Đang tải... (xem toàn văn)
https://1drv.ms/b/s!AmkCsf2WlV7n2CSmbhEI__B7JvqO?e=qfif37
JEE Main CHAPTERWISE SOLUTIONS 2019-2002 Chemistry All the 16 Question Papers of JEE Main Online 2019 (Jan & Apr Attempt) ARIHANT PRAKASHAN (Series), MEERUT Arihant Prakashan (Series), Meerut All Rights Reserved © Publisher No part of this publication may be re-produced, stored in a retrieval system or distributed in any form or by any means, electronic, mechanical, photocopying, recording, scanning, web or otherwise without the written permission of the publisher Arihant has obtained all the information in this book from the sources believed to be reliable and true However, Arihant or its editors or authors or illustrators don’t take any responsibility for the absolute accuracy of any information published and the damages or loss suffered there upon All disputes subject to Meerut (UP) jurisdiction only Administrative & Production Offices Regd Office ‘Ramchhaya’ 4577/15, Agarwal Road, Darya Ganj, New Delhi -110002 Tele: 011- 47630600, 43518550; Fax: 011- 23280316 Head Office Kalindi, TP Nagar, Meerut (UP) - 250002 Tele: 0121-2401479, 2512970, 4004199; Fax: 0121-2401648 Sales & Support Offices Agra, Ahmedabad, Bengaluru, Bareilly, Chennai, Delhi, Guwahati, Hyderabad, Jaipur, Jhansi, Kolkata, Lucknow, Meerut, Nagpur & Pune ISBN : 978-93-13195-31-3 Published by Arihant Publications (India) Ltd For further information about the books published by Arihant log on to www.arihantbooks.com or email to info@arihantbooks.com /@arihantpub Arihant Publications /arihantpub /arihantpub PREFACE JEE Main is a gateway examination for candidates expecting to seek admission in Bachelor in Engineering (BE), Bachelor of Technology (B.Tech) and Bachelor of Architecture (B.Arch) at Indian Institutes of Information Technology (IIITs), National Institutes of Technology (NITs), Delhi Technological University and other Centrally Funded Technical Institutes (CFTIs) JEE Main is also an examination which is like screening examination for JEE Advanced (The gateway examination to India's most reputed Technical Institutes, Indian Institutes of Technology— IITs) To make the students well-versed with the pattern as well as the level of the questions asked in the exam, this book contains Chapterwise Solutions of Questions asked in Last 18 Years’ Examinations of JEE Main (formerly known as AIEEE) Solutions to all the questions have been kept very detailed and accurate Along with the indication of level of the exam, this book will also teach you to solve the questions objectively in the examination To give the students a complete practice, along with Chapterwise Solutions, this book contains Practice Sets, based exactly on JEE Main Syllabus and Pattern By practicing these sets, students can attain efficiency in Time Management during the examination We hope this book would be highly beneficial for the students We would be grateful if any discrepancy or mistake in the questions or answers is brought to our notice so that these could be rectified in subsequent editions Publisher CONTENTS Some Basic Concepts of Chemistry 1-8 States of Matter 9-20 Atomic Structure 21-28 Chemical Bonding 29-40 Thermodynamics 41-56 Solutions 57-72 Equilibrium 73-90 Redox Reactions and Electrochemistry Chemical Kinetics and Surface Chemistry 91-107 108-128 10 Periodicity of Elements 129-134 11 Principles and Processes of Metallurgy 135-140 12 Hydrogen, s-and p-Block Elements 141-165 13 d-and f-Block Elements and Coordination Chemistry 166-191 14 Environmental Chemistry 192-198 15 General Organic Chemistry 199-214 16 Hydrocarbons and their Halogen Derivatives 215-238 17 Organic Compounds Containing Oxygen (Alcohols, Ethers, Aldehydes, Ketones, Carboxylic Acids and their Derivatives) 239-276 18 Organic Compounds Containing Nitrogen (Amines and Diazonium Salts) 277-293 19 Polymers and Biomolecules 294-307 20 Analytical Chemistry and Chemistry in Everyday Life 308-314 PRACTICE SETS for JEE MAIN Practice Set Practice Set Practice Set Practice Set Practice Set 315-320 321-326 327-331 332-336 337-342 SYLLABUS SECTION- A (Physical Chemistry) UNIT Some Basic Concepts in hemistry 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 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 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 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 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 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 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 ΔG o 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 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 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 10 Surface Chemistry Adsorption Physisorption and chemisorption and their characteristics, factors affecting adsorption of gases on solids- Freundlich and Langmuir adsorption isotherms, adsorption from solutions Catalysis Homogeneous and heterogeneous, activity and selectivity of solid catalysts, enzyme catalysis and its mechanism Colloidal state distinction among true solutions, colloids and suspensions, classification of colloids - lyophilic, lyophobic; multi molecular, macromole-cular and associated colloids (micelles), preparation and properties of colloids Tyndall effect, Brownian movement, electrophoresis, dialysis, coagulation and flocculation; Emulsions and their characteristics UNIT Chemical Kinetics Rate of a chemical reaction, factors affecting the rate of reactions concentration, temperature, pressure and catalyst; elementary and complex reactions, order and molecularity of reactions, rate law, rate constant and its units, differential and integral forms of zero and first order reactions, their characteristics and half - lives, effect of temperature on rate of reactions - Arrhenius theory, activation energy and its calculation, collision theory of bimolecular gaseous reactions (no derivation) SECTION- B (Inorganic Chemistry) UNIT 11 Classification of Elements and Periodicity in Properties UNIT 14 s - Block Elements 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 Group and 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 12 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 13 Hydrogen Position of hydrogen in periodic table, isotopes, preparation, properties and uses of hydrogen; physical and chemical properties of water and heavy water; Structure, preparation, reactions and uses of hydrogen peroxide; Classification of hydrides ionic, covalent and interstitial; Hydrogen as a fuel (Alkali and Alkaline Earth Metals) UNIT 15 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 16 d – and f – Block Elements Transition Elements General introduction, electronic configuration, occurrence and characteristics, general trends in properties of the first row transition elements - physical properties, ionization enthalpy, oxidation states, atomic radii, colour, catalytic behaviour, magnetic properties, complex formation, interstitial compounds, alloy formation; Preparation, properties and uses of K2 Cr2 O7 and KMnO4 Inner Transition Elements Lanthanoids Electronic configuration, oxidation states, chemical reactivity and lanthanoid contraction Actinoids Electronic configuration and oxidation states UNIT 17 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 18 Environmental Chemistry Environmental pollution Atmospheric, water and soil Atmospheric pollution Tropospheric and stratospheric Tropospheric pollutants : Gaseous pollutants Oxides of carbon, nitrogen and sulphur, hydrocarbons; their sources, harmful effects and prevention; Green house effect and Global warming; Acid rain; Particulate pollutants Smoke, dust, smog, fumes, mist; their sources, harmful effects and prevention Stratospheric pollution Formation and breakdown of ozone, depletion of ozone layer - its mechanism and effects Water pollution Major pollutants such as, pathogens, organic wastes and chemical pollutants their harmful effects and prevention Soil pollution Major pollutants such as: Pesticides (insecticides, herbicides and fungicides), their harmful effects and prevention Strategies to control environmental pollution SECTION- C (Organic Chemistry) UNIT 19 Purification & Characterisation of Organic Compounds Purification Crystallization, sublimation, distillation, differential extraction and chromatography principles and their applications Qualitative analysis Detection of nitrogen, sulphur, phosphorus and halogens Quantitative analysis (basic principles only) Estimation of carbon, hydrogen, nitrogen, halogens, sulphur, phosphorus Calculations of empirical formulae and molecular formulae; Numerical problems in organic quantitative analysis UNIT 20 Some Basic Principles of Organic Chemistry Tetravalency of carbon; Shapes of simple molecules hybridization (s and p); Classification of organic compounds based on functional groups: —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 21 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 Alkynes 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 22 Organic Compounds Containing Halogens General methods of preparation, properties and reactions; Nature of C—X bond; Mechanisms of substitution reactions Uses/environmental effects of chloroform, iodoform UNIT 23 Organic Compounds Containing Oxygen General methods of preparation, properties, reactions and uses Alcohols, Phenols and Ethers Alcohols Identification of primary, secondary and tertiary alcohols; mechanism of dehydration Phenols Acidic nature, electrophilic substitution reactions: halogenation, nitration and sulphonation, Reimer - Tiemann reaction Ethers Structure Aldehyde and Ketones Nature of carbonyl group; Nucleophilic addition to >C=O group, relative reactivities of aldehydes and ketones; Important reactions such as - Nucleophilic addition reactions (addition of HCN, NH 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 and factors affecting it 76 JEE Main Chapterwise Chemistry ∴ 12 + pOH = 14 pOH = 14 − 12 = [OH− ] = 10− in buffer solution For reaction Cd(OH)2 → Cd + + 2OH− S2 S Ksp = [Cd + ] [OH− ]2 Ksp = (S )(2S ) = 4S = 4(184 × 10− )3 Ksp = 24.9 × 10− 15 [Cd + ] = [Cd + ] = Ksp (d) 2NO(g ) N2 ( g ) + O ( g ) ∆ng = nproduct − nreactant = − = ∆ng = ⇒ So, K p = Kc - 10 20 mL of 0.1 M H2SO4 solution is added to 30 mL of 0.2 M NH 4OH solution The pH of the resultant mixture is [pK b ofNH 4OH = 4.7] (a) 9.3 Exp (a) 24.9 × 10− 15 H 2SO −2 ) = 24.9 × 10− 11 M [Cd + ] ⇒2.49 × 10− 10 M The expected solubility of Cd(OH)2 in a buffer solution of pH = 12 is 2.49 × 10− 10 M In which one of the following equilibria, [JEE Main 2019, 12 April Shift-II] K p ≠ Kc ? 2CO(g ) c c cc (a) 2C(s ) + O2 (g ) (b) 2HI(g ) (d) 5.2 (a) The reaction takes place when H2SO is added to NH4OH is as follows : [OH− ]2 (10 (c) 9.0 [ JEE Main 2019, Jan Shift-I] H2 (g ) + I2 (g ) (c) NO2 (g ) + SO2 (g ) NO(g ) + SO3 (g ) (d) 2NO(g ) N2 (g ) + O2 (g ) + 2NH 4OH → (NH )2SO + 2H 2O Strong acid = 24.9 × 10− 15 × 10 + ⇒ (b) 5.0 Weak base Millimoles at t = 20 × 0.1 = Millimoles at t = t 30 × 0.2 = Salt of strong acid + weak base 2 So, the resulting solution is a basic buffer [NH4OH + (NH4 )2 SO ] According to the Henderson’s equation, [(NH4 )2 SO ] pOH = pK b + log [NH4OH] + log = 4.7 = 47 ⇒ pH = 14 − pOH = 14 − 4.7 = 9.3 11 Consider the following reversible chemical reactions, Exp (a) Key Idea The relationship between Kp and Kc is ∆n g Kp = Kc (RT ) where, ∆ng = nproducts − nreactants If ∆ng = then Kp = Kc If ∆ng = + ve then Kp > Kc If ∆ng = − ve then Kp < Kc Consider the following equilibria reactions (a) 2C(s ) + O2 (g ) 2CO(g ) ∆ng = nproduct − nreactant = − (1) = ∆ng ≠ ⇒ So, K p ≠ Kc - (b) 2HI(g ) H2 ( g ) + I ( g ) ∆ng = nproduct − nreactant = − = ∆ng = ⇒ So, K p = Kc - (c) NO (g ) + SO (g ) NO(g ) + SO (g ) ∆ng = nproduct − nreactant = − = ∆ng = ⇒ So, K p = Kc - A 2( g ) + B ( g ) K1 K2 6AB ( g ) …(i) -2AB (g ) - 3A (g ) + 3B ( g ) …(ii) The relation between K and K is (b) K K = (d) K K = (a) K = K 13 (c) K = K 1− [ JEE Main 2019, Jan Shift-II] Exp (c) (i) A2 (g ) + B2 (g ) c AB(g ) K1 = [ AB]2 [ A2 ] [B2 ] (ii) AB(g ) A2 (g ) + 3B2 (g ); [ A2 ]3 [B2 ]3 1 = = 3, K2 = K [ AB]6 [ AB] [ A2 ][B2 ] c ⇒ K = K1−3 77 Equilibrium Kp for the following reactions KC at 300 K are, respectively (At 300 K, T = 24.62 dm atm mol −1 ) 12 The values of N ( g ) + O2 ( g ) N 2O4( g ) N ( g ) + 3H ( g ) =2NO(g ) =2NO (g ) =2NH (g ) The reaction involved is as follows : Ca(OH)2 + Na 2SO → CaSO + 2OH− + 2Na + × 1000 = 14 142 Millimoles 100 at t = Milimoles at t = t s (a) 1, 24.62 dm atm mol −1 , 606.0 dm atm mol −2 (b) 1, 24.62 dm atm mol −1 , 1.65 × 10−3 dm −6 atm −2 mol Exp (b) 86 [Limiting reagent] 28 1000 × = 0.28mol L−1 1000 100 = 0.28 M [OH– ] = Exp (b) 14 5.1 g NH4 SH is introduced in 3.0 L evacuated We know that, the relationship between K p and KC of a chemical equilibrium state (reaction) is Kp ∆n ∆n K p = KC (RT ) g ⇒ = (RT ) g KC flask at 327°C 30% of the solid NH SH decomposed to NH3 and H 2S as gases The Kp of the reaction at 327°C is (R = 0.082 atm mol −1 K −1 , molar mass of S = 32 g mol −1 , molar mass of N = 14 g mol −1 ) ∆ng = ΣnProducts − ΣnReactants c c c (i) N2 (g ) + O (g ) 2NO(g ) ⇒ (RT )2 − ( + 1) = (RT )0 = (ii) N2O (g ) 2NO (g ) ⇒ (RT )2 − = RT = 24.62 dm3 atm mol −1 –6 dm −2 atm mol The mass of calcium sulphate formed and the concentration of OH− in resulting solution, respectively, are : (Molar mass of Ca(OH)2 , Na 2SO4 and CaSO4 are74, 143 and 136 g mol −1 , respectively; K sp of Ca(OH)2 is [ JEE Main 2019, 10 Jan Shift-I] (a) 136 g, 0.28 mol L (b) 1.9 g, 0.28 mol L −1 (c) 13.6 g, 0.14 mol L −1 (d) 1.9 g, 0.14 mol L −1 (c) × 10−3 atm (d)1 × 10−4 atm Molar mass of NH4SH = 18 + 33 = 51g mol −1 sodium sulphate was dissolved in water and the volume was made upto 100 mL −1 (b) 242 atm Exp (b) A mixture of 100 mmol of Ca(OH)2 and g of 5.5 × 10 −6 ) (a) 242 × 10−4 atm [ JEE Main 2019, 10 Jan (Shift-I)] (iii) N2 (g ) + 3H2 (g ) 2NH3 (g ) ⇒(RT )2 − ( + 1) = (RT )− = (24.62 dm atm mol −1 )2 13 28 ∴ Molarity of OH– , [ JEE Main 2019, 10 Jan Shift-I] = 1.649 × 10 No of moles = (d)1, 4.1× 10−2 dm −3 atm −1 mol, 606 dm atm mol −2 −3 14 Weight Molecular mass 14 × 136 = g ∴ Mass of CaSO = 1000 No of moles of solute Also, Molarity = Volume of solution (in L) (c) 24.62 dm atm mol −1 , 606.0 dm atm −2 mol , 1.65 × 10−3 dm−6 atm −2 mol where, Number of moles of NH4SH introduced in the vessel Weight 5⋅1 mol = = = 01 Molar mass 51 NH4SH(s ) Active mass (mol L −1) KC = ⇒ NH3 (g ) + c Number of 0.1 moles at t = At t = t eq 011 ( − 0.03) 30% of 01 = 0.03 H2S(g ) 30% of 0.1 = 0.03 0.03 0.03 = 0.01 = 0.01 3 [NH3 ] [H2S] 0.01 × 0.01 = 10−4 (mol L −1) = [NH4HS(s )] ∆n g K p = KC (RT ) [where, ∆ng = Σnproduct − Σnreactant ] = − = 78 JEE Main Chapterwise Chemistry ∴ K p = KC (RT )2 Exp (a) = 10− × [0.082 × (273 + 327 )]2 atm2 = 0.242 atm2 15 Consider the reaction, N ( g ) + 3H ( g ) =2NH (g ) 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 p NH3 <> pNH ] = p + 3p = 4p Kp = Now, = Kp = ⇒ pNH = ⇒ pNH = p2NH pN × pH3 2 pNH 27 × p4 = × 44 pNH = pNH P 27 × 4 [Q P = p] 32 × × P × K p × 31/ 42 (b) 100 kJ mol − (d) − 100 kJ mol − [ JEE Main 2019, 11 Jan Shift-II] Exp (c) We know that, ∆G° = − 2.303 RT log K Also, given equilibrium is - – H3O + + OH [H+ ][OH− ] = 10−14 K = 10− 14 or ∴ ∆G° = − 2.303 × 8.314 JK –1mol −1 × 298K × log10−14 = 79881.8 J mol − = 18 Two solids dissociate as follows: 33 / × P × K1p/ 16 16 Given the equilibrium constant (K C ) of the reaction : Cu(s ) + Ag+ (aq ) → Cu2 + (aq ) + Ag(s ) ° of this reaction is10 × 1015 , calculate the E cell RT at 298 K 2.303 at 298 K = 0.059 V F (a) 0.4736 V (c) 0.4736 mV (a) − 80 kJ mol − (c) 80 kJ mol − - = 79.8 kJ ≈ 80 kJ mol − 32 × × P 44 × P × K1p/ H3O+ + OH − , the value of ∆G º at 298 K is approximately 17 For the equilibrium,2H2O 2H2O p × (3 p)3 pNH According to Nernst equation, 2.303 RT log Q Ecell = E °cell − nF 2.303 RT Given, = 0.059 V F 0.059 Ecell = E ° cell − log Q ∴ n At equilibrium, Ecell = 0.059 log KC E ° cell = n For the given reaction, n = Also, KC = 10 × 1015 [given] 0.059 ∴ log (10 × 1015 ) E° cell = = 0.472 V ≈ 0.473 V (b) 0.04736 mV (d) 0.04736 V [ JEE Main 2019, 11 Jan Shift-II] -B (g ) + C(g );K D(s ) -C( g ) + E ( g ); K A (s ) p1 p = x atm = y atm The total pressure when both the solids dissociate simultaneously is (a) x + y atm (b) x + y atm (d) 2( x + y ) atm (c) ( x + y ) atm [ JEE Main 2019, 12 Jan Shift-I] Exp (d) The equilibrium reaction for the dissociation of two solids is given as: A(s ) e B(g ) + C(g ) p1 p1 + p 79 Equilibrium At equilibrium K p1 = x = pB ⋅ pC = p1( p1 + p2 ) …(i) e Similarly, D(s) C(g ) + E(g ) At equilibrium p1 + p2 ⋅ p2 K p = y = pC ⋅ pE =( p1 + p2 ) p2 On adding Eq (i) and (ii), we get …(ii) K - 2C + D, the initial concentration ofB 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 [ JEE Main 2019, 12 Jan Shift-I] (b)16 (c) − 10 (c) × 10 (d) × 10− 11 M M Exp (c) Let the solubility of Ag 2CO is S Now, 0.1 M of AgNO is added to this solution after which let the solubility of Ag 2CO becomes S ′ ∴ [Ag + ] = S + 0.1 and [CO 23 − ] = S ′ Ksp = (S + 0.1)2 (S ′ ) Ksp = × 10 Given, Q K sp is very small, we neglectS ′ againstS in Eq (i) ∴ K sp = (0.1)2 S ′ or × 10− 12 = 0.01 S ′ or S ′ = × 10− 12 × 102 = × 10− 10 M Thus, molar solubility of Ag 2CO in 0.1 M AgNO is × 10− 10 M 21 Which of the following lines correctly show the temperature dependence of equilibrium constant, K , for an exothermic reaction? [JEE Main 2018] (d) In K A B T(K) (0, 0) For the given chemical reaction, A + 2B a0 a0 − x w 1.5a0 1.5a0 − x ×× ×× ×× C ×× ×× ×× 2C + D 2x .(i) − 12 Exp (d) At, t = t = t eq (b) × 10− 13 M [ JEE Main 2019, 12 Jan Shift-II] or …(iii) x + y= p1 + p2 Now, total pressure is given as pT = pB + pC + pE = p1 + ( p1 + p2 ) + p2 …(iv) = ( p1 + p2 ) On substituting the value of p1 + p2 from Eq (iii) to Eq (iv), we get pT =2 x + y (a) solubility of Ag2CO3 in 0.1 M AgNO3 is (a) × 10− 12 M K p1 + K p = x + y = p1( p1 + p2 ) + p2 ( p1 + p2 ) = ( p1 + p2 )2 19 In a chemical reaction, A + 2B 20 If K sp of Ag2CO3 is × 10− 12, the molar x [x = degree of dissociation] Given, at equilibrium [ A] =[B] a0 − x = 15 a0 − x x = 0.5a0 ∴ [A] = a0 − x = a0 − 0.5a0 = 0.5a0 [B] = 15 a0 − x = 15 a0 − × 0.5a0 = 0.5a0 [C] = x = × 0.5a0 = a0 [D] = x = 0.5a0 [C]2 [D] Now, K = [ A] [B]2 Now, substituting the values in above equation, we get (a )2 × (0.5a0 ) K= =4 (0.5a0 ) × (0.5a0 ) D (a) A and B (c) C and D (b) B and C (d) A and D Exp (a) From thermodynamics, −∆H° ∆S ° …(i) + lnk = 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 80 JEE Main Chapterwise Chemistry Exp (c) So, lines A and B in the graph represent temperature dependence of equilibrium constant K for an exothermic reaction as shown below A ln K B T(K) (0, 0) Its given that the final volume is 500 mL and this final volume was arrived when 50 mL of M Na2SO4 was added to unknown Ba2 + solution So, we can interpret the volume of unknown Ba2 + solution as 450 mL i.e 450mL + 50mL → 500mL Ba 2+ solution 22 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 1.0 × 10−7 and that of S2− from HS− ions is 1.2 × 10−13 then the concentration of S2− ions in aqueous solution is : [JEE Main 2018] (a) × 10−8 (b) × 10−20 (c) × 10−21 (d) × 10−19 Exp (b) Given [H2S] = 010 M [HCl] = 020 M So, [H+ ] = 020 M H2S H+ + HS− , K1 = 1.0 × 10−7 HS -H − + + + S2 − = 1.2 × 10−20 K × [H2S] [S2 − ] = [H+ ]2 = 1.2 × 10 [Ba 2+ ] = × 10−9 M Remember This is the concentration of Ba + ions in final solution Hence, for calculating the [Ba 2+ ] in original solution we have to use M1V1 = M V2 as M1 × 450 = 10−9 × 500 so, [according to the final equation] 1.2 × 10−20 × 0.1 M = (0.2M)2 −20 [Ba 2+ ][0.1] = × 10−10 or K = K1 × K = 10 × 10−7 × 1.2 × 10−13 Now From this we can calculate the concentration of SO24 − ion in the solution via M1V1 = M V2 × 50 = M × 500 (as 1M Na2SO4 is taken into consideration) = 01 M M2 = 10 Now for just precipitation, Ionic product = Solubility product (Ksp ) i.e [Ba 2+ ][SO 24 − ] = Ksp of BaSO So, × 10−13 + S2 − , K = 12 -2H −1 × × 10 M × 10−2 M = × 10−20 M 23 An aqueous solution contains an unknown concentration of Ba + When 50 mL of a M solution of Na 2SO is added, BaSO just begins to precipitate The final volume is 500 mL The solubility product of BaSO is × 10−10 What is the original concentration of Ba 2+ ? [JEE Main 2018] (a) × 10−9 M (b) × 10−9 M (c)11 × 10−9 M (d)10 × 10−10 M BaSO solution Given Ksp of BaSO = × 10−10 It means for, H2S Na 2SO solution M1 = 1.1 × 10−9 M 24 Which of the following are Lewis acids? [JEE Main 2018] (a) PH3 and BCl (b) AlCl and SiCl (c) PH3 and SiCl (d) BCl and AlCl Exp (d) / Lewis acids are defined as, ‘‘Electron deficient compounds which have the ability to accept atleast one lone pair.’’ The compound given are PH3 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 81 Equilibrium BCl3 Incomplete octet with following orbital picture 2p 2s 1s Vacant p- orbital BUsed in bond formation with Cl having one electron each from B and Cl Hence, vacant p-orbital of B can accept one lone pair thus it can be considered as Lewis acid AlCl -Similar condition is visible in AlCl as well i.e Al ( Valence orbital only) = 3s 3p Vacant p-orbital 3d vacant Used in bond formation with Cl Hence this compound can also be considered as Lewis acid SiCl - 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 is visible in following reaction Cl Cl + H2O Si Cl Cl Cl H O Si Cl H Cl Cl Lone pair acceptance in d-orbital Cl Si Cl OH + HCl 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 and AlCl 25 Which of the following salts is the most basic in aqueous solution? (a) Al(CN)3 (c) FeCl [JEE Main 2018] (b) CH3COOK (d) Pb(CH3COO)2 Exp (b) Among the given salts FeCl3 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(CH3 COO)2 are the salts of weak acid and weak base CH3 COOK 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 - (Strong base) (Weak acid) 26 An alkali is titrated against an acid with methyl orange as indicator, which of the following is a correct combination? [JEE Main 2018] 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 Exp (c) 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 3.9 –4.5 Yellow orange Weak base have the pH range greater than 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 range and reaches to end point near 3.9 where yellow orange colour of methyl orange changes to Pinkish red resulting to similar change in colour of solution as well 27 pK a of a weak acid (HA) and pK b of a weak base (BOH) are 3.2 and 3.4, respectively The pH of their salt (AB) solution is [JEE Main 2017 (Offline)] (a) 7.2 (b) 6.9 (c) 7.0 (d) 1.0 Exp (b) For a salt of weak acid and weak base, 1 pH = + pK a − pK b 2 Given, pK a (HA) = 3.2, pK a (BOH) = 3.4 1 pH = + (3.2) − (3.4) ∴ 2 = + 1.6 − 1.7 = 6.9 82 JEE Main Chapterwise Chemistry 28 The equilibrium constant at 298 K for a reaction, A + B C + D is 100 If the initial concentrations of all the four species were M each, then equilibrium concentration of D (in mol L−1 ) will be [JEE Main 2016 (Offline)] - (a) 0.818 (b) 1.818 (c) 1.182 (d) 0.182 Exp (b) A At equilibrium 1− x Keq = or ∴ O2 ( g ) SO3 ( g ), if K P = K C (RT )x where the symbols have usual meanings then, the value of x is (assuming ideality) [JEE Main 2014] SO2 ( g ) + + B - 1− x C D + 1+ x 1+ x [C ][D] (1 + x)(1 + x) (1 + x)2 = = [ A][B] (1 − x)(1 − x) (1 − x)2 + x 100 = 1 − x or e (b) − (a) −1 Initially at t = or 30 For the reaction, 10 = 1+ x 1− x 10 − 10 x = + x ⇒ 10 − = x + 10 x = 0.818 = 11 x ⇒ x = 11 [D] = + x = + 0.818 = 1818 29 The standard Gibbs energy change at 300K for the reaction, A B + C is 2494 J At a given time, the composition of the 1 reaction mixture is[ A] = ,[ B] = and[C ] = 2 The reaction proceeds in the [R = 8.314 JK / mol, e = 2.718] [JEE Main 2015] # (a) forward direction because Q > Kc (b) reverse direction because Q > Kc (c) forward direction because Q < Kc (d) reverse direction because Q < Kc Exp (b) Given, ∆G° = 2494.2J [ B][C] × (1 /2 ) = =4 Q= [ A]2 1 2 ∴ We know, ∆G = ∆G ° + RT ln Q = 24942 + 8.314 × 300 ln = 28747.27 J = positive value Also, we have Q ∆G = RT ln K if ∆G is positive, Q > K Therefore, reaction shifts in reverse direction (c) (d) Exp (b) By using this formula, Equilibrium constant, ∆n g K P = Kc (RT ) where ∆ng = (no of moles of products) − (no of moles of reactants) For the given reaction, SO ( g ) + O (g ) SO (g ) ∆ng = − + 1 = − = − 31 How many litres of water must be added to L of an aqueous solution of HCl with a pH of to create an aqueous solution with pH of 2? (a) 0.1 L (c) 2.0 L (b) 0.9 L [JEE Main 2014] (d) 9.0 L Exp (d) ∴ pH = [H+ ] = 10−1 = 0.1M ∴ [H+ ] = 10−2 = 0.01M pH = For dilution of HCl, M1V1 = M V2 0.1 × = 0.01 × V2 V2 = 10 L Volume of water to be added = 10 − 1= L 32 The equilibrium constant (K c ) for the reaction N ( g ) + O2 ( g ) → NO ( g ) at temperature T is × 10−4 The value of K c for the reaction 1 NO( g ) → N ( g ) + O2 ( g ) at the same 2 temperature is [AIEEE 2012] (a) 0.02 (b) 2.5 ×102 (c) × 10−4 (d) 50.0 83 Equilibrium Exp (d) 34 A vessel at 1000 K contains CO2 with a N2 (g ) + O (g ) → 2NO(g ) Kc = [NO]2 = × 10−4 [N2 ][O ] For the reverse reaction, 2NO(g ) → N2 (g ) + O (g ) [N ][O ] 1 104 Kc ′ = = 22 = = Kc [NO] × 10−4 On dividing the reverse reaction by 1 NO(g ) → N2 ( g ) + O ( g ) 2 [N ]1/ [O ]1/ Kc ′ ′ = = Kc ′ [NO] = pressure of 0.5 atm Some of the CO2 is converted into CO on the addition of graphite If the total pressure at equilibrium [AIEEE 2011] is 0.8 atm, the value of K p is (a) 1.8 atm (b) atm Exp (a) (a) × 10−1 (b) × 10−3 (c) × 10−5 (d) × 10−7 Exp (c) H Q = H+ + Q − [H+ ] = K aC by Ostwald’s dilution law [H+ ] = 10−pH = 10−3 M C = 0.1 M 10−3 = K a × 0.1 Total pressure of CO and CO gases pCO + pCC = ptotal 0.5 − ( p + p) = 0.8, p = 0.3 atm pCO = 0.5 − 0.3 = 0.2 atm pCO 0.6 × 0.6 = 18 atm 0.2 = 35 The K sp for Cr(OH)3 is 1.6 × 10−30 The molar solubility of this compound in water is [AIEEE 2011] (a) 1.6 × 10−30 (b) 1.6 × 10−30 (c) 1.6 × 10−30 / 27 (d)1.6 × 10−30 / 27 Exp (c) Let molar solubility of Cr(OH)3 = s mol L−1 Cr(OH)3 (s ) Cr 3+ (aq ) + 3OH− (aq ) s 3s −30 Ksp = 1.6 × 10 K a = 10−5 = [Cr 3+ ] [OH− ]3 = (s )(3s )3 = 27 s Alternate method HQ + 1H + Q− 0 Initial concentration 0.1 M Given, pH = 3, this suggests [H+ ] = 10−3 M at equilibrium Thus, [Q − ] at equilibrium = 10−3 M Hence, [HQ ] = 0.1M - 10 M -3 [Q10−3 M > 10 ] = × 10−10 M 37 Three reactions involving H 2PO–4 are given below (a) 1.2 × 10−10 g (c) 6.2 × 10−5 g Exp (b) [AgBr] = [Ag + ] = 0.05 M [AIEEE 2010] Ksp [AgBr] = [Ag+ ] [Br − ] I H 3PO4 + H 2O → H 3O+ + H 2PO4– II H 2PO4– + H 2O → HPO42– + H 3O+ III H 2PO4– + OH → H 3PO4 + O – [Br − ] = ⇒ 2– In which of the above does H 2PO–4 act as an acid? (a) II only (c) III only (b) 1.2 × 10−9 g (d) 50 × 10−8 g (b) I and II (d) I only Exp (a) Only in reaction (II), H2PO 4– , gives H+ to H2O Thus, behaves as an acid Ksp (AgBr ) [Ag+ ] 5.0 × 10-13 = 10−11 M [mol L −1] 0.05 Moles of KBr needed to precipitate AgBr = [Br − ] × V = 10−11 mol L −1 × 1L = 10−11 mol = Therefore, amount of KBr needed to precipitate AgBr = 10−11 mol × 120 g mol −1 = 1.2 × 10−9 g 40 At 25°C, the solubility product of Mg(OH) is aqueous solution, the ionisation constants for carbonic acid are [AIEEE 2010] K = 4.2 × 10−7 and K = 4.8 × 10−11 1.0 × 10−11 At which pH, will Mg 2+ ions start precipitating in the form of Mg(OH) from a solution of 0.001 M Mg 2+ ions? [AIEEE 2010] Select the correct statement for a saturated 0.034 M solution of the carbonic acid (a) 38 In (a) The concentration of CO 23 − is 0.034 M Mg(OH)2 − + HCO −3 + 2− K1 >> K ∴ [H+ ] = [HCO −3 ] K2 = [H+ ][CO 23 ] [HCO 3− ] So, [CO 23 − ] = K = 4.8 × 10−11 K1 = 4.2 × 10−7 K = 4.8 × 10−11 (d) r Mg 2+ + 2OH− Ksp = [Mg 2+ ] [OH– ]2 [OH− ] = Ksp [Mg 2+ ] = × 10−11 0.001 −1 pOH = − log [OH ] = − log [10−4 ] Exp (c) r H + HCO ; r H + CO ; (c) 11 Exp (b) (b) The concentration of CO 23 − is greater than that of HCO −3 (c) The concentration of H+ and HCO −3 are approximately equal (d) The concentration of H+ is double that ofCO 23 − H2CO (b) 10 ∴ pOH = and pH = 10 pH = 14 − pOH = 14 − = 10 41 Solid Ba(NO3 )2 is gradually dissolved in a 1.0 × 10−4 M Na 2CO3 solution At what concentration of Ba 2+ , will a precipitate begin to form? K sp for BaCO3 = 5.1 × 10−9 ) [AIEEE 2009] (a) 4.1 × 10−5 M (c) 8.1 × 10−8 M (b) 5.1 × 10−5 M (d) 8.1 × 10−7 M 85 Equilibrium Exp (b) Na CO a2Na [Na CO ] = BaCO + + [CO 23 − ] = 2+ CO 23 − −4 × 10 Since, strong acid has a weak conjugate base Thus, the basic strength of the conjugate bases of the given acid is opposite to their acidic strength.The basic strength of the conjugate bases of the given acids follows the order HSO −3 < H2O H3O+ > HSO −4 > HCO −3 Ksp (BaCO ) ]= [CO 23 − ] [Ba 2+ ] = 44 The equilibrium constants K p and K p for 5.1 × 10-9 [Ba 2+ ] = 5.1 × 10−5 M Hence, at 5.1 × 10−5 M concentration of Ba 2+ , a precipitate will begin to form 42 For the following three reactions I, II and III, equilibrium constants are given [AIEEE 2008] I CO ( g ) + H2O( g ) II CH4 ( g ) + H2O( g ) III CH4 (g) + 2H2O( g) a CO ( g ) + H ( g ); K a CO ( g ) +3H ( g ); K a CO ( g) 2 (a) : 36 Which of the following relations is correct? (b) K K = K (d) K K 23 = K 12 Exp (c) As equation ‘III’ is obtained on adding equation ‘I’ and equation ‘II’, so K = K1 ⋅ K III HSO–4 (c) : (d) : Let the total pressure at equilibrium for the given two reactions are p1 and p2 , respectively For first reaction, X a2 Y Initial mol At equilibrium (1 − α ) 2α Total moles = (1 − α ) + 2α [Here, α is the degree of dissociation] = (1 + α ) 1− α px = p1 1+ α [Q Partial pressure = (Total no of moles) × (Total pressure)] 2α py = p1 1+ α II H 3O+ IV HSO3F Which one of the following is the correct sequence of their acid strength? [AIEEE 2008] (a) IV < II < III < I (b) II < III < I < IV (c) I < III < II < IV (d) III < I < IV < II a Exp (a) 43 Four species are listed below I HCO–3 (b) : + 4H2 (g ); K (a) K K = K (c) K = K K the reactions X 2Y and Z P +Q , respectively are in the ratio of : If the degree of dissociation of X and Z be equal, then the ratio of total pressure at these equilibria is [AIEEE 2008] a × 10−4 K p1 = 2α p1 1 + α 1− α p1 1 + α = 4α − α2 P1 = 4α P1 [Qα 2 2 Exp (a) Exp (c) Change in volume affects number of moles per unit volume and move in the direction which undo the change N2 ( g ) + O ( g ) 2NO(g ) Number of moles of reactants and products are equal r 68 Which one of the following species acts as both Bronsted acid and base? [AIEEE 2002] (a) H2 PO –2 (b) HPO 23 – r r 71 pH of 0.005 M calcium acetate (pK a of CH 3COOH = 4.74) is (a) 7.04 (b) 9.37 [AIEEE 2002] (c) 9.26 (d) 8.37 Exp (a) Calcium acetate is a salt of weak acid and weak base 0.005 M calcium acetate, (CH3COO)2 Ca (CH3COO)2 Ca → Ca 2+ + 2CH3COO − (c) HPO 24 − (d) All of the above Exp (c) According to Bronsted Lowry concept of acids and bases, an acid is a proton donor whereas a base is a proton acceptor Thus, according to this concept, HPO 24 − can easily donate a proton and also accepts a proton on reaction with water HPO 24 − + H2O PO 34 − + H3O+ HPO 24 − NaCl is salt of strong acid and strong base Its not the case of buffer NaCl + H2O NaOH + HCl Aqueous NaCl, itself exact neutral solution HCl + H2O H3O+ + Cl − makes solution acidic r + H O r H PO 2 2− + OH− (2 × 0.005 = 0.01) 0.005 M ∴ [CH3COO − ] = 0.01 M CH3COO − + H2O r CH COOH + OH pK a log C + 2 log 0.01 = + 2.37 + = + 2.37 − 2.30 = 7.06 pH = + – Alkaline .. .JEE Main CHAPTERWISE SOLUTIONS 2019-2002 Chemistry All the 16 Question Papers of JEE Main Online 2019 (Jan & Apr Attempt) ARIHANT PRAKASHAN (Series) , MEERUT Arihant Prakashan (Series) ,... Published by Arihant Publications (India) Ltd For further information about the books published by Arihant log on to www.arihantbooks.com or email to info@arihantbooks.com /@arihantpub Arihant. .. length is C 2− (option-a) 30 JEE Main Chapterwise Chemistry Among the following, the molecule expected to be stabilised by anion formation is C , O2 , NO, F2 [JEE Main 2019, April Shift-I] (b)