JEE ADVANCED Comprehensive Chemistry 2019 JEE ADVANCED 2019 Comprehensive Chemistry McGraw Hill Education (India) Private Limited CHENNAI McGraw Hill Education Offices Chennai New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto McGraw Hill Education (India) Private Limited Published by McGraw Hill Education (India) Private Limited, 444/1, Sri Ekambara Naicker Industrial Estate, Alapakkam, Porur,Chennai - 600 116, Tamil Nadu, India Comprehensive Chemistry—JEE Advanced Copyright © 2018 by McGraw Hill Education (India) Private Limited No Part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication This edition can be exported from India only by the publishers McGraw Hill Education (India) Private Limited D102542 22 21 20 19 18 ISBN (13) : 978-93-87572-58-4 ISBN (10) : 93-87572-58-7 Information contained in this work has been obtained McGraw Hill Education (India), from sources believed to be reliable However, neither, McGraw Hill nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India) nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information This work is published with the understanding that McGraw Hill Education (India) and its authors are supplying information but are not attempting to render engineering or other professional services If such services are required, the assistance of an appropriate professional should be sought Printers, 312 EPIP, HSIDC, Kundli, Sonepat, Haryana Cover Design: Neeraj Dayal Visit us at: www.mheducation.co.in A Word to the Reader &RPSUHKHQVLYH &KHPLVWU\ -(( $GYDQFHG is based on the prescribed syllabus Divided into 34 chapters that covers the entire gamut of the subject (Physical Chemistry, Inorganic Chemistry & �������������������������������������������������������������������������������������������������������������������� This is followed by MCQs, Linked-Comprehension Type, Assertion-Reason and Matrix-Match type questions � ���������������������������������������������������������������������������������������� Some key features ∑ IUPAC recommendations and SI units used throughout ∑� ���������������������������������������������������������������������������������� What’s special? ∑ Over 3000 MCQs with one correct choice and completely solved ∑ Around 800 MCQs with more than one correct choice and fully solved ∑ More than 170 Linked comprehension type questions ∑ About 370 Assertion- Reason type questions & over 120 matrix-match questions ∑ Two model test papers with solutions ∑ JEE Advanced chemistry papers of 2012, 2013, 2014 and 2015 given with complete solutions Tips for studying Chemistry The three branches of Chemistry, viz., Physical, Inorganic and Organic are equally important Physical Chemistry is less diverse compared to Organic and Inorganic The applications to different problems in this are also straightforward Given below are some important topics of Physical, Organic and Inorganic that require special attention: Physical: ∑ Bohr’s theory of atomic structure, quantum numbers and orbitals ∑ MO approach to diatomic molecules, concepts of hybridization/VSEPR theory ∑ Van der Waals equation of state and its application to the behaviour of real gases ∑ Crystal systems, packing of atoms, ionic solids, density of crystals and imperfection ∑ Colligative properties of non-electrolytic and electrolytic solutions ∑ Electrolysis, conductance and galvanic cells ∑ Rate laws, effect of catalyst and temperature on the rate of reaction ∑ pH of salt solutions and solubility product ∑ Le-Chatelier principle, relation between Kp and Kc ∑ Thermochemical calculations and criterion of spontaneity ∑ Radioactive decay Organic and Inorganic ∑ Boron and its compounds ∑ Silicates and silicones vi A Word to the Reader ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ Oxoacids of P, S and halogens Interhalogens and compounds of noble gases Transition elements, coordination compounds and lanthanides Important compounds such as H2O2, NaHCO3, Na2CO3, KMnO4 and K2CrO7 Quantitive analysis of salts Isomerism including optical isomerism Inductance and resonance effects on acidity and basicity of acids and bases Factors affecting SN1 and SN2 reactions Reactions involving rearrangement Bromination and hydrogenation of cis- and trans-alkenes, debromination of dibromobutane Characteristic reaction of aldehede, ketones and carboxylic acids derivatives Reactions with Grignard reagent and those of diazonium salt Carbohydrates and polymers Quantitative analysis of organic compounds All the above topics are adequately covered in this book, supported by plenty of practice problems 1HZ )HDWXUH An interactive CD of 10 full length Mock Papers with Answer Key and Complete Solutions We wish the aspirants all the best in their endeavours THE PUBLISHERS Syllabus 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 ��������������������������������������������������������������������������������������������������������������������������� 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 vapourization; Second law of thermodynamics; Entropy; Free energy; Criterion of spontaneity Chemical equilibrium: Law of mass action; Equilibrium constant, Le Chatelier’s principle (effect of ��������������� ������������ ���� ����������� ������������ ������ ������o 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 ������������������������������������������������������������������������������������������������������������������������� 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:���������������������������������������������������������������������������������������������a, b, g), 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 �������������������������������������������������������������������������������������������������������������������� examples) Nuclear chemistry: Radioactivity: isotopes and isobars; Properties of a, b and g rays; Kinetics of radioactive decay ����������������������������������������������������������������������������������������������������������������������������� 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 viii Syllabus 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 ����������������������� Transition elements (3d series):� ����������� �������� ����������������� ���������� ������� ���� ������ ������������� ������� (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+��������������������������������������������������������������� 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); ��������� ������� ���������� ���� ������ �������� ��� ��������� ����������� ��� ��������� ���� ����������� ������� ���������� ���� 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); Reimer-Tiemann reaction, Kolbe reaction Characteristic reactions of the following (including those mentioned above): Alkyl halides: rearrangement reactions ������������������������������������������������������������������������������������������������������������������������ 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 Syllabus ix 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:� ��������������� ������ ���� ��������������� ��������� ���� ���������� ����������� ����������� ���������� 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: ������������������������������������������������ Practical organic chemistry: ������������������������������������������������������������������������������������� 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 Chemistry Paper I—2017 CI.3 SECTION • This section contains SIX questions of matching type • This section contains TWO tables (each having columns and rows) • Based on each table, there are THREE questions • Each question has FOUR options (a), (b), (c) and (d) ONLY ONE of these four options is correct Answer Q 13 Q 14 and Q 15 by appropriately matching the information given in the three columns of the following table The wave function yn, l, m is a mathematical function whose value depends upon spherical polar coordinates (r, q, f ) of the electron and characterized by the quantum numbers n, l and ml Here r is distance from nucleus, q is coaltitude and f is azimuth In the mathematical functions given in the table, Z is atomic number and a0 is bohr redius Column I Column Column3 Z ∝ e− Zr / a0 (P) a0 I 1s orbital (i) y n , l , m II 2s orbital (ii) One radial node (Q) Probability density at nucleus ∝ 1/a03 III 2pz orbital (iii) y n,l , m Z ∝ re− ( Zr / a0 ) Cosq a0 (R) Probability density is maximum at nucleus IV 3dz2 orbital (iv) xy-plane is a nodal plane (S) E nergy needed to excite electron from n =2 state to n = state is 23/32 times the energy needed to excite electron from n = state to n = state 13 For He+ ion, the only INCORRECT combination is: 16 The only CORRECT Combination in which the reaction proceeds through radical mechanism is: (a) (I), (i), (R) (b) (II), (ii), (Q) (a) (II), (iii), (R) (b) (III), (ii), (P) (c) (I), (i), (S) (d) (I), (iii), (R) 14 For the given orbital in column 1, the only CORRECT combination for any hydrogen-like species is: (c) (IV), (i), (Q) (d) (I), (ii), (R) 17 For the synthesis of benzoic acid, the only CORRECT combination is: (c) (II), (i), (S) (d) (I), (iv), (Q) 18 The only CORRECT combination that gives two different carboxylic acide is: (a) (I), (ii), (S) (b) (IV), (iv), (R) (c) (III), (iii), (P) (d) (II), (ii), (P) 15 For hydrogen atom, the only CORRECT combination is: (a) (II), (i), (Q) (b) (I), (iv), (R) (c) (I), (i), (p) (d) (I), (iv), (S) (a) (III), (IV), (R) (b) (IV), (ii), (P) (a) (IV), (iii), (Q) (b) (I), (i), (S) (c) (III), (iii), (P) (d) (II), (iv), (R) CI.4 Comprehensive Chemistry—JEE Advanced ANSWERS (c) (b), (d) 11 ( ) 16 ( ) (a), (b), (c) ( ) 12 ( ) 17 (c) (a), (b), (c) ( ) 13 ( ) 18 (c) Hints and Solutions ( ) ( ) 14 ( ) () 10 () 15 () Thus, n = The molecular orbitals formed from p orbitals of halogens along with the occupancy of 10 electrons are shown in the figure The electronic configuration of Co2+ confirms this fact 2+ 27Co 3d ¯ ¯ ¯ ¯ ¯ ¯ ¯ 4s 4p s*p Thus, the compound X is [CO(H2O)6]Cl2 ¯ ¯ ¯ ¯ ¯ ¯ p ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ The Conversion of X into Y may be represented by the chemical equation Õ*p ¯ ¯ p ¯ 4[Co(H2O)6] Cl2 + NH4Cl + 20 NH3 + O2 → [Co(NH3)6] Cl3 + 26 H2O (Y) Since NH3 is a strong ligand, it is able to couple the unpaired electrons in Co3+ ion It is Õp sp The highest occupied molecular orbital (HOMO) is ∏*p orbitals and the lowest unoccupied orbital (LUMO) is s*p On descending the halogen group, the energy difference between ∏* (i.e HOMO) and s* (i.e LUMO) is decreased This is responsible for changing colour from yellow to violet This is due to the electronic excitation from ∏*p orbital to s*p orbital Thus, the choices (b) and (c) are correct Supported by zero spin magnetic moment Cobalt, in the complex Y is d2sp3 hybridized The reaction of X with excess of HCl at room temperature gives blue-coloured complex Z The reaction corresponding to this is: The compound X is MCl2 6H2O, i.e [M(H2O)6] Cl2 The treatment of X with NH4Cl in the presence of air gives : complex Y This shows that M2+ in X is oxidised to M3+ in Y The metal ion may be that of cobalt since Co(II) complexes are readly oxidised to Co(III) complexes in the presence of air This is also supported by the spin only magnetic moment of X Given value is 3.87 bohr magneton corresponds to unpaired electrons: The above reaction is an equilibrium reaction and is weakly endothermic, i.e DrH° = positive At room temperature it lies more towards Z giving blue-coloured complex At 0°C, the equilibrium lies more towards complex X giving pink colour of the complex n(n + 2) = 3.8 i.e n (n + 2) 15 The compound Z involves sp3 hybridization and is thus tetrahedral in shape This is due to the fact that Cl is a weak ligand and it is unable to couple the unpaired electrons in Co2+ ion Chemistry Paper I—2017 CI.5 Since work done is represented by the area under P–V curve, it follows that the work done under adiabatic condition will be lesser than that done under isothermal condition This is supported by the spin-only magnetic moment of the complex which requires the presence of unpaired electrons Addition of AgNO3 to the complex will precipitate three equivalents of silver chloride (c) Since the external pressure is equal to pressure of the gas after the gas has gone expansion and it is maximum pressure which the external pressure can have Thus, work done on the gas will be maximum It is shown in the figure Thus, the choices (a), (b) and (c) are correct (a) For free expansion, pext = Thus, w = – pext DV = For an isothermal expansion, DT = Since the internal energy of an ideal gas depends only on temperature, it follows that DU = from the first law of thermodynamics (DU = q + w), it follows that q = (p1, V1) p1 w max Thus, for an isothermal free expansion, we have w = ; q = and DU = (p2, V2) For adiabatic free expansion, q = as no heat is allowed to enter or leave from the system Since w = 0, it follows that DU = (d) Expansion under isothermal conditions, T1 = T2, Thus, for both isothermal and adiabatic free expansions, q = 0, w = and DU = Thus, DU < (b) Since in adiabatic expansion, work is done at the expense of internal energy, the temperature T2 in the final state will be lesser in adiabatic expansion as compared to that in isothermal expansion Since p ∝ T, it follows that p2 in adiabatic expansion will be lesser than that in isothermal expansion The variations of P versus V in the two expansions are shown in the figure Thus, DU = Under adiabatic conditions, T2 < T1, Thus, the choices (a), (b) and (c) are correct (a) Since the pressure PL is larger than that predicted by Raoult’s law (represented by broken line in the figure), larger number of molecules of L are escaping from solution when compared from ideal solution Z pL ¬Raoult's Law Isothermal Adiabatic V1 Vđ V2 xM xL ¬ This is possible when the molecular attractions L M are weaker than those existing in L L and M M (d) The point Z represents vapour pressure of pure liquid L where xL =1 The Raoult’s law is applicable when xL →1 CI.6 Comprehensive Chemistry—JEE Advanced Addition of bromine across double bond is antiaddition The cis-alkene yields only racemic mixture whereas trans-alkene produces meso compound 5 The compound is named either as benzene derivative or by common name Thus, the IUPAC names are: – Chloro – – Methylbenzone and – Chlorotoluene (i) (ii) (a) The strength of oxoacids of chlorine follows the order: HClO4 > HClO3 > HCIO2 > HClO The copounds M and N are identical whereas O and P are enantiomers The compounds M and O and N and P are diastereomers The strength of conjugate bases follow the reverse order: – The choices (b) and (d) are correct Valence electrons – Lewis structure – Prediction : (b) Molecule – ClO4 < ClO3 < ClO2 < ClO O—H | : O — Cl – O : HClO4 1+7+4×6=32 pairs of electrons around Cl, sp3 | hybridization O : : : : : : : : (c) The reaction between Cl2 and H2O is Cl + H2O → HCl + HClO (d) ClO–4 is stabilized due to resonance structures : : : : HClO + + = 14 : Cl — O — H 4-pairs of electrons around Cl, sp3 hybridization Chemistry Paper I—2017 CI.7 From the expression G = K (a/l), we get k= Gl 120cm –7 –1 = (5 × 10–7S) = 600 × 10 S × cm a 1cm For infinite diluted solution, weak acid is considered to be completed ionized Hence, weak acid may be considered as strong acid of PH = (i.e [H] = 10–4M) Now Λ∞m = k c = 600 × 10 –7 Scm –1 10 –4 mol dm –3 = 600 ×10–3 S cm–1 dm–3 mol–1 = 600×10–3 S cm–1 (10 cm)3 mol–1 = 600 S cm2 mol–1 = ×102 S cm2 mol–1 Thus, the value of Z is Valence electrons rewis structure [TeBr6]2– + × + = 50 1 [BrF2]+ + × – = 20 2 : N | : S —F 0 : : : : : + + × = 32 F : : SNF3 Lone pairs : Species F : : – [XeF3] + × + = 30 3 Total lone pairs 10 Aromatic compounds have (4n + 2)p electrons The following compounds satisfy this requirement = where N = 4, a = 400 pm and m is mass/atom of the solid From this, use get m = Pa (8 g cm –3 )(400 × 10 –10 cm)3 = N n = 0 n = 1 n = 1 n = 1 n = Number of atoms in 256g is The number of aromatic compounds is = 1.28 × 10–22g N1 = 11 The expression of density is N r = 3m a 256 g = ×1024 1.28 × 10 –22 g Thus, the value of N is 12 Species Molecular electronic configuration H2 (s1s) He 2 (s1s) (s 1s) + Li2 * Conclusion Diamagnetic Paramagnetic KK (s2s) Diamagnetic KK(s2s) (s 2s) Diamagnetic B2 KK(s2s) (s 2s) (∏2px) (∏2py) Paramagnetic Diamagnetic N2 KK(s2s)2 (s*2s)2 (∏2px)2 (∏2py)2 KK(s2s)2 (s*2s)2 (∏2px)2 (∏2py)2 (s2py)2 Diamagnetic KK(s2s)2 (s*2s)2 (s2pz)2 (∏2px)2 (∏2py)2 (∏*2px)2 (∏*2py)2 Paramagnetic KK(s2s) (s 2s) (s2pz) (∏2px) (∏2py) (∏ 2px) (∏ 2py) Diamagnetic Be2 C2 O–2 F2 2 2 * * * 2 There are diamagnetic species 2 * * CI.8 Comprehensive Chemistry—JEE Advanced 13 The choice (d) is incorrect Is orbital (I in column 1) has no directional characteristics (iii in column contains cos q) DE1 = R∝hc – n1 n2 14 The choice (d) is correct 2s orbital (II in column 1) has one radial node ( = n – l– 1) (ii in column 2) and the plot of yn, l, m (r) has the characteristic of P shown in column 1 = R∝hc – 16 = 15 The choice (d) is correct Is orbital (I in column 1) has wave function dependence shown in (i) column The option S of column is also correct as shown in the following Transition from n = to n = Transition from n = to n = R hc 16 ∝ 1 DE2 = R∝hc – = R hc 36 36 ∝ DE (3 / 16) R∞he 27 = = DE2 (8 / 36) R∞he 32 Answer Q 15, Q 17 and Q 18 by appropriately matching the information given in the three columns of the following table Columns 1, 2, contain starting materials reaction conditions, and type of reactions respectively Column (I) Toluene (II) Acetophenone (III) Benzaldehyde Column (i) NaOH/Br2 (P) Condensation (iii) (CH3CO)2O/ (R) Substitution (iv) NaOH/CO2 (S) Haloform (ii) Br2/h2 CH3COOK (IV) Phenol Column (Q) Carboxylation 16 The combination Br2/h2 (Choice ii in column 2) produces Br radicals Of the two choices (b) and (d) which include (ii), the choice (d) is correct as bromination of tolune (choice I in column 1) involves substitution (choice S in column 3) in the methyl group as shown in the following 17 The choice (c) is correct combination as C6H5COCH3 (autophenone, choice II in column 1) undergoes haloform reaction (choice S in column 3) with the use of reagents NaOH/Br (choice ii in column 2) 18 The choice (c) is correct Heating of aromatic aldehyde (choice III in column 1) with acetic anhydride in the presence of sodium acetate (Choice ii in column 2) gives a, b- unsaturated corboxylic acid this is known as perkin reaction —CHO + (CH3CO)2O NaOAc D —CH = CHCOOH cinnamic acid Cinnamic acid exists in two geometrical isomers due to the presence of a double bond JEE Advanced 2017: Paper–II (Chemistry) SECTION – I This section contains SEVEN questions Each question has FOUR options (a), (b), (c) and (d) ONE or More Than One of these four option(s) is/are correct 1 The order of basicity among the following compounds is (I) (II) (III) (IV) (a) IV > I > II > III (b) IV > II > III > I (c) I > IV > III > II (d) II > I > IV > III The major product of the following reaction is (a) Au metal and NaCN [aq] in the presence of air (b) Zn metal and NaOH [aq] (c) Fe metal and conc HNO3 (d) Cu metal and conc HNO3 The standard state Gibbs free energies of formation of C (graphite) and C (diamond) at T = 298 K are DfG° (C, graphite) = kJ mol–1 DfG° (C, diamond) = 2.9 kJ mol-1 The standard state means that the pressure should be bar, and substance should be pure at a given temperature The conversion of C (graphite) to C (diamond) reduces its volume by × 10–6 m3 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 (Useful information: J = kg m2 s–1, 9pa = 1kg m–1 s–2, bar = 105 pa) (a) 14501 bar (b) 29001 bar (c) 1450 bar (d) 58001 bar The order of the oxidation state of the phosphorus atom in H3PO2, H3PO4, H3PO3 and H4P2O6 is (a) (c) (b) (d) Which of the following combination will produce H2 gas? (a) H3PO4 > H4P2O6 > H3PO3 > H3PO2 (b) H3PO2 > H3PO3 > H4P2O6 > H3PO4 (c) H3PO4 > H3PO2 > H3PO3 > H4P2O6 (d) H3PO3 > H3PO2 > H3PO4 > H4P2O6 For the following cell Zn (s) | ZnSO4 (aq) || CuSO4 (aq) | Cu(s) when the concentration of Zn2+ ions is 10 times the concentration of Cu2+ ions, the expression of DG of the cell reaction is CII.2 Comprehensive Chemistry—JEE Advanced (a) 2.303 RT – (2.2V) F (b) (1.1V) F (c) – (2.2V) F (d) 2.303 RT + (1.1V) F (Where F is Faraday constant, R is gas constant, T is temperature, E°cell = 1.1V) Pure water freezes at 273 k and bar The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution Use the freezing point depression constant of water as 2K kg mol–1 The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T) (Motor mass of ethanol is 46 g mol–1) Among the following, the option representing changes in the freezing point is (b) (c) (a) (d) SECTION This section contains SEVEN questions Each question has FOUR options (a), (b), (c) and (d) ONE OR MORE THAN ONE of these four options is (are) correct In a bimolecular reaction, the steric factor P was experimentally determined to be 4.5 The correct option(s) among the following is (are): (a) The activation energy of the reaction is unaffected by the value of steric factor (b) Since P = 4.5, the reaction will not proved unless and effective catalyst is used (c) The value of frequency factor predicted by Arrhenius equation is higher than that determined experimentally (d) Experimentally determined value of frequency factor is higher than that predicted by Arrhenius equation For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant K in terms of change in entropy is described by (a) With increase in temperature, the value of K for exothermic reaction decreases because favourable change in entropy of the surroundings decreases (b) With increases in temperature, the value of K for exothermic reaction decreases because the entropy change of the system is positive (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 endothermic reaction increases because unfavourable change in entropy of the surroundings decreases 10 The correct statement(s) about surface properties is (are): Chemistry Paper II—2017 CI.3 (a) Adsorption is accompained by decrease in enthalpy and decrease in entropy of the system (b) Cloud is an emulsion type of colloid in which liquid is dispersed phase and gas is dispersion medium (c) The critical temperature of ethane and nitrogen are 563K and 126K, respectively The adsorption of ethane will be more than that of nitrogen on same amount of activated charcoal at a given temperature (d) Brownian motion of colloidal particles does not depend on the size of the particle but depends on viscosity of the solution 11 Among the following the correct statement(s) is (are): (a) I and III follow SN1 mechanism (b) Compound IV configuration (c) The correct order of reactivity for I, III and IV is IV > I > II undergoes inversion of (d) I and II follows SN2 mechanism 13 The option(s) with only amphoteric oxide is (are): (a) ZnO, Al2O3, PbO, PbO2 (b) Cr2O3, CrO, SnO, PbO (c) Cr2O3, BeO, SnO, SnO2 (a) AlCl3 has the three-centre two electron bonds in its dimeric structure (b) BH3 has the three-centre two-electron bonds in its dimeric structure (d) NO, B2O3, PbO, SnO2 14 Compounds P and R upon ozonolysis produce Q and S, respectively The molecular formula of Q and S is C8H8O Q and undergoes Cannizzaro reaction but not haloform reaction, whereas S undergoes haloform reaction but not Cannizzaro reaction (c) Al (CH3)3 has the three-centre two-electron bonds in its dimeric structure 2 P (i) (ii ) Zn / H 2O → (C H O ) 8 (d) The Lewis acidity of BCl3 is greater than that of AlCl3 12 For the following compounds, the correct statement(s) with respect to nucleophilic substitution reaction is(are) (i ) O / CH Cl O / CH Cl S (ii ) Zn / H 2O (C8 H 8O ) 2 (ii) R → The option(s) with suitable combination of P and R, respectively, is (are) (I) (II) (b) (III) (IV) Q (a) (c) (d) CII.4 Comprehensive Chemistry—JEE Advanced SECTION-3 This section contains TWO paragraphs Based on each paragraph, there are TWO questions Each question has FOUR options (a), (b), (c) and (d) Only ONE of these options is correct 17 The reactions Q to R and R → S are Paragraph Upon mixing KClO3 in the presence of catalytic amount of MnO2, a gas W is formed Excess amount of W reacts with phosphorus to give X The reaction of X with pure HNO3 gives Y and Z 15 W and X are, respectively (a) O3 and P4O6 (a) Dehydration and Friedel-Crafts acylation (b) Friedel-Crafts alkylation, dehydration and Friedel-Crafts acylation (c) Friedel-Crafts alkylation and Friedel-Crafts acylation (b) O2 and P4O6 (c) O2 and P4O10 (d) O3 and P4O10 16 Y and Z are, respectively (a) N2O5 and HPO3 (b) N2O5 and H3PO4 (c) N2O4 and HPO3 (d) N2O4 and H3PO3 Paragraph (d) Aromatic sulfonation and Friedel-Crafts acylation 18 The product S is (a) (b) The reaction of compound P with CH3MqBr (excess) in (C2H5)2O followed by addition of H2O gives Q The compound Q on treatment with H2SO4 are 0°C gives R The reaction of R with CH3COCl in the presence of anhydrous AlCl3 in CH2Cl2 followed by treatment with H2O produces compound S [Et in compound P is ethyl group] (c) (d) ANSWERS (a) (a) 11 (b), (c), (d) 16 (a) (b) (c) 12 (a), (b), (d) 17 (c) (b) (a), (b) 13 (a), (c) 18 (c) (a) (a), (d) 14 (b), (c) (a) 10 (a), (c) 15 (c) Chemistry Paper II—2017 CI.5 HINTS & SOLUTIONS The molecule IV have maximum basicity This may be attributed to stability of its conjugate acid due to larger number of resonating structures The maximum basicity may be explained on the basis that the basic site = NH is attached to two pi-electrons donors which enhances that electron density on the basic site The next lesser basic is the molecule I This involves lesser number of resonating structures in its conjugate acid C (graphite) → C (diamond) DfG° 0 2.9 kJ mol–1 DrG° = DfG° (C, diamond) – DfG° (C, graphite) = 2.9 kJ mol–1 – D = 2.9 kJ mol–1 = 2900 J mol–1 Since DG = DH – T DS, we have DG = D(U + PV) – T D S = DU + D (PV) – T D S Since DU = q + w = q – pDV and at constant temperature, DS = q/T, we get DG = (q – p D V) + (p D V + V D p) – q = V D p or Dp = DG 2900 J mol –1 = = 1450 ×106 Pa V × 10 –6 m3 mol –1 = 14500 bar The molecule I has one pi-electrons doner –NH2 group and one electron-releasing methyl group and thus expected to be lesser basic than molecule IV The least basic is the molecule III The electron pair on nitrogen is delocalised over the whole molecule to make it aromatic Its lone pair of electrons on nitrogen is not available to make it basic Thus, the order of basicity is IV > I > II > III The reactions are: i.e p2 = p1 = 14500 bar i.e p2 = 14501 bar ; (Since p1 = 1bar) It x is the oxidation state of phosphorus, then we have H3PO4 3(+1)+ x + (–2) = ⇒ x = + H4P2O6 (+1) + 2x + (–2) = ⇒ x = + H3PO3 H3PO2 (+1) + x + 3(–2) = ⇒ x = + 3(+1) + x + (–2) = ⇒ x = + Thus, the choice (a) gives the correct order of oxidation state of phosphorus The cell reaction is Cu2+ (aq) + Zn(s) = Cu(s) = Zn2+ (aq) The cell potential is (Minor) (Major) (more steric hindrance) Zn metal produces H2 gas with NaOH (aq) Zn + NaOH → Na2 ZnO2 + H2 The other reaction: Au + NaCN + 2H2O + O2 → Na [Au (CN)2] + NaOH Fe+ HNO3 → Fe (NO3)3 + 3NO2 + 3H2O (Conc.) Cu + HNO3 → Cu (NO3)2 + 2NO2 + 2H2O (Conc.) Ecell = E°cell – [ Zn ] / C ° RT ln 2F [Cu ] / C ° = 1.1V – 2.303RT 10 log 1 2F = 1.1V – 2.303RT 2F Since DG = –n F Ecell, we get DG 2.303RT – = 1.1V – 2F 2F i.e DG = 2.303 RT– (2.2V) F Note: It is desireable to write cell reaction to which DG stands for Vapour pressure of solution increases with increase in temperature Thus, either choice (b) or choice (c) is correct CII.6 Comprehensive Chemistry—JEE Advanced Amount of ethanol, n = m 34.5 g = = 0.75 mol M 46 g mol –1 Molality of solution, n = n 0.75mol = = 1.5 mol kg–1 m1 0.5kg Depression in freezing point is – DTf = Kfm (2K kg mol–1) (1.5 mol kg–1) = 3K Freezing point of solution Tf = T*f – DTf = 273K – 3K = 270 K The choice (c) satisfies this requirement Note: It is assumed that ethanol is nonvolatile The activation energy of the reaction is not affected by steric factor (choice a) Since steric factor is more than one, the frequency factor (i.e frequency of colliding molecules) determined experimentally will be larger that computed from Arrhenius equation In fact Zexpt = PZ1 where Z’ is frequency factor computed from Arrhenius equation (Choice d) As per Le-Chatelier principle, if the temperature of a reaction at equilibrium is increased, the reaction shifts in a direction so as to decrease the temperature This decrease in temperature is achieved by releasing less heat in an exothermic reaction and absorbing more heat in an endothermic reaction Considering an exothermic reaction Reactant Product; DrH = negative which may be written as Reactant Product + q To decrease the value of q1 reaction moves towards reactant side causing a decrease in equilibrium constant [Product] decrease K= = = decrease [Reactant] increase The heat released in an exothermic reaction causes an increase in entropy of the surroundings Assuming q does not change significantly, the entropy change ( = q/T) in the surroundings decreases due to increase in temperature Since the entropy change is positive, this is referred to as favourable change in entropy and its value decreases with increase in temperature (Choice a) For an endothermic reaction Reactant Product; DrH = positive which may be written as Reactant + q Product With the increase in temperature, absorption of q increases thereby equilibrium is shifted towards product side causing an increase in equilibrium constant K= [Product] increase = = increase [Reactant] decrease The absorbed heat is supplied by the surroundings, the entropy change of the surroundings will be negative and is referred to as unfavourable change in entorpy Its value becomes more negative and thus a decrease in entropy of the surroundings will be observed (choice d) The change in equilibrium constant with change in temperature is independent of the entropy change of the reaction It depends only on the enthalpy of reaction through the expression d ln k ° ∆ r H ° = dT RT 10 Adsorption involves attractions between adsorbent and adsorbate and thus involves a decrease in enthalpy on adsorption The adsorbate is more ordered after adsorption and is thus accompanied with decrease in entropy More easily liquefiable gas ethane (which has a higher critical temperature) involves larger intermolecular attractions and thus will exhibit larger adsorption on the same amount of adsorbent Cloud belongs to areosol colloid Brownian motion depends on colloidal particle Thus, the choices (a) and (c) are correct 11 The structures of given dimers are: It has two 3c-2e bonds It has two 3c-2e bonds Chemistry Paper II—2017 CI.7 No 3c-2e bonds BCl3 is a stronger Lewis acid than AlCl3 due to the presence of more localized P orbital in boron 12 The main characteristics of SN1 mechanism are: (i) Steric hindrance for the incoming nucleophile (ii) Polar solvent accelerates the substitution (iii) Chiral starting material ends with the racemization of the products (iv) Intermediate is carbocation more stable the carbocation, the faster the Sn1 mechanism The main characteristics of Sn2 mechanism are (i) Lesser steric hindrance for the incoming nucleophile (ii) Aportic solvent accelerates the substitution (iii) Involves Walden inversion – stereochemistry around carbon atom is inverted Primary and secondary halides are likely to undergo SN2 mechanism Secondary and tertiary halides are likely to undergo SN2 mechanism Choice (a) is correct as intermediate cations are stable Choice (b) is correct as it can follow SN2 mechanism Choice (c) is incorrect as compound (III) is expected to have maximum reactivity Choice (d) is also correct as the compounds I and II can follow SN2 mechanism 13 The amphoteric oxides in the given oxides are: ZnO, Al2O3, PbO, PbO2, Cr2O3, BeO, SnO and SnO2 14 The ozonolysis products from P and R of choice (a) are (Q) Gives Cannizzaro reaction due to CHO group with no a-hydrogen (S) The molecular formula of S is not C8H8O The ozonolysis products from P and R of choice (b) are (Q) Gives Cannizzaro reaction due to –CHO group with No a-hydrogen (S) Gives holoform reaction due to COCH3 Group The ozonolysis products from P and R of choice (c) are (Q) Gives Cannizzaro reaction due to –CHO group with no a-hydrogen (S) Gives haloform reaction due to –COCH3 group The ozonolysis products from P and R of choice (d) are (Q) Gives Cannizzaro reaction due to –CHO group with no a-hydrogen (S) The molecular formula is not C8H8O Solutions (15 and 16) Thus, the choices (a) and (c) are correct The reactions are CrO is a basic oxide MnO2 → 2KCl + 3O 2KClO3 NO is a neutral oxide B2O3 is an acidic oxide (W) P4 + 5O2 → P4O10 (X) P4O10 + HNO3 → 2N2O5 + HPO3 (Y) (Z) CII.8 Comprehensive Chemistry—JEE Advanced 18 The given reactions are: 17 The conversion of Q to R involves Friedel-Crafts alkylation and dehydration The conversion of R to S involves Friedel-Crafts acylation 18 The S is compound of choice (c) ... 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