride.
(ii) K3[Cr(CN)6 ] : potassium hexacyanochromate (III)
The metal displaces Ca and is chelated, mobilised and usually
excreted. Less then 5% CaNa 2EDTA is absorbed in the
gastrointestinal tract and it possibly increases the absorption of
Pb present in the tract. Therefore, it is not recommended for oral
use. It is usually given intravenously.
Topic 2 Bonding and Important Property
of Coordination Compounds
1.
M 2+ + mH2O → [ M (H2O)m ]2+
(d 6 )
4. Complete removal of both the axial ligands (along the z-axis)
from an octahedral complex leads to the following splitting pattern.
Here, H2O is a weak field ligand and it should give a high spin
paramagnetic (µ = 4.90 BM) complex ion.
dx2–y 2
We know, µ = n ( n + 2 ) BM = 4.90 BM (n = 4)
[∴n = Number of unpaired electrons in the complex]
ns0
dxy
Energy
(n–1)d6
np0
M 2+ =
(d6)
dz2
t2g
H2O H2O H2O H2O
n=4
M2+ is sp3 hybridised, i.e. it
is a tetrahedral complex ion
[M(H2O)4]2+ [∴ M = 4]
dxzdyz
The single electron in the dx2 − y2 orbital is being repelled by four
ligands, while the electron in the dz2 orbital is only being repelled
by two ligands. Thus, the energy of the dx2 − y2 increases relative
to that of dz2 . A more stable arrangement arises when both the eg
electrons pair up and occupy the lower energy dz2 orbital. This
leaves the dx2 − y2 orbital empty.
Thus, four ligands can now approach along + x , − x, + y and − y
directions without any difficulty as dx2 − y2 orbital is empty.
However, ligands approaching along + z and −z directions meet
very strong repulsive forces from filled dz2 orbitals. Thus, only
four ligands succeed in bonding to the metal. A square planar
complex is formed, the attempt to form an octahedral complex
being unsuccessful.
(n −1) d -orbitals in tetrahedral field.
t2
+0.4 ∆ t
–0.6 ∆ t
e
CFSE = [(− 0.6) × 3 + (0.4 ) × 3 ]∆ t = − 0.6 ∆ t
Number of
electrons in e set.
Number of
electrons in t2 set.
So, the geometry and the crystal field stabilisation energy of the
complex is tetrahedral and − 0.6 ∆ t respectively.
2. The d-electron configuration [Ru(en) 3]Cl 2 and [ Fe(H2O)6 ]Cl 2
respectively are t62geg0 and t24geg2.
5.
Key Idea Crystal field splitting occurs due to the presence
of ligands in a definite geometry. In octahedral complexes
the energy of two, eg orbitals will increase by (0.6) ∆ o and
that of three t2g will decrease by (0.4) ∆ o .
The complex ion that will lose its crystal field stabilisation
energy upon oxidation of its metal to +3 state is[ Fe(phen)3 ]2+ .
[Ru(en)3 ]Cl2
Ru = 4d series
en = bidentate ligand (strong field ligand)
C.N. = 6
Oxidation number = +2
Ru 2+ = [Kr] 4 d6 5s0
−e−
[Fe(phen)3 ]2+ → [Fe(phen)3 ]3+
In [Fe(phen)3 ]2+ , electronic configuration of Fe2+ is 3d6 4 s0.
Phenanthrene is a strong field symmetrical bidentate ligand. The
splitting of orbital in Fe2+ is as follows:
Fe2+ =
eg
[Fe(H 2O) 6 ]Cl2 =
eg
0.6∆0
(n–1)d
(3d)
eg
∆0 t 4 , e 2
2g g
0.4∆0
+0.6 ∆o
Fe2+
–0.4 ∆o
t2g
CFSE = 6 × −0.4 ∆ o = −2.4 ∆ o .
t2g
Coordination Compounds
[Cr(H2O)6 ]2+ ,
(n = 5),
The splitting of orbital and arrangement of electrons in Fe3+ is as
follows :
+0.6 ∆o
Fe3+
–0.4 ∆o
t2g
CFSE = 5 × −0.4 ∆ o = −2. 0 ∆ o
Fe upon oxidation of its metal to +3 state lose its CFSE from
−2.4 ∆ o to −2.0∆ o .
6.
Key Idea Crystal field stabilisation energy (CFSE) for
octahedral complexes = (−0.4 x + 0.6 y)∆ o
where, x = number of electrons occupying t2g orbital.
y = number of electrons occupying eg orbital.
CFSE for tetrahedral complexes
= (−0.6x + 0.4 y)∆ t
where, x = number of electrons occupying e orbital.
y = number of electrons occupying t orbital.
In [Fe(H2O)6 ]Cl2, H2O is a weak field ligand, so it is a high spin
(outer orbital) octahedral complex of Fe2+ .
eg
Fe2+(3d6) =
t2g
∴ CFSE = (−0.4 x + 0.6 y)∆ o
= [ −0.4 × 4 + 0.6 × 2 ]∆ o = − 0.4 ∆o
In K2[NiCl 4 ], Cl − is a weak field ligand, so it is a high spin
tetrahedral complex of Ni2+ .
t
Ni2+(3d6) =
e
µ 2 = 5(5 + 2)
= 35 = 5.92 BM
So, µ 1 ≈ µ 2.Thus, statement (c) is correct.
(d) [Ni(NH3 )4 (H2O)2 ]2+ is also a high-spin octahedral complex
of Ni2+ (3d 8 , n = 2)
µ = 2(2 + 2) = 8 = 2.83 BM
Thus, statement (d) is correct.
eg
2+
8.
Key Idea The wavelength (λ ) of light absorbed by the
complexes is inversely proportional to its ∆ 0 CFSE
(magnitude). ∆ 0 (CFSE) ∝ 1 / λ
The complexes can be written as:
I. [CoCl(NH3 )5 ]2+ ≡ [Co(NH3 )5 (Cl)]2+ ]
II. [Co[NH3 ]5 H2O]3+ ≡ [Co(NH3 )5 (H2O)]3+
III. [Co(NH3 )5 ]3+ ≡ [Co(NH3 )5 (NH3 )]3+
So, the differentiating ligands in the octahedral complexes of
Co (III) in I, II and III are Cl s, H2O and NH3 respectively. In the
spectrochemical series, the order of this power for crystal field
splitting is Cl − < H2O < NH3.
So, the crystal field splitting energy (magnitude) order will be
∆CFSE
(I) < ∆CFSE
(II) < ∆CFSE
(III)
0
0
0
and the order of wavelength (λ ) of light absorbed by the
complexes will be
1

Q Energy (∆CFSE
)∝
λ (I) > λ (II) > λ (III)
0
λ 

9. The degenerate orbitals of [Cr(H2O)6 ]3+ are dxz and d yz.
Electronic configuration of Cr3 + is 3d 5 4 s1. The five d-orbitals in
an isolated gaseous atom or ion have same energy, i.e. they are
degenerate. This degeneracy has been removed due to the ligand
electron–metal electron repulsions in the octahedral complex to
yield three orbitals of lower energy, t2g set and two orbitals of
higher energy, eg set.
(e g )
∴CFSE = (−0.6 × 4 + 0.4 × 4 )∆ t = − 0.8∆ t
dx 2–y2 dz2
7. The explanation of given statements are as follows :
(a) Ruby, a pink or blood-red coloured gemstone belongs to
corundum (Al 2O3 , alumina) system which has trigonal
crystalline lattice containing the repeating unit of
Al 2O3 – Cr 3+ . So, ruby does not belong to beryl lattice
(Be3Al 2Si6O18 ).
Thus, statement (a) is incorrect.
(b) [Co(Cl)(NH3 )5 ]2+ is a low spin octahedral complex of Co3+ .
It absorbs low energy yellow light and high energy
complementary violet light will be shown off. Thus,
statement (b) is correct.
(c) [Fe(H2O)6 ]2+ and [Cr(H2O)6 ]2+ are the high-spin octahedral
complexes of Fe2+ (3d6 , n = 4 ) and Cr 2+ (3d 5 , n = 5) ions
and weak field ligand, H2O respectively. So, spin-only
magnetic moment = n(n + 2) of the complexes.
[Fe(H2O)6 ]2+ ,
(n = 4 ),
µ 1 = 4 (4 + 2)
= 24 = 4.89 BM
271
Free metal ion
∆0
(t2g)
dxy dxz dyz
10. [ Fe(H2O)6 ]2 ⇒ It will form 2 cationic species. i.e.
II
I. (i)As [ Fe(H2O)6 ]2+ ⇒ High spin octahedral complex of Fe2+ .
Fe2+ : 3d6 , x = 4 (unpaired electrons)
µ = 4 (4 + 2) BM = 4.9 BM
III
or (ii) as[ Fe(H2O)6 ]3+ = High spin octahedral complex of Fe3+ .
Fe3 + : 3d5 , x = 5, µ = 5 (5 + 2) = 5.92 BM
[H2O is a neutral weak field ligand]
So, [ Fe(H2O)6 ]2+ will be the cationic specie, µ = 4.9 BM.
[ Fe(CN)6 ] will have two anionic complexes
II
II. (i) [ Fe(CN)6 ]4– ⇒ Low spin, octahedral complex of Fe2+ .
As CN− is a strong ligand it will pair up the electrons.
272
Coordination Compounds
The
octahedral
homoleptic
complex
sp3d 2-hybridisation in the complex, i.e.
d2sp3
Fe2+(3d6)
sp3d2-hybridisation
n=0, µ=0
Mn2+=
III
or, (ii) [ Fe(CN)6 ]3– ⇒ Low spin octahedral complex of Fe3+ .
3d
n=1, µ=√1(1+2)
=1.73 BM
Metal Configuration Number of
ions
unpaired
electrons
Thus, the calculated spin only magnetic moments (BM) of the
anionic and cationic species of [ Fe(H2O)6 ]2 and [ Fe(CN)6 ]
respectively are 4.9 and 0.
11. cis-[Pt(Cl)2 (NH3 )2 ] is known as cis-platin. It is a σ-bonded
organo-metallic compound and is used as an anti-tumor agent in
the treatment of cancer.
Cl
12.
Key Idea In presence of strong field ligands, ∆ 0 > p, for
fourth electron it is more energetically favourable to occupy
t2 g orbital with configuration t24geg0and form low spin complexes.
The correct order of the spin only magnetic moment of metal ions
in the given low-spin complexes is V2+ > Cr 2+ > Ru 3+ > Fe2+ .
All the given complexes possess strong field ligands
(CN, NH3 ). Hence, readily form low spin complexes.
No. of
unpaired
electrons
Orbital
splitting
eg
[V(CN)6]4–
V
2+
3
0
t2g eg
3
t2g
eg
[Cr(NH3)6]2+
Cr2+
4
t2g eg0
2
t2g
3
Fe2+
(d6 ) = t2g4eg2
4
4.9
Cr2+
(d 4 ) = t23geg1
4
4.9
t23geg2
5
5.9
(d 3 ) = t23geg0
3
3.9
5
2+
(d ) =
Therefore, Co 2+ and V 2+ contains same value of magnetic
moment (3.9 BM).
15.
In K3 [Co(CN)6 ], Co have +3 oxidation state and electronic
configuration of Co3 + is [Ar ] 18 3d 6.
3d6
4p0
4s0
Co3+=
As, CN − is a strong field ligands so it pairs up the de− s
∴
3d6
[Co(CN)6]3–=
Inner orbital
complex
XX
XX
XX
XX
XX
XX
d2sp3
-hybridised
(6e− pairs donated
by 6 CN− ligands)
In an octahedral complex, the metal is at the centre of the
octahedron and the ligands are at the six corners. The lobes of the
eg orbitals (dx2 − y2 and dz2 ) point along the axes x , y and z under
the influence of an octahedral field, the d- orbitals split as follow.
eg
[Ru(NH3)6]3+ Ru3+
(d 7 ) = t25geg2
V2+
Cl
Complex Oxidation Configuration
state
Co2+
Spin only
Magnetic
moment (in
BM) = n (n + 2)
3.9
Mn
Pt
H3 N
4d
complexes. In[M (H2O)6 ]Cl2, M exist in +2 oxidation state. The
arrangement of electrons in the given metal ions are as follows:
(Not in options)
II
4p
14. As H2O is a weak field ligand. It readily forms high spin
[CN − is an anionic strong field ligand]
So,the anionic species is [Fe(CN)6 ]4− , µ = 0
H3 N
4s
Thus, 5 unpaired electrons are present in the complex which
suggest the presence of a weak ligand like NCS − .
d2sp3
Fe3+ ⇒ (3d5)
suggests
t25g eg0
dx2-y2, dz 2
1
t2g
eg
[Fe(CN)6]4–
Fe2+
t26g eg0
0
t2g
13. The magnetic moment of the magnitude 5.9 BM suggest the
presence of 5 unpaired electrons in Mn(II). This can be cross
verified by putting the value (5) of unpaired electrons in the
formula, µ = n(n + 2) BM
Thus, the valence electronic configuration of Mn(II) in the
complex is
Mn2+=
3d
4s
4p
4d
Average
d-orbitals
energy of
in free ion the d-orbitals
in a spherical
crystal field
dxy, dyz, dxz
Splitting of d-orbitals
in an octahedral
crystal field
As the d-orbitals, i.e. dx2 − y2 and dz2 are vacant. Hence, these
both orbitals are directly facing the ligands in K3 [Co(CN)6 ].
16. Mn 2 (CO)10 is an organometallic compound due to the
presence of MnC bond. The metal-carbon bond in
organometallic compounds possess both σ and-π character.
The MC σ bond is formed by the donation of lone pair of
electrons from the carbonyl carbon into a vacant orbital of the
metal. The MC π-bond is formed by the donation of pair of
Coordination Compounds
Here, C2O24 − is a bidentate ligand,
electrons from a filled d-orbital of metal into vacant
antibonding π * orbital of CO. The M  L bonding creates a
synergic effect which strengthens the bond between CO and
the metal.
¾
O¾
CO
CO
CO
CO
Mn
Mn
CO
CO CO
¾
¾
M (Metal)
M
C2O24 −
So, total number of sites offered by
and H2O ligands
around Th(IV) = Coordination number of Th (IV)
CO
written as:
= 4 × 2(by C2O24 −) + 2 × 1(by H2O) = 10
19. (A) Co is present in vitamin B12 (iii) having molecular formula,
II
O
C63H88 CoN14O14P.
C
CO
(B) Zn is present in carbonic anhydrase (iv) in which three
histidine units and the —OH group coordinate with one Zn
(II) ion.
OC——Co——–Co——CO
OC
C¾O
¾
O ¾
and H2 O is a monodentate ligand, H2O
CO
17. The structure of Co2 (CO)8 (a polynuclear metal carbonyl) can be
OC
C¾O
¾
The structure of Mn 2(CO)10 is shown below :
CO
273
CO
C
(C) Rh is present in Wilkinson catalyst (i) having molecular
formula [(Ph 3P)3 RhCl] .
O
Total number of bridging CO ligands = 2
and the Co Co bond = 1
(D) Mg is present in chlorophyll (ii) having molecular formula
II
18. Coordination number is defined as the total number of ligands to
C55H70O6N4 Mg (chlorophyll-b).
which the metal is directly attached.
20. The difference in the number of unpaired electrons of different metal ions in their high spin and low spin octahedral complexes are given in
the table below :
Metal ion
Mn
Fe
Ni
Number of e− in high spin complex ( n1 )
2+
3d
2+
4s
3d
3d
4d
4d
n1=5
n2=1
3d
2+
4p
Number of e− in low spin complax ( n2 )
4s
4p
3d
4d
4d
n1=4
n2=0
4s
n2 − n1
4s
4p
5 −1 = 4
4s
4p
4−0 = 4
4p
4d
n1=2
Ni 2+ does not form low spin octahedral complexes.
Co2+
3d
4s
4p
III
III
CoCl3 + 2en → [Co(en)2Cl2]Cl
4s
4d
4d
∴ n1=3
∴ n1=1
21. According to the situation given in question, reactions are as
follows:
3d
4p
••

CH2 — N H2
en = ethylene-1, 2-diamine, 

CH2 — NH2 
••

3 −1 = 2
274
Coordination Compounds
∆(CFSE) is measured with help of wavelength of the colour
absorbed by the given coordination compound, as
c
∆ O = hν = h ×
λ
Both the complexes contain three unpaired electrons. Therefore,
both are paramagnetic.
Cl
en
Cl
Co
Cl
III
en
Cl
en
en Cl
Co
Cl
‘B’ Optically inactive
(trans-form)
(Green)
‘A’ Optically active
(cis-form)
(Violet)
27. Fluoride ions help in making teeth enamel harder by converting
i.e.
Hydroxy
apatite
to
[3Ca 3 (PO4 )2 ⋅ Ca(OH)2 ]
[3Ca 3 (PO4 )2 ⋅ CaF2 ] i.e., Fluorapatite (Harder teeth enamel) via
following reaction:
22. Wilkinson’s catalyst is a σ-bonded organometallic compound
[(Ph 3P)3 RhCl] . It is commercially used for hydrogenation of
alkenes and vegetable oils (unsaturated).
IUPAC name Chloridotris (triphenylphosphene) rhodium (I).
[3Ca 3 (PO4 ) 2 ⋅ Ca(OH) 2 ] + 2F − → [3Ca 3 (PO4 ) 2 ⋅ CaF2 ] + 2OH−
From
drinking
water
28. Molarity (M) =
23. In homoleptic complexes, the metal atom/ion is linked to only
∴ Number of moles of complex
one type of ligand. Assuming, ligands are neutral, the octahedral
complexes of M 3+ can be,
[ M (L1 )6 ]3+ ,[ M (L2 )6 ]3+ and [ M (L3 )6 ]3+
(I)
(II)
Green
λ Absorption
L3
So,
λ III > λLI1 > λLII2
∴∆°absorption
: ∆LII2
>
∆LI1
>
∆LIII3
=
(III)
Blue
We know, ligand strength ∝ ∆°absorption
∴2 Cl − are present outside the square brackets, i.e. in ionisation
sphere. Thus, the formula of complex is
So, the increasing order of the ligand strength will be,
L3 < L1 < L2
[Co(H2 O) 5 Cl]Cl 2 ⋅H2 O.
29.
4 

Q ∆t = ∆o

9 
So, the octahedral complexes (a, b, c) have higher ∆ o values than
that of tetrahedral, K2[CoCl 4 ].
Now, for the complexes, a, b and c,
the magnitude of ∆ o ∝ ligand strength, which is based on their
positions in the spectrochemical series.
Cl − < H2O < NH3 < CN−
We know,
∆t < ∆o
Hence, K3[Co(CN)6 ] will have the highest ∆ value.
25. The spin only magnetic moment (µ) (in BM) is given by
µ(in BM) = n(n + 2)
where, n = number of unpaired electrons
The highest value of n in transition metal complex is 5 in its
d 5-configuration.
∴
µ = 5(5 + 2)BM = 5.916BM
26. ‘A’ absorbs yellow light of less energy and emits violet light of
high energy (complementary colour) because H2O is a weak
field ligand. But in case of ‘B’, due to presence of strong field
ligand (NH3 ), it absorbs high energy violet light and emits low
energy complementary yellow colour.
. × 1022
12
= 0.02 moles
6.02 × 1023
∴Number of Cl − present in ionisation sphere
Number of moles of ions precipitated 0.02
=
=2
=
0.01
Number of moles of complex
1
[QEnergy (∆, CFSE) ∝ ]
λ
except K 2[CoCl 4], which is a tetrahedral complex of Co (II)
(sp3-hybridised).
Molarity × volume (in mL) 01
. × 100
=
= 0.01 mole
1000
1000
Number of moles of ions precipitate =
Red (wavelength)
24. All of the complex given are the octahedral complexes of Co (III)
Number of moles of solute
Volume of solution (in L)
Complex ion
Electronic
configuration of
metal ion
[Cr(H2O)6 ]2+
Cr 2+ ; [Ar] 3 d 4
;4
[Fe(H2O)6 ]2+
Fe 2+ ; [Ar] 3 d6
; 4
[Mn(H2O)6 ]2+
Mn2+ ; [Ar] 3 d 5
;5
[CoCl 4 ]2−
Co 2+ ; [Ar] 3 d 7
;3
Number of unpaired
electrons (n)
30.
Compounds
Hybridisation Unpaired
electron(s)
Ni(CO)4
[NiCl 4 ]
2−
[Co(NH3 )4 Cl 2 ]Cl
sp3
Magnetic
character
No
Diamagnetic
3
two
Paramagnetic
3 2
No
Diamagnetic
sp
sp d
3 2
three
Paramagnetic
Na 2O2
—
No
Diamagnetic (O22 − )
CsO2
—
One
Paramagnetic
Na 3[CoF6 ]
sp d
O−2 (superoxide
ion is
paramagnetic)
32. This problem is based on conceptual mixing of properties of
lithium oxide and preparation, properties of coordination
compounds. To answer this question, keep in mind that on
adding acid, ammine complexes get destroyed.
(a) Li 2O + KCl → 2LiCl + K2O
This is wrong equation, since a stronger base K2O cannot be
generated by a weaker base Li 2O.
(b) [CoCl(NH3 )5 ]+ + 5H+ → Co2+ ( aq) + 5 NH4+ + Cl −
+2
26
+2
[Ar]3d 5
5
35 BM
3
15 BM
4
24 BM
weak ligand
Q : [V(H2O)6 ]2+
[Ar]
weak ligand
R :[Fe(H2O)6 ]2+
[Ar]3d6
Thus, order of spin-only magnetic moment = Q < R < P
35. In the given complex, NiCl 2 {P (C2H5 )2 (C6H5 )}2 nickel is in
+ 2 oxidation state and the ground state electronic configuration
of Ni 2+ ions in free gaseous state is
4s0
3d 8
4p0
2+
Ni
For the given four coordinated complex to be paramagnetic, it
must possess unpaired electrons in the valence shell. To satisfy
this condition, four lone pairs from the four ligands occupies the
four sp3-hybrid orbitals as :
Therefore, geometry of paramagnetic complex must be
tetrahedral. On the otherhand, for complex to be diamagnetic,
there should not be any unpaired electrons in the valence shell.
This condition can be fulfilled by pairing electrons of
3d-orbitals against Hund’s rule as
λ
L1
L2
L3
L4
Absorbed light
Red
Green
Yellow
Blue
Wavelength of absorbed light decreases.
∴ Increasing order of energy of wavelengths absorbed reflect
greater extent of crystal field splitting, hence, higher field
strength of the ligand.
Energy blue (L4 ) > green (L 2 ) > yellow (L 3 ) > red (L 1 )
∴
L4 > L2 > L3 > L1 in field strength of ligands.
34.
36. For a diamagnetic complex, there should not be any unpaired
electron in the valence shell of central metal.
In K3[Fe(CN)6], Fe (III) has d 5-configuration (odd electrons),
hence it is paramagnetic.
In [Co(NH3)6]Cl3, Co (III) has d 6-configuration in a strong
ligand field, hence all the electrons are paired and the complex is
diamagnetic.
In Na3[Co(ox)3], Co (III) has d 6-configuration and oxalate being
a chelating ligand, very strong ligand and all the six electrons
remains paired in lower t2g level, diamagnetic.
In [Ni(H2O)6]Cl2, Ni (II) has 3d 8 -configuration and H2O is a
weak ligand, hence
3d
sp3d 2
:
:
:
:
:
In the presence of strong ligand (as CN− , CO, NH3 , en) electrons
are paired and electrons from ligands are filled in available inner
orbitals
dsp2
The above electronic arrangement gives dsp2-hybridisation and
therefore, square planar geometry to the complex.
:
PLAN Spin only magnetic moment have the formula n( n + 2) BM,
where N is the number of unpaired electrons. In the presence of
weak ligand (as H 2O, Cl − , F − ) there is no pairing of electrons,
and electrons donated by ligands are filled in outer vacant
orbitals.
:
L4, according to wavelength of their absorbed light, then use of
the following relation to answer the question.
Ligand field strength ∝ Energy of light absorbed
1
∝
Wavelength of light absorbed
:
3d 8
33. Arrange the complex formed by different ligands L 1 , L 2 , L 3 and
:
Ni2+
:
(d) The 4th reaction is incorrect. It can be correctly
represented as
2CuSO4 + 10KCN → 2K3[Cu(CN)4 ]
+ 2K2SO4 + (CN)2 ↑
:
sp3
(c) [Mg (H2O)6 ]2+ + EDTA4− → [ Mg( EDTA)]2+ + 6H2O
This is wrong, since the formula of complex must be
[Mg(EDTA)]2+ as EDT.
:
:
Ni2+
OH−
Excess
Magnetic
moment
23
Unpaired
electrons
+3
275
E.C.
26
:
This is correct. All ammine complexes can be destroyed by
adding H ⊕ . Hence, on adding acid to [CoCl(NH3 )5 ], it gets
converted to Co2+ ( aq)+ NH+4 and Cl − .
P : [FeF6 ]3 −
O.N.
∴ In MnO−4 , Mn has + 7 oxidation state having no electron in
d-orbitals.
It is considered that higher the oxidation state of metal, greater is
the tendency to occur L →M charge transfer, because ligand is
able to donate the electron into the vacant d-orbital of metal.
Since, charge transfer is laporate as well as spin allowed,
therefore, it shows colour.
Time saving Technique There is no need to check all the four
options. Just find out the oxidation state of metal ion. If oxidation
state is highest and ligand present there is of electron donating
nature, gives LMCT, which shows more intense colour.
Atomic
number of
31. KMnO4 → K + + MnO4−
Complex
Coordination Compounds
Paramagnetic
276
Coordination Compounds
NiCl 2 (PPh 3 )2 , Ni 2+ has 3d 8 -configuration. Due to weak ligand
field, Ni is sp3-hybridised and complex is tetrahedral.
In K2[Pt(CN)4], Pt(II) has d 8 -configuration and CN– is a strong
ligand, hence all the eight electrons are spin paired . Therefore,
complex is diamagnetic.
In [Zn(H2O)6](NO3)2, Zn (II) has 3d 10configuration with all the
ten electrons spin paired, hence diamagnetic.
37. Magnetic moment = 2.83 BM indicates that there is two unpaired
48.
Cu 2+ + CN− → CuCN ↓
CuCN + 3CN− → [Cu(CN)4 ]3 −
49. Fe in [Fe(H2O)6 ]2+ has maximum (four) unpaired electrons, has
highest paramagnetism.
electrons.
u = n(n + 2) BM = 8 BM = 2.82 BM
−
8
2−
In [NiCl 4 ] , Ni has d configuration and Cl is a weak ligand :
3d
:
Ni
:
:
:
2+
4s
4p
3
sp -hybridisation
6
38. In Cr(CO)6 : 3d , has no unpaired electrons, zero magnetic
moment.
39. CuF2 : Cu 2+ has 3d 9 -configuration, allowed d-d transition,
hence, coloured.
40. In Ni(CO)4, Ni is sp3-hybridised while in [Ni(CN)4 ] 2− , Ni 2+ is
dsp2-hybridised.
41. Greater the extent of dπ - pπ back bonding, smaller will be the
50. In Ni (CO)4, Ni has 3d 10-configuration, diamagnetic. In
Ni (CN)4 ]2− , Ni has 3d 8 -configuration but due to strong ligand
field, all the d-electrons are spin paired giving
dsp2-hybridisation, diamagnetic.
In [NiCl 4 ]2− , Ni has 3d 8 -configuration and there is two unpaired
electrons (weak chloride ligand do not pair up d - electrons),
hence paramagnetic.
51. Salt with least number of unpaired electrons in d - orbital of
central metal will show lowest degree of paramagnetism
Mn 2+ (3d 5, 5 unpaired electrons)
Cu 2+ (3d 9 , 1 unpaired electrons)
Fe2+ (3d6, 4 unpaired electrons)
Ni 2+ (3d 8 , 2 unpaired electrons)
Hence, CuSO4 ⋅ 5H2O has lowest degree of paramagnetism.
52. (a) In [FeCl 4 ]− , oxidation number of Fe atom = + 3
Electronic configuration of Fe in ground state = 3d6 4 s2
bond order of CO bond in metal carbonyls. In Fe(CO)5, there is
maximum number of valence shell electrons (d-electrons),
greatest chances of pπ - dπ back bonding, lowest bond order of
CO bond.
Electronic configuration of Fe3+ = 3d 5 4 s0 4 p0
Cl
42. In CO, bond order = 3. In metal carbonyls like Fe(CO)5, due to
dπ - pπ back-bonding, bond order of CO decreases slightly
therefore, bond length increases slightly.
43. In Hg [Co(SCN)4 ], Co2+ has 3d 7 configuration. SCN − produces
weak ligand field, no pairing of electrons in d-orbitals occurs
against Hund’s rule, hence :
Co2+ :
µ = 3 (3 + 2) BM = 15 BM
3d7
Cl Cl Cl
sp 3-hybridisation
Thus, [FeCl 4] − has tetrahedral geometry.
(b) [Co(en)(NH3 )2 Cl 2 ]+ have three geometrical isomers. Thus,
statement (b) is incorrect.
NH3
Cl
NH3
en
Co3+
Cl
en
Co3+
Cl
Tetrahedral
sp3-hybridisation
3d 8
4s
Under influence of weak ligand field
In all other complexes, hybridisation at central metal is dsp2 and
complexes have square planar geometries.
45. In 1 L solution, there will be 0.01 mole of each [Co(NH3 )5 SO4 ]
Br and [Co(NH3 )5 Br]SO4. Addition of excess of AgNO3 will
give 0.01 mole of AgBr. Addition of excess of BaCl 2 will give
0.01 mole of BaSO4.
46. In MnO−4 , Mn + 7 has 3d 0 configuration.
by CO. Hence, Ni is sp3-hybridised and complex is tetrahedral. In
Co3+
NH3
Cl
NH3
(c) Fe 3+ in [FeCl 4 ]− is sp3-hybridised with 5 unpaired electrons.
(higher
spin-only
magnetic
moment
in
While
= n(n + 2 ) = 5.92 BM ).
Co3+
[Co(en)(NH3 )2 Cl 2 ]+ is d 2sp3-hybridised with zero unpaired
electrons
(low
spin-only
magnetic
moment
= n(n + 2 ) = 0 BM).
Thus, the statement (c) is correct.
(d) Co3+ [Co(en)(NH3 )2 Cl 2 ]+
eg
∆0>P
Co3+ : [Ar]3d6
10
47. In Ni (CO)4, Ni is in 3d state due to strong ligand field produced
Cl
en
Cl
NH3
44. [NiCl 4 ]2 − : Ni 2+ (3d 8 )
NH3
t2g
Coordination Compounds
Co3+ in [Co(en)(NH3 )2 Cl 2 ]+ is d 2sp3-hybridised and has
octahedral geometry with 0 unpaired electron. Thus,
statement (d) is incorrect.
53.
SnCl2 + Cl
–
Tin chloride
(Q)
55. Statement wise explanation is
Statement (a) [Co(en)(NH3 )3 H2O ]3+ have following 2
geometrical isomers.
NH3
Sn
Cl
Cl
Cl
–
SnCl3
(X )
en
Me
N
Me
Me
(3°amine)
(Q)
(Y )
Cl
Cl
+2
+4
[ M ( AA )b3 c] type complex
Hence, this is correct statement.
Statement (b) If bidentate ligand ‘en’ is replaced by two
cyanide ligands then [Co(NH3 )3 (H2O)(CN)2 ]+ is formed.
It is [ Ma3 b2 c ] type complex which has following 3 geometrical
isomers.
+1
NH3
SnCl4 + 2CuCl
(Q)
(Z )
N
Oxidation
Z is oxidised product and oxidation state of Sn is +4 in Z
compound. Structure of SnCl 4 (Z ) is
Cl
Sn
Cl
NH3
NH3
C
N
C
NH3
NH3
C
N
Valence shell having 8 electrons
in complex formation
Ni(3d8,4s2)
4p
[Fe(CO)5]
3d
4s
4p
H2O
NH3
C
NH3
(Mer)
Hence, this statement is also correct.
Statement (c) Co metal has [ Ar ]3d 7 4 s2 configuration while in
[Co(en)(NH3 )3 (H2O )]3+ it is in +3 oxidation state. Thus, Co 3+
has [Ar]3d6 configuration.
3d
Rearrangement
dsp3
4s
3d
Co3+ =
As en is a strong ligand, so pairing will occur
[Ni(CO)4]
Rearrangement
CN
Fac with respect to NH3 and
Mer with respect to —CN
Co
(i) Statement (a) The total number of valence shell electrons at
metal centre in Fe(CO)5 or Ni(CO)4 is 8 instead of 16 as
shown below
4s
NH3
H2O
H2O
(Fac)
N
NH3
C
Co
54. Statement wise explanation is
3d
N
Co
Cl
Cl
Thus, options (a, d) are correct.
Fe(3d6,4s2)
NH3
NH3
Mer
H2O
Fac
Sn
Reduction
+2
Co
NH3
Y complex has coordinate bond in between nitrogen and Sn metal.
SnCl2 + 2CuCl2
H2O
en
Co
SnCl −3 has (3σ + 1lp) and exist in pyramidal structure.
SnCl2 × NMe3
NH3
NH3
sp3 (pyramidal)
SnCl2+Me3N
277
sp3
Hence, this statement is incorrect.
(ii) Statement (b) Carbonyl complexes are predominantly low
spin complexes due to strong ligand fields. Hence, this
statement is correct.
(iii) Statement (c) For central metal lowering of oxidation state
results to increase in electron density on it. This in turn
results to increase in extent of synergic bonding. Thus, we
can say ‘‘metal carbonyl bond strengthens, when oxidation
state of metal is lowered’’.
Hence, it is a correct statement.
(iv) Statement (d) Increase in positive charge on metal (i.e.,
increase in oxidation state) results to decrease in synergic
bonding strength.
This in turn makes C—O bond stronger instead of weaker.
Hence this statement is also incorrect.
4s
Due to the presence of all paired electrons it show diamagnetic
behaviour rather than paramagnetic.
Hence, this statement is incorrect.
Statement (d) According to CFT, absorption of light by
coordination complexes depends upon CFSE i.e., crystal field
splitting energy (∆ 0 )as
1
∆0 ∝
λ
Among the complexes given [Co (en) (NH3)4 ]3+ has more ∆ 0
value as compared to complex [Co(en) (NH3)3(H2O) ]3+ . Thus,
[Co (en) (NH3)3(H2O) ]3+ absorbs the light at longer wavelength
for d-d transition.
Hence, this statement is also correct.
Note : For any complex, the value of ∆ 0 can be calculated via the
difference or gap between eg and t2g values.
278
Coordination Compounds
61. For, P i.e. dsp2, It is seen in [Ni(CN)4 ]2 −
Excess NH 4 OH / NH 4 Cl
56. [Co(H2O) 6 ] Cl 2 → Co(NH3 ) 6 ]Cl 3
O 2 (Air)
Pink (X)
Ni [Ar]3d 8 4 s2
Y
2+
[Co(H2 O) 6 ]
+ 4Cl
−
2−
(Excess)
X
→ [CoCl 4 ]
Ni2+ [Ar]3d 8
blue Z
3d
(a) Since NH3 is moderately strong ligand, hybridisation of
cobalt in Y is d 2sp3.
(b) Cobalt is sp3-hybridised in [CoCl 4 ]2− .
(c) [Co(NH 3 ) 6 ]Cl 3 + 3AgNO3 (aq) → 3AgCl ↓
Y
(d) [CoCl 4 ]2− + 6H2O q
Blue
[Co(H2O)6 ] 2+ + 4Cl − ; ∆H < 0
Cl (a)
Cl
N1
en
N2
(b )
Mn
en
Cl (a)
dsp2-hybridisation
3d
4s
4p
Ni2+
CN– CN– CN– CN–
Cl (b)
Structure : Square planar
So correct match for P is 6.
For Q i.e., sp3
Mn
N
4p
as CN − is a strong ligand so when it approaches towards central
metal pairing of unpaired electrons takes place.
Thus, in[Ni(CN)4 ]2 −
Pink
57. The structure of cis-[ Mn (en )2 Cl 2 ] complex is
4s
Ni2+
N4
3
Bond angles (Mn—N and Mn—Cl bond in cis positions)
Cl (a) —— Mn —— N(1)
Cl (a) —— Mn —— N(2)
Cl (a) —— Mn —— N(4)
Cl (b) —— Mn —— N1
Cl (b) —— Mn —— N3
Cl (b) —— Mn —— N4
Number of cis Cl—Mn—N = 6
58. In the complex [Fe(H2O)5 NO]SO4, Fe is in +1 oxidation state
It is seen in [FeCl4 ]2 − and Ni(CO)4
Fe − [Ar]3d6 4 s2
Fe 2+ − [Ar]3d6
3d
4s
4p
Fe2+
As Cl − is a weak ligand so when it approaches towards central
metal pairing of unpaired electrons does not take place.
Thus, in[FeCl4 ]2 −
because NO is in +1 state. Also NO is a strong ligand, complex
has 3d 7 -configuration at Fe(I) as :
sp3 hybridisation
3d
4s
4p
Cl–
Cl– Cl– Cl–
4s
4p
Fe2+
3d7
Structure-Tetrahedral
Likewise in Ni(CO)4
Ni [Ar]3d 8 4 s2
Strong ligand field
3d
Ni
Three unpaired electrons
59. A is diamagnetic, square planar complex because of strong
ligand field of CN− .
Ni(CN)42 –(Ni2+) :
As CO is a strong ligand, hence when it approaches towards
central metal atom pairing of unpaired electron of central atom
takes place.
Thus, in Ni(CO)4
sp3 hybridisation
dsp2
Diamagnetic
3d
B is paramagnetic, tetrahedral complex because of weak ligand
field of Cl − .
NiCl42 –(Ni2+) :
sp
3
Paramagnetic
60. Described in 2, A has dsp2 hybridisation while B has
sp3-hybridisation of Ni.
4s
4p
CO
CO CO CO
Ni
Structure Tetrahedral
So, for Q-4 and 5 are correct match.
For R i.e., sp3d 2
It is seen in [FeF6]4−
Fe [Ar]3d6 4 s2
Fe 2+ [Ar]3d6
Coordination Compounds
3d
4s
Coordination compounds of [MA4 B2 ] type show geometrical
isomerism. Molecular orbital electronic configuration (MOEC)
for various coordination compound can be drawn using VBT as
4d
4p
Fe2+
As F − is a weak field ligand hence, when it approaches towards
central metal atom, pairing of its electrons does not take place.
Thus, in[FeF6 ]4 −
A. MO EC for [Cr(NH3 )4 Cl 2 ]Cl is
3d
4s
4p
F–
F– F– F–
NH3NH3 NH3 NH3Cl Cl
Number of unpaired electrons (n) = 3
Magnetic properties = paramagnetic
Geometrical isomers of [Cr(NH3 )4 Cl 2 ]+ are
4d
Fe2+
F– F–
Structure : Octahedral
So, 1 is the correct match for R.
For S i.e., d 2 sp 3
Cl
H 3N
It is seen in [Ti(H2O)3 Cl3 ] and[Cr(NH3 )6 ]3+
2
Ti [Ar]3d 4 s
4s
B.
4p
4s
NH3
Cr
H2O H2O H2O Cl– Cl– Cl–
Structure Octahedral
Likewise in [Cr(NH3)6]3+
Cr [Ar]3d 5 4 s1
NH3
H3N
Cl
4s
C. MOEC of [Pt(en)(NH3 )Cl]NO3 is
××
××
en
en
×× ××
××
NH3 Cl
n=0
Magnetic property = diamagnetic
Ionisation isomers are [Pt(en)2 (NH3 )Cl]NO3
and [Pt(en)2 NH3 (NO3 )]Cl
×× ××
4p
Cr3+
Here, NH 3 is also a weak field ligand so due to its approach no
pairing takes place in Cr.
Thus, In[Cr(NH3 )6 ]3+
NO3
4p
NH3 NH3 NH3 NH3
So for, S-2 and 3 are the correct match.
PLAN This problem is based on concept of VBT and magnetic
properties of coordination compound.
Draw VBT for each coordination compound.
If unpaired electron is present then coordination compound will
be paramagnetic otherwise diamagnetic.
×× ×× ××
n=0
Magnetic property = Diamagnetic
Geometrical isomers are
H3N
Cr3+–
××
NH3NH3 NH3 NH3NO3NO3
d2sp3 hybridisation
4s
Trans
D. MOEC of [Co(NH3 )4 (NO3 )2 ]NO3
Cr 3+ [Ar]3d 3 4 s0
3d
+
n=1
Magnetic properties = paramagnetic
Ionisation isomers of [Ti(H2O)5 Cl](NO3 )2 are
4p
Ti3+
62.
and
NH3
[Ti(H2O)5 Cl](NO3 )2 and [Ti(H2O)5 (NO3 )]Cl(NO3 )
d2sp3 hybridisation
NH3 NH3
Cr
NH3
Here, both H2O and Cl are weak ligands
So, in[Ti(H2O)3 Cl3 ]
3d
H3 N
Cl
Cis
Ti3+
3d
Cl
+
H 3N
2
Ti3+ [Ar]3d 1
3d
4p
4s
×× ×× ×× ××
×× ××
sp3d2 hybridisation
3d
279
NH3
NH3
CO
NH3
H3N
NO3
Trans
NO3
and
CO
NH3
NH3
NO3
NH3
Cis
Thus, magnetic property and isomerism in given coordination
compound can be summarised as
(P) [Cr(NH3 )4 Cl 2 ]Cl → Paramagnetic and exhibits cis-trans
isomerism (3)
(Q) [Ti(H2O)5 Cl](NO3 )2 → Paramagnetic and exhibits
ionisation isomerism (1)
Coordination Compounds
(R) [Pt(en)(NH3 )Cl]NO3 → Diamagnetic and
ionisation isomerism (4)
(S) [Co(NH3 )4 (NO3 )2 ]NO3 → Diamagnetic and
cis-trans isomerism (2)
∴ P → 3, Q → 1, R → 4, S → 2
Hence, (b) is the correct choice.
exhibits
3d
d2sp3
exhibits
Spin only magnetic moment (µ s ) = 1 (1 + 2) BM
= 3 BM
Hence, difference in spin only magnetic moment
63. (A) [Co(NH3 )4 (H2O)2 ]Cl 2:Co2+ , 3d 7
show geometrical isomerism, paramagnetic.
(B) Pt(NH3 )2 Cl 2:Pt 2+ has d 8 -configuration with all paired
electrons. Show geometrical isomerism, diamagnetic.
(C) [Co(H2O)5 Cl]Cl : Co2+ , 3d 7
Cannot show geometrical isomerism, paramagnetic.
(D) [Ni(H2O)6 ]Cl 2: Ni 2+ , 3d 8 , weak ligand, has two unpaired
electrons. Paramagnetic but cannot show geometrical
isomerism.
64. Complex part is cationic, named first : [Co(NH3 )6 ]Cl 3 :
hexaammine cobalt (III) chloride.
−
65. False : Cyanide (CN ) is a strong ligand, brings about pairing of
3d electrons.
= 35 − 3
≈ 4 BM
(II)
Et3P
C
O
CH2COO –
The structure of a chelate of a divalent Co2+ with EDTA is
shown as
H2C H2C CH2
CH2
CO
H 2C
O
Potassium ferrocyanide
N
—
No unpaired electron, diamagnetic
Has one unpaired electron, paramagnetic
66. False : Lobes of 3dx2 − y2 orbitals lies in X Y plane on the X and Y
coordinate axes, therefore electron density of dx2 − y2 orbital in
X Y plane is non-zero.
2
3+
H3C
C
C
5
4s
3d 5(n-5)
4d
sp3d2
Spin only magnetic moment (µ s ) = 5 (5 + 2) BM = 35 BM
In case of CN− ligand, carbon is the donor atom , it produces
strong ligand field and forms low spin complex as
5
[Fe(CN)6 ] : Fe (3d )
3
O
CO
O
N
N
H
N
N
O
4p
4
Co
CO
C
CH 3
Ni
H3C
[ Fe(SCN)6 ] : Fe (3d ) =
CH2
Each N has four N  Co  O bonds thus total eight N  Co  O
bonds.
70.
H
H
O
O
67. When S is donor atom of SCN−, it produces weak ligand field
and forms high spin complex as
N
1
CO
O
3+
2
N—CH2CH2—N
OOCCH2
3−
acetylbromidodicarbonylbis
(triethylphosphine)iron (II)
PLAN EDTA is a multidentate ligand as it can donate six pairs of
electrons – two pair from the two nitrogen atoms and four pair
from the four terminal oxygens of the COO − groups.
–
CH COO –
OOCCH2
Potassium ferrocyanide
O
C  CH 3
Fe
Et3P
69.
O
Br
C
68.
–
3−
4p
4s
==
280
H
C
CH 3
O
Ni-DMG complex
Oxidation state of Ni is +2 and hybridisation is dsp2.
µ = 0 (no unpaired electron) hence, diamagnetic.
71. The spin-only magnetic moment (µ) of the complex is 1.73 BM.
It indicates that nucleus of complex, chromium ion has one
unpaired electron. So, the ligand NO is unit positively charged.
Coordination Compounds
281
In both A and B, hybridisation of chromium is d 2sp3 and
magnetic moment : µ = n (n + 2) BM = 0
K2[Cr(NO) (CN)4 (NH3 )]
potassium amminetetracyanonitrosoniumchromate (I)
Cr+1 :
(3d6, strong ligand, no unpaired electron)
3d 5
4s0
Cr(I) “under influence of strong ligand field”.
NH3
74.
3+
H3N
NC
NH3
d 2sp3
H3N
CN
NC
d 2sp3-octahedral
2–
CN
dsp2-square planar
CO
Cr
NC
NC
NH3
NH3
NO
2–
Ni
Co
Octahedral
CN
Ni
NH3
OC
CN
CO
CO
sp3-tetrahedral
Octahedral geometry
75. A has no water molecules of crystallisation.
72. [NiCl 4 ] 2 − : Ni 2+ (3d 8 ), weak ligand field.
Hence, A is [Cr(H2O)6 ]Cl 3.
Both B and C loses weight with concentrated H2SO4, therefore,
both B and C have some water molecules of crystallisation.
Moreover, weight loss with C is just double of the same with B
indicates that number of water molecules of crystallisation of C
is double of the same for B. Therefore, B has one and C has two
water molecules of crystallisation.
B = [Cr(H2O)5 Cl]Cl 2 ⋅ H2O, C = [Cr(H2O)4 Cl 2 ]Cl ⋅ 2H2O
Ni2+(3d 8) :
sp3
µ = n (n + 2) ΒΜ = 8 ΒM
[Ni(CN)4 ]2 − : Ni 2+ (3d 8 ) , strong ligand field.
Ni2+(3d 8) :
dsp2
Square planar
µ = 0 (no unpaired electron)
73. In complexes A and B, one halide (Cl − or Br − ) is outside
coordination sphere, i.e. complexes are :
[Cr(NH3 )4 Br2 ]Cl and [Cr(NH3 )4 BrCl]Br
A gives white precipitate AgCl with excess of AgNO3 which
dissolve in excess ammonia. Therefore, A must be
[Cr(NH3 )4 Br2 ]Cl.
B gives a pale yellow precipitate with excess of AgNO3, which
dissolve in concentrated ammonia solution. Therefore,
precipitate is AgBr and complex B is [Cr(NH3 )4 ClBr]Br.
76.
(i) [Ti(NO3 )4 ]: Ti 4+ (3d 0 ) No d-electron, no d-d transition
possible, colourless.
(ii) [Cu(NCCH3 )]BF4 : Cu + (3d 10 ) All d-orbitals are completely
filled, no d-d transition possible, colourless.
(iii) [Cr(NH3 )6 ]Cl 3 : Cr 3+ (3d 3 ) Complex
has
allowed
d-d-transitions from t2g to eg level, hence coloured.
(iv) K3[VF6 ]: V3+ (3d 2 ) Complex has allowed d-d-transitions
from t2g to eg level, hence coloured.
77. I− is a strong reducing agent, reduces Cu 2+ to Cu + and
precipitate out as stable CuI.
19
Extraction of Metals
Objective Questions I (Only one correct option)
7. Match the refining methods Column I with metals Column II.
1. The processes of calcination and roasting in metallurgical
industries, respectively, can lead to
Column I
(Refining Methods)
(2020 Main, 4 Sep II)
(a) global warming and acid rain
(b) global warming and photochemical smog
(c) photochemical smog and ozone layer depletion
(d) photochemical smog and global warming
2. An Ellingham diagram provides information about
(2020 Main, 5 Sep I)
(a) the kinetics of the reduction process.
(b) the pressure dependence of the standard electrode potentials
of reduction reactions involved in the extraction of metals.
(c) the temperature dependence of the standard Gibbs energies
of formation of some metal oxides.
(d) the conditions of pH and potential under which a species is
thermodynamically stable.
3. Calamine, malachite, magnetite and cryolite, respectively,
are
(a) ZnCO3 , CuCO3 , Fe2 O3 , Na 3 AlF6
(b) ZnSO4 , CuCO3 , Fe2 O3 , AlF3
(c) ZnSO4 , Cu(OH)2 , Fe3 O4 , Na 3 AlF6
(2019 Adv.)
(d) ZnCO3, CuCO3 ⋅ Cu(OH)2, Fe3O4, Na 3AlF6
4. The correct statement is
(a) leaching of bauxite using concentrated NaOH solution
gives sodium aluminate and sodium silicate.
(2019 Main, 12 April II)
(b) the hall-heroult process is used for the production of
aluminium and iron.
(c) pig iron is obtained from cast iron.
(d) the blistered appearance of copper during the metallurgical
process is due to the evolution of CO2.
5. The idea of froth floatation method came from a person X
and this method is related to the process Y of ores. X and Y ,
respectively, are
(2019 Main, 12 April I)
(a)
(b)
(c)
(d)
fisher woman and concentration
washer woman and concentration
fisher man and reduction
washer man and reduction
6. The correct statement is
(a)
(b)
(c)
(d)
(a)
(b)
(c)
(d)
Column II
(Metals)
I.
Liquation
(A)
Zr
II.
Zone refining
(B)
Ni
III.
Mond process
(C)
Sn
IV.
van Arkel method
(D)
Ga
I- (C) ; II-(D); III-(B) ; IV-(A)
I- (B) ; II-(C); III-(D) ; IV-(A)
I- (C) ; II-(A); III-(B) ; IV-(D)
I- (B) ; II-(D); III-(A) ; IV-(C)
8. The one that is not a carbonate ore is
(a) siderite
(b) calamine
(c) malachite
(2019 Main, 10 April I)
(2019 Main, 9 April II)
(d) bauxite
9. Assertion For the extraction of iron, haematite ore is used.
Reason Heamatite is a carbonate ore of iron.
(2019 Main, 9 April II)
(a) Only the reason is correct.
(b) Both the assertion and reason are correct explanation for the
assertion.
(c) Both the assertion and reason are correct and the reason is the
correct explanation for the assertion.
(d) Only the assertion is correct.
10. The ore that contains the metal in the form of fluoride is
(2019 Main, 9 April I)
(a) magnetite (b) sphalerite
(c) malachite
11. The Mond process is used for the
(d) cryolite
(2019 Main, 8 April II)
(a) purification of Ni
(b) extraction of Mo
(c) purification of Zr and Ti
(d) extraction of Zn
12. With respect to an ore, Ellingham diagram helps to predict the
feasibility of its
(a) electrolysis
(c) vapour phase refining
(2019 Main, 8 April I)
(b) zone refining
(d) thermal reduction
13. The pair that does not require calcination is
(2019 Main, 12 Jan II)
(2019 Main, 10 April II)
zone refining process is used for the refining of titanium.
zincite is a carbonate ore.
sodium cyanide cannot be used in the metallurgy of silver.
aniline is a froth stabiliser.
(a) ZnO and MgO
(c) ZnCO3 and CaO
(b) ZnO and Fe2O3 ⋅ xH2O
(d) Fe2O3 and CaCO3 ⋅ MgCO3
14. In the Hall-Heroult process, aluminium is formed at the cathode.
The cathode is made out of
(a) platinum
(c) pure aluminium
(2019 Main, 12 Jan I)
(b) carbon
(d) copper
Extraction of Metals 283
22. From the following statements regarding H2 O2 choose the
15. The reaction that does not define calcination is
∆
(a) Fe2O3 ⋅ XH2O → Fe2O3 + XH2O
(2019 Main, 11 Jan II)
∆
(b) ZnCO3 →
ZnO + CO2
∆
(c) CaCO3 ⋅ MgCO3 →
CaO + MgO + 2CO2
∆
(d) 2Cu2S + 3O2 →
2Cu2O + 2SO2
Column A
A.
B.
C.
D.
Column B
Metals
Zinc
Copper
Iron
Aluminium
P.
Q.
R.
S.
aqueous solution of its salts is
(b) A - R; B- S; C - P; D- Q
(d) A - R; B- S; C - Q; D- P
17. The electrolytes usually used in the electroplating of gold and
silver, respectively, are
(2019 Main, 10 Jan II)
(a) [Au(OH) 4 ] − and [Ag(OH) 2 ] − (b) [Au(NH3 ) 2 ] + and [Ag(CN) 2 ] −
−
(c) [Au(CN) 2 ] and [Ag(CN) 2 ]
−
−
(d) [Au(CN) 2 ] and [AgCl 2 ]
18. Hall-Heroult’s process is given by
−
(2019 Main, 10 Jan I)
Coke, 1673 K
(a) ZnO + C → Zn + CO
(b) Cr2O3 + 2Al → Al 2O3 + 2Cr
(c) 2Al 2O3 + 3C → 4Al + 3CO2
is
(2019 Main, 9 Jan II)
+O 2
4Cu
→
O
2Cu 2
∆Gº (kJ/mol)
2C
(a)
(b)
(c)
(d)
+O
→2
2
26. In the cyanide extraction process of silver from argentite ore,
the oxidising and reducing agents used are
(a)
(b)
(c)
(d)
(2012)
O2 and CO respectively
O2 and Zn dust respectively
HNO3 and Zn dust respectively
HNO3 and CO respectively
(a)
(b)
(c)
(d)
(2011)
II, III in haematite and III in magnetite
II, III in haematite and II in magnetite
II in haematite and II, III in magnetite
III in haematite and II, III in magnetite
(a) electrolytic reduction
(b) roasting followed by reduction with carbon
(c) roasting followed by reduction with another metal
(d) roasting followed by self-reduction
O3
3 Al 2
→ 2/
500ºC 800ºC
Temperature (ºC)
2000ºC
(2019 Main, 9 Jan I)
(b) azurite
(d) copper pyrites
21. Which one of the following ores is best concentrated by froth
floatation method?
(2016 Main)
(b) Galena
(d) Magnetite
(b) oxygen
(d) argon
(2008, 3M)
29. Extraction of zinc from zinc blende is achieved by
CO
20. The ore that contains both iron and copper is
(a) Siderite
(c) Malachite
(2013 Adv.)
(b) Ag, Cu and Sn
(d) Al, Cu and Pb
dilute aqueous solution of NaCN in the presence of
At 800°C, Cu can be used for the extraction of Zn from ZnO
At 1400°C, Al can be used for the extraction of Zn from ZnO
At 500°C, coke can be used for the extraction of Zn from ZnO
Coke cannot be used for the extraction of Cu from Cu 2O
(a) malachite
(c) dolomite
(a) Ag, Cu and Pb
(c) Ag, Mg and Pb
(a) nitrogen
(c) carbon dioxide
l+O 2
4/3 A
–1050
25. Sulphide ores are common for the metals
28. Native silver metal forms a water soluble complex with a
–300
O2
2Zn+
(2014 Main)
(b) Ca
(d) Cr
magnetite, respectively, are
19. The correct statement regarding the given Ellingham diagram
–600
(a) Ag
(c) Cu
27. Oxidation states of the metal in the minerals haematite and
(d) Cu2+ (aq) + H2 (g ) → Cu(s) + 2H+ (aq)
nO
→ 2Z
Al, which of the following statements is false?
(2015 Main)
(a) CO and CO2 are produced in this process
(b) Al 2O3 is mixed with CaF2 which lowers the melting point of the
mixture and brings conductivity
(c) Al 3+ is reduced at the cathode to form Al
(d) Na 3AlF6 serves as the electrolyte
24. The metal that cannot be obtained by electrolysis of an
(2019 Main, 11 Jan I)
(a) A - P; B- Q; C - R; D- S
(c) A - Q; B- R; C - S; D- P
(2015 Main)
23. In the context of the Hall-Heroult process for the extraction of
16. Match the ores ( Column A ) with the metals (Column B).
Ores
Siderite
Kaolinite
Malachite
Calamine
incorrect statement.
(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
30. Which ore contains both iron and copper?
(a) Cuprite
(c) Chalcopyrite
(2007, 3M)
(2005, 1M)
(b) Chalcocite
(d) Malachite
31. The methods chiefly used for the extraction of lead and tin
from their ores are respectively
(2004, 1M)
(a) self-reduction and carbon reduction
(b) self-reduction and electrolytic reduction
(c) carbon reduction and self-reduction
(d) cyanide process and carbon reduction
32. In the process of extraction of gold,
O2
Roasted gold ore + CN– + H2O → [ X ] + HO
[ X ] + Zn → [Y ] + Au
284 Extraction of Metals
Identify the complexes [X] and [Y].
(2003, 1M)
(a) X = [Au(CN)2 ] − , Y = [Zn(CN)4 ] 2−
(c) X = [Au(CN)2 ] − , Y = [Zn(CN)6 ] 4−
(d) X = [Au(CN)4 ] − , Y = [ Zn(CN)4 ] 2−
43. In metallurgy of iron, when limestone is added to the blast
33. Anhydrous ferric chloride is prepared by
(2002)
(a) heating hydrated ferric chloride at a high temperature
in a stream of air
(b) heating metallic iron in a stream of dry chlorine gas
(c) reaction of ferric oxide with hydrochloric acid
(d) reaction of metallic iron with hydrochloric acid
34. Which of the following process is used in extractive
metallurgy of magnesium?
(2002, 3M)
Fused salt electrolysis
Self-reduction
Aqueous solution electrolysis
Thermite reduction
smelting process in the extraction of copper is
(2001, 1M)
36. Electrolytic reduction of alumina to aluminium by
37. The chemical process in the production of steel from
(2000, 1M)
(a) reduction
(b) oxidation
(c) reduction followed by oxidation
(d) oxidation followed by reduction
38. In the commercial electrochemical process for aluminium
extraction, the electrolyte used is
(1999, 2M)
(a) Al(OH)3 in NaOH solution
(b) an aqueous solution of Al 2 (SO4 )3
(c) a molten mixture of Al 2O3 and Na 3AlF6
(d) a molten mixture of AlO(OH) and Al(OH)3
(a) as a catalyst
(b) to make the fused mixture very conducting
(c) to increase the temperature of the melt
(d) to decrease the rate of oxidation of carbon at the anode
(1985, 1M)
(b) heated ferric oxide
(d) heated aluminium oxide
41. In the alumino-thermite process, aluminium acts as
(a) an oxidising agent
(c) a reducing agent
(b) a flux
(d) a solder
(b) H3PO4
(d) HNO3
(1982)
Objective Questions II
(One or more than one correct option)
(2020 Adv.)
(a) Hydrated Al 2O3 precipitates, when CO2 is bubbled through a
solution of sodium aluminate.
(b) Addition of Na 3AlF6 lowers the melting point of alumina.
(c) CO2 is evolved at the anode during electrolysis.
(d) The cathode is a steel vessel with a lining of carbon.
46. Extraction of copper from copper pyrite (CuFeS 2 ) involves
(2016 Adv.)
(a) crushing followed by concentration of the ore by froth-floatation
(b) removal of iron as slag
(c) self reduction step to produce ‘blister copper’ following
evolution of SO2
(d) refining of ‘blister copper’ by carbon reduction
47. Copper is purified by electrolytic refining of blister copper.
The correct statement(s) about this process is/are
(2015 Adv.)
(a) impure Cu strip is used as cathode
(b) acidified aqueous CuSO4 is used as electrolyte
(c) pure Cu deposits at cathode
(d) impurities settle as anode-mud
48. Upon heating with Cu 2 S, the reagent(s) that give copper
metal is/are
(a) CuFeS2
(2014 Adv.)
(b) CuO
(c) Cu2O
extraction of
amount in the electrolytic reduction of alumina dissolved in
fused cryolite (Na 3 AlF6 ) is
(1993, 1M)
(a) heated cupric oxide
(c) heated stannic oxide
(a) H2SO4
(c) HCl
(d) CuSO4
49. The carbon-based reduction method is not used for the
39. The major role of fluorspar (CaF2 ) which is added in small
40. Hydrogen gas will not reduce
44. Iron is rendered passive by treatment with concentrated
(2000, 1M)
(a) in the presence of NaCl
(b) in the presence of fluorite
(c) in the presence of cryolite which forms a melt with lower melting
temperature
(d) in the presence of cryolite which forms a melt with higher
melting temperature
haematite ore involve
(1982)
(b) gangue
(d) calcium carbonate
extraction of aluminium from bauxite?
(b) FeSiO3
(d) Cu2S + FeO
Hall-Heroult process is carried out
furnace, the calcium ion ends up in
(a) slag
(c) metallic calcium
45. Which among the following statement(s) is(are) true for the
35. The chemical composition of ‘slag’ formed during the
(a) Cu2O + FeS
(c) CuFeS2
(1983)
(a) electrovalent and covalent
(b) electrovalent and coordinate covalent
(c) electrovalent, covalent and coordinate covalent
(d) covalent and coordinate covalent
(b) X = [Au(CN)4 ] 3 – ,Y = [Zn (CN)4 ] 2–
(a)
(b)
(c)
(d)
42. Type of bonds present in CuSO4 ⋅ 5H2O are only
(1983, 1M)
(a)
(b)
(c)
(d)
(2013 Adv.)
tin from SnO2
iron from Fe2O3
aluminium from Al 2O3
magnesium from MgCO3 , CaCO3
50. Extraction of metal from the ore cassiterite involves
(a)
(b)
(c)
(d)
(2011)
carbon reduction of an oxide ore
self-reduction of a sulphide ore
removal of copper impurity
removal of iron impurity
51. Addition of high proportions of manganese makes steel
useful in making rails
(a) gives hardness to steel
(b) helps the formation of oxides of iron
(c) can remove oxygen and sulphur
(d) can show highest oxidation state of + 7
(1998)
Extraction of Metals 285
52. Of the following, the metals that cannot be obtained by
electrolysis of the aqueous solution of their salts are
(a) Ag
(d) Al
(b) Mg
(e) Cr
(c) Cu
53. In the electrolysis of alumina, cryolite is added to
S ° [ C( s )] = 6.0 J K −1 mol −1
(1990, 1M)
S ° [ CO2 ( g )] = 210.0 JK −1 mol −1
(1986, 1M)
Assume that, the enthalpies and the entropies are temperature
independent.
(2020 Adv.)
(a) lower the melting point of alumina
(b) increase the electrical conductivity
(c) minimise the anode effect
(d) remove impurities from alumina
Match the Columns
59. Match the anionic species given in Column I that are present in
the ore (s) given in Column II.
Assertion and Reason
Column I
Read the following questions and answer as per the direction
given below :
(a) Statement I is correct; Statement II is correct; Statement II
is the correct explanation of Statement I.
(b) Statement I is correct; Statement II is 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 true.
54. Statement I Al(OH)3 is amphoteric in nature.
Statement II Al—O and O—H bonds can be broken with
(1998)
equal ease in Al(OH)3 .
Passage Based Questions
Passage
Copper is the most noble of the first row transition metals
and occurs in small deposits in several countries. Ores of copper
include chalcanthite (CuSO4 ⋅ 5H2 O), atacamite (Cu 2 Cl(OH)3 ),
cuprite (Cu 2 O), copper glance (Cu 2 S) and malachite
(Cu 2 (OH)2 CO3 ). However, 80% of the world copper production
comes from the ore chalcopyrite (CuFeS2 ). The extraction of
copper from chalcopyrite involves partial roasting, removal of iron
and self-reduction.
(2010)
55. Partial roasting of chalcopyrite produces
(a) Cu 2S and FeO
(c) CuS and Fe2O3
(b) Cu 2O and FeO
(d) Cu 2O and Fe2O3
56. Iron is removed from chalcopyrite as
(a) FeO
(b) FeS
(c) Fe 2O3
(d) FeSiO3
(a) S
(b) O
(c) S
2−
(d) SO2
the data given below to determine the minimum temperature
(in K) at which the reduction of cassiterite by coke would
take place.
−1
At 298 K : ∆ f H ° [ SnO2 ( s )] = − 5810
. kJ mol ,
S ° [ Sn ( s )] = 52.0 JK −1 mol −1
Carbonate
p.
Siderite
B.
Sulphide
q.
Malachite
C.
Hydroxide
r.
Bauxite
D.
Oxide
s.
Calamine
t.
Argentite
corresponding product(s) given in Column II.
Column I
(2009)
Column II
A. Cu + dil. HNO3
p.
NO
B. Cu + conc. HNO3
q.
NO2
C. Zn + dil. HNO3
r.
N2 O
D. Zn + conc. HNO3
s.
Cu(NO3 )2
Zn(NO3 )2
t.
61. Match the conversions in Column I with the type(s) of
reaction(s) given in Column II.
(2008, 6M)
Column I
Column II
A.
PbS → PbO
p.
Roasting
B.
CaCO3 → CaO
q.
Calcination
C.
ZnS → Zn
r.
Carbon reduction
D.
Cu 2 S → Cu
s.
Self-reduction
62. Match the extraction processes listed in Column I with metals
listed in Column II.
(2006, 6M)
Column II
A.
Self-reduction
p.
Lead
B.
Carbon reduction
q.
Silver
C.
Complex formation and
displacement by metal
r.
Copper
D.
Decomposition of iodide
s.
Boron
58. Tin is obtained from cassiterite by reduction with coke. Use
S ° [ SnO2 ( s )] = 56.0 J K −1 mol −1
A.
Column I
Numerical Answer Type Questions
∆ f H ° [(CO 2 )( g )] = − 394.0 kJ mol −1
Column II
60. Match each of the reactions given in Column I with the
57. In self-reduction, the reducing species is
2−
(2015 Adv.)
63. Each entry in Column X is in some way related to the entries in
Columns Y and Z. Match the appropriate entries.
Column X
Column Y
(1988, 3M)
Column Z
A. Invar
p. Co, Ni
m. Cutlery
B. Nichrome
q. Fe, Ni
n. Heating element
o. Watch spring
C. Stainless steel r. Fe, Cr, Ni
286 Extraction of Metals
64. Match the following choosing one item from Column X and
the appropriate item from Column Y.
Column X
Al
p.
Calamine
B.
Cu
q.
Cryolite
C.
Mg
r.
Malachite
D.
Zn
s.
Carnalite
metal centre in CuSO4 ⋅ 5H2O.
white photographic film. Also give reason, why the solution
of sodium thiosulphate on acidification turns milky white and
give balance equation of this reaction.
(2005, 2M)
81. A1 and A2 are two ores of metal M . A1 on calcination gives
black precipitate, CO2 and water.
65. Match the following metals listed in Column I with extraction
processes listed in Column II.
Column I
(1979, 2M)
A1
Column II
A.
Silver
p.
Fused salt electrolysis
B.
Calcium
q.
Carbon reduction
C.
Zinc
r.
Carbon monoxide reduction
D.
Iron
s.
Amalgamation
E.
Copper
t.
Self-reduction
n
atio
lcin
a
C
Dil. H
Cl
KI
Black solid + CO2 + H2O
; A2
Roasting
I2 + ppt.
Metal + gas
K 2Cr2O7
+ H2SO4
Green colour
Identify A1 and A2 .
(2004, 4M)
82. Which of the two, anhydrous or hydrated AlCl3 is more
soluble in diethyl ether? Justify using the concepts of bonding
in not more than 2 or 3 sentences.
(2003)
Fill in the Blanks
66. Silver jewellery items tarnish slowly in the air due to their
reaction with…………
(2009, 2M)
80. Write balanced chemical equation for developing a black and
Column Y
A.
79. Give the number of water molecule (s) directly bonded to the
(1983, 2M)
83. Write the balanced chemical reactions involved in the
extraction of lead from galena. Mention oxidation state of
lead in litharge.
(2003, 2M)
(1997)
67. In the extractive metallurgy of zinc, partial fusion of ZnO
with coke is called …… and reduction of the ore to the molten
metal is called …… (smelting, calcining, roasting, sintering).
84. Write the balanced chemical equation for developing
photographic films.
(2000)
(1988, 1M)
85. Write the chemical reactions involved in the extraction of
68. Silver chloride is sparingly soluble in water because its lattice
silver from argentite.
(2000, 2M)
Work out the following using chemical equations.
In moist air, copper corrodes to produce a green layer on the
surface.
(1998)
When the ore haematite is burnt in air with coke around
2000°C along with lime, the process not only produces steel
but also produces a silicate slag, that is useful in making
building materials such as cement. Discuss the same and
show through balanced chemical equations.
(1998, 4M)
Give balance equation for the reaction of aluminium with
aqueous sodium hydroxide.
(1997)
Write a balanced equation for the reaction of argentite with
KCN and name the products in the solution.
(1996)
Give reasons for the following
“Although aluminium is above hydrogen in the
electrochemical series, it is stable in air and water.”
energy greater than ……….. energy.
69. Galvanisation of iron denote coating with ……
70. Cassiterite is an ore of ……
(1987)
86.
(1983)
(1980, 1M)
71. In the thermite process …… is used as a reducing agent.
87.
(1980, 1M)
72. In the basic Bessemer process for the manufacture of steel,
the lining of the converter is made up of … . The slag formed
consists of ……
(1980, 1M)
88.
73. AgCl dissolve in excess of KCN solution to give …………
complex compound.
(1980)
True/False
74. Cu + disproportionate to Cu 2+ and elemental copper in
solution.
89.
90.
(1991)
75. Silver chloride is more soluble in very concentrated sodium
chloride solution than in pure water.
(1984, 1M)
76. Dilute HCl oxidises metallic Fe to Fe2+ .
(1983, 1M)
77. Silver fluoride is fairly soluble in water.
(1982)
Subjective Questions
78. Give the coordination number of Al in the crystalline state of
AlCl3.
(2009, 2M)
(1994, 1M)
91. Complete the following reaction :
Sn + 2KOH + 4H2 O → ...... + .......
(1994)
92. Give briefly the isolation of magnesium from sea water by the
Dow’s process.
Give equations for the steps involved.
(1993, 3M)
93. Complete and balance the following reaction :
Copper reacts with HNO3 to give NO and NO2 in the molar
ratio of 2:1
(1992)
Cu + HNO3 → KK + NO + NO2 + KK
Extraction of Metals 287
94. Write balanced equation for the extraction of “Copper from
98. Each of the following statement is true, only under some
copper pyrites by self reduction.”
(1990, 2M)
95. Give balanced equations for the extraction of “Silver from
silver glance by cyanide process.”
(1988, 1M)
96. Answer the following questions briefly
specific conditions. Write the condition for each
subquestion in not more than 2 sentences.
(i) Metals can be recovered from their ores by chemical methods
(i) What is the actual reducing agent of haematite in blast furnace?
(ii) Give the equation for the recovery of lead from galena by air
reduction.
(iii) Why is sodium chloride added during electrolysis of fused
anhydrous magnesium chloride?
(iv) Zinc, not copper is used for the recovery of metallic silver from
complex [Ag(CN)2 ]– , explain.
(v) Why is chalcocite roasted and not calcinated during recovery of
copper?
(1987, 5M)
97. Write balanced chemical equation for the following
“Gold is dissolved in aqua regia.”
(ii) High purity metals can be obtained by zone refining method.
(1984, 2M)
99. Give reason for the following in one or two sentences :
“Silver bromide is used in photography.”
(1983)
100. State the conditions under which the preparation of
alumina from aluminium is carried out. Give the necessary
equations which need not be balanced.
(1983, 2M)
101. Write the chemical equations involved in the extraction of
lead from galena by self reduction process.
(1979, 2M)
102. Write balanced equation involved in the preparation of tin
metal from cassiterite.
(1979)
(1987)
Answers
1.
5.
9.
13.
17.
21.
25.
29.
33.
37.
41.
45.
49.
53.
(a)
(b)
(d)
(a)
(c)
(b)
(a)
(b)
(b)
(a)
(c)
(a,b,c,d)
(c,d)
(a,b)
2.
6.
10.
14.
18.
22.
26.
30.
34.
38.
42.
46.
50.
54.
(c)
(d)
(d)
(b)
(c)
(a)
(b)
(c)
(a)
(c)
(c)
(a,b,c)
(a,d)
(b)
3.
7.
11.
15.
19.
23.
27.
31.
35.
39.
43.
47.
51.
55.
(d)
(a)
(a)
(d)
(b)
(d)
(d)
(a)
(b)
(b)
(a)
(b,c,d)
(a,c)
(b)
4.
8.
12.
16.
20.
24.
28.
32.
36.
40.
44.
48.
52.
56.
(a)
(d)
(d)
(d)
(d)
(b)
(b)
(a)
(c)
(d)
(d)
(b,c,d)
(b,d)
(d)
57. (c)
58. (935)
59. A → p, q, s; B → t; C → q; D → r
60. A → p, s; B → q, s; C → r, t; D → q, t
51. A → p; B → q; C → r, s; D → p, s
62. A → p, r; B → p; C → q; D → s
63. A → q, o; B → p, n; C → r, m
64. A → q; B → r; C → s; D → p
65. A → s; B → p; C → q; D → q, r; E → t
66. (H 2S)
67. (Sintering, Smelting)
68. (Hydration) 69. (Zn)
72. (Lime, calcium phosphate)
70. (Sn)
71. (Al)
73. K [Ag(CN) 2 ]
74. (T)
78. (6)
76. (T)
83. (2)
75. (T)
79. (4)
77. (T)
Hints & Solutions
1. Due to industrial process, CO2 release from calcination and SO2
release from roasting respectively which is responsible for global
warming and acid rain.
Calcination It involves heating where the volatile matter escapes
leaving behind the metal oxide. In this process, the ore is heated below
its melting point in the absence of air or in the limited supply of air.
∆
e.g. CaCO3 (s) 
→ CaO(s) +
CO2 ↑
(Carbon dioxide)
Responsible for
global warming
Roasting In this process the ore is heated in a regular supply of air
in a furnace at a temperature below the melting point of the metal.
e.g. 2ZnS + 3O2 → 2ZnO +
2.
2SO2 ↑
(Sulphur dioxide)
Responsible for acid rain
Ellingham diagram provides information about temperature
dependence of the standard Gibbs energies of formation of some
metal oxides.
It predicts the feasibility of thermal reduction of the ore. The
criterion of feasibility is that at given temperature, Gibbs
energy of the reaction must be negative.
3. ZnCO3-Calamine (zinc ore)
CuCO3 ⋅ Cu(OH)2-Malachite (copper ore)
Fe3O4-Magnetite (iron ore)
Na 3AlF6-Cryolite (aluminium ore)
Thus, option (d) is correct.
4. The correct statement is ‘‘leaching of bauxite using
concentrated NaOH solution gives sodium aluminate and
sodium silicate’’. Bauxite usually contains SiO2, iron oxides
and titanium oxide (TiO2 ) as impurities. Concentration is
carried out by digesting the powdered ore with a concentrated
solution of NaOH at 473-523 K and 35-36 bar pressure. Al2O3
is leached out as sodium aluminate (and SiO2 too as sodium
silicate) leaving the impurities behind.
Al 2O3 (s) + 2NaOH(aq) + 3H2O(l ) → 2Na[Al(OH)4 ](aq)
288 Extraction of Metals
The aluminate in solution is neutralised by passing CO2 gas and
hydrated Al 2O3 is precipitated. Here, the solution is seeded with
freshly prepared samples of hydrated Al 2O3 which induces
precipitation.
2Na[Al(OH)4 ](aq) + CO2 (g ) →
Al 2O3 ⋅ xH2O(s) + 2NaHCO3 (aq)
The sodium silicate remains in the solution and hydrated alumina
is filtered, dried and heated to give back pure Al 2O3.
1470 K
Al 2O3 ⋅ xH2O(s)  → Al 2O3 (s) + xH2O(g )
5. The idea of froth floatation method came from a person ‘washer
woman’ ( X ) and this method is related to the process
concentration (Y ) of ores.
This method is based upon the preferential wetting properties
with the frothing agent (collector) and water.
6. The explanation of given statements are as follows:
(a) Zone refining process is used for the refining of B, Ga, In, Si
and Ge.
Ti is refined by van Arkel method.
Thus, statement (a) is incorrect.
(b) Zincite (ZnO) is an oxide ore of Zn.
Thus, statement (b) is incorrect.
(c) NaCN is used in the hydrometallurgy of silver. It is known as
Mc. Arthur Forrest process.
The reactions occuring during the process are as follows:
Ag2S + 4NaCN → 2Na[Ag(CN)2 ] + Na 2S
4Na 2S + 2H2O + 5O2 → 2Na 2SO4 + 4NaOH + 2S
2Na[Ag(CN)2 ] + Zn → Na 2[Zn(CN)4 ] + 2Ag
Thus, statement (c) is incorrect.
(d) Aniline and cresol help in stabilising the froth in froth
floatation process.
Thus, statement (d) is correct.
8. Bauxite is not a carbonate ore. Its chemical formula is Al 2O3 or
AlOx(OH)3 − 2x , where 0 < x < I. Chemical formula of other ores
given in options are as follows:
Siderite-FeCO3
Calamine-ZnCO3
Malachite-CuCO3 ⋅ Cu(OH)2
9. Only assertion is correct and reason is incorrect. Haematite is not
a carbonate ore. It is an oxide ore, i.e. Fe2O3. Cast iron is
extracted chiefly from its oxide ore (haematite) by heating in the
presence of coke and limestone in a blast furnace.
10. Cryolite ore (Na 3AlF6, sodium hexafluoroaluminate) contain
fluorine while other given options such as malachite
(Cu 2 (CO)3 (OH)2 ), sphalerite ((Zn,Fe)S) and bauxite (Al 2O3 )
does not contain fluorine.
11. Mond process is used in the purification of Ni. It is a vapour
phase refining process.
It is based on the principle that Ni is heated in the presence of
carbon monoxide to form nickel tetracarbonyl, which is a
volatile complex. This complex is then decomposed by
subjecting it to a higher temperature (450-470 K) to obtain pure
nickel metal.
Crude nickel (s) + 4 CO (g)
(Impure)
330-350 K
[Ni (CO)4] (g) 450-470 K
(Volatile compound)
Crude nickel (s) + 4CO(g)
330-350 K
(Impure)
450-470 K
Ni(CO)4(g)
(Volatile compound)
Recycled
4CO(g) +
feasibility of its thermal reduction. It is a graph representation of
Gibbs energy change versus absolute temperature.
0
(Crude
metal)
500-600 K
MI4(g)
1700-1800 K
(Volatile compound)
Recycled
2I2(g) +
–100
∆Gº/kJ mol–1 of O2
–300
–400
–500
–600
–700
+O 2
2CO
+O 2
2Zn
2CO 2
O
2Zn
–800
–900
–1000
2Cu 2O
4Cu+O 2
2FeO
O
2
+
2Fe
–200
4/3
O2
Al+
2 Mg
–1100
C+O2
2C+O
2
CO2
2CO
l +O 3
2/3 A 2
O
2Mg
+O 2
–1200
Ni(s)
(Pure)
IV. van Arkel method
M (s) + 2I2 (g)
Ni (s)
Pure + 4 CO (g)
12. With respect to an ore, Ellingham diagram helps to predict the
7. Refining of crude metals results pure metals and its impurities
get separated out.
I. Liquation In this method low melting metals like Sn, Pb, Bi
and Hg can be made to flow down through a sloping surface
leaving behind the higher melting impurities on the hearth.
II. Zone refining The basic principle of the method is,
impurities are more soluble in the molten metal than in the
solid state of the metal. This method is useful to produce
semiconductors and ultra-pure metals like B, Ga, In,Si and Ge.
III. Mond process
Recycled
M(s)
(Ultra-pure)
Here, M = Zr, Hf, Ti
Hence, the correct matching is
I→ (C), II→(D), III →(B), IV → (A).
0°C
273 K
400°C
673 K
800°C
1200°C 1600°C 2000°C
1073 K
1473 K
1873 K
2273 K
Temperature
Gibbs energy (∆Gº) versus T plots (schematic)
for the formation of some oxides (Ellingham diagram)
Generally, the diagram consists of plots of ∆G ° versus T for the
formation of oxides of elements
2xM (s) + O2 (g ) →
2M x O(s)
Thermal reduction product
In this reaction, amount of gas decreases thus, randomness
decreases. Hence, ∆S becomes negative. Therefore, the value of
Extraction of Metals 289
free energy increases with increase in temperature. There is a
point in a curve below which ∆G is negative. So, M xO is stable.
Above this point, M xO will decompose on its own.
18. Hall-Heroult’s process is an electro-reduction process by which
pure alumina (Al 2O3 ) is reduced to crude Al.
In this process, electrolysis of a fused mixture of Al 2O3,
Na 3[AlF6 ] (cryolite) and CaF2 (fluorspar) is carried out at carbon
cathode and graphite anode.
The overall reaction is represented as:
2Al 2O3 + 3C → 4Al + 3CO2
13. The hydroxide, hydrated oxides and carbonate ores, after
concentration, are subjected to calcination. In the process, the
ore is heated below its melting point in the limited supply or
absence of air. As the result, these are converted into their oxides.
So, among the given options, the options having either carbonates
(e.g. ZnCO3 and CaCO3 ⋅ MgCO3) or hydrated oxide (e.g.
Fe2O3 ⋅ xH2O), require calcination while pair of option (a), i.e.
ZnO and MgO does not require calcination.
19. From the Ellingham diagram, we can say that any oxide with
lower value of ∆G ° is more stable than a oxide with higher value
of ∆G °. We can also predict that the oxide placed higher in the
diagram can be reduced by the element involved in the formation
of its oxide placed lower at that temperature in the diagram.
It is happening in case of ZnO for its reduction by Al at 1400°C.
14. In the Hall - Heroult’s process, aluminium in formed at the
cathode. The cathode is made out of carbon. In this method,
Al 2O3 is melted with cryolite, Na 3[AlF6 ] and electrolysed in a
graphite lined steel tank, which serves as the cathode. The anode
is also made of graphite.
The cell runs continuously and at intervals molten aluminium is
drained from the bottom of the cell and more bauxite is added.
The electrolytic reactions are as follows:
At cathode
Al3+ + 3e− → Al
At anode
C(s) + O 2− (melt) → CO(g)+ 2e−
2−
20. The formulae of the given ores are as follows:
Malachite
Copper pyrites
Dolomite
Azurite
21. Sulphide ores are concentrated by froth floatation method e.g.
Galena (PbS)
−
C(s) + 2O (melt) → CO2(g)+ 4 e
15. Calcination is one of the pyrometallurgical process, like roasting
by which a concentrated ore gets converted into its oxide.
In calcination, a hydrated carbonate or bicarbonate ore or a
hydrated ore is heated at lower temperature (compared to roasting)
in absence of air to give its oxide as in options (a), (b) and (c).
Here, volatile non-metallic oxides like H2O, CO2, are also
produced.
Roasting is valid mainly for sulphide ores like option (d), where
SO2 gets liberated. In this reaction, calcination cannot be used.
22. H2O2 acts as an oxidising as well as reducing agent, because
oxidation number of oxygen in H2O2 is −1. So, it can be oxidised
to oxidation state 0 or reduced to oxidation state –2.
H2O2 decomposes on exposure to light. So, it has to be stored in
plastic or wax linked glass bottles in dark for the prevention of
exposure. It also has to be kept away from dust.
23. (a) In Hall-Heroult process for extraction Al, carbon anode is
oxidised to CO and CO2.
(b) When Al 2O3 is mixed with CaF2, it lowers the melting point
of the mixture and brings conductivity.
(c) Al 3+ is reduced at cathode to form Al.
(d) Here, Al 2O3 is an electrolyte, undergoing the redox process.
Na 3AlF6 although is an electrolyte but serves as a solvent, not
electrolyte.
16. The correct match is: A → R; B → S; C → Q; D → P.
(A) Siderite is an ore of iron with molecular formula FeCO 3 (R).
(B) Kaolinite is an ore of aluminium with molecular formula
Al 2Si 2(OH)2O 5 (S).
(C) Malachite is an ore of copper with molecular formula
CuCO 3 ⋅ Cu(OH)2 (Q).
(D) Calamine is an ore of zinc with molecular formulaZnCO 3 (P).
24. Higher the position of element in the electrochemical series more
difficult is the reduction of its cations.
If Ca 2+ (aq) is electrolysed, water is reduced in preference to it.
Hence, it cannot be reduced electrolytically from their aqueous
solution.
Ca 2+ ( aq) + H2O → Ca 2+ + OH− + H2 ↑
17. Electroplating is a process of coating one metal or metal object
with a very thin layer of another metal typically applying a direct
electric current.
Electrolytes used in the electroplating of gold and silver are
given in the table below:
Process
(a)
(b)
Gold
plating
Silver
plating
25.
Pure metal block
Article to acts an anode
Electrolyte
be plated
by which
out acts as electroplating (aqueous solution)
cathode
will be done
Article
Au(s)
I
Na[A u(CN)2 ]
(Sodium
auro-cyanide)
Article
Ag(s)
: CuCO3 ⋅ Cu(OH)2
: CuFeS2
: CaMg(CO3 )2
: Cu 3 (CO)3 (OH)2
I
Na[A g(CN)2 ]
(Sodium argento
cyanide)
Element
Ores
Ag
Ag 2 S
Name
Cu
CuFeS2
Pb
Sn
PbS
SnO2
Galena
Cassiterite
Mg
MgCO3 ⋅ CaCO3
Dolomite
Al
Al 2 O3 ⋅ xH2 O
Argentite
Copper pyrites
Bauxite
26. The reactions involved in extraction of silver by cyanide process
are
Ag2S + CN− + O2 → [Ag(CN)2 ]− + SO2
…(i)
290 Extraction of Metals
[Ag (CN)2 ]− + Zn → [Zn (CN)4 ]2− + Ag
…(ii)
In reaction (i), sulphide is oxidised to SO2 by oxygen. In the
reaction (ii), silver ion (Ag+ ) is reduced to Ag by Zn. Therefore,
O2 is oxidising agent and Zn is reducing agent.
27. Haematite is Fe2O3, in which oxidation number of iron is III.
Magnetite is Fe3O4 which is infact a mixed oxide (FeO ⋅ Fe2O3),
hence iron is present in both II and III oxidation state.
28. A water soluble complex with silver and dilute aqueous solution
of NaCN is Na[Ag(CN)2 ]. In the cyanide process, the native
silver is crushed and treated with aqueous NaCN solution and
aerated.
4Ag + 8NaCN + 2H2 O + O2 → 4Na [Ag(CN)2 ]
+ 4NaOH
29. Zinc blende contain ZnS which is first roasted partially and then
subjected to reduction with carbon
ZnS + O2 → ZnO + SO2 Roasting
∆
ZnO + C → Zn + CO ↑ Carbon reduction
30. Chalcopyrite contain both iron and copper.
31. Lead is mainly extracted by self-reduction process while tin is
extracted by carbon reduction method.
Au + 2CN− → [Au(CN)2 ]−
32.
X
2Au(CN)−2 + Zn → [Zn(CN)4 ]2 − + 2Au
Y
33. Heating iron in stream of dry chlorine gas gives FeCl 3 in
anhydrous form. In all other cases (a and c) hydrated FeCl 3 is
obtained while in (d), FeCl 2 is formed.
34. Mg is extracted by electrolysis of molten MgCl 2 .
35. Iron present in copper pyrite is removed by forming FeSiO3 as
slag.
(b) The electrolysis of alumina by Hall-Heroult’s process is
carried by using a fused mixture of alumina and cryolite
(Na2AlF6 ) along with minor quantities of aluminium fluoride
and fluorspar. The addition of cryolite and fluorspar
increases the electrical conductivity of alumina and lowers
the fusion temperature. Hence, the statement (b) is true.
(c) At anode, alumina reacts with fluorine to liberate oxygen and
evolved oxygen reacts with carbon to form carbon dioxide
and hence, statement (c) is true.
(d) Steel cathode with carbon lining and graphite anode are used
and hence,the statement (d) is true.
46. CuFeS2 (copper pyrite) is converted into copper into following
steps:
Step I Crushing (grinding ) followed by concentration by
froth-floatation process.
Step II Roasting of ore in the presence of SiO2 which removes iron as
slag (FeSiO3).
2CuFeS2 + O2 → Cu 2S + 2FeS + SO2
2FeS + 3O2 → 2SO2 + 2FeO
FeO + SiO2 → FeSiO3 (Slag)
Step III Self-reduction in Bessemer converter
2Cu 2S + 3O2 → 2Cu 2O + 2SO2
2Cu 2O + Cu 2S → 6Cu + SO2
Copper obtained is blister copper (98% pure).
Step IV Refining of blister copper is done by electrolysis
Impure copper—Anode
Pure copper— Cathode
At anode
: Cu → Cu 2+ + 2e−
At cathode
: Cu 2+ + 2e− → Cu
Carbon-reduction method is not used. Thus, (d) is incorrect.
47. (a) is wrong statement. Impure copper is set as anode where
36. Cryolite is added to alumina in order to lower the melting point.
37. Haematite ore contain Fe2 O3 which is reduced by CO in the blast
furnace as
Fe2 O3 + CO → Fe + CO2
38. Al 2 O3 mixed with cryolite Na 3[AlF6 ] is fused and electrolysed in
copper is oxidised to Cu 2+ and goes into electrolytic
solutions.
(b) CuSO4 is used as an electrolyte in purification process.
(c) Pure copper is deposited at cathode as:
(At cathode)
Cu 2+ + 2e– → Cu :
(d) Less active metals like Ag, Au etc settle down as anode mud.
the extraction of Al.
39. Fluorspar (CaF2 ) improve the electrical conductivity during
electrolytic reduction of alumina.
40. Al itself is a very strong reducing agent.
41. In thermite welding, Al acts as a reducing agent
2Al + Fe2 O3 → Al 2 O3 + 2Fe + Heat
42. The actual representation of CuSO4 ⋅ 5H2 O (blue vitriol) is
[Cu(H2 O)4 ]SO4 ⋅ H2 O and it has covalent, ionic and coordinate
covalent bonds.
43. Ca 2 + end up in CaSiO3 (slag).
44. Iron is rendered passive by concentrated HNO3 due to formation
of a thick protective layer of Fe3O4 .
45. (a) 2Na[Al(OH)4 ](aq) + CO2 → Na 2CO3
+ H2O + 2Al(OH)3 ↓ or Al 2O3.2H2O (ppt.)
Hence, the statement (a) is true.
1100 °C
48. (b) 4CuO → 2Cu 2O + O2
∆
2Cu 2O + Cu 2S → 6Cu + SO2
∆
(c) Cu 2S + 2Cu 2O → 6Cu + SO2
720 ° C
(d) CuSO4 → CuO + SO2 +
1
O2
2
1100 ° C
4CuO → 2Cu 2O + O2
∆
2Cu 2O + Cu 2S → 6Cu + SO2
Reaction is believed to proceed as
Cu 2S r 2Cu + + S2−
2Cu 2O r 4 Cu + + 2O2−
S + 2O2− → SO2 + 6e−
o
6Cu + + 6e− → 6Cu ; Ecell
= 0.52
2−
Extraction of Metals 291
Here, copper sulphide is reduced to copper metal. Solidified
copper has blistered appearance due to evolution of SO2 and thus
obtained copper is known as blister copper.
57. Cu 2S + 2Cu 2O → 6Cu + SO2
In Cu 2S, sulphur is S2− and in SO2, sulphur is in +4 state.
Hence, S2− is acting as reducing agent.
Other compounds which give Cu are
58. Tin is obtained from cassiterite by reduction with coke and the
1100 °C
(i) CuO as 4CuO → 2Cu 2O + O2
balanced chemical reaction is
SnO2 (s) + C (s) → Sn (s) + CO2 (g )
Standard enthalpy of reaction,
∆
2Cu 2O + Cu 2S → 6Cu + SO2
720 °C
(ii) CuSO4 as CuSO4 → CuO + SO2 +
1
O2
2
∆H R°n = (∆H °f CO2 (g )) − (∆H f° SnO2 (s))
∆
∆H R°n = − 394 − (− 581) ⇒187 kJ
4CuO → 2Cu 2O + O2
∆
°
°
Standard entropy of reaction, ∆S R°n = ∆S Products
− ∆S Reactants
2Cu 2O + Cu 2S → 6Cu + SO2
While CuFeS2 will not give Cu on heating. The heating in the
presence of O2 gives Cu 2S and FeS with the evolution of SO 2.
∆S R°n = [ S ° (Sn (s)) + S ° (CO2 (g ))] −
[ S ° (SnO2 (s)) + S ° (C(s))]
49. Al has greater affinity for oxygen, hence oxide is not reduced by
∆S R°n = [ 52 + 210 ] − [ 56 + 6 ] ⇒ 200 JK −1 mol −1
carbon. MgO and CaO (formed in the calcination from
carbonates) are stable species and not reduced by carbon.
We know that, ∆H ° = T∆S °
∆H ° 187 × 1000
∴
T =
=
⇒ 935 K
∆S °
200
For the reaction to be spontaneous, the temperature should be
greater than 935 K.
1300 °C
During Smelting SnO2 + C → Sn + CO
∆ 4Fe + 3 CO
2Fe2O3 + 3 C →
2
50. The important ore of tin is cassiterite (SnO2). Tin is extracted
from cassiterite ore by carbon reduction method in a blast furnace.
SnO2 + 2C 
→ Sn + 2CO
The product often contain traces of iron which is removed by
blowing air through the melt to oxidise to FeO which then floats
to the surface.
59. Siderite = FeCO3, Malachite = CuCO3 ⋅ Cu(OH)2
Bauxite = Al 2O3 ⋅ 2H2O2 consisting some Al (OH)3
Calamine = ZnCO3, Argentite = Ag2S
60. A. 3Cu + 8HNO3 → 3Cu(NO3 )2 + 2NO + 4H2 O
Dil.
2Fe + O2 
→ 2FeO
B. Cu + 4HNO3 → Cu(NO3 )2 + 2NO2 + 2H2 O
Conc.
51. Addition of manganese to iron improve hardness of steel as well
as remove oxygen and sulphur.
C. 4Zn + 10HNO3 → 4Zn(NO3 )2 + N2 O + 5H2 O
Dil.
52. Magnesium and aluminium are both highly electropositive, more
electropositive than water cannot be obtained by electrolysis of
aqueous solution of their salts.
53. Alumina (Al 2 O3 ) has very high melting point and it is poor
D. Zn + 4HNO3 → Zn(NO3 )2 + 2NO2 + 2H2 O
Conc.
O2
61. A. PbS → PbO + SO2 , roasting
B. CaCO3 → CaO + CO2 ↑; calcination
conductor of electricity. Both these factors posses difficulty in
electrolysis of molten alumina.
Cryolite, Na 3AlF6, when mixed with alumina, lowers melting
point as well as improve electrical conductivity, hence helps in
electrolysis of Al 2 O3.
54. Al(OH)3 is amphoteric
Al(OH)3 + 3HCl → AlCl 3 + 3H2 O
Base
Al(OH)3 + NaOH → Na[Al(OH)4 ]
Acid
C. ZnS → Zn, can be done by carbon reduction
or self reduction
D. Cu 2 S → Cu, roasting followed by self reduction
62.
Extraction
methods
A.
B.
C.
Self reduction
Carbon reduction
Complex formation
and displacement
by metal
D.
Decomposition of
iodide
High charge and small size of Al 3+ makes Al—O and O—H
bonds equally ionisable.
55. 2CuFeS2 + O2 → Cu 2S + 2FeS + SO2 ↑
2Cu 2S + 3O2 → 2Cu 2O + 2SO2 ↑
2FeS + 3O2 → 2FeO + 2SO2 ↑
56. FeO + SiO2 → FeSiO3 (Slag)
Metals
extracted
r. Copper, (P) Lead
p. Lead
q. Silver : Ag 2 S + NaCN
→ Na[Ag(CN)2 ]
Na[Ag(CN)2 ] + Zn
→ Na 2 [Zn(CN)4 ] + Ag
s. Boron :
∆
3
BI 3 → B + I2
2
292 Extraction of Metals
63. Column X
Column Y
Column Z
Invar
Fe, Ni
Watch spring
Nichrome
Co, Ni
Heating element
Stainless steel
Fe, Cr, Ni
Cutlery
Column X (Metals)
64.
A.
Al
Column Y
(Ores)
q.
76. True Iron is more electropositive than hydrogen
Fe + 2HCl → FeCl 2 + H2 ↑
77. True : Solubility of silver halides decreases down in the group
Solubility :
AgF > AgCl > AgBr > AgI
78. In crystalline state, AlCl 3 has rock-salt like structure with
coordination number of Al = 6.
Cryolite
79. Four, the complex has formula [Cu(H2 O)4 ] SO4 ⋅ H2 O
80. (a) 2AgBr + C6H4 (OH)2 → 2Ag + 2HBr + C6 H4 O2
B.
Cu
r.
Malachite
C.
Mg
s.
Carnalite
D.
Zn
p.
Calamine
65. A. Silver is extracted by amalgamation process
Hydroquinone
(developer)
Quinone
AgBr + 2Na 2 S2 O3 → Na 3[Ag(S2 O3 )2 ] + NaBr
(b) Na 2 S2 O3 + 2H+ → 2Na + + H2 SO3 + S ↓
Colloidal
sulphur
Distillation
Ag + Hg → Ag(Hg) → Ag( s) + Hg( v ) ↑
Amalgam
B. Calcium is extracted by electrolysis of fused CaCl 2 .
C. Zinc is extracted by carbon reduction method
ZnO + C → Zn + CO
D. Iron is extracted by both carbon reduction method and CO
reduction methods
Fe2 O3 + 3C → 2Fe + 3CO
Fe2 O3 + 3CO → 2Fe + 3CO2
E. Copper is extracted by self reduction methods
Cu 2 S + O2 → Cu 2 O + SO2
Cu 2 O + Cu 2 S → Cu + SO2
66. H 2S Ag 2 S (black) is formed on the surface.
∆
67. ZnO + C →
Zn + CO = Smelting
81. A1 is basic copper carbonate (Cu(OH)2 ⋅ CuCO3 ) while A2 is
Cu 2S. The confirmatory reactions are :
∆
(i) CuCO3 ⋅ Cu(OH)2 → CuO + CO2 ↑ + H2O
Black
HCl
(ii) CuCO3 ⋅ Cu(OH)2 → CuCl 2 + CO2 ↑ + H2O
KI
CuCl 2 → Cu 2 I2 ↓ + KCl + I2
D
Roasting
A2 → Cu 2 O + SO2 ↑
Cu 2 S + Cu 2 O → Cu + SO2 ↑
SO2 is a reducing gas that gives green colour with acidified
K 2 Cr2 O7 as
3SO2 + K 2 Cr2 O7 + H2 SO4 → K 2 SO4 + Cr2 (SO4 )3
Green
sintering
+ 4H2 O
68. Hydration energy Energy required to break the crystal lattice
during dissolving process comes from hydration. If lattice
energy is very high and hydration energy is low, salt becomes
sparingly soluble.
69. Zn Galvanisation involves coating of iron with zinc metal in
order to prevent if from rusting.
70. Sn Cassiterite is an ore of tin.
71. Al Aluminium reduces Fe2 O3 to Fe.
72. Lime, calcium phosphate In basic Bessemer process, the
Bessemer converter is lined with lime but in acid Bessemer
process, it is lined with silica. In basic Bessemer process,
phosphorus is slagged off as calcium phosphate :
P4 + 5O2 → P4 O10
6CaO + P4 O10 → 2Ca 3(PO4 )2
Thomas slag
73. K [Ag(CN)2 ] : AgCl + 2KCN → K[Ag(CN)2 ] + KCl
74. True: Cu + is unstable
H+
2Cu + → Cu 2 + + Cu
75. True Complex (Na[AgCl 2 ]) formation increases solubility of
otherwise sparingly soluble AgCl.
82. Anhydrous AlCl 3 is more soluble in diethyl ether as the oxygen
atom of ether donate its lone-pair of electrons to the vacant
orbital of Al in electron deficient AlCl 3. In case of hydrated
AlCl 3, Al is not electron deficient as oxygen of water molecule
has already donated its lone-pair of electrons to compensate
electron deficiency of Al.
Cl
Cl


••
••
Cl — Al ← OEt 2
Cl — Al ← OH2
••
••


Cl
Cl
hydrated
anhydrous
83. The reactions involved in the extraction of lead from galena
(PbS) by self reduction are
2PbS + 3O2 → 2PbO + 2SO2
PbS + 2PbO → 3Pb + SO2
PbS + 2O2 → PbSO4 (side reaction)
PbSO4 + PbS → 2Pb + 2SO2
In litharge (PbO), the oxidation state of Pb is +2
84. The common photographic film is coated with AgBr and during
developing of photographic film, the unreacted AgBr is removed
by Na 2S2O3 as
AgBr + 2Na 2S2O3 → Na 3[Ag(S2O3 )2 ] + NaBr
Extraction of Metals 293
85.
4NaCN + Ag2S → 2Na[Ag(CN)2 ] + Na 2S
96. (i) Carbon monoxide :
2Na[Ag(CN)2 ] + Zn → Na 2[Zn(CN)4 ] + 2Ag
86. 2Cu + H2O + CO2 + O2 → Cu(OH)2 ⋅ CuCO3
(ii)
Green
(basic copper carbonate)
87.
2PbO + PbS → 3Pb + SO2
(iii) To improve electrical conductivity of melt.
(iv) A metal which is much more electropositive than Ag can only
replace Ag + completely from [Ag(CN)2 ] − as
2000 ° C
C + O2 → CO
3CO + Fe2O3 → 2Fe + 3CO2
CaCO3 → CaO + CO2
Zn + 2[Ag(CN)2 ]− → [Zn(CN)4 ]2− + 2Ag
CaO + SiO2 → CaSiO3
(v) Chalcocite is a sulphide ore of copper, during roasting, SO2 is
liberated, which is not possible in calcination.
Slag
H2O
88. Al + NaOH → NaAlO2+ 32 H2
89. 4KCN + Ag2S →
97. 2Au + 3HNO3 + 11HCl → 2HAuCl 4 + 6H2O + 3NOCl
2K[Ag(CN)2 ] + K2S
98.
Potassium
dicyanoargentate (I)
90. Due to formation of protective, inert layer of Al 2O3 on surface.
91. Sn + 2KOH + 4H2O → K2[Sn(OH)6 ] + 2H2
92. Sea water (contain MgCl 2) + Ca(OH)2
→ Mg(OH)2 ↓ + CaCl 2 ( aq )
(i) Mg(OH)2 + 2HCl → MgCl 2 + 2H2O
Heat to
→ MgCl 2 ( s )
Dryness
electrolysis
Fusion
NaCl
(ii) MgCl 2 (s) → Mg2+ + 2Cl − → Mg
(At cathode)
93. 7Cu + 20HNO3 → 7Cu(NO3 )2 + 4NO + 2NO2 + 10H2O
94.
2CuFeS2
Copper pyrite
Flux
95.
(Al 3+ , Mg2+ ) and they are obtained by electrolytic
reduction of their molten salt.
(ii) Metals like Ge is required in high purity, can be readily
melted and can easily crystallise out from the melt form.
99. AgBr is sensitive to visible light.
1
Br2
2
A photographic plate coated with AgBr, when exposed to
light, gets blackened due to the above reaction.
hν
AgBr → Ag +
100. Al + NaOH(aq) → NaAlO2 + H2
∆
Al(OH)3 → Al 2 O3 + H2 O
Roasting
101. In the first step, galena is heated in presence of O2
(limited quantity) in a reverberatory furnace, where PbS is
partially oxidised to PbO :
Slag
2Cu 2O + Cu 2S → 6Cu + SO2
(i) If the metal is moderately electropositive, e.g. Fe, Sn, Pb or
Cu, they can be obtained from their ore by chemical
reduction methods.
However, if the metal is highly electropositive, e.g. Al, Mg
etc., no reducing agent exist for reduction of their ions
NaAlO2 + CO2 ( aq) → Na 2 CO3 + Al(OH)3 ↓
+ O2 → Cu 2S + 2FeS + SO2
2Cu 2S + 3O2 → 2Cu 2O + 2SO2
2FeS + 3O2 → 2FeO + 2SO2
FeS + SiO2 → FeSiO3
C + O2 → CO
CO + Fe2O3 → CO2 + Fe
2PbS + 3O2 → 2PbO + 2SO2
Bessemerisation
Ag2S + 2NaCN → 2AgCN + Na 2S
AgCN + NaCN → Na[Ag(CN)2 ]
2Na[Ag(CN)2 ] + Zn → Na 2[Zn(CN)4 ] + 2Ag
3
PbS + 2 O2 → PbO + SO2
In the second step, more PbS is added and heated in absence
of O2 , where the following self reduction takes place
PbS + 2PbO → 3Pb + SO2
102. SnO2 + 2C → Sn + 2CO( g ), Carbon reduction method.
20
Qualitative Analysis
Objective Questions I (Only one correct option)
1. A colourless aqueous solution contains nitrates of two
metals, X and Y. When it was added to an aqueous solution of
NaCl, a white precipitate was formed. This precipitate was
found to be partly soluble in hot water to give a residue P and
a solution Q. The residue P was soluble in aqueous NH 3 and
also in excess sodium thiosulphate.The hot solution Q gave a
yellow precipitate with KI. The metals X and Y ,
respectively, are
(2020 Adv.)
(a) Ag and Pb
(c) Cd and Pb
(b) Ag and Cd
(d) Cd and Zn
generates compound Y . Reaction of Y with NaOH gives X .
Further, the reaction of X with Y and water affords
compound Z. Y and Z respectively, are
(2020 Main, 6 Sep II)
(b) SO3 and NaHSO3
(d) S and Na2 SO3
3. Among (A) - (D), the complexes that can display
geometrical isomerism are
+
(A) [Pt(NH3 )3 Cl]
(2020 Main, 8 Jan II)
(B) [Pt(NH3 )Cl5 ]−
(C) [Pt(NH3 )2Cl(NO2 )] (D) [Pt(NH3 )4ClBr]2+
(a) (D) and (A)
(c) (A) and (B)
(b) (C) and (D)
(d) (B) and (C)
4. An organic compound X showing the following solubility
profile is
(2019 Main, 8 April I)
Water
5% HCl
X
10% NaOH
10% NaHCO3
(a) o -toluidine
(c) m-cresol
Insoluble
S2 O23 − →
Ag
X
(Clear solution)
+
→
Y
(White ppt.)
With time
→
Z
(Black ppt.)
(b) [Ag(S2O3 )3 ]5− , Ag2SO3 , Ag2S
(c) [Ag(SO3 )2 ]3− , Ag2S2O3 , Ag
(d) [Ag(SO3 )3 ]3− , Ag2SO4 , Ag
organic compound gave 141 mg of AgBr. The percentage of
bromine in the compound is (atomic mass Ag = 108,
Br = 80)
(2015 Main)
(a) 24
(b) 36
(c) 48
precipitates as a sulphide is
(a) Fe (III)
(c) Mg (II)
(2013 Adv.)
(b) Al (III)
(d) Zn(II)
9. Passing H2 S gas into a mixture of Mn 2+ , Ni 2+ , Cu 2+ and
Hg 2+ ions in an acidified aqueous solution precipitates
(a) CuS and HgS
(c) MnS and NiS
(b) MnS and CuS
(d) NiS and HgS
(2011)
10. A solution of a metal ion when treated with KI gives a red
precipitate which dissolves in excess KI to give a colourless
solution. Moreover, the solution of metal ion on treatment
with a solution of cobalt (II) thiocyanate gives rise to a deep
blue crystalline precipitate. The metal ion is
(2007, 3M)
Soluble
(c) Cu2+
(d) Co2+
Insoluble
(d) 60
8. Upon treatment with ammoniacal H 2S , the metal ion that
(b) Hg2+
11. MgSO4 on reaction with NH4 OH and Na 2 HPO4 forms a
white crystalline precipitate. What is its formula?
precipitate ‘X ’ is obtained, which is soluble in excess of
NaOH. Compound ‘X ’ when heated strongly gives an oxide
which is used in chromatography as an adsorbent. The metal
‘M ’ is
(2018 Main)
(c) Al
(2016 Adv.)
(a) [Ag(S2O3 )2 ]3− , Ag2S2O3 , Ag2S
(a) Pb2+
(b) oleic acid
(d) benzamide
(b) Ca
Ag
+
Insoluble
5. When metal ‘M ’ is treated with NaOH, a white gelatinous
(a) Zn
species X, Y and Z, respectively, are
7. In Carius method of estimation of halogens 250 mg of an
2. Reaction of an inorganic sulphite X with dilute H 2SO 4
(a) SO2 and Na2 SO3
(c) SO2 and NaHSO3
6. In the following reaction sequence in aqueous solution, the
(d) Fe
(a) Mg(NH4 )PO4
(b) Mg3 (PO4 )2
(c) MgCl 2 ⋅ MgSO4
(d) MgSO4
(2006)
12. CuSO4 decolourises on addition of KCN, the product is
(a) [Cu(CN)4 ]2 −
(b) Cu2 + get reduced to form [Cu(CN)4 ]3 −
(c) Cu(CN)2
(d) CuCN
(2006, 3M)
Qualitative Analysis 295
13. A solution when diluted with H2 O and boiled, it gives a
white precipitate. On addition of excess NH4 Cl / NH4 OH,
the volume of precipitate decreases leaving behind a white
gelatinous precipitate. Identify the precipitate which
dissolves in NH4 OH / NH4 Cl.
(2006, 3M)
(a) Zn(OH)2
(b) Al(OH)3
(c) Mg(OH)2
(d) Ca(OH)2
which on addition of excess of KI convert into orange colour
solution. The cation of metal nitrate is
(2005, 1M)
(b) Bi 3+
(d) Pb2+
(c) Sn 2+
15. (NH4 )2 Cr2 O7 on heating gives a gas which is also given by
(2004, 1M)
(a) Heating NH4NO2
(c) Mg3N2 + H2O
(b) Heating NH4NO3
(d) Na(comp. ) + H2O2
16. A sodium salt of an unknown anion when treated with
MgCl 2 gives white precipitate only on boiling. The anion is
(2004, 1M)
(a) SO2–
4
(b) HCO–3
(d) NO−3
(c) CO2−
3
[Fe(H2 O)5 (NO)+ ] SO4 . The oxidation state of iron is
(1987, 1M)
(a) 1
(b) 2
[Y ] + K 2 Cr2 O7 + H2 SO4 → green solution
(a) SO23− , SO2
2−
(c) S , H2S
(2003, 1M)
(b) Cl − , HCl
(d)
CO23− ,
CO2
The aqueous solution on treatment with silver nitrate gives a
white precipitate. The saturated aqueous solution also
dissolves magnesium ribbon with evolution of a colourless
gas Y. Identify X and Y.
(2002, 3M)
(b) X = Cl 2 , Y = CO2
(d) X = H2 , Y = Cl 2
19. An aqueous solution of a substance gives a white precipitate
on treatment with dilute hydrochloric acid, which dissolves
on heating. When hydrogen sulphide is passed through the
hot acidic solution, a black precipitate is obtained. The
substance is a
(2000, 1M)
(a) Hg2+
2 salt
(b) Cr 2+ salt
(c) Ag+ salt
(1986, 1M)
(a) Bi 3+ , Sn 4+
(b) Al 3+ , Hg2+
(c) Zn 2+ , Cu2+
(d) Ni 2+ , Cu2+
24. The compound insoluble in acetic acid is
(a) calcium oxide
(c) calcium oxalate
(1986, 1M)
(b) calcium carbonate
(d) calcium hydroxide
25. The ion that cannot be precipitated by both HCl and H2 S is
(a) Pb2–
(c) Ag+
26. For