43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry [16 ed.] 8194767733, 9788194767732

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Disha Publication feels proud to release its thoroughly Revised & Updated Edition of the Book “43 Years IIT-JEE Advanced + JEE Main Chapter-wise & Topic-wise Solved Paper CHEMISTRY”. This is the 16th Edition which speaks volumes about the Quality & Hard Work that has gone in building this book. It is an integrated book, which contains Chapter-wise & Topic-wise collection of past JEE Advanced (including1978 - 2012 IIT-JEE & 2013 2020 JEE Advanced) questions from 1978 to 2020 and past JEE Main from2013 – 2020 (including all Online & Offline Papers). •

The unique feature of this new edition is the division of questions into

31 chapters as per NCERT. With this new feature this book has become the 1st to adopt NCERT Chapterisation among JEE Advanced Solved Papers. •

Each chapter divides the questions into 2 - 4 topics which are further

divided into 10 categories of questions – 1.

MCQ 1 option correct

2.

Integer Answer

3.

Numeric Answer

4.

Fill in the Blanks

5.

True/False

6.

MCQ more than 1 option correct

7.

Multiple Matching

8.

Passage Based

9.

Assertion-Reason

10. Subjective Questions.

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•

All the Screening and Mains papers of IIT-JEE have been incorporated

in the book. All online papers of JEE Main including the 16 papers of 2020 & 16 papers of 2019 phase I & II have been incorporated in the book.

IIT-JEE/ JEE Advanced Papers • 1978 – 2012 IIT JEE Screening & Main • 2013 – 2020 JEE Advanced Papers 1 & 2

•

JEE Main (Online + Offline) Papers JEE Main 2013 – 5 (4 + 1) papers JEE Main 2014 – 4 (3 + 1) papers JEE Main 2015 – 3 (2 + 1) papers JEE Main 2016 – 3 (2 + 1) papers JEE Main 2017 – 3 (2 + 1) papers JEE Main 2018 – 4 (3 + 1) papers JEE Main 2019 – 16 Papers JEE Main 2020 – 16 Papers

Detailed solution of each and every question has been provided for

100% conceptual clarity of the student. Well elaborated detailed solutions with user friendly language provided at the end of the book.•

Solutions

have been given with enough diagrams, proper reasoning to bring

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conceptual clarity. •

The students are advised to attempt questions of a

topic immediately after they complete a topic in their class/ school/ home. The book contains around 4000 Milestone Problems in Chemistry. If utilised properly this book can take your JEE preparation to the next level. Although all efforts are made to ensure quality but some errors might have crept in. We would invite our readers to send us these errors so that we can incorporate the corrections in the upcoming editions. All the Best Disha Experts

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Class – XI 1.

Some Basic Concepts of Chemistry Topic 1 : Measurement , Mole Concept and

Percentage Composition Topic 2 : Stoichiometry, Equivalent Concept, Neutralization and Redox Titration 2.

Structure of Atom Topic 1 : Different Atomic Models that leads to

Bohr Model Topic 2 : Advancement towards Quantum Mechanical Model of Atom Topic 3 : Quantum Mechanical Model

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3.

Classification of Elements and Periodicity in

Properties Topic 1 : Periodic Classification Topic 2 : Periodic Properties 4.

Chemical Bonding and Molecular Structure Topic 1 : Lewis Approach to Chemical Bonding ,

Ionic Bond and Bond Parameters Topic 2 : VSEPR Theory and Hybridisation Topic 3 : VBT, MOT and Hydrogen Bonding 5.

States of Matter Topic 1 : Intermolecular Forces, , Gas Laws and

Ideal Gas Equation Topic 2 : Kinetic Theory of Gases and Molecular Speeds

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Topic 3 : Deviation from Ideal Gas Behaviour, Liquification of Gases and Liquid State 6.

Thermodynamics Topic 1 : Thermodynamics Topic 2 : Thermochemistry

7.

Equilibrium Topic 1 : Chemical Equlibrium Topic 2 : Ionic Equilibrium

8.

Redox Reactions Topic 1 : Oxidation and Reduction Reactions Topic 2 : Oxidation Number Topic 3 : Redox Reactions and Electrode

Processes 9.

Hydrogen

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Topic 1 : Preparation and properties of Hydrogen and Hydrides Topic 2 : Preparation and Properties of H2O and D2O Topic 3 : Preparation and Properties of H2O2 10. The s-Block Elements Topic 1 : Group -1 Elements (Alkali Metals) Topic 2 : Group -2 Elements (Alkaline Earth Metals) 11. The p-Block Elements (Group-13 and 14) Topic 1 : Group-13 Elements (Boron Family) Topic 2 : Group-14 Elements (Carbon Family) 12. Organic Chemistry — Some Basic Principles & Techniques

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Topic 1 : Classification and Nomenclature of Organic Compounds Topic 2 : Isomerism in Organic Compounds Topic 3 : Concept of Reaction Mechanism, Purification and Analysis of Organic Compounds 13. Hydrocarbons Topic 1 : Alkanes Topic 2 : Alkenes Topic 3 : Alkynes Topic 4 : Aromatic Hydrocarbons 14. Environmental Chemistry Topic 1 : Air Pollution Topic 2 : Water and Soil Pollution

Class – XII 15. The Solid State

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Topic 1 : Different types, Crystal Structures and Properties of Solids Topic 2 : Cubic Cystems, Bragg’s Equation and Imperfection in Solids 16. Solutions Topic 1 : Solution and Vapour Pressure of Liquid Solutions Topic 2 : Colligative Properties of Solutions 17. Electrochemistry Topic 1 : Conductance of Electrolytic Solution and Electrolysis Topic 2 : Cells, Nernst Equation, Commercial Cells and Corrosion 18. Chemical Kinetics and Nuclear Chemistry Topic 1 : Rate of Reactions, Order of Reactions and Half Life Period

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Topic 2 : Effect of Temperature and Catalyst on Rate of Reactions, Collosion Theory of Chemical Reactions Topic 3 : Nuclear Chemistry 19. Surface Chemistry Topic 1 : Adsorption Topic 2 : Catalysis and Theories of Catalysis Topic 3 : Colloids and Emulsions 20. General Principles and Processes of Isolation of Elements Topic 1 : Occurance of Metals and Metallurgical Processes Topic 2 : Purification and Uses of Metals 21. The p-Block Elements (Group 15, 16, 17 & 18)

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Topic 1 : Group-15 Elements (Nitrogen Family) Topic 2 : Group-16 Elements (Oxygen Family) Topic 3 : Group-17 Elements (Halogen Family) Topic 4 : Group-18 Elements (Noble Gases) 22. The d-and f-Block Elements Topic 1 : d-Block Elements Topic 2 : f-Block Elements 23. Co-ordination Compounds Topic 1 : Important Terms, Coordination Number, Nomenclature and Isomerism of Coordination Compounds Topic 2 : Bonding, Stability and Application of Coordination Compounds 24. Haloalkanes and Haloarenes Topic 1 : Preparation and Properties of Haloalkanes

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Topic 2 : Preparation and Properties of Haloarenes Topic 3 : Polyhalogen Compounds 25. Alcohols, Phenols and Ethers Topic 1 : Preparation and Properties of Alcohols Topic 2 : Preparation of Phenols Topic 3 : Preparation and Properties of Ethers 26. Aldehydes, Ketones and Carboxylic Acids Topic 1 : Preparation and Properties of Carbonyl Compounds Topic 2 : Preparation and Properties of Carboxylic Acids and its derivatives 27. Amines Topic 1 : Cyanides, Isocyanides, Nitriles and Nitro Compounds Topic 2 : Aliphatic and Aromatic Amines Topic 3 : Diazonium Salts

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28. Biomolecules Topic 1 : Carbohydrates and Lipids Topic 2 : Amino Acids and Proteins Topic 3 : Nucleic Acids, Enzymes, Vitamins and Hormones 29. Polymers Topic 1 : Classification of Polymers Topic 2 : Preparation, Properties and Uses of Polymers 30. Chemistry in Everyday Life Topic 1 : Drugs Classification, therapuetic action of drugs and Chemicals in Food Topic 2 : Cleansing Agents 31. Analytical Chemistry Topic 1 : Analysis of Inorganic Compounds Topic 2 : Analysis of Organic Compounds

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Hints & Solutions Class – XI 1. Some Basic Concepts of Chemistry 2. Structure of Atom 3. Classification of Elements and Periodicity in Properties 4. Chemical Bonding and Molecular Structure 5. States of Matter 6. Thermodynamics 7. Equilibrium 8. Redox Reactions 9. Hydrogen 10. The s-Block Elements 11. The p-Block Elements (Group-13 and 14) 12. Organic Chemistry — Some Basic Principles & Techniques 13. Hydrocarbons 14. Environmental Chemistry

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Class – XII 15. The Solid State 16. Solutions 17. Electrochemistry 18. Chemical Kinetics and Nuclear Chemistry 19. Surface Chemistry 20. General Principles and Processes of Isolation of Elements 21. The p-Block Elements (Group 15, 16, 17 & 18) 22. The d-and f-Block Elements 23. 24. 25. 26. 27. 28. 29.

Co-ordination Compounds Haloalkanes and Haloarenes Alcohols, Phenols and Ethers Aldehydes, Ketones and Carboxylic Acids Amines Biomolecules Polymers

30. Chemistry in Everyday Life 31. Analytical Chemistry

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1.

A solution of two components containing n1 moles of the 1st component and n2 moles of the 2nd component is prepared. M1 and M2 are the molecular weights of component 1 and 2 respectively. If d is the density of the solution in g mL–1, C2 is the molarity and x2 is the mole fraction of the 2nd component, then C2 can be expressed as: [Main Sep. 06, 2020 (I)]

(a) C2 = (b) C2 = (c) C2 = (d) C2 = 2.

The average molar mass of chlorine is 35.5 g mol–1. The ratio of 35Cl to 37Cl in naturally occrring chlorine is close to: [Main Sep. 06, 2020 (II)]

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(a) (b) (c) (d) 3. (a) (b) (c)

(d) 4.

(a) (b) (c) (d) 5.

(a) (b) (c)

4:1 3:1 2:1 1:1 Amongst the following statements, that which was not proposed by Dalton was: [Main Jan. 07, 2020 (I)] Chemical reactions involve reorganization of atoms. These are neither created nor destroyed in a chemical reaction. All the atoms of a given element have identical properties including identical mass. Atoms of different elements differ in mass. When gases combine or reproduced in a chemical reaction they do so in a simple ratio by volume provided all gases are at the same T & P. Matter consists of indivisible atoms. 8 g of NaOH is dissolved in 18 g of H2O. Mole fraction of NaOH in solution and molality (in mol kg–1) of the solution respectively are : [Main Jan. 12, 2019 (II)] 0.2, 22.20 0.2, 11.11 0.167, 11.11 0.167, 22.20 The amount of sugar (C12H22O11) required to prepare 2 L of its 0.1 M aqueous solution is: [Main Jan. 10, 2019 (II)] 136.8 g 17.1 g 68.4 g

(d) 34.2 g 6. The ratio of mass percent of C and H of an organic compound (CXHYOZ) is 6 : 1. If one molecule of the above compound (CXHYOZ) contains half as much oxygen as required to burn one molecule of

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compound CXHY completely to CO2 and H2O. The empirical formula of compound CXHYOZ is : [Main 2018] (a) (b) (c) (d) 7.

C3H6O3 C2H4O C3H4O2 C2H4O3 An unknown chlorohydrocarbon has 3.55% of chlorine. If each molecule of the hydrocarbon has one chlorine atom only, chlorine atoms present in 1g of chlorohydrocarbon are: (Atomic wt. of Cl = 35.5u; Avogadro constant = 6.023 × 1023 mol–1 ) [Main Online April 16, 2018] (a) 6.023 × 109 (b) 6.023 × 1023 (c) 6.023 × 1021 (d) 6.023 × 1020 8. The most abundant elements by mass in the body of a healthy human adult are : [Main 2017] Oxygen (61.4%) ; Carbon (22.9%), Hydrogen (10.0%) ; and Nitrogen (2.6%). The weight which a 75 kg person would gain if all 1H atoms are replaced by 2H atoms is (a) 15 kg (b) 37.5 kg (c) 7.5 kg (d) 10 kg 9. What quantity (in mL) of a 45% acid solution of a mono-protic strong acid must be mixed with a 20% solution of the same acid to produce 800 mL of a 29.875% acid solution? [Main Online April 9, 2017] (a) 320 (b) 325 (c) 316 (d) 330 10. 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

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(a) (b) (c) (d) 11.

(a) (b) (c) (d) 12.

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: [Main 2016] C4H8 C4H10 C3H6 C3H8 5 L of an alkane requires 25 L of oxygen for its complete combustion. If all volumes are measured at constant temperature and pressure, the alkane is : [Main Online April 9, 2016] Isobutane Ethane Butane Propane The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3– Na+ (Mol. wt. 206. What would be the maximum uptake of Ca2 + ions by the resin when expressed in mole per gram resin ? [Main 2015]

(a) (b) (c) (d) 13.

Choose the incorrect formula out of the four compounds for an element X below :[Main Online April 11, 2015] (a) X2O3 (b) X2Cl3 (c) X2(SO4)3 (d) XPO4 14. Dissolving 120 g of a compound of (mol. wt. 60) in 1000 g of water gave a solution of density 1.12 g/mL. The molarity of the solution is:

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[Main Online April 9, 2014] (a) (b) (c) (d) 15.

1.00 M 2.00 M 2.50 M 4.00 M A gaseous hydrocarbon gives upon combustion 0.72 g of water and 3.08 g. of CO2. The empirical formula of the hydrocarbon is : [Main 2013] (a) C2H4 (b) C3H4 (c) C6H5 (d) C7H8 16. 10 mL of 2(M) NaOH solution is added to 200 mL of 0.5 (M) of NaOH solution. What is the final concentration ? [Main Online April 25, 2013] (a) 0.57 (M) (b) 5.7 (M) (c) 11.4 (M) (d) 1.14 (M) 17. The density of 3M solution of sodium chloride is 1.252 g mL–1. The molality of the solution will be : (molar mass, NaCl = 58.5 g mol–1) [Main Online April 22, 2013] (a) 260 m (b) 2.18 m (c) 2.79 m (d) 3.00 m 18. Which has maximum number of atoms? [2003S] (a) 24g of C (12) (b) 56g of Fe (56) (c) 27g of Al (27)

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(d) 108g of Ag (108) 19.

How many moles of electron weigh one kilogram? [2002S]

(a) (b) (c) (d) 20. (a) (b) (c) (d) 21. (a) (b) (c) (d) 22. (a) (b) (c) (d) 23. (a) (b) (c)

If two compounds have the same empirical formula but different molecular fomulae they must have [1987 - 1 Mark] different percentage composition different molecular weight same viscosity same vapour density The largest number of molecules is in [1979] 36 g of water 28 g of carbon monoxide 46 g of ethyl alcohol 54 g of nitrogen pentoxide The total number of electrons in one molecule of carbon dioxide is [1979] 22 44 66 88 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] 1:4 1:8 7 : 32

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(d) 3 : 16 24. 27 g of Al will react completely with how many grams of oxygen? [1978] (a) 8 g (b) 16 g (c) 32 g (d) 24 g 25. A compound was found to contain nitrogen and oxygen in the ratio 28 g and 80 g respectively. The formula of compound is [1978] (a) NO (b) N2O3 (c) N2O5 (d) N2O4 26.

27.

28.

29. 30.

The ratio of the mass percentages of ‘C & H’ and ‘C & O’ of a saturated acyclic organic compound ‘X’ are 4 : 1 and 3 : 4 respectively. Then, the moles of oxygen gas required for complete combustion of two moles of organic compound ‘X’ is _________. [Main Sep. 02, 2020 (II)] The molarity of HNO3 in a sample which has density 1.4 g/mL and mass percentage of 63% is ______ . (Molecular Weight of HNO3 = 63) [Main Jan. 09, 2020 (I)] If the value of Avogadro number is 6.023 × 1023 mol–1 and the value of Boltzmann constant is 1.380 × 10–23 J K–1, then the number of significant digits in the calculated value of the universal gas constant is [Adv. 2014] Calculate the molarity of water if its density is 1000 kg/m3. [2003 - 2 Marks] . What The composition of a sample of Wurtzite is percentage of the iron is present in the form of Fe (III)? [1994 - 2 Marks]

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31.

A compound contains 28 percent of nitrogen and 72 percent of metal by weight. 3 atoms of metal combine with 2 atoms of N. Find the atomic weight of metal. [1980]

32.

The weight of 1 × 1022 molecules of CuSO4.5H2O is ............... [1991 - 1 Mark] 33. The modern atomic mass unit is based on ........................... [1980] 34. The total number of electrons present in 18 mL of water is ....................... [1980]

35.

36.

(b)

(c)

37.

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 cm3/g. If the virus is considered to be a single particle, find its molar mass. [1999 - 3 Marks] (a) One litre of a sample of hard water contains 1 mg of CaCl2 and 1 mg of MgCl2. Find the total hardness in terms of parts of CaCO3 per 106 parts of water by weight. A sample of hard water contains 20 mg of Ca++ ions per litre. How many milli-equivalent of Na2CO3 would be required to soften 1 litre of the sample? 1 g of Mg is burnt in a closed vessel which contains 0.5 g of O2. (i) Which reactant is left in excess? (ii) Find the weight of the excess reactants? (iii) How may milliliters of 0.5 N H2SO4 will dissolve the residue in the vessel. [1980] A hydrocarbon contains 10.5g of carbon per gram of hydrogen. 1 litre of the vapour of the hydrocarbon at 127°C and 1 atmosphere pressure weighs 2.8g. Find the molecular formula.

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[1980] 38. Find [1980] (i) The total number of neutrons and (ii) The total mass of neutron in 7 mg of 14C. (Assume that mass of neutron = mass of hydrogen atom)

1.

The ammonia (NH3) released on quantitative reaction of 0.6 g urea (NH2CONH2) with sodium hydroxide (NaOH) can be neutralized by: [Main Jan. 07, 2020 (II)] (a) 200 mL of 0.4 N HCl (b) 200 mL of 0.2 N HCl (c) 100 mL of 0.2 N HCl (d) 100 mL of 0.1 N HCl 2. For a reaction, N2(g) + 3H2(g) → 2 NH3(g); identify dihydrogen (H2) as a limiting reagent in the following reaction mixtures. [Main April 9, 2019 (I)] (a) 56 g of N2 + 10 g of H2 (b)35 g of N2 + 8 g of H2 (c) 28 g of N2 + 6 g of H2 (d) 14 g of N2 + 4 g of H2 3. A 10 mg effervescent tablet containing sodium bicarbonate and oxalic acid releases 0.25 mL of CO2 at T = 298.15K and P = l bar. If molar volume of CO2 is 25.0 L under such condition, what is the percentage of sodium bicarbonate in each tablet ? [Main Jan. 11, 2019 (I)] –1 [Molar mass of NaHCO3 = 84 g mol ] (a) 0.84 (b) 33.6 (c) 16.8 (d) 8.4

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4.

A sample of NaClO3 is converted by heat to NaCl with a loss of 0.16 g of oxygen. The residue is dissolved in water and precipitated as AgCl. The mass of AgCl (in g) obtained will be: (Given: Molar mass of AgCl = 143.5 g mol–1 ) [Main Online April 15, 2018 (I)] (a) 0.35 (b) 0.54 (c) 0.41 (d) 0.48 5. 1 gram of a carbonate (M2CO3) on treatment with excess HCl produces 0.01186 mole of CO2. The molar mass of M2CO3 in g mol–1 is : [Main 2017] (a) 1186 (b) 84.3 (c) 118.6 (d) 11.86 6. The amount of arsenic pentasulphide that can be obtained when 35.5 g arsenic acid is treated with excess H2S in the presence of conc. HCl (assuming 100% conversion) is : [Main Online April 9, 2016] (a) 0.25 mol (b) 0.50 mol (c) 0.333 mol (d) 0.125 mol 7. 3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06N) 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: [Main 2015] (a) 42 mg (b) 54 mg

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(c) 18 mg (d) 36 mg 8.

A sample of a hydrate of barium chloride weighing 61 g was heated until all the water of hydration is removed. The dried sample weighed 52 g. The formula of the hydrated salt is : (atomic mass, Ba = 137 amu, Cl = 35.5 amu) [Main Online April 10, 2015]

(a) BaCl2.4H2O (b) BaCl2.3H2O (c) BaCl2.H2O (d) BaCl2.2H2O 9.

Mixture X = 0.02 mol of [Co(NH3)5SO4]Br and 0.02 mol of [Co(NH3)5Br]SO4 was prepared in 2 litre of solution.[2003S] 1 litre of mixture X + excess AgNO3 →Y. 1 litre of mixture X + excess BaCl2 →Z No. 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 10. 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 neutralize 10 mL of this solution is [2001S] (a) 40 mL (b) 20 mL (c) 10 mL (d) 4 mL 11. In the standardization of Na2S2O3 using K2Cr2O7 by iodometry, the equivalent weight of K2Cr2O7 is [2001S] (a) (molecular weight)/2 (b) (molecular weight)/6

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(c) (molecular weight)/3 (d) same as molecular weight 12. The normality of 0.3 M phosphorous acid (H3PO3) is, [1999 - 2 Marks] (a) 0.1 (b) 0.9 (c)0.3 (d) 0.6 13. The equivalent weight of MnSO4 is half of its molecular weight when it is converted to : [1988 - 1 Mark] (a) Mn2O3 (b) MnO2 (c) (d) 14. (a) (b) (c) (d) 15. (a) (b) (c) (d) 16.

(a) (b) (c) (d)

In which mode of expression, the concentration of a solution remains independent of temperature? [1988 - 1 Mark] Molarity Normality Formality Molality A molal solution is one that contains one mole of a solute in: [1986 - 1 Mark] 1000 g of the solvent one litre of the solvent one litre of the solution 22.4 litres of the solution If 0.50 mole of BaCl2 is mixed with 0.20 mol of Na3PO4, the maximum number of moles of Ba3(PO4)2 that can be formed is [1981 - 1 Mark] 0.70 0.50 0.20 0.10

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17. M is molecular weight of KMnO4. The equivalent weight of KMnO4 when it is converted into K2MnO4 is [1980] (a) M (b) M/3 (c) M/5 (d) M/7 18. 2.76 g of silver carbonate on being strongly heated yields a residue weighing [1979] (a) 2.16 g (b) 2.48 g (c) 2.32 g (d) 2.64 g 19.

The minimum number of moles of O2 required for complete combustion of 1 mole of propane and 2 moles of butane is ______. [Main Sep. 05, 2020 [I)] 20. The volume, in mL, of 0.02 M K2Cr2O7 solution required to react with 0.288 g of ferrous oxalate in acidic medium is _________. (Molar mass of Fe = 56 g mol–1) [Main Sep. 05, 2020 (II)] 21. A 20.0 mL solution containing 0.2 g impure H2O2 reacts completely with 0.316 g of KMnO4 in acid solution. The purity of H2O2 (in %) is __________ (mol. wt. of H2O2 = 34; mol. wt. of KMnO4 = 158)

[Main Sep. 04, 2020 (I)] 22. The mass of ammonia in grams produced when 2.8 kg of dinitrogen quantitatively reacts with 1 kg of dihydrogen is __________. [Main Sep. 04, 2020 (I)] 23. A 100 mL solution was made by adding 1.43 g of The normality of the solution is 0.1 N. The value of x is ___________. [Main Sep. 04, 2020 [II)] (The atomic mass of Na is 23 g/ mol)

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24.

The mole fraction of glucose (C6H12O6) in an aqueous binary solution is 0.1. The mass percentage of water in it, to the nearest integer, is ______. [Main Sep. 03, 2020 (I)] 25. 6.023 × 1022 molecules are present in 10 g of a substance 'x'. The molarity of a solution containing 5 g of substance 'x' in 2 L solution is __________ × 10–3. [Main Sep. 03, 2020 [II)] 26. The volume (in mL) of 0.1 N NaOH required to neutralise 10 mL of 0.1 N phosphinic acid is _________. [Main Sep. 03, 2020 [II)] 27. 10.30 mg of O2 is dissolved into a liter of sea water of density 1.03 g/mL. The concentration of O2 in ppm is _____. [Main Jan. 09, 2020 (II)] 28. Ferrous sulphate heptahydrate is used to fortify foods with iron. The amount (in grams) of the salt required to achieve 10 ppm of iron in 100 kg of wheat is ______.Atomic weight: Fe = 55.85; S = 32.00; O = 16.00 [Main Jan. 08, 2020 (I)] 29. NaClO3 is used, even in spacecrafts, to produce O2. The daily consumption of pure O2 by a person is 492 L at 1 atm, 300 K. How much amount of NaClO3, in grams, is required to produce O2 for the daily consumption of a person at 1 atm, 300 K? ________. NaClO3(s) + Fe(s) → O2(g) + NaCl (s) + FeO(s) R = 0.082 Latm mol–1 K–1 [Main Jan. 08, 2020 (II)] 30. In the chemical reaction between stoichiometric quantities of KMnO4 and KI in weakly basic solution, what is the number of moles of I2 released for 4 moles of KMnO4 consumed? [Adv. 2020] 31. The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by NiCl2.6H2O to form a stable coordination compound. Assume that both the reactions are 100% complete. If 1584 g of ammonium sulphate and 952 g of NiCl2.6H2O are used in the preparation, the combined weight (in grams) of gypsum and the nickel-ammonia coordination compound thus produced is________.

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(Atomic weights in g mol–1: H = 1, N = 14, O = 16, S = 32, Cl = 35.5, Ca = 40, Ni = 59) [Adv. 2018] 32. Galena (an ore) is partially oxidized 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 contents undergo self-reduction. The weight (in kg) of Pb produced per kg of O2 consumed is______. (Atomic weights in g mol–1: O = 16, S = 32, Pb = 207) [Adv. 2018] 33. How many millilitres of 0.5 M H2SO4 are needed to dissolve 0.5 g of copper(II) carbonate? [1999 - 3 Marks] 34. 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 filterate is titrated with (M/10) KIO3 solution in presence of 6M HCl till all I– ions are converted into ICl. It requires50 mL. of (M/10) KIO3 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 + 3H2O) [1992 - 4 Marks] 35. A 1.0 g sample of Fe2O3 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 upto 100.0 mL. An aliquot of 25.0 mL of this solution requires 17.0 mL of 0.0167 M solution of an oxidant for titration. Calculate the number of electrons taken up by the oxidant in the reaction of the above titration. [1991 - 4 Marks] 36. Calculate the molality of 1 litre solution of 93% H2SO4 (weight/volume). The density of the solution is 1.84 g/mL. [1990 - 1 Marks] 37. A sample of hydrazine sulphate (N2H6SO4) 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

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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. [1988 - 3 Marks] Reaction : 4Fe3+ + N2H4 → N2 + 4Fe2+ + 4H+ + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O. 38. Hydroxylamine reduces iron (III) according to the equation: 2NH2OH + 4 Fe3+ → N2O(g) ↑ + H2O + 4 Fe2+ + 4H+ Iron (II) thus produced is estimated by titration with a standard permanganate solution. The reaction is : + 5 Fe2+ + 8H+ → Mn2+ + 5 Fe3+ + 4H2O A 10 mL. sample of hydroxylamine solution was diluted to 1 litre. 50 mL. of this diluted solution was boiled with an excess of iron (III) solution. The resulting solution required 12 mL. of 0.02 M KMnO4 solution for complete oxidation of iron (II). Calculate the weight of hydroxylamine in one litre of the original solution. (H = 1, N = 14, O = 16, K = 39, Mn = 55, Fe = 56) [1982 - 4 Marks] 39. A 1.00 g sample of H2O2 solution containing X per cent H2O2 by weight requires X mL of a KMnO4 solution for complete oxidation under acidic conditions. Calculate the normality of the KMnO4 solution. [1981 - 3 Marks] 40. 4.215 g of a metallic carbonate was heated in a hard glass tube and the CO2 evolved was found to measure 1336 mL at 27°C and 700 mm pressure. What is the equivalent weight of the metal? [1979] 41. What weight of AgCl will be precipitated when a solution containing 4.77 g of NaCl is added to a solution of 5.77 g of AgNO3? [1978] 42. Igniting MnO2 converts it quantitatively to Mn3O4. A sample of pyrolusite is of the following composition : MnO2 80%, SiO2 and

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other inert constituents 15%, rest being water. The sample is ignited in air to constant weight. What is the percentage of Mn in the ignited sample? [1978] [O = 16, Mn = 54.9] 43.

3 g of a salt of molecular weight 30 is dissolved in 250 g of water. The molality of the solution is ............... . [1983 - 1 Mark]

44.

Read the following statement and explanation and answer as per the options given below : Statement(S) : In the titration of Na2CO3 with HCl using methyl orange indicator, the volume required at the equivalence point is twice that of the acid required using phenolphthalein indicator. explanation(E) : Two moles of HCl are required for the complete neutralization of one mole of Na2CO3. [1991 - 2 Marks] Both S and E are true, and E is the correct explanation of S. Both S and E are true, but E is not the correct explanation of S. S is true but E is false. S is false but E is true.

(a) (b) (c) (d) 45.

Hydrogen peroxide solution (20 mL) reacts quantitatively with a solution of KMnO4 (20 mL) acidified with dilute H2SO4. 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 H2SO4. Write the balanced equations involved in the reactions and calculate the molarity of H2O2. [2001 - 5 Marks]

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46.

47.

48.

49.

50.

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 to decolourise the blue starch-iodine complex. Calculate the molarity of the sodium thiosulphate solution. [1998 - 5 Marks] A 3.00 g sample containing Fe3O4, Fe2O3 and an inert impure substance, is treated with excess of KI solution in presence of dilute H2SO4.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 Na2S2O3 solution to reduce the iodine present. A 50 mL of the diluted solution, after complete extraction of the iodine requires 12.80 mL of 0.25 M KMnO4 solution in dilute H2SO4 medium for the oxidation of Fe2+. Calculate the percentages of Fe2O3 and Fe3O4 in the original sample. [1996 - 5 Marks] –2 8.0575 × 10 kg of Glauber’s salt is dissolved in water to obtain 1 dm3 of a solution of density 1077.2 kg m–3. Calculate the molarity, molality and mole fraction of Na2SO4 in the solution. [1994 - 3 Marks] Upon mixing 45.0 mL. of 0.25 M lead nitrate solution with 25.0 mL of 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 the species left behind in the final solution. Assume that lead sulphate is completely insoluble. [1993 - 3 Marks] 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 1.5g of the same sample requires 150 mL. of (M/10) HCl for complete neutralisation. Calculate the % composition of the components of the mixture. [1992 - 5 Marks]

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51.

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

ions

on titration with 0.02 M KMnO4 in presence of H2SO4 consumes 22.6 mL. of the oxidant. The resultant solution is neutralized with Na2CO3, acidified with dil. acetic acid and treated with excess KI. The liberated iodine requires 11.3 mL of 0.05 M Na2S2O3 solution for in the complete reduction. Find out the molar ratio of Cu2+ to compound. Write down the balanced redox reactions involved in the above titrations. [1991 - 5 Marks] 52. A mixture of H2C2O4 (oxalic acid) and NaHC2O4 weighing 2.02 g was dissolved in water and solution made upto one litre. Ten millilitres of the solution required 3.0 mL. of 0.1 N sodium hydroxide solution for complete neutralization. In another experiment, 10.0 mL. of the same solution, in hot dilute sulphuric acid medium. require 4.0 mL. of 0.1 N potassium permanganate solution for complete reaction. Calculate the amount of H2C2O4 and NaHC2O4 in the mixture. [1990 - 5 Marks] 53. A solid mixture (5.0 g) consisting of lead nitrate and sodium nitrate was heated below 600ºC 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 - 4 Marks] 54. An equal volume of a reducing agent is titrated separately with 1M KMnO4 in acid neutral and alkaline media. The volumes of KMnO4 required are 20 mL. in acid, 33.4 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 reactions. Find out the volume of 1M K2Cr2O7 consumed; if the same volume of the reducing agent is titrated in acid medium. [1989 - 5 Marks] 55. A sugar syrup of weight 214.2 g contains 34.2 g of sugar (C12H22O11). Calculate : (i) molal concentration and (ii) mole fraction of sugar in the syrup.

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56. (i)

[1988 - 2 Marks] What is the weight of sodium bromate and molarity of solution necessary to prepare 85.5 mL of 0.672 N solution when the halfcell reaction is + 6H+ + 6e– → Br– + 3H2O

(ii) What would be the weight as well as molarity if the half-cell reaction is : + 12H+ + 10e– → Br2 + 6H2O [1987 - 5 Marks] 57. Five mL of 8N nitric acid, 4.8 mL of 5N hydrochloric acid and a certain volume of 17M sulphuric acid are mixed together and made upto 2litre. Thirty mL. of this acid mixture exactly neutralise 42.9 mL of sodium carbonate solution containing one gram of Na2CO3.10H2O in 100 mL. of water. Calculate the amount in gram of the sulphate ions in solution. [1985 - 4 Marks] 58. 2.68 × 10–3 moles of a solution containing an ion An+ require 1.61 × for the oxidation of An+ to in acid 10–3 moles of medium. What is the value of n? [1984 - 2 Marks] 59. The density of a 3 M sodium thiosulphate solution (Na2S2O3) is 1.25 g per mL. Calculate (i) the percentage by weight of sodium thiosulphate, (ii) the mole fraction of sodium thiosulphate and (iii) the molalities of Na+ and S2O32– ions. [1983 - 5 Marks] 60. 4.08 g of a mixture of BaO and an 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 neutralization. Identify the metal M. [1983 - 4 Marks] (At. wt. H = 1, C = 12, O = 16, Cl = 35.5, Ba = 138) 61. (i) A sample of MnSO4.4H2O is strongly heated in air. The residue is Mn3O4.

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(ii) The residue is dissolved in 100 mL of 0.1 N FeSO4 containing dilute H2SO4. (iii) The solution reacts completely with 50 mL of KMnO4 solution. (iv) 25 mL of the KMnO4 solution used in step (iii) requires 30 mL of 0.1 N FeSO4 solution for complete reaction. Find the amount of MnSO4.4H2O present in the sample. [1980] 62. A mixture contains NaCl and unknown chloride MCl. (i) 1 g of this is dissolved in water. Excess of acidified AgNO3 solution is added to it. 2.567 g of white ppt. is formed. (ii) 1 g of original mixture is heated to 300°C. Some vapours come out which are absorbed in acidified AgNO3 solution, 1.341 g of white precipitate was obtained. Find the molecular weight of unknown chloride. [1980] 63. 5 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 an electric spark. After the explosion, the volume of the mixed gases remaining was 25 mL. On adding a concentrated solution of potassium hydroxide, the volume further diminished to 15 mL of the residual gas being pure oxygen. All volumes have been reduced to N.T.P. Calculate the molecular formula of the hydrocarbon gas. [1979] 64. One gram of an alloy of aluminium and magnesium when treated with excess of dil. HCl forms magnesium chloride, aluminium chloride and hydrogen. The evolved hydrogen, collected over mercury at 0°C has a volume of 1.20 litres at 0.92 atm. pressure. Calculate the composition of the alloy. [H = 1, Mg = 24, Al = 27] [1978]

Topic-1 : Measurement, Mole Concept and Percentage Composition

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1. (c) 2. (b) 3. (c) 4. (c) 5. (c) 6. (d) 7. (d) 8. (c) 9. (c) 10. (N) 11. (d) 12. (b) 13. (b) 14. (b) 15. (d) 16. (a) 17. (c) 18. (a) 19. (d) 20. (b) 21. (a) 22. (a) 23. (c) 24. (d) 25. (c) 26. (5) 27. (14.00) 28. (4) 29. (55.55) 30. (15.05) 31. (24) Topic-2 : Measurement, Mole Concept and Percentage Composition 1. (c) 2. (a) 3. (d)

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4. (d) 5. (b) 6. (d) 7. (c) 8. (d) 9. (a) 10. (a) 11. (b) 12. (d) 13. (b) 14. (d) 15. (a) 16. (d) 17. (a) 18. (a) 19. (18) 20. (50) 21. (85) 22. (3400) 23. (10) 24. (47) 25. (25) 26. (10) 27. (10) 28. (4.96) 29. (2130) 30. (6) 31. (2992) 32. (6.47) 33. (8.09) 34. (85) 35. (6.0) 36. (10.43) 37. (6.5) 38. (39.6) 39. (0.58)

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40. (12.15) 41. (4.87) 42. (59.33) 44. (b)

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1.

(a) (b) (c) (d) 2. (a) (b) (c) (d) 3.

The difference between the radii of 3rd and 4th orbits of Li2+ is ∆R1. The difference between the radii of 3rd and 4th orbits of He+ is ∆R2. Ratio ∆R1 : ∆R2 is : [Main Sep. 05, 2020 (I)] 8:3 3:8 2:3 3:2 The region in the electromagnetic spectrum where the Balmer series lines appear is : [Main Sep. 04, 2020 (I)] Visible Microwave Infrared Ultraviolet The shortest wavelength of H atom in the Lyman series is λ1.The longest wavelength in the Balmer series is He+ is : [Main Sep. 04, 2020 (II)]

(a) (b) (c)

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(d) 4.

For the Balmer series in the spectrum of H atom,

the correct

statements among (I) to (IV) are: [Main Jan. 08, 2020 (I)] (I) As wavelength decreases, the lines in the series converge (II) The integer n1 is equal to 2 (III)The lines of longest wavelength corresponds to n2 = 3 (IV) The ionization energy of hydrogen can be calculated from wave number of these lines (a) (I), (III), (IV) (b) (I), (II), (III) (c) (I), (II), (IV) (d) (II), (III), (IV) 5. The radius of the second Bohr orbit, in terms of the Bohr radius, a0, in Li2+ is: [Main Jan. 08, 2020 (II)] (a) (b) (c) (d) 6. (a) (b) (c) (d) 7.

(a) (b) (c)

Among the following, the energy of 2s orbital is lowest in: [Main April 12, 2019 (II)] K H Li Na The ratio of the shortest wavelength of two spectral series of hydrogen spectrum is found to be about 9. The spectral series are : [Main April 10, 2019 (II)] Lyman and Paschen Balmer and Brackett Brackett and Pfund

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(d) Paschen and Pfund 8. For any given series of spectral lines of atomic hydrogen, let be the difference in maximum and minimum frequencies in cm–1. The ratio ∆ Lyman/∆ Balmeris : [Main April 9, 2019 (I)] (a) 4 : 1 (b) 9 : 4 (c) 5 : 4 (d) 27 : 5 9. What is the work function of the metal if the light of wavelength 4000Å generates photoelectrons of velocity 6×105 ms-1 from it ? [Main Jan. 12, 2019 (I)] –31 (Mass of electron=9×10 kg Velocity of light = 3×108ms–1 Planck’s constant=6.626×10–34 Js Charge of electron =1.6×10–19 JeV–1) (a) 0.9 eV (b) 3.1 eV (c) 2.1 eV (d) 4.0 eV 10. Heat treatment of muscular pain involves radiation of wavelength of about 900 nm. Which spectral line of H-atom is suitable for this purpose ? [Main Jan. 11, 2019 (I)] 5 –1 –34 8 –1 [RH = 1 × 10 cm . h = 6.6 × 10 Js, c = 3 × l0 ms ] (a) Paschen, ∞ → 3 (b) Paschen, 5 → 3 (c) Balmer, ∞ → 2 (d) Lyman, ∞ → 1 11. The ground state energy of hydrogen atom is –13.6 eV. The energy of second excited state of He+ ion in eV is: [Main Jan. 10, 2019 (II)] (a) – 54.4 (b) – 3.4 (c) – 6.04 (d) – 27.2 12. Which of the following statements is false? [Main Online April 16, 2018] (a) Splitting of spectral lines in electrical field is called Stark effect

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(b) Frequency of emitted radiation from a black body goes from a lower wavelength to higher wavelength as the temperature increases (c) Photon has momentum as well as wavelength (d) Rydberg constant has unit of energy 13. Ejection of the photoelectron from metal in the photoelectric effect experiment can be stopped by applying 0.5V when the radiation of 250 nm is used. The work function of the metal is : [Main Online April 15, 2018 (I)] (a) 4 eV (b) 5.5 eV (c) 4.5 eV (d) 5 eV 14. The radius of the second Bohr orbit for hydrogen atom is : (Plank's const. h = 6.6262 × 10–34 Js ; mass of electron = 9.1091 × 10–31 kg ; charge of electron e = 1.60210 × 10–19 C ; permittivity of vaccum ∈0 = 8.854185 × 10–12 kg–1 m–3 A2) [Main 2017] (a) 1.65Å (b) 4.76Å (c) 0.529Å (d) 2.12Å 15. If the shortest wavelength in Lyman series of hydrogen atom is A, then the longest wavelength in Paschen series of He+ is : [Main Online April 8, 2017] (a) (b) (c) (d) 16.

A stream of electrons from a heated filaments was passed 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: [Main 2016]

(a)

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(b) (c) meV (d) 2meV 17. Which of the following is the energy of a possible excited state of hydrogen ? [Main 2015] (a) –3.4 eV (b) +6.8 eV (c) +13.6 eV (d) –6.8 eV 18. If m and e are the mass and charge of the revolving electron in the orbit of radius r for hydrogen atom, the total energy of the revolving electron will be: [Main Online April 12, 2014] (a) (b) (c) (d) 19.

If λ0 and λ be threshold wavelength and wavelength of incident light, the velocity of photoelectron ejected from the metal surface is: [Main Online April 11, 2014]

(a) (b) (c) (d) 20.

Energy of an electron is given by E = – 2.178 × Wavelength of light required to excite an electron in an hydrogen atom from level n = 1 to n = 2 will be: [Main 2013]

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(h = 6.62 × 10 –34 Js and c = 3.0 × 108 ms–1) (a) 1.214 × 10–7 m (b) 2.816 × 10.–7 m (c) 6.500 × 10–7 m (d) 8.500 × 10–7 m 21. The wave number of the first emission line in the Balmer series of H-Spectrum is : [Main Online April 22, 2013] (R = Rydberg constant) : (a) (b) (c) (d) 22.

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [a0 is Bohr radius] : [2012]

(a)

(b)

(c)

(d) 23.

Given that the abundances of isotopes 54Fe, 56Fe and respectively, the atomic mass of Fe is

57

Fe are 5%, 90% and 5%, [2009S]

(a) (b) (c) (d)

55.85 55.95 55.75 56.05

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24. (a) (b) (c) (d) 25.

(a) (b) (c) (d) 26. (a) (b) (c) (d) 27. (a) (b) (c) (d) 28.

The radius of which of the following orbit is same as that of the first Bohr’s orbit of hydrogen atom? [2004S] + He (n = 2) Li2+ (n = 2) Li2+ (n = 3) Be3+ (n = 2) Rurtherford's experiment, which established the nuclear model of the atom, used a beam of [2002S] -particles, which impinged on a metal foil and got absorbed -rays, which impinged on a metal foil and ejected electrons helium atoms, which impinged on a metal foil and got scattered helium nuclei, which impinged on a metal foil and got scattered Which of the following does not characterise X-rays? [1992 - 1 Mark] The radiation can ionise gases It causes ZnS to fluorescence Deflected by electric and magnetic fields Have wavelengths shorter than ultraviolet rays The wavelength of a spectral line for an electronic transition is inversely related to : [1988 - 1 Mark] the number of electrons undergoing the transition the nuclear charge of the atom the difference in the energy of the energy levels involved in the transition the velocity of the electron undergoing the transition. The triad of nuclei that is isotonic is [1988 - 1 Mark]

(a)

,

,

(b)

,

,

(c)

,

,

(d)

,

,

29.

The ratio of the energy of a photon of 2000Å wavelength radiation to that of 4000Å radiation is : [1986 - 1 Mark] (a) ¼

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(b) 4 (c) ½ (d) 2 30. Rutherford’s alpha particle scattering experiment eventually led to the conclusion that : [1986 - 1 Mark] (a) mass and energy are related (b) electrons occupy space around the nucleus (c) neutrons are buried deep in the nucleus (d) the point of impact with matter can be precisely determined. 31. Electromagnetic radiation with maximum wavelength is : [1985 - 1 Mark] (a) ultraviolet (b) radiowave (c) X-ray (d) infrared 32. The radius of an atomic nucleus is of the order of : [1985 - 1 Mark] –10 –13 (a) 10 cm (b) 10 cm –15 (c) 10 cm (d) 10–8 cm 33. Bohr model can explain : [1985 - 1 Mark] (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 34. Which electronic level would allow the hydrogen atom to absorb a photon but not to emit a photon? [1984 - 1 Mark] (a) 3s (b) 2p (c) 2s (d) 1s 35. 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 - 1 Mark] (a) e, p, n, α (b) n, p, e, α (c) n, p, α, e (d) n, α, p, e 36. Rutherford’s scattering experiment is related to the size of the

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[1983 - 1 Mark] (a) (b) (c) (d) 37. (a) (b) (c) (d) 38. (a) (b) (c) (d)

nucleus atom electron neutron Rutherford’s experiment on scattering of α-particles showed for the first time that the atom has [1981 - 1 Mark] electrons protons nucleus neutrons The number of neutrons in dipositive zinc ion with mass number 70 is [1979] 34 36 38 40

39. 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] 40.

Wavelength of high energy transition of H–atoms is 91.2nm. Calculate the corresponding wavelength of He atoms. [2003 - 2 Marks] 41. Calculate the wave number for the shortest wavelength transition in the Balmer series of atomic hydrogen. [1996 - 1 Mark]

42.

According to Bohr’s theory, the electronic energy of hydrogen atom in the nth Bohr’s orbit is given byEn =

J. Calculate the longest wavelength

of light that will be needed to remove an electron from the third Bohr orbit of the He+ ion. [1990 - 3 Marks]

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43.

Calculate the wavelength in Angstrom of the photon that is emitted when an electron in the Bohr orbit, n = 2 returns to the orbit, n = 1 in the hydrogen atom. The ionization potential of the ground state hydrogen atom is 2.17 × 10–11 erg per atom. [1982 - 4 Marks] 44. The energy of the electron in the second and the 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 radiation when the electron drops from the third to the second orbit. [1981 - 3 Marks]

45.

The light radiations with discrete quantities of energy are called ............... . [1993 - 1 Mark] 46. Elements of the same mass number but of different atomic numbers are known as ............... . [1983 - 1 Mark] 47. Isotopes of an element differ in the number of ............. in their nuclei. [1982 - 1 Mark] 48. The mass of a hydrogen atom is ............. kg. [1982 - 1 Mark] 49.

In a given electric field, β-particles are deflected more than α-particles in spite of α-particles having larger charge. [1993 - 1 Mark] 50. Gamma rays are electromagnetic radiations of wavelengths of 10–6 cm to 10–5 cm. [1983 - 1 Mark]

51.

(a) (b) (c) (d) 52.

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 - 2 Marks] – 3.4 eV – 4.2 eV – 6.8 eV – 1.5 eV The sum of the number of neutrons and proton in the isotope of hydrogen is :

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[1986 - 1 Mark] (a) (b) (c) (d) 53. (a) (b) (c) (d) 54. (a) (b) (c) (d) 55.

6 2 4 3 When alpha particles are sent through a thin metal foil, most of them go straight through the foil because : [1984 - 1 Mark] alpha particles are much heavier than electrons alpha particles are positively charged most part of the atom is empty space alpha particle move with high velocity Many elements have non-integral atomic masses because: [1984 - 1 Mark] they have isotopes their isotopes have non-integral masses their isotopes have different masses the constitutents, neutrons, protons and electrons, combine to give fractional masses An isotone of is : [1984 - 1 Mark]

(a) (b) (c) (d)

56. 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 [Adv. 2019] (I) (II) (III)

List–I Radius of the nth orbit Angular momentum of the electron in the nth orbit Kinetic energy of the electron in the nth orbit

List–II (P) ∝ n−2 (Q) ∝ n−1 (R) ∝ n0

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(IV)

Potential energy of the electron in the nth orbit

(S) ∝ n1 (T) ∝ n2 (U) ∝ n1/2 Which of the following options has the correct ombination considering List–I and List–II? (a) (II), (R) (b) (II), (Q) (c) (I), (P) (d) (I), (T) 57. 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. [Adv. 2019] (I) (II) (III) (IV)

List–I Radius of the nth orbit Angular momentum of the electron in the nth orbit Kinetic energy of the electron in the nth orbit Potential energy of the electron in the nth orbit

List–II (P) ∝ n–2 (Q) ∝ n–1 (R) ∝ n0

(S) ∝ n1 (T) ∝ n2 (U) ∝ n1/2 Which of the following options has the correct combination considering List–I and List–II? (a) (III), (S) (b) (IV), (Q) (c) (III), (P) (d) (IV), (U) 58.

Calculate the energy required to excite one litre 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 - 4 Marks] 59. Consider the hydrogen atom to be a proton embedded in a cavity of radius a0 (Bohr 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

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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 - 2 Marks] 60. Iodine molecule dissociates into atoms after absorbing light of 4500 . 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 - 2 Marks] 61. Find out the number of waves made by a Bohr electron in one complete revolution in its 3rd orbit.[1994 - 3 Marks] 62. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+ spectrum? [1993 - 3 Marks] 63. Estimate the difference in energy between Ist and 2nd Bohr 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 λ = 3.0 × 10–8m? Which hydrogen atom-like species does this atomic number correspond to ? [1993 - 5 Marks] 64. The electron energy in hydrogen atom is given byE = (–21.7 × 10–12)/n2 ergs. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength (in cm) of light that can be used to cause this transition? [1984 - 3 Marks] 65. 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]

1.

The figure that is not a direct manifestation of the quantum nature of atoms is : [Main Sep. 02, 2020 (I)]

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(a)

(b)

(c)

(d)

2.

The de Broglie wavelength of an electron in the 4th Bohr orbit is: [Main Jan. 09, 2020 (I)]

(a) (b) (c) (d) 3.

2πa0 4πa0 6πa0 8πa0 If p is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength λ, then for 1.5p 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): [Main April 8, 2019 (II)] (a) (b)

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(c) (d) 4.

(a) (b) (c) (d) 5. (a) (b)

If the de Broglie wavelength of the electron in nth Bohr orbit in a hydrogenic atom is equal to 1.5 πa0 (a0 is Bohr radius), then the value of n/z is : [Main Jan. 12, 2019 (II)] 0.40 1.50 1.0 0.75 The de-Broglie's wavelength of electron present in first Bohr orbit of 'H' atom is: [Main Online April 15, 2018 (II)] 4 × 0.529Å 2π × 0.529Å

(c) (d) 0.529Å 6.

At temperature T, the average kinetic energy of any particle is

. The de

Broglie wavelength follows the order : (a) (b) (c) (d) 7.

(a) (b) (c) (d) 8. (a) (b)

[Main Online April 11, 2015] Visible photon > Thermal neutron > Thermal electron Thermal proton > Thermal electron > Visible photon Thermal proton > Visible photon > Thermal electron Visible photon > Thermal electron > Thermal neutron The de-Broglie wavelength of a particle of mass 6.63 g moving with a velocity of 100 ms–1 is: [Main Online April 12, 2014] –33 10 m 10–35 m 10–31 m 10–25 m The de Broglie wavelength of a car of mass 1000 kg and velocity 36 km/hr is : [Main Online April 23, 2013] –34 6.626 × l0 m 6.626 × 10–38 m

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(c) 6.626 × 10–31 m (d) 6.626 × 10– 30 m 9. The wavelength associated with a golf ball weighing 200 g and moving at a speed of 5 m/h is of the order [2001S] –10 (a) 10 m (b) 10–20 m (c)10–30 m (d) 10–40 m 10. Which of the following relates to photons both as wave motion and as a stream of particles? [1992 - 1 Mark] (a) Inference (b) E = mc2 (c) Diffraction (d) E = hv

11. The atomic masses of ‘He’ and ‘Ne’ are 4 and 20 a.m.u., 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 [Adv. 2013] 12.

The work function of sodium metal is 4.41 × 10–19 J. If photons of wavelength 300 nm are incident on the metal, the kinetic energy of the ejected electrons will be (h = 6.63 × 10–34 J s; c = 3 × 108 m/s) _______ × 10–21 J. [Main Sep. 02, 2020 (II)]

13. 14.

Wave functions of electrons in atoms and molecules are called ............... . [1993 - 1 Mark] The uncertainty principle and the concept of wave nature of matter were proposed by ............... and ............... respectively. (Heisenberg, Schrodinger, Maxwell, de Broglie) [1988 - 1 Mark]

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15.

Find the velocity (ms–1) of electron in first Bohr’s orbit of radius a0. Also find the de Broglie’s wavelength (in m). Find the orbital angular momentum of 2p orbital of hydrogen atom in units of h / 2π. [2005 - 2 Marks] –1 16. A ball of mass 100 g is moving with 100 ms . Find its wavelength. [2004 - 1 Mark] 17. The Schrodinger wave equation for hydrogen atom is [2004 - 2 Marks] Where a0 is Bohr’s radius. If the radial node in 2s be at r0, then find r0 in terms of a0.

1.

In the sixth period, the orbitals that are filled are : [Main Sep. 05, 2020 (I)]

(a) (b) (c) (d) 2. (a) (b) (c) (d) 3.

(a) (b) (c) (d) 4.

6s, 4f, 5d, 6p 6s, 5d, 5f, 6p 6s, 5f, 6d, 6p 6s, 6p, 6d, 6f The correct statement about probability density (except at infinite distance from nucleus) is : [Main Sep. 05, 2020 (II)] It can be zero for 1s orbital It can be negative for 2p orbital It can be zero for 3p orbital It can never be zero for 2s orbital Consider the hypothetical situation where the azimuthal quantum number, l, takes values 0, 1, 2, ..... n + 1, where n is the principal quantum number. Then, the element with atomic number : [Main Sep. 03, 2020 (II)] 9 is the first alkali metal 13 has a half-filled valence subshell 8 is the first noble gas 6 has a 2p-valence subshell The number of subshells associated with n = 4 and m = – 2 quantum numbers is :

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[Main Sep. 02, 2020 (II)] (a) (b) (c) (d) 5.

8 2 16 4 The number of orbitals associated with quantum numbers n = 5, ms = +

is:

[Main Jan. 07, 2020 (I)] (a) 11 (b) 25 (c) 50 (d) 15 6.

The electrons are more likely to be found : [Main April 12, 2019 (I)]

(a) (b) (c) (d) 7.

in the region a and c in the region a and b only in the region a only in the region c The graph between |ψ|2 and r (radial distance) is shown below. This represents : [Main April 10, 2019 (I)]

(a) 3s orbital

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(b) 2s orbital (c) 1s orbital (d) 2p orbital 8. The isoelectronic set of ions is [Main April 10, 2019 (I)] (a) N , O , F and Na (b) N3 –, Li+, Mg2+ and O2– (c) F –, Li+, Na+ and Mg2+ (d) Li+, Na+, O2 – and F – 9. The quantum number of four electrons are given below : I. n = 4, l = 2, ml = –2, ms = – 1/2 II. n = 3, l = 2, ml = 1, ms = + 1/2 III. n = 4, l = 1, ml = 0, ms = + 1/2 IV. n = 3, l = 1, ml = 1, ms = – 1/2 The correct order of their increasing energies will be : 3–

2–

–

+

[Main April 8, 2019 (I)] (a) (b) (c) (d) 10. (a) (b) (c) (d) 11.

IV < III < II < I I < II < III < IV IV < II < III < I I < III < II < IV The total number of orbitals associated with the principal quantum number 5 is : [Main Online April 9, 2016] 20 25 10 5 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 is4π r2dr. The qualitative sketch of the dependence of P on r is [Adv. 2016]

(a)

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(b)

(c)

(d)

12. (a) (b) (c) (d) 13.

If the principal quantum number n = 6, the correct sequence of filling of electrons will be :[Main Online April 10, 2015] ns → (n – 2) f → np → (n –1)d ns → (n – 2) f →(n –1)d → np ns → np → (n – 1)d → (n–2)f ns → (n–1)d → (n–2)f → np The correct set of four quantum numbers for the valence electrons of rubidium atom (Z = 37) is: [Main 2014]

(a) (b) (c) (d)

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14.

(a) (b) (c) (d) 15.

In an atom how many orbital(s) will have the quantum numbers; n = 3, l = 2 and ml = + 2 ? [Main Online April 9, 2013] 5 3 1 7 The number of radial nodes of 3s and 2p orbitals are respectively [2005S]

(a) (b) (c) (d) 16.

(a) (b) (c) (d) 17.

(b) (c) (d) 18. (a) (b) (c) (d) 19. (a) (b) (c)

2, 0 0, 2 1, 2 2, 1 If the nitrogen atom has electronic configuration 1s7, it would have energy lower than that of the normal ground state configuration 1s22s22p3, because the electrons would be closer to the nucleus. Yet 1s7 is not observed because it violates. [2002S] Heisenberg uncertainty principle Hund's rule Pauli exclusion principle Bohr postulate of stationary orbits The quantum numbers +1/2 and –1/2 for the electron spin represent [2001S] (a) rotation of the electron in clockwise and anticlockwise direction respectively rotation of the electron in anticlockwise and clockwise direction respectively magnetic moment of the electron pointing up and down respectively two quantum mechanical spin states which have no classical analogue The electronic configuration of an element is 1s2, 2s2 2p6, 3s2 3p6 3d5, 4s1. This represents its [2000S] excited state ground state cationic form anionic form The number of nodal planes in a px orbital is [2000S] one two three

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(d) zero 20. The electrons, identified by quantum numbers n and l,(i) n = 4, l = 1, (ii) n = 4, l = 0, (iii) n = 3, l = 2, and (iv) n = 3, l = 1 can be placed in order of increasing energy, from the lowest to highest, as [1999 - 2 Marks] (a) (iv) < (ii) < (iii) < (i) (b) (ii) < (iv) < (i) < (iii) (c) (i) < (iii) < (ii) < (iv) (d) (iii) < (i) < (iv) < (ii) 21. For a d-electron, the orbital angular momentum is [1997 - 1 Mark] (a) (b) (c) (d) 22.

The orbital angular momentum of an electron in 2s orbital is: [1996 - 1 Mark]

(a)

(b)

Zero

(c) (d) 23.

A 3p orbital has :

[1995S] two non spherical nodes two spherical nodes one spherical & one non spherical node one spherical and two non spherical nodes The correct set of quantum numbers for the unpaired electron of chlorine atom is : [1989 - 1 Mark] n l m (a) 2 1 0 (b) 2 1 1 (c) 3 1 1 (d) 3 0 0 25. The correct ground state electronic configuration of chromium atom is :

(a) (b) (c) (d) 24.

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[1989 - 1 Mark] (a) (b) (c) (d) 26. (a) (b) (c) (d) 27.

[Ar] 3d 4s [Ar] 3d4 4s2 [Ar] 3d6 4s0 [Ar] 4d5 4s1 The outermost electronic configuration of the most electronegative element is [1988 - 1 Mark] 2 3 ns np ns2 np4 ns2 np5 ns2 np6 The orbital diagram in which the Aufbau principle is violated is : [1988 - 1 Mark] 2s 2p 5

1

(a) (b) (c) (d) 28.

29.

(a) (b) (c) (d) 30. (a) (b)

Which one of the following sets of quantum numbers represents an impossible arrangement? [1986 - 1 Mark] n l ml ms (a) 3 2 –2 ½ (b) 4 0 0 ½ (c) 3 2 –3 ½ (d) 5 3 0 –½ Correct set of four quantum numbers for the valence (outermost) electron of rubidium (Z = 37) is : [1984 - 1 Mark] 5, 0, 0, +½ 5, 1, 0, +½ 5, 1, 1, +½ 6, 0,0, +½ Any p-orbital can accommodate upto [1983 - 1 Mark] four electrons six electrons

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(c) two electrons with parallel spins (d) two electrons with opposite spins 31. The principal quantum number of an atom is related to the [1983 - 1 Mark] (a) (b) (c) (d)

size of the orbital spin angular momentum orbital angular momentum orientation of the orbital in space

32. Not considering the electronic spin, the degeneracy of the second excited st00ate (n = 3) of H atom is 9, while the degeneracy of the second excited state of H– is [Adv. 2015] 33. In an atom, the total number of electrons having quantum numbers n = 4, | ml | = 1 and ms =

is

[Adv. 2014]

34. The maximum number of electrons that can have principal quantum number, n = 3, and spin quantum

is [2011]

35.

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 - 2 Marks]

36.

The outermost electronic configuration of Cr is ...............

37.

The ............... .

and

[1994 - 1 Mark] orbitals of atom have identical shapes but differ in their

38.

[1993 - 1 Mark] When there are two electrons in the same orbital, they have ............. spins. [1982 - 1 Mark]

39.

The electron density in the XY plane in

orbital is zero. [1986 - 1 Mark]

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40.

The energy of the electron in the 3d-orbital is less than that in the 4s-orbital in the hydrogen atom. [1983 - 1 Mark] 41. The outer electronic configuration of the ground state chromium atom is 3d44s2. [1982 - 1 Mark]

42.

The ground state energy of hydrogen atom is –13.6 eV. Consider an electronic state ψ of He+ whose energy, azimuthal quantum number and magnetic quantum number are –3.4 eV, 2 and 0, respectively. Which of the following statement(s) is (are) true from the state ψ? [Adv. 2019] (a) It is a 4d state (b) It has 3 radial nodes (c) It has 2 angular nodes (d) The nuclear charge experienced by the electron in this state is less than 2e, where e is the magnitude of the electronic charge 43. Ground state electronic configuration of nitrogen atom can be represented by [1999 - 3 Marks] (a) (b) (c) (d) 44.

Which of the following satement(s) is (are) correct? [1998 - 2 Marks] (Atomic Number of

(a) The electronic configuration of Cr is [Ar] 3d 4s . 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 ofAg = 47) (d) The oxidation state of nitrogen in HN3 is – 3. 5

1

(Qs. 45-47) are based on the table, having 3 columns and 4 rows. Each question has four options (A), (B), (C) and (D). Only one of these four options is correct. By appropriately matching the information given in the three columns of the following table.

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The wave function,

is a mathematical function whose value depends upon

spherical polar coordinates (r, θ, φ) of the electron and characterized by the quantum numbers 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. [Adv. 2017] Column Column 2 1 (I) 1s (i) orbital

Column 3 (P)

(II) 2s (ii) One radial node orbital

(Q)Probability

density

at

(III)2pz (iii) orbital

(R)Probability density maximum at nucleus

is

(IV)

(S) Energy needed to excite electron from n = 2 state to

(iv) xy-plane is a nodal plane orbital

nucleus

n = 4 state is

45. (a) (b) (c) (d) 46. (a) (b)

times the

energy needed to excite electron from n = 2 state to n = 6 state For the given orbital in Column 1, the only CORRECT combination for any hydrogen-like species is (I) (ii) (S) (IV) (iv) (R) (II) (ii) (P) (III) (iii) (P) For hydrogen atom, the only CORRECT combination is (I) (i) (S) (II) (i) (Q)

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(c) (I) (i) (P) (d) (I) (iv) (R) 47. For He+ ion, the only INCORRECT combination is (a) (I) (i) (R) (b) (II) (ii) (Q) (c) (I) (iii) (R) (d) (I) (i) (S) 48. Match the entries in Column I with the correctly related quantum number(s) in Column II. [2008 - 6M] Column I Column II (A)Orbital angular momentum of the electron in(p)Principal quantum a number hydrogen-like atomic orbital (B)A hydrogen-like one-electron wave function(q)Azimuthal quantum obeying Pauli principle number (C)Shape, size and orientation of hydrogen- like(r) Magnetic quantum atomic orbitals number (D)Probability density of electron at the nucleus(s) Electron spin in hydrogen-like atom quantum number 49. According to Bohr’s theory, [2006 - 6M] En = Total energy, Kn = Kinetic energy, Vn = Potential energy, rn = Radius of nth orbit Match the following : Column I Column II (A)Vn / Kn = ? (p)0 (B) If radius of nth orbit ,(q)–1 x=? (C)Angular momentum lowest orbital (D)

in(r) –2 (s) 1

The hydrogen-like species Li2+ is in a spherically symmetric state S1 with one radial node. Upon absorbing light the ion undergoes transition to a state S2. The state S2

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50.

has one radial node and its energy is equal to the ground state energy of the hydrogen atom. [2010] The state S1 is :

(a) (b) (c) (d) 51. (a) (b) (c) (d) 52. (a) (b) (c) (d)

1s 2s 2p 3s Energy of the state S1 in units of the hydrogen atom ground state energy is : 0.75 1.50 2.25 4.50 The orbital angular momentum quantum number of the state S2 is : 0 1 2 3

53.

Give reasons why the ground state outermost electronic configuration of silicon is : [1985 - 2 Marks] 3s 3p 3s 3p 3 and not

Topic-1 : Different Atomic Models that Leads to Bohr Model 1. (c) 2. (a) 3. (c) 4. (b) 5. (c) 6. (a) 7. (a) 8. (b) 9. (c) 10. (a) 11. (c)

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12. (b) 13. (c) 14. (d) 15. (d) 16. (b) 17. (a) 18. (d) 19. (c) 20. (a) 21. (a) 22. (c) 23. (b) 24. (d) 25. (d) 26. (c) 27. (c) 28. (a) 29. (d) 30. (b) 31. (b) 32. (b) 33. (b) 34. (d) 35. (d) 36. (a) 37. (c) 38. (d) 39. (4) 40. (22.8) 41. (27419) 42. (2055) 43. (220) 44. (660) 49. (True) 50. (False) 51. (a, d) 52. (b, d) 53. (a, c) 54. (a, c) 55. (b, d) 56. (d)

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57. (c) Topic-2 : Advancement Towards Quantum Mechanical Model of Atom 1. (d) 2. (d) 3. (d) 4. (d) 5. (b) 6. (d) 7. (a) 8. (b) 9. (a) 10. (d) 11. (5) 12. (222) Topic-3 : Quantium Mechanical Model of Atom 1. (a) 2. (c) 3. (b) 4. (b) 5. (b) 6. (a) 7. (b) 8. (a) 9. (c) 10. (b) 11. (d) 12. (b) 13. (a) 14. (c) 15. (a) 16. (c) 17. (d) 18. (b) 19. (a) 20. (a)

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21. (a) 22. (b) 23. (c) 24. (c) 25. (a) 26. (c) 27. (b) 28. (c) 29. (a) 30. (d) 31. (a) 32. (3) 33. (6) 34. (9) 35. (10) 39. (False) 40. (True) 41. (False) 42. (a, c) 43. (a, d) 44. (a, b, c) 45. (c) 46. (a) 47. (c) 48. (A) - (q); (B) - (p), (q), (r, s); (C) - (p, q, r); (D) - (p, q, r) 49. (A) - (r); (B) - (q); (C) - (p); (D) - (s) 50. (b) 51. (c) 52. (b)

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1. (a) (b) (c) (d) 2. (a) (b) (c) (d) 3. (a) (b) (c) (d)

The atomic number of the element unnilennium is : [Main Sep. 03, 2020 (I)] 109 102 108 119 The group number, number of valence electrons, and valency of an element with atomic number 15, respectively, are : [Main April 12, 2019 (I)] 16, 5 and 2 15, 5 and 3 16, 6 and 3 15, 6 and 2 The element with Z=120 (not yet discovered) will be an/a: [Main Jan. 12, 2019 (I)] Inner-transition metal Alkaline earth metal Alkali metal Transition metal

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4.

(b) (c) (d) 5.

(a) (b) (c) (d) 6. (a) (b) (c) (d)

7.

Similarity in chemical properties of the atoms of elements in a group of the periodic table is most closely related to: [Main Online April 12, 2014] (a) atomic numbers atomic masses number of principal energy levels number of valence electrons Which of the following has the maximum number of unpaired electrons? [1996 - 1 Mark] Mg2+ Ti3+ V3+ Fe2+ The statement that is not correct for the periodic classification of element is [1992 - 1 Mark] The properties of elements are the periodic functions of their atomic numbers Non-metallic elements are lesser in number than metallic elements The first ionisation energies of elements along a period do not vary in a regular manner with increase in atomic number For transition elements the d-subshells are filled with electrons monotonically with increase in atomic number.

The atomic number of Unnilunium is ______ . [Main Sep. 06, 2020 (II)]

8.

The statements that are true for the long form of the periodic table are : [1988 - 1 Mark] (a) it reflects the sequence of filling the electrons in the order of subenergy level s, p, d and f.

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(b) it helps to predict the stable valency states of the elements (c) it reflects trends in physical and chemical properties of the elements (d) it helps to predict the relative ionicity of the bond between any two elements.

1. (a) (b) (c) (d) 2.

The set that contains atomic numbers of only transition elements, is : [Main Sep. 06, 2020 (I)] 37, 42, 50, 64 21, 25, 42, 72 9, 17, 34, 38 21, 32, 53, 64 The correct order of the ionic radii of O2–, N3–, F–, Mg2+, Na+ and Al3+ is : [Main Sep. 05, 2020 (II)]

(a) (b) (c) (d) 3.

(a) (b) (c) (d) 4.

The elements with atomic numbers 101 and 104 belong to, respectively : [Main Sep. 04, 2020 (I)] Group 11 and Group 4 Actinoids and Group 6 Actinoids and Group 4 Group 6 and Actinoids The process that is NOT endothermic in nature is : [Main Sep. 04, 2020 (II)]

(a) (b)

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(c)

(d)

5.

The ionic radii of O2–, F–, Na+ and Mg2+ are in the order: [Main Sep. 04, 2020 (I)] – 2– + 2+ (a) F > O > Na > Mg (b) O2– > F– > Na+ > Mg2+ (c) Mg2+ > Na+ > F– > O2– (d) O2– > F– > Mg2+ > Na+ 6. Among the statements (I - IV), the correct ones are : [Main Sep. 03, 2020 (II)] (I) Be has smaller atomic radius compared to Mg. (II) Be has higher ionization enthalpy than Al. (III)Charge/radius ratio of Be is greater than that of Al. (IV)Both Be and Al form mainly covalent compounds. (a) (I), (II) and (IV) (b) (I), (III) and (IV) (c) (II), (III) and (IV) (d) (I), (II) and (III) 7. The five successive ionization enthalpies of an element are 800, 2427, 3658, 25024 and 32824 kJ mol–1. The number of valence electrons in the element is : [Main Sep. 03, 2020 (II)] (a) 5 (b) 4 (c) 3 (d) 2 8. In general, the property (magnitudes only) that shows an opposite trend in comparison to other properties across a period is : [Main Sep. 02, 2020 (I)] (a) Ionization enthalpy (b) Electronegativity (c) Electron gain enthalpy (d) Atomic radius

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9.

Three elements X, Y and Z are in the 3rd period of the periodic table. The oxides of X, Y and Z, respectively, are basic, amphoteric and acidic. The correct order of the atomic numbers of X, Y and Z is : [Main Sep. 02, 2020 (II)] (a) Z < Y < X (b) X < Y < Z (c) X < Z < Y (d) Y < X < Z 10. B has a smaller first ionization enthalpy than Be. Consider the following statements: (I) it is easier to remove 2p electron than 2s electron (II) 2p electron of B is more shielded from the nucleus by the inner core of electrons than the 2s electrons of Be (III)2s electron has more penetration power than 2p electron (IV) atomic radius of B is more than Be (atomic number B = 5, Be = 4) The correct statements are: [Main Jan. 09, 2020 (I)] (a) (I), (II) and (IV) (b) (II), (III) and (IV) (c) (I), (II) and (III) (d) (I), (III) and (IV) 11. The acidic, basic and amphoteric oxides, respectively, are: [Main Jan. 09, 2020 (I)] (a) Na2O, SO3, Al2O3 (b) Cl2O, CaO, P4O10 (c) N2O3, Li2O, Al2O3 (d) MgO, Cl2O, Al2O3 12. The first and second ionisation enthalpies of a metal are 496 and 4560 kJ mol–1, respectively. How many moles of HCl and H2SO4, respectively, will be needed to react completely with 1 mole of the metal hydroxide? [Main, Jan. 09, 2020 (II)]

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(a) (b) (c) (d) 13.

1 and 1 2 and 0.5 1 and 2 1 and 0.5 The first ionization energy (in kJ/mol) of Na, Mg, Al and Si respectively, are: [Main Jan. 08, 2020 (I)] (a) 496, 737, 577,786 (b) 496, 577,737, 786 (c) 786,737,577,496 (d) 496, 577,786, 737 14. The increasing order of the atomic radii of the following elements is: [Main Jan. 08, 2020 (II)] (i) C (ii) O (iii) F (iv) Cl (v) Br (a) (ii) < (iii) < (iv) < (i) < (v) (b) (iv) < (iii) < (ii) < (i) < (v) (c) (iii) < (ii) < (i) < (iv) < (v) (d) (i) < (ii) < (iii) < (iv) < (v) 15. The electron gain enthalpy (in kJ/mol) of fluorine, chlorine, bromine and iodine, respectively, are: [Main Jan. 07, 2020 (I)] (a) –296, –325, –333 and –349 (b) –349, –333, –325 and –296 (c) –333, –349, –325 and –296 (d) –333, –325, –349 and –296 16. Within each pair of elements F & Cl, S & Se, and Li & Na, respectively, the elements that release more energy upon an electron gain are: [Main Jan. 07, 2020 (II)] (a) Cl, Se and Na (b) Cl, S and Li

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(c) F, S and Li (d) F, Se and Na 17. In comparison to boron, berylium has: (a) (b) (c) (d) 18. (a) (b) (c) (d) 19.

(a) (b) (c) (d) 20.

(a) (b) (c) (d) 21.

(a) (b) (c)

[Main April 12, 2019 (II)] lesser nuclear charge and lesser first ionisation enthalpy. greater nuclear charge and lesser first ionisation enthalpy. greater nulear charge and greater first ionisation enthalpy. lesser nuclear charge and greater first ionisation enthalpy. The correct order of the atomic radii of C, Cs, Al, and S is : [Main Jan. 11, 2019 (I)] C < S < A1 < Cs S < C < Cs < Al S < C < Al < Cs C < S < Cs < Al The correct option with respect to the Pauling electronegativity values of the elements is: [Main Jan. 11, 2019 (II)] Te > Se Ga < Ge Si < Al P>S The 71st electron of an element X with an atomic number of 71 enters into the orbital: [Main Jan. 10, 2019 (II)] 6p 4f 5d 6s In general, the properties that decrease and increase down a group in the periodic table, respectively, are: [Main Jan. 9, 2019 (I)] atomic radius and electronegativity. electron gain enthalpy and electronegativity. electronegativity and atomic radius.

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(d) electronegativity and electron gain enthalpy. 22. The correct order of electron affinity is: [Main Online April 15, 2018 (II)] (a) O > F > Cl (b) F > O > Cl (c) F > Cl > O (d) Cl > F > O 23. Both lithium and magnesium display several similar properties due to the diagonal relationship; however, the one which is incorrect is : [Main 2017] (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 24. The group having isoelectronic species is : [Main 2017] 2– – + 2+ (a) O , F , Na , Mg (b) O– , F–, Na, Mg+ (c) O2– , F–, Na , Mg2+ (d) O– , F–, Na+ , Mg2+ 25. Consider the following ionization enthalpies of two elements ‘A’ and ‘B’. [Main Online April 8, 2017] Element Ionization enthalpy (kJ/mol) 1st 2nd 3rd A 899 1757 14847 B 737 1450 7731 Which of the following statements is correct ? (a) Both ‘A’ and ‘B’ belong to group–1 where ‘B’ comes below ‘A’. (b) Both ‘A’ and ‘B’ belong to group–1 where ‘A’ comes below ‘B’. (c) Both ‘A’ and ‘B’ belong to group–2 where ‘B’ comes below ‘A’. (d) Both ‘A’ and ‘B’ belong to group–2 where ‘A’ comes below ‘B’. 26. Which of the following atoms has the highest first ionization energy? [Main 2016] (a) K

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(b) Sc (c) Rb (d) Na 27. The following statements concern elements in the periodic table. Which of the following is true ? [Main Online April 10, 2016] (a) For Group 15 elements, the stability of +5 oxidation state increases down the group (b) Elements of Group 16 have lower ionization enthalpy values compared to those of Group 15 in the corresponding periods. (c) The Group 13 elements are all metals (d) All the elements in Group 17 are gases. 28. The ionic radii (in Å) of N3–, O2– and F– are respectively : [Main 2015] (a) 1.71, 1.40 and 1.36 (b) 1.71, 1.36 and 1.40 (c) 1.36, 1.40 and 1.71 (d) 1.36, 1.71 and 1.40 29. In the long form of the periodic table, the valence shell electronic configuration of 5s25p4 corresponds to the element present in : [Main Online April 10, 2015] (a) Group 16 and period 6 (b) Group 17 and period 6 (c) Group 16 and period 5 (d) Group 17 and period 5 30. Which of the following series correctly represents relations between the elements from X to Y? X→Y [Main Online April 11, 2014] (a) 3Li → 19K Ionization enthalpy increases (b) 9F → 35Br Electron gain enthalpy (negative sign) increases (c) 6C → 32Ge Atomic radii increases (d) 18Ar → 54Xe Noble character increases 31. Which of the following arrangements represents the increasing order (smallest to largest) of ionic radii of the given species O2–, S2–, N3–, P3–?

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[Main Online April 12, 2014] (a) (b) (c) (d) 32. (a) (b) (c) (d) 33. (a) (b) (c) (d) 34.

(a) (b) (c) (d) 35. (a) (b) (c) (d) 36. (a) (b) (c)

O N2O5 > SO3 Na2O > MgO > Al2O3 2–

3–

2–

3–

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(d) K2O > CaO > MgO 37. Amongst H2O, H2S, H2Se and H2Te, the one with the highest boiling point is [2000S] (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 38. Which has most stable +2 oxidation state : [1995S] (a) Sn (b) Pb (c) Fe (d) Ag 39. Amongst the following elements (whose electronic configurations are given below), the one having the highest ionization energy is : [1990 - 1 Mark] 2 1 (a) [Ne] 3s 3p (b) [Ne] 3s23p3 (c) [Ne] 3s23p2 (d) [Ne] 3d104s24p3 40. Which one of the following is the strongest base? (a) AsH3 (b) NH3 [1989 - 1 Mark] (c) PH3 (d) SbH3 41. Which one of the following is the smallest in size? (a) N3– (b) O2– [1989 - 1 Mark] – (c) F (d) Na+ 42. The first ionisation potential of Na, Mg, Al and Si are in the order [1988 - 1 Mark]

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(a) (b) (c) (d) 43.

(a) (b) (c) (d) 44. (a) (b) (c) (d) 45. (a) (b) (c) (d) 46. (a) (b) (c) (d) 47.

(b) (c)

Na < Mg > Al < Si Na > Mg > Al > Si Na < Mg < Al < Si Na > Mg > Al < Si The electronegativity of the following elements increases in the order [1987 - 1 Mark] C, N, Si, P N, Si, C, P Si, P, C, N P, Si, N, C The first ionisation potential in electron volts of nitrogen and oxygen atoms are respectively given by [1987 - 1 Mark] 14.6, 13.6 13.6, 14.6 13.6, 13.6 14.6, 14.6 Atomic radii of fluorine and neon in Ångstorm units are respectively given by [1987 - 1 Mark] 0.72, 1.60 1.60, 1.60 0.72, 0.72 None of these values The element with the highest first ionization potential is [1982 - 1 Mark] boron carbon nitrogen oxygen The correct order of second ionisation potential of carbon, nitrogen, oxygen and fluorine is [1981 - 1 Mark] (a) C > N > O > F O>N>F>C O>F>N>C

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(d) F > O > N > C 48.

The 1st, 2nd, and the 3rd ionization enthalpies, I1, I2, and I3, 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? [Adv. 2020]

49.

Among the following, the number of elements showing only one nonzero oxidation state is : O, Cl, F, N, P, Sn, Tl, Na, Ti [2010]

50.

On Mulliken scale, the average of ionization potential and electron affinity is known as ............... . [1985 - 1 Mark] 51. The energy released when an electron is added to a neutral gaseous atom is called .......... of the atom. [1982 - 1 Mark]

52. 53. 54.

The basic nature of the hydroxides of group 13 (Gr. III B) decreases progressively down the group. [1993 - 1 Mark] The decreasing order of electron affinity of F, Cl, Br isF > Cl > Br. [1993 - 1 Mark] In group IA, of alkali metals, the ionisation potential decreases on moving down the group. Therefore, lithium is a poor reducing agent.

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[1987 - 1 Mark] 55.

The option(s) with only amphoteric oxides is(are) [Adv. 2017]

(a) (b) (c) (d)

Cr2O3, BeO, SnO, SnO2 Cr2O3, CrO, SnO, PbO NO, B2O3, PbO, SnO2 ZnO, Al2O3, PbO, PbO2

56.

Ionic radii of [1999 - 3 Marks]

(a) (b) (c) (d) 57.

(a) (b) (c) (d)

Ti4+ < Mn7+ 35 Cl– < 37Cl– K+ > Cl– P3+ > P5+ Sodium sulphate is soluble in water whereas barium sulphate is sparingly soluble because : [1989 - 1 Mark] the hydration energy of sodium sulphate is more than its lattice energy the lattice energy of barium sulphate is more than its hydration energy the lattice energy has no role to play in solubility the hydration energy of sodium sulphate is less than its lattice energy.

58.

Read the following statement and explanation and answer as per the options given below : ASSERTION : The first ionization energy of Be is greater than that of B. [2000S] REASON : 2p orbital is lower in energy than 2s (a) If both assertion and reason are CORRECT, and reason is the CORRECT explanation of the assertion. (b) If both assertion and reason are CORRECT, but reason is NOT the CORRECT explanation of the assertion.

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(c) If assertion is CORRECT, but reason is INCORRECT. (d) If assertion is INCORRECT, but reason is CORRECT. 59. Arrange the following in : (i) Arrange the following ions in order of their increasing radii : Li+, Mg2+, K+, Al3+. [1997 - 1 Mark] (ii) Increasing order of basic character : MgO, SrO, K2O, NiO, Cs2O [1991 - 1 Mark] 3– + – 2– 2+ (iii) Increasing order of ionic size : N , Na , F , O , Mg [1991 - 1 Mark] (iv) Increasing size : Cl–, S2–, Ca2+, Ar [1986 - 1 Mark] (v) Increasing first ionization potential : Mg, Al, Si, Na [1985 - 1 Mark] (vi) Increasing acidic property : ZnO, Na2O2, P2O5, MgO [1985 - 1 Mark] (vii) Decreasing ionic size : Mg2+, O2–, Na+, F– [1985 - 1 Mark] 60. The first ionization energy of carbon atom is greater than that of boron atom whereas, the reverse is true for the second ionization energy. [1989 - 2 Marks]

Topic-1 : Periodic Classification 1. (a) 2. (b) 3. (b) 4. (a) 5. (d)

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6. (d) 7. (101) 8. (a, c, d) Topic-2 : Periodic Properties 1. (b) 2. (c) 3. (c) 4. (b) 5. (b) 6. (a) 7. (c) 8. (d) 9. (b) 10. (c) 11. (c) 12. (d) 13. (a) 14. (c) 15. (c) 16. (b) 17. (d) 18. (a) 19. (b) 20. (c) 21. (c) 22. (d) 23. (a) 24. (a) 25. (c)

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26. (b) 27. (b) 28. (a) 29. (c) 30. (c) 31. (a) 32. (c) 33. (d) 34. (a) 35. (b) 36. (a) 37. (a) 38. (b) 39. (b) 40. (b) 41. (d) 42. (a) 43. (c) 44. (a) 45. (a) 46. (c) 47. (c) 48. (9) 49. (2) 52. (False) 53. (False) 54. (False) 55. (a, d) 56. (d) 57. (a, b) 58. (c)

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1.

The dipole moments of CCl4, CHCl3 and CH4 are in the order: [Main Jan. 07, 2020 (I)]

(a) (b) (c) (d) 2.

(a) (b) (c) (d) 3.

CHCl3 < CH4 = CCl4 CCl4 < CH4 < CHCl3 CH4 < CCl4 < CHCl3 CH4 = CCl4 < CHCl3 Which of the following compounds contain(s) no covalent bond(s)? KCl, PH3, O2, B2H6, H2SO4 [Main 2018] KCl, B2H6, PH3 KCl, H2SO4 KCl KCl, B2H6 Which compound exhibits maximum dipole moment among the following? [Main Online April 11, 2015]

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(a)

(b)

(c)

(d)

4.

Which of these statements is not true? [Main Online April 19, 2014]

(a) (b) (c) (d) 5.

NO is not isoelectronic with O2 B is always covalent in its compounds In aqueous solution, the Tl+ ion is much more stable than Tl (III) LiAlH4 is a versatile reducing agent in organic synthesis. The correct order of bond dissociation energy among N2, O2, +

is

shown in which of the following arrangements? [Main Online April 11, 2014]

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(a) (b) (c) (d) 6. (a) (b) (c) (d) 7. (a) (b) (c) (d) 8.

(a) (b) (c) (d) 9. (a) (b) (c) (d) 10. (a) (b)

Which one of the following molecules is polar ? [Main Online April 9, 2013] XeF4 IF5 SbF5 CF4 The geometry of H2S and its dipole moment are [1999 - 2 Marks] angular and non-zero angular and zero linear and non-zero linear and zero The critical temperature of water is higher than that of O2 because the H2O molecule has [1997 - 1 Mark] fewer electrons than O2 two covalent bonds V-shape dipole moment. Which contains both polar and non-polar bonds? [1997 - 1 Mark] NH4Cl HCN H2O2 CH4 Which one is most ionic : [1995S] P2O5 CrO3

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(c) MnO (d) Mn2O7 11. Pick out the isoelectronic structures from the following; [1993 - 1 Mark] II. H3O+ III. NH3 IV I. (a) (b) (c) (d) 12.

(a) (b) (c) (d) 13. (a) (b) (c) (d) 14.

(a) (b) (c) (d) 15.

(b) (c) (d)

I and II III and IV I and III II, III and IV The cyanide ion, CN– and N2 are isoelectronic. But in contrast to CN–, N2 is chemically inert, because of [1992 - 1 Mark] low bond energy absence of bond polarity unsymmetrical electron distribution presence of more number of electrons in bonding orbitals The molecule which has zero dipole moment is : [1989 - 1 Mark] CH2Cl2 BF3 NF3 ClO2 The bond between two identical non-metal atoms has a pair of electrons : [1986 - 1 Mark] unequally shared between the two transferred fully from one atom to another with identical spins equally shared between them The types of bonds present in CuSO4.5H2O are only (a) electrovalent and covalent [1983 - 1 Mark] electrovalent and coordinate covalent electrovalent, covalent and coordinate covalent covalent and coordinate covalent

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16. (a) (b) (c) (d) 17. (a) (b) (c) (d) 18. (a) (b) (c) (d) 19.

(a) (b) (c) (d) 20. (a) (b) (c) (d) 21.

(a) (b)

Carbon tetrachloride has no net dipole moment because of [1983 - 1 Mark] its planar structure its regular tetrahedral structure similar sizes of carbon and chlorine similar electron affinities of carbon and chlorine The compound with no dipole moment is [1982 - 1 Mark] methyl chloride carbon tetrachloride methylene chloride chloroform The ion that is isoelectronic with CO is [1982 - 1 Mark] – CN O2+ O2– N2+ If a molecule MX3 has zero dipole moment, the sigma bonding orbitals used by M (atomic number < 21) are [1981 - 1 Mark] pure p sp hybrid sp2 hybrid sp3 hybrid Which of the following is soluble in water [1980] CS2 C2H5OH CCl4 CHCl3 The total number of electrons that take part in forming the bond in N2 is [1980] 2 4

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(c) 6 (d) 10 22. Which of the following compounds are covalent? [1980] (a) (b) (c) (d) 23.

H2 CaO KCl Na2S Element X is strongly electropositive and element Y is strongly electronegative. Both are univalent. The compound formed would be [1980] + – (a) X Y (b) X –X + (c) X – Y (d) 24. The octet rule is not valid for the molecule [1979] (a) CO2 (b) H2O (c) O2 (d) CO 25. The compound which contains both ionic and covalent bonds is [1979] (a) CH4 (b) H2 (c) KCN (d) KCl 26. Consider the following compounds in the liquid form: O2, HF, H2O, NH3, H2O2, CCl4, CHCl3, C6H6, C6H5Cl. When a charged comb is brought near their flowing stream, how many of them show deflection as per the following figure? [Adv. 2020]

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27.

Among B2H6, B3N3H6, N2O, N2O4, H2S2O3, H2S2O8, the total number of molecules containing covalent bond between two atoms of the same kind is _______ [Adv. 2019]

28.

The dipole moment of KCl is 3.336 × 10–29 Coulomb meters 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 - 2 Marks]

29.

The two types of bonds present in

31.

The dipole moment of CH3F is greater than that of

are covalent and ........ [1994 - 1 Mark] 30. There are ............. π bonds in a nitrogen molecule. [1982 - 1 Mark]

[1993 - 1 Mark] 32. The presence of polar bonds in a poly-atomic molecule suggests that the molecule has non-zero dipole moment. [1990 - 1 Mark] 33. All molecules with polar bonds have dipole moment. [1985 - ½ Mark]

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34.

Linear overlap of two atomic p-orbitals leads to a sigma bond. [1983 - 1 Mark]

35.

Each of the following options contains a set of four molecules. Identify the option(s) were all four molecules posses permanent dipole moment at room temperature. [Adv. 2019] BF3, O3, SF6, XeF6 NO2, NH3, POCl3, CH3Cl SO2, C6H5Cl, H2Se, BrF5 BeCl2, CO2, BCl3, CHCl3 The molecules that will have dipole moment are [1992 - 1 Mark] 2, 2-dimethylpropane trans-2-pentene cis-3-hexene 2,2,3,3-tetramethylbutane

(a) (b) (c) (d) 36. (a) (b) (c) (d) 37.

Read the following Assertion and Reason and answer as per the options given below : [1998 - 2 Marks] Assertion : The electronic structure of O3 is Reason :

(a) (b) (c) (d) 38.

structure is not allowed because octet around O

cannot be expanded. If both assertion and reason are correct, and reason is the correct explanation of the assertion. If both assertion and reason are correct, but reason is not the correct explanation of the assertion. If assertion is correct but reason is incorrect. If assertion is incorrect but reason is correct. Read the following Assertion and Reason and answer as per the options given below :

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(a) (b) (c) (d)

[1998 - 2 Marks] Assertion : LiCl is predominantly a covalent compound. Reason : Electronegativity difference between Li and Cl is too small. If both assertion and reason are correct, and reason is the correct explanation of the assertion. If both assertion and reason are correct, but reason is not the correct explanation of the assertion. If assertion is correct but reason is incorrect. If assertion is incorrect but reason is correct.

39. Write the Lewis dot structure of the following : O3, COCl2 40.

(i)

[1986 - 1 Mark] Write the Lewis dot structural formula for each of 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 H3O+ :

; (ii)

; (iii) CN–; (iv) NCS– [1983 - 1 × 4 = 4 Marks]

1.

The compound that has the largest H – M – H bond angle (M = N, S, C), is : [Main Sep. 05, 2020 (II)] (a) H2O (b) NH3 (c) H2S

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(d) CH4 2. The molecule in which hybrid MOs involve only one d-orbital of the central atom is : [Main Sep. 04, 2020 (II)] (a) [Ni(CN)4]2– (b) BrF5 (c) XeF4 (d) [CrF6]3– 3. If AB4 molecule is a polar molecule, a possible geometry of AB4 is : [Main Sep. 02, 2020 (I)] (a) Square pyramidal (b) Tetrahedral (c) Rectangular planar (d) Square planar 4. The shape/ structure of [XeF5]– and XeO3F2, respectively, are : [Main Sep. 02, 2020 (II)] (a) pentagonal planar and trigonal bipyramidal (b) octahedral and square pyramidal (c) trigonal bipyramidal and pentagonal planar (d) trigonal bipyramidal and trigonal bipyramidal 5. The molecular geometry of SF6 is octahedral. What is the geometry of SF4 (including lone pair(s) of electrons, if any)? [Main Sep. 02, 2020 (II)] (a) Tetrahedral (b) Trigonal bipyramidal (c) Pyramidal (d) Square planar 6. The correct statement about ICl5 and ICl–is 4 : (a) both are is isostructural. [Main April 8, 2019 (II)] (b) ICl5 is trigonal bipyramidal and ICl is tetrahedral. (c) ICl5 is square pyramidal and ICl4– is tetrahedral. (d) ICl5 is square pyramidal and ICl4– is square planar. – 4

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7. (a) (b) (c) (d) 8.

The ion that has sp3d2 hydridisation for the central atom, is: [Main April 8, 2019 (II)] [ICl4]– [ICl2]– [IF6]– [BrF2]– Total number of lone pair of electrons in

ion is : [Main 2018]

(a) (a) (c) (d) 9. (a) (b) (c) (d) 10. (a) (b) (c) (d) 11. (a) (b) (c) (d) 12.

3 6 9 12 The incorrect geometry is represented by __________. NF3- trigonal planar [Main Online April 16, 2018] BF3- trigonal planar AsF5- trigonal bipyramidal H2O- bent Identify the pair in which the geometry of the species is T-shape and square pyramidal, respectively [Main Online April 15, 2018 (I)] – ICl2and ICl5 IO3– and IO2F–2 ClF3 and IO–4 XeOF2 and XeOF4 sp3d2 Hybridisation is not displayed by : [Main Online April 8, 2017] BrF5 SF6 [CrF6]3– PF5 The species in which the N atom is in a state of sp hybridisation is : [Main 2016]

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(a) (b) (c) (d) 13. (a) (b) (c) (d) 14.

The group of molecules having identical shape is : [Main Online April 9, 2016] PCl5, IF5, XeO2F2 BF3, PCl3, XeO3 SF4, XeF4, CCl4 ClF3, XeOF2, XeF3+ ion in CaC2 are: The number and type of bonds in [Main Online April 9, 2014]

(a) (b) (c) (d) 15. (a) (b) (c) (d) 16.

One σ bond and one π-bond One σ bond and two π-bond Two σ bond and two π-bond Two σ bond and one π-bond The shape of IF–6 is :

[Main Online April 23, 2013]

Trigonally distorted octahedron Pyramidal Octahedral Square antiprism The species having pyramidal shape is : [2010]

(a) SO3 (b) BrF3 (c) (d) OSF2 17. Which species has the maximum number of lone pair of electrons on the central atom? [2005S] − (a) [ClO3]

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18.

(b) XeF4 (c) SF4 (d) [I3]− Which of the following are isoelectronic and isostructural? CO32–, ClO3–, SO3

NO3–, [2003S]

(a) NO , CO (b) SO3 ,NO (c) ClO , CO (d) CO32–, SO3 19. Specify the coordination geometry around and hybridisation of N and B atoms in a 1 : 1 complex of BF3 and NH3 [2002S] 3 3 (a) N : tetrahedral, sp ; B : tetrahedral, sp (b) N : pyramidal, sp3; B : pyramidal, sp3 (c) N : pyramidal, sp3; B : planar, sp2 (d) N : pyramidal, sp3; B : tetrahedral, sp3 20. The correct order of hybridization of the central atom in the following species NH3, [PtCl4]2 –, PCl5 and BCl3 is [2001S] 2 3 2 3 3 2 3 2 (a) dsp , dsp , sp , sp (b) sp , dsp , sp d, sp 2 2 3 3 (c) dsp , sp , sp , dsp (d) dsp2, sp3, sp2, dsp3 , and 21. The hybridisation of atomic orbitals of nitrogen in – 3 – 3

2– 3 2– 3

– 3

are [2000S] (a) (b) (c) (d) 22. (a) (b) (c) (d) 23.

sp, sp and sp respectively sp, sp2 and sp3 respectively sp2, sp and sp3 respectively sp2, sp3 and sp respectively Molecular shapes of SF4, CF4 and XeF4 are 3

2

[2000S]

the same, with 2, 0 and 1 lone pairs of electrons respectively the same, with 1, 1 and 1 lone pairs of electrons respectively different, with 0, 1 and 2 lone pairs of electrons respectively different, with 1, 0 and 2 lone pairs of electrons respectively The geometry and the type of hybrid orbital present about the central atom in BF3 is

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[1998 - 2 Marks] (a) (b) (c) (d) 24. (a) (b) (c) (d) 25.

linear, sp trigonal planar, sp2 tetrahedral, sp3 pyramidal, sp3. Which one of the following compounds has sp2 hydridization? [1997 - 1 Mark] CO2 SO2 N2O CO Among the following species, identify the isostructural pairs. NF3, , BF3, H3O+, HN3 [1996 - 1 Mark]

(a) (b) (c) (d) 26.

The type of hybrid orbitals used by the chlorine atom in

(a) (b) (c) (d) 27.

sp sp2 sp none of these The compound in which formation is :

is

[1992 - 1 Mark] 3

uses its sp3 hybrid orbitals for bond [1989 - 1 Mark]

(a) (b) (c) (d) 28.

H OOH (H2N)2 O (CH3)3 OH CH3 HO The species in which the central atom uses sp2 hybrid orbitals in its bonding is

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[1988 - 1 Mark] (a) PH3 (b) NH3 (c) (d) SbH3 29. The hybridisation of sulphur in sulphur dioxide is : [1986 - 1 Mark] (a) (b) (c) (d)

sp sp3 sp2 dsp2

30.

The sum of the number of lone pairs of electrons on each central atom in the following species is [TeBr6]2-, [BrF2]+, SNF3 and [XeF3]– (Atomic numbers: N = 7, F = 9, S = 16, Br = 35, Te = 52,Xe = 54) [Adv. 2017] , N2O, , O3, 31. Among the triatomic molecules/ions, BeCl2, SCl2,

,

and

XeF2,

the

total

number

of

linear

molecule(s)/ion(s) where the hybridization of the central atom does not have contribution from the d-orbital(s) is [Atomic number : S = 16, Cl = 17, I = 53 and Xe = 54] [Adv. 2015] 32. A list of species having the formula XZ4 is given below. XeF4, SF4, SiF4, BF4–, BrF4–, [Cu(NH3)4]2+, [FeCl4]2–, [CoCl4]2– and [PtCl4]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 [Adv. 2014] 33. Based on VSEPR theory, the number of 90 degree F–Br–F angles in BrF5 is [2010] 34.

The shape of [CH3]+ is ............... .

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35.

[1990 - 1 Mark] The angle between two covalent bonds is maximum in ................. . (CH4, H2O, CO2) [1981 - 1 Mark]

36. 37. 38.

39. (a) 40. (a) (b) (c)

sp2 hybrid orbitals have equal s and p character. [1987 - 1 Mark] In benzene, carbon uses all the three p-orbitals for hybridisation. [1987 - 1 Mark] SnCl2 is a non-linear molecule. [1985 - ½ Mark] The compound(s) with TWO lone pairs of electrons on the central atom is(are [Adv. 2016] BrF5 (b) ClF3 (c) XeF4 (d) SF4 The linear structure is assumed by : [1991 - 1 Mark] SnCl2 NCO– CS2

(d) 41.

CO2 is isostructural with :

[1986 - 1 Mark]

(a) (b) (c) (d)

HgCl2 SnCl2 C2H2 NO2

42.

Draw the structure of XeF4 and OSF4 according to VSEPR theory, clearly indicating the state of hybridisation of the central atom and lone pair of electrons (if any) on the central atom. [2004 - 2 Marks]

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43.

Using VSEPR theory, draw the shape of PCl5 and BrF5. [2003 - 2 Marks]

44.

Interpret the non-linear shape of H2S molecule and non-planar shape of PCl3 using valence shell electron pair repulsion (VSEPR) theory. (Atomic numbers : H = 1, P = 15, S = 16, Cl = 17.) [1998 - 4 Marks] 45. Using the VSEPR theory, identify the type of hybridization and draw the structure of OF2. What are the oxidation states of O and F ? [1994 - 3 Marks]

1.

The potential energy curve for the H2 molecule as a function of internuclear distance is : [Main Sep. 05, 2020 (I)] (b)

(a)

(c) 2.

(d) The structure of PCl5 in the solid state is: [Main Sep. 05, 2020 (I)]

(a) (b) (c) (d) 3.

tetrahedral [PCl4] and octahedral [PCl6] square planar [PCl4]+ and octahedral [PCl6]– square pyramidal trigonal bipyramidal Of the species, NO, NO+, NO2+ and NO–, the one with minimum bond strength is : [Main Sep. 03, 2020 (I)] (a) NO+ +

–

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(b) NO (c) NO2+ (d) NO– 4. If the magnetic moment of a dioxygen species is 1.73 B.M, it may be: [Main Jan. 09, 2020 (I)] – + (a) O2 or O2 (b) O2 or O2+ (c) O2 or O2– (d) O2, O2– or O2+ 5. The bond order and the magnetic characteristics of CN– are: [Main Jan. 07, 2020 (II)] (a)

, diamagnetic

(b) 3, diamagnetic (c) 3, paramagnetic (d)

, paramagnetic

6.

Among the following, the molecule expected to be stabilised by anion formation is: [Main April 9, 2019 (I)] C2, O2, NO, F2 (a) C2 (b) F2 (c) NO (d) O2 7. Among the following species, the diamagnetic molecule is: [Main April 9, 2019 (II)] (a) NO (b) CO (c) B2 (d) O2 , O2 8. Among the following molecules/ions, Which one is diamagnetic and has the shortest bond length? [Main April 8, 2019 (II)] (a) O2

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(b) (c) (d) 9.

Two pi and half sigma bonds are present in: [Main Jan. 10, 2019 (I)]

(a) (b) (c) (d) 10.

(a) (b) (c) (d) 11.

O N2 O2 N2+ According to molecular orbital theory, which of the following is true with respect to Li2+ and Li2– ? [Main Jan. 9, 2019 (I)] + – Li2 is unstable and Li2 is stable Li2+ is stable and Li2– is unstable Both are stable Both are unstable According to molecular orbital theory, which of the following will not be a viable molecule? [Main 2018] + 2

(a) (b) (c) (d) 12. (I) (II) H – N - - -N - - -N In hydrogen azide, the bond orders of bonds (I) and (II) are _________. [Main Online April 15, 2018 (I)] (a) I < 2, II > 2 (b) I > 2, II > 2 (c) I > 2, II < 2 (d) I < 2, II < 2

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13.

Which of the following best describes the diagram of molecular orbital? [Main Online April 15, 2018 (II)]

(a) (b) (c) (d) 14.

A bonding π orbital A non-bonding orbital An antibonding σ orbital An antibonding π orbital Which of the following is paramagnetic ? [Main Online April 8, 2017] (a) NO+ (b) CO (c) (d) B2 15. Which of the following species is not paramagnetic ? [Main 2017] (a) (b) (c) (d) 16.

NO CO O2 B2 After understanding the assertion and reason, choose the correct option. [Main Online April 10, 2015] Assertion : In the bonding molecular orbital (MO) of H2, electron density is increased between the nuclei. Reason : The bonding MO is ΨA + ΨB, which shows destructive interference of the combining electron waves. (a) Assertion is incorrect, reason is correct. (b) Assertion is correct, reason is incorrect. (c) Assertion and reason are correct and reason is the correct explanation for the assertion.

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(d) Assertion and reason are correct, but reason is not the correct explanation for the assertion. 17. Which one of the following properties is not shown by NO? [Main 2014] (a) It is diamagnetic in gaseous state (b) It is neutral oxide (c) It combines with oxygen to form nitrogen dioxide (d) Its bond order is 2.5 18. Which one of the following molecules is paramagnetic? [Main Online April 19, 2014] (a) N2 (b) NO (c) C O (d) O3 19. Assuming 2s-2p mixing is NOT operative, the paramagnetic species among the following is [Adv. 2014] (a) Be2 (b) B2 (c) C2 (d) N2 20. In which of the following pairs of molecules/ions, both the species are not likely to exist ? [Main 2013] (a) (b) (c) (d) 21.

Which one of the following molecules is expected to exhibit diamagnetic behaviour ? [Main 2013] (a) C2 (b) N2 (c) O2

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(d) S2 22. Which of the following is the wrong statement (a) ONCl and ONO– are not isoelectronic. [Main 2013] (b) O3 molecule is bent (c) Ozone is violet-black in solid state (d) Ozone is diamagnetic gas. and 23. Stability of the species Li2,

increases in the order of : [Main 2013]

(a) (b) (c) (d) 24.

(a) (b) (c) (d) 25.

Bond order normally gives idea of stability of a molecular species. All the molecules viz. H2, Li2 and B2 have the same bond order yet they are not equally stable. Their stability order is [Main Online April 22, 2013] H2 > B2 > Li2 Li2 > H2 > B2 Li2 > B2 > H2 B2 > H2 > Li2 In which of the following ionization processes the bond energy has increased and also the magnetic behaviour has changed from paramagnetic to diamagnetic ? [Main Online April 9, 2013]

(a) (b) (c) (d)

26. Assuming that Hund’s rule is violated, the bond order and magnetic nature of the diatomic

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molecule B2 is [2010] (a) (b) (c) (d) 27.

1 and diamagnetic 0 and dimagnetic 1 and paramagnetic 0 and paramagnetic The species having bond order different from that in CO is

(a) (b) (c) (d) 28.

NO NO+ CN– N2 Among the following, the paramagnetic compound is

[2007] –

[2007] (a) (b) (c) (d) 29.

(a) (b) (c) (d) 30. (a) (b) (c)

Na2O2 O3 N2O KO2 According to molecular orbital theory which of the following statement about the magnetic character and bond order is correct regarding [2004S] Paramagnetic and Bond order < O2 Paramagnetic and Bond order > O2 Diamagnetic and Bond order < O2 Diamagnetic and Bond order > O2 Which of the following molecular species has unpaired electron(s) ? [2002S] N2 F2

(d) 31.

Identify the least stable ion amongst the following : [2002S]

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(a) (b) (c) (d) 32. (a) (b) (c) (d) 33.

Li– Be– B– C– The common features among the species CN–, CO and NO+ are [2001S] bond order three and isoelectronic bond order three and weak field ligands bond order two and π−acceptors isoelectronic and weak field ligands The correct order of increasing C — O bond length of CO, CO , CO2, is < CO2 < CO

(a) (c) CO < 34.

(a) (b) (c) (d) 35. (a) (b) (c) (d) 36. (a) (b) (c)

[1999 - 2 Marks] < CO2

(b) CO2
ion-ion (b) ion-dipole > ion-ion > dipole-dipole (c) ion-dipole > dipole-dipole > ion-ion (d) ion-ion > ion-dipole > dipole-dipole 5. 0.5 moles of gas A and x moles of gas B exert a pressure of 200 Pa in a container of volume 10 m3 at 1000 K. Given R is the gas constant in JK–1 mol–1, x is: [Main Jan. 9, 2019 (I)]

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(a) (b) (c) (d) 6.

(a) (b) (c) (d) 7.

(a) (b) (c) (d) 8.

(a) (b) (c) (d) 9.

An open vessel at 27°C is heated until two fifth of the air (assumed as an ideal gas) in it has escaped from the vessel. Assuming that the volume of the vessel remains constant, the temperature at which the vessel has been heated is : [Main Jan. 12, 2019 (II)] 500 ºC 500 K 750 °C 750 K Assuming ideal gas behaviour, the ratio of density of ammonia to that of hydrogen chloride at same temperature and pressure is: (Atomic wt. of Cl = 35.5u) [Main Online April 16, 2018] 1.46 1.64 0.46 0.64 At 300 K, the density of a certain gaseous molecule at 2 bar is double to that of dinitrogen (N2) at 4 bar. The molar mass of gaseous molecule is :[Main Online April 9, 2017] 28 g mol–1 56 g mol–1 112 g mol–1 224 g mol–1 Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure pi and temperature T1 are connected through a narrow tube

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of negligible volume. The temperature of one of the bulbs is then raised to T2. The final pressure pf is : [Main 2016]

(a) (b) (c) (d) 10. (a) (b) (c) (d) 11. (a) (b) (c) (d) 12.

The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is : [Main 2015] London force hydrogen bond ion - ion interaction ion - dipole interaction When does a gas deviate the most from its ideal behaviour? [Main Online April 11, 2015] At low pressure and low temperature At low pressure and high temperature At high pressure and low temperature At high pressure and high temperature Sulphur dioxide and oxygen were allowed to diffuse through a porous partition. 20 dm3 of SO2 diffuses through the porous partition in 60 seconds. The volume of O2 in dm3 which diffuses under the similar condition in 30 seconds will be (atomic mass of sulphur = 32 u):

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[Main Online April 19, 2014] (a) (b) (c) (d) 13. (a) (b) (c) (d) 14. (a) (b) (c) (d) 15.

7.09 14.1 10.0 28.2 The ratio of the rate of diffusion of helium and methane under identical condition of pressure and temperature will be [2005S] 4 2 1 0.5 Positive deviation from ideal behaviour takes place because of [2003S] Molecular interaction between atoms and PV/nRT > 1 Molecular interaction between atoms and PV/nRT < 1 Finite size of atoms and PV/nRT > 1 Finite size of atoms and PV/nRT < 1 Which of the following volume (V) - temperature (T) plots represents the behaviour of one mole of an ideal gas at one atmospheric pressure ? [2002S]

(a)

(b)

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(c)

(d)

16.

(a) (b) (c) (d) 17.. (a) (b) (c) (d) 18.

(a) (b) (c) (d) 19.

At 100°C and 1 atm, if the density of liquid water is 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 litre of steam at that temperature is [2000S] 3 6 cm 60 cm3 0.6 cm3 0.06 cm3 A gas will approach ideal behaviour at [1999 - 2 Marks] low temperature and low pressure. low temperature and high pressure. high temperature and low pressure. high temperature and high pressure. X mL of H2 gas effuses through a hole in a container in 5 seconds. The time taken for the effusion of the same volume of the gas specified below under identical conditions is : [1996 - 1 Mark] 10 seconds : He 20 seconds : O2 25 seconds : CO 55 seconds : CO2 One mole of N2O4(g) at 300 K is kept in a closed container under one atmosphere. It is heated to 600 K when 20% by mass of N2O4 (g) decomposes to NO2(g).The resultant pressure is : [1996 - 1 Mark]

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(a) (b) (c) (d) 20. (a) (b) (c) (d) 21. (a) (b) (c) (d) 22. (a) (b) (c) (d) 23.

(a) (b) (c) (d) 24. (a) (b)

1.2 atm 2.4 atm 2.0 atm 1.0 atm At constant volume, for a fixed number of moles of a gas the pressure of the gas increases with rise in temperature due to [1992 - 1 Mark] Increase in average molecular speed Increased rate of collisions amongst molecules Increase in molecular attraction Decrease in mean free path The density of neon will be highest at [1990 - 1 Mark] S.T.P. 0ºC, 2 atm 273ºC, 1 atm. 273ºC, 2 atm. The rate of diffusion of methane at a given temperature is twice that of a gas X. The molecular weight of X is [1990 - 1 Mark] 64.0 32.0 4.0 8.0 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 - 1 Mark] at the centre of the tube. near the hydrogen chloride bottle. near the ammonia bottle. throughout the length of the tube. Rate of diffusion of a gas is : [1985 - 1 Mark] directly proportional to its density. directly proportional to its molecular weight.

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(c) directly proportional to the square root of its molecular weight. (d) inversely proportional to the square root of its molecular weight. 25. Equal weights of methane and hydrogen are mixed in an empty container at 25ºC. The fraction of the total pressure exerted by hydrogen is : [1984 - 1 Mark] (a) (b) (c) (d) 26.

Equal weights of methane and oxygen are mixed in an empty container at 25ºC. The fraction of the total pressure exerted by oxygen is [1981 - 1 Mark]

(a) (b) (c) (d) 27. (a) (b) (c) (d)

The temperature at which a real gas obeys the ideal gas laws over a wide range of pressure is [1981 - 1 Mark] Critical temperature Boyle temperature Inversion temperature Reduced temperature

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28.

A closed tank has two compartments A and B, both filled with oxygen (assumed to be ideal gas). The partition separating the two compartments is fixed and is a perfect heat insulator (Figure 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 (Figure 2), the volume (in m3) of the compartment A after the system attains equilibrium is_______ [Adv. 2018]

Figure 1

Figure 2 29. The diffusion coefficient of an ideal gas is proportional to its mean free path and mean speed. The absolute temperature of 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 [Adv. 2016] 30. To an evacuated vessel with movable piston under external pressure of 1 atm, 0.1 mol of He and 1.0 mol 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] 31.

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.

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[Main Sep. 06, 2020 (I)] 32. 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 = d0 at which the electron-electron repulsion and the nucleus-nucleus repulsion energies are absent? As reference, the potential energy of H atom is taken as zero when its electron and the nucleus are infinitely far apart. [Adv. 2020] 23 –1 Use Avogadro constant as 6.023 × 10 mol .

33. The degree of dissociation is 0.4 at 400 K and 1.0 atm for the PCl3 + Cl2. Assuming ideal behaviour of all gaseous reaction PCl5 gases, calculate the density of equilibrium mixture at 400 K and 1.0 atmosphere. (Relative atomic mass of P = 31.0 and Cl = 35.5) [1998 - 3 Marks] 34. For the reaction, N2O5(g) → 2NO2(g) + 0.5 O2(g), calculate the mole fraction of 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. [1998 - 3 Marks] 35. 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 - 4 Marks]

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36. A straight glass tube has two inlets X and Y at two ends. The length of tube is 200 cm. HCl gas through inlets X and NH3 gas through inlet Y are allowed to enter the tube at the same time. What fumes appear at point P inside the tube. Find distance of P from X. [1980] 37. Calculate density of NH3 at 30°C and 5 atm pressure. [1978] 38.

The value of PV for 5.6 litres of an ideal gas is .............. RT, at N.T.P. [1987 - 1 Mark] 39. The rate of diffusion of gas is ............... proportional to both ............... and square root of molecular mass. [1986 - 1 Mark] 40. Cp – Cv for an ideal gas is ............... . [1984 - 1 Mark] 41.

A mixture of ideal gases is cooled upto liquid helium temperature (4.22 K) to form an ideal solution. [1996 - 1 Mark]

42.

Refer to the figure given : [2006 - 5M; –1]

Which of the following statements is wrong?

(a) For gas A, a = 0 and Z will linearly depend on pressure (b) For gas B, b = 0 and Z will linearly depend on pressure

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(c) Gas C is a real gas and we can find ‘a’ and ‘b’ if intersection data is given (d) All van der Waal gases will behave like gas C and give positive slope at high pressure 43. According to Graham’s law, at a given temperature the ratio of the rates of diffusion rA/rB of gases A and B is given by [1998 - 2 Marks] 1/2 (a) (PA/PB) (MA/MB) (b) (MA/MB) (PA/PB)1/2 (c) (PA/PB) (MB/MA)1/2 (d) (MA/MB) (PB/PA)1/2 (Where P and M are pressures and molecular weights of gases A and B respectively.) 44. 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 - 1 Mark] (a) 1 : 2 (b) 1 : 1 (c) 1 : 16 (d) 15 : 16 45. If a gas is expanded at constant temperature : [1986 - 1 Mark] (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 46. When an ideal gas undergoes unrestrained expansion, no cooling occurs because the molecules : [1984 - 1 Mark] (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

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47.

Match the type of interaction in column A with the distance dependence of their interaction energy in column B : [Main Sep. 02, 2020 (II)] A B (I) ion-ion (A) (II)

dipole-dipole

(B)

(III) London dispersion

(C) (D)

(a) (b) (c) (d)

(I)-(B), (II)-(D), (III)-(C) (I)-(A), (II)-(B), (III)-(D) (I)-(A), (II)-(B), (III)-(C) (I)-(A), (II)-(C), (III)-(D)

X and Y are two volatile liquids with molar weights of 10 g mol–1 and 40 g mol–1 respectively. 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 atmosphere 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.

48. The value of d in cm (shown in the figure), as estimated from Graham’s law, is [Adv. 2014] (a) 8

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(b) (c) (d) 49. (a) (b) (c) (d)

12 16 20 The experimental value of d is found to be smaller than the estimate obtained using Graham’s law. This is due to [Adv. 2014] Larger mean free path for X as compared to that of Y Larger mean free path for Y as compared to that of X Increased collision frequency of Y with the inert gas as compared to that of X with the inert gas Increased collision frequency of X with the inert gas as compared to that of Y with the inert gas

50. 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. [2002 - 5 Marks] (a) Determine (i) molecular weight, (ii) molar volume, (iii) compression factor (Z) of the vapour and (iv) which forces among the gas molecules are dominating, the attractive or the repulsive? (b) If the vapour behaves ideally at 1000 K, determine the average translational kinetic energy of a molecule. 51. One mole of nitrogen gas at 0.8 atm takes 38 s to diffuse through a pinhole, whereas one mole of an unknown compound of xenon with flourine at 1.6 atm takes 57 s to diffuse through the same hole. Calculate the molecular formula of the compound. [1999 - 5 Marks] 52. 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 degrees at the same volume, the pressure increases by 10%. Calculate the temperature t and volume V. (Molecular weight of the gas = 120.) [1999 - 5 Marks]

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53. 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 300K. Determine the molar mass of the gas. [1998 - 3 Marks] 54. A mixture of ethane (C2H6) and ethene (C2H4) occupies 40 litres at 1.00 atm and at 400 K. The mixture reacts completely with 130 g of O2 to produce CO2 and H2O. Assuming ideal gas behaviour, calculate the mole fractions of C2H4 and C2H6 in the mixture. [1995 - 4 Marks] ), which is 55. The composition of the equilibrium mixture ( attained at 1200°C, is determined by measuring the rate of effusion through a pin–hole. It is observed that at 1.80 mmHg pressure, the mixture effuses 1.16 times as fast as krypton effuses under the same conditions. Calculate the fraction of the chlorine molecules dissociated into atoms. (Relative atomic mass of Kr = 84.) [1995 - 4 Marks] 3 56. A 20.0 cm 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 cm3. A further contraction of 14.0 cm3 occurs when the residual gas is treated with KOH solution. Find out the composition of the gaseous mixture in terms of volume percentage. [1995 - 4 Marks] 57. 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 - 2 Marks] 58. 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 27º C, the weight of the full cylinder reduces to 23.2 kg. Find out the volume of the gas in cubic meters used up at the normal usage conditions, find the final pressure inside the cylinder. Assume LPG to be n-butane with normal boiling point of 0ºC. [1994 - 3 Marks]

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59. At 27ºC, hydrogen is leaked through a tiny hole into a vessel for 20 minutes. Another unknown gas at the same temperature and pressure as that of H2 is leaked through the same hole for 20 minutes. After the effusion of the gases the mixture exerts a pressure of 6 atmosphere. The hydrogen content of the mixture is 0.7 mole. If the volume of the container is 3 litres, what is the molecular weight of the unknown gas? [1992 - 3 Marks] 60. At room temperature the following reactions proceed nearly to completion : [1992 - 4 Marks] 2NO + O2 → 2NO2 → N2O4 The dimer, N2O4, solidifies at 262 K. A 250 mL flask and a 100 mL. flask are separated by a stop-cock. 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 at 220K. 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). 61. Calculate the volume occupied by 5.0 g of acetylene gas at 50ºC and 740 mm pressure. [1991 - 2 Marks] 62. A spherical balloon of 21 cm diameter is to be filled up with hydrogen at N.T.P. from a cylinder containing the gas at 20 atmospheres at 27ºC. If the cylinder can hold 2.82 litres of water, calculate the number of balloons that can be filled up. [1987 - 5 Marks] 63. Oxygen is present in 1 litre flask at a pressure of 7.6 × 10–10 mm of Hg. Calculate the number of oxygen molecules in the flask at 0ºC. [1983 - 2 Marks] 64. When 2 g of a gas A is introduced into an evaluated flask kept at 25ºC, 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 weightsMA : MB. [1983 - 2 Marks]

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65. 1 litre of mixture of CO and CO2 is taken. The mixture is passed through a tube containing red hot charcoal. The volume now becomes 1.6 litre. The volumes are measured under the same conditions. Find the composition of mixture by volume. [1980] 66. 3.7 g of a gas at 25°C occupied the same volume as 0.184g of hydrogen at 17°C and at the same pressure. What is the molecular weight of the gas? [1979]

1.

Identify the correct labels of A, B and C in the following graph from the options given below:

Root mean square speed (Vrms); most probable speed (Vmp); average speed (Vav) [Main Jan. 07, 2020 (II)] (a) A – Vmp; B – Vrms; C – Vav (b) A – Vav; B – Vrms; C – Vmp (c) A – Vrms; B – Vmp; C – Vav (d) A – Vmp ; B – Vav ; C – Vrms 2. 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 [Adv. 2020]

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(a) (b) (c) (d) 3.

1:1:1 1 : 1 : 1.224 1 : 1.128 : 1.224 1 : 1.128 : 1 Points I, II and III in the following plot respectively correspond to (Vmp : most probable velocity) [Main April 10, 2019 (II)]

(a) (b) (c) (d) 4.

Vmp of N2 (300 K); Vmp of O2 (400 K); Vmp of H2 (300 K) Vmp of O2 (400 K); Vmp of N2 (300 K); Vmp of H2 (300 K) Vmp of N2 (300 K); Vmp of H2 (300 K); Vmp of O2 (400 K) Vmp of H2 (300 K); Vmp of N2 (300 K); Vmp of O2 (400 K) Initially, the root mean square (rms) velocity of N2 molecules at certain temperature is u. If this temperature is doubled and all the nitrogen molecules dissociate into nitrogen atoms, then the rms velocity will be : [Main Online April 10, 2016] (a) 2u

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(b) 14 u (c) 4u (d) u/2 5. Which of the following is not an assumption of the kinetic theory of gases? [Main Online April 10, 2015] (a) Gas particles have negligible volume. (b) A gas consists of many identical particles which are in continual motion. (c) At high pressure, gas particles are difficult to compress. (d) Collisions of gas particles are perfectly elastic. 6. The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is 1 : 4. The ratio of number of their molecules is: [Main 2014] (a) 1 : 4 (b) 7 : 32 (c) 1 : 8 (d) 3 : 16 7. The temperature at which oxygen molecules have the same root mean square speed as helium atoms have at 300 K is: Atomic masses: He = 4 u, O = 16 u) [Main Online April 9, 2014] (a) 300 K (b) 600 K (c) 1200 K (d) 2400 K 8. For gaseous state, if most probable speed is denoted by C*, average and mean square speed by C, then for a large number of speed by molecules the ratios of these speeds are : [Main 2013] (a) C* : : C = 1.225 : 1.128 : 1 (b) C* : : C = 1.128 : 1.225 : 1 (c) C* : : C = 1 : 1.128 : 1.225 (d) C* : : C = 1 : 1.225 : 1.128

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9.

(a) (b) (c) (d) 10.

By how many folds the temperature of a gas would increase when the root mean square velocity of the gas molecules in a container of fixed volume is increased from 5 × 104 cm/s to 10 × 104 cm/s ? [Main Online April 9, 2013] Two Three Six Four The root mean square velocity of one mole of a monoatomic gas having molar mass M is ur.m.s.. The relation between the average kinetic energy (E) of the gas and ur.m.s. is [2004S] (a) (b) (c) (d) 11. The root mean square velocity of an ideal gas at constant pressure varies with density (d) as [2001S] 2 d d

(a) (b) (c) (d) 1/ 12.

The rms velocity of hydrogen is

times the rms velocity of

nitrogen. If T is the temperature of the gas, then [2000S] (a) (b) (c) (d) 13.

T(H2)= T(N2) T(H2) > T(N2) T(H2) < T(N2) T(N2) T(H2) = The ratio between the root mean square speed of H2 at 50 K and that of O2 at 800 K is,

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[1996 - 1 Mark] (a) (b) (c) (d) 14.

4 2 1 1/4 Longest mean free path stands for : [1995S]

(a) (b) (c) (d) 15. (a) (b) (c) (d) 16. (a) (b) (c) (d) 17. (a) (b) (c) (d)

H2 N2 O2 Cl2 According to kinetic theory of gases, for a diatomic molecule [1991 - 1 Mark] the pressure exerted by the gas is proportional to mean velocity of the molecule the pressure exerted by the gas is proportional to the root mean velocity of the molecule the root mean square velocity of the molecule is inversely proportional to the temperature the mean translational kinetic energy of the molecule is proportional to the absolute temperature. The average velocity of an ideal gas molecule at 27ºC is 0.3 m/sec. The average velocity at 927ºC will be: [1986 - 1 Mark] 0.6 m/sec 0.3 m/sec 0.9 m/sec 3.0 m/sec Helium atom is two times heavier than a hydrogen molecule. At 298 K, the average kinetic energy of a helium atom is [1982 - 1 Mark] two times that of a hydrogen molecule. same as that of a hydrogen molecule. four times that of a hydrogen molecule. half that of a hydrogen molecule.

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18. (a) (b) (c) (d)

The ratio of root mean square velocity to average velocity of a gas molecule at a particular temperature is [1981 - 1 Mark] 1.086 : 1 1 : 1.086 2 : 1.086 1.086 : 2

19. 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] 20. The average velocity of gas molecules is 400 m/sec. Calculate its rms velocity at the same temperature. [2003 - 2 Marks] 21.

Eight gram each of oxygen and hydrogen at 27ºC will have the total kinetic energy in the ratio of ............... . [1989 - 1 Mark] 22. The total energy of one mole of an ideal monoatomic gas at 27ºC is ............... calories. [1984 - 1 Mark]

23. 24.

Kinetic energy of a molecule is zero at 0ºC. [1985 - ½ Mark] A gas in a closed container will exert much higher pressure due to gravity at the bottom than at the top. [1985 - ½ Mark]

25. Which of the following statement(s) is (are) correct regarding the root mean square speed (Urms) of a molecule in a gas at equilibrium? [Adv. 2019] (a) Eav at a given temperature does not depend on its molecular mass

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(b) Urms is inversely proportional to the square root of its molecular mass (c) Urms is doubled when its temperature is increased four times (d) Eav is doubled when its temperature is increased four times 26. According to kinetic theory of gases [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 27.

(a) (b) (c) (d)

Read the following statement and explanation and answer as per the options given below : Assertion : The pressure of a fixed amount of an ideal gas is proportional to its temperature Reason : Frequency of collisions and their impact both increase in proportion to the square root of temperature. [2000S] If both assertion and reason are CORRECT, and reason is the CORRECT explanation of the assertion. If both assertion and reason are CORRECT, but reason is NOT the CORRECT explanation of the assertion. If assertion is CORRECT, but reason is INCORRECT. If assertion is INCORRECT, but reason is CORRECT.

28. A gas bulb of 1 litre capacity contains 2.0 × 1021 molecules of nitrogen exerting a pressure of 7.57 × 103 Nm–2. Calculate the root mean square (r.m.s) speed and the temperature of the gas molecules. If the ratio of the most probable speed to the root mean square speed is 0.82, calculate the most probable speed for these molecules at this temperature. [1993 - 4 Marks] 29. The average velocity at T1K, and the most probable velocity at T2K of CO2 gas is 9.0 × 104 cm sec–1. Calculate the value of T1 and T2.

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[1990 - 4 Marks] 30. Calculate the root mean square velocity of ozone kept in a closed vessel at 20ºC and 82 cm mercury pressure. [1985 - 2 Marks] 31. Calculate the average of kinetic energy, in Joules of the molecules in 8.0 g of methane at 27ºC. [1982 - 2 Marks]

1.

Consider the following table : Gas a/(k Pa dm6mol–1) b/(dm3mol–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 : [Main April 10, 2019 (I)] (a) Gas C will occupy more volume than gas A; gas B will be more compressible than gas D (b) Gas C will occupy lesser volume than gas A; gas B will be lesser 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 lesser volume than gas A; gas B will be more compressible than gas D 2. Consider the van der Waals constants, a and b, for the following gases, Gas Ar Ne Kr Xe 6 –2 a/ (atm dm mol ) 1.3 0.2 5.1 4.1 b/ (10–2 dm3 mol–1) 3.2 1.7 1.0 5.0 Which gas is expected to have the highest critical temperature? [Main April 9, 2019 (I)] (a) Kr

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(b) Ne (c) Xe (d) Ar 3. At a given temperature T, gases Ne, Ar, Xe and Kr are found to deviate from ideal gas behaviour. Their equation of state is given as at T. Here, b is the van der Waals constant. Which gas will exhibit steepest increase in the plot of Z (compression factor) vs P? [Main April 9, 2019 (II)] (a) Xe (b) Kr (c) Ne (d) Ar 4. At very high pressures, the compressibility factor of one mole of a gas is given by : [Main Online April 9, 2016] (a) (b) (c) (d) 5.

The qualitative sketches I, II and III given below show the variation of surface tension with molar concentration of three different aqueous solutions of KCl, CH3OH and CH3(CH2)11OSO3– Na+ at room temperature. The correct assignment of the sketches is [Adv. 2016]

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(a) I : KCl II : CH3OH III : CH3(CH2)11OSO3– Na+ (b) I : CH3(CH2)11OSO3– Na+ II : CH3OH III : KCl (c) I : KCl II : CH3(CH2)11OSO3– Na+ III : CH3OH (d) I : CH3OH II : KCl III : CH3(CH2)11OSO3– Na+ 6. If Z is a compressibility factor, van der Waals equation at low pressure can be written as: [Main 2014] (a) (b) (c) (d) 7.

van der Waals equation for a gas is stated as, . [Main Online April 9, 2014]

This equation reduces to the perfect gas equation, (a) (b) (c) (d) 8.

when,

temperature is sufficient high and pressure is low. temperature is sufficient low and pressure is high. both temperature and pressure are very high. both temperature and pressure are very low. For one mole of a van der Waal’s gas when b = 0 and T = 300 K, the PV vs, 1/V plot is shown below. The value of the van der Waal’s

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constant a (atm. liter2 mol–2) is : [2012]

(a) (b) (c) (d) 9.

1.0 1.5 4.5 3.0 The term that corrects for the attractive forces present in a real gas in the van der Waals equation is [2009 - 3M; –1] (a) nb (b) (c)

(d) – nb 10. When the temperature is increased, surface tension of water [2002S] (a) (b) (c) (d) 11.

increases decreases remains constant shows irregular behaviour The compressibility of a gas is less than unity at STP. Therefore, [2000S]

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(a) (b) (c) (d) 12.

Vm > 22.4 litres Vm < 22.4 litres Vm = 22.4 litres Vm = 44.8 litres The compressibility factor for an ideal gas is [1997 - 1 Mark]

(a) 1.5 (b) 1.0 (c) 2.0 (d) 13.

∞ Arrange the van der Waals constant for the gases: [1995S]

I II III IV. (a) (b) (c) (d) 14.

(a) (b) (c) (d) 15. (a) (b)

C6H6(g) A. 0.217 C6H5.CH3(g) B. 5.464 Ne(g) C. 18.000 H2O(g) D. 24.060 I-A, II-D, III-C, IV-B I-D, II-A, III-B, IV-C I-C, II-D, III-A, IV-B I-B, II-C, III-A, IV-D The values of van der Waals constant ‘a’ for the gases O2, N2, NH3 and CH4 are 1.360, 1.390, 4.170 and 2.253 L2 atm mol–2 respectively. The gas which can most easily be liquified is : [1989 - 1 Mark] O2 N2 NH3 CH4 In van der Waals equation of state for a non-ideal gas, the term that accounts for intermolecular forces is [1988 - 1 Mark] (V – b) RT

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(c) (d) (RT)–1 16. Calculate the pressure exerted by one mole of CO2 gas at 273 K if the van der Waal's constant a = 3.592 dm6 atm mol–2. Assume that the volume occupied by CO2 molecules is negligible. [2000 - 2 Marks]

17.

In the van der Waal’s equation

the

constant ‘a’ reflects the actual volume of the gas molecules. [1993 - 1 Mark] 18. One mole of a monoatomic real gas satisfies the equationp(V – b) = RT where b is a constant. The relationship of interatomic potential V(r) and interatomic distance r for the gas is given by [Adv. 2015]

(a)

(b)

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(c)

(d)

19. (a) (b) (c) (d)

20.

A gas described by van der Waals equation – [2008- 1 Mark] behave similar to an ideal gas in the limit of large molar volumes behaves similar to an ideal gas is in limit of large pressures is characterised by van der Waals coefficients that are dependent on the identity of the gas but are independent of the temperature. has the pressure that is lower than the pressure exerted by the same gas behaving ideally Match gases under specified conditions listed in Column I with their properties/laws in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS. [2007] Column I Column II (A) Hydrogen gas (p) Compressibility (P = 200 atm, T = 273 K) factor ≠ 1 (B) Hydrogen gas (q) Attractive forces (P ~ 0, T = 273 K) are dominant (C) CO2 (P = 1 atm, T = 273 K) (r) PV = nRT (D) Real gas with very large (s) P(V – nb) = nRT

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molar volume 21.

Read the following statement and explanation and answer as per the options given below : Assertion : The value of van der Waals’constant ‘a’ is larger for ammonia than for nitrogen. Reason : Hydrogen bonding is present in ammonia. [1998 - 2 Marks] (a) If both assertion and reason are correct, and reason is the correct explanation of the assertion. (b) If both assertion and reason are correct, but reason is not the correct explanation of the assertion. (c) If assertion is correct but reason is incorrect. (d) If assertion is incorrect but reason is correct. 22. A graph is plotted between PVm along Y-axis and P alongX-axis, where Vm is the molar volume of a real gas. Find the intercept along Yaxis. [2004 - 2 Marks] 23. The compression factor (compressibility factor) for one mole of a van der Waals gas at 0°C and 100 atmospheric 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 - 5 Marks] 24. Using van der Waal’s equation, calculate the constant, ‘a’ when two moles of a gas confined in a four litre flask exerts a pressure of 11.0 atmospheres at a temperature of 300 K. The value of ‘b’ is 0.05 L mol– 1 . [1998 - 4 Marks] Topic-1 : Intermolecular Forces, Gas Laws and Ideal Gas Equation 1. (a)2(b) 3. (a)

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4. (d) 5. (d) 6. (b) 7. (c) 8. (c) 9. (a) 10. (b) 11 (c)

12. (b) 13. (b) 14. (c) 15. (c) 16. (c) 17. (c) 18. (b) 19. (b) 20. (a)

21. (b) 22. (a) 23. (b) 24. (d) 25. (b) 26. (a) 27. (b) 28. (2.22) 29. (4) 30. (7)

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31. (750) 32. (–5246.49) 33. (4.53) 34. (0.4) 35. (2.197) 36. (85.2) 37. (3.42) 41. (False) 42. (b) 43. (c) 44. (d) 45. (a, b) 46. (b) 47. (d) 48. (c) 49. (d) Topic-2 : Kinetic Theory of Gases and Molecular Speeds 1. (d) 2. (b) 3. (a) 4. (a) 5. (c) 6. (b) 7. (d) 8. (c) 9. (d) 10. (c) 11. (d)

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12. (c) 13. (c) 14. (a) 15. (d) 16. (a) 17. (b) 18. (a) 19. (4) 20. (434) 23. (False) 24. (False) 25. (a,b,c) 26. (a,b,c,d) 27. (a) Topic-3 : Deviation from Ideal Gas Behaviour, Liquification of Gases and Liquid State 1. (a) 2. (a) 3. (a) 4. (a) 5. (d) 6. (b)7. (a) 8. (b) 9. (b) 10. (b) 11. (b) 12. (b) 13. (c) 14. (c) 15. (c) 16. (0.99)

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17. (False) 18. (c) 19. (a, c)

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1. (a) (b) (c) (d) 2. (a) (b) (c) (d) 3. (A) (B) (C)

The true statement amongst the following is : [Main Jan. 09, 2020 (II)] Both ∆S and S are functions of temperature. Both S and ∆S are not functions of temperature. S is not a function of temperature but ∆S is a function of temperature. S is a function of temperature but ∆S is not a function of temperature. Five moles of an ideal gas at 1 bar and 298 K is expanded into vacuum to double the volume. The work done is : [Main Sep. 04, 2020 (II)] CV(T2 – T1) –RT (V2 – V1) –RT ln V1/V1 zero Among the following, the set of parameters that represents path functions, is: [Main April 9, 2019 (I)] q+w q w (D) H – TS

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(a) (b) (c) (d) 4.

(a) (b) (c) (d) 5.

(a) (b) (c) (d) 6.

(B) and (C) (B), (C) and (D) (A) and (D) (A), (B) and (C) 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) [Main April 8, 2019 (I)] Cyclic process : q = –w Adiabatic process : ∆U = –w Isochoric process: ∆U = q Isothermal process: q = – w 5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If CV = 28 J K–1 mol– 1 , calculate ∆U and ∆pV for this process. (R = 8.0 J K–1 mol–1) [Main April 8, 2019 (II)] ∆U = 14 kJ; ∆(pV) = 18 kJ ∆U = 14 kJ; ∆(pV) = 0.8 kJ ∆U = 14 kJ; ∆(pV) = 4 kJ ∆U = 14 kJ; ∆(pV) = 8.0 kJ For a diatomic ideal gas in a closed system, which of the following plots does not correctly describe the relation between various thermodynamic quantities? [Main Jan. 12, 2019 (I)]

(a)

(b)

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(c)

(d)

7. (a) (b) (c) (d) 8.

(a) (b) (c) (d) 9.

The process with negative entropy change is: [Main Jan. 10, 2019 (II)] Dissociation of CaSO4(s) to CaO(s) and SO3(g) Sublimation of dry ice Dissolution of iodine in water Synthesis of ammonia from N2 and H2 Which of the following lines correctly show the temperature dependence of equilibrium constant, K, for an exothermic reaction? [2018]

A and B B and C C and D A and D For which of the following processes, ∆S is negative? [Main Online April 16, 2018]

(a) (b) (c) (d)

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10.

∆U is equal to [2017]

(a) (b) (c) (d) 11.

Isochoric work Isobaric work Adiabatic work Isothermal work A gas undergoes change from state A to state B. In this process, the heat absorbed and work done by the gas is 5 J and 8 J, respectively. Now gas is brought back to A by another process during which 3 J of heat is evolved. In this reverse process of B to A : [Main Online April 9, 2017] (a) 10 J of the work will be done by the gas. (b) 6 J of the work will be done by the gas. (c) 10 J of the work will be done by the surrounding on gas. (d) 6 J of the work will be done by the surrounding on gas. 12. If 100 mole of H2O2 decomposes at 1 bar and 300 K, the work done (kJ) by one mole of O2(g) as it expands against 1 bar pressure is : [Main Online April 10, 2016] H2O(l) + O2(g) 2H2O2(l) (R = 83 JK–1 mol–1) (a) 124.50 (b) 249.00 (c) 498.00 (d) 62.25 13. A reaction at 1 bar is non–spontaneous at low temperature but becomes spontaneous at high temperature. Identify the correct statement about the reaction among the following : [Main Online April 9, 2016] (a) ∆H is negative while ∆S is positive (b) Both ∆H and ∆S are negative (c) ∆H is positive while ∆S is negative (d) Both ∆H and ∆S are positive. 14. 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 (∆Ssurr) in JK–1 is

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(1 L atm = 101.3 J) [Adv. 2016] (a) (b) (c) (d) 15.

5.763 1.013 –1.013 –5.763 A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37.0 ºC. As it does so, it absorbs 208 J of heat. The values of q and w for the process will be: [2013] (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 16. Which of the following statements/relationships is not correct in thermodynamic changes? [Main Online April 23, 2013] (a) ∆U = 0 (isothermal reversible expansion of a gas) (b) w = – nRT ln

(c) w = nRT ln

(isothermal reversible expansion of an ideal gas) (isothermal reversible expansion of an ideal gas)

(d) For a system of constant volume heat involved directly changes to internal energy. 17. When one mole of monoatomic ideal gas at T K undergoes adiabatic change under a constant external pressure of1 atm, volume changes from 1 litre to 2 litre. The final temperature in Kelvin would be [2005S] (a) (b) T + (c) T

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(d) T – 18.

A mono-atomic 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; –1]

(a) (b) 2 R (c) 0 (d) 19.

(a) (b) (c) (d) 20.

(a) (b) (c) (d) 21. (a) (b) (c) (d) 22.

Two moles of an ideal gas is expanded isothermally and reversibly from 1 litre to 10 litres at 300 K. The enthalpy change (in kJ) for the process is [2004S] 11.4 kJ – 11.4 kJ 0 kJ 4.8 kJ One mole of a non-ideal gas undergoes a change of state (2.0 atm, 3.0L, 95K) → (4.0 atm, 5.0 L, 245K) with a change in internal energy, ∆U = 30.0L atm. The change in enthalpy (∆H) of the process in L atm is [2002S] 40.0 42.3 44.0 not defined, because pressure is not constant Which one of the following statements is false? [2001S] Work is a state function. Temperature is a state function. Change in the state is completely defined when the initial and final states are specified. Work appears at the boundary of the system. In thermodynamics, a process is called reversible when

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[2001S] (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. 23.

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 graphs below. If the work done along the solid line path ws and that along the dotted line path is wd, then the integer closest to the ratio wd / ws is : [2010]

24.

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 increase 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 - 6M]

25. For a dimerization reaction, kJ mol–1, at 298 K, be ___________ J.

2A(g) → A2(g), J K–1 mol–1, then the

will

[Main Sep. 05, 2020 (II)]

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26.

The magnitude of work done by a gas that undergoes a reversible expansion along the path ABC shown in the figure is ________ [Main Jan. 08, 2020 (I)]

27.

At constant volume, 4 mol of an ideal gas when heated from 300 K to 500 K changes its internal energy by 5000 J. The molar heat capacity at constant volume is ______. [Main Jan. 08, 2020 (II)] 28. Tin is obtained from cassiterite by reduction with coke. Use the data given below to determine the minimum temperature (in K) at which the reduction of cassiterite by coke would take place. [Adv. 2020] At 298 K:

Assume that the enthalpies and the entropies are temperature independent. Two moles of 29. For the reaction, 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)

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[2006 - 6M] 30. A sample of argon gas at 1 atm pressure and 27°C expands reversibly and adiabatically from 1.25 dm3 to 2.50 dm3. Calculate the enthalpy change in this process. CV, m for argon is 12.48 JK–1 mol–1. [2000 - 4 Marks] 31. An athlete is given 100 g of glcuose (C6H12O6) of energy equivalent to 1560 kJ. He utilizes 50 percent of this gained energy in the event. In order to avoids 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/mole. [1989 - 2 Marks] 32. 33.

Enthalpy is an ................ property. [1997 - 1 Mark] A system is said to be ................ if it can neither exchange matter nor energy with the surroundings. [1993 - 1 Mark]

34.

Heat capacity of a diatomic gas is higher than that of a monoatomic gas. [1985 - ½ Mark] 35. First law of thermodynamics is not adequate in predicting the direction of a process. [1982 - 1 Mark]

36.

In thermodynamics, the P – V work done is given by

For a system undergoing a particular process, the work done is,

This equation is applicable to a [Adv. 2020] (a) system that satisfies the van der Waals equation of state.

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(b) process that is reversible and isothermal. (c) process that is reversible and adiabatic. (d) process that is irreversible and at constant pressure. 37. 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. [Adv. 2018]

The correct option(s) is (are) (a) qAc = ∆UBC and wAB = P2(V2 – V1) (b) wBC = P2(V2 – V1) and qBC = ∆HAC (c) ∆HCA < ∆UCA and qAC = ∆UBC (d) qBC = ∆HAC and ∆HCA > ∆UCA 38. An ideal gas is expanded from (P1, V1, T1) to (P2, V2, T2) under different conditions. The correct statement(s) among the following is (are) [Adv. 2017] (a) The work done on the gas is maximum when it is compressed irreversibly from (P2, V2) to (P1, V1) against constant pressure P1 (b) If the expansion is carried out freely, it is simultaneously both isothermal as well as adiabatic (c) 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 from V1 to V2 under isothermal conditions (d) 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

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39.

An ideal gas in a thermally insulated vessel at internal 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, [Adv. 2014]

(a) (b) (c) (d) 40.

q=0 T2 = T1 P2V2 = P1V1 P2V2γ = P1V1γ The reversible expansion of an ideal gas under adiabatic and isothermal conditions is shown in the figure. Which of the following statement(s) is (are) correct ? [2012 - II]

(a) (b) (c) (d) 41.

T1 = T2 T3 > T1 wisothermal > wadiabatic ∆Uisothermal > ∆Uadiabatic For an ideal gas, consider only P–V work in going from an 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. Which of the following choice(s) is (are) correct? [Take ∆S as change in entropy and w as work done].

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[2012]

(a) ∆SX→Z = ∆SX→Y + ∆SY→Z (b) wX→Z = wX→Y + wY→Z (c) wX→Y→Z = wX→Y (d) ∆SX→Y→Z = ∆SX→Y 42.

Among the following, the intensive property is (properties are) [2010]

(a) (b) (c) (d) 43.

molar conductivity electromotive force resistance heat capacity Among the following the state function(s) is (are) [2009]

(a) (b) (c) (d) 44.

Internal energy Irreversible expansion work Reversible expansion work Molar enthalpy Identify the intensive quantities from the following: [1993 - 1 Mark]

(a) (b) (c) (d)

Enthalpy Temperature Volume Refractive Index

45. Match the thermodynamic processes given under Column-I with the expressions given under Column-II. [Adv. 2015]

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Column-I

ColumnII (A) Freezing of water at 273 K and 1 atm (p)q = 0 (B) Expansion of 1 mol of an ideal gas into a vacuum(q)w = 0 under isolated conditions (C) Mixing of equal volumes of two ideal gases at constant (r) ∆Ssys < 0 temperature and pressure in an isolated container (D) Reversible heating of H2(g) at 1 atm from 300 K to(s) ∆U = 0 600 K, followed by reversible cooling to 300 K at 1 atm (t) ∆G = 0 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 [Adv. 2013]

46. 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 47. The pair of isochoric processes among the transformation of states is (a) K to L and L to M (b) L to M and N to K (c) L to M and M to N (d) M to N and N to K

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48.

Statement - 1 : There is a natural asymmetry between converting work to heat and converting heat to work. Statement - 2 : No process is possible in which the sole result is the absorption of heat form a reservoir and its complete conversion into work. [2008S] (a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (c) Statement-1 is True, Statement-2 is False (d) Statement-1 is False, Statement-2 is True 49. Read the following statement and explanation and answer as per the options given below : Assertion : The heat absorbed during the isothermal expansion of an ideal gas against vacuum is zero. Reason : The volume occupied by the molecules of an ideal gas is zero. [2000S] (a) If both assertion and reason are CORRECT, and reason is the CORRECT explanation of the assertion. (b) If both assertion and reason are CORRECT, but reason is NOT the CORRECT explanation of the assertion. (c) If assertion is CORRECT, but reason is INCORRECT. (d) If assertion is INCORRECT, but reason is CORRECT. 50.

An insulated container contains 1 mol of a liquid, molar volume 100 mL, at 1 bar. When liquid is steeply pressed to 100 bar, volume decreases to 99 mL. Find. ∆H and ∆U for the process. [2004 - 2 Marks] 51. Cv value of He is always 3R/2 but Cv value of H2 is 3R/2 at low temperature and 5R/2 at moderate temperature and more than 5R/2 at higher temperature. Explain in two to three lines. [2003 - 2 Marks]

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52. (a) (b) (c)

53.

1. (a) (b) (c) (d) 2. (1) (2) (3) (4) (a) (b)

Two moles of a perfect gas undergo the following processes: [2002 - 5 Marks] a reversible isobaric expansion from (1.0 atm, 20.0L) to (1.0 atm, 40.0 L); a reversible isochoric change of state from (1.0 atm, 40.0 L) to (0.5 atm, 40.0 L); 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? “The heat energy q, absorbed by a gas is ∆H”, is true at what condition(s). [1984 - 1 Mark]

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 : [Main Sep. 05, 2020 (II)] –780 kJ mol–1 780 kJ mol–1 –784 kJ mol–1 784 kJ mol–1 For one mole of an ideal gas, which of these statements must be true? [Main Sep. 04, 2020 (I)] U and H each depends only on temperature Compressibility factor z is not equal to 1 CP, m – CV, m = R dU = CVdT for any process (1) and (3) (2), (3) and (4)

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(c) (3) and (4) (d) (1), (3) and (4) 3. The intermolecular potential energy for the molecules A, B, C and D given below suggests that : [Main Sep. 04, 2020 (I)]

(a) (b) (c) (d) 4.

(a) (c) 5.

(a) (b) (c) (d)

A-D has the shortest bond length A-A has the largest bond enthalpy D is more electronegative than other atoms A-B has the stiffest bond If enthalpy of atomisation for Br2( ) is x kJ/mol and bond enthalpy for Br2 is y kJ/mol, the relation between them: [Main Jan. 09, 2020 (I)] is x = y (b) does not exist is x > y (d) is x < y The difference between ∆H and ∆U (∆H – ∆U), when the combustion of one mole of heptane (I) is carried out at a temperature T, is equal to : [Main April 10, 2019 (II)] – 4 RT – 3 RT 4RT 3 RT

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6.

The combustion of benzene (l) gives CO2 (g) and H2O (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) [Main 2018] (a) 4152.6 (b) –452.46 (c) 3260 (d) –3267.6 7. For which of the following reactions, ∆H is equal to ∆U ? [Main Online April 15, 2018 (I)] (a) (b) (c) (d) 8. Given C(graphite) + O2(g) → CO2(g) ; ∆rH° = –393.5 kJ mol–1 H2 (g) +

O2 (g) → H2O(l) ; ∆rH° = –285.8 kJ mol–1

CO2(g) + 2H2O(l) → CH4(g) + 2O2(g); ∆rH° = + 890.3 kJ mol–1 Based on the above thermochemical equations, the value of ∆rH° at 298 K for the reaction C(graphite) + 2H2(g) → CH4(g) will be : (a) (b) (c) (d) 9.

[Main 2017]

+ 74.8 kJ mol + 144.0 kJ mol–1 – 74.8 kJ mol–1 – 144.0 kJ mol–1 For a reaction, A(g) → A(l); ∆H= –3RT. The correct statement for the reaction is : [Main Online April 8, 2017] –1

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(a) (b) (c) (d) 10.

∆H = ∆U ≠ O ∆H = ∆U = O |∆H| < |∆U| |∆H| > |∆U| The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K are∆ƒG0 [C(graphite)] = 0 kJ mol–1 ∆ƒG0 [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 m3 mol–1. If C(graphite) is converted to C(diamond) isothermally atT = 298 K, the pressure at which C(graphite) is in equilibrium with C(diamond), is [Useful information : 1 J = 1 kg m2s–2; 1 Pa = 1 kg m–1 s–2; 1 bar = 105 Pa] [Adv. 2017] (a) 14501 bar (b) 58001 bar (c) 1450 bar (d) 29001 bar 11. 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 : [Main 2016] (a) –676.5 (b) – 110.5 (c) 110.5 (d) 676.5 12. The heat of atomization of methane and ethane are 360 kJ/mol and 620 kJ/mol, respectively. The longest wavelength of light capable of breaking the C – C bond is (Avogadro number = 6.02 × 1023, h = 6.62 × 10–34 J s) : [Main Online April 10, 2015] 4 (a) 2.48 × 10 nm (b) 1.49 × 103 nm (c) 2.48 × 103 nm (d) 1.49 × 104 nm 13. For complete combustion of ethanol,

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the amount of heat produced as measured in bomb calorimeter, is 1364.47 kJ mol–1 at 25 ºC. Assuming ideality the enthalpy of combustion, ∆cH, for the reaction will be: (R = 8.314 kJ mol–1) [Main 2014] (a) (b) (c) (d) 14.

The standard enthalpy of formation of NH3 is – 46.0 kJ/mol. If the enthalpy of formation of H2 from its atoms is – 436 kJ/mol and that of N2 is – 712 kJ/mol, the average bond enthalpy of N - H bond in NH3 is: [Main Online April 9, 2014] (a) – 1102 kJ/mol (b) – 964 kJ/mol (c) + 352 kJ/mol (d) + 1056 kJ/mol 15. For the process [Adv. 2014] H2O(l) → H2O(g) at T = 100°C and 1 atmosphere pressure, the correct choice is (a) ∆Ssystem > 0 and ∆Ssurroundings > 0 (b) ∆Ssystem > 0 and ∆Ssurroundings < 0 (c) ∆Ssystem < 0 and ∆Ssurroundings > 0 (d) ∆Ssystem < 0 and ∆Ssurroundings < 0 16. Given : (I)

(II)

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The molar enthalpy of vapourisation of water will be : [Main Online April 9, 2013] –1 (a) 241.8 kJ mol (b) 22.0 kJ mol–1 (c) 44.1 kJ mol–1 (d) 527.7 kJ mol–1 17. The standard enthalpies of formation of CO2(g), H2O(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[Adv. 2013-I] (a) +2900 kJ (b) –2900 kJ (c) –16.11 kJ (d) +16.11 kJ 18. The species which by definition has ZERO standard molar enthalpy of formation at 298 K is [2010] (a) Br2 (g) (b) Cl2 (g) (c) H2O (g) (d) CH4 (g) 19. For the process H2O(l) (1 bar, 373 K) → H2O(g) (1 bar, 373 K), the correct set of thermodynamic parameters is [2007] (a) ∆G = 0, ∆S = +ve (b) ∆G = 0, ∆S = –ve (c) ∆G = +ve, ∆S = 0 (d) ∆G = –ve, ∆S = +ve B is 20. The value of log10 K for a reaction A ∆rS°298K (Given : = 10 JK–1 mol–1 and R = 8.314 JK–1 mol–1; 2.303 × 8.314 × 298 = 5705) [2007] (a) 5

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(b) 10 (c) 95 (d) 100 21. The enthalpy of vapourization of liquid is 30 kJ mol–1 and entropy of vapourization is 75 J mol–1 K. The boiling point of the liquid at 1 atm is [2004S] (a) 250 K (b) 400 K (c) 450 K (d) 600 K 22. Which of the reaction defines ∆H°f ? [2003S] (a) (b) (c) (d) 23. The ∆H0f for CO2(g), CO(g) and H2O(g) are –393.5, –110.5 and –241.8 kJ mol–1 respectively. The standard enthalpy change (in kJ) for the reaction CO2(g) + H2(g) CO(g) + H2O(g) is [2000S] (a) 524.1 (b) 41.2 (c) –262.5 (d) –41.2 24. For which change ∆H ≠ ∆E : [1995S] (a) H2(g) + I2(g) → 2HI(g) (b) HCl + NaOH → NaCl (c) C(s) + O2(g) → CO2(g) (d) N2(g) + 3H2(g) → 2NH3(g) 25. The difference between heats of reaction at constant pressure and constant volume for the reaction :

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2C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O(l) at 25ºC in kJ is (a) (b) (c) (d)

[1991 - 1 Mark]

–7.43 +3.72 –3.72 +7.43

26.

The internal energy change (in J) when 90 g of water undergoes complete evaporation at 100°C is ________ . (Given : ∆Hvap for water at 373 K = 41 kJ/ mol, R = 8.314 JK–1 mol–1) [Main Sep. 02, 2020 (I)] 27. The heat of combustion of ethanol into carbon dioxids and water is – 327 kcal at constant pressure. The heat evolved (in cal) at constant volume and 27°C (if all gases behave ideally) is (R = 2 cal mol–1 K–1) __________. [Main Sep. 02, 2020 (II)] 28. For the reaction ; A(l) → 2B(g) ∆U = 2.1 kcal, ∆S = 20 cal K–1 at 300 K. Hence ∆G in kcal is. [Main Jan. 07, 2020 (I)] 29. The standard heat of formation (∆f H°298) of ethane (in kJ/mol), if the heat of combustion of ethane, hydrogen and graphite are –1560, – 393.5 and –286 kJ/mol, respectively is ________. [Main Jan. 07, 2020 (II)] 30. The surface of copper gets tarnished by the formation of copper oxide. N2 gas was passed to prevent the oxide formation during heating of copper at 1250 K. However, the N2 gas contains 1 mole % of water vapour as impurity. The water vapour oxidises copper as per the reaction given below: 2Cu(s) + H2O(g) → Cu2O(s) + H2(g) pH2 is the minimum partial pressure of H2 (in bar) needed to prevent the oxidation at 1250 K. The value of ln(pH2) is_____. (Given: total pressure = 1 bar, R (universal gas constant) = 8 J K–1 mol–1, ln (10) = 2.3. Cu (s) and Cu2O (s) are mutually immiscible. At 1250 K : 2 Cu(s) + ½O2(g) → Cu2O(s); ∆G° = –78,000 J mol–1

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H2(g) + ½O2(g) → H2O(g); ∆G° = –1,78,000J mol–1;

(G is the Gibbs energy) [Adv. 2018] 31. Diborane is a potential rocket fuel which undergoes combustion according to the reaction. [2000 - 2 Marks] B2O3 (s) + 3H2O (g) B2H6 (g) + 3 O2 (g) From the following data, calculate the enthalpy change for the combustion of diborane. ∆H = –1273 kJ mol–1 ∆H = – 286 kJ mol–1 ∆H = 44 kJ mol–1 ∆H = 36 kJ mol–1 32. 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 - 3 Marks] 33. From the following data, calculate the enthalpy change for the combustion of cyclopropane at 298 K. The enthalpy of formation of CO2(g), H2O(l) and propene(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 - 5 Marks] 34. Compute the heat of formation of liquid methyl alcohol in kilojoules per mole, using the following data. Heat of vaporization 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, 249kJ / mol. Average bond energies : C – H = 415kJ/mol, C – O = 365 kJ/mol, O – H = 463 kJ/mol [1997 - 5 Marks] 35. The standard molar enthalpies of formation of cyclohexane(l)and benzene(l) at 25°C are – 156 and + 49 kJ mol–1 respectively. The

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36.

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 - 2 Marks] The polymerisation of ethylene to linear polyethylene is represented by the reaction [1994 - 2 Marks]

where n has a large integral value. Given that the average enthalpies of bond dissociation for C = C and C– C at 298 K are + 590 and + 331 kJ mol–1, respectively, calculate the enthalpy of polymerisation per mole of ethylene at 298 K. 37. Determine the enthalpy change of the reaction. C3H8(g) + H2(g) → C2H6(g) + CH4(g), at 25º, using the given heat of combustion values under standard conditions: Compound H2(g) CH4(g) C2H6(g) C(graphite) ∆Hº (kJ/mol) – 285.8 –890.0 –1560.0 –393.5 The standard heat of formation of C3H8(g) is –103.8 kJ/mol. [1992 - 3 Marks] 38. A gas mixture of 3.67 litres of ethylene and methane on complete combustion at 25ºC produces 6.11 litres of CO2. Find out the amount of heat evolved on burning one litre of the gas mixture. The heats of combustion of ethylene and methane are –1423 and –891 kJ mol–1 at 25ºC. [1991 - 5 Marks] 39. The standard enthalpy of combustion at 25ºC of hydrogen, cyclohexene (C6H10) and cyclohexane (C6H12) are –241, –3800 and – 3920 kJ/mole respectively. Calculate the heat of hydrogenation of cyclohexene. [1989 - 2 Marks] 40. An intimate mixture of ferric oxide, Fe2O3, 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 : [1988 - 2 Marks]

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H f (Al2O3) = 399 kcal/mole; Hf (Fe2O3) = 199 kcal/mole; Density of Fe2O3 = 5.2 g/cc; Density of Al = 2.7 g/cc. 41. The standard molar heats 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 - 2 Marks] 42. The bond dissociation energies of gaseous H2, Cl2 and HCl are 104, 58 and 103 kcal/mole respectively. Calculate the enthalpy of formation of HCl gas. [1985 - 2 Marks] 43. 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 (25ºC). [1984 - 4 Marks] 44. The molar heats of combustion of C2H2(g), C(graphite) and H2(g) are 310.62 kcal, 94.05 kcal and 68.32 kcal, respectively. Calculate the standard heat of formation of C2H2(g). [1983 - 2 Marks] 45. The enthalpy for the following reaction (∆Hº) at 25ºC are given below : [1981 - 2 Marks] (i)

H2(g) +

O2(g) → OH(g)

10.06 kcal

(ii) H2(g) → 2H(g) 104.18 kcal (iii) O2(g) → 2O(g) 118.32 kcal Calculate the O–H bond energy in the hydroxyl radical. 46.

When Fe(s) is dissolved in aqueous hydrochloric acid in a closed vessel, the work done is ............ . [1997 - 1 Mark]

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47.

The heat content of the products is more than that of the reactants in an ................ reaction. [1993 - 1 Mark]

48.

Choose the reaction(s) from the following options, for which the standard enthalpy of reaction is equal to the standard enthalpy of formation [Adv. 2019]

(a) (b) (c) (d) 49.

(a) (b) (c) (d) 50. (a) (b) (c) (d)

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 [Adv. 2017] With increase in temperature, the value of K for exothermic reaction decreases because the entropy change of the system is positive With increase in temperature, the value of K for endothermic reaction increases because unfavourable change in entropy of the surroundings decreases With increase in temperature, the value of K for endothermic reaction increases because the entropy change of the system is negative With increase in temperature, the value of K for exothermic reaction decreases because favourable change in entropy of the surroundings decreases The following is (are) endothermic reaction(s): Combustion of methane [1999 - 3 Marks] Decomposition of water Dehydrogenation of ethane to ethylene Conversion of graphite to diamond

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51. Match the transformations in column I with appropriate options in column II [2011] Column-I Column-II (A) CO2(s) → CO2(g) (p) phase transition (B) CaCO3(s) → CaO(s) + CO2(g) (q) allotropic change (C) 2H• → H2(g) (r) ∆H is positive (D)

P(white, solid) → P(red, solid)

(s) (t)

∆S is positive ∆S is negative

52. In the following equilibrium N2O4 (g) 2NO2(g) When 5 moles of each is taken and the temperature is kept at 298 K, the total pressure was found to be 20 bar. [2004 - 2 Marks] Given : (i) Find ∆G of the reaction at 298 K. (ii) Find the direction of the reaction 53. 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 1, 2-pentadiene (C). The equilibrium was maintained at 175°C. Calculate ∆G° for the following equilibria : B

A

=?

From the calculated value of

B

C and

=? 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 - 10 Marks] 54. Show that the reaction

at 300 K, is

spontaneous and exothermic, when the standard entropy change is –

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55.

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 - 3 Marks] Anhydrous AlCl3 is covalent. From the data given below, predict whether it would remain covalent or become ionic in aqueous for solution. (Ionisation energy for Al = 5137 kJ mol–1; Al3+ = –4665 kJ mol–1;

for Cl– = – 381 kJ mol–1.)

[1997 - 2 Marks] 56. 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 In order to get the same calorific output, what should butane, be the rate of supply of butane and oxygen ? Assume that losses due to incomplete combustion, etc, are the same for both the fuels and the gases behave ideally. Heats of combustion : [1993 - 3 Marks] 57. Using the data (all values are in kcal mol at 25ºC) given below, calculate the bond energy of C–C and C–H bonds. [1990 - 5 Marks] ∆Hºcombustion(ethane) = –372.0 ∆Hºcombustion(propane) = –530.0 ∆HºC(s) → C(g) = 172.0 Bond energy of H–H = 104.0 ∆Hºf of H2O(l) = –68.0 ∆Hºf of CO2(g) = –94.0 –1

Topic-1 : Thermodynamics 1. (a) 2. (d)

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3. (a) 4. (b) 5. (c) 6. (a) 7. (d) 8. (a) 9. (b) 10. (c) 11. (d) 12. (a) 13. (d) 14. (c) 15. (a) 16. (c) 17. (a) 18. (b) 19. (c) 20. (c) 21. (a) 22. (c) 23. (2) 24. (9) 25. (–13538) 26. (48) 27. (6.25) 28. (935.00) 29. (557) 30.(115.41) 31. (319.1) 34. (True) 35. (True) 36. (a,b,c)37.(b, c) 38. (a,b,c) 39. (a,b,c) 40. (a,c,d) 41. (a, c)

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42. 43. 44. 45. 46. 47. 48. 49.

(a, b) (a, d) (b, d) A- (r, t); B -(p, q, s); C -(p, q, s); D- (p, q, s, t) (c) (b) (a) (b) Topic-2 : Thermochemistry

1. (c) 2. (d) 3. (d) 4. (c) 5. (a) 6. (d) 7. (b) 8. (c) 9. (d) 10. (a) 11. (b) 12. (b) 13. (a) 14. (c) 15. (b) 16. (c) 17. (c) 18. (b) 19. (a) 20. (b) 21. (b) 22. (b) 23. (b) 24. (d) 25. (a) 26. (189494)

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27. (–326400) 28. (–2.70) 29. (–192.5) 30. (–14.6) 31. (–2035) 32. (309.16) 33. (–2091.32) 34. (–266) 35. (–152) 36. (–72) 37. (–55.7) 38. (50.9) 39. (–121) 40. (3.94) 41. (–372.0) 42. (–22) 43. (41.104) 44. (54.2) 45. (101.19) 48. (a, c) 49. (b, d) 50. (b, c, d)51. A – p, r, s ; B – r, s ; C – t ; D – p, q, t

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1.

For the reaction [Main Sep. 06, 2020 (I)]

Fe2N(s) +

H2(g)

2 Fe(s) + NH3(g)

(a) Kc = Kp(RT) (b) Kc = Kp (c) Kc = Kp (d) Kc = Kp 2.

The variation of equilibrium constant with temperature is given below: [Main Sep. 06, 2020 (I)] Temperature Equilibrium Constant T1 = 25 °C K1 = 10 T2 = 100 °C K2 = 100 The values of ∆Hº , ∆Gº at T1 and ∆Gº at T2 (in kj mol–1) respectively, are close to

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[use R = 8.314 J K–1 mol–1] (a) 28.4, – 7.14 and – 5.71 (b) 0.64, – 7.14 and – 5.71 (c) 28.4, – 5.71 and – 14.29 (d) 0.64, – 5.71 and – 14.29 3. The value of Kc is 64 at 800 K for the reaction . The value of Kc for the following reaction is : [Main Sep. 06, 2020 (II)]

(a) (b) (c) (d) 4.

1/64 8 1/4 1/8 Consider the following reaction : [Main Sep. 05, 2020 (I)]

2NO2(g); DH0 = + 58 kJ N2O4(g) For each of the following cases ((i), (ii)), the direction in which the equilibrium shifts is : (i) Temperature is decreases (ii) Pressure is increased by adding N2 at constant T. (a) (i) towards product, (ii) towards product (b) (i) towards reactant, (ii) towards product (c) (i) towards reactant, (ii) no change (d) (i) towards product, (ii) no change the variation of the rate of the 5. For the equilibrium forward (a) and reverse (b) reaction with time is given by : [Main Sep. 04, 2020 (I)]

(a)

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(b)

(c)

(d)

6.

If the equilibrium constant for is

is

the equilibrium constant for

and that of is :

[Main Sep. 04, 2020 (II)] (a) (b) (c) (d) 7.

In the figure shown below reactant A (represented by square) is in equilibrium with product B (represented by circle). The equilibrium constant is: [Main Jan. 09, 2020 (II)]

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(a) 4 (b) 8 (c) 1 (d) 2 8. For the reaction, 2 SO3(g), 2SO2(g) + O2(g) DH = – 57.2 kJ mol–1 and Kc = 1.7×1016 Which of the following statement is INCORRECT ? [Main April 10, 2019 (II)] (a) The equilibrium constant is large suggestive of reaction going to completion and so no catalyst is required. (b) The equilibrium will shift in forward direction as the pressure increases. (c) The equilibrium constant decreases as the temperature increases. (d) The addition of inert gas at constant volume will not affect the equilibrium constant. 9. (a) (b) (c) (d) 10.

In which one of the following equilibria, Kp ≠ Kc ? [Main April 12, 2019 (II)] 2CO(g) 2C(s) + O2(g) H2(g) + I2(g) 2HI(g) NO(g) + SO3(g) NO2(g) + SO2(g) N2(g) + O2(g) 2NO(g) For the equilibrium

; the value of

298 K is approximately: [Main Jan. 11, 2019 (II)] (a) 100 kJ mol–1 (b) –80 kJ mol–1 (c) 80 kJ mol–1

Gº at

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(d) –100 kJ mol–1 11. Consider the following reversible chemical reactions:

The relation between K1 and K2 is:

[Main Jan. 9, 2019 (II)]

(a) K1K2 = (b) K2 = K31 (c) K2 = K–31 (d) K1K2= 3 12. In which of the following reactions, an increase in the volume of the container will favour the formation of products? [Main Online April 15, 2018 (I)] (a) (b) (c) (d) 13.

The following reaction occurs in the Blast Furnace where iron ore is reduced to iron metal : [Main Online April 9, 2017] 2 Fe(l) + 3 CO2(g) Fe2O3(s) + 3 CO(g) Using the Le Chatelier’s principle, predict which one of the following will not disturb the equilibrium ? (a) Removal of CO (b) Removal of CO2 (c) Addition of CO2 (d) Addition of Fe2O3 C + D is 14. The equilibrium constant at 298 K for a reaction A +B 100. If the initial concentration of all the four species were 1 M each, then equilibrium concentration of D (in mol L–1) will be : [Main 2016]

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(a) (b) (c) (d) 15.

(a) (b) (c) (d) 16.

1.818 1.182 0.182 0.818 A solid XY kept in an evacuated sealed container undergoes decomposition to form a mixture of gases X and Y at temperature T. The equilibrium pressure is 10 bar in the vessel. Kp for this reaction is: [Main Online April 10, 2016] 25 100 10 5 The following reaction is performed at 298 K. [Main 2015] 2NO(g) + O2(g)

2NO2(g)

The standard free energy of formation of NO(g) is 86.6KJ/mol at 298 K. What is the standard free energy of formation of NO2(g) at 298 K? (Kp = 1.6 × 1012) (a) 86600 – (b) 0.5[2 × 86,600 – R(298) ln(1.6 × 1012)] (c) R(298) ln(1.6 × 1012) – 86600 (d) 86600 + R(298) ln(1.6 × l012) water system at constant temperature 17. The increase of pressure on ice will lead to [Main Online April 11, 2015] (a) a decrease in the entropy of the system (b) an increase in the Gibb’s energy of the system (c) no effect on the equilibrium (d) a shift of the equilibrium in the forward direction 18. For the reaction

if

where the symbols have usual meaning then the value of x is (assuming ideality):

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[Main 2014] (a) –1 (b) (c) (d) 1 19. What happens when an inert gas is added to an equilibrium keeping volume unchanged? [Main Online April 12, 2014] (a) More product will form (b) Less product will form (c) More reactant will form (d) Equilibrium will remain unchanged 2C + D, initial concentration of B was 1.5 20. In reaction A + 2B times of [A], but at equilibrium the concentrations of A and B became equal. The equilibrium constant for the reaction is : [Main Online April 9, 2013] (a) 8 (b) 4 (c) 12 (d) 6 21. Solubility product constant (Ksp) of salts of types MX, MX2 and M3X 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 – [2008S] (a) MX > MX2 > M3X (b) M3X > MX2 > MX (c) MX2 > M3X > MX (d) MX > M3X > MX2 22.

The Haber’s process for the formation of NH3 at 298K is

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2NH3; ∆H = –46.0 kJ; Which of the following is the correct statement [2006 - 3M, –1] (a) The condition for equilibrium is

(b)

(c) (d) 23.

where G is Gibb's free energy per mole of the gaseous species measured at that partial pressure. On adding N2, the equilibrium will shift to forward direction because according to IInd law of thermodynamics, the entropy must increase in the direction of spontaneous reaction The catalyst will increase the rate of forward reaction by 2 times and that of backward reaction by 1.5 times None of these Consider the following equilibrium in a closed container [2002S] 2NO2(g) N2O4(g)

At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statements holds true regarding the equilibrium constant (Kp) and degree of dissociation ( )? (a) neither Kp nor changes (b) both Kp and change (c) Kp changes, but does not change (d) Kp does not change, but 24.

changes

At constant temperature, the equilibrium constant (Kp) for the 2NO2 is expressed by Kp = decomposition reaction N2O4

(4x2P)/(1−x2), where P = pressure, x = extent of decomposition. Which one of the following statements is true? [2001S] (a) Kp increases with increase of P (b) Kp increases with increase of x (c) Kp increases with decrease of x (d) Kp remains constant with change in P and x

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25.

(b) (c) (d) 26. (a) (b) (c) (d) 27.

For a sparingly soluble salt ApBq, the relationship of its solubility product (LS) with its solubility (S) is [2001S] p+q p q (a) LS = S .p .q LS = S p +q.pq.q p LS = S pq.p p.q q LS = S pq.( pq) p+q When two reactants, A & B are mixed to give productsC & D, the reaction quotient Q, at the initial stages of the reaction [2000S] is zero decreases with time is independent of time increases with time 2NH3(g) at 500°C, the For the reversible reaction, N2(g) + 3H2(g) when partial pressure is measured in value of Kp is atmospheres. The corresponding value of Kc, with concentration in mole litre–1, is [2000S]

(a) (b) (c) (d) 28.

For the chemical reaction 3X(g) + Y(g) X3Y at equilibrium is affected by

X3Y(g), the amount of [1999 - 2 Marks]

(a) temperature and pressure (b) temperature only

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(c) pressure only (d) temperature, pressure and catalyst 29. Amongst the following hydroxides, the one which has the lowest value of Ksp at ordinary temperature (about 25ºC) is [1990 - 1 Mark] (a) Mg(OH)2 (b) Ca(OH)2 (c) Ba(OH)2 (d) Be(OH)2 30. When equal volumes of the following solutions are mixed, precipitation of AgCl (Ksp = 1.8×10–10) will occur only with (a) 10–4 M (Ag+) and 10–4 M (Cl–) [1988 - 1 Mark] (b) 10–5 M (Ag+) and 10–5 M (Cl–) (c) 10–6 M (Ag+) and 10–6 M (Cl–) (d) 10–10 M (Ag+) and 10–10 M (Cl–) 31. An example of a reversible reaction is : [1985 - 1 Mark] (a) Pb(NO3)2aq + 2NaI(aq) → PbI2(s) + 2NaNO3(aq) (b) AgNO3(aq) + HCl(aq) → AgCl(s) + NaNO3(aq) (c) 2Na(s) + H2O(l) → 2NaOH(aq) + H2(g) (d) KNO3(aq) + NaCl(aq) → KCl(aq) + NaNO3(aq) 32. Pure ammonia is placed in a vessel at a temperature where its dissociation constant (α) is appreciable. At equilibrium: [1984 - 1 Mark] (a) Kp does not change significantly with pressure. (b) α does not change with pressure. (c) concentration of NH3 does not change with pressure. (d) concentration of hydrogen is less than that of nitrogen. 33. A liquid is in equilibrium with its vapour at its boiling point. On the average, the molecules in the two phases have equal : [1984 - 1 Mark] (a) inter-molecular forces (b) potential energy

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(c) total energy (d) kinetic energy 34. The precipitate of [1982 - 1 Mark] CaF2(Ksp = 1.7 × 10 ) –10

is obtained when equal volumes of the following are mixed (a) 10–4M Ca2+ + 10–4M F– (b) 10–2M Ca2+ + 10–3M F– (c) 10–5M Ca2+ + 10–3M F– (d) 10–3M Ca2+ + 10–5M F– 35. For the reaction : [1981 - 1 Mark] 2HI(g) H2(g) + I2(g) the equilibrium constant Kp changes with (a) total pressure (b) catalyst (c) the amounts of H2 and I2 present (d) temperature 36. The oxidation of SO2 by O2 to SO3 is an exothermic reaction. The yield of SO3 will be maximum if [1981 - 1 Mark] (a) temperature is increased and pressure is kept constant (b) temperature is reduced and pressure is increased (c) both temperature and pressure are increased (d) both temperature and pressure are reduced 37. Molten sodium chloride conducts electricitry due to the presence of [1981 - 1 Mark] (a) free electrons (b) free ions (c) free molecules (d) atoms of sodium and chlorine 38.

In 1 L saturated solution of AgCl [Ksp(AgCl) = 1.6 × 10–10], 0.1 mol of CuCl [Ksp(CuCl) = 1.0 × 10–6] is added. The resultant concentration of

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Ag+ in the solution is 1.6 × 10–x. The value of “x” is [2011] 39.

For a reaction 1.0 mol of X, 1.5 mol of Y and 0.5 mol of Z were taken in a 1 L vessel and allowed to react. At equilibrium, the concentration of Z was 1.0 mol L–1. The equilibrium constant of the reaction is

40.

41.

The value of x is ____________.

[Main Sep. 05, 2020 (II)] at 1000 K. At time t ', the temperature Consider the reaction 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. What is the ratio of the standard Gibbs energy of the reaction at 1000 K to that at 2000 K? [Adv. 2020]

For the following reaction, the equilibrium constant Kc at298 K is 1.6 × 1017. [Adv. 2019]

When equal volumes of 0.06 M Fe2+ (aq) and 0.2 M S2– (aq) solutions are mixed, the equilibrium concentration of Fe2+ (aq) is found to be Y × 10–17 M. The value of Y is

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42.

An aqueous solution of a metal bromide MBr2 (0.05M) is saturated with H2S. What is the minimum pH at which MS will precipitate? [1993 - 3 Marks] for concentration of saturatedH2S = 0.1 M and

, for H2S.

43. The equilibrium constant of the reaction 2AB(g) A2(g) + B2(g) at 100ºC is 50. If a one litre flask, containing one mole of A2 is connected to a two litre flask, containing two mole of B2, how many mole of AB will be formed at 373ºC? [1985 - 4 Marks] 44.

45.

46.

A ten-fold increase in pressure on the reaction, at equilibrium results in .... in KP. [1996 - 1 Mark] For a given reversible reaction at a fixed temperature, equilibrium constants Kp and Kc are related by........ [1994 - 1 Mark] Solubility of temperature.

sodium

hydroxide increases with increase in

[1985 - ½ Mark] 47. When a liquid and its vapour are at equilibrium and the pressure is suddenly decreased, cooling occurs. [1984 - 1 Mark] 2AB, is K, then for 48. If equilibrium constant for the reaction A2 + B2 ½ A2 + ½ B2, the equilibrium constant the backward reaction AB is 1/K. [1984 - 1 Mark] 49. The %yield of ammonia as a function of time in the reaction 2NH3(g), ∆H < 0 at (P, T1) is given below N2(g) + 3H2(g)

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If this reaction is conducted at (P, T2), with T2 > T1, the% yield of ammonia as a function of time is represented by [Adv. 2015]

(a)

(b)

(c)

(d)

50.

The thermal dissociation equilibrium of CaCO3(s) is studied under different conditions [Adv. 2013]

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CaCO3(s) CaO(s) + CO2(g). 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 catalyst, if any 51. The Ksp of Ag2CrO4 is 1.1 × 10–12 at 298 K. The solubility (in mol/L) of Ag2CrO4 in a 0.1 M AgNO3 solution is [Adv. 2013] –11 (a) 1.1 × 10 (b) 1.1 × 10–10 (c) 1.1 × 10–12 (d) 1.1 × 10–9 CO2(g) + H2(g) at a given 52. For the reaction CO(g) + H2O(g) temperature, the equilibrium amount of CO2(g) can be increased by [1998 - 2 Marks] (a) adding a suitable catalyst (b) adding an inert gas (c) decreasing the volume of the container (d) increasing the amount of CO(g). 53. For the reaction : [1991 - 1 Mark] PCl5(g) → PCl3(g) + Cl2(g) The forward reaction at constant temperature is favoured by (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 PCl5 at constant volume 54. The equilibrium : [1989 - 1 Mark] SO2(g) + Cl2(g) SO2Cl2(g) is attained at 25ºC in a closed container and an inert gas, helium is introduced. Which of the following statements are correct? (a) Concentration of SO2, Cl2 and SO2Cl2 do not change

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(b) More chlorine is formed (c) Concentration of SO2 is reduced (d) More SO2Cl2 is formed. 55. When NaNO3 is heated in a closed vessel, oxygen is liberated and NaNO2 is left behind. At equilibrium. [1986 - 1 Mark] (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 56. For the gas phase reaction : [1984 - 1 Mark] C2H6 (∆H = –32.7 kcal) C2H4 + H2 carried out in a vessel, the equilibrium concentration of C2H4 can be increased by : (a) increasing the temperature (b) decreasing the pressure (c) removing some H2 (d) adding some C2H6 Each question contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason). Each question has 5 choices (a), (b), (c) and (d) out of which ONLY ONE is correct. Mark your answer as 57. Statement -1 For every chemical reaction at equilibrium, standard Gibbs energy of reaction is zero. Statement -2 At constant temperature and pressure, chemical reactions are spontaneous in the direction of decreasing Gibbs energy. [2008] 58. Statement -1 The endothermic reactions are favoured at lower temperature and the exothermic reactions are favoured at higher temperature. Statement -2 When a system in equilibrium is disturbed by changing the temperature, it will tend to adjust itself so as to overcome the effect of change. [1991 - 2 Marks]

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(a) If both Statement -1 and Statement -2 are correct,and Statement -2 is the correct explanation of the Statement -2. (b) If both Statement -1 and Statement -2 are correct, but Statement -2 is not the correct explanation of the Statement -1. (c) If Statement -1 is correct but Statement -2 is incorrect. (d) If Statement -1 is incorrect but Statement -2 is correct. 59.

When 3.06 g of solid NH4HS is introduced into a two litre evacuated flask at 27° C, 30% of the solid decomposes into gaseous ammonia and hydrogen sulphide. (i) Calculate Kc and Kp for the reaction at 27°C. (ii) What would happen to the equilibrium when more solid NH4HS is introduced into the flask ? [1999 - 7 Marks] + Ag + 2NH3, Kc = 6.2 × 10–8 and Ksp of 60. Given : Ag(NH3) AgCl = 1.8 × 10–10 at 298 K. 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 - 5 Marks] 61. A sample of AgCl was treated with 5.00 mL of 1.5 M Na2CO3 solution to give Ag2CO3. The remaining solution contained 0.0026 g of Cl– per litre. Calculate the solubility product of AgCl (Ksp(Ag2CO3) = 8.2 × 10–12). [1997 - 5 Marks] 62. For the reaction the equilibrium costant, at 25°C , is . Calculate the silver ion concentration in a solution which was originally 0.10 molar in KCN and 0.03 molar in AgNO3. [1994 - 3 Marks] 63. At temperature T, a compound AB2 (g) dissociates according to the reaction [1994 - 4 Marks]

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with a degree of dissociation x which is small compared with unity. Deduce the expression for x in terms of the equilibrium constant Kp and the total pressure, P. 64. 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 : Hydrogen is introduced until the total pressure of the system is 8.5 atmosphere at equilibrium and 0.08 mole of methanol is formed. Calculate (i) Kp and Kc 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 - 5 Marks] 65. The solubility product (Ksp) of Ca(OH)2 at 25ºC is4.42 × 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 - 4 Marks] 66. The solubility product of Ag2C2O4 at 25ºC is 1.29 × 10–11 mol3 L–3. A solution of K2C2O4 containing 0.1520 mole in 500 mL water is shaken at 25ºC with excess of Ag2CO3 till the following equilibrium is reached : [1991 - 4 Marks] Ag2C2O4 + K2CO3 Ag2CO3 + K2C2O4 At equilibrium, the solution contains 0.0358 mole of K2CO3. Assuming the degree of dissociation of K2C2O4 and K2CO3 to be equal, calculate the solubility product of Ag2CO3. CH3OH(g) 67. For the reaction : CO(g) + 2H2(g) hydrogen gas is introduced into a five litre flask at 327ºC, containing 0.2 mole of CO(g) and a catalyst, until the pressure is 4.92 atm. At this point 0.1 mole of CH3OH(g) is formed. Calculate the equilibrium constant, Kp and Kc. [1990 - 5 Marks] 68. Freshly precipitated aluminium and magnesium hydroxides are stirred vigorously in a buffer solution containing 0.25 mole/L of ammonium chloride and 0.05 mole/L of ammonium hydroxide.

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Calculate the concentration of aluminium and magnesium ions in solution: Kb[NH4OH] = 1.80 × 10–5 [1989- 3 Marks] Ksp [Mg(OH)2] = 6 × 10–10 Ksp [Al(OH)3] = 6 × 10–32 69. The equilibrium constant Kp of the reaction : 2SO3(g) 2SO2(g) + O2(g) is 900 atm. at 800 K. A mixture containing SO3 and O2 having initial partial 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 - 3 Marks] 70. N2O4 is 25% dissociated at 37ºC and one atmosphere pressure. Calculate (i) Kp and (ii) the percentage dissociation at 0.1 atmosphere and 37ºC. [1988 - 4 Marks] 71. At a certain temperature equilibrium constant (Kc) is 16 for the reaction. [1987 - 5 Marks] SO3(g) + NO(g) SO2(g) + NO2(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(g) and NO2(g)? 72. The solubility of Mg(OH)2 in pure water is 9.57×10–3 g/litre. Calculate its solubility (in g/litre) in 0.02 M Mg(NO3)2 solution. [1986 - 5 Marks] 73. One mole of Cl2 and 3 moles of PCl5 are placed in a 100 litre vessel heated to 227ºC. The equilibrium pressure is 2.05 atmosphere. Assuming ideal behaviour, calculate the degree of dissociation for PCl5 and Kp for the reaction : PCl3(g) + Cl2(g). PCl5(g) [1984 - 6 Marks] ++ 74. A solution contains a mixture of Ag (0.10 M) and Hg2 (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? [Ksp : AgI = 8.5 × 10–17; Hg2I2 = 2.5 × 10–26]

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[1984 - 4 Marks] 75. One mole of nitrogen is mixed with three moles of hydrogen in a 4 litre container. If 0.25 per cent of nitrogen is converted to ammonia by the following reaction [1981 - 4 Marks] 2NH3(g), N2(g) + 3H2(g) calculate the equilibrium constant (Kc) in concentration units. What will be the value of Kc for the following equilibrium ? N2(g) +

1. (A) (B) (C) (D) (a) (b) (c) (d) 2. (a) (b) (c) (d) 3.

H2(g)

NH3(ag)

Arrange the following solutions in the decreasing order of pOH : [Main Sep. 06, 2020 (I)] 0.01 M HCl 0.01 M NaOH 0.01 M CH3COONa 0.01 M NaCl (A) > (C) > (D) > (B) (A) > (D) > (C) > (B) (B) > (C) > (D) > (A) (B) > (D) > (C) > (A) An acidic buffer is obtained on mixing : [Main Sep. 03, 2020 (I)] 100 mL of 0.1 M CH3COOH and 100 mL of 0.1 M NaOH 100 mL of 0.1 M HCl and 200 mL of 0.1 M NaCl 100 mL of 0.1 M CH3COOH and 200 mL of 0.1 M NaOH 100 mL of 0.1 M HCl and 200 mL of 0.1 M CH3COONa 100 mL of 0.1 M HCl is taken in a beaker and to it 100 mL of 0.1 M NaOH is added in steps of 2 mL and the pH is continuously measured. Which of the following graphs correctly depicts the change in pH? [Main Sep. 03, 2020 (II)]

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(a)

(b)

(c)

(d)

4.

The Ksp for the following dissociation is 1.6 × 10–5

Which of the following choices is correct for a mixture of 300 mL 0.134 M Pb(NO3)2 and 100 mL 0.4 M NaCl? [Main Jan. 09, 2020 (I)] (a) Not enough data provided (b) Q < Ksp (c) Q > Ksp (d) Q = Ksp

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5.

(a) (b) (c) (d) 6.

The solubility product of Cr(OH)3 at 298 K is 6.0 × 10–31. The concentration of hydroxide ions in a saturated solution of Cr(OH)3 will be: [Main Jan. 09, 2020 (II)] (2.22 × 10–31)1/4 (18 × 10–31)1/4 (18 × 10–31)1/2 (4.86 × 10–29)1/4 The stoichiometry and solubility product of a salt with the solubility curve given below is, respectively: [Main Jan. 08, 2020 (I)]

(a) (b) (c) (d) 7.

X2Y, 2 × 10–9 M3 XY2, 4 × 10–9 M3 XY2 , 1 × 10–9 M3 XY, 2 × l0–6 M3 For the following Assertion and Reason, the correct option is: Assertion: The pH of water increases with increase in temperature. Reason: The dissociation of water into H+ and OH– is an exothermic reaction. [Main Jan. 08, 2020 (II)] (a) Both assertion and reason are true, and the reason is the correct explanation for the assertion. (b) Both assertion and reason are false. (c) Both assertion and reason are true, but the reason is not the correct explanation for the assertion.

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(d) Assertion is not true, but reason is true. 8. 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 : [Main April 12, 2019 (I)] –19 (a) 3×10 (b) 12×10–21 (c) 3×10–22 (d) 12×10–23 9. The pH of a 0.02 M NH4Cl solution will be [given Kb(NH4OH)=10–5 and log 2 = 0.301] [Main April 10, 2019 (II)] (a) 2.65 (b) 4.35 (c) 4.65 (d) 5.35 10. 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? [Main April 9, 2019 (II)]

(a) (B) (b) (A) (c) (C) (d) (D)

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11.

If solubility product of Zr3(PO4)4 is denoted by Ksp and its molar solubility is denoted by S, then which of the following relation between S and Ksp is correct ? [Main April 8, 2019 (I)]

(a) (b) (c) (d) 12. (a) (b) (c) 13.

(a) (b) (c) (d) 14.

(a)

Which of the following salts is the most basic in aqueous solution? [Main 2018] Al(CN)3 CH3COOK FeCl3 (d) Pb(CH3COO)2 An aqueous solution contains 0.10 MH2S 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 : [Main 2018] –8 5 × 10 3 × 10–20 6 × 10–21 5 × 10–19 The minimum volume of water required to dissolve 0.1g lead (II) chloride to get a saturated solution (KSPof PbCl2 = 3.2 × 10–8; atomic mass of Pb = 207 u) is : [Main Online April 15, 2018 (I)] 1.798 L

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(b) 0.36 L (c) 17.95 L (d) 0.18 L 15. pKa of a weak acid (HA) and pKb of a weak base (BOH) are 3.2 and 3.4, respectively. The pH of their salt (AB) solution is [Main 2017] (a) 7.2 (b) 6.9 (c) 7.0 (d) 1.0 16. Addition of sodium hydroxide solution to a weak acid (HA) results in a buffer of pH 6. If ionisation constant of HA is 10–5, the ratio of salt to acid concentration in the buffer solution will be :

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[Main Online April 8, 2017] (a) 4 : 5 (b) 1 : 10 (c) 10 : 1 (d) 5 : 4 17. The conjugate base of hydrazoic acid is: [Main Online April 12, 2014] –3 (a) N (b) (c) (d) 18. In some solutions, the concentration of H3O+ remains constant even when small amounts of strong acid or strong base are added to them. These solutions are known as: [Main Online April 11, 2014] (a) Ideal solutions (b) Colloidal solutions (c) True solutions (d) Buffer solutions 19. How many litres of water must be added to 1 litre an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2 ? [Main 2013] (a) 0.1 L (b) 0.9 L (c) 2.0 L (d) 9.0 L 20. NaOH is a strong base. What will be pH of5.0 × 10–2 M NaOH solution ? (log 2 = 0.3) [Main Online April 22, 2013] (a) 14.00 (b) 13.70 (c) 13.00 (d) 12.70

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21.

(a) (c) 22.

(a) (b) (c) (d) 23.

(a) (b) (c) (d) 24.

(a) (b) (c) (d) 25. (a) (b) (c)

2.5 mL of (2/5) M weak monoacidic base (Kb = 1 × 10–12 at 25°) is titrated (2/15) M HCl in water at 25°C. The concentration of H+ at equivalence point is (Kw = 1 × 10–14 at 25°C) [2008S] –14 –7 3.7 × 10 M (b) 3.2 × 10 M –2 3.2 × 10 M (d) 2.7 × 10–2 M 0.1 mole of CH3NH2 (Kb = 5 ×10–4) is mixed with 0.08 mole of HCl and diluted to one litre. What will be the H+ concentration in the solution? [2005S] -2 8 × 10 M 8 × 10-11 M 1.6 × 10-11 M 8 × 10-5 M A 0.004 M solution of Na2SO4 is isotonic with 0.010 M solution of glucose at same temperature. The percentage dissociation of Na2SO4 is [2004S] 25% 50% 75% 85% A weak acid HX has the dissociation constant 1 × 10–5 M. It forms a salt NaX on reaction with alkali. The percentage hydrolysis of 0.1 M solution of NaX is [2004S] 0.0001 % 0.01 % 0.1% 0.15 % The set with correct order of acidity is [2001S] HClO < HClO2 < HClO3 < HClO4 HClO4 < HClO3 < HClO2 < HClO HClO < HClO4 < HClO3 < HClO2

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(d) HClO4 < HClO2 < HClO3 < HClO 26. The pH of 0.1 M solution of the following salts increases in the order. [1999 - 2 Marks] (a) NaCl < NH4Cl < NaCN < HCl (b) HCl < NH4Cl < NaCl < NaCN (c) NaCN < NH4Cl < NaCl < HCl (d) HCl < NaCl < NaCN < NH4Cl 27. The following acids have been arranged in the order of decreasing acid strength. Identify the correct order. ClOH (I), BrOH(II), IOH(III) [1996 - 1 Mark] (a) I > II > III (b) II > I >III (c) III > II > I (d) I > III > II 28. Which one is more acidic in aqueous solution. [1995S] (a) NiCl2 (b) FeCl3 (c) AlCl3 (d) BeCl2 29. The degree of dissociation of water at 25ºC is 1.9 × 10–7% and density is 1.0 g cm–3. The ionic constant for water is : [1995S] –10 (a) 1.0 × 10 (b) 1.0 × 10–14 (c) 1.0 × 10–16 (d) 1.0 × 10–8 30. Which of the following solutions will have pH close to 1.0? [1992 - 1 Mark] (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 31. The following equilibrium is established when hydrogen chloride is dissolved in acetic acid.

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HCl + CH3COOH

Cl– +

The set that characterises the conjugate acid-base pairs is [1992 - 1 Mark] (a) (HCl, CH3COOH) and (

) and (CH3COOH, Cl–)

(b) (HCl, (c) (

, Cl–)

, HCl) and (Cl–, CH3COOH)

(d) (HCl, Cl–) and (

, CH3COOH)

32. The reaction which proceeds in the forward direction is [1991 - 1 Mark] (a) Fe2O3 + 6HCl → 2FeCl3 + 3H2O (b) NH3 + H2O + NaCl → NH4Cl + NaOH (c) SnCl4 + Hg2Cl2 → SnCl2 + 2HgCl2 (d) 2CuI + I2 + 4K+ → 2Cu2+ + 4KI 33. Which one of the following is the strongest acid? [1989 - 1 Mark] (a) ClO3(OH) (b) ClO2(OH) (c) SO(OH)2 (d) SO2(OH)2 34. The pKa of acetylsalicyclic and (aspirin) is 3.5. The pH of gastric juice in human stomach is about 2-3 and the pH in the small intestine is about 8. Aspirin will be [1988 - 1 Mark] (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. 35. The compound whose 0.1 M solution is basic is : [1986 - 1 Mark] (a) ammonium acetate (b) ammonium chloride (c) ammonium sulphate (d) sodium acetate

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36.

The compound insoluble in acetic acid is : [1986 - 1 Mark]

(a) (b) (c) (d) 37.

calcium oxide calcium carbonate calcium oxalate calcium hydroxide The compound that is not a Lewis acid is : [1985 - 1 Mark]

(a) (b) (c) (d) 38.

BF3 AlCl3 BeCl2 SnCl4 The conjugate acid of

is : [1985 - 1 Mark]

(a) NH3 (b) NH2OH (c) (d) N2H4 39. The best indicator for detection of end point in titration of a weak acid and a strong base is : [1985 - 1 Mark] (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) 40. A certain weak acid has a dissociation constant of 1.0 × 10–4. The equilibrium constant for its reaction with a strong base is : [1984 - 1 Mark] –4 (a) 1.0 × 10 (b) 1.0 × 10–10 (c) 1.0 × 1010 (d) 1.0 × 1014 41. A certain buffer solution contains equal concentration of X– and HX. The Kb for X– is 10–10. The pH of the buffer is:

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(a) 4 (b) 7 [1984 - 1 Mark] (c) 10 (d) 14 42. At 90ºC, pure water has [H3O+] 10–6 mole litre–1. What is the value of Kw at 90ºC? [1981 - 1 Mark] (a) 10–6 (b) 10–12 (c) 10–14 (d) 10–8 43. Of the given anions, the strongest Bronsted base is [1981 - 1 Mark] (a) ClO– (b) Cl (c) Cl (d) Cl 44.

The pH of a 10–8 molar solution of HCl in water is [1981 - 1 Mark]

(a) (b) (c) (d) 45. (a) (b) (c) (d) 46.

8 –8 between 7 and 8 between 6 and 7 An acidic buffer solution can be prepared by mixing the solutions of [1981 - 1 Mark] ammonium acetate and acetic acid ammonium chloride and ammonioum hydroxide sulphuric acid and sodium sulphate sodium chloride and sodium hydroxide. 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

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added is shown in the figure below. What is the pKb of the base? The BH+ + A–. neutralization reaction is given by B + HA [Adv. 2020]

47.

Amongst the following, the total number of compounds whose aqueous solution turns red litmus paper blue is KCN, K2SO4, (NH4)2C2O4, NaCl, Zn(NO3)2, FeCl3, K2CO3, NH4NO3 and LiCN [2010] 48. The total number of diprotic acids among the following is: H3PO4, H2SO4, H3PO3, H2CO3, H2S2O7, H3BO3, H3PO2, H2CrO4 and H2SO3. [2010] 49. The dissociation constant of a substituted benzoic acid at 25°C is 1.0 × 10–4. The pH of a 0.01 M solution of its sodium salt is [2009 - 2 Marks] 50. 0.1 M NaOH is titrated with 0.1 M HA till the end point; Ka for HA is 5.6 × 10–6 and degree of hydrolysis is less compared to 1. Calculate pH of the resulting solution at the end point. [2004 - 2 Marks] 51. 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 (Ka=1.0×10–5)? [1997 - 2 Marks]

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52.

A soft drink was bottled with a partial prssure of CO2 of 3 bar over the liquid at room temperature. The partial pressure of CO2 over the solution approaches a value of 30 bar when 44g of CO2 is dissolved in 1 kg of water at room temperature. The approximate pH of the soft drink is ______ × 10–1. (First dissociation constant of H2CO3 = 4.0 × 10–7; log 2 = 0.3; density of the soft drink = 1g mL–1) [Main Sep. 05, 2020 (I)] 53. Two solutions, A and B, each of 100 L was made by dissolving 4g of NaOH and 9.8 g of H2SO4 in water, respectively. The pH of the resultant solution obtained from mixing 40 L of solution A and 10 L of solution. B is ________. [Main Jan. 07, 2020 (I)] 54. 3 g of acetic acid is added to 250 mL of 0.1 M HCl and the solution made up to 500 mL. To 20 mL of this solution

mL of 5 M NaOH is

added. The pH of the solution is ________. [Main Jan. 07, 2020 (II)] [Given: pKa of acetic acid = 4.75, molar mass of acetic acid = 60 g/mol, log 3 = 0.4771] Neglect any changes in volume. 55. If the solubility product of AB2 is 3.20 × 10–11M3, then the solubility of AB2 in pure water is ______ × 10–4 mol L–1. [Assuming that neither kind of ion reacts with water] [Main Sep. 06, 2020 (II)] 2+ 56. An acidified solution of 0.05 M Zn is saturated with 0.1 M H2S. What is the minimum molar concentration (M) of H+ required to prevent the precipitation of ZnS? Use Ksp (ZnS) = 1.25 × 10–22 and overall dissociation constant of H2S, KNET = K1K2 = 1 × 10–21. [Adv. 2020] 57. The solubility of a salt of weak acid (AB) at pH 3 is Y×10–3 mol L–1. The value of Y is________. [Adv. 2018] (Given that the value of solubility product of AB (Ksp) = 2×10–10 and the value of ionization constant of HB (Ka) = 1×10–8)

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58. 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 moles litre–1 and the pKa of H2SO3 is 1.92, estimate the pH of rain on that day. [2000 - 5 Marks] 59. What will be the resultant pH when 200mL of an aqueous solution of HCl (pH = 2.0) is mixed with 300 mL of an aqueous solution of NaOH (pH = 12.0) ? [1998 - 2 Marks] 60. What is the pH of a 0.50 M aqueous NaCN solution? pKb of CN– is 4.70. [1996 - 2 Marks] 61. Calculate the pH of an aqueous solution of 1.0 M ammonium formate assuming complete dissociation. (pKa of formic acid = 3.8 and pKb of ammonia = 4.8.) [1995 - 2 Marks] 62. The pH of blood stream is maintained by a proper balance of H2CO3 and NaHCO3 concentrations. What volume of 5M NaHCO3 solution should be mixed with a 10 mL sample of blood which is 2M in H2CO3 in order to maintain a pH of 7.4 ? Ka for H2CO3 in blood is 7.8 × 10–7. [1993 - 2 Marks] 63.

In the reaction I–+I2

, the Lewis acid is ........... .

[1997 - 1 Mark] 64. 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 - 1 Mark] in aqueous solution is ............. 65. The conjugate base of [1982 - 1 Mark] 66.

Aluminium chloride (AlCl3) is a Lewis acid because it can donate electrons. [1982 - 1 Mark]

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67.

(a) (b) (d) 68.

(a)

The initial rate of hydrolysis of methyl acetate (1M) by a weak acid (HA, 1M) is 1/100th of that of a strong acid (HX, 1M), at 25°C. The Ka of HA is [Adv. 2013] –4 1 × 10 1 × 10–5 (c) 1 × 10–6 1 × 10–3 Aqueous solutions of HNO3, KOH, CH3COOH and CH3COONa of identical concentrations are provided. The pair(s) of solutions which form a buffer upon mixing is(are) HNO3 and CH3COOH [2010] (b) KOH and CH3COONa

(c) HNO3 and CH3COONa (d) CH3COOH and CH3COONa 69.

A buffer solution can be prepared from a mixture of [1999 - 3 Marks]

(a) (b) (c) (d) 70.

sodium acetate and acetic acid in water sodium acetate and hydrochloric acid in water ammonia and ammonium chloride in water ammonia and sodium hydroxide in water Which of the following statements(s) is (are) correct? [1998 - 2 Marks] (a) The pH of 1.0 × 10–8 M solution of HCl is 8 (b) The conjugate base of H2

is HPO

(c) Autoprotolysis constant of water increases with temperature (d) When a solution of a weak monoprotic acid is titrated against a strong base, at half-neutralisation pointpH = (1/2) pKa.

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71.

Dilution processes of different aqueous solutions, with water, are given in LIST-I. The effects of dilution of the solutions on [H+] are given in LIST-II. (Note: Degree of dissociation (α) of weak acid and weak base is K+ > Cs+ > Rb+ Na+ > Li+ > K+ > Rb+ > Cs+ Na+ > Li+ > K+ > Cs+ > Rb+ Li+ > Na+ > K+ > Rb+ > Cs+ A metal on combution in excess of air forms X. X upon hydrolysis with water yields H2O2 and O2 along with another product. The metal is: [Main Jan. 12, 2019 (I)] Na Rb Mg Li The correct statement(s) among I to III with respect to potassium ions that are abundant within the cell fluidsis/are: [Main Jan. 12, 2019 (II)] They activate many enzymes II. They participate in the oxidation of glucose to produce ATP Along with sodium ions, they are responsible for the transmission of nerve signals I and II only I and III only I, II and III III only The metal that forms nitride by reacting directly with N2 of air, is: [Main Jan. 9, 2019 (II)] K Li Rb Cs Lithium aluminium hydride reacts with silicon tetrachloride to form: [Main Online April 15, 2018 (II)] LiCl, AlH3 and SiH4 LiCl, AlCl3 and SiH4 LiH, AlCl3 and SiCl2 LiH, AlH3 and SiH4 Which one of the following is an oxide ? [Main Online April 9, 2017] KO2 BaO2 SiO2 CsO2 The main oxides formed on combustion of Li, Na and K in excess of air are, respectively: [Main 2016] Li2O2, Na2O2 and KO2 Li2O, Na2O2 and KO2 Li2O, Na2O and KO2 LiO2, Na2O2 and K2O Which of the alkaline earth metal halides given below is essentially covalent in nature ? [Main Online April 11, 2015]

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(a) (b) (c) (d) 15.

SrCl2 CaCl2 BaCl2 MgCl2 Which of the following statements about Na2O2 is not correct? [Main Online April 11, 2014]

(a) It is diamagnetic in nature (b) It is derivative of H2O2 (c) Na2O2 oxidises Cr3+ to

in acid medium.

(d) It is the super oxide of sodium 16. The first ionisation potential of Na is 5.1 eV. The value of electron gain enthalpy of Na+ will be [Main 2013] (a) – 2.55 eV (b) – 5.1 eV (c) – 10.2 eV (d) + 2.55 eV 17. The solubility order for alkali metal fluoride in water is [Main Online April 22, 2013] (a) LiF < RbF < KF < NaF (b) RbF < KF < NaF < LiF (c) LiF > NaF > KF > RbF (d) LiF < NaF < KF < RbF 18. The metallic lustre exhibited by sodium is explained by (a) diffusion of sodium ions [1987 - 1 Mark] (b) oscillation of loose electrons (c) excitation of free protons (d) existence of body centered cubic lattice 19. The pair of compounds which cannot exist together in solution is : [1986 - 1 Mark] (a) NaHCO3 and NaOH (b) Na2CO3 and NaHCO3 (c) Na2CO3 and NaOH (d) NaHCO3 and NaCl 20. The oxide that gives hydrogen peroxide on treatment with a dilute acid is : [1985 - 1 Mark] (a) PbO2 (b) Na2O2 (c) MnO2 (d) TiO2 21. A solution of sodium metal in liquid ammonia is strongly reducing due to the presence of [1981 - 1 Mark] (a) sodium atoms (b) sodium hydride (c) sodium amide (d) solvated electrons

22.

The electrolysis of molten sodium hydride liberates ............. gas at the ............. . [1989 - 1 Mark]

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23.

Sodium dissolved in liquid ammonia conducts electricity because ............... . [1985 - 1 Mark]

24.

Sodium when burnt in excess of oxygen gives sodium oxide.

25.

The softness of group I-A metals increases down the group with increasing atomic number.

[1987 - 1 Mark] [1986 - 1 Mark]

26.

The compound(s) formed upon combustion of sodium metal in excess air is (are) [2009 - 5M, –1]

(a) Na2O2 (b) Na2O (c) NaO2 (d) NaOH 27. Highly pure dilute solution of sodium in liquid ammonia (a) shows blue colour [1998 - 2 Marks] (b) exhibits electrical conductivity (c) produces sodium amide (d) produces hydrogen gas. 28.

(a) (b) (c) (d)

This question contains STATEMENT-1 (Assertion) and STATEMENT-2 (Reason) and has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct. [2007] Statement-1 : Alkali metals dissolve in liquid ammonia to give blue solutions. because Statement-2 : Alkali metals is liquid ammonia give solvated species of the type [M(NH3)n]+ (M = alkali metals). Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 Statement-1 is True, Statement-2 is True; Statement-2 is not correct explanation for Statement-1 Statement-1 is True, Statement-2 is False Statement-1 is False, Statement-2 is True.

29.

Explain the difference in the nature of bonding in LiF and LiI. [1996 - 2 Marks]

30. Write down the balanced equations for the reactions when: (i) Potassium ferricyanide reacts with hydrogen peroxide in basic solution. [1989 - 1 Mark] (ii) Carbon dioxide is passed through a concentrated aqueous solution of sodium chloride saturated with ammonia. [1988 - 1 Mark] 31. Give reason of the following : Sodium carbonate is made by Solvay process but the same process is not extended to the manufacture of potassium carbonate. [1981 - 1 Mark]

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1.

(a) (b) (c) (d) 2. (a) (b) (c) (d) 3. (A) (B) (C) (D) (a) (b) (c) (d) 4. (a) (b) (c) (d) 5.

(a)

Among the sulphates of alkaline earth metals, the solubilities of BeSO4 and MgSO4 in water, respectively, are : [Main Sep. 06, 2020 (I)] poor and poor high and poor high and high poor and high An alkaline earth metal 'M' readily forms water soluble sulphate and water insoluble hydroxide. Its oxide MO is very stable to heat and does not have rock-salt structure. M is : [Main Sep. 04, 2020 (II)] Sr Ca Mg Be Among the statements (A)–(D), the correct ones are: [Main Jan. 09, 2020 (II)] Lithium has the highest hydration enthalpy among the alkali metals. Lithium chloride is insoluble in pyridine. Lithium cannot form ethynide upon its reaction with ethyne. Both lithium and magnesium react slowly with H2O. (A), (B) and (D) only (A), (C) and (D) only (B) and (D) only (A) and (D) only When gypsum is heated to 393 K, it forms: [Main Jan. 08, 2020 (I)] Anhydrous CaSO4 CaSO4  5 H2O CaSO4  0.5 H2O Dead burnt plaster A metal (A) on heating in nitrogen gas gives compound B. B on treatment with H2O gives a colourless gas which when passed through CuSO4 solution gives a dark blue-violet coloured solution. A and B respectively, are: [Main Jan. 08, 2020 (II)] Na and NaNO3

(b) Na and Na3N (c) Mg and Mg3N2 (d) Mg and Mg (NO3)2 6.

The correct sequence of thermal stability of the following carbonates is : [Main April 12, 2019 (I)]

(a) (b) (c) (d) 7.

BaCO3 < CaCO3 < SrCO3 < MgCO3 MgCO3 < CaCO3 < SrCO3 < BaCO3 MgCO3 < SrCO3 < CaCO3 < BaCO3 BaCO3 < SrCO3 < CaCO3 < MgCO3 The alloy used in the construction of aircrafts is : [Main April 10, 2019 (I)]

(a) Mg – Al

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(b) Mg – Zn (c) Mg – Sn (d) Mg – Mn 8. Magnesium powder burns in air to give: (a) (b) (c) (d) 9. (a) (b) (c) (d) 10. (a)

[Main April 9, 2019 (I)] Mg(NO3) 2 and Mg3N2 MgO and Mg3N2 MgO only MgO and Mg(NO3)2 The structures of beryllium chloride in the solid state and vapour phase, respectively, are: [Main April 9, 2019 (II)] chain and chain dimeric and dimeric chain and dimeric dimeric and chain The covalent alkaline earth metal halide (X = Cl, Br, I) is : [Main April 8, 2019 (II)] MgX2

(b) CaX2 (c) BeX2 (d) SrX2 11.

The amphoteric hydroxide is : [Main Jan. 11, 2019 (I)]

(a) (b) (c) (d) 12.

Be(OH)2 Ca(OH)2 Mg(OH)2 Sr(OH)2 The metal used for making X-ray tube window is:

(a) (b) (c) (d) 13.

Mg Na Be Ca The alkaline earth metal nitrate that does not crystallise with water molecules, is:

(a) (b) (c) (d) 14.

Mg (NO3)2 Sr (NO3)2 Ca (NO3)2 Ba (NO3)2 The commercial name for calcium oxide is :

[Main Jan. 10, 2019 (I)]

[Main Jan. 9, 2019 (I)]

(a) (b) (c) (d) 15. (a) (b) (c) (d)

[Main Online April 10, 2016] Quick lime Milk of lime Slaked lime Limestone The correct order of the solubility of alkaline–earth metal sulphates in water is : [Main Online April 9, 2016] Mg > Ca > Sr > Ba Mg > Sr > Ca > Ba Mg < Ca < Sr < Ba Mg < Sr < Ca < Ba

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16.

(a) (b) (c) (d) 17. (a)

Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy ? [Main 2015] BaSO4 SrSO4 CaSO4 BeSO4 The correct order of thermal stability of hydroxides is: [Main Online April 10, 2015] Ba(OH)2 < Ca(OH)2 < Sr(OH)2 < Mg(OH)2

(b) Mg(OH)2 < Sr(OH)2 < Ca(OH)2 < Ba(OH)2 (c) Mg(OH)2 < Ca(OH)2 < Sr(OH)2 < Ba(OH)2 (d) Ba(OH)2 < Sr(OH)2 < Ca(OH)2 < Mg(OH)2 18.

A sodium salt on treatment with MgCl2 gives white precipitate only on heating. The anion of the sodium salt is [2004S]

(a) (b) (c) (d) 19.

The set representing the correct order of first ionization potential is

[2001S] K > Na > Li Be > Mg > Ca B>C>N Ge > Si > C The following compounds have been arranged in order of their increasing thermal stabilities. Identify the correct order. [1996 - 1 Mark] K2CO3 (I) MgCO3 (II) CaCO3 (III) BeCO3 (IV) (a) I < II < III < IV (b) IV < II< III < I (c) IV < II < I < III (d) II < IV < III < I 21. Molecular formula of Glauber’s salt is : [1985 - 1 Mark] (a) MgSO4.7H2O

(a) (b) (c) (d) 20.

(b) CuSO4.5H2O (c) FeSO4.7H2O (d) Na2SO4.10H2O 22.

Calcium is obtained by

(a) (b) (c) (d) 23.

electrolysis of molten CaCl2. electrolysis of solution of CaCl2 in water. Reduction of CaCl2 with carbon. roasting of limestone. HCl is added to following oxides. Which one would give H2O2?

[1980]

[1980]

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(a) MnO2 (b) PbO2 (c) BaO2.8H2O (d) NO2 24.

A substance absorbs CO2 and violently reacts with water. The substance is [1978]

(a) (b) (c) (d)

CaCO3 CaO H2SO4 ZnO

25.

Ca2+ has a smaller ionic radius than K+ because it has ............

26.

Anhydrous MgCl2 is obtained by heating hydrated salt with ..........

[1993 - 1 Mark] [1980]

27.

MgCl2.6H2O on heating give anhydrous MgCl2. [1982 - 1 Mark]

28.

MgSO4 on reaction with NH4OH and Na2HPO4 forms a white crystalline precipitate. What is its formula? [2006 - 5M, –1]

(a) (b) (c) (d) 29.

The species that do not contain peroxide ions are

(a) (b) (c) (d)

PbO2 H2O2 SrO2 BaO2

[1992 - 1 Mark]

30. Match the following compounds (Column-I) with their uses (Column-II) : [Main Sep. 06, 2020 (II)] (I) (II) (III) (IV)

Column-I Ca(OH)2 NaCl CaSO4⋅ H2O CaCO3

(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)

(A) (B) (C) (D)

Column-II casts of statutes white wash antacid washing soda preparation

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(d) (I)-(C), (II)-(D), (III)-(B), (IV)-(A) 31. Match the following items in column I with the corresponding items in column II. [Main Jan. 11, 2019 (II)] Column-I Column-II (i) Na2CO3.10H2O (A) Portland cement ingredient (ii) Mg(HCO3)2 (B) Castner-Kellner process (iii) NaOH (C) Solvay process (iv) Ca3Al2O6 (D) Temporary hardness (a) (i) → (B); (ii) → (C); (iii) → (A); (iv) → (D) (b) (i) → (C); (ii) → (B); (iii) → (D); (iv) → (A) (c) (i) → (D); (ii) → (A); (iii) → (B); (iv) → (C) (d) (i) → (C); (ii) → (D); (iii) → (B); (iv) → (A)

32. Give reason of the following: BeCl2 can be easily hydrolysed. [1999 - 2 Marks] 33. Work out the following using chemical equation: Chlorination of calcium hydroxide produces bleaching powder. 34.

35.

36.

37.

38.

[1998 - 2 Marks] Arrange the following sulphates of alkaline earth metals in order of decreasing thermal stability : BeSO4, MgSO4, CaSO4, SrSO4 [1997 - 1 Mark] The crystalline salts of alkaline earth metals contain more water of crystallisation than the corresponding alkali metal salts. [1997 - 2 Marks] 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. Identify A, B, C and D. [1997 - 3 Marks] Mg3N2 when reacted with water gives off NH3 but HCl is not obtained from MgCl2 on reaction with water at room temperature. [1995 - 2 Marks] Complete and balance the following reactions : .

[1994 - 1 Mark] Give briefly the isolation of magnesium from sea water by the Dow process. Give equations for the steps involved. [1993 - 3 Marks] 40. Write down the balanced equations for the reactions when: Carbon dioxide is passed through a suspension of lime stone in water. [1991 - 1 Mark] 41. How will you prepare bleaching powder from slaked lime [1982 - 1 Mark]

39.

Topic-1 : Group-1 Elements (Alkali Metals) 1. (b)

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2. (d) 3. (d) 4. (b) 5. (c) 6. (a) 7. (d) 8. (b) 9. (c) 10. (b) 11. (b) 12. (c) 13. (b) 14. (d) 15. (d) 16. (b) 17. (d) 18. (b) 19. (a) 20. (b) 21. (d) 24. (False) 25. (True) 26. (a, b) 27. (a, b) 28. (b) Topic-2 : Group-2 Elements (Alkaline Earthmetals) 1. (c) 2. (d) 3. (b) 4. (c) 5. (c) 6. (b) 7. (a) 8. (b) 9. (c) 10. (c) 11. (a) 12. (c) 13. (d) 14. (a) 15. (a) 16. (d) 17. (c) 18. (a) 19. (b) 20. (b) 21. (d)

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22. 23. 24. 27. 28. 29. 30. 31.

(a) (c) (b) (False) (a) (a) (b) (d)

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1.

(a) (b) (c) (d) 2. (a) (b) (c) (d) 3.

(a) (b)

The reaction of H3N3B3Cl3(A) with LiBH4 in tetrahydrofuran gives inorganic benzene (B). Further, the reaction of (A) with (C) leads to H3N3B3(Me)3. Compounds (B) and (C) respectively, are: [Main Jan. 09, 2020 (II)] Borazine and MeBr Diborane and MeMgBr Boron nitride and MeBr Borazine and MeMgBr C60, an allotrope of carbon cantains: [Main April 9, 2019 (I)] 12 hexagons and 20 pentagons. 18 hexagons and 14 pentagons. 16 hexagons and 16 pentagons. 20 hexagons and 12 pentagons. Diborane (B2H6) reacts independently with O2 and H2O to produce, respectively; [Main April 8, 2019 (I)] B2O3 and H3BO3 B2O3 and [BH4]–

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(c) H3BO3 and B2O3 (d) HBO2 and H3BO3 4. The hydride that is NOT electron deficient is: [Main Jan. 11, 2019 (II)] (a) SiH4 (b) B2H6 (c) GaH3 (4) AlH3 5. The electronegativity of aluminium is similar to: [Main Jan. 10, 2019 (I)] (a) Carbon (b) Beryllium (c) Boron (d) Lithium 6. When metal ‘M’ is treated with NaOH, a white gelatinous 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 : [Main 2018] (a) Zn (b) Ca (c) Al (d) Fe 7. A group 13 element ‘X’ reacts with chlorine gas to produce a compound XCl3. XCl3 is electron deficient and easily reacts with NH3 to form Cl3X ← NH3 adduct, however, XCl3 does not dimerize.X is: [Main Online April 16, 2018] (a) B (b) Al (c) In (d) Ga 8. Identify the reaction which does not liberate hydrogen : [Main Online April 10, 2016] (a) Reaction of lithium hydride with B2H6. (b) Electrolysis of acidified water using Pt electrodes (c) Reaction of zinc with aqueous alkali (d) Allowing a solution of sodium in liquid ammonia to stand

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9.

The increasing order of atomic radii of the following Group 13 elements is [Adv. 2016] (a) A < Ga < In < T (b) Ga < A < In < T (c) A < In < Ga < T (d) A < Ga < T < In 10. How can the following reaction be made to proceed in forward direction? [2006 - 3M, –1] NaBO2 + Na[B(OH)4] + H2O B(OH)3 + NaOH (a) addition of borax (b) addition of cis -1,2-diol (c) addition of Na2HPO4 (d) addition of trans -1,2-diol 11. H3BO3 is: [2003S] (a) Monobasic and weak Lewis acid (b) Monobasic and weak Bronsted acid (c) Monobasic and strong Lewis acid (d) Tribasic and weak Bronsted acid 12. Which of the following statements about anhydrous aluminium chloride is correct? [1981 - 1 Mark] (a) it exists as AlCl3 molecules (b) it is not easily hydrolysed (c) it sublimes at 100ºC under vacuum (d) it is a strong Lewis base 13.

Three moles of B2H6 are completely reacted with methanol. The number of moles of boron containing product formed is [Adv. 2015] 14. The coordination number of Al in the crystalline state of AlCl3 is [2009]

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15.

Hydrogen gas is liberated by the action of aluminium with concentrated solution of .............. . [1987 - 1 Mark]

16.

All the Al–Cl bonds in Al2Cl6 are equivalent. [1989 - 1 Mark]

17. (a) (b) (c) (d) 18. (a) (b) (c) (d) 19. (a) (b) (c) (d) 20.

Among the following, the correct statement(s) is (are) [Adv. 2017] Al(CH3)3 has the three–centre two–electron bonds in its dimeric structure BH3 has the three–centre two–electron bonds in its dimeric structure AlCl3 has the three–centre two–electron bonds in its dimeric structure The Lewis acidity of BCl3 is greater than that of AlCl3 The crystalline form of borax has [Adv. 2016] tetranuclear [B4O5(OH)4]2– unit all boron atoms in the same plane equal number of sp2 and sp3 hybridized boron atoms one terminal hydroxide per boron atom The correct statement(s) for orthoboric acid is/are [Adv. 2014] It behaves as a weak acid in water due to self ionization. Acidity of its aqueous solution increases upon addition of ethylene glycol It has a three dimensional structure due to hydrogen bonding It is a weak electrolyte in water Match the following : [2006 - 6M]

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(A) (B) (C) (D)

Column I Bi3+ → (BiO)+ [AlO2]– → Al(OH)3 [SiO4]4– → [Si2O7]6– [B4O7]2– → [B(OH)3]

(p) (q) (r) (s)

Column II Heat Hydrolysis Acidification Dilution by water

This question contains STATEMENT-1 (Assertion/ Statement ) and STATEMENT-2 (Reason/Explanation) and has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct. (a) Statement-1 is True, Statement-2 isTrue; Statement-2 is a correct explanation for Statement-1 (b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (c) Statement-1 is True, Statement-2 is False (d) Statement-1 is False, Statement-2 is True. 21. Statement-1 : Boron always forms covalent bond. because Statement-2 : The small size of B3+ favours formation of covalent bond. [2007] 22. Statement-1 : In water, orthoboric acid behaves as a weak monobasic acid. because Statement-2 : In water, orthoboric acid acts as a proton donor. [2007] 23. Statement-1 : Al(OH)3 is amphoteric in nature Statement-2 : Al–O and O–H bonds can be broken with equal ease in Al(OH)3. [1998 - 2 Marks] 24.

AlF3 is insoluble in anhydrous HF but it becomes soluble in presence of little amount of KF. Addition of boron trifluoride to the resulting

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solution causes reprecipitation of AlF3. Explain with balanced chemical equations. [2004 - 2 Marks] 25. How is boron obtained from borax? Give chemical equations with reaction conditions. Write the structure of B2H6 and its reaction with HCl. [2002 - 5 Marks] 26. Compound (X) on reduction with LiAlH4 gives a hydride (Y) containing 21.72% hydrogen along with 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 - 5 Marks] 27. State the conditions under which the following preparation is carried out. Give the necessary equations which need not be balanced : Alumina from aluminium. [1983 - 1 Mark] 28. State with balanced equations what happens when : (i) Aluminium sulphide gives a foul odour when it becomes damp. Write a balanced chemical equation for the reaction. [1997 - 2 Marks] (ii) Reaction of aluminium with aqueous sodium hydroxide. [1997 - 1 Mark] 29. Give reason of the following : The hydroxides of aluminium and iron are insoluble in water. However, NaOH is used to separate one from the other. [1991 - 1 Mark] 30. Complete the following equations (no balancing is needed) + Al3+ → Al(OH)3 + .... (i) [1981 - 1 Mark] (ii) AlBr3 + K2Cr2O7 + H3PO4 → K3PO4 + AlPO4 + H2O +....+.... [1981 - 1 Mark]

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1. (a) (b) (c) (d) 2. (a) (b) (c) (d) 3. (a) (b) (c) (d) 4. (a) (b) (c) (d) 5. (a)

The correct statement among the following is : [Main April 12, 2019 (I)] (SiH3)3N is planar and less basic than (CH3)3N. (SiH3)3N is pyramidal and more basic than (CH3)3N. (SiH3)3N is pyramidal and less basic than (CH3)3N. (SiH3)3N is planar and more basic than (CH3)3N. The C – C bond length is maximum in : [Main April 12, 2019 (II)] graphite C70 C60 diamond The correct order of catenation is : [Main April 10, 2019 (I)] C > Sn > Si Ge C > Si > Ge Sn Si > Sn > C > Ge Ge > Sn > Si > C The element that does NOT show catenation is : [Main Jan. 12, 2019 (II)] Ge Si Sn Pb The chloride that CANNOT get hydrolysed is : [Main Jan. 11, 2019 (I)] PbCl4

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(b) CC14 (c) SnCl4 (d) 6. (a) (b) (c) (d) 7.

(a) (b) (c) (d) 8.

(a) (b) (c) (d) 9. (a) (b) (c) (d) 10. (a) (b)

SiCl4 Which of the following are Lewis acids? [Main 2018] PH3 and BCl3 AlCl3 and SiCl4 PH3 and SiCl4 BCl3 and AlCl3 The gas evolved on heating CaF2 and SiO2 with concentrated H2SO4, on hydrolysis gives a white gelatinous precipitate. The precipitate is: [Main Online April 9, 2014] hydrofluosilicic acid silica gel silicic acid calciumfluorosilicate Name of the structure of silicates in which three oxygen atoms of [SiO4]4– are shared. [2005S] Pyrosilicate Sheet silicate Linear chain silicate Three dimensional silicate (Me)2SiCl2 on hydrolysis will produce [2003S] (Me)2Si(OH)2 (Me)2Si = O –[–O–(Me)2Si–O–]n– Me2SiCl(OH) Which one of the following oxides is neutral? CO SnO2

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[1996 - 1 Mark] (c) ZnO (d) SiO2 11. Which of the following halides is least stable and has doubtful existence? [1996 - 1 Mark] (a) CI4 (b) GeI4 (c) SnI4 (d) PbI4 12. Moderate electrical conductivity is shown by [1982 - 1 Mark] (a) silica (b) graphite (c) diamond (d) carborundum 13. Lead pencil contains [1980] (a) Pb (b) FeS (c) Graphite (d) PbS 14.

The value of n in the molecular formula BenAl2Si6O18 is [2010]

15.

Compounds that formally contain Pb4+ are easily reduced to Pb2+. The stability of the lower oxidation state is due to ........... . [1997 - 1 Mark]

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16.

A liquid which is permanently supercooled is frequently called a ........... . [1997 - 1 Mark]

17.

One recently discovered allotrope of carbon (e.g., C60) is commonly known as.......... . [1994 - 1 Mark]

18. The hydrolysis of trialkylchlorosilane

yields ........... . [1994 - 1 Mark]

19. The hydrolysis of alkyl substituted chlorosilanes gives .............. . [1991 - 1 Mark] 20.

The tendency for catenation is much higher for C than for Si. [1993 - 1 Mark]

21.

Diamond is harder than graphite. [1993 - 1 Mark]

22.

Graphite is better lubricant on the moon than on the earth. [1987 - 1 Mark]

23.

Carbon tetrachloride is inflammable. [1985 - ½ Mark]

24.

Carbon tetrachloride burns in air when lighted to give phosgene. [1983 - 1 Mark]

25.

When PbO2 reacts with a dilute acid, it gives hydrogen peroxide. [1982 - 1 Mark]

26.

Choose the correct statement(s) among the following. is a reducing agent.

(a)

[Adv. 2020] (b) SnO2 reacts with KOH to form K2[Sn(OH)6]. (c) A solution of PbCl2 in HCl contains Pb2+ and Cl– ions.

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(d) The reaction of Pb3O4 with hot dilute nitric acid to give PbO2 is a redox reaction. 27. A tin chloride Q undergoes the following reactions (not balanced) [Adv. 2019]

(a) (b) (c) (d) 28.

(a) (b) (c) (d)

X is a mono anion having pyramidal geometry. Both Y and Z are neutral compounds. Choose the correct option(s). The oxidation state of the central atom in Z is +2 The central atom in Z has one lone pair of electrons The central atom in X is sp3 hybridized There is a coordinate bond in Y Under hydrolytic conditions, the compounds used for preparation of linear polymer and for chain termination, respectively, are [Adv. 2015] CH3SiCl3 and Si(CH3)4 (CH3)2SiCl2 and (CH3)3SiCl (CH3)2SiCl2 and CH3SiCl3 SiCl4 and (CH3)3SiCl

29. With respect to graphite and diamond, which of the statement(s) given below is (are) correct ? [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. 30. The material used in the solar cells contains [1993 - 1 Mark] (a) Cs (b) Si

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(c) Sn (d) Ti 31.

Match the items in Column I with its main use listed in Column II: [Main Online April 9, 2016] Column I Column II (A) Silica gel (i) Transistor (B) Silicon (ii) Ion– exchanger (C) Silicone (iii) Drying agent (D) Silicate (iv) Sealant

(a) (A)–(iii), (B)–(i), (C)–(iv), (D)–(ii) (b) (A)–(iv), (B)–(i), (C)–(ii), (D)–(iii) (c) (A)–(ii), (B)–(i), (C)–(iv), (D)–(iii) (d) (A)–(ii), (B)–(iv), (C)–(i), (D)–(iii) 32.

Match gases under specified conditions listed in Column I with their properties/laws in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS. [1995S] (A) (B) (C) (D)

Column I Explosive Artificial gem Self reduction Magnetic material

(p) (q) (r) (s) (t) (u) (v) (w)

Column II NaN3 Fe3O4 Cu Al2O3 Pb(N3)2 Fe2O3 Cu SiC

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33.

Each entry in column X is in some way related to the entries in column Y and Z. Match the appropriate entries : [Multiple Concepts, 1989 - 5 × 1 = 5 Marks]

X A . Mica B. Superphosphate C. Carbon fibres D. Rock salt E.

Carborundum

Y Z (a) Graphite crystallite (i) Abrasive (b) Cubic (ii) Insulator (c) Layer structure (iii) Fertilizer (d) Diamond structure (iv) Reinforced plastics (e) Bone ash (v) Preservative

Example : Yeast Fermentation Ethanol 34.

Assertion : Among the carbon allotropes, diamond is an insulator, whereas, graphite is a good conductor of electricity. Reason : Hybridization of carbon in diamond and graphite are sp3 and sp2, respectively.

[Main Online April 10, 2016] (a) Both assertion and reason are correct, but the reason is not the correct explanation for the assertion (b) Both assertion and reason are correct, and the reason is the correct explanation for the assertion (c) Both assertion and reason are incorrect (d) Assertion is incorrect statement, but the reason is correct. This question contains STATEMENT-1 (Assertion/ Statement ) and STATEMENT-2 (Reason/Explanation) and has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct. (a) Statement-1 is True, Statement-2 isTrue; Statement-2 is a correct explanation for Statement-1

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(b) (c) (d) 35.

Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 Statement-1 is True, Statement-2 is False Statement-1 is False, Statement-2 is True. Statement-1 : Pb+4 compounds are stronger oxidising agents than Sn4+ compounds [2008] Statement-2 : The higher oxidation states for the group 14 elements are more stable for the heavier members of the group due to 'inert pair effect'.

36.

Statement-1 : Between SiCl4 and CCl4, only SiCl4 reacts with water.

Statement-2 : SiCl4 is ionic and CCl4 is covalent. [2001S] 37.

Starting from SiCl4, prepare the following in steps not exceeding the number given in parentheses (give reactions only):

(i) Silicon (1) (ii) Linear silicone containing methyl groups only (4) (iii) Na2SiO3 (3) [2001 - 5 Marks] 38.

Draw the structure of a cyclic silicate, (Si3O9)6– with proper labelling. [1998 - 4 Marks]

39.

Complete and balance the following chemical reactions : [1994 - 1 Mark]

40.

State with balanced equations what happens when :

(i) SnCl4 + C2H5Cl + Na [1998 - 1 Mark]

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(ii) Write balanced equations for the preparation of crystalline silicon from SiCl4. [1990 - 1 Mark] 41.

Give reasons for the following :

(i) (SiH3)3N is a weaker base than (CH3)3N. [1995 - 2 Marks] (ii) The molecule of magnesium chloride is linear whereas that of stannous chloride is angular. [1987 - 1 Mark] (iii) Graphite is used as a solid lubricant; [1985 - 1 Mark] (iv) Solid carbon dioxide is known as dry ice. [1983 - 1 Mark] (v) Carbon acts as an abrasive and also as a lubricant. [1981 - 1 Mark] 42.

State with balanced equations, what happens when :

(i) Tin is treated with moderately concentrated nitric acid. (ii) Aluminium is reacted with hot concentrated caustic soda solution [1979]

Topic-1 : Group-13 Elements (Boron Family) 1. (d)2. (d) 3. (a) 4. (a) 5. (b) 6. (c) 7. (a) 8. (a) 9. (b)

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10. (b) 11. (a) 12. (c) 13. (6) 14. (6) 16. (False) 17. (a,b,d) 18. (a, c, d) 19. (b, d) 20. (A)-(q), (B)-(s), (C)-(p), (D)-(r) 21. (a) 22. (c) 23. (c) Topic-2 : Group-14 Elements (Carbon Family) 1. (a) 2. (d) 3. (b) 4. (d) 5. (b) 6. (b, d) 7. (d) 8. (b) 9. (c) 10. (a) 11. (d) 12. (b) 13. (c) 14. (3) 20. (True)

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21. (True) 22. (True) 23. (False) 24. (False) 25. (False) 26. (a, b)27. (c, d) 28. (b) 29. (b, d) 30. (b) 31. (a) 32. A-(t); B-(s); C-(v); D-(q) 33. A-(c)-(ii); B-(e)-(iii); C-(a)-(iv); D-(b)-(v); E-(d)-(i). 34. (b) 35. (c) 36. (c)

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1.

The IUPAC name of the following compound is: [Main Sep. 06, 2020 (II)]

(a) (b) (c) (d) 2.

2-nitro-4-hydroxymethyl-5-amino benzaldehyde 3-amino-4-hydroxymethyl-5-nitrobenzaldehyde 5-amino-4-hydroxymethyl-2-nitrobenzaldehyde 4-amino-2-formyl-5-hydroxymethyl nitrobenzene The IUPAC name of the following compound is : [Main Sep. 04, 2020 (I)]

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(a) (b) (c) (d) 3.

5-Bromo-3-methylcyclopentanoic acid 4-Bromo-2-methylcyclopentane carboxylic acid 3-Bromo-5-methylcyclopentanoic acid 3-Bromo-5-methylcyclopentane carboxylic acid The IUPAC name for the following compound is : [Main Sep. 02, 2020 (I)]

(a) 2, 5-dimethyl-5-carboxy-hex-3-enal (b) 2, 5-dimethyl-6-carboxy-hex-3-enal (c) 2, 5-dimethyl-6-oxo-hex-3-enoic acid (d) 6-formyl-2-methyl-hex-3-enoic acid 4. Which one of the following structures has the IUPAC name 3-ethynyl-2-hydroxy-4-methylhex-3-en-5-ynoic acid? [Adv. 2020] (a)

(b)

(c)

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(d)

5.

The IUPAC name for the following compound is : [Main April 12, 2019 (II)]

(a) (b) (c) (d) 6.

3-methyl-4-(3-methylprop-l-enyl)-l-heptyne 3,5-dimethyl-4-propylhept-6-en-l-yne 3-methyl-4-(1-methylprop-2-ynyl)-l-heptene 3,5-dimethyl-4-propylhept-1-en-6-yne The correct IUPAC name of the following compound is: [Main April 9, 2019 (I)]

(a) (b) (c) (d) 7.

5-chloro-4-methyl-1-nitrobenzene 2-chloro-1-methyl-4-nitrobenzene 3-chloro-4-methyl-1-nitrobenzene 2-methyl-5-nitro-1-chlorobenzene What is the IUPAC name of the following compound? [Main Jan. 10, 2019 (II)]

(a) 3-Bromo-1, 2-dimethylbut-1-ene

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(b) 3-Bromo-3-methyl-1, 2-dimethylprop-1-ene (c) 2-Bromo-3-methylpent-3-ene (d) 4-Bromo-3-methylpent-2-ene 8. The IUPAC name of the following compound is :

[Main Online April 15, 2018(I)] (a) (b) (c) (d) 9.

3-ethyl-4-methylhex-4-ene 4, 4-diethyl-3-methylbut-2-ene 4-methyl-3-ethylhex-4-ene 4-ethyl-3-methylhex-2-ene The IUPAC name of the following compound is : [Main Online April 8, 2017]

(a) (b) (c) (d) 10.

1, 1–dimethyl–2–ethylcyclohexane 2–ethyl–1,1–dimethylcyclohexane 1–ethyl–2,2–dimethylcyclohexane 2, 2–dimethyl–1–ethylcyclohexane The hydrocarbon with seven carbon atoms containing a neopentyl and a vinyl group is : [Main Online April 9, 2016] 2, 2–dimethyl–4– pentene 4, 4– dimethyl pentene isopropyl–2–butene 2, 2–dimethyl–3–pentene The correct IUPAC name of the following compound is: [Main Online April 19, 2014]

(a) (b) (c) (d) 11.

(a) 4 - methyl - 3 - ethylhexane

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(b) 3 - ethyl - 4 - methylhexane (c) 3, 4 - ethylmethylhexane (d) 4 - ethyl - 3 - methylhexane 12. In allene (C3H4), the type(s) of hybridization of the carbon atoms is (are): [Main Online April 11, 2014] (a) sp and sp3 (b) sp2 and sp (c) only sp2 (d) sp2 and sp3 13. The IUPAC name of the following compound is [2009]

(a) (b) (c) (d) 14. (a) (b) (c) (d) 15. (a)

4-Bromo-3-cyanophenol 2-Bromo-5-hydroxybenzonitrile 2- Cyano-4-hydroxybromobenzene 6-Bromo-3-hydroxybenzonitrile The IUPAC name of C6H5COCl is

[2006 - 3M, –1]

Benzene chloro ketone Benzoyl chloride Chloro phenyl ketone Benzene carbonyl chloride The compound which has one isopropyl group is : 2, 2, 3, 3-tetramethylpentane [1989]

(b) 2, 2-dimethylpentane (c) 2, 2, 3-trimethylpentane (d) 2-methylpentane 16. The IUPAC name of the compound [1987]

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(a) (b) (c) (d) 17.

CH2 = CH – CH (CH3)2 is 1, 1-dimethyl –2-propene 3-methyl–1-butene 2-vinylpropane 1-isopropylethylene The IUPAC name of the compound having the formula

is : [1984] (a) (b) (c) (d) 18.

3, 3, 3-Trimethyl-1-propene 1, 1, 1-Trimethyl-2-propene 3, 3-Dimethyl-1-butene 2, 2-Dimethyl-3-butene The IUPAC name of succinic acid is ............... .

[1990] 19. The kind of delocalization involving sigma bond orbitals is called ............... . [1994] 20. A ............... diol has two hydroxyl groups on ............... carbon atoms. [1986] 21.

The IUPAC name(s) of the following compound is (are)

[Adv. 2017] (a) (b) (c) (d)

1-chloro-4-methylbenzene 4-chlorotoluene 4-methylchlorobenzene 1-methyl-4-chlorobenzene

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22.

(i)

Write the IUPAC name for the following : [1991]

(ii) Give the IUPAC name of the following compound : [1990]

(Me = Methyl)

(iii) Write the IUPAC name of :

[1986] CH3CH2CH = CHCOOH

23.

Write the structural formula of 4-chloro-2-pentene. [1988]

1.

Which of the following compounds will show the maximum ‘enol’ content? [Main April 8, 2019 (II)] (a) CH3COCH2COOC2H5

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(b) CH3COCH2COCH3 (c) CH3COCH3 (d) CH3COCH2CONH2 2. In the following structure, the double bonds are marked as I, II, III and IV [Main Online April 9, 2017]

Geometrical isomerism is not possible at site (s) : (a) III (b) I (c) I and III (d) III and IV 3. The absolute configuration of

is :

[Main 2016] (a) (b) (c) (d) 4. (a) (b) (c) (d) 5.

(2S, 3S) (2R, 3R) (2R, 3S) (2S, 3R) Which of the following compounds will exhibit geometrical isomerism ? [Main 2015] 2 - Phenyl -1 - butene 1, 1 - Diphenyl - 1 - propene 1 - Phenyl - 2 - butene 3 - Phenyl -1 - butene The number of structural isomers for C6H14 is :

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[Main Online April 11, 2015] (a) (b) (c) (d) 6.

4 3 6 5 Which of the following pairs of compounds are positional isomers ? [Main Online April 11, 2015]

(a) and (b) and (c) and (d) and 7. (a) (b) (c) (d) 8.

Which one of the following acids does not exhibit optical isomerism? [Main Online April 12, 2014] Lactic acid Tartaric acid Maleic acid α-amino acids The number of isomers of C6H14 is

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[1987, 2007] (a) (b) (c) (d) 9.

4 5 6 7

(a) (b) (c) (d)

3 4 5 6

10.

The number of isomers for the compound with molecular formula C2BrClFI is [2001S]

The structure

shows

[1995S]: (a) (b) (c) (d) 11. (a) (b) (c) (d) 12. (a) (b) (c) (d) 13.

geometrical isomerism optical isomerism geometrical & optical isomerism tautomerism. Isomers which can be interconverted through rotation around a single bond are [1992] Conformers Diastereomers Enantiomers Positional isomers An isomer of ethanol is : [1986] methanol diethyl ether acetone dimethyl ether The compound which is not isomeric with diethyl ether is

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[1981] (a) (b) (c) (d) 14.

n-propyl methyl ether butan-1-ol 2-methylpropan-2-ol butanone Amongst the given options, the compound(s) in which all the atoms are in one plane in all the possible conformations (if any), is (are) [2011]

(a)

(b) (c) (d) 15.

(a) (b) (c) (d)

The correct statements(s) concerning the structures E,F and G is (are) – [2008]

E,F, and G are resonance structures E,F and E, G are tautomers F and G are geometrical isomers F and G are diastereomers

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16.

Tautomerism is exhibited by [1998]

(a)

CH = CH – OH

(b)

(c)

(d) 17.

Only two isomeric monochloro derivatives are possible for: [1986]

(a) (b) (c) (d) 18.

n-butane 2, 4-dimethylpentane benzene 2-methylpropane Write tautomeric forms for phenol. [1992]

1.

Which one of the following compounds possesses the most acidic hydrogen? [Main Sep. 03, 2020 (I)]

(a) (b)

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(c)

(d) 2.

Glycerol is separated in soap industries by : [Main Sep. 03, 2020 (I)]

(a) Fractional distillation (b) Differential extraction (c) Steam distillation (d) Distillation under reduced pressure 3. The increasing order of basicity for the following intermediates is (from weak to strong) [Main Jan. 09, 2020 (I)]

(a) (b) (c) (d) 4.

(iii) < (i) < (ii) < (iv) < (v) (v) < (i) < (iv) < (ii) < (iii) (v) < (iii) < (ii) < (iv) < (i) (iii) < (iv) < (ii) < (i) < (v) A flask contains a mixture of isohexane and 3-methylpentane. One of the liquids boils at 63° C while the other boils at 60° C. What is the best way to separate the two liquids and which one will be distilled out first? [Main Jan. 08, 2020 (I)] (a) fractional distillation, isohexane (b) simple distillation, 3-methylpentane

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(c) simple distillation, isohexane (d) fractional distillation, 3-methylpentane 5. The correct order of stability for the following alkoxides is:

[Main Jan. 07, 2020 (II)] (a) (B) > (A) > (C) (b) (C) > (B) > (A) (c) (C) > (A) > (B) (d) (B) > (C) > (A) 6.

The increasing order of nucleophilicity of the following nucleophiles is : [Main April 10, 2019 (II)]

(i) (ii) H2O (iii) (iv) OH– (a) (i) < (iv) < (iii) < (ii) (b) (ii) < (iii) < (iv) < (i) (c) (iv) < (i) < (iii) < (ii) (d) (ii) < (iii) < (i) < (iv) 7. The increasing order of basicity of the following compounds is [Main 2018] (i) (ii) (iii) (iv)

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(a) (b) (c) (d) 8.

(i) < (ii) < (iii) < (iv) (ii) < (i) < (iii) < (iv) (ii) < (i) < (iv) < (iii) (iv) < (ii) < (i) < (iii) The most polar compound among the following is : [Main Online April 16, 2018]

(a)

(b)

(c)

(d)

9.

(a) (b) (c) (d) 10.

Two compounds I and II are eluted by column chromatography(adsorption of I > II). Which one of the following is a correct statement? [Main Online April 15, 2018 (II)] II moves slower and has higher Rf value than I II moves faster and has higher Rf value than I I moves faster and has higher Rf value than II I moves slower and has higher Rf value than II Which of the following molecules is least resonance stabilized? [Main 2017]

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(a)

(b)

(c)

(d) 11. (a) (b) (c) (d) 12.

The distillation technique most suited for separating glycerol from pent-l-ye in the soap industry is : [Main 2016] Steam distillation. Distillation under reduced pressure. Simple distillation Fractional distillation For which of the following molecule significant [Main 2014]

(i)

(ii)

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(iii)

(iv)

(a) (b) (c) (d) 13.

(a)

Only (i) (i) and (ii) Only (iii) (iii) and (iv) In which of the following pairs A is more stable than B? [Main Online April 9, 2014] A B ,

(b) (c)

, ,

(d) 14.

A solution of ( – ) – 1 – chloro –1– phenylethane in toluene racemises slowly in the presence of a small amount of SbCl5, due to the formation of : [Main 2013] (a) carbanion (b) carbene

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(c) carbocation (d) free radical 15. The order of stability of the following carbocations : [Main 2013]

(a) (b) (c) (d) 16.

III > II > I II > III > I I > II > III III > I > II Which one of the following is most stable ? [Main Online April 9, 2013]

(a)

(b)

(c)

(d) 17.

In CH3CH2OH, the bond that undergoes heterolytic cleavage most readily is [1988]

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(a) (b) (c) (d) 18.

C—C C—O C—H O—H The Cl—C—Cl angle in 1,1,2,2-tetrachloroethene tetrachloromethane respectively will be about

and [1988]

(a) (b) (c) (d) 19.

120º and 109.5º 90º and 109.5º 109.5º and 90º 109.5º and 120º The bond between carbon atom (1) and carbon atom (2) in compound = CH2 involves the hybrids as

N≡

[1987] (a) (b) (c) (d) 20. (a) (b) (c) (d) 21. (a) (b) (c) (d) 22. (a)

sp and sp sp3 and sp sp and sp2 sp and sp Out of the following compounds, which will have a zero dipole moment? [1987] 1, 1-dichloroethylene cis-1, 2-dichloroethylene trans-1, 2-dichloroethylene None of these compounds The compound 1, 2-butadiene has [1983] only sp hybridized carbon atoms only sp2 hybridized carbon atoms both sp and sp2 hybridized carbon atoms sp, sp2 and sp3 hybridized carbon atoms Molecule in which the distance between the two adjacent carbon atoms is largest is [1981] Ethane 2

2

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(b) Ethene (c) Ethyne (d) Benzene

23.

The total number of contributing structures showing hyperconjugation (involving C−H bonds) for the following carbocation is [2011]

24.

In an estimation of bromine by Carius method, 1.6 g of an organic compound gave 1.88 g of AgBr. The mass percentage of bromine in the compound is ______. (Atomic mass, Ag = 108, Br = 80 g mol–1) [Main Sep. 06, 2020 (I)] 25.

The valence atomic orbitals on carbon in silver acetylide is ............... hybridized. [1990] 26. ................. ring is most strained. [1981] (Cyclopropane, Cyclobutane, Cyclopentane) 27. The compound having both sp and sp2 hybridized carbon atoms is ................. . [1981] (propene, propane, propadiene) 28. Among the given cations, ................. is most stable. [1981] (sec-butyl carbonium ion; tert-butyl carbonium ion; n-butyl carbonium ion)

29.

With respect to the compounds I-V, choose the correct statement(s).

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(a) The acidity of compound I is due to delocalization in the conjugate base. (b) The conjugate base of compound IV is aromatic. (c) Compound II becomes more acidic, when it has a –NO2 substituent. (d) The acidity of compounds follows the order I > IV > V > II > III. 30. The hyperconjugative stabilities of tert-butyl cation and2-butene, respectively, are due to [Adv. 2013] (a) σ → p (empty) and σ → π* electron delocalisations (b) σ → σ* and σ → π electron delocalisations (c) σ → p (filled) and σ → π electron delocalisations (d) p (filled) → σ* and σ → π* electron delocalisations 31. Which of the following molecules, in pure form, is (are) unstable at room temperature ? [2012] (a) (b)

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(c)

(d) 32.

Among the following compounds, the strongest acid is [1998]

(a) HC ≡ CH (b) C6H6 (c) C2H6 (d) CH3OH 33. What is

the

decreasing

order of ?

strength

of

the

bases [1993]

(a) (b) (c) (d) 34.

Dipole moment is shown by : [1986]

(a) (b) (c) (d) 35. (a)

1, 4-dichlorobenzene cis- 1, 2-dichloroethane trans -1, 2-dichloroethene trans- 1, 2-dichloro-2-pentene Resonance structures of a molecule should have : identical arrangement of atoms [1984]

(b) nearly the same energy content (c) the same number of paired electrons

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(d) identical bonding

36.

(i)

, where

is the dipole moment of a stable

is the mole conformer of the molecule, Z – CH2 – CH2 – Z and fraction of the stable conformer. Given : µobs = 1.0 D and x(Anti) = 0.82 Draw all the stable conformers of Z – CH2 – CH2 – Z and calculate the value of µ(Gauche). (ii) Draw the stable conformer of Y – CHD – CHD – Y (meso form), when Y = CH3 (rotation about C2 – C3) and Y = OH (rotation about C1 – C2) in Newmann projection. [2005] 37. Which of the following is more acidic and why ? [2004]

38. [2003] Write resonance structure of the given compound. 39. Which one is more soluble in diethyl ether - anhydrous AlCl3 or hydrous AlCl3? Explain in terms of bonding. [2003] 40. Give reasons for the following : . (i) CH2=CH – is more basic than [1994] (ii) Phenyl group is known to exert negative inductive effect. But each phenyl ring in biphenyl (C6H5 – C6H5) is more reactive than benzene towards electrophilic substitution. [1992]

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(iii) Carbon oxygen bond lengths in formic acid are 1.23Å and 1.36Å and both the carbon oxygen bonds in sodium formate have the same value i.e. 1.27Å. [1988] 41. n-butane, n-butanol, n-butyl chloride, isobutane in increasing order of boiling point. [1988] 42. For nitromethane molecule, write structure(s). (i) showing significant resonance stabilisation. [1986] (ii) indicating tautomerism. [1986] Topic-1 : Classifcation and Nomenclature of Organic Compounds 1. (c) 2. (b) 3. (c) 4. (d) 5. (d) 6. (b) 7. (d) 8. (d) 9. (b) 10. (b) 11. (b) 12. (b) 13. (b) 14. (b) 15. (d) 16. (b) 17. (c) 21. (a, b) Topic-2 : Isomerism in Organic Compounds

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1. (b) 2. (b) 3. (d) 4. (c) 5. (d) 6. (a) 7. (c) 8. (b) 9. (d) 10. (b) 11. (a) 12. (d) 13. (d) 14. (b, c) 15. (b,c,d) 16. (a,c,d) 17. (a, d) Topic-3 : Concepts of Reaction Mechanism in Organic Compounds and Purification 1. (d) 2. (d) 3. (c) 4.(a) 5. (b) 6. (d) 7. (c) 8. (c) 9. (b) 10. (d) 11. (b) 12. (d) 13. (d) 14. (c) 15. (d)

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16. (a) 17. (d) 18. (a) 19. (c) 20. (c) 21. (d) 22. (a) 23. (6) 24. (50) 29. (a, b, c) 30. (a) 31. (b, c) 32. (d) 33. (a) 34. (b, d) 35. (a,b,c)

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1.

Newman projections P, Q, R and S are shown below: [Adv. 2020]

Which one of the following options represents identical molecules?

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(a) P and Q (b) Q and S (c) Q and R (d) R and S 2. In the following skew conformation of ethane, H′ – C – C – H3 dihedral angle is : [Main April 12, 2019 (II)]

(a) 58o (b) 149o (c) 151o (d) 120o 3. At 300 K and 1 atmospheric pressure, 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 : [Main April 10, 2019 (I)] (a) C4H10 (b) C4H6 (c) C4H7Cl (d) C4H8 4. The major product obtained in the photo catalysed bromination of 2methylbutane is: [Main 2005, Online May 19, 2012; Online April 12, 2014] (a) 1-bromo-2-methylbutane (b) 1-bromo-3-methylbutane (c) 2-bromo-3-methylbutane (d) 2-bromo-2-methylbutane

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5.

Isomers of hexane, based on their branching, can be divided into three distinct classes as shown in the figure. [Adv. 2014]

The correct order of their boiling point is (a) I > II > III (b) III > II > I (c) II > III > I (d) III > I > II 6. The bond energy (in kcal mol–1) of a C–C single bond is approximately [2010] (a) 1 (b) 10 (c) 100 (d) 1000 7. On monochlorination of 2-methylbutane, the total number of chiral compounds formed is [2004S] (a) 2 (b) 4 (c) 6 (d) 8 8.

In the given conformation, if C2 is rotated about C2 – C3 bond anticlockwise by an angle of 120º then the conformation obtained is [2004S]

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(a) (b) (c) (d) 9.

fully eclipsed conformation partially eclipsed conformation gauche conformation staggered conformation Consider the following reaction [2002S]

Identify the structure of the major product 'X' (a)

(b)

(c)

(d) 10. When cyclohexane is poured on water, it floats, because: (a) cyclohexane is in ‘boat’ form [1997 - 1 Mark] (b) cyclohexane is in ‘chair’ form (c) cyclohexane is in ‘crown’ form (d) cyclohexane is less dense than water. 11. Which of the following will have least hindered rotation about carbon-carbon bond? [1987 - 1 Mark] (a) Ethane (b) Ethylene

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(c) Acetylene (d) Hexachloroethane 12. The compound with the highest boiling point is [1982 - 1 Mark] (a) (b) (c) (d) 13.

n-hexane n-pentane 2, 2-dimethylpropane 2-methylbutane Marsh gas mainly contains [1980]

(a) (b) (c) (d) 14.

C2H2 CH4 H2S CO The total number of stereoisomers that can exist for M is [Adv. 2015]

15.

The total number(s) of stable conformers with non-zero dipole moment for the following compound is (are) [Adv. 2014]

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16.

Photobromination of 2-methylpropane gives a mixture of1-bromo-2methylpropane and 2-bromo-2-methylpropane in the ratio of 9: 1. [1993 - 1 Mark]

17.

Which of the following reactions produce(s) propane as a major product ? [Adv. 2019]

(a) (b) (c) (d) 18.

Which of the given statement(s) about N, O, P and Q with respect to M is (are) correct ? [2012]

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(a) (b) (c) (d) 19.

M and N are non-mirror image stereoisomers M and O are identical M and P are enantiomers M and Q are identical In the Newman projection for 2,2-dimethylbutane

X and Y can respectively be [2010] (a) H and H (b) H and C2H5 (c) C2H5 and H 20.

(d) CH3 and CH3 ;

(isomeric products) Identify N and M [2006 - 5M, –1] (a) 6, 4 (b) 6, 6

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(c) 4, 4 (d) 3, 3 21.

Draw Newmann projection of relatively less stable staggered form of n-butane. The reason of low stability of this form is van der Waal’s repulsion, torsional strain, or both. [2004 - 2 Marks] 22. Optically active 2-iodobutane on treatment with NaI in acetone gives a product which does not show optical activity. Explain briefly. [1995 - 2 Marks] 23. n-Butane is produced by the monobromination of ethane followed by the Wurtz reaction. Calculate the volume of ethane at NTP required to produce 55 g n-butane, if the bromination takes place with 90 per cent yield and the Wurtz reaction with 85 per cent yield. [1989 - 3 Marks] 24. Give reason of the following : Methane does not react with chlorine in the dark. [1983 - 1 Mark]

1.

The major product of the following reaction is :

[Main Sep. 06, 2020 (I)]

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(a)

(b)

(c)

(d)

2. (a) (b) (c) (d) 3.

Which of the following compounds shows geometrical isomerism? [Main Sep. 06, 2020 (I)] 2-methylpent-2-ene 4-methylpent-2-ene 4-methylpent-1-ene 2-methylpent-1-ene The increasing order of the boiling point of the major products A, B and C of the following reactions will be: [Main Sep. 06, 2020 (II)]

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(a) (b) (c) (a) (b) (c) (d) 4.

B3>4 2>3>1>4 Which of the following compounds does not dissolve in conc. H2SO4 even on warming? [1983 - 1 Mark]

(a) (b) (c) (d) 8.

(a) (b) (c) (d)

ethylene benzene hexane aniline Among the following, the compound that can be most readily sulphonated is [1982] benzene nitrobenzene toluene chlorobenzene

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9. (a) (b) (c) (d)

The bond order of individual carbon-carbon bonds in benzene is [1981] one two between one and two one and two, alternately

10.

Among the following, the number of aromatic compound(s) is [Adv. 2017]

11.

The bond dissociation energy needed to form the benzyl radical from toluene is........than the formation of the methyl radical from methane. [1994 - 1 Mark] and ........ 12. Kolbe electrolysis of potassium succinate gives [1993 - 1 Mark]

13.

................. ring is most strained. [1981]

(Cyclopropane, Cyclobutane, Cyclopentane) 14.

An electron donating substituent in benzene orients the incoming electrophilic group to the meta position. [1987]

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15.

Choose the correct option(s) that give(s) an aromatic compound as the major product. [Adv. 2019]

(a)

(b)

(c) (d) 16.

Among the following, reaction(s) which gives (give)tert-butyl benzene as the major product is (are) [Adv. 2016]

(a)

(b)

(c)

(d) 17. The major product U in the following reactions is

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[Adv. 2015]

(a)

(b)

(c)

(d) 18.

Among P, Q, R and S, the aromatic compound(s) is/are [Adv. 2013-I]

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(a) (b) (c) (d) 19. (a) (b) (c) (d) 20. (a) (b) (c) (d)

P R Q S Toluene, when treated with Br2/Fe, gives p-bromotoluene as the major product because CH3 group is para directing [1999 - 3 Marks] is meta directing activates the ring by hyperconjugation deactivates the ring An aromatic molecule will [1999] have 4n π electrons have (4n + 2) π electrons be planar be cyclic

21. Give reasons for the following : (i) tert-Butylbenzene does not give benzoic acid on treatment with acidic KMnO4. [2000 - 1 Mark] (ii) Toluene reacts with bromine in the presence of light to give benzyl bromide while in presence of FeBr3 it gives pbromotoluene. Give explanation for the above observations. [1996 - 2 Marks] (iii) Although benzene is highly unsaturated, normally it does not undergo addition reaction.[1983 - 1 Mark] 22. Write the structural formula of the major product in each of the following cases : (i)

+

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[1997 - 1 Mark] (ii) [1994 - 1 Mark] (iii) [1994 - 1 Mark] (iv)

+ (CH3)2CHCH2Cl

[1992 - 1 Mark] 23. Show the steps to carry out the following transformations. (i) Ethylbenzene → benzene [1998 - 2 Marks] (ii) Ethylbenzene → 2- phenylpropionic acid. [1998 - 3 Marks] 24. Give reason of the following : Normally, benzene gives electrophilic substitution reaction rather than electrophilic addition reaction although it has double bonds. [1994] 25. Arrange the following in : benzene, toluene, methoxybenzene, chlorobenzene in increasing order of reactivity towards sulphonation with fuming sulphuric acid. [1988] 26. Write down the reactions involved in the preparation of the following, using the reagents indicated against it in parenthesis. Ethylbenzene from benzene [C2H5OH, PCl5, anhydrous AlCl3]. [1984 - 2 Marks] Topic-1 : Alkanes 1. (c) 2. (b) 3. (b)

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4. (d) 5. (b) 6. (c) 7. (b) 8. (c) 9. (b) 10. (d) 11. (a) 12. (a) 13. (b) 14. (2) 15. (3) 16. 17. (a, c) 18. (a, b, c) 19. (b, d) 20. (a) 21. 22. 23. 24. Topic-2 : Alkenes 1. (d) 2. (b) 3. (a) 4. (d) 5. (b, c) 6. (d) 7. (a) 8. (b) 9. (d) 10. (d) 11. (b) 12. (c) 13. (d) 14. (c) 15. (b)

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16. (d) 17. (c) 18. (d) 19. (d) 20. (d) 21. (c) 22. (d) 23. (a) 24. (b) 25. (a) 26. (b) 27. (a) 28. (b) 29. (c) 30. (a) 31. (a) 32. (c) 33. (a) 34. (a) 35. (b) 36. -7 37. 38. 39. 40. (b, d) 41. (b, d) 42. (d) 43. (b) 44. (b) 45. (c) 46. (a) 47. 48. (i) 49. 50. (a) 51. 52.

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53. 55. 57. 59.

54. 56. 58. 60. Topic-3 : Alkynes

1. (a) 2. (c) 3. (a) 4. (d) 5. (c) 6. (d) 7. (d) 8. (a) 9. (d) 10. (b) 11. (d) 12. (a) 13. (a) 14. (c) 15. (8) 16. (13) 17. 18. 19. 20. 21. (b, c) 22. (a) 23. (c) 24. (d) 25. (b) 26. 27. 28. 29. 30.

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31. 32. 33. 34.

35. Topic-4 : Aromatic Hydrocarbons

1. (c) 2. (a) 3. (a) 4. (a) 5. (b) 6. (c) 7. (c) 8. (c) 9. (c) 10. (4) 11. 12. 13. 14. 15. (b, c) 16. (b, c, d) 17. (b) 18. (a, b, c, d) 19. (a,c) 20. (b,c,d) 21. (i) 22. (i) 23. 24. 25. 26.

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1. (a) (b) (c) (d) 2. (a) (b) (c) (d) 3. (1) (2) (3) (4)

The processes of calcination and roasting in metallurgical industries, respectively, can lead to : [Main Sep. 04, 2020 (II)] Global warming and photochemical smog Global warming and acid rain Photochemical smog and ozone layer depletion Photochemical smog and global warming Thermal power plants can lead to : [Main Sep. 03, 2020 (I)] Acid rain Blue baby syndrome Ozone layer depletion Eutrophication The incorrect statement(s) among (1) - (4) regarding acid rain is (are) : [Main Sep. 03, 2020 (II)] It can corrode water pipes. It can damage structures made up of stone. It cannot cause respiratory ailments in animals It is not harmful for trees

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(a) (b) (c) (d) 4. (a) (b) (c) (d) 5. (1) (2) (3) (4) (5) (a) (b) (c) (d) 6. (a) (b) (c) (d) 7. (a) (b) (c) (d)

(1), (3) and (4) (3) only (1), (2) and (4) (3) and (4) The statement that is not true about ozone is : [Main Sep. 02, 2020 (I)] in the stratosphere, CFCs release chlorine free radicals (Cl) which reacts with O3 to give chlorine dioxide radicals. in the atmosphere, it is depleted by CFCs. in the stratosphere, it forms a protective shield against UV radiation. it is a toxic gas and its reaction with NO gives NO2. Among the gases (a)–(e), the gases that cause greenhouse effect are: [Main Jan. 08, 2020 (I)] CO2 H2O CFCs O2 O3 (1), (2), (3) and (4) (1), (2), (3) and (5) (1) and (4) (1), (3), (4) and (5) The correct set of species responsible for the photochemical smog is : [Main April 12, 2019 (I)] N2, NO2 and hydrocarbons CO2, NO2, SO2 and hydrocarbons NO, NO2, O3 and hydrocarbons N2, O2, O3 and hydrocarbons The primary pollutant that leads to photochemical smog is: [Main April 12, 2019 (II)] acrolein nitrogen oxides ozone sulphur dioxide

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8. (a) (b) (c) (d) 9. (a) (b) (c) (d) 10. (a) (b) (c) (d) 11. (a) (b) (c) (d) 12. (a) (b) (c) (d) 13.

The regions of the atmosphere, where clouds are formed and where we live, respectively, are : [Main April 10, 2019 (I)] Troposphere and Stratosphere Stratosphere and Troposphere Troposphere and Troposphere Stratosphere and Stratosphere Air pollution that occurs in sunlight is : [Main April 10, 2019 (II)] reducing smog acid rain oxidizing smog fog Excessive release of CO2 into the atmosphere results in: [Main April 9, 2019 (I)] global warming polar vortex formation of smog depletion of ozone The layer of atmosphere between 10 km to 50 km above the sea level is called as: [Main April 9, 2019 (II)] troposphere thermosphere stratosphere mesosphere Which is wrong with respect to our responsibility as a human being to protect our environment ? [Main April 8, 2019 (I)] Restricting the use of vehicles Avoiding the use of floodlighted facilities Setting up compost tin in gardens. Using plastic bags. The molecule that has minimum/no role in the formation of photochemical smog, is: [Main Jan. 12, 2019 (I)]

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(a) (b) (c) (d) 14.

N2 CH2= O O3 NO The compound that is NOT a common component of photochemical smog is : [Main Jan. 12, 2019 (II)] (a) O3 (b) H3C –

–OONO2

(c) CH2 = CHCHO (d) CF2Cl2 15. The upper stratosphere consisting of the ozone layer protects us from the sun’s radiation that falls in the wavelength region [Main Jan. 12, 2019 (II)] (a) 200 – 315 nm (b) 400 – 550 nm (c) 0.8 – 1.5 nm (d) 600 – 750 nm 16. Peroxyacetyl nitrate (PAN), an eye irritant, is produced by : [Main Jan. 11, 2019 (I)] (a) Classical smog (b) Acid rain (c) Organic waste (d) Photochemical smog 17. Taj Mahal is being slowly disfigured and discoloured. This is primarily due to : [Main Jan. 11, 2019 (II)] (a) Global warming (b) Acid rain (c) Water pollution (d) Soil pollution 18. The higher concentration of which gas in air can cause stiffness of flower buds? [Main Jan. 11, 2019 (II)]

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(a) NO2 (b) CO2 (c) SO2 (4) CO 19. The reaction that is NOT involved in the ozone layer depletion mechanism in the stratosphere is: [Main Jan. 10, 2019 (II)] (a) (b) (c) (d) 20. (a) (b) (c) 21. (a) (b) (c) (d) 22. (a) (b) (c) (d) 23. (a) (b)

Which of the following is a set of green house gases ? [Main Online April 9, 2017] CH4, O3, N2, SO2 O3, N2, CO2, NO2 O3, NO2, SO2, Cl2 (d) CO2, CH4, N2O, O3 Identify the pollutant gases largely responsible for the discoloured and lustreless nature of marble of the Taj Mahal. [Main Online April 8, 2017] O3 and CO2 CO2 and NO2 SO2 and NO2 SO2 and O3 Which one of the following substances used in dry cleaning is a better strategy to control environmental pollution ? [Main Online April 10, 2016] Sulphur dioxide Carbon dioxide Nitrogen dioxide Tetrachloroethylene Photochemical smog consists of excessive amount of X, in addition to aldehydes, ketones, peroxyacetyl nitrate (PAN), and so forth X is : [Main Online April 10, 2015] CO CH4

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(c) O3 (d) CO2 24. Global warming is due to increase of: (a) (b) (c) (d) 25. (a) (b)

(c) (d) 26. (a) (b) (c) (d)

[Main Online April 12, 2014] methane and nitrous oxide in atmosphere methane and CO2 in atmosphere methane and O3 in atmosphere methane and CO in atmosphere Which of the following statements about the depletion of ozone layer is correct? [Main Online April 11, 2014] The problem of ozone depletion is less serious at poles because NO2 solidifies and is not available for consuming CIO• radicals. The problem of ozone depletion is more serious at poles because ice crystals in the clouds over poles act as catalyst for photochemical reactions involving the decomposition of ozone of Cl• and CIO• radicals. Freons, chlorofluorocarbons, are inert. Chemically, they do not react with ozone in stratosphere. Oxides of nitrogen also do not react with ozone in stratosphere. The gas leaked from a storage tank of the Union Carbide plant in Bhopal gas tragedy was : [Main 2013] Methyl isocyanate Methylamine Ammonia Phosgene

27. Assertion : Ozone is destroyed by CFCs in the upper stratosphere. Reason : Ozone holes increase the amount of UV radiation reaching the earth. [Main April 8, 2019 (I)] (a) Assertion and reason are incorrect. (b) Assertion and reason are both correct, and the reason is the correct explanation for the assertion.

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(c) Assertion and reason are correct, but the reason is not the explanation for the assertion. (d) Assertion is false, but the reason is correct.

1.

(a) (b) (c) (d) 2. (a) (b) (c) (d) 3. (a) (b) (c) (d) 4. (a) (b)

The presence of soluble fluoride ion upto 1 ppm concentration in drinking water, is : [Main Sep. 06, 2020 (I)] harmful for teeth harmful to skin harmful to bones safe for teeth The condition that indicates a polluted environment is: [Main Sep. 05, 2020 (I)] eutrophication 0.03% of CO2 in the atmosphere BOD value of 5 ppm pH of rain water to be 5.6 Biochemical Oxygen Demand (BOD) is the amount of oxygen required (in ppm): [Main Jan. 09, 2020 (II)] for sustaining life in a water body. by bacteria to break-down organic waste in a certain volume of a water sample. for the photochemical breakdown of waste present in 1 m3 volume of a water body. by anaerobic bacteria to breakdown inorganic waste present in a water body. The maximum prescribed concentration of copper in drinking water is: [Main April 8, 2019 (II)] 5 ppm 0.05 ppm

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(c) 0.5 ppm (d) 3 ppm 5. Water samples with BOD values of 4 ppm and 18 ppm, respectively, are : [Main Jan. 12, 2019 (I)] (a) Clean and Clean (b) Highly polluted and Clean (c) Clean and Highly polluted (d) Highly polluted and Highly polluted 6. The concentration of dissolved oxygen (DO) in cold water can go upto : [Main Jan. 11, 2019 (I)] (a) 14 ppm (b) 8 ppm (c) 10 ppm (d) 16 ppm 7. Water filled in two glasses A and B have BOD values of 10 and 20, respectively. The correct statement regarding, them, is: [Main Jan. 10, 2019 (I)] (a) B is more polluted than A. (b) A is suitable for drinking, whereas B is not. (c) Both A and B are suitable for drinking. (d) A is more polluted than B. 8. A water sample has ppm level concentration of the following metals: Fe = 0.2; Mn = 5.0; Cu = 3.0; Zn = 5.0. The metal that makes the water sample unsuitable for drinking is: [Main Jan. 9, 2019 (I)] (a) Cu (b) Mn (c) Fe (d) Zn 9. Which of the following conditions in drinking water causes methemoglobinemia? [Main Jan. 9, 2019 (II)] (a) > 50 ppm of lead (b) > 50 ppm of chloride

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(c) > 50 ppm of nitrate (d) > 100 ppm of sulphate 10. The pH of rain water, is approximately: [Main Jan. 9, 2019 (II)] (a) (b) (c) (d) 11.

5.6 7.5 7.0 6.5 The recommended concentration of fluoride ion in drinking water is up to 1ppm as fluoride ion is required to make teeth enamel harder by converting [3Ca3(PO4)2. Ca(OH)2] to : [Main 2018] (a) [CaF2] (b) [3(CaF2).Ca(OH)2] (c) [3Ca3(PO4)2.CaF2] (d) [3{(Ca(OH)2}.CaF2] 12. Biochemical oxygen demand(BOD) value can be a measure of water pollution caused by the organic matter. Which of the following statements is correct? [Main Online April 15, 2018 (II)] (a) Polluted water has BOD value higher than 10 ppm (b) Aerobic bacteria decreases the BOD value (c) Anaerobic bacteria increases the BOD value (d) Clean water has BOD value higher than 10 ppm 13. A water sample has ppm level concentration of following anions F– = 10; SO2–4 = 100; NO–3= 50 the anion/anions that make/makes the water sample unsuitable for drinking is/are : [Main 2017] – (a) only NO3 (b) both SO2–4 and NO–3 (c) only F– (d) only SO2–4 14. The concentration of fluoride, lead, nitrate and iron in a water sample from an underground lake was found to be 1000 ppb, 40 ppb, 100 ppm

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(a) (b) (c) (d) 15. (a) (b) (c) (d) 16. (a) (b) (c) (d)

and 0.2 ppm, respectively. This water is unsuitable for drinking due to high concentration of : [Main 2016] Nitrate Iron Fluoride Lead BOD stands for : [Main Online April 9, 2016] Biochemical Oxidation Demand Biological Oxygen Demand Biochemical Oxygen Demand Bacterial Oxidation Demand Addition of phosphate fertilisers to water bodies causes: [Main Online April 11, 2015] increase in amount of dissolved oxygen in water deposition of calcium phosphate increase in fish population enhanced growth of algae

Topic-1 : Air Pollution 1. (b) 2. (a) 3. (d) 4. (a) 5. (b) 6. (c) 7. (b) 8. (c) 9. (c) 10. (a) 11. (c)

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12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27.

(d) (a) (d) (a) (d) (b) (c) (c) (d) (c) (d) (c) (a) (b) (a) (c) Topic-2 : Water and Soil Pollution

1. (d) 2. (a) 3. (b) 4. (d) 5. (c) 6. (c) 7. (a) 8. (b) 9. (c) 10. (a) 11. (c) 12. (a) 13. (c)

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14. (a) 15. (c) 16. (d)

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1. (a) (b) (c) (d) 2. (a) (b) (c) (d) 3. (a) (b) (c) (d)

‘X’ melts at low temperature and is a bad conductor of electricity in both liquid and solid state. X is: [Main Jan. 09, 2020 (I)] Zinc sulphide Mercury Silicon carbide Carbon tetrachloride The ratio of number of atoms present in a simple cubic, body centered cubic and face centered cubic structure are, respectively : [Main April 12, 2019 (II)] 8:1:6 1:2:4 4:2:1 4:2:3 Which primitive unit cell has unequal edge lengths (a b c) and all axial angles different from 90°? [Main Jan. 10, 2019 (I)] Triclinic Hexagonal Monoclinic Tetragonal

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4.

All of the following share the same crystal structure except. [Main Online April 15, 2018 (II)] (a) RbCl (b) NaCl (c) CsCl (d) Li Cl 5. In a monoclinic unit cell, the relation of sides and angles are respectively: [Main Online April 12, 2014] (a) a = b ≠ c and α = β = γ = 90° (b) a ≠ b ≠ c and α = β = γ = 90° (c) a ≠ b ≠ c and β = γ = 90° ≠ α (d) a ≠ b ≠ c and α ≠ β ≠ γ ≠ 90° 6. Which of the following exists as covalent crystals in the solid state? [Main 2013] (a) Iodine (b) Silicon (c) Sulphur (d) Phosphorus 7. Copper crystallises in fcc with a unit length of 361pm. What is the radius of copper atom? [Main Online April 25, 2013] (a) 157 pm (b) 128 pm (c) 108 pm (d) 181 pm 8. In which of the following crystals alternate tetrahedral voids are occupied? [2005S] (a) NaCl (b) ZnS (c) CaF2 (d) Na2O

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9.

An element with molar mass 2.7 × 10–2 kg mol–1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7 × 103 kg m–3, the radius of the element is approximately ___________ × 10–12 m (to the nearest integer). [Main Sep. 03, 2020 (I)]

10.

Match the crystal system/unit cells mentioned in Column I with their characteristic features mentioned in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS. [2007] Column I Column II (A) Simple cubic and (p) have these parameters, face-centered cubic a = b = c and α = β = γ parameters (B) cubic and (q) are two crystal systems rhombohedral (C) cubic and tetragonal (r) have only two crystallographic angles of 90° (D) hexagonal and (s) belong to same crystal monoclinic system

1.

A crystal is made up of metal ions ‘M1’ and ‘M2’ and oxide ions. Oxide ions. form a ccp lattice structure. The cation ‘M1’ occpies 50% of octahedral voids and the cation ‘M2’ occupies 12.5% of tetrahedral voids of oxide lattice. The oxidation numbers of ‘M1’ and ‘M2’ are, respectively : [Main Sep. 06, 2020 (II)] +2, +4 +1, +3 +3, +1 +4, +2 A diatomic molecule X2 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

(a) (b) (c) (d) 2.

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number of molecules present in 200 g of X2 is :

[Main Sep. 05, 2020 (I)] (Avogadroconstant (NA) = 6 × 10 mol ) (a) 40 NA (b) 8 NA (c) 4 NA (d) 2 NA 3. An element crystallises in a face-centred cubic (fcc) unit cell with cell edge a. The distance between the centres of two nearest octahedral voids in the crystal lattice is : [Main Sep. 05, 2020 (II)] 23

–1

(a) (b) a (c) (d) 4.

Which of the following compounds is likely to show both Frenkel and Schottky defects in its crystalline form? [Main Jan. 08, 2020 (II)] (a) AgBr (b) CsCl (c) KBr (d) ZnS 5. A solid having density of 9 × 103 kg m–3 forms face centred cubic pm. What is the molar mass of the crystals of edge length solid? [Main Jan. 11, 2019 (I)] 23 –1 [Avogadro constant 6 × 10 mol ,  3] (a) 0.0432 kg mol–1 (b) 0.0216 kg mol–1

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(c) 0.0305 kg mol–1 (d) 0.4320 kg mol–1 6. 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: [Main Jan. 10, 2019 (II)] (a) hcp lattice – A,

Tetrahedral voids – B

(b) hcp lattice – A,

Tetrahedral voids – B

(c) hcp lattice – B,

Tetrahedral voids – A

(d) hcp lattice – B,

Tetrahedral voids – A

7.

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] [Main Jan. 9, 2019 (II)] (a) (b) (c) (d) 8.

Which type of ‘defect’ has the presence of cations in the interstitial sites? [2018] (a) Schottky defect (b) Vacancy defect (c) Frenkel defect (d) Metal deficiency defect

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9.

Which of the following arrangements shows the schematic alignment of magnetic moments of antiferromagnetic substance? [Main Online April 15, 2018 (I)]

(a) (b) (c) (d) 10.

A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is 'a', the closest approach between two atoms in metallic crystal will be: [Main 2017] (a) 2a (b)

(c) (d) 11. (a) (b) (c) (d) 12. (a) (b) (c) (d) 13. (a)

Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 4.29Å. The radius of sodium atom is approximately : [Main 2015] 5.72Å 0.93Å 1.86Å 3.22Å The correct statement for the molecule, CsI3 is [Main 2014]: It is a covalent molecule. It contains Cs+ and I3–ions. It contains Cs3+ and I– ions. It contains Cs+, I– and lattice I2 molecule. The appearance of colour in solid alkali meta halides is generally due to: [Main Online April 11, 2014] Schottky defect

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(b) Frenkel defect (c) Interstitial position (d) F-centres 14. In a face centered cubic lattice atoms A are at the corner points and atoms B at the face centered points. If atom B is missing from one of the face centered points, the formula of the ionic compound is: [Main Online April 9, 2014] (a) AB2 (b) A5B2 (c) A2B3 (d) A2B5 15. An element having an atomic radius of 0.14 nm crystallizes in an fcc unit cell. What is the length of a side of the cell ? [Main Online April 9, 2013] (a) 0.56 nm (b) 0.24 nm (c) 0.96 nm (d) 0.4 nm 16. Which one of the following statements about packing in solids is incorrect ? [Main Online April 22, 2013] (a) Coordination number in bcc mode of packing is 8. (b) Coordination number in hcp mode of packing is 12. (c) Void space in hcp mode of packing is 32%. (d) Void space is ccp mode of packing is 26%. 17. In a face centred cubic lattice, atoms of A form the corner points and atoms of B form the face centred points. If two atoms of A are missing from the corner points, the formula of the ionic compound is : [Main Online April 23, 2013] (a) AB3 (b) AB4 (c) A2B5 (d) AB2 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

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A+ is [Adv. 2013]

(a) (c) (b) (d) 19.

104 pm 183 pm 125 pm 57 pm A compound MpXq has cubic close packing (ccp) arrangement of X. Its unit cell structure is shown below. The empirical formula of the compound is [2012 - I]

(a) (b) (c) (d) 20.

MX MX2 M 2X M5X14 The packing efficiency of the two-dimensional square unit cell shown below is : [2010]

(a) (b) (c) (d)

39.27% 68.02% 74.05% 78.54%

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21.

(a) (b) (c) (d) 22.

(a) (b) (c) (d) 23. (a) (b) (c) (d) 24. (a) (b) (c) (d) 25.

A substance AxBy crystallizes in a face centred cubic (FCC) 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 AxBy [2002S] AB3 A4B3 A3B Compostion cannot be specified In a solid ‘AB’ having the NaCl structure, ‘A’ atoms occupy the corners of the cubic unit cell. If all the face-centered atoms along one of the axes are removed, then the resultant stoichiometry of the solid is [2001S] AB2 A2B A4B3 A3B4 The coordination number of a metal crystallizing in a hexagonal close-packed structure is [1999 - 2 Marks] 12 4 8 6 CsBr has bcc structure with edge length 4.3. The shortest inter ionic distance between Cs+ and Br– is [1995S] 3.72 1.86 7.44 4.3 Consider an ionic solid MX with NaCl structure. Construct a new structure (Z) whose unit cell is constructed from the unit cell of MX

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following the sequential instructions given below. Neglect the charge balance. [Adv. 2018] (i) Remove all the anions (X) except the central one (ii) Replace all the face centered cations (M) by anions (X) (iii) Remove all the corner cations (M) (iv) Replace the central anion (X) with cation (M) The value of

in Z is____.

26.

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 [Adv. 2017] 27. The number of hexagonal faces that are present in a truncated octahedron is [2011] 28.

A metal crystallises into two cubic phases, face centered 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 - 3 Marks] 29. Chromium metal crystallizes with a body centred cubic lattice. The length of the unit cell edge is found to be 287 pm. Calculate the atomic radius. What would be the density of chromium in g/cm3? [1997 - 3 Marks] 30. Sodium metal crystallizes in body centred cubic lattice with the cell edge, a = 4.29Å. What is the radius of sodium atom? [1994 - 2 Marks]

31.

The cubic unit cell structure of a compound containing cation M and anion X is shown below. When compared to the anion, the cation has smaller ionic radius. Choose the correct statement(s). [Adv. 2020]

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(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 to 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. 32. The CORRECT statement(s) for cubic close packed (ccp) three dimensional structure is (are) [Adv. 2016] (a) The number of the nearest neighbours of an atom present in the topmost layer is 12 (b) The efficiency of atom packing is 74% (c) The number of octahedral and tetrahedral voids per atom are 1 and 2, respectively times the radius of the atom (d) The unit cell edge length is 33. 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 [Adv. 2015] (a) (b) (c)

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(d) 34. (a) (b) (c) (d) 35. (a) (b) (c) (d)

The correct statement(s) regarding defects in solids is (are) [2009S] Frenkel defect is usually favoured by a very small difference in the sizes of cation and anion Frenkel defect is a dislocation defect Trapping of an electron in the lattice leads to the formation of F-centre Schottky defects have no effect on the physical properties of solids Which of the following statement(s) is (are) correct? [1998 - 2 Marks] The coordination number of each type of ion in CsCl crystal is 8. A metal that crystallizes in bcc structure has a coordination number of 12. A unit cell of an ionic crystal shares some of its ions with other unit cells. The length of the unit cell in NaCl is 552 pm. (rNa+ = 95 pm; rCl–=181 pm).

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 third layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be 'r'. Answer the following questions 36. The number of atoms in the HCP unit cell is [2008] (a) 4

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(b) 6 (c) 12 (d) 17 37. The volume of this hcp unit cell is – [2008] (a) (b) (c) (d) 38.

The empty space in this hcp unit cell is [2008]

(a) (b) (c) (d)

74% 47.6% 32% 26%

Read the following statement (Assertion) and explanation (Reason) and answer each question as per the options given below : (a) If both assertion and reason are correct, and reason is the correct explanation of the assertion. (b) If both assertion and reason are correct, but reason is not the correct explanation of the assertion. (c) If assertion is correct but reason is incorrect. (d) If assertion is incorrect but reason is correct. 39. Assertion : In any ionic solid [MX] with Schottky defects, the number of positive and negative ions are same. Reason : Equal number of cation and anion vacancies are present. [2001S] 40.

The edge length of unit cell of a metal having molecular weight 75 g/mol is 5Å which crystallizes in cubic lattice. If the density is 2 g/cc then find the radius of metal atom(NA = 6 × 1023). Give the answer in pm

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41.

42.

(a) (b) 43.

44.

[2006 - 6M] In face centred cubic (fcc) crystal lattice, edge length is 400 pm. Find the diameter of greatest sphere which can be fit into the interstitial void without distortion of lattice. [2005 - 2 Marks] A compound AB has rock salt type structure. The formula weight of AB is 6.023 Y amu, and the closest A – B distance is Y1/3 nm, where Y is an arbitrary number. [2004 - 2 Marks] Find the density of lattice If the density of lattice is found to be 20 kg m–3, then predict the type of defect. You are given marbles of diameter 10 mm. They are to be placed such that their centres are lying in a square bound by four lines each of length 40 mm. What will be the arrangements of marbles in a plane so that maximum number of marbles can be placed inside the area? Sketch the diagram and derive expression for the number of molecules per unit area. [2003 - 2 Marks] The figures given below show the location of atoms in three crystallographic planes in an FCC lattice. Draw the unit cell for the corresponding structure and identify these planes in your diagram. [2000 - 3 Marks] 45.A

46.

metallic element crystallises into a lattice containing a sequence of layers of ABABAB..... Any packing of spheres leaves out voids in the lattice. What percentage by volume of this lattice is empty space? [1996 - 3 Marks] The density of mercury is 13.6 g/mL. Calculate approximately the diameter of an atom of mercury assuming that each atom is occupying a cube of edge length equal to the diameter of the mercury atom. [1983 - 3 Marks]

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Topic-1 : Periodic Classification 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

(d) (b) (a) (d) (c) (b) (b) (b) (143) A-p, s; B-p,q; C-q; D-q, r. Topic-2 : Periodic Properties

1. (a) 2. (c) 3. (a) 4. (a) 5. (c) 6. (d) 7. (d) 8. (c) 9. (d) 10. (d) 11. (c) 12. (b) 13. (d) 14. (d) 15. (d) 16. (c)

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17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39.

(b) (a) (b) (d) (a) (d) (a) (a) (3) (2) (8) (1.259) (124.27) (1.86) (a,c) (b, c, d) (a) (b, c) (a,c,d) (b) (a) (d) (a)

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1.

(a) (b) (c) (d) 2.

A set of solutions is prepared using 180g of water as a solvent and 10g of different non-volatile solutes A, B and C. The relative lowering of vapour pressure in the presence of these solutes are in the order [Given, molar mass of A = 100g mol–1; B = 200g mol–1; C = 10,000g mol–1] [Main Sep. 06, 2020 (II)] B>C>A C>B>A A>B>C A>C>B Henry’s constant (in kbar) for four gases α, β, γ and δ in water at 298 K is given below : [Main Sep. 03, 2020 (I)]

(density of water = 103 kg m–3 at 298 K) This table implies that : (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

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(d) The pressure of a 55.5 molal solution of δ is 250 bar 3. An open beaker of water in equilibrium with water vapour is in a sealed container. When a few grams of glucose are added to the beaker of water, the rate at which water molecules : [Main Sep. 02, 2020 (I)] (a) leaves the vapour increases (b) leaves the solution increases (c) leaves the solution decreases (d) leaves the vapour decreases 4. A graph of vapour pressure and temperature for three different liquids X, Y, and Z is shown below:

The following inferences are made: [Main Jan. 08, 2020 (I)] (A) X has higher intermolecular interactions compared to Y. (B) X has lower intermolecular interactions compared to Y. (C) Z has lower intermolecular interactions compared to Y. The correct inference(s) is/are: (a) (A) and (C) (b) (A) (c) (B) (d) (C) 5. At 35 °C, the vapour pressure of CS2 is 512 mm Hg and that of acetone is 344 mm Hg. A solution of CS2 in acetone has a total vapour pressure of 600 mm Hg. The false statement amongst the following is: [Main Jan. 07, 2020 (I)] (a) Raoult’s law is not obeyed by this system (b) a mixture of 100 mL CS2 and 100 mL acetone has a volume < 200 mL (c) CS2 and acetone are less attracted to each other than to themselves

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(d) heat must be absorbed in order to produce the solution at 35 °C 6. Two open beakers one containing a solvent and the other containing a mixture of that solvent with a non volatile solute are together sealed in a container. Over time: [Main Jan. 07, 2020 (II)] (a) the volume of the solution increases and the volume of the solvent decreases (b) the volume of the solution decreases and the volume of the solvent increases (c) the volume of the solution and the solvent does not change (d) the volume of the solution does not change and the volume of the solvent decreases 7. 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 : [Main April 12, 2019 (I)] –2 (a) 13.88×10 (b) 13.88×10–1 (c) 13.88 (d) 13.88×10–3 8. What would be the molality of 20% (mass/mass) aqueous solution of Kl? (molar mass of Kl = 166 g mol–1) [Main April 9, 2019 (II)] (a) 1.08 (b) 1.35 (c) 1.48 (d) 1.51 9. The vapour pressures of pure liquids A and B are 400 and 600 mmHg, respectively at 298K. 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: [Main April 8, 2019 (I)] (a) 450 mmHg, 0.4, 0.6 (b) 500 mmHg, 0.5, 0.5 (c) 450 mmHg, 0.5, 0.5

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(d) 500 mmHg, 0.4, 06 10. The solubility of N2 in water at 300 K and 500 torr partial pressure is 0.01 g L–1.The solubility (in g L–1)at 750 torr partial pressure is : [Main Online April 9, 2016] (a) 0.0075 (b) 0.005 (c) 0.02 (d) 0.015 11. A solution at 20 °C is composed of 1.5 mol of benzene and 3.5 mol of toluene. If the vapour pressure of pure benzene and pure toluene at this temperature are 74.7 torr and 22.3 torr, respectively, then the total vapour pressure of the solution and the benzene mole fraction in equilibrium with it will be, respectively : [Main Online April 10, 2015] (a) 35.8 torr and 0.280 (b) 38.0 torr and 0.589 (c) 30.5 torr and 0.389 (d) 30.5 torr and 0.480 12. For an ideal solution of two components A and B, which of the following is true? [Main Online April 19, 2014] (a) ∆Hmixing < 0 (zero) (b) ∆Hmixing > 0 (zero) (c) A – B interaction is stronger than A – A and B – B interactions (d) A– A, B – B and A – B interactions are identical. 13. The molarity of a solution obtained by mixing 750 mL of 0.5(M) HCl with 250 mL of 2(M) HCl will be : [Main 2013] (a) 0.875 M (b) 1.00 M (c) 1.75 M (d) 0.975 M 14. Vapour pressure of pure benzene is 119 torr and that of toluene is 37.0 torr at the same temperature. Mole fraction of toluene in vapour phase which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene 0.50, will be : [Main Online April 23, 2013] (a) 0.137

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(b) (c) (d) 15. (a) (b) (c) (d) 16.

(a) (b) (c) (d) 17. (a) (b) (c) (d) 18.

(b) (c) (d)

0.237 0.435 0.205 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] 1.78 M 2.00 M 2.05 M 2.22 M 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 at 298 K and 5 atm pressure is [2009] –4 4.0× 10 4.0 × 10–5 5.0 × 10– 4 4.0 × 10–6 For a dilute solution, Raoult’s law states that : [1985 - 1 Mark] the lowering of vapour pressure is equal to the mole fraction of solute. the relative lowering of vapour pressure is equal to the mole fraction of solute. the relative lowering of vapour pressure is proportional to the amount of solute in solution. the vapour pressure of the solution is equal to the mole fraction of solvent. An azeotropic solution of two liquids has boiling point lower than either of them when it [1981 - 1 Mark] (a) shows negative deviation from Raoult’s law shows no deviation from Raoult’s law shows positive deviation from Raoult’s law is saturated

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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 xA and xB, respectively, has vapour pressure of 22.5 Torr. The value of xA/xB in the new solution is________. (given that the vapour pressure of pure liquid A is 20 Torr at temperature T) [Adv. 2018] 20. The mole fraction of a solute in a solution is 0.1. At 298 K, molarity of this solution is the same as its molality. Density of this solution at 298 K is 2.0 g cm–3 . The ratio of the molecular weights of the solute and solvent,

, is

[Adv. 2016] 21. A compound H2X with molar weight of 80g 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 [Adv. 2014] 22. 29.2% (w/w) HCl stock solution has a density of 1.25 g 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 of 0.4 M HCl is : [2012] 23.

At 300 K, the vapour pressure of a solution containing 1 mole of nhexane and 3 moles of n-heptane is 550 mm of Hg. At the same temperature, if one more mole of n-heptane is added to this solution, the vapour pressure of the solution increases by 10 mm of Hg. What is the vapour pressure in mm Hg of n-heptane in its pure state ________? [Main Sep. 04, 2020 (I)] 24. 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

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25.

26.

27.

28.

29.

have the total vapor pressures of 0.3 and 0.4 bar, respectively. What is the vapor pressure of pure liquid B in bar? [Adv. 2020] The mole fraction of urea in an aqueous urea solution containing 900 g of water is 0.05. If the density of the solution is 1.2 g cm–3, the molarity of urea solution is ___ (Given data: Molar masses of urea and water are 60 g mol–1 and 18 g mol–1, respectively) [Adv. 2019] –1 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 46.0 Torr. Find the mole fraction of benzene in the vapour above the solution. [1996 - 3 Marks] The vapour pressure of pure benzene is 639.7 mm of mercury and the vapour of a solution of a solute in benzene at the same temperature is 631.9 mm of mercury. Calculate the molality of the solution. [1981 - 3 Marks] 0.5 g of fuming H2SO4 (Oleum) is diluted with water. This solution is completely neutralized by 26.7 mL of 0.4 N NaOH. Find the percentage of free SO3 in the sample of oleum. [1980] For a solution formed by mixing liquids L and M, the vapour pressure of L plotted against the mole fraction of M in solution is shown in the following figure. Here xL and xM represent mole fractions of L and M, respectively, in the solution. The correct statement(s) applicable to this system is (are) [Adv. 2017]

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(a) (b) (c) (d) 30. (a) (b) (c) (d) 31. (a) (b) (c) (d)

The point Z represents vapour pressure of pure liquid M and Raoult's law is obeyed from xL= 0 to xL= 1 The point Z represents vapour pressure of pure liquid L and Raoult's law is obeyed when xL → 1 The point Z represents vapour pressure of pure liquid M and Raoult's law is obeyed when xL → 0 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. Mixture(s) showing positive deviation from Raoult’s law at 35 °C is (are) [Adv. 2016] carbon tetrachloride + methanol carbon disulphide + acetone benzene + toluene phenol + aniline Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statement(s) is(are) [Adv. 2013] ∆G is positive ∆Ssystem is positive ∆Ssurroundings = 0 ∆H = 0

32.

The vapour pressure of a dilute aqueous solution of glucose (C6H12O6) is 750 mm of mercury at 373 K. Calculate (i) molality and (ii) mole fraction of the solution. [1989 - 3 Marks] 33. The vapour pressure of ethanol and methanol are 44.5 mm and 88.7 Hg respectively. An ideal solution is formed at the same temperature by 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 - 4 Marks]

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34.

'Two volatile and miscible liquids can be separated by fractional distillation into pure component', is true under what conditions? [1984 - 1 Mark] 35. An organic compound CxH2yOy was burnt with twice the amount of oxygen needed for complete combustion to CO2 and H2O. The hot gases when cooled to 0ºC and 1 atm. pressure, measured 2.24 liters. The water collected during cooling weighed 0.9 g. The vapour pressure of pure water at 20ºC is 17.5 mm Hg and is lowered by 0.104 mm when 50 g of the organic compound is dissolved in 1000 g of water. Give the molecular formula of the organic compound. [1983 - 5 Marks] 36. A bottle of commercial sulphuric acid (density 1.787 g/mL) is labelled as 86 percent by weight. What is the molarity of the acid? What volume of the acid has to be used to make 1 litre of 0.2 M H2SO4? [1979] 37. 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 acid be diluted in order to prepare a 1.5 N solution? [1978]

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 oC. The osmotic pressure of the solution is : [Main April 12, 2019 (II)] (R = 0.08206 L atm K–1 mol–1) (a) 8.2 atm (b) 2.46 atm (c) 4.92 atm (d) 1.64 atm 2. 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

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pure water at this temperature is 35 mmHg, lowering of vapour pressure will be : [Main April 10, 2019 (I)] –1 (molar mass of urea = 60 g mol ) (a) 0.027 mmHg (b) 0.028 mmHg (c) 0.017 mmHg (d) 0.031 mmHg 3. Molal depression constant for a solvent is 4.0 K kg mol–1. The depression in the freezing point of the solvent for 0.03 mol kg–1 solution K2SO4 is: [Main April 9, 2019 (II)] (Assume complete dissociation of the electrolyte) (a) 0.18 K (b) 0.24 K (c) 0.12 K (d) 0.36 K 4. K2HgI4 is 40% ionised in aqueous solution. The value of its van’t Hoff factor (i) is : [Main Jan. 11, 2019 (II)] (a) 1.6 (b) 1.8 (c) 2.0 (d) 2.2 5. 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 Kb and Kf is: [Main Jan. 10, 2019 (II)] (a) Kb = 1.5 Kf (b) Kb = Kf (c) Kb = 0.5 Kf (d) Kb = 2 Kf 6. For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point? [Main 2018] (a) [Co(H2O)6]Cl3

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(b) [Co(H2O)5Cl]Cl2.H2O (c) [Co(H2O)4Cl2]Cl. 2H2O (d) [Co(H2O)3Cl3].3H2O 7. The freezing point of benzene decreases by 0.45°C when 0.2g 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 : [Main 2017] –1 (Kf for benzene = 5.12 K kg mol ) (a) 64.6% (b) 80.4% (c) 74.6% (d) 94.6% 8. 5 g of Na2SO4 was dissolved in x g of H2O. The change in freezing point was found to be 3.82 °C. If Na2SO4 is 81.5% ionised, the value of x [Main Online April 8, 2017] –1 (Kf for water = 1.86°C kg mol ) is approximately : (molar mass of S = 32 g mol–1 and that of Na = 23 g mol–1) (a) 15 g (b) 25 g (c) 45 g (d) 65 g 9. Pure water freezes at 273 K and 1 bar. The addition of34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 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] Among the following, the option representing change in the freezing point is [Adv. 2017]

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(a)

(b)

(c)

(d)

10.

(a) (b) (c) (d)

18 g glucose (C6H12O6) is added to 178.2 g water. The vapour pressure of water (in torr) for this aqueous solution is: [Main 2016, 2006] 752.4 759.0 7.6 76.0

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11.

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 (g mol–1) of the substance is : [Main 2015] (a) 128 (b) 488 (c) 32 (d) 64 12. Determination of the molar mass of acetic acid in benzene using freezing point depression is affected by : [Main Online April 11, 2015] (a) partial ionization (b) dissociation (c) complex formation (d) association 13. Consider separate solutions of 0.500 M C2H5OH(aq), 0.100 M Mg3 (PO4)2 (aq), 0.250 M KBr(aq) and 0.125 M Which statement is true about these solutions, Na3PO4(aq) at assuming all salts to be strong electrolytes? [Main 2014] (a) They all have the same osmotic pressure. (b) 0.100 M Mg3(PO4)2(aq) has the highest osmotic pressure. (c) 0.125 M Na3PO4(aq) has the highest osmotic pressure. (d) 0.500 M C2H5OH(aq) has the highest osmotic pressure. 14. The observed osmotic pressure for a 0.10 M solution of Fe(NH4)2(SO4)2 at 25 °C is 10.8 atm. The expected and experimental (observed) values of van’t Hoff factor (i) will be respectively: (R = 0.082 L atm K–1 mol–1) [Main Online April 19, 2014] (a) 5 and 4.42 (b) 4 and 4.00 (c) 5 and 3.42 (d) 3 and 5.42 15. 12 g of a nonvolatile solute dissolved in 108 g of water produces the relative lowering of vapour pressure of 0.1. The molecular mass of

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the solute is : [Main Online April 9, 2013] (a) (b) (c) (d) 16.

(a) (b) (c) (d) 17.

(a) (b) (c) (d) 18.

(a) (b) (c) (d) 19.

(a)

80 60 20 40 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 Kb = 0.76 K kg mol–1) [2012] 724 740 736 718 The freezing point (in °C) of a solution containing 0.1 gof K3[Fe(CN)6] (Mol. wt. 329) in 100 g of water (Kf = 1.86 Kkg mol–1) is [2011] –2 –2.3 × 10 –5.7 × 10–2 –5.7 × 10–3 –1.2 × 10–2 When 20 g of naphthoic acid (C11H8O2) is dissolved in 50 g of benzene (Kf = 1.72 K kg mol–1), a freezing point depression of 2K is observed. The van't Hoff factor (i) is [2007] 0.5 1 2 3 The elevation in boiling point of a solution of 13.44 g of CuCl2 in 1 kg of water using the following information will be (Molecular weight of CuCl2 = 134.4 and Kb = 0.52 K molal−1) [2005S] 0.16

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(d) 20. (a) (b) (c) (d) 21. (a) (b) (c) (d) 22.

(a) (b) (c) (d) 23. (a) (b) (c) (d) 24. (a) (b)

(b) 0.05 (c) 0.1 0.2 During depression of freezing point in a solution, the following are in equililbrium [2003S] liquid solvent, solid solvent liquid solvent, solid solute liquid solute, solid solute liquid solute, solid solvent The molecular weight of benzoic acid in benzene as determined by depression in freezing point method corresponds to: [1996 - 1 Mark] ionization of benzoic acid. dimerization of benzoic acid. trimerization of benzoic acid. solvation of benzoic acid. 0.2 molal acid HX is 20% ionised in solution. Kf = 1.86 K molality–1. The freezing point of the solution is : [1995S] – 0.45 – 0.90 – 0.31 – 0.53 The freezing point of equimolal aqueous solutions will be highest for : [1990 - 1 Mark] C6H5NH3Cl (aniline hydrochloride) Ca(NO3)2 La(NO3)3 C6H12O6 (glucose) Which of the following 0.1 M aqueous solutions will have the lowest freezing point? [1989 - 1 Mark] Potassium sulphate Sodium chloride

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(c) Urea (d) Glucose 25. When mercuric iodide is added to the aqueous solution of potassium iodide then [1987 - 1 Mark] (a) freezing point is raised. (b) freezing point is lowered. (c) freezing point does not change. (d) boiling point does not change. 26.

The elevation of boiling point of 0.10 m aqueous CrCl3.xNH3 solution is two times that of 0.05 m aqueous CaCl2 solution. The value of x is ______. [Assume 100% ionisation of the complex and CaCl2, coordination number of Cr as 6, and that all NH3 molecules are present inside the coordination sphere] [Main Sep. 06, 2020 (I)] 27. 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 [Kf of water = 1.86 K kg mol–1] [Adv. 2015] 2+ – 28. MX2 dissociates into M 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 [Adv. 2014] 29.

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) [Main Sep. 04, 2020 (II)]

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30.

If 250 cm3 of an aqueous solution containing 0.73 g of a protein A is isotonic with one litre of another aqueous solution containing 1.65 g of a protein B, at 298 K, the ratio of the molecular masses of A and B is __________ × 10–2 (to the nearest integer). [Main Sep. 03, 2020 (II)] 31. How much amount of NaCl should be added to 600 g of water (r = 1.00 g/mL) to decrease the freezing point of water to –0.2°C? _______. (The freezing point depression constant for water = 2 K kg mol–1) [Main Jan. 09, 2020 (I)] 32. A cylinder containing an ideal gas (0.1 mol of 1.0 dm3) is in thermal equilibrium with a large volume of 0.5 molal aqueous solution of ethylene glycol at its freezing point. If the stoppers S1 and S2 (as shown in the figure) are suddenly withdrawn, the volume of the gas in litres after equilibrium is achieved will be _______. (Given, Kf (water) = 2.0 K kg mol–1, R = 0.08 dm3 atm K–1 mol–1) [Main Jan. 09, 2020 (II)]

33. On dissolving 0.5 g of a non-volatile non-ionic solute to 39 g of benzene, its vapour pressure decreases from 650 mm Hg to 640 mm Hg. 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 and5.12 K kg mol-1, respectively) [Adv. 2019] 34. 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 solutions of NaCl in these solvents. NaCl completely dissociates in both the solvents. [Adv. 2018]

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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 dimerization in these solvents. If the degree of dimerization is 0.7 in solvent Y, the degree of dimerization in solvent X is______. 35. 75.2 g of C6H5OH (phenol) is dissolved in a solvent ofKf = 14. If the depression in freezing point is 7 K then find the % of phenol that dimerises. [2006 - 6M] 3 –3 36. To 500 cm of water, 3.0 × 10 kg of acetic acid is added. If 23% of acetic acid is dissociated, what will be the depression in freezing point? Kf and density of water are 1.86 K kg–1 mol–1 and 0.997 g cm–3, respectively. [2000 - 3 Marks] 37. 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 Kf for water is 1.86 K kg mol–1. Calculate the vapour pressure of this solution at 298 K. [1998 - 4 Marks] 38. Addition of 0.643 g of a compound to 50 mL. of benzene (density : 0.879 g/mL.) lowers the freezing point from 5.51ºC to 5.03ºC. If Kf for benzene is 5.12 K kg mol–1, calculate the molecular weight of the compound. [1992 - 2 Marks] 39. The degree of dissociation of calcium nitrate in a dilute aqueous solution, containing 7.0 g. of the salt per 100 g of water at 100ºC is 70%. If the

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40.

41.

vapour pressure of water at 100ºC is 760 mm, calculate the vapour pressure of the solution. [1991 - 4 Marks] 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 the solution is 600 mm Hg. What is the molecular weight of the solid substance? [1990 - 3 Marks] Given that

is the depression in freezing point of the solvent in a

solution of a non-volatile solute of molality, m, the quantity is equal to ........... . [1994 - 1 Mark] 42. (a) (b) (c) (d)

In the depression of freezing point experiment, it is found that the [1999 - 3 Marks] vapour pressure of the solution is less than that of pure solvent vapour pressure of the solution is more than that of pure solvent only solute molecules solidify at the freezing point only solvent molecules solidify at the freezing

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. Application of colligative properties are very useful in day-to-day life. One of its example 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 (Kf water) = 1.86 K kg mol–1 Freezing point depression constant of ethanol (Kf ethanol) = 2.0 K kg mol–1 Boiling point elevation constant of water (Kbwater)

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= 0.52 K kg mol–1 Boiling point elevation constant of ethanol (Kbethanol) = 1.2 K kg mol–1 Standard freezing point of water = 273 K 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 solution to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. 43. The freezing point of the solution M is [2008 - 3 Marks] (a) 268.7 K (b) 268.5 K (c) 234.2 K (d) 150.9 K 44. The vapour pressure of the solution M is [2008 - 3 Marks] (a) 39.3 mm Hg (b) 36.0 mm Hg (c) 29.5 mm Hg (d) 28.8 mm Hg 45. 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 [2008 - 3 Marks] (a) 380.4 K (b) 376.2 K (c) 375.5 K (d) 354.7 K 46.

1.22g of benzoic acid is dissolved in 100 g of acetone and 100 g of benzene separately. Boiling point of the solution in acetone increases by 0.17º C, while that in the benzene increases by 0.13º C; Kb for

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acetone and benzene is1.7 K kg mol–1 and 2.6 K kg mol–1 respectively. Find molecular weight of benzoic acid in two cases and justify your answer. [2004 - 2 Marks] 47. Match the following if the molecular weights of X, Y and Z are same. [2003 - 2 Marks] Boiling Point Kb X 100 0.63 Y 27 0.53 Z 253 0.98 48. Nitrobenzene is formed as the major product along with a minor product in the reaction of benzene with a hotmixture of nitric acid and sulphuric acid. The minorproduct 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 and determine its molecular and structural formula. (Molal boiling point elevation constant of benzene is2.53 K kg mol–1.) [1999 - 10 Marks] 49. What weight of the non-volatile solute, urea (NH2 – CO – NH2) needs to be dissolved in 100g of water, in order to decrease the vapour pressure of water by 25%? What will be the molality of the solution? [1993 - 3 Marks] Topic-1 : Solution and Vapour Pressure of Liquid Solutions 1. (c) 2. (d) 3. (a) 4. (c) 5. (b) 6. (a) 7. (c)

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8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31.

(d) (d) (d) (b) (d) (a) (b) (c) (a) (b) (c) (19) (9) (8) (8) (600) (0.02) (2.98) (0.735) (0.156) (3.84) (b, d) (a, b) (b, c, d) Topic-2 : Colligative Properties of Solutions

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

(c) (c) (d) (b) (d) (d) (d) (c) (c) (a) (d)

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12. (d) 13. (a) 14. (a) 15. (c) 16. (a) 17. (a) 18. (a) 19. (a) 20. (a) 21. (b) 22. (a) 23. (d) 24. (a) 25. (a) 26. (5.0) 27. (1) 28. (2) 29. (167) 30. (177) 31. (1.75) 32. (2.18) 33. (1.03) 34. (0.05) 35. (75) 36. (0.228) 37. (23.44) 38. (156.056) 39. (746.3) 40. (65.25) 42. (a, d) 43. (d) 44. (b) 45. (b)

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1.

The variation of molar conductivity with concentration of an electrolyte (X) in aqueous solution is shown in the given figure. [Main Sep. 05, 2020 (II)]

The electrolyte X is : (a) HCl (b) NaCl (c) KNO3 (d) CH3COOH 2. Let CNaCl and

be the conductances (in S) measured for

saturated aqueous solutions of NaCl and BaSO4, respectively, at a temperature T. Which of the following is false? [Sep. 03, 2020 (I)] (a) Ionic mobilities of ions from both salts increase with T

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(b) (c) (d) 3.

for T2 > T1 for T2 7> T1 at a given T The equation that is incorrect is: [Main Jan. 07, 2020 (II)]

4.

The decreasing order of electrical conductivity of the following aqueous solutions is: [Main April 12, 2019 (II)] 0.1 M Formic acid (A), 0.1 M Acetic acid (B), 0.1 M Benzoic acid (C). (a) A > C > B (b) C > B > A (c) A > B > C (d) C > A > B 5. Consider the statements S1 and S2 : S1 : Conductivity always increases with decrease in the concentration of electrolyte. S2 : Molar conductivity always increases with decrease in the concentration of electrolyte. The correct option among the following is : [Main April 10, 2019 (I)] (a) Both S1 and S2 are wrong (b) S1 is wrong and S2 is correct (c) Both S1 and S2 are correct (d) S1 is correct and S2 is wrong 6. A solution of Ni (NO3)2 is electrolysed between platinum electrodes using 0.1 Faraday electricity. How many mole of Ni will be deposited

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at the cathode? [Main April 9, 2019 (II)] (a) (b) (c) (d) 7.

0.05 0.20 0.15 0.10 The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount of PbSO4 electrolyzed in g during the process is : (Molar mass of PbSO4 = 303 g mol–1) [Main Jan. 9, 2019 (I)] (a) 22.8 (b) 15.2 (c) 7.6 (d) 11.4 8. Molar conductivity of aqueous solution of sodium stearate, which behaves as a strong electrolyte is recorded at varying concentration (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) [Adv. 2019]

(a)

(b)

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(c)

(d)

9.

(a) (b) (d) 10.

(a) (b) (c) (d) 11.

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 u) [Main 2018] 6.4 hours 0.8 hours (c) 3.2 hours 1.6 hours When an electric current is passes through acidified water, 112 mL of hydrogen gas at N.T.P. was collected at the cathode in 965 seconds. The current passed, in ampere, is : [Main Online April 15, 2018 (I)] 2.0 0.1 0.5 1.0 Given [Main 2017]

Among the following, the strongest reducing agent is

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(a) (b) (c) (d) 12.

(a) (b) (c) 13.

(a) (b) (c) (d) 14.

(a) (b) (c) (d) 15.

(a) (b) (c) (d)

Cr Mn2+ Cr3+ Cl– What will occur if a block of copper metal is dropped into a beaker containing a solution of 1M ZnSO4? [Main Online April 9, 2016] The copper metal will dissolve with evolution of oxygen gas The copper metal will dissolve with evolution of hydrogen gas No reaction will occur (d) The copper metal will dissolve and zinc metal will be deposited. 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 amu) [Main 2015] 2g 127 g 0g 63.5 g A variable, opposite external potential (Eext) is applied to the cell Zn|Zn2+ (1 M) || Cu2+ (1 M) | Cu, of potential 1.1 V. When Eext < 1.1 V and Eext > 1.1 V, respectively electrons flow from : [Main Online April 10, 2015] anode to cathode in both cases cathode to anode and anode to cathode anode to cathode and cathode to anode cathode to anode in both cases Resistance of 0.2 M solution of an electrolyte is 50 Ω. The specific conductance of the solution is 1.4 S m–1. The resistance of 0.5 M solution of the same electrolyte is 280 Ω. The molar conductivity of 0.5 M solution of the electrolyte in S m2 mol–1 is: [Main 2014] –4 5 × 10 5 × 10–3 5 × 103 5 × 102

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16.

The equivalent conductance of NaCl at concentration C and at infinite and respectively. The correct relationship dilution are and is given as: between (Where the constant B is positive) [Main 2014] (a) (b) (c) (d) 17.

The standard electrode potentials

of four metals A, B, C

and D are – 1.2 V, 0.6 V, 0.85 V and – 0.76 V, respectively. The sequence of deposition of metals on applying potential is: [Main Online April 9, 2014] (a) A, C, B, D (b) B, D, C, A (c) C, B, D, A (d) D, A, B, C 18. Given Fe3+ (aq) + e– → Fe2+ (aq); E° = + 0.77 V Al3+ (aq) + 3e– → Al(s); E° = – 1.66 V Br2(aq) + 2e– → 2Br–; E° = + 1.09 V Considering the electrode potentials, which of the following represents the correct order of reducing power? [Main Online April 11, 2014] 2+ – (a) Fe < Al < Br (b) Br– < Fe2+ < Al (c) Al < Br– < Fe2+ (d) Al < Fe2+ < Br– 19. How many electrons would be required to deposit 6.35 g of copper at the cathode during the electrolysis of an aqueous solution of copper sulphate? (Atomic mass of copper = 63.5 u, NA = Avogadro’s constant): [Main Online April 12, 2014]

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(a) (b) (c) (d) 20.

(a) (b) (c) (d) 21.

A solution of copper sulphate (CuSO4) is electrolysed for 10 minutes with a current of 1.5 amperes. The mass of copper deposited at the cathode (at. mass of Cu = 63u) is: [Main Online April 25, 2013] 0.3892 g 0.2938 g 0.2398 g 0.3928 g AgNO3(aq.) 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 [2011]

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(a) (b) (c) (d) 22.

(P) (Q) (R) (S) Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C mol–1) [2008S] 4 4 (a) 9.65 × 10 sec (b) 19.3 × 10 sec 4 (c) 28.95 × 10 sec (d) 38.6 × 104 sec 23. The correct order of equivalent conductance at infinite dilution of LiCl, NaCl and KCl is [2001S] (a) LiCl > NaCl > KCl (b) KCl > NaCl > LiCl (c) NaCl > KCl > LiCl (d) LiCl > KCl > NaCl 24. The electric charge for electrode deposition of one gram equivalent of a substance is : [1984 - 1 Mark] (a) one ampere per second. (b) 96,500 coloumbs per second. (c) one ampere for one hour. (d) charge on one mole of electrons. 25. Faraday’s laws of electrolysis are related to the [1983 - 1 Mark] (a) atomic number of the reactants. (b) atomic number of the anion.

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(c) equivalent weight of the electrolyte. (d) speed of the cation. 26. The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt electrodes. The distance between the electrodes is 120 cm with an area of cross section of 1 cm2. 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 cm2 mol–1 . The value of Z is [Adv. 2017] 27. 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 the difference in their pKa values, pKa(HX) HY (0.10 M). If – pKa(HY), is (consider degree of ionization of both acids to be 1.1 V, Zn dissolves at Zn electrode and Cu deposits at Cu electrode (c) If Eext < 1.1 V, Zn dissolves at anode and Cu deposits at cathode (d) If Eext = 1.1 V, no flow of e– or current occurs 3. Given that the standard potentials (E0) of Cu2 + /Cu and Cu+/Cu are 0.34 V and 0.522 V respectively, the E0 of Cu2 +/Cu + is: [Main Jan. 07, 2020 (I)]

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(a) (b) (c) (d) 4.

+ 0.182 V + 0.158 V – 0.182 V – 0.158 V The standard Gibbs energy for the given cell reaction in kJ mol–1 at 298 K is: [Main April 9, 2019 (I)] 2+ 2+ Zn(s) + Cu (aq) → Zn (aq) + Cu(s), E° = 2 V at 298 K (Faraday’s constant, F = 96000 C mol–1) (a) – 384 (b) 384 (c) 192 (d) – 192 5. Calculate the standard cell potential (in V) of the cell in which following reaction takes place : 2+ Fe (aq) + Ag+ (aq) → Fe3+ (aq) + Ag(s) Given that [Main April 8, 2019 (II)] =xV;

=yV;

=zV

(a) x – z (b) x – y (c) x + 2y – 3z (d) x + y – z 6. For the cell Zn(s) | Zn2+(aq) || Mx+ (aq) | M(s), different half cells and their standard electrode potentials are given below: Mx+(aq)/ Au3+(aq)/ Ag+(aq)/ Fe3+ (aq)/ Fe2+ (aq)/ M(s) Au(s) Ag(s) Fe2+(aq) Fe(s) 1.40 0.80 0.77 –0.44 If

= – 0.76V, which cathode will give a maximum

° value of Ecell per electron transferred?

[Main Jan. 11, 2019 (I)] (a) Ag+/Ag

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(b) Fe3+/Fe2+ (c) Au3+/Au (d) Fe2+/Fe 7. If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium constant (K) for the reaction at 300 K is approximately (R = 8JK–1 mol–1, F= 96000 C mol–1) [Main Jan. 9, 2019 (II)] (a) e (b) e–160 (c) e320 (d) e160 8. What is the standard reduction potential (E°) for Fe3+ → Fe ? Given that : [Main Online April 8, 2017] = – 0.47V Fe2+ + 2e– → Fe; –80

Fe3+ +e– → Fe2+;

= + 0.77V

(a) – 0.057 V (b) + 0.057 V (c) + 0.30 V (d) – 0.30 V 9. 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; E0 (cell) = 1.1 V] [Adv. 2017] (a) 1.1 F (b) 2.303RT – 2.2F (c) 2.303RT + 1.1F (d) – 2.2F 10. Galvanization is applying a coating of: [Main 2016]

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(a) (b) (c) (d) 11.

Cu Zn Pb Cr Identify the correct statement :

[Main Online April 10, 2016] (a) Corrosion of iron can be minimized by forming a contact with another metal with a higher reduction potential (b) Iron corrodes in oxygen free water (c) Corrosion of iron can be minimized by forming an impermeable barrier at its surface (d) Iron corrodes more rapidly in salt water because its electrochemical potential is higher 12. For the following electrochemical cell at 298 K, Pt(s) | H2(g, 1 bar) | H+ (aq, 1 M) || M 4+ (aq), M 2+ (aq) | Pt(s) .

Ecell = 0.092 V when

[Adv. 2016] Given :

2.303

The value of x is (a) –2 (b) –1 (c) 1 (d) 2 13. Given below are the half-cell reactions:

The E° for

will be: [Main 2014]

(a) (b) (c) (d)

–2.69 V; the reaction will not occur –2.69 V; the reaction will occur –0.33 V; the reaction will not occur –0.33 V; the reaction will occur

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14.

Given :

Based on the data given above, strongest oxidising agent will be : [Main 2013] (a) C l (b) Cr3+ (c) Mn2+ (d) MnO–4 15. Consider the following cell reaction: [2011] 2Fe(s) + O2(g) + 4H+(aq) → 2Fe2+(aq) + 2H2O(l); E° = 1.67V At [Fe2+] = 10–3 M, P(O2) = 0.1 atm and pH = 3, the cell potential at 25ºC is (a) 1.47 V (b) 1.77 V (c) 1.87 V (d) 1.57 V 16. The rusting of iron takes place as follows [2005S] + – H2O(l) ; E° = +1.23 V 2H + 2e + ½O2 2+ − Fe(s) ; E° = −0.44 V Fe + 2e Calculate ∆G° for the net process (a) −322 kJ mol−1 (b) −161 kJ mol−1 (c) −152 kJ mol−1 (d) −76 kJ mol−1 17. The emf of the cell [2004S] 2+ 2+ Zn | Zn (0.01 M) | | Fe (0.001 M) | Fe at 298 K is 0.2905 then the value of equilibrium constant for the cell reaction is (a) (b)

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(c) 18.

(d) In the electrolytic cell, flow of electrons is from [2003S]

(a) (b) (c) (d) 19.

Cathode to anode in solution Cathode to anode through external supply Cathode to anode through internal supply Anode to cathode through internal supply 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 : [2002S] E° = 1.51 V E° = 1.38 V E° = 0.77 V

Cl2(g) + 2e– → 2Cl –(aq.)

E° = 1.40 V

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Identify the only incorrect statement regarding the quantitative estimation of aqueous Fe(NO3)2 (a) MnO4–can be used in aqueous HCl (b) Cr2O2–7 can be used in aqueous HCl (c) MnO4–can be used in aqueous H2SO4 (d) Cr2O2–7 can be used in aqueous H2SO4 20. Saturated solution of KNO3 is used to make ‘salt−bridge’ because [2001S] + (a) velocity of K is greater than that of (b) velocity of is greater than that of K+ are nearly the same (c) velocities of both K+ and (d) KNO3 is highly soluble in water 21. For the electrochemical cell, M | M + || X – | X, E°(M + / M) = 0.44V and E°(X / X –) = 0.33V. From this data one can deduce that [2000S] – (a) M + X → M + + X is the spontaneous reaction (b) M + + X – → M + X is the spontaneous reaction (c) Ecell = 0.77V (d) Ecell = – 0.77 V 22. A 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 reduction potential of Z > Y > X, then, [1999 - 2 Marks] (a) Y will oxidize X and not Z (b) Y will oxidize Z and not X (c) Y will oxidize both X and Z (d) Y will reduce both X and Z 23. The standard reduction potentials of Cu2+ | Cu and Cu2+ | Cu+ are 0.337 V and 0.153 respectively. The standard electrode potential of Cu+ |Cu half cell is [1997 - 1 Mark] (a) 0.184 V (b) 0.827 V

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(c) 0.521 V (d) 0.490 V 24. A dilute aqueous solution of Na2SO4 is electrolyzed using platinum electrodes.The products at the anode and cathode are: [1996 - 1 Mark] (a) O2, H2 (b) (c) O2,Na (d) 25.

The standard oxidation potentials, Eº, for the half reactions are as [1988 - 1 Mark] 2+ – Zn = Zn + 2e ; Eº = +0.76 V Fe = Fe2+ + 2e–; Eº = +0.41 V The EMF for the cell reaction : Fe2+ + Zn → Zn2+ + Fe (a) –0.35 V (b) +0.35 V (c) +1.17 V (d) –1.17 V 26. A solution of sodium sulphate in water is electrolysed using inert electrodes. The products at the cathode and anode are respectively [1987 - 1 Mark] (a) H2, O2 (b) O2, H2 (c) O2, Na (d) O2, SO2 27. The reaction : [1985 - 1 Mark] + – ½ H2(g) + AgCl(s) → H (aq) + Cl (aq) + Ag(s) occurs in the galvanic cell (a) Ag | AgCl(s) | KCl (soln) | AgNO3 (soln) | 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

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28. A solution containing one mole per litre of each Cu(NO3)2; AgNO3; Hg2(NO3)2; is being electrolysed by using inert electrodes. The values of standard electrode potentials in volts (reduction potentials) are : [1984 - 1 Mark] + ++ Ag/Ag = +0.80, 2Hg/Hg2 = +0.79 Cu/Cu++ = +0.34, Mg/Mg++ = –2.37 With increasing voltage, the sequence of deposition of metals on the cathode will be : (a) Ag, Hg, Cu, Mg (b) Mg, Cu, Hg, Ag (c) Ag, Hg, Cu (d) Cu, Hg, Ag 29. The standard reduction potentials at 298 K for the following half reactions are given against each [1981 - 1 Mark] 2+ Zn(s) –0.762 Zn (aq) + 2e 3+ Cr(s) –0.740 Cr (aq) + 2e + H2(g) 0.000 2H (aq) + 2e 2+ 3+ Fe (aq) 0.770 Fe (aq) + 2e which is the strongest reducing agent? (a) Zn(s) (b) Cr(s) (c) H2(g) (d) Fe2+ (aq) 30. For the electrochemical cell, [Adv. 2018] Mg(s) | Mg (aq, 1 M) || Cu (aq, 1 M) | Cu(s) the standard emf of the cell is 2.70 V at 300 K. When the concentration of Mg2+ is changed to x M, the cell potential changes to 2.67 V at 300 K. The value of x is_______. 2+

2+

(given,

, where F is the Faraday constant and R is the

gas constant, ln 10=2.30)

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31. All the energy released from the reaction X → Y, ∆rG° = –193 kJ mol–1 is used for oxidizing M+ as M+ → M3+ + 2e–, E° = – 0.25 V Under standard conditions, the number of moles of M+ oxidized when one mole of X is converted to Y is [F = 96500 C mol–1] [Adv. 2015] 32. An oxidation-reduction reaction in which 3 electrons are transferred (in V) is ______ has a ∆G0 of 17.37 kJ mol–1 at 25ºC. The value of × 10–2. (1 F = 96,500 C mol–1) 33.

[Main Sep. 05, 2020 (I)] The Gibbs energy change (in J) for the given reaction at [Cu2+] = [Sn2+] = 1 M and 298 K is :

Take F = 96500 C mol–1) [Main Sep. 02, 2020 (I)] 34.

For the disproportionation reaction at 298 K, ln K

(where K is the equilibrium constant) is _______ × 10–1. Given [Main Sep. 02, 2020 (II)] ;

;

35.

What would be the electrode potential for the given half cell reaction at pH=5? __________. + 4e–; = 1.23V 2H2O O2 + (R = 8.314 J mol–1 K–1; Temp = 298 K; oxygen under std. atm. pressure of 1 bar) [Main Jan. 08, 2020 (I)] 36. For an electrochemical cell

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Sn(s)|Sn2+ (aq, 1M) ||Pb2+ (aq, 1M)|Pb(s) the ratio

when this cell

attains equilibrium is _______. [Main Jan. 08, 2020 (II)] 37. Consider an electrochemical cell: A(s) | An+ (aq, 2 M) |B2n+ (aq, 1 M) | B(s). The value of ∆H° for the cell reaction is twice that of ∆G° at 300 K. If the emf of the cell is zero, the ∆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.) [Adv. 2018] 38. 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 conc. of CuSO4 in the cell with higher emf value is 0.5 M. Find out the conc. of CuSO4 in the other cell (2.203 RT/F = 0.06). [2003 - 2 Marks] 2+ 39. The standard reduction potential for Cu | Cu is +0.34 V. Calculate the reduction potential at pH = 14 for the above couple. Ksp of Cu(OH)2 is 1.0×10–19 [1996 - 3 Marks] 40. 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 , assuming that the only reaction that

25°C. Calculate occurs is

. (Given

.) [1995 - 4 Marks]

41.

The more ............... the standard reduction potential, the ............... is its ability to displace hydrogen from acids. [1986 - 1 Mark]

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42.

The dependence of electrode potential for the electrode Mn+ / M with concentration under STP conditions is given by the expression : [1993 - 1 Mark]

43.

In a galvanic cell, the salt bridge [Adv. 2014]

(a) (b) (c) (d) 44.

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 ion in an aqueous solution, E° is + 0.96V. For the reduction of Values of E° for some metal ions are given below 2+ V (aq) + 2e– → V E° = – 1.19 V Fe3+ (aq) + 3e– → Fe E° = – 0.04 V 3+ – Au (aq) + 3e → Au E° = + 1.40 V 2+ – Hg (aq) + 2e → Hg E° = + 0.86 V – The pair(s) of metals that is (are) oxidized by NO3 in aqueous solution is (are) [2009] (a) V and Hg (b) Hg and Fe (c) Fe and Au (d) Fe and V 45. The standard reduction potential values of three metallic cations, X, Y and Z are 0.52,– 3.03 and – 1.18 V respectively. The order of reducing power of the corresponding metals is [1998 - 2 Marks] (a) Y > Z > X (b) X > Y > Z (c) Z > Y > X (d) Z > X > Y

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46.

The standard reduction potential data at 25°C is given below : [Adv. 2013] 3+ 2+ 2+ E°(Fe , Fe ) = + 0.77 V; E°(Fe , Fe) = – 0.44 V; E°(Cu2+, Cu) = + 0.34 V; E°(Cu+, Cu) = + 0.52 V E°[O2(g) + 4H+ + 4e– → 2H2O] = + 1.23 V; E°[O2(g) + 2H2O + 4e– → 4OH–] = + 0.40 V E°(Cr3+, Cr) = – 0.74 V; E°(Cr2+, Cr) = – 0.91 V Match E° of the redox pair in List I with the values given in List II and select the correct answer using the code given below the lists: List I List II P. E°(Fe3+, Fe) 1. – 0.18 V + – 4H + 4OH ) 2. – 0.4 V Q. E°(4H2O 2+ + R. E°(Cu + Cu → 2Cu ) 3. – 0.04 V 3+ 2+ S. E°(Cr , Cr ) 4. – 0.83 V Codes : P Q R S (a) 4 1 2 3 (b) 2 3 4 1 (c) 1 2 3 4 (d) 3 4 1 2

PASSAGE : I The electrochemical cell shown below is a concentration cell. M | M2+ (saturated solution of a sparingly soluble salt, MX2) || M2+ (0.001 mol dm–3) | M. The emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V. [2012] –1 47. The value of ∆G (kJ mol ) for the given cell is (take 1F = 96500 C mol–1) (a) –5.7 (b) 5.7 (c) 11.4 (d) –11.4

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48. The solubility product (Ksp; mol3 dm–9) of MX2 at 298 K based on the information available for 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 PASSAGE : II The concentration of potassium ions inside a biological cell is atleast 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] 49. For the above cell (a) (b) (c) (d) 50. 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 reactions 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

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respect to normal hydrogen electrode) values. Using this data, obtain the correct explanations to questions given. [2007] – – I2 + 2e → 2I E° = 0.54 Cl2 + 2e– → 2Cl– E° = 1.36 Mn3+ + e– → Mn2+ E° = 1.50 Fe3+ + e– → Fe2+ E° = 0.77 O2 + 4H+ + 4e– → 2H2O E° = 1.23 51. 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) Mn2+ is oxidised by chlorine 52. While Fe3+ is stable, Mn3+ is not stable in acid solution because (a) O2 oxideses Mn2+ to Mn3+ (b) O2 oxideses both Mn2+ to Mn3+ and Fe2+ to Fe3+ (c) Fe3+ oxideses H2O to O2 (d) Mn3+ oxideses H2O to O2 53. Sodium fusion extract, obtained from aniline, on treatment with iron (II) sulphate and H2SO4 in 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 PASSAGE : IV Tollen’s test is given by aldehydes. Ag+ + e–

→ Ag; E°red = +0.800 V

C6H12O6 + H2O → C6H12O7 + 2H+ + 2e– ; E°ox = –0.05V Gluconic acid

[Ag(NH3)2]+ + e–

→ Ag + 2NH3; E°red = 0.373V

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Given 54.

&

Calculate (ln K) for [2006 - 5M, –2]

(a) 55.6 (b) 29.6 (c) 66 (d) 58.38 55. On adding NH3, pH of the solution increases to 11 then, identify the effect on potential of half-cell [2006 - 5M, –2] (a) Eox increased from E°ox by 0.65 V (b) Eox decreased from E°ox by 0.65 V (c) Ered increased from E°red by 0.65 V (d) Ered decreased from E°red by 0.65 V 56. NH3 is used in this reaction rather than any other base. Select the correct statement out of the following [2006 - 5M, –2] + + (a) [Ag(NH3)2] is a weaker oxidizing agent than Ag (b) to dissolve the insoluble silver oxide formed under the reaction conditions (c) Ag precipitates gluconic acid as its silver salt (d) NH3 changes the standard reduction potential of [Ag(NH3)2]+ 57.

(a) For the reaction AgCl (s) Given :

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Write the cell representation of above reaction and calculate

at

298 K. Also find the solubility product of AgCl. (b) If 6.539 × 10–2 g of metallic zinc is added to 100 mL saturated solution of AgCl. Find the value of log10

.

How many moles of Ag will be precipitated in the above reaction. Given that [2005 - 6 Marks] – + Ag + e → Ag ; E° = 0.80 V ; Zn2+ + 2e– → Zn ; E° = –0.76 V (It was given that Atomic mass of Zn = 65.39) 58. Find the equilibrium constant for the reaction, at 298 K given :

= 0.15 V ;

= – 0.40 V,

= –0.42 V 59.

[2004 - 4 Marks] The standard potential of the following cell is 0.23 V at 15°C and 0.21 V at 35°C. [2001 - 10 Marks]

(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 theAg+(aq) / Ag(s) couple is 0.80 V at 25°C. 60. 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 : 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 increase or decrease with time?

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[2000 - 2 Marks] 61. A cell, Ag | Ag ||Cu |Cu, initiallly contains 1 M Ag+ and 1 M Cu2+ ions. Calculate the change in the cell potential after the passage of 9.65 A of current for 1 h. [1999 - 6 Marks] 62. Find the solubility product of a saturated solution of Ag2CrO4 in water at 298 K if the emf of the cell Ag | Ag+ (satd. Ag2CrO4 soln.) || Ag+ (0.1M) | Ag is 0.164 V at 298 K. [1998 - 6 Marks] 63. Calculate the equilibrium constant for the reaction,2Fe3+ + 3I– +

2+

–

64.

2Fe2+ + I3 . The standard reduction potentials in acidic conditions are 0.77 V and 0.54 Vrespectively for Fe3+ | Fe2+ and I3– | I– couples. [1998 - 3 Marks] Calculate the equilibrium constant for the reaction [1997 - 2 Marks]

(given

)

65.

Although aluminium is above hydrogen in the electrochemical series, it is stable in air and water. Explain. [1994 - 1 Mark] 66. The Edison storage cells is represented as Fe(s) | FeO(s) | KOH(aq) | Ni2O3(s) | Ni(s) The half-cell reactions are : – 2e– 2NiO(s) + 2OH ; Ni2O3(s) + H2O(l) + Eº = + 0.40V – Fe(s) + 2OH– ; Eº = – 0.87V FeO(s) + H2O(l) + 2e (i) What is the cell reaction ? (ii) What is the cell e.m.f ? 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 Ni2O3? [1994 - 4 Marks] + 67. The standard reduction potential of the Ag /Ag electrode at 298 K is 0.799 V. Given that for AgI, Ksp = 8.7 × 10–17, evaluate the potential of

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the Ag+/Ag electrode in a saturated solution of AgI. Also calculate the standard reduction potential of the I–/ AgI/Ag electrode. [1994 - 3 Marks] 68. The standard reduction potential for the half-cell is 0.78 V. (i) Calculate the reduction potential in 8 M 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 concentration. [1993 - 2 Marks] 69. An aqueous solution of NaCl on electrolysis gives H2(g), Cl2(g) and NaOH according to the reaction : 2Cl–(aq) + 2H2O = 2OH–(aq) + H2(g) + Cl2(g). A direct current of 25 amperes with a current efficiency of 62% is passed through 20 litres of NaCl solution (20% by weight). Write down the reactions taking place at the anode and the cathode. How long will it take to produce 1 kg of Cl2? What will be the molarity of the solution with respect to hydroxide ion? (Assume no loss due to evaporation.) [1992 - 3 Marks] 70. For the galvanic cell. [1992 - 4 Marks] Ag | AgCl(s), KCl(0.2 M) || KBr (0.001M), 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 25ºC. [Ksp(AgCl) = 2.8 × 10–10; Ksp(AgBr) = 3.3 × 10–13] 71. Zinc granules are added in excess to a 500 mL. of 1.0 M nickel nitrate solution at 25ºC until the equilibrium is reached. If the standard reduction potential of Zn2+ | Zn and Ni2+ | Ni are –0.75 V and –0.24 V respectively, find out the concentration of Ni2+ in solution at equilibrium. [1991 - 2 Marks] ++ 72. The standard reduction potential of Cu /Cu and Ag+/Ag electrodes are 0.337 and 0.799 volt respectively. Construct a galvanic cell using these electrodes so that its standard e.m.f. is positive. For what concentration of Ag+ will the e.m.f. of the cell, at 25ºC, be zero if the concentration of Cu++ is 0.01 M?

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[1990 - 3 Marks] 73. The standard reduction potential at 25ºC of the reaction, 2H2O + 2e– H2 + 2OH– is –0.8277V. Calculate the equilibrium constant for the H3O+ + OH– at 25ºC. reaction 2H2O [1989 - 3 Marks] 74. A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of 10–6 M hydrogen ions. The EMF of the cell is 0.118 V at 25ºC. Calculate the concentration of hydrogen ions at the positive electrode. [1988 - 2 Marks] 75. The EMF of a cell corresponding to the reaction : Zn(s) + 2H+(aq) → Zn2+ + (0.1 M) + H2 (g) (1 atm.) is 0.28 volt at 25ºC. Write the half-cell reactions and calculate the pH of the solution at the hydrogen electrode. = –0.76 volt;

=0 [1986 - 4 Marks]

76.

Consider the cell [1982 - 2 Marks]

Zn | Zn (aq) (1.0 M) || Cu (aq) (1.0 M) | Cu. The standard reduction potentials are : +0.350 volts for 2e– + Cu2+ (aq) → Cu and –0.763 volts for 2e– + Zn2+ (aq) → Zn (i) Write down the cell reaction. (ii) Calculate the emf of the cell. (iii) Is the cell reaction spontaneous or not? 77. (a) 19 g of molten SnCl2 is electrolysed for some time. Inert electrodes are used. 0.119 g of Sn is deposited at the cathode. No substance is lost during the electrolysis. Find the ratio of the weights of SnCl2: SnCl4 after electrolysis. (b) A hot solution of NaCl in water is electrolysed. Iron electrodes are used. Diaphragm cell is not used. Give equations for all the chemical reactions that take place during electrolysis. (c) Find the charge in coulombs of 1 gram ion of N3–. 2+

2+

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[1980]

Topic-1 : Conductance of Electrolytic Solution and Electrolysis 1. (d) 2. (N) 3. (d) 4. (a) 5. (b) 6. (a) 7. (c) 8. (d) 9. (c) 10. (d) 11. (a) 12. (c) 13. (d) 14. (c) 15. (a) 16. (c) 17. (c) 18. (d) 19. (c) 20. (b) 21. (d) 22. (b) 23. (b) 24. (d) 25. (c) 26. (6) 27. (3)

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28. 29. 30. 31. 32. 33. 34. 36. 37. 38. 39. 40.

(11) (60) (13.32) (0.154) (19.06) (125.09) (27171.96) (a) (i)-(s); (ii)-(r); (iii)-(q); (iv)-(p) (b) (d) (d) Topic-2 : Nernst Equation, Commercial Cells and Correction

1. (d) 2. (b) 3. (b) 4. (a) 5. (c) 6. (a) 7. (d) 8. (a) 9. (b) 10. (b) 11. (c) 12. (d) 13. (a) 14. (d) 15. (d) 16. (a) 17. (b)

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18. (c) 19. (a) 20. (c) 21. (b) 22. (a) 23. (c) 24. (a) 25. (b) 26. (a) 27. (c) 28. (c) 29. (a) 30. (10) 31. (4) 32. (–6) 33. (96500) 34. (144) 35. (1.52) 36. (2.15) 37. (–11.62) 38. (0.05) 39. (–0.22) 40.(0.792) 42. (False) 43. (a) 44. (a, b, d) 45. (a) 46. (d) 47. (d) 48. (b) 49. (b)

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50. 51. 52. 53. 54. 55. 56.

(c) (c) (d) (a) (d) (a) (b)

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1. Consider the following reactions A → P1 ; B → P2 ; C → P3 ; D → P4, The order of the above reactions are (i), (ii), (iii), and (iv), respectively. The following graph is obtained when log [rate] vs. log[conc.] are plotted : [Main Sep. 06, 2020 (I)]

Among the following, the correct sequence for the order of the reactions is: (a) (iv) > (i) > (ii) > (iii) (b) (i) > (ii) > (iii) > (iv) (c) (iii) > (i) > (ii) > (iv) (d) (iv) > (ii) > (i) > (iii)

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2.

(a) (b) (c) (d) 3. (a) (b) (c) (d) 4.

A flask contains a mixture of compounds A and B. Both compounds decompose by first-order kinetics. The half-lives for A and B are 300 s and 180 s, respectively. If the concentrations of A and B are equal initially, the time required for the concentration of A to be four times that of B (in s) is: (Use ln 2 = 0.693) [Main Sep. 05, 2020 (I)] 180 900 300 120 It is true that : [Main Sep. 03, 2020 (I)] A second order reaction is always a multistep reaction A zero order reaction is a multistep reaction A first order reaction is always a single step reaction A zero order reaction is a single step reaction For the reaction

which statement is correct ? [Main Sep. 03, 2020 (II)]

(a) (b) (c) (d) 5.

The results given in the below table were obtained during kinetic studies of the following reaction : 2A + B → C + D [Main Sep. 02, 2020 (II)]

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X and Y in the given table are respectively : (a) 0.4, 0.4 (b) 0.4, 0.3 (c) 0.3, 0.4 (d) 0.3, 0.3 6. For the following reactions

ks and ke, are, respectively, the rate constants for substitution and elimination, and

=

, the correct option is

.

[Main Jan. 07, 2020 (II)] (a) (b) (c) (d) 7.

> A and ke(A) > ke(B) A > B and ke(B) > ke(A) B > A and ke(B) > ke(A) A > B and ke(A) > ke(B) For the reaction B

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2H2(g) + 2NO(g) → N2(g) + 2H2O(g) the observed rate expression is, rate = kf[NO]2 [H2]. The rate expression for the reverse reaction is: [Main Jan. 07, 2020 (II)] 2 (a) kb[N2][H2O] (b) kb[N2][H2O]2/[NO] (c) kb[N2][H2O] (d) kb[N2][H2O]2/[H2] 8. NO2 required for a reaction is produced by the decomposition of N2O5 in CCl4 as per the equation, The initial concentration of N2O5 is 3.00 mol L–1 and it is 2.75 mol L–1 after 30 minutes. The rate of formation of NO2 is : [Main April 12, 2019 (II)] (a) 4.167×10–3 mol L–1 min–1 (b) 1.667×10–2 mol L–1 min–1 (c) 8.333×10–3 mol L–1 min–1 (d) 2.083×10–3 mol L–1 min–1 9. The given plots represents the variation of the concentration of a reactant R with time for two different reactions (i) and (ii). The respective orders of the reactions are: [Main April 9, 2019 (I)]

(a) (b) (c) (d) 10.

1, 0 1, 1 0, 1 0, 2 For a reaction scheme

, if the rate of formation of

B is set to be zero then the concentration of B is given by :

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[Main April 8, 2019 (II)] (a) (k1 – k2) [A] (b) k1k2 [A] (c) (k1 + k2) [A] (d) 11.

(a) (b) (c) (d) 12.

[A] Decomposition of X exhibits a rate constant of 0.05 µg/year. How many years are required for the decomposition of 5 µg of X into 2.5 µg? [Main Jan. 12, 2019 (I)] 50 25 20 40 The reaction 2X 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 : [Main Jan. 11, 2019 (II)]

(a) (b) (c) (d) 13.

9.0 h 12.0 h 18.0 h 7.2 h For an elementary chemical reaction, A2

2A,

is:

the expression for

[Main Jan. 10, 2019 (II)] (a) (b) (c) (d)

k1[A2] – k – 1[A] 2k1[A2] – k– 1[A]2 k1[A2] + k – 1[A]2 2k1[A2] – 2k– 1[A]2 2

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14.

(a) (b) (c) (d) 15.

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 : [Main 2018] 2 3 1 0 If 50% of a reaction occurs in 100 seconds and 75% of the reaction occurs in 200 seconds, the order of this reaction is: [Main Online April 16, 2018]

(a) (b) (c) (d) 16.

2 3 Zero 1 Decomposition of H2O2 follows a first order reaction. In fifty minutes the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be: [Main 2016] (a) 2.66 L min–1 at STP (b) 1.34 × 10–2 mol min–1 (c) 6.96 × 10–2 mol min–1 (d) 6.93 × 10–4 mol min–1 17. The rate law for the reaction below is given by the expression k [A] [B] A + B → Product If the concentration of B is increased from 0.1 to 0.3 mole, keeping the value of A at 0.1 mole, the rate constant will be: [Main Online April 10, 2016] (a) 3k (b) 9k (c) k/3

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(d) k 18. Higher order (>3) reactions are rare due to : [Main 2015] (a) shifting of equilibrium towards reactants due to elastic collisions (b) loss of active species on collision (c) low probability of simultaneous collision of all the reacting species (d) increase in entropy and activation energy as more molecules are involved 19. A + 2B → C, the rate equation for this reaction is given as Rate = k[A] [B] [Main Online April 11, 2015] If the concentration of A is kept the same but that of B is doubled what will happen to the rate itself ? (a) halved (b) the same (c) doubled (d) quadrupled 20. For the non-stoichiometric reaction 2A + B → C + D, the following kinetic data were obtained in three separate experiments, all at 298 K.

The rate law for the formation of C is: [Main 2014] (a) (b)

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(c) (d) 21.

The half-life period of a first order reaction is 15 minutes. The amount of substance left after one hour will be: [Main Online April 9, 2014]

(a)

of the original amount

(b)

of the original amount

(c)

of the original amount

(d)

of the original amount

22.

(a) (b) (c) (d) 23.

For the elementary reaction M → N, the rate of 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 [Adv. 2014] 4 3 2 1

The rate constant of a zero order reaction is 2.0 × 10–2 mol L–1 s–1. If the concentration of the reactant after 25 seconds is 0.5 M. What is the initial concentration ? [Main Online April 23, 2013] (a) 0.5 M (b) 1.25 M (c) 12.5 M (d) 1.0 M 24. 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

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shown in the figure. The overall order of the reaction is [Adv. 2013]

(a) (b) (c) (d) 25.

(a) (c) 26.

(a) (b) (c) (d) 27.

2 3 0 1 Under the same reaction conditions, initial concentration of 1.386 mol dm–3 of a substance becomes half in 40 seconds and 20 seconds through first order and zero order kinetics, respectively. Ratio (k1/k0) of the rate constant for first order (k1) and zero order (k0) of the reaction is – [2008] 0.5 mol–1 dm3 (b) 1.0 mol dm–3 1.5 mol dm–3 (d) 2.0 mol–1 dm3 Consider a reaction 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] 0 1 2 3 Which one of the following statement for order of reaction is not correct? [2005S] (a) Order can be determined experimentally (b) Order of reaction is equal to sum of the powers of concentration terms in differential rate law.

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28.

(a) (b) (c) (d) 29.

(a) (b) (c) (d) 30.

(a) (b) (c) (d) 31.

(c) It is not affected with the stoichiometric coefficient of the reactants (d) Order cannot be fractional. The reaction, A → Product, follows first order kinetics. In 40 minutes the concentration of A changes from 0.1 to 0.025 M. The rate of reaction, when concentration of A is 0.01 M is [2004S] –4 –1 1.73 ×10 M min 3.47 ×10–5 M min–1 3.47 ×10–4 M min–1 1.73 ×10–5 M min–1 In a first order reaction the concentration of reactant decreases from 800 mol/dm3 to 50 mol/dm3 in 2 × 104 sec. The rate constant of reaction in sec–1 is: [2003S] 4 2 × 10 3.45 × 10–5 1.386 × 10–4 2 × 10–4 2NH3(g). The rate Consider the chemical reaction, N2(g) + 3H2(g) of this reaction can be expressed in terms of time derivative of concentration of N2(g), H2(g) or NH3(g). Identify the correct relationship amongst the rate expressions. [2002S] Rate = –d[N2]/dt = –1/3d[H2]/dt = 1/2d[NH3]/dt Rate = –d[N2]/dt = –3d[H2]/dt = 2d [NH3]/dt Rate = d[N2]/dt =1/3d[H2]/dt = 1/2d [NH3]/dt Rate = –d[N2]/dt = –d[H2]/dt = d [NH3]/dt If ‘I’ is the intensity of absorbed light and ‘C’ is the concentration of AB*, the rate of AB for the photochemical process, AB + hv formation of AB* is directly proportional to [2001S] (a) C

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(b) I (c) I2 (d) C.I 32. The rate constant for the reaction, 2N2O5 → 4NO2 + O2, is 3.0 × 10–5 sec–1 . If the rate is 2.40 × 10–5 mol litre-1 sec-1, then the concentration of N2O5 (in mol litre-1) is [2000S] (a) 1.4 (b) 1.2 (c) 0.04 (d) 0.8 33. The specific rate constant of a first order reaction depends on the [1983 - 1 Mark] (a) concentration of the reactant (b) concentration of the product (c) time (d) temperature 34. The rate constant of a reaction depends on (a) temperature [1981 - 1 Mark] (b) initial concentration of the reactants (c) time of reaction (d) extent of reaction 35.

In dilute aqueous H2SO4, the complex diaquodio-xalatoferrate(II) is oxidized by MnO4 –. For this reaction, the ratio of the rate of change of [H+] to the rate of change of [MnO4–] is [Adv. 2015] 36. An organic compound undergoes first-order decomposition. The time taken for its decomposition to1/8 and 1/10 of its initial concentration are t1/8 and t1/10 respectively. What is the value of

(log102 = 0.3)

× 10 ?

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[2012] 37. The concentration of R in the reaction R → P was measured as a function of time and the following data is obtained:

The order of reaction is [2010] 38.

If 75% of a first order reaction was completed in 90 minutes, 60% of the same reaction would be completed in approximately (in minutes) ___________. (Take : log 2 = 0.30; log 2.5 = 0.40) [Main Sep. 04, 2020 (I)] 39. During the nuclear explosion, one of the products is 90Sr with half life of 6.93 years. If 1 g 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. [Main Jan. 07, 2020 (I)] 40. Consider the kinetic data given in the following table for the reaction A + B + C → product. [Adv. 2019] Experiment

[A]

[B]

[C]

Rate of reaction

No.

(mol dm –3) 0.1

(mol dm –3) 0.1

(mol dm –3)

1

(mol dm –3) 0.2

2

0.2

0.2

0.1

6.0 × 10 –5

3

0.2

0.1

0.2

1.2 × 10 –4

4

0.3

0.1

0.1

9.0 × 10 –5

6.0 × 10 –5

The rate of the reaction for [A] = 0.15 mol dm–3, [B] = 0.25 mol dm–3 and [C] = 0.15 mol dm–3 is found to be Y ×10–5 moldm–3 s–1. The value of Y is _____ 41. The decomposition reaction

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is started in a closed cylinder under isothermal isochoric condition at an initial pressure of 1 atm. After Y×103s, 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 __ [Adv. 2019] –1 –1 42. The rate of a first-order reaction is 0.04 mol litre s at 10 minutes and 0.03 mol litre–1 s–1 at 20 minutes after initiation. Find the half-life of the reaction. [2001 - 5 Marks] 43. The gas phase decomposition of dimethyl ether follows first order kinetics. The reaction is carried out in a constant volume container at 500°C and has a half life of 14.5 minutes. Initially, only dimethyl ether is present at a pressure of 0.40 atmosphere. What is the total pressure of the system after 12 minutes? Assume ideal gas behaviour. [1993 - 4 Marks]

44.

For the reaction N2(g) + 3H2(g) → 2NH3(g), 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 ........ kg h–1. [1994 - 1 Mark] 45. The hydrolysis of ethyl acetate in ............... medium is a ............... order reaction. [1986 - 1 Mark] 46. The rate of chemical change is directly proportional to ............... . [1985 - 1 Mark] 47.

For a first order reaction, the rate of the reaction doubles as the concentration of the reactant (s) doubles. [1986 - 1 Mark]

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48. Which of the following plots is(are) correct for the given reaction? ([P]0 is the initial concentration of P) [Adv. 2020]

(a)

(b)

(c)

(d)

49.

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

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and Pt, respectively. Initially, only A is present with concentration [A]0, and t1/3 is the time required for the partial pressure of A to reach 1/3rd of its initial value. The correct option(s) is (are) (Assume that all these gases behave as ideal gases) [Adv. 2018]

(a)

(b)

(c)

(d)

50.

For the first order reaction [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 to 99.6% completion in eight half-life duration 51. The following statement(s) is (are) correct : [1999 - 3 Marks] (a) A plot of log Kp versus 1/T is linear P (b) A plot of log [X] versus time is linear for a first order reaction, X (c) A plot of P versus 1/T is linear at constant volume (d) A plot of P versus 1/V is linear at constant temperature

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52.

For a first order reaction, [1998 - 2 Marks]

(a) the degree of dissociation is equal to (1–e–kt) (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 t1/2 of the reaction (d) the pre-exponential factor in the Arrhenius equation has the dimension of time, T–1. 53. The rate law for the reaction : [1988 - 1 Mark] RCl + NaOH (aq.) → ROH + NaCl is given by, Rate = k1 [RCl]. The rate of the reaction will be (a) doubled on doubling the concentration of sodium hydroxide. (b) halved on reducing the concentration of alkyl halide to one half. (c) increased on increasing the temperature of the reaction. (d) unaffected by increasing the temperature of the reaction. 54.

At constant temperature and volume, X decomposes as [2005 - 4 Marks] 2X(g) → 3Y(g) + 2Z(g); PX is the partial pressure of X.

(i) (ii) (iii) (iv) 55.

What is the order of reaction with respect to X? Find the rate constant. Find the time for 75% completion of the reaction. Find the total pressure when pressure of X is 700 mm of Hg. For the given reactions, A + B → Products, following data were obtained. [2004 - 2 Marks]

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(a) (b) 56.

57.

58.

[A0] [B0] R0 (mol L–1 s–1) 1. 0.1 0.2 0.05 2. 0.2 0.2 0.10 3. 0.1 0.1 0.05 Write the rate law expression Find the rate constant The vapour pressure of the two miscible liquids (A) and (B) are 300 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 polymerizing into a completely insoluble solid. The polymerization follows first-order kinetics. After 100 minutes, 0.525 mole of a solute is dissolved which arrests the polymerization completely. The final vapour pressure of the solution is 400 mm of Hg. Estimate the rate of constant of the polymerization reaction. Assume negligible volume change on mixing and polymerization and ideal behaviour for the final solution. [2001 - 10 Marks] is 4.5 × 10–3 The rate constant for an isomerisation reaction, min–1. If the initial concentration of A is 1 M, calculate the rate of the reaction after 1 h. [1999 - 4 Marks] nB, with time, is presented in The progress of the reaction, A figure given below. Determine

(i) the value of n (ii) the equilibrium constant, K and (iii) the initial rate of conversion of A. [1994 - 3 Marks]

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59. The decomposition of N2O5 according to the equation : 2N2O5(g) → 4NO2(g) + O2(g) [1991 - 6 Marks] 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 and on complete decomposition, the total pressure is 584.5 mm of Hg. Calculate the rate constant of the reaction. 60. A first order reaction has k = 1.5 × 10–6 per second at 200ºC. If the reaction is allowed to run for 10 hours, what percentage of the initial concentration would have changed in the product? What is the half life of this reaction? [1987 - 5 Marks] 61. While studying the decomposition of gaseous N2O5, 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 - 2 Marks] 62. Rate of a reaction A + B → products, is given below as a function of different initial concentrations of A and B : [1982 - 4 Marks] [A] (mol/l) [B] (mol/l) Initial rate (mol / l/min) 0.01 0.01 0.005 0.02 0.01 0.010 0.01 0.02 0.005 Determine the order of the reaction with respect to A and with respect to B. What is the half-life of A in the reaction?

1.

The rate constant (k) of a reaction is measured at different temperatures (T), and the data are plotted in the given figure. The activation energy of the reaction in kJ mol–1 is : (R is gas constant) [Main Sep. 05, 2020 (II)]

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(a) (b) (c) (d) 2.

2/R 1/R R 2R For following reactions:

it was found that the Ea is decrease by 30 kJ/mol in the presence of catalyst. If the rate remains unchanged, the activation energy for catalysed reaction is (Assume pre exponential factor is same): [Main Jan. 09, 2020 (I)] (a) 75 kJ/mol (b) 105 kJ/mol (c) 135 kJ/mol (d) 198 kJ/mol 3. The rate of a certain biochemical reaction at physiological temperature (T) occurs 106 times faster with enzyme than without. The change in the activation energy upon adding enzyme is: [Main Jan. 08, 2020 (I)] (a) –6(2.303)RT (b) –6RT (c) +6(2.303)RT (d) +6RT 4.

Consider the following plots of rate constant versus

for four

different reactions. Which of the following orders is correct for the

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activation energies of these reactions? [Main Jan. 08, 2020 (II)]

(a) (b) (c) (d) 5.

Eb > Ea > Ed > Ec Ea > Ec > Ed > Eb Ec > Ea > Ed > Eb Eb > Ed > Ec > Ea For the reaction of H2 with I2, the rate constant is 2.5×10–4 dm3 mol–1 s– 1 at 327 oC and 1.0 dm3 mol–1 s–1 at 527 oC. The activation energy for the reaction, in kJ mol–1 is : (R = 8.314J K–1 mol–1) [Main April 10, 2019 (II)] (a) 166 (b) 150 (c) 72 (d) 59 6. Consider the given plot of enthalpy of the following reaction between A and B, A + B → C + D. Identify the incorrect statement. [Main April 9, 2019 (II)]

(a) (b) (c) (d)

Activation enthalpy to form C is 5kJ mol–1 less than that to form D. C is the thermodynamically stable product. D is kinetically stable product. Formation of A and B from C has highest enthalpy of activation.

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7.

If a reaction follows the Arrhenius equation, the plot lnk vs

gives straight line with a gradient (– y) unit. The energy required to activate the reactant is: [Main Jan. 11, 2019 (I)] (a) y/R unit (b) y unit (c) yR unit (d) – y unit 8. Two reactions R1 and R2 have identical pre-exponential factors. Activation energy of R1 exceeds that of R2 by10 kJ mol–1. If k1 and k2 are rate constants for reactions R1 and R2 respectively at 300 K, then ln(k2/k1) is equal to : (R = 8.314 J mol–1K–1) [Main 2017] (a) 8 (b) 12 (c) 6 (d) 4 9. The rate of a reaction quadruples when the temperature changes from 300 to 310 K. The activation energy of this reaction is : (Assume activation energy and pre-exponential factor are independent of temperature; ln 2 = 0.693; R = 8.314 J mol–1 K–1) [Main Online April 9, 2017] –1 (a) 107.2 kJ mol (b) 53.6 kJ mol–1 (c) 26.8 kJ mol–1 (d) 214.4 kJ mol–1 10. The rate coefficient (k) for a particular reactions is1.3 × 10–4 M–1 s–1 at 100 °C, and 1.3 × 10–3 M–1 s–1 at 150 °C. What is the energy of activation (Ea) (in kJ) for this reaction? (R = molar gas constant = 8.314 JK–1 mol–1) [Main Online April 12, 2014] (a) 16 (b) 60

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(c) 99 (d) 132 11. The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy of the reaction will be : (R = 8.314 JK–1 mol–1 and log 2 = 0.301) [Main 2013] (a) 53.6 kJ mol –1 (b) 48.6 kJ mol–1 (c) 58.5 kJ mol–1 (d) 60.5 kJ mol–1 12. The reaction X → Y is an exothermic reaction. Activation energy of the reaction for X into Y is 150 kJ mol–1. Enthalpy of reaction is 135 kJ mol–1. The activation energy for the reverse reaction, Y → X will be : [Main Online April 22, 2013] (a) 280 kJ mol–1 (b) 285 kJ mol–1 (c) 270 kJ mol–1 (d) 15 kJ mol–1 13. Plots showing the variation of the rate constant (k) with temperature (T) are given below. The plot that follows Arrhenius equation is [2010] (a)

(b)

(c)

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(d) 14.

For a first order reaction A→P, the temperature (T) dependent rate constant (k) was found to follow the equation log k = – (2000) . The pre-exponential factor A and the activation energy Ea, respectively, are [2009]

(a) (b) (c) (d) 15.

(a) (b) (c) (d) 16. (a) (b) (c) (d)

1.0 × 10 s and 9.2 kJ mol 6.0 s–1 and 16.6 kJ mol–1 1.0 × 106 s–1 and 16.6 kJ mol–1 1.0 × 106 s–1 and 38.3 kJ mol–1 The rate constant, the activation energy and the Arrhenius parameter of a chemical reaction at 25°C are 3.0 × 10–4s–1, 104.4 kJ mol–1 and 6.0 × 1014 s–1 respectively. The value of the rate constant as T → ∞ is, [1996 - 1 Mark] 2.0 × 1018 s–1 6.0 × 1014 s–1 infinity 3.6 × 1030 s–1 A catalyst is a substance which [1983 - 1 Mark] increases the equilibrium concentration of the product changes the equilibrium constant of the reaction shortens the time to reach equilibrium supplies energy to the reaction 6

–1

–1

17.

The rate of a reaction decreased by 3.555 times when the temperature was changed from 40ºC to 30ºC. The activation energy (in kJ mol–1) of the reaction is ______. Take; R = 8.314 J mol–1 K–1 In 3.555 = 1.268 [Main Sep. 06, 2020 (II)] 18. The number of molecules with energy greater than the threshold energy for a reaction increases five fold by a rise of temperature from

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27°C to 42°C. Its energy of activation in J/ mol is ___________. (Take ln 5 = 1.6094; R = 8.314 J mol–1K–1) [Main Sep. 04, 2020 (II)] 19.

A sample of milk splits after 60 min. at 300 K and after 40 min. at 400 K when the population of lactobacillus acidophilus in it doubles. The activation energy (in kJ/mol) for this process is closest to _____.

(Given, R = 8.3 J mol–1 K–1, ln

= 0.4, e–3 = 4.0) [Main Jan. 09, 2020 (II)]

20.

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 ∆G° (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) [Adv. 2018] 21. A hydrogenation reaction is carried out at 500 K. If 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 - 3 Marks] 22. At 380°C, the half-life period for the first order decomposition of is 360 min. The energy of activation of the reaction is 200 kJ

23.

. Calculate the time required for 75% decomposition at 450°C. [1995 - 4 Marks] In the Arrhenius equation for a certain reaction, the value of A and Ea (activation energy) are 4 × 1013 sec–1 and 98.6 kJ mol–1 respectively. If the reaction is of first order, at what temperature will its half-life period be ten minutes? [1990 - 3 Marks]

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24.

In the Arrhenius equation, k = A exp (–Ea/RT), A may be termed as the rate constant at .............. . [1997 - 1 Mark]

25.

The rate of an exothermic reaction increases with increasing temperature. [1990 - 1 Mark] 26. Catalyst does not affect the energy of activation in a chemical reaction. [1989 - 1 Mark] 27. Catalyst makes a reaction more exothermic. [1987 - 1 Mark]

28.

For a reaction, A P, the plots of [A] and [P] with time at temperatures T1 and T2 are given below. [Adv. 2018]

If T2 > T1, the correct statement(s) is (are) (Assume ∆H° and ∆S° are independent of temperature and ratio of lnK at T1 to lnK at T2 is greater than

(b) (c) (d) 29.

. Here H, S, G and K are enthalpy,

entropy, Gibbs energy and equilibrium constant, respectively.) (a) ∆H° < 0, ∆S° < 0 ∆G° < 0, ∆H° > 0 ∆G° < 0, ∆S° < 0 ∆G° < 0, ∆S° > 0 In a bimolecular reaction, the steric factor P was experimentally determined to be 4.5. The correct option(s) among the following is(are) [Adv. 2017]

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(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) Since P = 4.5, the reaction will not proceed unless an effective catalyst is used (d) The value of frequency factor predicted by Arrhenius equation is higher than that determined experimentally 30. According to the Arrhenius equation, [JEE Adv. 2016] (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. 31. A catalyst : [1984 - 1 Mark] (a) increases the average kinetic energy of reacting molecules (b) decreases the activation energy (c) alters the reaction mechanism (d) increases the frequency of collisions of reacting species 32.

Read the following assertion and statement and answer as per the options given below : Assertion : For each ten degree rise of temperature the specific rate constant is nearly doubled. Statement : Energy-wise distribution of molecules in a gas is an experimental function of temperature. [1989 - 2 Marks] (a) If both assertion and statement are correct and statement is an explanation of assertion.

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(b) If assertion is correct and statement is wrong, statement is not an explanation of assertion. (c) If assertion is wrong and statement is correct, statement is not an explanation of assertion. (d) If both assertion and statement are wrong and statement is not explanation of assertion. 33.

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 - 5 Marks] 34. The rate constant for the first order decomposition of a certain reaction is described by the equation [1997 - 5 Marks]

(i) What is the energy of activation for this reaction? (ii) At what temperature will its half-life period be 256 minutes? 35. From the following data for the reaction between A and B. [1994 - 5 Marks]

Calculate (i) the order of the reaction with respect to A and with respect to B, (ii) the rate constant at 300K (iii) the energy of activation, and (iv) the pre-exponential factor , requires activation energy of 70kJ mol– 36. A first order reaction 1 . When a 20% solution of A was kept at 25°C for 20 minutes, 25% decomposition took place. What will be the percent 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 - 4 Marks]

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37.

Two reactions (i) A → products, (ii) B → products, follows first order kinetics. The rate of the reaction : (i) is doubled when the temperature is raised from 300K to 310K. The half life for this reaction at 310K is 30 minutes. 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 300K. [1992 - 3 Marks] 38. A first order reaction is 50% complete in 30 minutes at 27ºC and in 10 minutes at 47ºC. Calculate the reaction rate constant at 27ºC and the energy of activation of the reaction in kJ/mole. [1988 - 3 Marks]

1.

Bombardment of aluminium by α-particle leads to its artificial disintegration in two ways, (i) and (ii) as shown. Products X, Y and Z respectively are, [2011]

(a) (b) (d) (d) 2.

proton, neutron, positron neutron, positron, proton proton, positron, neutron positron, proton, neutron A positron is emitted from

(a) (b) (c) 3.

. The ratio of the atomic mass and

atomic number of the resulting nuclide is [2007] 22/10 22/11 23/10 (d) 23/12 23 Na is the more stable isotope of Na. Find out the process by which can undergo radioactive decay [2003S]

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(a) β– emission (b) α emission (c) β+ emission (d) K electron capture 4.

The number of neutrons accompanying the formation of from the absorption of a slow neutron by

and

, followed by

nuclear fission is, [1999 - 2 Marks] (a) (b) (c) (d) 5.

0 2 1 3 is a stable isotope,

is expected to disintegrate by [1996 - 1 Mark]

(a) (b) (c) (d) 6.

α-emission β-emission positron emission proton emission The half-life period of a radioactive element is 140 days. After 560 days, one gram of the element will reduced to : [1986 - 1 Mark]

(a)

g

(b)

g

(c) (d)

g g

7.

The radiations from a naturally occurring radioactive substance, as seen after deflection by a magnetic field in one direction, are : [1984 - 1 Mark] (a) definitely alpha rays (b) definitely beta rays

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(c) both alpha and beta rays (d) either alpha or beta rays 8. If uranium (mass number 238 and atomic number 92) emits an αparticle, the product has mass no. and atomic no. [1981 - 1 Mark] (a) 236 and 92 (b) 234 and 90 (c) 238 and 90 (d) 236 and 90 9.

A closed vessel with rigid walls contains 1 mol of air at 298 K. Considering complete decay of

10.

11.

to

the ratio

of the final pressure to the initial pressure of the system at 298 K is [Adv. 2015] 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 X belongs in the periodic table? [2012] The number of neutrons emitted when nuclear fission to

12.

and 1 mol of

and

undergoes controlled

is

[2010] The total number of α and β particles emitted in the nuclear reaction [2009]

13.

is known to undergo radioactive decay to form

by

emitting alpha and beta particles. A rock initially contained 68 × 10–6 . If the number of alpha particles that it would emit during g of

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its radioactive decay of

to

in three half-lives is Z × 1018,

then what is the value of Z ? 14.

[Adv. 2020] and One of the hazards of nuclear explosion is the generation of its subsequent incorporation in bones. This nuclide has a half-life of 28.1 years. Suppose one microgram was absorbed by a new-born will remain in his bones after 20 years? child, how much [1995 - 2 Marks]

15.

A radioactive nucleus decays emitting one alpha and two beta particles; the daughter nucleus is ............... of the parent. [1989 - 1 Mark] on 16. The number of neutrons in the parent nucleus which gives beta emission is ............... . [1985 - 1 Mark]

17.

An element

undergoes an α-emission followed by two

successive β-emissions. The element formed is .............. [1982 - 1 Mark] 18.

In β-emission from a nucleus, the atomic number of the daughter element decreases by one. [1990 - 1 Mark]

19.

In the decay sequence, [Adv. 2019]

x1, x2, x3 and x4 are particles/radiation emitted by the respective isotopes. The correct option(s) is (are) (a) Z is an isotope of uranium (b) x1 will deflect towards negatively charged plate

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(c) x3 is γ-ray (d) x2 is β-ray 20. 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 (are) [Adv. 2016] – (a) β –decay (β emission) (b) orbital or K–electron capture (c) neutron emission (d) β+–decay (positron emission) 21. In the nuclear transmutation [Adv. 2013] +X→ +Y (X, Y) is (are) (a) (γ, n) (b) (p, D) (c) (n, D) (d) (γ, p) 22. Decrease in atomic number is observed during [1998 - 2 Marks] (a) (b) (c) (d) 23.

alpha emission beta emission positron emission electron capture. Nuclear reactions accompanied with emission of neutron(s) are : [1988 - 1 Mark] + → (a) (b)

+

(c)

→

(d)

→ + +

→

+

Several short-lived radioactive species have been used to determine the age of wood or animal fossils. One of the most interesting substances is 6C14

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(half-life 5760 years) which is used in determining the age of carbonbearing materials (e.g. wood, animal fossils, etc.). Carbon-14 is produced by the bombardment of nitrogen atoms present in the upper atmosphere with neutrons (from cosmic rays). Thus carbon-14 is oxidised to CO2 and eventually ingested by plants and animals. The death of plants or animals put an end to the intake of C14 from the atmosphere. After this the amount of C14 in the dead tissues starts decreasing due to its disintegration as per the following reaction : The C14 isotope enters the biosphere when carbon dioxide is taken up in plant photosynthesis. Plants are eaten by animals, which exhale C14 as CO2. Eventually, C14 participates in many aspects of the carbon cycle. The C14 lost by radioactive decay is constantly replenished by the production of new isotopes in the atmosphere. In this decay-replenishment process, a dynamic equilibrium is established whereby the ratio of C14 to C12 remains constant in living matter. But when an individual plant or an animal dies, the C14 isotope in it is no longer replenished, so the ratio decreases as C14 decays. So, the number of C14 nuclei after time t (after the death of living matter) would be less than in a living matter. The decay constant can be calculated using the following formula,

The intensity of the cosmic rays have remain the same for 30,000 years. But since some years the changes in this are observed due to excessive burning of fossil fuel and nuclear tests. 24. Why do we use the carbon dating to calculate the age of the fossil? [2006 - 5M, –2] (a) Rate of exchange of carbon between atmosphere and living is slower than decay of C14 (b) It is not appropriate to use C14 dating to determine age (c) Rate of exchange of C14 between atmosphere and living organism is so fast that an equilibrium is set up between the intake of C14 by organism and its exponential decay (d) none of the above

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25. (a) (b) (c) (d) 26.

(a)

What should be the age of the fossil for meaningful determination of its age? [2006 - 5M, –2] 6 years 6000 years 60,000 years can be used to calculate any age A nuclear explosion has taken place leading to increase in concentration of C14 in nearby areas. C14 concentration is C1 in nearby areas and C2 in areas far away. If the age of the fossil is determined to be T1 and T2 at the respective places then [2006 - 5M, –2] The age of the fossil will increase at the place where explosion has taken place and

(b) The age of the fossil will decrease at the place where explosion has taken place and (c) The age of fossil will be determined to be same (d) 27.

(a) (b) (c) (d)

Statement-1 : 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-2 : Proton-proton electrostatic repulsions begin to overcome attractive forces involving protons and neutrons in heavier nuclides. [2008] Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 Statement-1 is True, Statement-2 is False Statement-1 is False, Statement-2 is True

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28.

(a) (b) (c) (d) 29.

Assertion : Nuclide

Al is less stable than

Ca

[1998 - 2 Marks] Reason : Nuclides having odd number of protons and neutrons are generally unstable. If both assertion and reason are correct, and reason is the correct explanation of the assertion. If both assertion and reason are correct, but reason is not the correct explanation of the assertion. If assertion is correct but reason is incorrect. If assertion is incorrect but reason is correct. Complete and balance the following reactions. (i) [2004 - 1 Mark]

(ii) [2005 - 1 Mark] + ..............

(iii) 30.

31.

[2005 - 1 Mark] 64 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 - 5 Marks] is radioactive and it emits α and β particles to form

.

Calculate the number of α and β particles emitted in this conversion. An ore of

is found to contain

and

in the weight is 4.5 × 109 years.

ratio of 1:0.1. The half-life period of Calculate the age of the ore.

[2000 - 5 Marks] 32. Write a balanced equation for the reaction of N with α-particle. [1997 - 1 Mark] 227 33. Ac has a half-life of 21.8 years with respect to radioactive decay. The decay follows two parallel paths, one leading to 227Th and the other to 223Fr. The percentage yields of these two daughter nuclides 14

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are 1.2 and 98.8 respectively. What are the decay constants (λ) for each of the separate paths? [1996 - 2 Marks] to in a sample of water is8.0 × 10–18 : 1. 34. The nucleidic ratio, Tritium undergoes decay with a half life period of 12.3 years. How many tritium atoms would 10.0 g of such a sample contain 40 years after the original sample is collected? [1992 - 4 Marks] 35. An experiment requires minimum beta activity product at the rate of , which is 346 beta particles per minute. The half life period of a beta emitter is 66.6 hours. Find the minimum amount of required to carry out the experiment in 6.909 hours. [1989 - 5 Marks] disintegrates to give

36.

37.

as the final product. How many

alpha and beta particles are emitted during this process? [1986 - 2 Marks] Radioactive decay is a first order process. Radioactive carbon in wood sample decays with a half life of 5770 years. What is the rate constant (in years–1) for the decay? What fraction would remain after 11540 years? [1984 - 3 Marks]

Topic-1 : Rate of Reactions, Order of Reactions and Half Life Period 1. (d) 2. (b) 3. (b) 4. (c) 5. (c) 6. (b) 7. (d) 8. (b)

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9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31.

(a) (d) (a) (c) (d) (a) (d) (d) (d) (c) (c) (d) (c) (b) (d) (d) (a) (d) (d) (c) (c) (a) (b)

32. (d) 33. (d) 34. (a) 35. (8) 36. (9) 37. (0)

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38. (60) 39. (23.03) 40.(6.75) 41. (2.30) 42. (24.14) 43. (0.749) 48. (a) 49. (a,d) 50. (a,b,d) 51. (a,b,d) 52. (a,d) 53. (b,c) Topic-2 : Effect of Temperature and Catalyst on Rate of Reactions, Collosion Theory of Chemical Reactions 1. (d) 2. (a) 3. (a) 4. (c) 5. (a) 6. (a) 7. (b) 8. (d) 9. (a) 10. (b) 11. (a) 12. (b) 13. (a)

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14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 28. 29. 30. 31. 32.

(d) (b) (c) (100) (84297.48) (– 3.98) (– 8500) (100) (20.34) (311.35) (a,c) (a,b) (b,c,d) (b,c) (a) Topic-2 : Nuclear Chemistry

1. (a) 2. (c) 3. (a) 4. (d) 5. (b) 6. (d) 7. (d) 8. (b) 9. (9) 10. (8) 11. (3) 12. (8) 13. (1.20) 14. (0.061)

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18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28.

(False) (a,b,d) (b,d) (a,b) (a,c,d) (a,d) (c) (b) (a) (a) (c)

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1.

Adsorption of a gas follows Freundlich adsorption isotherm. If x is the mass of the gas adsorbed on mass m of the adsorbent, the correct plot of

versus p is : [Main Sep. 05, 2020 (II)]

(a)

(b)

(c)

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(d)

2.

(1) (2) (3) (4) (a) (b) (c) (d) 3.

Amongst the following statements regarding adsorption, those that are valid are : [Main Sep. 02, 2020 (II)] ∆H becomes less negative as adsorption proceeds. On a given adsorbent, ammonia is adsorbed more than nitrogen gas. On adsorption, the residual force acting along the surface of the adsorbent increases. With increase in temperature, the equilibrium concentration of adsorbate increases. (4) and (1) (2) and (3) (1) and (2) (3) and (4) A mixture of gases O2, H2 and CO are taken in a closed vessel containing charcoal. The graph that represents the correct behaviour of pressure with time is: [Main Jan. 09, 2020 (II)]

(a)

(b)

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(c)

(d)

4.

Adsorption of a gas follows Freundlich adsorption isotherm. x is the mass of the gas adsorbed on mass m of the adsorbent. The plot of versus log p is shown in the given graph.

is proportional to : [Main April 8, 2019 (I)]

(a) (b) (c) (d) 5. H2 33

p2/3 p3/2 p3 p2 Given gas critical temperature/K for CH4 CO2 SO2 is 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?

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[Main Jan. 12, 2019 (I)] (a) (b) (c) (d) 6.

SO2 CH4 CO2 H2 If x gram of gas is adsorbed by m gram of adsorbent at pressure P, the plot of log

versus log P is linear. The slope of the plot is: (n and k

are constants and n > 1 ) [Main Online April 15, 2018 (II)] (a) log k (b) (c) 2k (d) n 7. Adsorption of a gas on a surface follows Freundlich adsorption isotherm. Plot of log

versus log p gives a straight line with slope

equal to 0.5, then : (

is the mass of the gas adsorbed per gram of adsorbent) [Main Online April 9, 2017]

(a) (b) (c) (d) 8.

(a) (b) (c) (d)

Adsorption is independent of pressure. Adsorption is proportional to the pressure. Adsorption is proportional to the square root of pressure. Adsorption is proportional to the square of pressure. 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] Only 1/n appears as the slope. log (1/n) appears as the intercept. Both k and 1/n appear in the slope term. 1/n appears as the intercept.

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9.

(a) (b) (c) (d) 10. (a) (b) (c) (d) 11. (a) (b) (c) (d) 12. (a) (b) (c) (d) 13. (a) (b) (c) (d)

A particular adsorption process has the following characteristics : (i) It arises due to van der Waals forces and (ii) it is reversile. Identify the correct statement that describes the above adsorption process : [Main Online April 9, 2016] Adsorption is monolayer. Adsorption increases with increase in temperature. Enthalpy of adsorption is greater than 100 kJ mol–1 Energy of activation is low. The following statements relate to the adsorption of gases on a solid surface. Identify the incorrect statement among them : [Main Online April 10, 2015] Enthalpy of adsorption is negative Energy appears as heat On adsorption, the residual forces on the surface are increased Entropy of adsorption is negative Methylene blue, from its aqueous solution, is adsorbed on activated charcoal at 25°C. For this process, the correct statement is [Adv. 2013] The adsorption requires activation at 25°C The adsorption is accompanied by a decrease in enthalpy The adsorption increases with increase of temperature The adsorption is irreversible Adsorption of gases on solid surface is generally exothermic because [2004S] enthalpy is positive entropy decreases entropy increases free energy increases Rate of physisorption increases with [2003S] decrease in temperature increase in temperature decrease in pressure decrease in surface area

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14.

The mass of gas adsorbed, x, per unit mass of adsorbate, m, was measured at various pressures, p. A graph between

and log p

gives a straight line with slope equal to 2 and the intercept equal to 0.4771. The value of

at a pressure of 4 atm is : (Given log 3 =

0.4771) [Main Sep. 02, 2020 (I)] 15. 20% of surface sites are occupied by N2 molecules. The density of surface site is 6.023 × 1014 cm–2 and total surface area is 1000 cm2. The catalyst is heated to 300 K while N2 is completely desorbed into a pressure of 0.001 atm and volume of 2.46 cm3. Find the number of active sites occupied by each N2 molecule. [2005 - 4 Marks] 16.

For Freundlich adsorption isotherm, a plot of log (x/m) (y-axis) and log p (x-axis) gives a straight line. The intercept and slope for the line is 0.4771 and 2, respectively. The mass of gas, adsorbed per gram of adsorbent if the initial pressure is 0.04 atm, is ______ × 10–4g. (log 3 = 0.4771) [Main Sep. 06, 2020 (II)]

17.

(a) (b) (c) (d) 18.

When O2 is adsorbed on a metallic surface, electron transfer occurs from the metal to O2. The true statement(s) regarding this adsorption is(are) [Adv. 2015] O2 is physisorbed Heat is released Occupancy of π*2ρ of O2 is increased Bond length of O2 is increased The given graphs/data I, II, III and IV represent general trends observed for different physisorption and chemisorption processes

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under mild conditions of temperature and presure. Which of the following choice(s) about I, II, III and IV is (are) correct? [2012]

(a) (b) (c) (d) 19.

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 The correct statement(s) pertaining to the adsorption of a gas on a solid surface is (are)

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[2011] (a) Adsorption is always exothermic (b) Physisorption may transform into chemisorption at high temperature (c) Physisorption increases with increasing temperature but chemisorption decreases with increasing temperature (d) Chemisorption is more exothermic than physiosorption, however it is very slow due to higher energy of activation 20.

1 g of charcoal adsorbs 100 mL 0.5 M CH3COOH to form a monolayer, and thereby the molarity of CH3COOH reduces to 0.49. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal = 3.01 × 102 m2/g. [2003 - 2 Marks]

1.

Which of the following is not an example of heterogeneous catalytic reaction? [Main Jan. 10, 2019 (I)] Ostwald’s process Combustion of coal Hydrogenatoin of vegetable oils Haber’s process Identify the correct statement regarding enzymes [2004] Enzymes are specific biological catalysts that cannot be poisoned Enzymes are normally heterogeneous catalysts that are very specific in their action Enzymes are specific biological catalysts that can normally function at very high temperatures (T~1000K) Enzymes are specific biological catalysts that possess well-defined active sites

(a) (b) (c) (d) 2. (a) (b) (c) (d)

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3.

Match the catalysts to the correct processes : [2015]

(a) (b) (c) (d)

Catalyst Process (A) TiCl4 (i) Wacker process (B) PdCl2 (ii) Ziegler - Natta polymerization (C) CuCl2 (iii) Contact process (D) V2O5 (iv) Deacon's process (A) - (ii), (B) - (iii), (C) - (iv), (D) - (i) (A) - (iii), (B) - (i), (C) - (ii), (D) - (iv) (A) - (iii), (B) - (ii), (C) - (iv), (D) - (i) (A) - (ii), (B) - (i), (C) - (iv), (D) - (iii)

1.

Kraft temperature is the temperature:

(a) (b) (c) (d) 2.

[Main Sep. 06, 2020 (I)] below which the aqueous solution of detergents starts freezing. below which the formation of micelles takes place. above which the aqueous solution of detergents starts boiling. above which the formation of micelles takes place. Identify the correct molecular picture showing what happens at the critical micellar concentration (CMC) of an aqueous solution of a non-polar tail, • water). surfactant ( polar head, [Main Sep. 05, 2020 (I)]

(a) (D) (b) (B)

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(c) (A) (d) (C) 3. A sample of red ink (a colloidal suspension) is prepared by mixing eosin dye, egg white, HCHO and water. The component which ensures stabilitiy of the ink sample is : [Main Sep. 04, 2020 (II)] (a) Egg white (b) Water (c) HCHO (d) Eosin dye 4. Tyndall effect is observed when : [Main Sep. 03, 2020 (I)] (a) The diameter of dispersed particles is much larger than the wavelength of light used. (b) The diameter of dispersed particles is much smaller than the wavelength of light used. (c) The refractive index of dispersed phase is greater than that of the dispersion medium. (d) The diameter of dispersed particles is similar to the wavelength of light used. 5. An ionic micelle is formed on the addition of : [Main Sep. 03, 2020 (II)] (a) liquid diethyl ether to aqueous NaCl solution (b) excess water to liquid

excess water to liquid

(c) (d) sodium stearate to pure toluene

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6. (a) (b) (c) (d) 7.

(a) (b) (c) (d) 8.

Which of the following is used for the preparation of colloids ? [Main Sep. 02, 2020 (I)] Ostwald Process Van Arkel Method Bredig's Arc Method Mond Process The size of a raw mango shrinks to a much smaller size when kept in a concentrated salt solution. Which one of the following processes can explain this? [Main Sep. 02, 2020 (II)] Osmosis Dialysis Diffusion Reverse osmosis As per Hardy-Schulze formulation, the flocculation values of the following for ferric hydroxide sol are in the order: [Main Jan. 08, 2020 (I)]

(a) K3[Fe(CN)6] < K2CrO4 < KBr = KNO3 = AlCl3 (b) K3[Fe(CN)6] < K2CrO4 < AlCl3 < KBr < KNO3 (c) AlCl3 > K3[Fe(CN)6] > K2CrO4 > KBr = KNO3 (d) K3[Fe(CN)6] > AlCl3 > K2CrO4 > KBr > KNO3 9. (a) (b) (c) (d) 10. (a) (b)

Peptization is a : [Main April 12, 2019 (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 The correct option among the following is : [Main April 10, 2019 (II)] Colloidal medicines are more effective because they have small surface area. Addition of alum to water makes it unfit for drinking

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(c) Colloidal particles in lyophobic sols can be precipitated by electrophoresis. (d) Brownian motion in colloidal solution is faster if the viscosity of the solution is very high. 11. The aerosol is a kind of colloid in which: [Main April 9, 2019 (I)] (a) solid is dispersed in gas (b) gas is dispersed in solid (c) gas is dispersed in liquid (d) liquid is dispersed in water 12. Among the colloids cheese (C), milk (M) and smoke (S), the correct combination of the dispersed phase and dispersion medium, respectively is : [Main Jan. 11, 2019 (II)] (a) C : liquid in solid; M : liquid in solid; S : solid in gas (b) C : liquid in solid; M : liquid in liquid; S : solid in gas (c) C : solid in liquid; M : liquid in liquid; S : gas in solid (d) C : solid in liquid; M : solid in liquid; S : solid in gas 13. Haemoglobin and gold sol are examples of: [Main Jan. 10, 2019 (II)] (a) positively and negatively charged sols, respectively (b) positively charged sols (c) negatively charged sols (d) negatively and positively charged sols, respectively 14. For coagulation of arsenious sulphide sol, which one of the following salt solution will be most effective? [Main Jan. 9, 2019 (II)] (a) BaCl2 (b) A1Cl3 (c) NaCl (d) Na3PO4 15.

Which of the following statements about colloids is false? [Main Online April 15, 2018 (I)]

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(a) When silver nitrate solution is added to potassium iodide solution, a negatively charged colloidal solution is formed (b) Freezing point of colloidal solution is lower than true solution at same concentration of a solute (c) Colloidal particles can pass through ordinary filter paper (d) When excess of electrolyte is added to colloidal solution, colloidal particle will be precipitated 16. The Tyndall effect is observed only when following conditions are satisfied : [Main 2017] (i) The diameter of the dispersed particles is much smaller than the wavelength of the light used. (ii) The diameter of the dispersed particle is not much smaller than the wavelength of the light used. (iii) The refractive indices of the dispersed phase and dispersion medium are almost similar in magnitude. (iv) The refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude. (a) (i) and (iv) (b) (ii) and (iv) (c) (i) and (iii) (d) (ii) and (iii) 17. Among the following, correct statement is : [Main Online April 8, 2017] (a) Brownian movement is more pronounced for smaller particles than for bigger–particles. (b) Sols of metal sulphides are lyophilic. (c) Hardy Schulze law states that bigger the size of the ons, the greater is its coagulating power. (d) One would expect charcoal to adsorb chlorine more than hydrogen sulphide. 18. Gold numbers of some colloids are : Gelatin : 0.005 – 0.01, Gum arabic : 0.15 – 0.25; Oleate : 0.04 – 1.0; Starch : 15 – 25. Which among these is a better protective colloid ?

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[Main Online April 10, 2016] (a) (b) (c) (d) 19.

Gelatin Starch Oleate Gum arabic Under ambient conditions, which among the following surfactants will form micelles in aqueous solution at lowest molar concentration? [Main Online April 11, 2015]

(a) CH3–(CH2)8–COO – Na+ (b) CH3(CH2)11

(CH3)3Br–

(c) CH3–(CH2)13–OSO–3Na+ (d) CH3(CH2)15 20.

(CH3)3Br–

The coagulating power of electrolytes having ions Na+, Al3+ and Ba2+ for arsenic sulphide sol increases in the order [Main 2013]

(a) Al3+ < Ba2+ < Na+ (b) Na+ < Ba2+ < Al3+ (c) Ba2+ < Na+ < Al3+ (d) Al3+ < Na+ < Ba2+ 21.

(a) (b) (c) (d) 22. (a)

The migration of dispersion medium under the influence of an electric potential is called : [Main Online April 9, 2013] Cataphoresis Electroosmosis Electrophoresis Sedimentation Smoke is an example of : [Main Online April 23, 2013] Solid dispersed in solid

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(b) Gas dispersed in liquid (c) Solid dispersed in gas (d) Gas dispersed in solid 23. Among the electrolytes Na2SO4, CaCl2, Al2(SO4)3 and NH4Cl, the most effective coagulating agent for Sb2S3 sol is [2009S] (a) Na2SO4 (b) CaCl2 (c) Al2(SO4)3 (d) NH4Cl 24. Among the following, the surfactant that will form micelles in aqueous solution at the lowest molar concentration at ambient condition is : [2008S] + – (a) CH3(CH2)15N (CH3)3Br (b) CH3(CH2)11OSO–3Na+ (c) CH3(CH2)6COO–Na+ (d) CH3(CH2)11N+ (CH3)3Br– 25.

Lyophilic sols are [2005S]

(a) Irreversible sols (b) They are prepared from inorganic compound (c) Coagulated by adding electrolytes (d) Self−stabilizing 26.

The flocculation value of HCl for arsenic sulphide sol. is 30 m mol L– 1 . If H2SO4 is used for the flocculation of arsenic sulphide, the amount, in grams, of H2SO4 in 250 mL required for the above purpose is  [Main Jan. 07, 2020 (II)] (molecular mass of H2SO4 = 98 g/mol)

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27. (a) (b)

(c) (d) 28. (a) (b) (c) (d)

The correct statement(s) about surface properties is (are) [Adv. 2017] Adsorption is accompanied by decrease in enthalpy and decrease in entropy of the system 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 on same amount of activated charcoal at a given temperature Cloud is an emulsion type of colloid in which liquid is dispersed phase and gas is dispersion medium Brownian motion of colloidal particles does not depend on the size of the particles but depends on viscosity of the solution. Choose the correct reason(s) for the stability of the lyophobic colloidal particles. [2012] Preferential adsorption of ions on their surface from the solution. Preferential adsorption of solvent on their surface from the solution. Attraction between different particles having opposite charges on their surface. Potential difference between the fixed layer and the diffused layer of opposite charges around the colloidal particles.

Read the following statement (Assertion) and explanation (Reason) and answer the question as per the options given below : (a) If both assertion and reason are correct, and reason is the correct explanation of the assertion. (b) If both assertion and reason are correct, but reason is not the correct explanation of the assertion. (c) If assertion is correct but reason is incorrect. (d) If assertion is incorrect but reason is correct. 29. Assertion : Micelles are formed by surfactant molecules above the critical micellar concentration (CMC).

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Reason : The conductivity of a solution having surfactant molecules decreases sharply at the CMC. [2007]

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1.

The INCORRECT statement is : [Main Sep. 06, 2020 (I)]

(a) (b) (c) (d) 2.

(a) (b) (c) (d) 3. (a)

bronze is an alloy of copper and tin. cast iron is used to manufacture wrought iron. german silver is an alloy of zinc, copper and nickel. brass is an alloy of copper and nickel. For a reaction, 4 M(s) + nO2(g) → 2M2On(s), the free energy change is plotted as a function of temperature. The temperature below which the oxide is stable could be inferred from the plot as the point at which : [Main Sep. 06, 2020 (II)] the slope change from negative to positive the free energy change shows a change from negative to positive value the slop changes from positive to negative the slop changes from positive to zero An Ellingham diagram provides information about: [Main Sep. 05, 2020 (I)] the conditions of pH and potential under which a species is thermodynamically stable.

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(b) the temperature dependence of the standard Gibbs energies of formation of some metal oxides. (c) the pressure dependence of the standard electrode potentials of reduction reactions involved in the extraction of metals. (d) the kinetics of the reduction process. 4. Among statements (1) - (4), the correct ones are : [Main Sep. 04, 2020 (I)] (1) Lime stone is decomposed to CaO during the extraction of iron from its oxides. (2) In the extraction of silver, silver is extracted as an anionic complex. (3) Nickel is purified by Mond’s process. (4) Zr and Ti are purified by Van Arkel method. (a) (1), (2), (3) and (4) (b) (1), (3) and (4) only (c) (2), (3) and (4) only (d) (3) and (4) only 5. According to the following diagram, A reduces BO2 when the temperature is: [Main Jan. 09, 2020 (I)]

(a) (b) (c) (d) 6.

< 1400 °C > 1400 °C > 1200 °C but < 1400 °C < 1200 °C Among the reactions (a) - (d), the reaction(s) that does/do not occur in the blast furnace during the extraction of iron is/are: [Main Jan. 08, 2020 (II)] (a) CaO + SiO2 → CaSiO3 (b) 3Fe2O3 + CO → 2Fe3O4 + CO2

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(c) FeO + SiO2 → FeSiO3 (d) FeO → Fe + (a) (b) (c) (d) 7. (a) (b) (c) (d) 8. (a) (b) (c) (d) 9. (a) (b) (c) (d) 10. (a) (b) (c) (d) 11.

O2

(a) (a) and (d) (c) and (d) (d) The refining method used when the metal and the impurities have low and high melting temperatures, respectively, is: [Main Jan. 07, 2020 (II)] liquation vapour phase refining zone refining distillation 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 : [Main April 12, 2019 (I)] fisher woman and concentration washer woman and concentration fisher man and reduction washer man and reduction The correct statement is : [Main April 10, 2019 (II)] aniline is a froth stabilizer. zincite is a carbonate ore. sodium cyanide cannot be used in the metallurgy of silver. zone refining process is used for the refining of titanium. The ore that contains the metal in the form of fluoride is: [Main April 9, 2019 (I)] cryolite malachite magnetite sphalerite The one that is not a carbonate ore is: [Main April 9, 2019 (II)]

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(a) (b) (c) (d) 12. (a)

malachite calamine siderite bauxite The Mond process is used for the : purification of Ni [Main April 8, 2019 (II)]

(b) extraction of Mo (c) purification of Zr and Ti (d) extraction of Zn 13. In the Hall-Heroult process, aluminium is formed at the cathode. The cathode is made out of: [Main Jan. 12, 2019 (I)] (a) Pure aluminium (b) Carbon (c) Copper (d) Platinum 14. Hall-Heroult’s process is given by: [Main Jan. 10, 2019 (I)] (a) Cu2+ (aq) + H2(g) → Cu(s) + 2H+ (aq) (b) Cr2O3 + 2Al → Al2O3 + 2Cr (c) 2Al2O3 + 3C → 4Al + 3CO2 Zn + CO (d) ZnO + C 15. The ore that contains both iron and copper is: [Main Jan. 9, 2019 (I)] (a) (b) (c) (d) 16.

copper pyrites malachite dolomite azurite Calamine, malachite, magnetite and cryolite, respectively, are [Adv. 2019] (a) ZnCO3, CuCO3.Cu(OH)2, Fe3O4, Na3A1F6 (b) ZnSO4, Cu(OH)2, Fe3O4, Na3AlF6 (c) ZnSO4, CuCO3, Fe2O3, AlF3

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(d) ZnCO3, CuCO3, Fe2O3, Na3AlF6 17. In the leaching method, bauxite ore is digested with a concentrated solution of NaOH that produces 'X'. When CO2 gas is passed through the aqueous solution of 'X', a hydrated compound 'Y' is precipitated. 'X' and 'Y' respectively are. [Main Online April 15, 2018 (II)] (a) Na[Al(OH)4] and Al2(CO3)3·xH2O (b) Al(OH)3 and Al2O3·xH2O (c) NaAlO2 and Al2(CO3)3·xH2O (d) Na[Al(OH)4] and Al2O3·xH2O 18. Which one of the following ores is best concentrated by froth floatation method? [Main 2016] (a) Galena (b) Malachite (c)Magnetite (d) Siderite 19. Extraction of copper by smelting uses silica as an additive to remove : [Main Online April 10, 2016] (a) Cu2O (b) FeS (c) FeO (d) Cu2S 20. In the context of the Hall - Heroult process for the extraction of Al, which of the following statements is false ? [Main 2015] (a) Al3+ is reduced at the cathode to form Al (b) Na3AlF6 serves as the electrolyte (c) CO and CO2 are produced in this process (d) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity 21. Calamine is an ore of : [Main Online April 11, 2015] (a) zinc (b) aluminium(c) iron

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(d) copper 22. The form of iron obtained from blast furnace is: [Main Online April 9, 2014] (a) Steel (b) Cast Iron (c) Pig Iron (d) Wrought Iron 23. In Goldschmidt aluminothermic process which of the following reducing agents is used: [Main Online April 22, 2013] (a) calcium (b) coke (c) Al-powder (d) sodium 24. Sulfide ores are common for the metals [Adv. 2013] (a) Ag, Cu and Pb (b) Ag, Mg and Pb (c) Ag, Cu and Sn (d) Al, Cu and Pb 25. In the cyanide extraction process of silver from argentite ore, the oxidising and reducing agents used are (a) O2 and CO respectively [2012] (b) O2 and Zn dust respectively (c) HNO3 and Zn dust respectively (d) HNO3 and CO respectively 26. Oxidation states of the metal in the minerals haematite and magnetite, respectively, are [2011] (a) II, III in haematite and III in magnetite (b) II, III in haematite and II in magnetite (c) II in haematite and II, III in magnetite (d) III in haematite and II, III in magnetite 27. Extraction of zinc from zinc blende is achieved by

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(a) electrolytic reduction [2007] (b) roasting followed by reduction with carbon (c) roasting followed by reduction with another metal (d) roasting followed by self-reduction 28. Which ore contains both iron and copper? [2005S] (a) Cuprite (b) Chalcocite (c) Chalcopyrite (d) Malachite 29. In the process of extraction of gold, [2003S] Roasted gold ore Identify the complexes [X] and [Y] (a) X = [Au(CN)2]–, Y = [Zn(CN)4]2– (b) X = [Au(CN)4]3–, 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– 30. Which of the following process is used in the extractive metallurgy of magnesium? [2002S] (a) fused salt electrolysis (b) self reduction (c) aqueous solution electrolysis (d) thermite reduction 31. Electrolytic reduction of alumina to aluminium by Hall-Heroult process is carried out [2000S] (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

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(d) in the presence of cryolite which forms a melt with higher melting temperature 32. The chemical processes in the production of steel from haematite ore involve [2000S] (a) reduction (b) oxidation (c) reduction followed by oxidation (d) oxidation followed by reduction 33. The chemical composition of ‘slag’ formed during the smelting process in the extraction of copper is [2001S] (a) Cu2O + FeS (b) FeSiO3 (c) CuFeS2 (d) Cu2S + FeO 34.

(a) (b) (c) (d) 35. (a) (b) (c) (d) 36. (a) (b)

In the commercial electrochemical process for aluminium extraction the electrolyte used is [1999 - 2 Marks] Al(OH)3 in NaOH solution an aqueous solution of Al2(SO4)3. a molten mixture of Al2O3 and Na3AlF6 a molten mixture of AlO(OH) and Al(OH)3 In the alumino-thermite process, aluminium acts as [1983 - 1 Mark] an oxidizing agent a flux a reducing agent a solder In the metallurgy of iron, when limestone is added to the blast furnace, the calcium ion ends up in [1982 - 1 Mark] slag gangue

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(c) metallic calcium (d) calcium carbonate 37. Copper can be extracted from [1978] (a) (b) (c) (d)

Kupfernical Dolomite Malachite Galena

38.

In 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) [1988 - 1 Mark] 39. In the basic Bessemer process for the manufacture of steel, the lining of the converter is made of ........ . The slag formed consists of ...... [1980] 40. In the thermite process ........ is used as reducing agent. [1980] 41. Casseterite is ore of ........... [1980]

42. (a) (b) (c) (d) 43.

Which among the following statement(s) is(are) true for the extraction of aluminium from bauxite? [Adv. 2020] Hydrated Al2O3 precipitates, when CO2 is bubbled through a solution of sodium aluminate. Addition of Na3AlF6 lowers the melting point of alumina. CO2 is evolved at the anode during electrolysis. The cathode is a steel vessel with a lining of carbon. The cyanide process of gold extraction involves leaching out gold from its ore with CN– in the presence of Q in water to form R. Subsequently, R is treated with T to obtain Au and Z. Choose the correct option(s) [Adv. 2019]

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(a) (b) (c) (d) 44. (a) (b) (c) (d) 45. (a) (b) (c) (d) 46. (a) (b) (c) (d) 47. (a) (b) (c) (d) 48.

(a) (b)

Q is O2 T is Zn Z is [Zn(CN)4]2– R is [Au(CN)4]– Extraction of copper from copper pyrite (CuFeS2) involves [Adv. 2016] crushing followed by concentration of the ore by froth-flotation removal of iron as slag self-reduction step to produce ‘blister copper’ following evolution of SO2 refining of ‘blister copper’ by carbon reduction Upon heating with Cu2S, the reagent(s) that give copper metal is/are [Adv. 2014] CuFeS2 CuO Cu2O CuSO4 The carbon-based reduction method is NOT used for the extraction of [Adv. 2013] Tin from SnO2 Iron from Fe2O3 Aluminium from Al2O3 Magnesium from MgCO3.CaCO3 Extraction of metal from the ore cassiterite involves carbon reduction of an oxide ore [2011] self-reduction of a sulphide ore removal of copper impurity removal of iron impurity The major role of fluorspar (CaF2), which is added in small quantities in the electrolytic reduction of alumina dissolved in fused cryolite (Na3AlF6), is [1993 - 1 Mark] as a catalyst to make the fused mixture very conducting

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(c) to lower the temperature of the melt (d) to decrease the rate of oxidation of carbon at the anode. 49. Of the following, the metals that cannot be obtained by electrolysis of the aqueous solution of their salts are : [1990 - 1 Mark] (a) Ag (b) Mg (c) Cu (d) Al (e) Cr. 50. In the electrolysis of alumina, cryolite is added to : [1986 - 1 Mark] (a) lower the melting point of alumina (b) increase the electrical conductivity (c) minimise the anode effect (d) remove impurities from alumina 51.

Match the anionic species given in Column-I that are present in the ore(s) given in Column-II. [Adv. 2015] (A) (B) (C) (D)

52.

Column-I Carbonate Sulphide Hydroxide Oxide

(p) (q) (r) (s) (t)

Column-II Siderite Malachite Bauxite Calamine Argentite

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

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(D) Cu2S → Cu 53.

(s)

self reduction

Match the extraction processes listed in Column I with metals listed in Column II : [2006 - 6M]

Column I Column II (A) Self reduction (p) Lead (B) Carbon reduction (q) Silver (C) Complex formation and (r) Copper displacement by metal (D) Decomposition of iodide (s) Boron 54. Match the following choosing one item from column X and the appropriate item from column Y: [Multiple concept, 1986 - ½×8 = 4 Marks] X Y (i) Lewis acid (a) K electron capture (ii) Philosopher’s wool (b) Zinc ore (iii) Electrophile (c) HCHO (iv) Preservative (d) (v)

Electron emission

(e) Small proton to neutron ratio (vi) Bronsted acid (f) SO3 (vii) Black jack (g) BF3 (viii) X-ray emission (h) ZnO 55. Match the following, choosing one item from column X and the appropriate item from column Y. [1983 - 2 Marks] X Y (i) Al (a) Calamine (ii) Cu (b) Cryolite (iii) Mg (c) Malachite (iv) Zn (d) Carnallite

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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 (CuSO4.5H2O), atacamite (Cu2Cl(OH)3), cuprite (Cu2O), copper glance (Cu2S) and malachite (Cu2(OH)2CO3). 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] 56. Partial roasting of chalcopyrite produces (a) Cu2S and FeO (b) Cu2O and FeO (c) CuS and Fe2O3 (d) Cu2O and Fe2O3 57. Iron is removed from chalcopyrite as (a) FeO (b) FeS (c) Fe2O3 (d) FeSiO3 58. In self-reduction, the reducing species is (a) S (b) O2– (c) S2– (d) SO2 59.

Some reactions of two ores, A1 and A2 of the metal M are given below. [2004 - 4 Marks]

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Identify A1, A2, M, C, D, and G, and explain using the required chemical reactions. 60. Write down reactions involved in the extraction of Pb. What is the oxidation number of lead in litharge? [2003 - 2 Marks] 61. 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 - 4 Marks] 62. State with balanced equations what happens when : (i) Write balanced equations for the extraction of copper from copper pyrites by self-reduction. [1990 - 2 Marks] (ii) Write balanced equations for the extraction of silver from silver glance by cyanide process. [1988 - 1 Mark] 63. Give the equations for the recovery of lead from Galena by air reduction. [1987 - 1 Mark] 64. What is the actual reducing agent of haematite in blast furnace? (1987 - 1 Mark) 65. Give reasons for the following : (i) Why is chalcocite roasted and not calcinated during recovery of copper? [1987 - 1 Mark] (ii) Metals can be recovered from their ores by chemical methods. [1984 - 1 Mark] 66. Give balanced equations for extraction of silver from its sulphide ore [1982 - 2 Marks] 67. Give balanced equations for the extraction of aluminium from bauxite by electrolysis.

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[1982 - 2 Marks] 68. Write the matching pairs: [1980] Bleaching agent Aluminium Smelling salt Carbon Cryolite Tin Bell metal Ammonium carbonate Fluorspar Ammonium phosphate Fertilizer Calcium Anthracite Chlorine Examples : Bleaching agent – Chlorine Smelling salt – Ammonium carbonate 69. (a) Write the chemical equations involved in the extraction of lead from galena by self reduction process. (b) Match the following extraction processes with the appropriate metals listed below : (i) Silver (A) Fused salt electrolysis (ii) Calcium (B) Carbon reduction (iii) Zinc (C) Carbon monoxide reduction (iv) Iron (D) Amalgamation (v) Copper (E) Self reduction [1979]

1. (a) (b) (c) (d) 2. (a)

The element that can be refined by distillation is : [Main Sep. 06, 2020 (II)] nickel zinc tin gallium Boron and silicon of very high purity can be obtained through : [Main Sep. 05, 2020 (II)] liquation

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(b) zone refining (c) vapour phase refining (d) electrolytic refining 3. Cast iron is used for the manufacture of : [Main Sep. 02, 2020 (II)] (a) (b) (c) (d) 4.

wrought iron and pig iron pig iron, scrap iron and steel wrought iron, pig iron and steel wrought iron and steel The purest form of commercial iron is: [Main Jan. 07, 2020 (I)]

(a) (b) (c) (d) 5. (a) (b) (c) (d) 6. (a) (b) (c) (d) 7.

pig iron wrought iron cast iron scrap iron and pig iron The metal that cannot be obtained by electrolysis of an aqueous solution of its salts is: [Main 2014] Ag Ca Cu Cr Copper is purified by electrolytic refining of blister copper. The correct statement(s) about this process is(are) [Adv. 2015] Impure Cu strip is used as cathode Acidified aqueous CuSO4 is used as electrolyte Pure Cu deposits at cathode Impurities settle as anode-mud Match the refining methods (Column I) with metals (Column II). [Main April 10, 2019 (I)] Column I Column II (Refining methods) (Metals)

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(a) (b) (c) (d) 8. 9.

(I) Liquation (a) Zr (II) Zone Refining (b) Ni (III) Mond Process (c) Sn (IV) Van Arkel Method (d) Ga (I) – (c); (II) – (a); (III) – (b); (IV) – (d) (I) – (b); (II) – (c); (III) – (d); (IV) – (a) (I) – (c); (II) – (d); (III) – (b); (IV) – (a) (I) – (b); (II) – (d); (III) – (a)l (IV) – (c) Magnesium oxide is used for the lining of steel making furnace. [1987 - 1 Mark] High purity metals can be obtained by zone refining method. [1984 - 1 Mark] Topic-1 : Occurrence of Metals and Metallurgical Processes 1. (d) 2. (b) 3. (b)

4. (a) 5. (b) 6. (c) 7. (a) 8. (b) 9. (a) 10. (a) 11. (d) 12.(a) 13. (b) 14. (c)

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15. (a) 16. (a) 17. (d)18. (a) 19. (c) 20. (b) 21. (a) 22. (c) 23.(c) 24. (a) 25. (b) 26. (d) 27. (b) 28. (c) 29. (a) 30. (a) 31. (c) 32. (c) 33. (b) 34. (c) 35. (c) 36. (a) 37. (c) 42. (a, b, c, d) 43. (a, b, c) 44. (a, b, c) 45. (b, c, d) 46. (c, d)

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47. (a, c, d) 48. (b, c) 49. (b, d) 50. (a, b) 51. A-(p, q, s), B-(t), C-(q, r), D-(r) 52. (A) - p; (B) - q; (C) - p, r; (D) - p, s 53. (A)- p, r; (B) - p, r; (C) - q; (D) - s 55. (c) 56. (b) 57. (d) 58. (c) Topic-2 : Purification and Uses of Metal 1. (b) 2. (b) 3. (d) 4. (b) 5. (b) 6. (b, c, d) 7. (c)

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1. (a) (b) (c) (d) 2. (a) (b) (c) (d) 3. (a) (b) (c) (d) 4.

The correct statement with respect to dinitrogen is: [Main Sep. 06, 2020 (I)] N2 is paramagnetic in nature. it can combine with dioxygen at 25 °C. liquid dinitrogen is not used in cryosurgery. it can be used as an inert diluent for reactive chemicals. The reaction of NO with N2O4 at 250 K gives : [Main Sep. 06, 2020 (II)] N2O NO2 N2O3 N2O5 Reaction of ammonia with excess Cl2 gives : [Main Sep. 05, 2020 (II)] NH4Cl and N2 NH4Cl and HCl NCl3 and NH4Cl NCl3 and HCl On heating, lead (II) nitrate gives a brown gas (A). The gas (A) on cooling changes to a colourless solid/liquid (B). (B) on heating with

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(a) (b) (c) (d) 5. (a) (b) (c) (d) 6. (a) (b) (c) (d) 7.

(a) (b) (c) (d) 8.

(a) (b)

NO changes to a blue solid (C). The oxidation number of nitrogen in solid (C) is : [Main Sep. 04, 2020 (I)] +5 +2 +3 +4 Aqua regia is used for dissolving noble metals (Au, Pt, etc.). The gas evolved in this process is : [Main Sep. 03, 2020 (I)] NO N2O5 N2 N2O3 In a molecule of pyrophosphoric acid, the number of P – OH, P = O and P – O – P bonds/ moiety(ies) respectively are : [Main Sep. 03, 2020 (I)] 2, 4 and 1 3, 3 and 3 4, 2 and 0 4, 2 and 1 On heating compound (A) gives a gas (B) which is a constituent of air. This gas when treated with H2 in the presence of a catalyst gives another gas (C) which is basic in nature. (A) should not be : [Main Sep. 02, 2020 (I)] NaN3 Pb(NO3)2 (NH4)2Cr2O7 NH4NO2 White phosphorus on reaction with concentrated NaOH solution in an inert atmosphere of CO2 gives phosphine and compound (X). (X) on acidification with HCl gives compound (Y). The basicity of compound (Y) is: [Main Jan. 08, 2020 (II)] 2 1

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(c) 4 (d) 3 9. The number of pentagons in C60 and trigons (triangles) in white phosphorous, respectively, are : [Main April 10, 2019 (II)] (a) 20 and 3 (b) 12 and 4 (c) 12 and 3 (d) 20 and 4 10. The correct order of the oxidation states of nitrogen in NO, N2O, NO2 and N2O3 is: [Main April 9, 2019 (I)] (a) NO2 < NO < N2O3 < N2O (b) NO2 < N2O3 < NO < N2O (c) N2O < N2O3 < NO < NO2 (d) N2O < NO < N2O3 < NO2 11. The pair that contains two P–H bonds in each of the oxoacids is: [Main Jan. 10, 2019 (II)] (a) H4P2O5 and H4P2O6 (b) H3PO2 and H4P2O5 (c) H3PO3 and H3PO2 (d) H4P2O5 and H3PO3 12. Good reducing nature of H3PO2 is attributed to the presence of: [Main Jan. 9, 2019 (II)] (a) Two P — OH bonds (b) One P — H bond (c) Two P — H bonds (d) One P — OH bond 13. The compound that does not produce nitrogen gas by the thermal decomposition is : [Main 2018] (a) Ba(N3)2 (b) (NH4)2Cr2O7 (c) NH4NO2 (d) (NH4)2SO4

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14.

The number of P – O bonds in P4O6 is: [Main Online April 15, 2018 (II)]

(a) (b) (c) (d) 15.

9 6 12 18 A metal ‘M’ reacts with nitrogen gas to afford ‘M3N’. ‘M3N’ on heating at high temperature gives back ‘M’ and on reaction with water produces a gas ‘B’. Gas ‘B’ reacts with aqueous solution of CuSO4 to form a deep blue compound. ‘M’ and ‘B’ respectively are : [Main Online April 8, 2017] Li and NH3 Ba and N2 Na and NH3 Al and N2 The order of the oxidation state of the phosphorus atom in H3PO2, H3PO4, H3PO3 and H4P2O6 is [Adv. 2017] H3PO3 > H3PO2 > H3PO4 > H4P2O6 (b) H3PO4 > H3PO2 > H3PO3 > H4P2O6 H3PO4 > H4P2O6 > H3PO3 > H3PO2 (d) H3PO2 > H3PO3 > H4P2O6 > H3PO4 The pair in which phosphorus atoms have a formal oxidation state of + 3 is : [Main 2016] Orthophosphorous and hypophosphoric acids Pyrophosphorous and pyrophosphoric acids Orthophosphorous and pyrophosphorous acids Pyrophosphorous and hypophosphoric acids Which of the following compound has a P–P bond ? [Main Online April 11, 2015] H4P2O5 (HPO3)3 H4P2O6 H4P2O7

(a) (b) (c) (d) 16.

(a) (c) 17. (a) (b) (c) (d) 18. (a) (b) (c) (d)

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19. (a) (b) (c) (d) 20.

(a) (b) (c) (d) 21. (a) (b) (c) (d) 22.

(a) (b) (c) (d) 23. (a) (b) (c)

Which one of the following does not have a pyramidal shape? [Main Online April 11, 2014] (CH3)3N (SiH3)3 N P(CH3)3 P(SiH3)3 The product formed in the reaction of SOCl2 with white phosphorous is [Adv. 2014] PCl3 SO2Cl2 SCl2 POCl3 Concentrated nitric acid, upon long standing, turns yellow brown due to the formation of [Adv. 2013] NO NO2 N2O N2O4 The reaction of white phosphorus with aqueous NaOH gives phosphine along with another phosphorus containing compound. The reaction type; the oxidation states of phosphorus in phosphine and the other product are respectively [2012] redox reaction; – 3 and – 5 redox reaction; + 3 and + 5 disproportionation reaction; – 3 and + 5 disproportionation reaction; – 3 and + 3 Which ordering of compounds is according to the decreasing order of the oxidation state of nitrogen? [2012] HNO3, NO, NH4Cl, N2 HNO3, NO, N2, NH4 Cl HNO3, NH4Cl, NO, N2

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24.

(d) NO, HNO3, NH4Cl, N2 Extra pure N2 can be obtained by heating

(a) NH3 with CuO (b) NH4NO3 (c) (NH4)2Cr2O7 (d) Ba(N3)2 25. The reaction of P4 with X leads selectively to P4O6. The X is (a) (b) (c) (d) 26.

(a) (b) (c) (d) 27. (a) (b) (c) (d) 28. (a) (b) (c) (d) 29.

[2011]

[2009]

Dry O2 A mixture of O2 and N2 Moist O2 O2 in the presence of aqueous NaOH The percentage of π-character in the orbitals forming P – P bonds in P4 is [2007] 25 33 50 75 When PbO2 reacts with conc. HNO3, the gas evolved is [2005S] NO2 O2 N2 N2O Blue liquid which is obtained on reacting equimolar amounts of two gases at −30°C is? [2005S] N2O N2O3 N2O4 N2O5 Which is the most thermodynamically stable allotropic form of phosphorus?

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[2005S] (a) (b) (c) (d) 30. (a) (b) (c) (d) 31.

red white black yellow For H3PO3 and H3PO4, the correct choice is:

[2003S]

H3PO3 is dibasic and reducing H3PO3 is dibasic and non-reducing H3PO4 is tribasic and reducing H3PO3 is tribasic and non-reducing Ammonia can be dried by [2000S]

(a) (b) (c) (d) 32. (a) (b) (c) (d) 33.

(a) (b) (c) (d) 34. (a) (b) (c) (d)

conc. H2SO4 P4O10 CaO anhydrous CaCl2 The number of P – O – P bonds in cyclic metaphosphoric acid is [2000S] zero two three four In compounds of type ECl3, where E = B, P, As or Bi, the angles Cl – E– Cl for different E are in the order [1999 - 2 Marks] B > P = As = Bi B > P > As > Bi B < P = As = Bi B < P < As < Bi On heating ammonium dichromate, the gas evolved is [1999 - 2 Marks] oxygen ammonia nitrous oxide nitrogen

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35. (a) (b) (c) (d) 36. (a) (b) (c) (d) 37. (a) (b) (c) (d) 38. (a) (b) (c) (d) 39. (a) (b) (c) (d) 40. (a) (b) (c) (d)

One mole of calcium phosphide on reaction with excess water gives [1999 - 2 Marks] one mole of phosphine two moles of phosphoric acid two moles of phosphine one mole of phosphorus pentoxide In P4O10 each P atom is linked with .......... O atoms [1995S] 2 3 4 5 Amongst the trihalides of nitrogen which one is least basic? [1987 - 1 Mark] NF3 NCl3 NBr3 NI3 Which of the following oxides of nitrogen is a coloured gas? [1987 - 1 Mark] N2O NO N2O5 NO2 The bonds present in N2O5 are : (1986 - 1 Mark) only ionic covalent and coordinate only covalent covalent and ionic Nitrogen dioxide cannot be obtained by heating : [1985 - 1 Mark] KNO3 Pb(NO3)2 Cu(NO3)2 AgNO3

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41.

Ammonia can be dried by [1980]

(a) (b) (c) (d) 42.

(a) (b) (c) (d) 43. (a) (b) (c) (d) 44.

Conc. H2SO4 P2O5 Anhydrous CuSO4 none White P reacts with caustic soda. The products are PH3 and NaH2PO2. This reaction is an example of [1980] Oxidation Reduction oxidation and reduction Neutralisation The reddish brown coloured gas formed when nitric oxide is oxidised by air is [1979] N2O5 N2O4 NO2 N2O3 The total number of lone pairs of electrons in N2O3 is

[Adv. 2015] 45. Among the following, the number of compounds than can react with PCl5 to give POCl3 is [2011] O2, CO2, SO2, H2O, H2SO4, P4O10 46.

In P4O10, the number of oxygen atoms bonded to each phosphorus atom is ............... . [1992 - 1 Mark] 47. The basicity of phosphorous acid (H3PO3) is ............... . [1990 - 1 Mark] 48. .............. phosphorus is reactive because of its highly strained tetrahedral structure.

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[1987 - 1 Mark] 49.

The lowest possible oxidation state of nitrogen is [1980]

50.

Nitric oxide, though an odd electron molecule, is diamagnetic in liquid state. [1991 - 1 Mark] 51. The H-N-H bond angle in NH3 is greater than the H-As-H bond angle is AsH3. [1984 - 1 Mark] 52. Red phosphorus is less volatile than white phosphorus because the former has a tetrahedral structure. [1982 - 1 Mark]

53. (a) (b) (c) (d) 54.

(a) (c) 55.

(a) (b) (c) (d) 56.

Based on the compounds of group 15 elements, the correct statement(s) is (are) [Adv. 2018] Bi2O5 is more basic than N2O5 NF3 is more covalent than BiF3 PH3 boils at lower temperature than NH3 The N–N single bond is stronger than the P–P single bond The compound(s) which generate(s) N2 gas upon thermal decomposition below 300 °C is (are) [Adv. 2018] NH4NO3 (b) (NH4)2Cr2O7 Ba(N3)2 (d) Mg3N2 The nitrogen containing compound produced in the reaction of HNO3 with P4O10 [Adv. 2016] can also be prepared by reaction of P4 and HNO3 is diamagnetic contains one N-N bond reacts with Na metal producing a brown gas The nitrogen oxide(s) that contain(s) N-N bond(s) is(are) [2009]

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(a) (b) (c) (d) 57.

(a) 58. (a) (b) (c) (d) 59. (a) (b) (c) (d) 60. (a) (b) (c) (d) 61. (a) (b) (c) (d)

N2O N2O3 N2O4 N2O5 A solution of colourless salt H on boiling with excess NaOH produces a non-flammable gas. The gas evolution ceases after sometime. Upon addition of Zn dust to the same solution, the gas evolution restarts. The colourless salt (s) H is (are) [2008] NH4NO3 (b) NH4NO2 (c) NH4Cl (d) (NH4)2SO4 Ammonia, on reaction with hypochlorite anion, can form [1999 - 3 Marks] NO NH4Cl N2H4 HNO2 White phosphorus (P4) has [1998 - 2 Marks] six P-P single bonds four P-P single bonds four lone pairs of electrons PPP angle of 60° Sodium nitrate decomposes above 800° C to give [1998 - 2 Marks] N2 O2 NO2 Na2O Nitrogen(I) oxide is produced by : [1989 - 1 Mark] thermal decomposition of ammonium nitrate disproportionation of N2O4 thermal decomposition of ammonium nitrite interaction of hydroxylamine and nitrous acid.

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62.

Match the following, choosing one item from column X and the appropriate item from column Y. [1983 - 2 Marks] X Y (i) Haber (a) Activation energy (ii) Graham (b) Diffusion of gases (iii) Arrhenius (c) Octet rule (iv) Lewis (d) Ammonia synthesis

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. Hybridization easily explains the ease of sigma donation capability of NH3 and PH3. Phosphine is a flammable gas and is prepared from white phosphorous. [2008] 63. Among the following, the correct statement is (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 64. 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

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65. (a) (b) (c) (d) 66.

(a) (b) (c) (d) 67.

White phosphorus on reaction with NaOH gives PH3 as one of the products. This is a dimerization reaction disproportionation reaction condensation reaction precipitation reaction Statement-1 : Although and are known, the pentahalides of nitrogen have not been observed Statement-2 : Phosphorus has lower electronegativity than nitrogen. [1994 - 2 Marks] Statement-1 is True, Statement-2 isTrue; Statement-2 is a correct explanation for Statement-1 Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 Statement-1 is True, Statement-2 is False Statement-1 is False, Statement-2 is True. Write the structures of isostructural? Justify your answer.

and

.

Are they

[2005 - 2 Marks] ? 68. How many grams of CaO are required to neutralize 852 g of molecule. Draw structure of [2005 - 2 Marks] 69. Give reason(s) why elemental nitrogen exists as a diatomic molecule whereas elemental phosphorus as a tetraatomic molecule. [2000 - 2 Marks] 70. The Haber process can be represented by the following scheme;

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Identify A, B, C, D and E. [1999 - 5 Marks] 71. Reaction of phosphoric acid with Ca5(PO4)3F yields a fertilizer “triple superphosphate”. Represent the same through balanced chemical equation. [1998 - 2 Marks] 72. A soluble compound of a poisonous element M, when heated with Zn/H2SO4, 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 - 2 Marks] 73. Draw the structure of P4O10 and identify the number of single and double P—O bonds. [1996 - 3 Marks] 74. Complete and balance the following chemical reactions : (i) Red phosphorus is reacted with iodine in presence of water. [1992 - 1 Mark] P + I2 + H2O → ........... + ........... . (ii) Anhydrous potassium nitrate is heated with excess of metallic potassium. [1992 - 1 Mark] KNO3(s) + K(s) → ........... + ........... . 75. Arrange the following in : [1991 - 1 Mark] Increasing order of extent of hydrolysis :

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CCl4, MgCl2, AlCl3, PCl5, SiCl4 76. Write down the balanced equations for the reactions when: (i) a mixture of potassium chlorate, oxalic acid and sulphuric acid is heated; [1985 - 1 Mark] (ii) ammonium sulphate is heated with a mixture of nitric oxide and nitrogen dioxide. [1985 - 1 Mark] 77. Write down the resonance structures of nitrous oxide. [1985 - 2 Marks] OR Write the two resonance structures of N2O that satisfy the octet rule. [1990 - 1 Mark] 78. State with balanced equations what happens when : (i) P4O10+ PCl5 [1998 - 1 Mark] (ii) Phosphorus is treated with concentrated nitric acid. [1997 - 1 Mark] OR Manufacture of phosphoric acid from phosphorus. [1997 - 1 Mark] (iii) Elemental phosphorus reacts with conc. HNO3 to give phosphoric acid. [1991 - 1 Mark] (iv) Nitrogen is obtained in the reaction of aqueous ammonia with potassium permanganate. [1991 - 1 Mark] (v) Sodium nitrite is produced by absorbing the oxides of nitrogen in aqueous solution of washing soda. [1991 - 1 Mark] (vi) Aqueous solution of sodium nitrate is heated with zinc dust and caustic soda solution. [1990 - 1 Mark] (vii) Write balanced equations for the preparation of ammonium sulphate from gypsum, ammonia and carbon dioxide. [1990 - 1 Mark] (viii)Write balanced equations for the preparation of phosphine from CaO and white phosphorus.

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[1990 - 2 Marks] (ix) Hypophosphorous acid is heated. [1989 - 1 Mark] (x) Phosphorus reacts with nitric acid to give equimolar ratio of nitric oxide and nitrogen dioxide. [1988 - 1 Mark] (xi) Dilute nitric acid is slowly reacted with metallic tin. [1987 - 1 Mark] (xii) White phosphorous (P4) is boiled with a strong solution of sodium hydroxide in an inert atmosphere. [1982/87 - 1 Mark] 79. Give reasons for the following : (i) The experimentally determined N – F bond length in NF3 is greater than the sum of the single covalent bond radii of N and F. [1995 - 2 Marks] (ii) Ammonium chloride is acidic in liquid ammonia solvent. [1991 - 1 Mark] (iii) Phosphine has lower boiling point than ammonia. [1989 - 1 Mark] (iv) H3PO3 is a dibasic acid. [1989 - 1 Mark] (v) Orthophosphorus acid is not tribasic acid. [1987 - 1 Mark] (vi) A bottle of liquor ammonia should be cooled before opening the stopper. [1983 - 1 Mark] (vii) Orthophosphoric acid, H3PO4, is tribasic, but phosphorous acid, H3PO3, is dibasic. [1982 - 1 Mark] 80. Give structural formula for the following : (i) Phosphorous acid, H3PO3 [1981 - 1 Mark] (ii) Pyrophosphoric acid, H4P2O7 [1981 - 1 Mark]

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1.

(a) (b) (c) (d) 2. (a) (b) (c) (d) 3.

(a) (b) (c) (d) 4. (a) (b) (c) (d) 5.

(a) (b)

Reaction of an inorganic sulphite X with dilute H2SO4 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: [Main Sep. 06, 2020 (II)] SO2 and Na2SO3 SO3 and NaHSO3 SO2 and NaHSO3 S and Na2SO3 If the boiling point of H2O is 373 K, the boiling point of H2S will be : [Main Sep. 03, 2020 (I)] less than 300 K equal to 373 K more than 373 K greater than 300 K but less than 373 K The number of bonds between sulphur and oxygen atoms in and the number of bonds between sulphur and sulphur atoms in rhombic sulphur, respectively, are: [Main Jan. 08, 2020 (I)] 4 and 6 8 and 8 8 and 6 4 and 8 The oxoacid of sulphur that does not contain bond between Sulphur atoms is : [Main April 10, 2019 (I)] H2S4O6 H2S2O3 H2S2O7 H2S2O4 In KO2 , the nature of oxygen species and the oxidation state of oxygen atom are, respectively: [Main Online April 15, 2018 (II)] Superoxide and –1 Superoxide and –1/2

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(c) Peroxide and –1/2 (d) Oxide and –2 6. The number of S=O and S–OH bonds present in peroxodisulphuric acid and pyrosulphuric acid respectively are : [Main Online April 8, 2017] (a) (2 and 2) and (2 and 2) (b) (2 and 4) and (2 and 4) (c) (4 and 2) and (2 and 4) (d) (4 and 2) and (4 and 2) 7. Identify the incorrect statement : [Main Online April 10, 2016] (a) The S–S–S bond angles in the S8 and S6 rings are the same. (b) Rhombic and monoclinic sulphur have S8 molecules. (c) S2 is paramagnetic like oxygen (d) S8 ring has a crown shape. 8. In the following reaction sequence in aqueous solution, the species X, Y and Z, respectively, are [Adv. 2016]

(a) (b) (c) (d) 9.

[Ag(S2O3)2]3–, Ag2S2O3, Ag2S [Ag(S2O3)3]5–, Ag2SO3, Ag2S [Ag(SO3)2]3–, Ag2S2O3, Ag [Ag(SO3)3]3–, Ag2SO4, Ag Aqueous solution of Na2S2O3 on reaction with Cl2 gives –

(a) Na2S4O6 (b) NaHSO4 (c) NaCl (d) NaOH 10. Which of the following is not oxidized by O3 ? (a) (b) (c) (d)

[2008]

[2005S]

KI FeSO4 KMnO4 K2MnO4

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11.

The acid having O – O bond is [2004S]

(a) (b) (c) (d) 12.

(b) (c) (d) 13. (a) (b) (c) (d) 14. (a) (b) (c) (d) 15. (a) (b) (c) (d) 16. (a) (b)

H2S2O3 H2S2O6 H2S2O8 H2S4O6 The number of S − S bonds in sulphur trioxide trimer (S3O9) is [2001S] (a) three two one zero Sodium thiosulphate is prepared by [1996 - 1 Mark] reducing Na2SO4 solution with H2S boiling Na2SO3 solution with S in alkaline medium neutralising H2S2O3 solution with NaOH boiling Na2SO3 solution with S in acidic medium Hydrolysis of one mole of peroxodisulphuric acid produces two moles of sulphuric acid [1996 - 1 Mark] two moles of peroxomonosulphuric acid one mole of sulphuric acid and one mole of peroxomonosulphuric acid one mole of sulphuric acid, one mole of peroxomonosulphuric acid and one mole of hydrogen peroxide. H2SO4 cannot be used to prepare HBr from NaBr as it : [1995S] reacts slowly with NaBr oxidises HBr reduces HBr disproportionates HBr The compound which gives off oxygen on moderate heating is : [1986 - 1 Mark] cupric oxide mercuric oxide

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(c) zinc oxide (d) aluminium oxide 17. A gas that cannot be collected over water is : [1985 - 1 Mark] (a) (b) (c) (d) 18.

N2 O2 SO2 PH3 Which of the following is coloured [1980]

(a) (b) (c) (d)

NO N2O SO3 None

19.

The total number of compounds having at least one bridging oxo group among the molecules given below is______. N2O3, N2O5, P4O6, P4O7, H4P2O5, H5P3O10, H2S2O3, H2S2O5 [Adv. 2018]

20.

The amount of water produced (in g) in the oxidation of 1 mole of rhombic sulphur by conc.HNO3 to a compound with the highest oxidation state of sulphur is ____ (Given data : Molar mass of water = 18g mol–1) [Adv. 2019]

21.

The lead chamber process involves oxidation of SO2 by atomic oxygen under the influence of ............... as catalyst. [1992 - 1 Mark]

22.

The correct statement(s) about O3 is(are)

(a) O—O bond lengths are equal (b) Thermal decomposition of O3 is endothermic

[Adv. 2013-II]

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(c) O3 is diamagnetic in nature (d) O3 has a bent structure 23. Which of the following halides react(s) with AgNO3(aq) to give a precipitate that dissolves in Na2S2O3(aq)? [2012] (a) HCl (b) HF (c) HBr (d) HI PASSAGE -1 Upon heating KClO3 in the presence of catalytic amount of MnO2, a gas W is formed. Excess amount of W reacts with white phosphorus to give X. The reaction of X with pure HNO3 gives Y and Z. [Adv. 2017] 24. W and X are, respectively (A) O3 and P4O6 (B) O2 and P4O6 (C) O2 and P4O10 (D) O3 and P4O10 25. Y and Z are, respectively (A) N2O3 and H3PO4 (B) N2O5 and HPO3 (C) N2O4 and HPO3 (D) N2O4 and H3PO3 PASSAGE - 2 The reactions of Cl2 gas with cold-dilute and hot-concentrated NaOH in water give sodium salts of two (different) oxoacids of chlorine, P and Q, respectively. The Cl2 gas reacts with SO2 gas, in 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. [Adv. 2013] 26. P and Q, respectively, are the sodium salts of (a) Hypochlorus and chloric acids (b) Hypochlorus and chlorus acids

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(c) (d) 27. (a) (b)

Chloric and perchloric acids Chloric and hypochlorus acids R, S and T respectively, are SO2Cl2, PCl5 and H3PO4 SO2Cl2, PCl3 and H3PO3

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(c) SOCl2, PCl3 and H3PO2 (d) SOCl2, PCl5 and H3PO4 28.

(a) (b) (c) (d)

Assertion: Nitrogen and oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen. Reason: The reaction between nitrogen and oxygen requires high temperature. [Main 2015] The assertion is incorrect, but the reason is correct Both the assertion and reason are incorrect Both assertion and reason are correct, and the reason is the correct explanation for the assertion Both assertion and reason are correct, but the reason is not the correct explanation for the assertion

29.

Identify the missing compounds. Give the equation from A to B and A to C. [2005 - 4 Marks] 30. Identify the following: [2003 - 4 Marks] Also mention the oxidation state of S in all the compounds. 31. 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 - 4 Marks] 32. In the following equation, [1999 - 6 Marks] C + 2D A + 2B + H2O (A = HNO2, B = H2SO3, C = NH2OH). Identify D. Draw the structures of A, B, C and D.

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33.

Thionyl chloride can be synthesized by chlorinating SO2 using PCI5. 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 with2, 2 – dimethoxypropane. Discuss all this using balanced chemical equations. [1998 - 6 Marks]

34.

PbS

A + PbS

Pb + SO2; Identify A and B.

[1991 - 2 Marks] 35. Write the two resonance structures of ozone which satisfy the octet rule. [1991 - 1 Mark] 36. Mention the products formed in the following : Sulphur dioxide gas, water vapour and air are passed over heated sodium chloride. [1986 - 1 Mark] 37. Arrange the following in : CO2, N2O5, SiO2, SO3 in the order of increasing acidic character. [1988 - 1 Mark] 38. What happens when : (i) hydrogen sulphide is bubled through an aqueous solution of sulphur dioxide. [1985 - 1 Mark] (ii) tin is treated with concentrated nitric acid. [1985 - 1 Mark] (iii) Pb3O4 is treated with nitric acid. [1985 - 1 Mark] 39. Show with equations, how the following compound is prepared (equations need not be balanced) : sodium thiosulphate from sodium sulphite. [1982 - 1 Mark] 40. State with balanced equations what happens when : Sulphur is precipitated in the reaction of hydrogen sulphide with sodium bisulphite solution. [1991 - 1 Mark]

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41. Give reasons for the following : (i) Sulphur dioxide is a more powerful reducing agent in an alkaline medium than in acidic medium. [1992 - 1 Mark] (ii) Valency of oxygen is generally two whereas sulphur shows valency of two, four and six. [1988 - 1 Mark] (iii) Sulphur melts to a clear mobile liquid at 119ºC, but on further heating above 160ºC, it becomes viscous. [1981 - 1 Mark]

1.

Arrange the following bonds according to their average bond energies in descending order: C – Cl,C – Br, C – F, C – I [Main Jan. 08, 2020 (II)] (a) C – F > C – Cl > C – Br > C – I (b) C – Br > C – I > C – C1 > C – F (c) C – I > C – Br > C – Cl > C – F (d) C – Cl > C – Br > C – I > C – F 2. In the following reactions, products (A) and (B), respectively, are: [Main Jan. 07, 2020 (II)] NaOH + Cl2 → (A) + side products (hot and conc.) Ca(OH)2 + Cl2 → (B) + side products (dry) (a) NaClO3, and Ca(OCl)2 (b) NaClO3 and Ca(ClO3)2 (c) NaOCl and Ca(OCl)2 (d) NaOCl and Ca(ClO3)2 3. A colorless 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 aq. NH3 and also in excess sodium thiosulfate. The hot solution Q gave a yellow precipitate with KI. The metals X and Y, respectively, are [Adv. 2020]

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(a) (b) (c) (d) 4. (a) (b) (c) (d) 5.

(a)

Ag and Pb Ag and Cd Cd and Pb Cd and Zn HF has highest boiling point among hydrogen halides, because it has: [Main April 9, 2019 (II)] strongest van der Waals’ interactions lowest ionic character strongest hydrogen bonding lowest dissociation enthalpy Chlorine on reaction with hot and concentrated sodium hydroxide gives : [Main Jan. 12, 2019 (II)] Cl– and

(b) Cl– and ClO– (c) ClO3– and (d) Cl– and 6. (a) (b) (c) (d) 7. (a) (b) (c) (d) 8. (a)

The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are : [Main 2017] – – ClO and ClO3 ClO–2and ClO–3 Cl– and ClO– Cl– and ClO–2 Aqueous solution of which salt will not contain ions with the electronic configuration 1s22s22p63s23p6? [Main Online April 10, 2016] NaF KBr NaCl CaI2 Which among the following is the most reactive? [Main 2015] I2

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(b) IC1 (c) Cl2 (d) Br2 9. Chlorine water on standing loses its colour and forms : [Main Online April 11, 2015] (a) HCl only (b) HCl and HClO2 (c) HCl and HOCl (d) HOCl and HOCl2 10. Shapes of certain interhalogen compounds are stated below. Which one of them is not correctly stated? [Main Online April 11, 2014] (a) IF7 : pentagonal bipyramid (b) BrF5 : trigonal bipyramid (c) BrF3 : planar T-shaped (d) ICI3 : planar dimeric 11. Electron gain enthalpy with negative sign of fluorine is less than that of chlorine due to : [Main Online April 9, 2013] (a) High ionization enthalpy of fluorine (b) Smaller size of chlorine atom (c) Smaller size of fluorine atom (d) Bigger size of 2p orbital of fluorine 12. Which one of the following species is not a pseudohalide? [1997 - 1 Mark] – (a) CNO (b) RCOO– (c) OCN– (d) NNN– 13. KF combines with HF to form KHF2. The compound contains the species. [1996 - 1 Mark] (a) K+, F– and H+ (b) K+, F– and HF (c) K+ and [HF2]–

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(d) [KHF]+ and F– 14. Which of the following statements is correct for CsBr3? (a) It is a covalent compound. [1996 - 1 Mark] (b) It contains Cs3+ and Br – ions. (c) It contains Cs+ and Br3–ions (d) It contains Cs+, and Br – and lattice Br2 molecule 15. Bromine can be liberated from potassium bromide solution by the action of [1987 - 1 Mark] (a) Iodine solution (b) Chlorine water (c) Sodium chloride (d) Potassium iodide 16. Chlorine acts as a bleaching agent only in presence of [1983 - 1 Mark] (a) dry air (b) moisture (c) sunlight (d) pure oxygen 17. HBr and HI reduce sulphuric acid, HCl can reduce KMnO4 and HF can reduce [1981 - 1 Mark] (a) H2SO4 (b) KMnO4 (c) K2Cr2O7 (d) none of the above 18. A solution of KBr is treated with each of the following. Which one would liberate bromine [1980] (a) Cl2 (b) HI (c) I2 (d) SO2 19. Which of the following is most stable to heat

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[1980] (a) (b) (c) (d)

HCl HOCl HBr HI

20. The number of Cl = O bonds in perchloric acid is, "______." [Main Sep. 06, 2020 (I)] 21. Reaction of Br2 with Na2CO3in 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] 22.

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 ______. [Main Jan. 07, 2020 (I)]

23.

Solubility of iodine in water is greatly increased by the addition of iodide ions because of the formation of ........ [1994 - 1 Mark] 24. ............... acid gives hypo ............... ion. [1988 - 1 Mark] (hydrobromic, hypobromous, perbromic, bromide, bromite, perbromate) 25. The increase in the solubility of iodine in an aqueous solution of potassium iodide is due to the formation of ............. . [1982 - 1 Mark] 26. ................. is a weak acid. (HF, HCl, HI) [1981 - 1 Mark] 27. Iodine reacts with hot NaOH solution. The products are NaI and ....... [1980]

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28.

HBr is a stronger acid than HI because of hydrogen bonding. [1993 - 1 Mark] 29. In aqueous solution, chlorine is a stronger oxidizing agent than fluorine. [1984 - 1 Mark] 2+ 30. Dil. HCl oxidizes metallic Fe to Fe . [1983 - 1 Mark] 31. (a) (b) (c) (d) 32.

(a) (b) (c) (d) 33.

(a) (b) (c) (d) 34.

With respect to hypochlorite, chlorate and perchlorate ions, choose the correct statement(s). The hypochlorite ion is the strongest conjugate base. The molecular shape of only chlorate ion is influenced by the lone pair of electrons of Cl. The hypochlorite and chlorate ions disproportionate to give rise to identical set of ions. The hypochlorite ion oxidizes the sulfite ion. The colour of the X2 molecules of group 17 elements changes gradually from yellow to violet down the group. This is due to [Adv. 2017] The physical state of X2 at room temperature changes from gas to solid down the group Decrease in ionization energy down the group Decrease in π* - σ* gap down the group Decrease in HOMO-LUMO gap down the group The correct statement(s) about the oxoacids, HClO4 and HClO, is (are) [Adv. 2017] 3 The central atom in both HClO4 and HClO is sp hybridized HClO4 is more acidic than HClO because of the resonance stabilization of its anion HClO4 is formed in the reaction between Cl2 and H2O The conjugate base of HClO4 is weaker base than H2O The compounds used as refrigerant are [1990 - 1 Mark]

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(a) (b) (c) (d) (e)

NH3 CCl4 CF4 CF2Cl2 CH2F2

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 - II] 35. Bleaching powder contains a salt of an oxoacid as one of its components. The anhydride of that oxoacid is (a) Cl2O (b) Cl2O7 (c) ClO2 (d) Cl2O6 36. 25 mL of household solution was mixed with 30 mL of 0.50 M KI and 10 mL of 4N acetic acid. In the titration of the liberated iodine, 48 mL of 0.25 N Na2S2O3 was used to reach the end point. The molarity of the household bleach solution is (a) 0.48 M (b) 0.96 M (c) 0.24 M (d) 0.024 M 37.

(a) (b) (c) (d)

Statement-1 : F atom has less electron affinity than Cl atom. Statement-2 : Additional electrons are repelled more effectively by 3p electrons in Cl atom than by 2p electrons in F atom [1998 - 2 Marks] Statement-1 is True, Statement-2 isTrue; Statement-2 is a correct explanation for Statement-1 Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 Statement-1 is True, Statement-2 is False Statement-1 is False, Statement-2 is True.

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38.

Give an example of oxidation of one halide by another halogen. Explain the feasibility of the reaction. [2000 - 2 Marks] 39. Complete the following chemical equations : (a) (b) Justify the formation of the products in the above reactions. [1996 - 2 Marks] 40. Gradual addition of KI solution to Bi(NO3)3solution initially produces a dark brown precipitate which dissolves in excess of KI to give a clear yellow solution. Write chemical equations for the above reactions. [1996 - 2 Marks] 41. Complete and balance the following chemical reactions : (iii) NH3 + NaOCl → ........ + ........ [1993 - 1 Mark] 42. Mention the products formed in the following : (i) Chlorine gas is bubbled through a solution of ferrous bromide. [1986 - 1 Mark] (ii) Iodine is added to a solution of stannous chloride. [1986 - 1 Mark] 43. Arrange the following in : (i) HOCl, HOClO2, HOClO3, HOClO in increasing order of thermal stability. [1988 - 1 Mark] (ii) increasing bond strength [1986 - 1 Mark] HCl, HBr, HF, HI 44. State with balanced equations what happens when : (i) Sodium iodate is added to a solution of sodium bisulphite. [1990 - 1 Marks] (ii) Sodium chlorate reacts with sulphur dioxide in dilute sulphuric acid medium. [1989 - 1 Mark]

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(iii) Sodium bromate reacts with fluorine in presence of alkali. [1989 - 1 Mark] (iv) Iodate ion reacts with bisulphite ion to liberate iodine. [1988 - 1 Mark] (v) Sodium iodate is treated with sodium bisulphite solution. [1982 - 1 Mark] 45. Give reasons for the following : (i) Bond dissociation energy of F2 is less than that of Cl2. [1992 - 1 Mark] (ii) Fluorine cannot be prepared from fluorides by chemical oxidation. [1985 - 1 Mark] (iii) Anhydrous HCl is a bad conductor of electricity but aqueous HCl is a good conductor; [1985 - 1 Mark] (iv) In the preparation of hydrogen iodide from alkali iodides, phosphoric acid is preferred to sulphuric acid [1982 - 1 Mark] 46. Write balanced equation involved in the preparation of (i) Anhydrous aluminium chloride from alumina. (ii) Bleaching powder from slaked lime. (iii) Tin metal from cassiterite (iv) Chlorine from sodium chloride. [1979] 47. Account for the following. Limit your answer to two sentences (i) Hydrogen bromide cannot be prepared by action of concentrated sulphuric acid or 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]

1.

The reaction in which the hybridisation of the underlined atom is affected is : [Main Sep. 04, 2020 (II)]

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(a) (b) (c) (d) 2. (a) (b) (c) (d) 3.

(a) (b) (c) (d) 4.

(a) (b) (c) (d) 5. (a) (b) (c) (d) 6. (a)

The noble gas that does NOT occur in the atmosphere is : [Main April 10, 2019 (II)] He Kr Ne Ra The type of hybridisation and number of lone pair(s) of electrons of Xe in XeOF4, respectively, are: [Main Jan. 10, 2019 (I)] 3 2 sp d and 1 sp3d and 2 sp3d2 and 2 sp3d and 1 In XeO3F2, the number of bond pair(s), π-bond(s) and lone pair(s) on Xe atom respectively are: [Main Online April 15, 2018 (II)] 5, 3, 0 5, 2, 0 4, 2, 2 4, 4, 0 Which intermolecular force is most responsible in allowing xenon gas to liquefy? [Main Online April 9, 2016] Instantaneous dipole–induced dipole Ion–dipole Ionic Dipole – dipole Which one has the highest boiling point ? [Main 2015] Kr

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(b) Xe (c) He (d) Ne 7. The geometry of XeOF4 by VSEPR theory is : [Main Online April 10, 2015] (a) pentagonal planar (b) octahedral (c) square pyramidal (d) trigonal bipyramidal 8. Which of the following xenon-oxo compounds may not be obtained by hydrolysis of xenon fluorides? [Main Online April 12, 2014] (a) XeO2F2 (b) XeOF4 (c) XeO3 (d) XeO4 9. Under ambient conditions, the total number of gases released as products in the final step of the reaction scheme shown below is [Adv. 2014]

(a) (b) (c) (d) 10. (a) (b) (c) (d)

0 1 2 3 The shape of XeO2F2 molecule is trigonal bipyramidal square planar tetrahedral see-saw

[2012 ]

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11. (a) (b) (c) (d) 12.

Total number of lone pair of electrons with central atom in XeOF4 is [2004S] 0 1 2 3 At 143 K, the reaction of XeF4 with O2F2, produces a xenon compound Y. The total number of lone pair(s) of electron present on the whole molecule of Y is ______ [Adv. 2019]

13.

The unbalanced chemical reactions given in List I show missing reagent or condition (?) which are provided in List II. Match List I with List II and select the correct answer using the code given below the lists [Adv. 2013-II] List I List II 1. NO P. PbO2 + H2SO4 PbSO4 + O2 + other product 2. I2 Q. Na2S2O3 + H2O NaHSO4 + other product N2 + other product 3. Warm R. N2H4 Xe + other product 4. Cl2 S. XeF2 Codes : P Q R S (a) 4 2 3 1 (b) 3 2 1 4 (c) 1 4 2 3 (d) 3 4 2 1 14. 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.

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[2010] Column-I (A) (CH3)2SiCl2 (B) XeF4 (C) Cl2 (D) VCl5

(p) (q) (r) (s) (t)

Column-II Hydrogen halide formation Redox reaction Reacts with glass Polymerization O2 formation

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 given 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. 15. Argon is used in arc welding because of its [2007] (a) low reactivity with metal (b) ability to lower the melting point of metal (c) flammability (d) high calorific value 16. The structure of XeO3 is [2007] (a) linear (b) planar (c) pyramidal (d) T-shaped 17. XeF4 and XeF6 are expected to be [2007] (a) oxidizing (b) reducing

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(c) unreactive (d) strongly basic 18. (i) (ii) (iii) (iv) (v) 19.

Write balanced equations for the reactions of the following compounds with water : [2002 - 5 Marks] Al4C3 CaNCN BF3 NCl3 XeF4 Draw the molecular structures of XeF2, XeF4 and XeO2F2 indicating the location of lone pair(s) of electrons. [2000 - 3 Marks] Topic-1 : Group-15 Elements (Nitrogen Family)

1. (d) 2. (c) 3. (d) 4. (c) 5. (a) 6. (d) 7. (b) 8. (b) 9. (b) 10. (d) 11. (b) 12. (c) 13. (d) 14. (c) 15. (a) 16. (c) 17. (c) 18. (c)

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19. (b) 20. (a) 21. (b) 22. (c) 23. (b) 24. (d) 25. (b) 26. (d) 27. (b) 28. (b) 29. (c) 30. (a) 31. (c) 32. (c) 33. (b) 34. (d) 35. (c) 36. (c) 37. (a) 38. (d) 39. (b) 40. (a) 41. (d) 42. (c) 43. (c) 44. (8) 45. (4) 50. (True) 51. (True) 52. (False) 53. (a,b,c) 54. (b, c) 55. (b, d) 56. (a,b,c) 57. (a, b) 58. (c)

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59. (a,c,d) 60. (a,b,d) 61. (a, d) 62. (i) (d) (ii)(b) 64. (c) 65. (b) 66. (b)

(iii) (a) (iv) (c) 63.(c)

Topic-2 : Group-16 Elements (Oxygen Family) 1. (c) 2. (a) 3. (b) 4. (c) 5. (b) 6. (d) 7. (a) 8. (a) 9. (b) 10. (c) 11. (c) 12. (d) 13. (b) 14. (c) 15. (b) 16. (b) 17. (c) 18. (d) 19. (6) 20. (288) 22. (a,c,d) 23. (a,c,d) 24. (c) 25. (b) 26. (a) 27. (a) 28. (c)

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Topic-3 : Group-17 Elements (Halogen Family) 1. (a) 2. (a) 3. (a) 4. (c) 5. (a) 6. (c) 7. (a) 8. (b) 9. (c) 10. (b) 11. (c) 12. (b) 13. (c) 14. (c) 15. (b) 16. (b) 17. (d) 18. (a) 19. (a) 20. (3) 21. (5) 22. (1.67) 28. (False) 29. (False) 30. (True) 31. (a,b,d) 32. (c, d) 33. (a,b,d) 34. (a,d) 35. (a) 36. (c) 37. (c) Topic-4 : Group-18 Elements (Nobel Gases)

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1. (d) 2. (d) 3. (a) 4. (a) 5. (a) 6. (b) 7. (c) 8. (d) 9. (c) 10. (d) 11. (b) 12. (19) 13. (d) 14. (A) – (p), (s); (B) – (p), (q), (r), (t); (C) – (p), (q); (D) – (p), (q) 15. (a) 16. (c) 17. (a)

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1. (1) (2) (3) (a) (b) (c) (d) 2. (a) (b) (c) (d) 3.

The incorrect statement(s) among (1) - (3) is (are) : [Main Sep. 04, 2020 (II)] W(VI) is more stable than Cr(VI). in the presence of HCl, permanganate titrations provide satisfactory results. some lanthanoid oxides can be used as phosophorus. (2) and (3) only (1) and (2) only (2) only (1) only The incorrect statement is : [Main Sep. 03, 2020 (II)] Manganate and permanganate ions are tetrahedral In manganate and permanganate ions, the π-bonding takes place by overlap of p-orbitals of oxygen and d-orbitals of manganese Manganate and permanganate ions are paramagnetic Manganate ion is green in colour and permanganate ion is purple in colour The third ionization enthalpy is minimum for:

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[Main Jan. 08, 2020 (I)] (a) (b) (c) (d) 4.

Co Fe Ni Mn The atomic radius of Ag is closest to: [Main Jan. 07, 2020 (I)]

(a) (b) (c) (d) 5.

Au Ni Cu Hg 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 H2SO4 to give a pungent gas Z. X, Y, and Z, respectively, are : [Main April 12, 2019 (II)] (a) KMnO4, K2MnO4 and Cl2 (b) K2MnO4, KMnO4 and SO2 (c) K3MnO4, K2MnO4 and Cl2 (d) K2MnO4, KMnO4 and Cl2 6. Consider the hydrated ions of Ti2+, V2+, Ti3+, and Sc3+. The correct order of their spin-only magnetic moments is: [Main April 10, 2019 (I)] 2+ 2+ 3+ 3+ (a) V < Ti < Ti < Sc (b) Sc3+ < Ti3+ < Ti2+ < V2+ (c) Ti3+ < Ti2+ < Sc3+ < V2+ (d) Sc3+ < Ti3+ < V2+ < Ti2+ 7. The INCORRECT statement is : [Main April 10, 2019 (II)] (a) the gemstone, ruby, has Cr3+ ions occupying the octahedral sites of beryl. (b) the spin-only magnetic moment of [Ni(NH3)4(H2O)2]2+ is 2.83 BM. (c) the color of [CoCl(NH3)5]2+ is violet as it absorbs the yellow light.

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(d) the spin-only magnetic moments of [Fe(H2O)6]2+ and [Cr(H2O)6]2+ are nearly similar. 8. The statement that is INCORRECT about the interstitial compounds is : [Main April 8, 2019 (II)] (a) they are chemically reactive. (b) they are very hard. (c) they have metallic conductivity. (d) they have high melting points. 9. The element that usually does NOT show variable oxidation states is: [Main Jan. 11, 2019 (I)] (a) Cu (b) Ti (c) Sc (d) V 10.

In the above sequence of reactions, A and D, respectively, are: [Main Jan. 11, 2019 (II)] (a) KI and KMnO4 (b) MnO2 and KIO3 (c) KIO3 and MnO2 (d) KI and K2MnO4 11. The transition element that has lowest enthalpy of atomisation is: [Main Jan. 9, 2019 (II)] (a) Fe (b) Cu (c) V (d) Zn 12. When XO2 is fused with an alkali metal hydroxide in presence of an oxidizing agent such as KNO3, a dark green product is formed which

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(a) (b) (c) (d) 13. (i) (ii) (a) (b) (c) (d) 14.

(a) 15. (a) (b) (c) (d) 16. (a) (b) (c) (d) 17.

disproportionates in acidic solution to afford a dark purple solution. X is: [Main Online April 16, 2018] Mn Cr V Ti In the following reactions, ZnO is respectively acting as a/an: [Main 2017] ZnO + Na2O → Na2ZnO2 ZnO + CO2 → ZnCO3 base and acid base and base acid and acid acid and base Which of the following ions does not liberate hydrogen gas on reaction with dilute acids ? [Main Online April 9, 2017] Ti2+ (b) V2+ (c) Cr2+ (d) Mn2+ Which of the following combination will produce H2 gas? [JEE Adv. 2017] Fe metal and conc. HNO3 Cu metal and conc. HNO3 Zn metal and NaOH(aq) Au metal and NaCN(aq) in the presence of air The reaction of zinc with dilute and concentrated nitric acid, respectively, produces: [Main 2016] NO and N2O NO2 and N2O N2O and NO2 NO2 and NO Which of the following compounds is metallic and ferromagnetic? [Main 2016]

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(a) (b) (c) (d) 18. (a) (b) (c) (d) 19.

(a) (b) (c) (d) 20.

VO2 MnO2 TiO2 CrO2 The transition metal ions responsible for colour in ruby and emerald are, respectively : [Main Online April 10, 2016] 3+ 3+ Co and Cr Co3+ and Co3+ Cr3+ and Cr3+ Cr3+ and Co3+ When concentrated HCl is added to an aqueous solution of CoCl2, its colour changes from reddish pink to deep blue. Which complex ion gives blue colour in this reaction? [Main Online April 11, 2015] 2– [CoCl4] [CoCl6]3– [CoCl6]4– [Co(H2O)6]2+ The equation which is balanced and represents the correct product(s) is: [Main 2014]

(a) (b) (c)

(d) 21.

Which series of reactions correctly represents chemical reactions related to iron and its compound? [Main 2014]

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(a)

(b)

(c) (d) 22.

(a) (b) (c) (d) 23. (a) (b) (c) (d) 24.

(a) (b) (c) (d) 25.

Which one of the following exhibits the large number of oxidation states? [Main Online April 12, 2014] Ti (22) V (23) Cr (24) Mn (25) Which of the following is not formed when H2S reacts with acidic K2Cr2O7 solution?[Main Online April 9, 2014] CrSO4 Cr2(SO4)3 K2SO4 S Which of the following arrangements does not represent the correct order of the property stated against it ? [Main 2013] V2+ < Cr2+ < Mn2+ < Fe2+ : paramagnetic behaviour Ni2+ < Co2+ < Fe2+ < Mn2+ : ionic size Co3+ < Fe3+ < Cr3+ < Sc3+ : stability in aqueous solution Sc < Ti < Cr < Mn : number of oxidation states Four successive members of the first row transition elements are listed below with atomic numbers. Which one of them is expected to value ? have the highest

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[Main 2013] (a) (b) (c) (d) 26.

(a) (b) (c) (d) 27.

(a) (b) (c) (d) 28. (a) (b) (c) (d) 29. (a) (c) 30.

Cr(Z = 24) Mn(Z = 25) Fe(Z = 26) Co(Z = 27) When a small amount of KMnO4 is added to concentrated H2SO4, a green oily compound is obtained which is highly explosive in nature. Compound may be : [Main Online April 23, 2013] MnSO4 Mn2O7 MnO2 Mn2O3 The element with which of the following outer electron configuration may exhibit the largest number of oxidation states in its compounds : [Main Online April 9, 2013] 5 2 3d 4s 3d84s2 3d74s2 3d64s2 The colour of light absorbed by an aqueous solution of CuSO4 is: [2012] orange-red blue-green yellow violet Among the following, the coloured compound is [2008] CuCl (b) K3 [Cu (CN)4] CuF2 (d) [Cu (CH3CN)4]BF4 Native silver metal forms a water soluble complex with a dilute aqueous solution of NaCN in the presence of [2008]

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(a) (b) (c) (d) 31. (a) (b) (c) (d) 32.

(b) (c) (d) 33.

(a) (b) (c) (d) 34.

nitrogen oxygen carbon dioxide argon CuSO4 decolourises on addition of KCN, the product formed is [2006 - 3M, –1] Cu2+ get reduced to form [Cu(CN)4]3– [Cu(CN)4]2– CuCN Cu(CN)2 Which pair of compounds is expected to show similar colour in aqueous medium? [2005S] (a) FeCl2 and CuCl2 VOCl2 and CuCl2 VOCl2 and FeCl2 FeCl2 and MnCl2 (NH4)2Cr2O7 on heating liberates a gas. The same gas will be obtained by [2004S] heating NH4NO2 heating NH4NO3 treating H2O2 with NaNO2 treating Mg3N2 with H2O in alkaline medium is The product of oxidation of I– with [2004S]

(a) (b) I2 (c) IO– (d) 35. When MnO2 is fused with KOH, a coloured compound is formed, the product and its colour is:

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[2003S] (a) K2MnO4, purple green (b) KMnO4 , purple (c) Mn2O3, brown (d) Mn3O4, black 36. (a) (b) (c) (d) 37. (a) (b) (c) (d) 38. (a) (b) (c) (d) 39.

Anhydrous ferric chloride is prepared by [2002S] heating hydrated ferric chloride at a high temperature in a stream of air heating metallic iron in a stream of dry chlorine gas reaction of metallic iron with hydrochloric acid reaction of metallic iron with nitric acid In the dichromate anion, [1999 - 2 Marks] 4 Cr – O bonds are equivalent 6 Cr – O bonds are equivalent all Cr – O bonds are equivalent all Cr – O bonds are nonequivalent Which of the following compounds is expected to be coloured? [1997 - 1 Mark] Ag2SO4 CuF2 MgF2 CuCl. The number of moles of KMnO4 that will be needed to react with one mole of sulphite ion in acidic solution is [1997 - 1 Mark]

(a) (b)

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(c) (d) 1 40. (a) (b) (c) (d) 41.

(a) (b) (c) (d) 42. (a) (b) (c) (d) 43.

(a)

Ammonium dichromate is used in some fireworks. The green coloured powder blown in the air is [1997 - 1 Mark]

CrO3 Cr2O3 Cr CrO(O2) An aqueous solution of FeSO4, Al2(SO4)3 and chrome alum is heated with excess of Na2O2 and filtered. The materials obtained are : [1996 - 1 Mark] a colourless filtrate and a green residue a yellow filtrate and a green residue a yellow filtrate and a brown residue a green filtrate and a brown residue Which compound does not dissolve in hot, dilute HNO3? [1996 - 1 Mark] HgS PbS CuS CdS Which compound is formed when excess of KCN is added to aqueous solution of copper sulphate? [1996 - 1 Mark] Cu(CN)2

(b) K2 [Cu(CN)4] (c) K [Cu(CN)2] (d) K3[Cu(CN)4] 44.

Which pair gives Cl2 at room temperature? [1995S]

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(a) HCl(conc) + KMnO4 (b) NaCl + H2SO4(conc) (c) NaCl + MnO2 (d) NaCl + HNO3(conc) (a) Cu(CN)2 (b) K2 [Cu(CN)4] (c) K [Cu(CN)2] (d) K3[Cu(CN)4] 45.

Which one is solder ? [1995S]

(a) (b) (c) (d) 46.

(a) (b) (c) (d) 47. (a) (b) (c) (d) 48. (a)

Cu & Pb Zn & Cu Pb & Sn Fe & Zn Zinc-copper couple that can be used as a reducing agent is obtained by : [1984 - 1 Mark] mixing zinc dust and copper gauze zinc coated with copper copper coated with zinc zinc and copper wires welded together Iron is rendered passive by treatment with concentrated [1982 - 1 Mark] H2SO4 H3PO4 HCl HNO3 Sodium thiosulphate is used in photography because of its [1981 - 1 Mark] reducing behaviour

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(b) oxidising behaviour (c) complex forming behaviour (d) reaction with light 49. How many unpaired electrons are present in Ni2+? [1981 - 1 Mark] (a) (b) (c) (d) 50.

0 2 4 8 Which of the following dissolve in hot conc. NaOH solution [1980]

(a) (b) (c) (d) 51.

Fe Zn Cu Ag One of the constituent of German silver is [1980]

(a) (b) (c) (d) 52.

Ag Cu Mg Al Which of the following is the weakest base [1980]

(a) (b) (c) (d) 53.

NaOH Ca(OH)2 KOH Zn(OH)2 When same amount of zinc is treated separately with excess of sulphuric acid and excess of sodium hydroxide, the ratio of volume of hydrogen evolved is [1979] (a) 1 : 1 (b) 1 : 2

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(c) 2 : 1 (d) 9 : 4 54.

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% H2O2, a blue alcohol layer was obtained. The blue color is due to the formation of a chromium (VI) compound ‘X’. What is the number of oxygen atoms bonded to chromium through only single bonds in a molecule of X? 55. In neutral or faintly alkaline solution, 8 moles of permanganate anion quantitatively oxidize thiosulphate anions to produce X moles of a sulphur containing product. The magnitude of X is [Adv. 2016] 56. Consider the following list of reagents: [Adv. 2014] Acidified K2Cr2O7, alkaline KMnO4, CuSO4, H2O2, Cl2, O3, FeCl3, HNO3 and Na2S2O3. The total number of reagents that can oxidise aqueous iodide to iodine is 57.

The oxidation number of Mn in the product of alkaline oxidative fusion of MnO2 is [2009 - 4 Marks]

58.

The sum of the total number of bonds between chromium and oxygen atoms in chromate and dichromate ions is _____. [Main Jan. 09, 2020 (II)] 59. Consider the following reactions: NaCl + K2Cr2O7 + H2SO4 (Conc.) → (A) + Side products (A) + NaOH → (B) + Side products (B) + H2SO4 (dilute) + H2O2 → (C) + Side products The sum of the total number of atoms in one molecule each of (A), (B) and (C) is 

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[Main Jan. 07, 2020 (II)]

60.

Silver jewellery items tarnish slowly in the air due to their reaction with ............... . [1997 - 1 Mark] 61. The salts ............... and ............... are isostructural. (FeSO4.7H2O, CuSO4.5H2O, MnSO4.4H2O, ZnSO4.7H2O)

63. 64.

[1988 - 1 Mark] 62. Silver chloride is sparingly soluble in water because its lattice energy is greater than ........... energy. [1987 - 1 Mark] Galvanization of iron denotes coating with ............... . [1983 - 1 Mark] by ................. . Mn2+ can be oxidised to Mn

(SnO2, PbO2, BaO2) 65.

[1981 - 1 Mark] AgCN dissolves in excess KCN solution to give the complex compound ........... [1980]

66.

Dipositive zinc exhibits paramagnetism due to loss of two electrons from 3d-orbital of neutral atom. [1987 - 1 Mark] 67. Silver chloride is more soluble in very concentrated sodium chloride solution than in pure water. [1984 - 1 Mark] 68. Silver fluoride is fairly soluble in water. [1982 - 1 Mark]

69. Fusion of MnO2 with KOH in presence of O2 produces a salt W. Alkaline solution of W upon electrolytic oxidation yields another salt

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X. The manganese containing ions present in W and X, respectively are Y and Z. Correct statement(s) is (are) [ Adv. 2019] (a) In both Y and Z, p-bonding occurs between p-orbitals of oxygen and dorbitals of manganese (b) In aqueous acidic solution, Y undergoes dispro-portionation reaction to give Z and MnO2 (c) Both Y and Z are coloured and have tetrahedral shape (d) Y is diamagnetic in nature while Z is paramagnetic 70. Consider the following reactions (unbalanced) Zn + hot conc. H2SO4 G+R+X [Adv. 2019] Zn + conc. NaOH T + Q G + H2S + NH4OH Z (a precipitate) + X + Y Choose the correct option(s) (a) The oxidation state of Zn in T is + 1 (b) Bond order of Q is 1 in its ground state (c) Z is dirty white in colour (d) R is a V-shaped molecule 71. Fe3+ is reduced to Fe2+ by using [Adv. 2015] (a) H2O2 in presence of NaOH (b) Na2O2 in water (c) H2O2 in presence of H2SO4 (d) Na2O2 in presence of H2SO4 72. The correct statement(s) about Cr2+ and Mn3+ is (are) [Atomic numbers of Cr = 24 and Mn = 25] [Adv. 2015] 2+ (a) Cr is a reducing agent (b) Mn3+ is an oxidizing agent (c) Both Cr2+ and Mn3+ exhibit d 4 electronic configuration (d) When Cr2+ is used as a reducing agent, the chromium ion attains d 5 electronic configuration

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73. The pair(s) of reagents that yield paramagnetic speciesis/are [Adv. 2014] (a) Na and excess of NH3 (b) K and excess of O2 (c) Cu and dilute HNO3 (d) O2 and 2-ethylanthraquinol 74. For the given aqueous reactions, which of the statement (s) is (are) true? excess

KI

+

K3[Fe(CN)6]

brownish-

yellow solution ZnSO4

white precipitate + Na2S2O3

colourless solution (a) The first reaction is a redox reaction. [2012] (b) (c) (d) 75.

White precipitate is Zn3[Fe(CN)6]2. Addition of filtrate to starch solution gives blue colour. White precipitate is soluble in NaOH solution. The equilibrium [2011]

in aqueous medium at 25°C shifts towards the left in the presence of (a) (b) Cl– (c) SCN– (d) CN– 76. Reduction of the metal centre in aqueous permanganate ion involves [2011]

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(a) (b) (c) (d) 77.

3 electrons in neutral medium 5 electrons in neutral medium 3 electrons in alkaline medium 5 electrons in acidic medium Addition of high proportions of manganese makes steel useful in making rails of railroads, because manganese [1998 - 2 Marks]

(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. 78. Which of the following alloys contain(s) Cu and Zn? [1993 - 1 Mark] (a) (b) (c) (d) 79. (a) (b) (c) (d) (e) 80. (a) (b) (c) (d)

Bronze Brass Gun metal Type metal The aqueous solutions of the following salts will be coloured in the case of [1990 - 1 Mark] Zn(NO3)2 LiNO3 Co(NO3)2 CrCl3 Potash alum Potassium manganate (K2MnO4) is formed when [1988 - 1 Mark] chlorine is passed into aqueous KMnO4 solution manganese dioxide is fused with potassium hydroxide in air formaldehyde reacts with potassium permanganate in presence of a strong alkali potassium permanganate reacts with conc. sulphuric acid

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81. Match the catalysts (Column I) with products (Column II). [Main April 9, 2019 (I)] Column I Column II Catalyst Product (A) V2O5 (i) Polyethylene (B) TiCl4/ Al(Me) 3 (ii) ethanol (C) PdCl2 (iii) H2SO4 (D) Iron Oxide (iv) NH3 (a) (A)-(iii); (B)-(iv); (C)-(i); (D)-(ii) (b) (A)-(ii); (B)-(iii); (C)-(i); (D)-(iv) (c) (A)-(iii); (B)-(i); (C)-(ii); (D)-(iv) (d) (A)-(iv); (B)-(iii); (C)-(ii); (D)-(i) 82. Match each of the reactions given in Column I with the corresponding product(s) given in Column II. [2009] Column I Column II (A) Cu + dil HNO3 (p) NO (B) Cu + conc HNO3 (q) NO2 (C) Zn + dil HNO3 (r) N2O (D) Zn + conc HNO3 (s) Cu(NO3)2 (t) Zn(NO3)2 83. Match the following, choosing one item from column X and one from column Y. [Multiple Concepts, 1982 - 3 Marks] X Y (i) Hg2Cl2 (a) cassiterite (ii) (NaPO3)n (b) lunar caustic (c) producer gas (iii) (iv) SnO2 (v) KCl.MgCl2.6H2O (vi) AgNO3

(d) water softener (e) brown ring test (f) carnallite

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(vii) CO + N2

(g) calomel

When a metal rod M is dipped into an aqueous colourless concentrated solution of compound N, the solution turns light blue. Addition of aqueous NaCl to the blue solution gives a white precipitate O. Addition of aqueous NH3 dissolves O and gives an intense blue solution. [2011] 84. The metal rod M is (a) Fe (b) Cu (c) Ni (d) CO 85. The compound N is (a) AgNO3 (b) Zn(NO3)2 (c) Al(NO3)3 (d) Pb(NO3)2 86. The final solution contains

Read the following statement-1(Asseration/Statement) and Statement -2 (Reason/Explanation) and answer as per the options given below : (a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

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(c) (d) 87.

(a) (b) (c) (d) 88.

89.

Statement-1 is True, Statement-2 is False Statement-1 is False, Statement-2 is True For the following Assertion and Reason, the correct option is: Assertion: For hydrogenation reactions, the catalytic activity increases from Group 5 to Group 11 metals with maximum activity shown by Group 7-9 elements. Reason: The reactants are most strongly adsorbed on group 7-9 elements. [Main Jan. 08, 2020 (II)] The assertion is true, but the reason is false. Both assertion and reason are false. Both assertion and reason are true and the reason is the correct explanation for the assertion. Both assertion and reason are true but the reason is not the correct explanation for the assertion. Statement-1 : Zn2+ is diamagnetic. Statement-2 : Two electrons are lost from 4s orbital to form Zn2+. [1998 - 2 Marks] Statement-1 : To a solution of potassium chromate if a strong acid is added it changes its colour from yellow to orange. Statement-2 : The colour change is due to the oxidation of potassium chromate. [1988 - 2 Marks]

90.

Identify (A), (B) and MCl4. Also explain colour difference between MCl4 and (A). [2005 - 4 Marks]

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91.

Write the chemical reaction involved in developing of a black and white photographic film. An aqueous Na2S2O3 solution is acidified to give a milky white turbitity. Identify the product and write the balanced half chemical reaction for it. [2005 - 4 Marks] 92. (i) Write the chemical reactions involved in the extraction of metallic silver from argentite. (ii) Write the balanced chemical equation for developing photographic films. [2000 - 4 Marks] 93. Write the chemical reaction associated with the 'brown ring test'. [2000 - 2 Marks] 94. Work out the following using chemical equations [1998 - 2 Marks] In moist air copper corrodes to produce a green layer on the surface. 95. Compare qualitatively the first and second ionisation potentials of copper and zinc. Explain the observation. [1996 - 2 Marks] 96. The acidic, aqueous solution of ferrous ion forms a brown complex in ,by the following two steps. Complete and the presence of balance the equations : [1993 - 2 Marks]

97.

Mention the products formed when zinc oxide is treated with excess of sodium hydroxide solution. [1986 - 1 Mark] 98. What happens when : (i) aqueous ammonia is added dropwise to a solution of copper sulphate till it is in excess. [1985 - 1 Mark] (ii) CrCl3 solution is treated with sodium hydroxide and then with hydrogen peroxide. [1985 - 1 Mark]

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99.

State the conditions under which the following preparation is carried out. Potassium permanganate from manganese hydroxide. Give the necessary equations which need not be balanced. [1983 - 1 Mark] 100. Give reasons for the following : (i) CrO3 is an acid anhydride. [1999 - 2 Marks] 2– 2– (ii) The species [CuCl4] exists while [CuI4] does not. [1992 - 1 Mark] (iii) The colour of mercurous chloride, Hg2Cl2, changes from white to black when treated with ammonia. [1988 - 1 Mark] (iv) Zinc and not copper is used for the recovery of metallic silver from complex [Ag(CN)2]–. Explain. [1987 - 1 Mark] (v) Most transition metal compounds are coloured. [1986 - 1 Mark] (vi) Silver bromide is used in photography. [1983 - 1 Mark] 101. State with balanced equations what happens when : (i) Write balanced equations for the reaction of zinc with dilute nitric acid. [1997 - 1 Mark] (ii) Write a balanced equation for the reaction of argentite with KCN and name the products in solution. [1996 - 1 Mark] (iii) [1994 - 1 Mark] (iv) [1994 - 1 Mark] (v)

+........ [1993 - 1 Mark]

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(vi) [1993 - 1 Mark] (vii) Potassium dichromate and concentrated hydrochloric acid are heated together. [1992 - 1 Mark] (viii) Na2CO3 is added to a solution of copper sulphate. [1992 - 1 Marks] CuSO4 + Na2CO3 + H2O → ........... + Na2SO4 + ........... (ix) Copper reacts with HNO3 to give NO and NO2 in molar ratio of 2 : 1. [1992 - 1 Marks] Cu + HNO3 → ........... + NO + NO2 + ........... . (x) Potassium permanganate is added to a hot solution of manganous sulphate. [1990 - 1 Mark] (xi) Iron reacts with cold dilute nitric acid. [1990 - 1 Mark] (xii) A mixture of potassium dichromate and sodium chloride is heated with concentrated H2SO4. [1990 - 1 Mark] (xiii) Write balanced equations for the extraction of copper from copper pyrites by self-reduction. [1990 - 2 Marks] (xiv) Cobalt(II) solution reacts with KNO2 in acetic acid medium. [1989 - 1 Mark] (xv) Silver chloride is treated with aqueous sodium cyanide and the product thus formed is allowed to react with zinc in alkaline medium. [1989 - 1 Mark] (xvi) Write balanced equations for the extraction of silver from silver glance by cyanide process. [1988 - 1 Mark] (xvii) Gold is dissolved in aqua regia. [1987 - 1 Mark] (xviii) Potassium permanganate is reacted with warm solution of oxalic acid in the presence of sulphuric acid.

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[1987 - 1 Mark] (xix) potassium ferrocyanide is heated with concentrated sulphuric acid; [1985 - 1 Mark] (xx) potassium permanganate interacts with manganese dioxide in presence of potassium hydroxide; [1985 - 1 Mark] (xxi) aqueous solution of potassium chromate and acid are mixed. [1984 - 2 Marks] (xxii) aqueous solution of potassium manganate and acid are mixed. [1984 - 2 Marks] (xxiii) aqueous solution of ferric sulphate and potassium iodide are mixed. [1984 - 2 Marks] (xxiv) sulphur dioxide gas is bubbled through an aqueous solution of copper sulphate in presence of potassium thiocyanate. [1982 - 1 Mark] 102. Complete the following equation (no balancing is needed): SO2 +

+ .... —→

+ Mn2+ + ....

[1981 - 1 Mark] 103. A solution of FeCl3 in water gives a brown precipitate on standing. [1980] 104. State with balanced equations, what happens when (i)

Silver is treated with hot concentrated sulphuric acid.

(ii) Ammonium dichromate is heated. (iii) Hydrogen sulphide is passed through a solution of potassium permanganate acidified with dilute sulphuric acid. [1979] 105. A white amorphous powder (A) on heating yields a colourless, noncombustible gas (B) and a solid (C). The latter compound assumes a yellow colour on heating and changes to white on cooling. ‘C’ dissolves in dilute acid and the resulting solution gives a white precipitate on adding K4[Fe(CN)6] solution.

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‘A’ dissolves in dilute HCl with the evolution of gas, which is identical in all respects with ‘B’. The gas ‘B’ turns lime water milky, but the milkiness disappears with the continuous passage of gas. The solution of ‘A’, as obtained above, gives a white precipitate (D) on the addition of excess of NH4OH and passing H2S. Another portion of the solution gives initially a white precipitate (E) on the addition of sodium hydroxide solution, which dissolves on futher addition of the base. Identify the compounds A, B, D, and E. [1979] 106. A certain inorganic compound (A) on heating loses its water of crystallisation. On further heating, a blackish brown powder (B) and two oxides of sulphur (C and D) are obtained. The powder (B) on boiling with hydrochloric acid gives a yellow solution (E). When H2S is passed in (E) a white turbidity (F) and an apple green solution (G) are obtained. The solution (E) on treatment with thiocyanate ions gives a blood red coloured compound (H). Identify compounds from (A) to (H). [1978]

1. (a) (b) (c) (d) 2. (a) (b) (c) (d)

The lanthanoid that does NOT show + 4 oxidation state is: [Main Sep. 06, 2020 (I)] Dy Ce Eu Tb Mischmetal is an alloy consisting mainly of: [Main Sep. 06, 2020 (II)] lanthanoid metals actinoid and transition metals lanthanoid and actinoid metals actinoid metals

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3.

The correct electronic configuration and spin-only magnetic moment (BM) of Gd3+(Z = 64), respectively, are: [Main Sep. 05, 2020 (I)] 7 (a) [Xe] 4f and 8.9 (b) [Xe] 4f 7 and 7.9 (c) [Xe] 5f 7 and 8.9 (d) [Xe] 5f 7 and 7.9 4. The electronic configurations of bivalent europium and trivalent cerium are: (atomic number : Xe = 54, Ce = 58, Eu = 63) [Main Jan. 09, 2020 (I)] (a) [Xe] 4f 2 and [Xe] 4f 7 (b) [Xe] 4f 7 and [Xe] 4f 1 (c) [Xe] 4f 7 6s2 and [Xe] 4f 2 6s2 (d) [Xe] 4f 4 and [Xe] 4f 9 5. The maximum number of possible oxidation states of actinoides are shown by: [Main April 9, 2019 (II)] (a) Nobelium (No) and lawrencium (Lr) (b) Actinium (Ac) and thorium (Th) (c) Berkelium (Bk) and californium (Cf) (d) Neptunium (Np) and plutonium (Pu) 6. The lanthanide ion that would show colour is : [Main April 8, 2019 (I)] 3+ (a) Gd (b) Sm3+ (c) La3+ (d) Lu3+ 7. The correct order of atomic radii is : [Main Jan. 12, 2019 (II)] (a) N > Ce > Eu > Ho (b) Ho > N > Eu > Ce (c) Ce > Eu > Ho > N (d) Eu > Ce > Ho > N

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8. (a) (b) (c) (d)

The effect of lanthanoid contraction in the lanthanoid series of elements by and large means: [Main Jan. 10, 2019 (I)] increase in both atomic and ionic radii decrease in atomic radii and increase in ionic radii decrease in both atomic and ionic radii increase in atomic radii and decrease in ionic radii Topic-1 : d-Block Elements

1. (c) 2. (c) 3. (b) 4. (a) 5. (a) 6. (b) 7. (a) 8. (a) 9. (c) 10. (b) 11. (d) 12. (a) 13. (d) 14. (d) 15. (c) 16. (c) 17. (d) 18. (c) 19. (a) 20. (b) 21. (d) 22. (d)

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23. (a) 24. (a) 25. (d) 26. (b) 27. (a) 28. (a) 29. (c) 30. (b) 31. (a) 32. (b) 33. (a) 34. (a) 35. (a) 36. (b) 37. (b) 38. (b) 39. (a) 40. (b) 41. (c) 42. (a) 43. (d) 44. (a) 45. (c) 46. (b) 47. (d) 48. (c) 49. (b) 50. (b) 51. (b) 52. (d) 53. (a)

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54. (4) 55. (6) 56. (7) 57. (6) 58. (12) 59. (18) 66. (False) 67. (True) 68. (True) 69. (a, b, c) 70. (b, c, d) 71. (a, b) 72. (a, b, c) 73. (a, b, c) 74. (a, c, d) 75. (b, c, d) 76. (a, d) 77. (a, c) 78. (b, c) 79. (c, d) 80. (b, c) 81. (c) 82. (A) –p, s ; (B) –q, s ; (C) –r, t ; (D) –q, t 84. (b) 85. (a) 86. (c) 87. (a) 88. (b) 89. (c)

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Topic-2 : f-Block Elements 1. (c) 2. (a) 3. (b) 4. (b) 5. (d) 6. (b) 7. (d) 8. (c)

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1.

Consider the complex ions, trans-[Co(en)2Cl2]+ (A) and cis[Co(en)2Cl2]+ (B). The correct statement regarding them is : [Main Sep. 05, 2020 (II)] (a) both (A) and (B) cannot be optically active. (b) (A) can be optically active, but (B) cannot be optically active. (c) both (A) and (B) can be optically active. (d) (A) cannot be optically active, but (B) can be optically active. 2. The complex that can show optical activity is : [Main Sep. 03, 2020 (I)] 3– (a) trans-[Cr(Cl2)(ox)2] (b) trans-[Fe(NH3)2(CN)4]– (c) cis-[Fe(NH3)2(CN)4]– (d) cis-[CrCl2(ox)2]3– (ox = oxalate) 3. Complex A has a composition of H12O6Cl3Cr. If the complex on treatment with conc. H2SO4 loses 13.5% of its original mass, the correct molecular formula of A is : [Given : atomic mass of Cr = 52 amu and Cl = 35 amu] [Main Sep. 03, 2020 (II)] (a) [Cr(H2O)6]Cl3 (b) [Cr(H2O)3Cl3] ⋅ 3H2O

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(c) [Cr(H2O)5Cl]Cl2 ⋅ H2O (d) [Cr(H2O)4Cl2]Cl ⋅ 2H2O 4. The one that is not expected to show isomerism is : [Main Sep. 02, 2020 (II)] 2+ (a) [Ni(NH3)4(H2O)2] (b) [Ni(en)3]2+ (c) [Ni(NH3)2Cl2] (d) [Pt(NH3)2Cl2] 5. Complex X of composition Cr(H2O)6Cln has a spin only magnetic moment of 3.83 BM. It reacts with AgNO3 and shows geometrical isomerism. The IUPAC nomenclature of X is: [Main Jan. 09, 2020 (I)] (a) Hexaaqua chromium (III) chloride (b) Tetraaquadichlorido chromium (IV) chloride dihydrate (c) Dichloridotetraaqua chromium (IV) chloride dihydrate (d) Tetraaquadichlorido chromium (III) chloride dihydrate 6. The isomer(s) of [Co(NH3)4Cl2] that has/have a Cl–Co–Cl angle of 90°, is/are : [Main Jan. 09, 2020 (II)] (a) meridional and trans (b) cis and trans (c) trans only (d) cis only 7. The complex that can show fac- and mer- isomers is: [Main Jan. 08, 2020 (I)] + (a) [Co(NH3)4Cl2] (b) [Pt(NH3)2Cl2] (c) [CoCl2(en)2] (d) [Co(NH3)3(NO2)3] 8. Among (a) - (d), the complexes that can display geometrical isomerism are: [Main Jan. 08, 2020 (II)] + (1) [Pt(NH3)3Cl] (2) [Pt(NH3)Cl5]– (3) [Pt(NH3)2Cl(NO2)]

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(4) (a) (b) (c) (d) 9. (a) (b) (c) (d) 10.

(a) (b) (c) (d) 11. (A) (B)

[Pt(NH3)4ClBr]2+ (2) and (3) (4) and (1) (3) and (4) (1) and (2) The IUPAC name of the complex [Pt (NH3)2Cl(NH2CH3)]Cl is: [Main Jan. 07, 2020 (I)] Diamminechlorido (methanamine) platinum (II)chloride Diammine (methanamine) chlorido platinum (II)chloride Diamminechlorido (aminomethane) platinum (II)chloride Bisammine (methanamine) chlorido platinum (II)chloride The number of possible optical isomers for the complexes MA2B2 with sp3 and dsp2 hybridized metal atom, respectively, is: Note: A and B are unidentate neutral and unidentate monoanionic ligands, respectively. [Main Jan. 07, 2020 (II)] 0 and 2 2 and 2 0 and 0 0 and 1 Among the statements (A)-(D), the incorrect ones are: [Main Jan. 07, 2020 (II)] Octahedral Co(III) complexes with strong field ligands have very high magnetic moments. When 0 < P, the d-electron configuration of Co(III) in an octahedral complex is

.

(C) Wavelength of light absorbed by [Co(en)3]3+ is lower than that of [CoF6]3–. (D) If the 0 for an octahedral complex of Co(III) is 18,000 cm–1, the t for its tetrahedral complex with the same ligand will be 16,000 cm–1. (a) (A) and (D) only (b) (C) and (D) only (c) (A) and (B) only (d) (B) and (C) only

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12.

(a) (b) (c) (d) 13.

The coordination numbers of Co and Al in [Co(Cl)(en)2]Cl and K3[Al(C2O4)3], respectively, are : (en = ethane-1, 2-diamine) [Main April 12, 2019 (II)] 5 and 3 3 and 3 6 and 6 5 and 6 The maximum possible denticities of a ligand given below towards a common transition and inner-transition metal ion, respectively, are: [Main April 9, 2019 (II)]

(a) (b) (c) (d) 14.

8 and 6 6 and 8 6 and 6 8 and 8 The coordination number of Th in K4[Th(C2O4)4(H2O)2] is: (C2O42– = oxalato) [Main Jan. 11, 2019 (II)] (a) 14 (b) 6 (c) 8 (d) 10 15. A reaction of cobalt(III) chloride and ethylenediamine 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? [Main Jan. 10, 2019 (II)] (a) Geometrical isomers (b) Coordination isomers (c) Linkage isomers (d) Ionisation isomers 16. Consider the following reaction and statements: [Main 2018] + – [Co(NH3)4Br2] + Br → [Co(NH3)3Br3] + NH3 (I) Two isomers are produced if the reactant complex ion is a cis-isomer.

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(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: (a) (I) and (II) (b) (I) and (III) (c) (III) and (IV) (d) (II) and (IV) 17. The total number of possible isomers for square-planar [Pt(Cl)(NO2) (NO3)(SCN)]2– is: [Main Online April 15, 2018 (II)] (a) 16 (b) 12 (c) 8 (d) 24 18. On treatment of 100 mL of 0.1 M solution of CoCl3 . 6H2O with excess AgNO3; 1.2 × 1022 ions are precipitated. The complex is : [Main 2017] (a) [Co(H2O)4 Cl2]Cl.2H2O (b) [Co(H2O)3Cl3].3H2O (c) [Co(H2O)6]Cl3 (d) [Co(H2O)5Cl]Cl2.H2O 19. Which one of the following complexes shows optical isomerism? [Main 2016] (a) trans [Co(en)2Cl2]Cl (b) [Co(NH3)4Cl2]Cl (c) [Co(NH3)3Cl3] (d) cis[Co(en)2Cl2]Cl (en = ethylenediamine) 20. Which of the following is an example of homoleptic complex? [Main Online April 10, 2016] (a) [Co(NH3)6]Cl3

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(b) [Pt(NH3)2Cl2] (c) [Co(NH3)4Cl2] (d) [Co(NH3)5Cl]Cl2 21. The number of geometric isomers that can exist for square planar complex [Pt (Cl) (py) (NH3) (NH2OH)]+ is (py = pyridine) : [Main 2015] (a) 4 (b) 6 (c) 2 (d) 3 22. The correct statement on the isomerism associated with the following complex ions, [Main Online April 10, 2015] 2+ (A) [Ni(H2O)5NH3] , (B) [Ni(H2O)4(NH3)2]2+ and (C) [Ni(H2O)3(NH3)3]2+is : (a) (A) and (B) show only geometrical isomerism (b) (A) and (B) show geometrical and optical isomerism (c) (B) and (C) show geometrical and optical isomerism (d) (B) and (C) show only geometrical isomerism 23. An octahedral complex with molecular composition M.5 NH3.Cl.SO4 has two isomers, A and B. The solution of A gives a white precipitate with AgNO3 solution and the solution of B gives white precipitate with BaCl2 solution. The type of isomerism exhibited by the complex is: [Main Online April 19, 2014] (a) Linkage isomerism (b) Ionisation isomerism (c) Coordinate isomerism (d) Geometrical isomerism 24. Which of the following complex species is not expected to exhibit optical isomerism ? [Main 2013] 3+ (a) [Co(en)3] (b) [Co(en)2 Cl2]+

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(c) [Co(NH3)3 Cl3] (d) [Co(en) (NH3)2 Cl2]+ 25. Type of isomerism which exists between [Pd(C6H5)2(SCN)2] and [Pd(C6H5)2(NCS)2] is : [Main Online April 9, 2013] (a) Linkage isomerism (b) Coordination isomerism (c) Ionisation isomerism (d) Solvate isomerism 26. As per IUPAC nomenclature, the name of the complex [Co(H2O)4(NH3)2]Cl3 is : [2012] (a) Tetraaquadiaminecobalt (III) chloride (b) Tetraaquadiamminecobalt (III) chloride (c) Diaminetetraaquacoblat (II) chloride (d) Diamminetetraaquacobalt (III) chloride 27. The correct structure of ethylenediaminetetraacetic acid (EDTA) is [2010] (a) (b) (c)

(d)

28.

The ionisation isomer of [Cr(H2O)4Cl(NO2)]Cl is

(a) [Cr(H2O)4(O2N)]Cl2 (b) [Cr(H2O)4Cl2](NO2) (c) [Cr(H2O)4Cl(ONO)]Cl

[2010]

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(d) [Cr(H2O)4Cl2(NO2)].H2O 29. The IUPAC name of [Ni (NH3)4] [NiCl4] is (a) (b) (c) (d) 30. (a) (d) 31. (a) (b) (c) (d) 32. (a) (b) (c) (d)

[2008]

Tetrachloronickel (II) - tetraamminenickel (II) Tetraamminenickel (II) - tetrachloronickel (II) Tetraamminenickel (II) - tetrachloronickelate (II) Tetrachloronickel (II) - tetrachloronickelate (0) Ans. (C) Which kind of isomerism is exhibited by octahedral Co(NH3)4Br2Cl? [2005S] Geometrical and Ionization (b) Geometrical and Optical (c) Optical and Ionization Geometrical only The complex ion which has no ‘d’ electron in the central metal atom is [2001S] − [MnO4] [Co(NH3)6]3+ [Fe(CN)6]3 – [Cr(H2O)6]3+ Which of the following is an organometallic compound? [1997 - 1 Mark] Lithium methoxide Lithium acetate Lithium dimethylamide Methyl lithium

33. The total number of coordination sites in ethylenediaminetetraacetate (EDTA4–) is ______. [Main Sep. 05, 2020 (I)] 34. 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)2Cl2] complex is ___ (en = NH2CH2CH2NH2) [Adv. 2019]

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35.

The number of geometric isomers possible for the complex [CoL2Cl2]– (L = H2NCH2CH2O–) is [Adv. 2016] 36. Among the complex ions, [Co(NH2–CH2–CH2–NH2)2Cl2]+, [CrCl2(C2O4)2]3–, [Fe(H2O)4(OH)2]+, [Fe(NH3)2(CN)4]–, [Co(NH2–CH2–CH2–NH2)2(NH3)Cl]2+ and 2+ [Co(NH3)4(H2O)Cl] , the number of complex ion(s) that show(s) cistrans isomerism is [Adv. 2015] 37. In the complex acetylbromidodicarbonylbis (triethylphosphine) iron (II), the number of Fe–C bond(s) is [Adv. 2015] 4– 38. EDTA is ethylenediaminetetraacetate ion. The total number of N— Co—O bond angles in [Co(EDTA)]1–complex ion is [Adv. 2013] 39. 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(H2O)5Cl]Cl2, as silver chloride is close to [2011] 40. Total number of geometrical isomers for the complex [RhCl(CO) (PPh3) (NH3)] is [2010] 41. The number of water molecule(s) directly bonded to the metal centre in CuSO4.5H2O is [2009 - 4 Marks] 42.

The IUPAC name of [Co(NH3)6] Cl3 is ............

[1994 - 1 Mark]

43. The pair(s) of coordination complexes/ions exhibiting the same kind of isomerism is(are) [Adv. 2013] (a) [Cr(NH3)5Cl]Cl2 and [Cr(NH3)4Cl2]Cl (b) [Co(NH3)4Cl2]+ and [Pt(NH3)2(H2O)Cl]+ (c) [CoBr2Cl2]2– and [PtBr2Cl2]2–

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(d) [Pt(NH3)3(NO3)]Cl and [Pt(NH3)3Cl]Br 44. The compound(s) that exhibit(s) geometrical isomerism is (are) [2009] (a) (b) (c) (d)

[Pt(en)Cl2] [Pt(en)2]Cl2 [Pt(en)2Cl2]Cl2 [Pt(NH3)2Cl2]

Read the following statement-1(Asseration/Statement) and Statement -2 (Reason/Explanation) and answer as per the options given below : (a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (c) Statement-1 is True, Statement-2 is False (d) Statement-1 is False, Statement-2 is True 45.

Statement-1 : The geometrical isomers of the complex [M(NH3)4Cl2] are optically inactive. Statement-2 : Both geometrical isomers of the complex [M(NH3)4Cl2] possess axis of symmetry. [2008 - 2 Marks]

46.

A, B, and C are three complexes of chromium (III) with the empirical formula H12O6Cl3Cr. All the three complexes have water and chloride ion as ligands. Complex A does not react with concentrated H2SO4, whereas complexes B and C lose 6.75% and 13.5% of their original mass, respectively, on treatment with concentrated H2SO4. Identify A, B and C. [1999 - 6 Marks] 47. Write the formulae of the following complexes : (i) Pentaamminechlorocobalt(III) [1997 - 1 Mark] (ii) Lithium tetrahydroaluminate(III). [1997 - 1 Mark] 48. Write down the IUPAC names of the following compounds :

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(i)

[Cr(NH3)5CO3]Cl

[1996 - 1 Mark]

(ii) [1995 - 1 Mark] (iii) [1995 - 1 Mark]

1.

The species that has a spin-only magnetic moment of 5.9 BM, is : (Td = tetrahedral) [Main Sep. 06, 2020 (I)]

(a) [Ni(CN)4]2– (square planar) (b) [NiCl4]2– (Td) (c) Ni(CO)4(Td) (d) [MnBr4]2– (Td) 2.

For a d4 metal ion in an octahedral field, the correct electronic configuration is : [Main Sep. 06, 2020 (II)]

(a) (b) (c) (d) 3. (a) (b) (c) (d)

The values of the crystal field stabilization energies for a high spin d6 metal ion in octahedral and tetrahedral fields, respectively, are : [Main Sep. 05, 2020 (I)] – 0.4 ∆o and – 0.6 ∆t – 2.4 ∆o and – 0.6 ∆t – 1.6 ∆o and – 0.4 ∆t – 0.4 ∆o and – 0.27 ∆t

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4. (a) (b) (c) (d) 5.

The pair in which both the species have the same magnetic moment (spin only) is : [Main Sep. 04, 2020 (I)] 2+ 2+ [Cr(H2O)6] and [Fe(H2O)6] [Co(OH)4]2– and [Fe(NH3)6]2+ [Mn(H2O)6]2+ and [Cr(H2O)]2+ [Cr(H2O)6]2+ and [CoCl4]2– The Crystal Field Stabilization Energy (CFSE) of is : [Main Sep. 04, 2020 (II)]

(a) (b) (c) (d) 6.

The one that can exhibit highest paramagnetic behaviour among the following is : [Main Sep. 04, 2020 (II)] gly = glycinato; bpy = 2, 2'-bipyridine (a) [Pd(gly)2] (b) [Fe(en)(bpy)(NH3)2]2+ (c) [Co(OX)2(OH)2]– (∆0 > P) (d) [Ti(NH3)6]3+ 7. The electronic spectrum of [Ti(H2O)6]3+ shows a single broad peak with a maximum at 20,300 cm–1. The crystal field stabilization energy (CFSE) of the complex ion, in kJ mol–1, is : (1 kJ mol–1 = 83.7 cm–1) [Main Sep. 03, 2020 (I)] (a) 145.5 (b) 242.5 (c) 83.7 (d) 97 8. The d-electron configuration of [Ru(en)3]Cl2 and [Fe(H2O)6]Cl2, respectively are: [Main Sep. 03, 2020 (II)]

(a)

and

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(b)

and

(c)

and

(d)

and

9.

Consider that a d6 metal ion (M2+) 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 stabilization energy of the complex is : [Main Sep. 02, 2020 (I)] (a) octahedral and (b) tetrahedral and –0.6 ∆t (c) octahedral and –1.6 ∆0 (d) tetrahedral and 10. For octahedral Mn(II) and tetrahedral Ni(II) complexes, consider the following statements: [Main Sep. 02, 2020 (I)] (I) both the complexes can be high spin. (II) Ni(II) complex can very rarely below spin. (III)with strong field ligands, Mn(II) complexes can be low spin. (IV) aqueous solution of Mn(II) ions is yellow in color. The correct statements are : (a) (I) and (II) only (b) (I), (III) and (IV) only (c) (I), (II) and (III) only (d) (II), (III) and (IV) only 11. [Pd(F)(Cl)(Br)(I)]2– has n number of geometrical isomers. Then, the spin-only magnetic moment and crystal field stabilisation energy [CFSE] of [Fe(CN)6]n–6, respectively, are: [Note : Ignore the pairing energy] [Main Jan. 09, 2020 (I)] (a) 2.84 BM and –1.6 D0 (b) 5.92 BM and 0 (c) 1.73 BM and –2.0 D0 (d) 0 BM and –2.4 D0

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12.

The correct order of the spin-only magnetic moments of the following complexes is: [Main Jan. 09, 2020 (II)] (A) [Cr(H2O)6]Br2 (B) Na4[Fe(CN)6] (C) Na3[Fe(C2O4)3] ( 0 > P) (D) (Et4N)2[CoCl4] (a) (C) > (A) > (D) > (B) (b) (C) > (A) > (B) > (D) (c) (A) > (D) > (C) > (B) (d) (B) (A) > (D) > (C) 13. The correct order of the calculated spin-only magnetic moments of complexes (A) to (D) is: [Main Jan. 08, 2020 (II)] (A) Ni(CO)4 (B) [Ni(H2O)6]Cl2 (C) Na2[Ni(CN)4] (D) PdCl2(PPh3)2 (a) (A) ≈ (C) < (B) ≈ (D) (b) (C) < (D) < (B) < (A) (c) (C) ≈D) < (B) < (A) (d) (A) ≈ (C) ≈ (D) < (B) 14. The theory that can completely/properly explain the nature of bonding in [Ni(CO)4] is: [Main Jan. 07, 2020 (I)] (a) Werner’s theory (b) Molecular orbital theory (c) Crystal field theory (d) Valence bond theory 15. The compound used in the treatment of lead poisoning is : [Main April 12, 2019 (II)] (a) D-penicillamine (b) desferrioxime B (c) cis-platin (d) EDTA

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16.

The degenerate orbitals of [Cr(H2O) 6]3+ are:

[Main April 9, 2019 (I)]

(a) (b) (c) (d) 17.

The correct order of the spin-only magnetic moment of metal ions in the following low-spin complexes, [V(CN)6]4–, [Fe(CN)6]4–, [Ru(NH3)6]3+, and [Cr(NH3)6]2+, is [Main April 8, 2019 (I)] 2+ 3+ 2+ 2+ (a) Cr > Ru > Fe > V (b) V2+ > Cr2+ > Ru3+ > Fe2+ (c) V2+ > Ru3+ > Cr2+ > Fe2+ (d) Cr2+ > V2+ > Ru3+ > Fe2+ 18. Mn2(CO)10 is an organometallic compound due to the presence of : [Main Jan. 12, 2019 (I)] (a) Mn–C bond (b) Mn–Mn bond (c) Mn–O bond (d) C–O bond 19. The magnetic moment of an octahedral homoleptic Mn(II) complex is 5.9 BM. The suitable ligand for this complex is : [Main Jan. 12, 2019 (II)] (a) Ethylenediamine (b) CN– (c) NCS– (d) CO 20. The number of bridging CO ligand(s) and Co-Co bond(s) in Co2(CO)8, respectively are:[Main Jan. 11, 2019 (II)] (a) 2 and 1 (b) 2 and 0 (c) 0 and 2 (d) 4 and 0

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21.

Wilkinson catalyst is: [Main Jan. 10, 2019 (I)]

(a) (b) (c) (d) 22.

[(Ph3P)3IrCl] [(Et3P)3RhCl] [(Ph3P)3RhCl] [(Et3P)3IrCl] The complex that has highest crystal field splitting energy (∆) is: [Main Jan. 9, 2019 (II)] (a) [Co(NH3)5(H2O)]Cl3 (b) K2[CoCl4] (c) [Co(NH3)5Cl]Cl2 (d) K3[Co(CN)6] 23. In Wilkinson's catalyst, the hybridisation of central metal ion and its shape are respectively: [Main Online April 16, 2018] (a) sp3d, trigonal bipyramidal (b) d 2sp3, octahedral (c) dsp2, square planar (d) sp3, tetrahedral 24. [Co2(CO)8] displays : [Main Online April 9, 2017] (a) one Co – Co bond, six terminal CO and two bridging CO (b) one Co – Co bond, four terminal CO and four bridging CO (c) no Co – Co bond, six terminal CO and two bridging CO (d) no Co – Co bond, four terminal CO and four bridging CO 25. The pair having the same magnetic moment is: [At. No.: Cr = 24, Mn = 25, Fe = 26, Co = 27][Main 2016] (a) [Mn(H2O)6]2+ and [Cr(H2O)6]2+ (b) [CoCl4]2– and [Fe(H2O)6]2+ (c) [Cr(H2O)6]2+ and [CoCl4]2– (d) [Cr(H2O)6]2+ and [Fe(H2O)6]2+ 26. The geometries of the ammonia complexes of Ni2+ , Pt2+ and Zn2+ respectively, are [JEE Adv. 2016]

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(a) (b) (c) (d) 27. (a) (b) (c) (d) 28. (a) (b) (c) (d) 29.

(a) (b) (c) (d) 30.

octahedral, square planar and tetrahedral square planar, octahedral and tetrahedral tetrahedral, square planar and octahedral octahedral, tetrahedral and square planar Which of the following compounds is not colored yellow? [Main 2015] (NH4)3[As(Mo3O10)4] BaCrO4 Zn2[Fe(CN)6] K3[Co(NO2)6] The color of KMnO4 is due to : [Main 2015] L → M charge transfer transition σ - σ* transition M → L charge transfer transition d – d transition Which molecule/ion among the following cannot act as a ligand in complex compounds ? [Main Online April 10, 2015] CH4 CO CN– Br– The octahedral complex of a metal ion M3+ 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: [Main 2014]

(a) (b) (c) (d)

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31.

(a) (b) (c) (d)

An octahedral complex of Co3+ is diamagnetic. The hybridisation involved in the formation of the complex is: [Main Online April 9, 2014] sp3d2 dsp2 d2sp3 sp3d

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32.

(a) (b) (c) (d) 33.

(a) (b) (c) (d) 34.

Consider the coordination compound, [Co(NH3)6]Cl3. In the formation of this complex, the species which acts as the Lewis acid is: [Main Online April 11, 2014] 3+ [Co(NH3)6] Cl– Co3+ NH3 Nickel (Z = 28) combines with a uninegative monodentate ligand to form a diamagnetic complex [NiL4]2–. The hybridisation involved and the number of unpaired electrons present in the complex are respectively: [Main Online April 19, 2014] 3 sp , two dsp2, zero dsp2, one sp3, zero The structure of which of the following chloro species can be explained on the basis of dsp2 hybridisation ? [Main Online April 25, 2013]

(a) (b) (c) (d) 35.

(a) (b) (c) (d) 36.

The magnetic moment of the complex anion [Cr(NO) (NH3) (CN)4]2– is : [Main Online April 23, 2013] 5.91 BM 3.87 BM 1.73 BM 2.82 BM Consider the following complex ions, P, Q and R. P = [FeF6]3–, Q = [V(H2O)6]2+ and R = [Fe(H2O)6]2+

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The correct order of the complex ions, according to their spin-only magnetic moment values (in B.M.) is [Adv. 2013] (a) R < Q < P (b) Q < R < P (c) R < P < Q (d) Q < P < R 37. NiCl2 {P(C2H5)2(C6H5)}2 exhibits temperature depend-ent magnetic behaviour (paramagnetic/diamagnetic). The coordination geometries of Ni2+ in the paramagnetic and diamagnetic states are respectively [2012] (a) tetrahedral and tetrahedral (b) square planar and square planar (c) tetrahedral and square planar (d) square planar and tetrahedral 38. Among the following complexes (K−P) K3[Fe(CN)6] (K), [Co(NH3)6]Cl3 (L), Na3[Co(oxalate)3] (M), the [Ni(H2O)6]Cl2 (N), K2[Pt(CN)4] (O) and [Zn(H2O)6](NO3)2 (P) the diamagnetic complexes are [2011] (a) K, L, M, N (b) K, M, O, P (c) L, M, O, P (d) L, M, N, O 39. Geometrical shapes of the complexes formed by the reaction of Ni2+ with Cl– , CN– and H2O, respectively, are [2011] (a) octahedral, tetrahedral and square planar (b) tetrahedral, square planar and octahedral (c) square planar, tetrahedral and octahedral (d) octahedral, square planar and octahedral 40. The complex showing a spin-only magnetic moment of2.82 B.M. is : [2010] (a) [Ni(CO)4]

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(b) [NiCl4]2– (c) [Ni(PPh3)4] (d) [Ni(CN)4]2– 41. The spin only magnetic moment value (in Bohr magneton units) of Cr(CO)6 is [2009] (a) 0 (b) 2.84 (c) 4.90 (d) 5.92 42. Both [Ni (CO)4] and [Ni (CN)4]2– are diamagnetic. The hybridisations of nickel in these complexes, respectively, are [2008] 3 3 3 2 (a) sp , sp (b) sp , dsp (c) dsp2, sp3 (d) dsp2, sp2 43. Among the following metal carbonyls, the C–O bond order is lowest in [2007] + (a) [Mn(CO)6] (b) [Fe(CO)5] (c) [Cr(CO)6] (d) [V(CO)6]– 44. then the formation constant of [Ag(NH3)2]+ is [2006 - 3M, –1] (a) (b) (c) (d) 45.

6.8 × 10 1.08 × 10–5 1.08 × 10–6 6.8 × 10–5 The spin magnetic moment of cobalt in the compound Hg[Co(SCN)4] is [2004S] –6

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(a) (b) (c) (d) 46.

The species having tetrahedral shape is

(a) (b) (c) (d) 47. (a)

[PdCl4] [Ni(CN)4]2– [Pd(CN)4]2– [NiCl4]2– The geometry of Ni(CO)4 and Ni(PPh3)2Cl2 are both square planar

[2004S] 2–

[1999 - 2 Marks] (b) tetrahedral and square planar, respectively (c) both tetrahedral (d) square planar and tetrahedral, respectively 48. Amongst [Ni(CO)4], [Ni(CN)4]2– and [ (a) [Ni(CO)4] and [

]

[1991 - 1 Mark] ] are diamagnetic and [Ni(CN)4]2– is

paramagnetic (b) [

] and [Ni(CN)4]2– are diamagnetic and [Ni(CO)4] is

paramagnetic (c) [Ni(CO)4] and [Ni(CN)4]2–

are

diamagnetic

and

[

]is

paramagnetic (d) [Ni(CO)4] is diamagnetic and [

] and [Ni(CN)4]2– are

paramagnetic 49. Amongst the following, the lowest degree of paramagnetism per mole of the compound at 298 K will be shown by [1988 - 1 Mark] (a) MnSO4.4H2O

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(b) CuSO4.5H2O (c) FeSO4.6H2O (d) NiSO4.6H2O 50.

Among the species given below, the total number of diamagnetic species is_______. H atom, NO2 monomer, O2– (superoxide), dimeric sulphur in vapour phase, Mn3O4, (NH4)2[FeCl4],(NH4)2[NiCl4], K2MnO4, K2CrO4 [Adv. 2018] 3+ – 51. For the octahedral complexes of Fe 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] [Adv. 2015] 52.

Considering that the magnetic moment (in BM) of [Ru(H2O)6]2+ would be _______. [Main Sep. 05, 2020 [II]] 53. The volume (in mL) of 0.125 M AgNO3 required to quantitatively precipitate chloride ions in 0.3 g of [Co(NH3)6]Cl3 is _______. [Main Jan. 08, 2020 (I)] = 267.46 g/mol = 169.87 g/mol 54. Complexes (ML5) of metals Ni and Fe have ideal square pyramidal and trigonal bipyramidal geometries, respectively. The sum of the 90°, 120° and 180° L-M-L angles in the two complexes is ________. [Main Jan. 08, 2020 (II)] 55.

The type of magnetism exhibited by [Mn(H2O)6]2+ ion is ............. [1994 - 1 Mark]

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56.

Both potassium ferrocyanide and potassium ferricyanide are diamagnetic. [1989 - 1 Mark]

57. Choose the correct statement(s) among the following: (a) [FeCl4]– has tetrahedral geometry. [Adv. 2020] (b) [Co(en)(NH3)2Cl2] has 2 geometrical isomers. (c) [FeCl4]– has higher spin-only magnetic moment than [Co(en) (NH3)2Cl2]+. (d) The cobalt ion in [Co(en)(NH3)2Cl2]+ has sp3d2 hybridization. 58. The correct option(s) regarding the complex [Co(en)(NH3)3(H2O)]3+ (en = H2NCH2CH2NH2) is (are) (a) It has two geometrical isomers [Adv. 2018] (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+ 59. The correct statement(s) regarding the binary transition metal carbonyl compounds is (are) (Atomic numbers: Fe = 26,Ni = 28) [Adv. 2018] (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 60. Addition of excess aqueous ammonia to a pink coloured aqueous solution of MCl2.6H2O(X) and NH4Cl 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 +

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results in the formation of a blue coloured complex Z. The calculated spin only magnetic moment of X and Z is 3.87 B.M., whereas it is zero for complex Y. Among the following options, which statement(s) is (are) correct? [Adv. 2017] (a) Addition of silver nitrate to Y gives only two equivalents of silver chloride (b) The hybridization of the central metal ion in Y is d2sp3 (c) Z is a tetrahedral complex (d) When X and Z are in equilibrium at 0°C, the colour of the solution is pink 61. If the bond length of CO bond in carbon monoxide is 1.128Å, then what is the value of CO bond length in Fe(CO)5? [2006 - 5M, –1] (a) 1.15Å (b) 1.128Å (c) 1.13Å (d) 1.118Å 62. In nitroprusside ion the iron and NO exist as FeII and NO+ rather than FeIII and NO. These forms can be differentiated by [1998 - 2 Marks] (a) estimating the concentration of iron (b) measuring the concentration of CN– (c) measuring the solid state magnetic moment (d) thermally decomposing the compound. 63. Among the following ions which one has the highest paramagnetism? [1993 - 1 Mark] 3+ (a) [Cr(H2O)6] (b) [Fe(H2O)6]2+ (c) [Cu(H2O)6]2+ (d) [Zn(H2O)6]2+ 64.

Match the metals (column I) compound(s)/enzyme(s) (column II) :

with

the

coordination

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[Main Jan. 11, 2019 (I)] (column I) Metals

(column II) Coordination compound(s)/enzyme(s) (A) Co (i) Wilkinson catalyst (B) Zn (ii) Chlorophyll (C) Rh (iii)Vitamin B12 (D) Mg (iv) Carbonic anhydrase (a) (A)-(iii); (B)-(iv); (C)-(b) (A)-(i); (B)-(ii); (C)(i); (D)-(ii) (iii); (D)-(iv) (c) (A)-(ii); (B)-(i); (C)-(d) (A)-(iv); (B)-(iii); (C)(iv); (D)-(iii) (i); (D)-(ii) 65. Match each set of hybrid orbitals from LIST-I with complex(es) given in LIST-II [Adv. 2018] List-I ListII 2 (A) dsp (p) [FeF6]4– (B) sp3 (q) [Ti(H2O)3Cl3] (C) sp3d2 (r) [Cr(NH3)6]3+ (D) d2sp3 (s) [FeCl4]2– (t) [Ni(CO)4] (w) [Ni(CN)4]2– The correct option is (a) A-t; B-s, w ; C-q, r; D-p (b) A-t, w; B-s; C- q; D-p, q (c) A-w; B-s, t ; C-p; D-q, r (d) A-s, w; B-t, w ; C-p, q; D-r 66. Match each coordination compound in List-I with an appropriate pair of characteristics from List- II and select the correct answer using the code given below the lists.

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{en = H2NCH2CH2NH2; atomic numbers : Ti = 22; Cr = 24; Co = 27; Pt = 78} [Adv. 2014] List-I ListII (A) [Cr(NH3)4Cl2]Cl (p) Paramagnetic and exhibits ionisation isomerism (B) [Ti(H2O)5Cl](NO3)2 (q) Diamagnetic and exhibits cistrans isomerism (C) [Pt(en)(NH3)Cl]NO3 (r) Paramagnetic and exhibits cistrans isomerism (D) [Co(NH3)4(NO3)2]NO3(s) Diamagnetic and exhibits ionisation isomerism Code: A (a) (s) (b) (r) (c) (q) (d) (p) 67. Match the Column II.

B C (q) (r) (p) (s) (p) (r) (r) (s) complexes in

D (p) (q) (s) (q) Column I with their properties listed in [2007]

(A) (B) (C) (D)

Column I [Co(NH3)4(H2O)2] Cl2 [Pt(NH3)2Cl2] [Co(H2O)5Cl]Cl [Ni(H2O)6]Cl2

(p) (q) (r) (s)

Column II geometrical isomers paramagnetic diamagnetic metal ion with +2 oxidation state

The coordination number of Ni2+ is 4.

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68. (a) (b) (c) (d) 69. (a) (b) (c) (d) 70. (a) (b) (c) (d)

The IUPAC name of A and B are [2006 - 5M, –2] Potassium tetracyanonickelate (II), potassium tetrachloronickelate (II) Tetracyanopotassiumnickelate (II), tetrachloropotassiumnickelate (II) Tetracyanonickel (II), tetrachloronickel (II) Potassium tetracyanonickel (II), potassium tetrachloronickel (II) Predict the magnetic nature of A and B [2006 - 5M, –2] Both are diamagnetic A is diamagnetic and B is paramagnetic with one unpaired electron A is diamagnetic and B is paramagnetic with two unpaired electrons Both are paramagnetic The hybridization of A and B are [2006 - 5M, –2] dsp2, sp3 sp3, sp3 dsp2, dsp2 sp3d2, d2sp3

Read the following statement-1(Asseration/Statement) and Statement -2 (Reason/Explanation) and answer as per the options given below : (a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (c) Statement-1 is True, Statement-2 is False (d) Statement-1 is False, Statement-2 is True 71.

Statement-1 : [Fe(H2O)5NO]SO4 is paramagnetic. Statement-2 : The Fe in [Fe(H2O)5NO]SO4 has three unpaired electrons. [2008]

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72. What are (A) and (B)? Give IUPAC name of (A). Find the spin only magnetic moment of (B). [2005 - 4 Marks] 73. Nickel chloride, when treated with dimethylgyloxime in presence of ammonium hydroxide, a bright red precipitate is obtained. Answer the following. [2004 - 4 Marks] (a) Draw the structure of the complex showing H-bonds (b) Give oxidation state of nickel and its hybridisation (c) Predict the magnetic behaviour of the complex 74. Write the IUPAC nomenclature of the given complex along with its hybridisation and structure. [2003 - 4 Marks] 75. Deduce the structure of [NiCl4] and [Ni(CN)4] considering the hybridization of the metal ion. Calculate the magnetic moment (spin only) of the species. [2002 - 5 Marks] 76. A metal complex having composition Cr(NH3)4Cl2Br 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 hybridization of chromium in each. Calculate their magnetic moments (spin–only value). [2001 - 5 Marks] 2-

77.

Draw the structures of

2-

,

and

.

Write the hybridisation of atomic orbitals of the transition metal in each case. [2000 - 4 Marks]

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78.

A compound of vanadium has a magnetic moment of 1.73 BM. Work out the electronic configuration of the vanadium ion in the compound. [1997 - 2 Marks] 79. Identify the complexes which are expected to be coloured. Explain [1994 - 2 Marks] (i) [Ti(NO3)4] (ii) [Cu(NCCH3)4]+ BF4– (iii) [Cr(NH3)6]3+3Cl– (iv) K3 [VF6] Topic-1 : Important Terms Coordination Number, Nomenclature andIsomerism of Coordination Compounds 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

(d) (d) (d) (c) (d) (d) (d) (c) (a) (c) (a) (d) (b) (d) (a) (b) (b) (d) (d) (a)

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21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 43. 44. 45.

(d) (d) (b) (c) (a) (d) (c) (b) (c) (a) (a) (d) (6) (6) (5) (6) (3) (8) (6) (3) (4) (b, d) (c, d) (b) Topic-2 : Bonding, Stability and Application of Coordination Compounds

1. (d) 2. (a) 3. (a) 4. (a) 5. (b)

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6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39.

(c) (d) (c) (b) (c) (c) (c) (d) (b) (d) (a) (b) (a) (c) (a) (c) (d) (c) (a) (d) (a) (c) (a) (a) (b) (c) (c) (b) (a) (c) (b) (c) (c) (b)

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40. (b) 41. (a) 42. (b) 43. (d) 44. (b) 45. (c) 46. (d) 47. (c) 48. (c) 49. (b) 50. (1) 51. (4) 52. (0) 53. (26.92) 54. (20) 56. (False) 57. (a, c) 58. (a,b,d) 59. (b, c) 60. (b,c,d) 61. (a, c) 62. (c) 63. (b) 64. (a) 65. (c) 66. (b) 67. (A): (p), (q) and (s); (B): (p), (r) and (s); (C): (q) and (s); (D): (q) and (s) 68. (a) 69. (c) 70. (a) 71. (a)

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1.

The decreasing order of reactivity of the following organic molecules towards AgNO3 solution is : [Main Sep. 04, 2020 (I)]

(a) (b) (c) (d) 2.

(C) > (D) > (A) > (B) (A) > (B) > (D) > (C) (A) > (B) > (C) > (D) (B) > (A) > (C) > (D) Which of the following compounds will form the precipitate with aq. AgNO3 solution most readily? [Main Sep. 04, 2020 (II)]

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(a)

(b) (c)

(d) 3.

Among the following compounds, which one has the shortest C – Cl bond? [Main Sep. 04, 2020 (II)]

(a)

(b) (c) H3C – Cl (d) 4.

The mechanism of SN1 reaction is given as :

A student writes general characteristics based on the given mechanism as : [Main Sep. 03, 2020 (I)]

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(1) The reaction is favoured by weak nucleophiles. would be easily formed if the substituents are bulky. (2) (3) The reaction is accompanied by racemization. (4) The reaction is favoured by non-polar solvents. Which observations are correct? (a) (1) and (2) (b) (1) and (3) (c) (1), (2) and (3) (d) (2) and (4) 5. The major product in the following reaction is : [Main Sep. 03, 2020 (II)]

(a) (b) (c) (d) 6.

Which of the following compounds will show retention in configuration on nucleophic substitution by OH– ion ? [Main Sep. 02, 2020 (I)]

(a)

(b)

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(c) (d) 7.

The major product obtained from E2-elimination of 3-bromo-2fluoropentane is : [Main Sep. 02, 2020 (II)]

(a) (b) (c) (d) 8.

Consider the reaction sequence given below : [Main Sep. 02, 2020 (II)]

Which of the following statements is true ? (a) Changing the base from OH to OR will have no effect on reaction (2). (b) Changing the concentration of base will have no effect on reaction (1).

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(c) Doubling the concentration of base will double the rate of both the reactions. (d) Changing the concentration of base will have no effect on reaction (2). 9. The decreasing order of reactivity towards dehydrohalogenation (E1) reaction of the following compounds is: [Main Jan. 08, 2020 (I)] (A) (B) (C) (D) (a) (b) (c) (d) 10.

D>B>C>A B>D>A>C B>D>C>A B>A>D>C Consider the following reactions:

(1) (CH3)3CCH(OH)CH3 (2) (CH3)2CHCH(Br)CH3 (3) (CH3)2CHCH(Br)CH3 (4) – CH2–CHO Which of these reaction(s) will not produce Saytzeff product? [Main Jan. 07, 2020 (I)] (a) (1), (3) and (4) (b) (4) only (c) (3) only (d) (2) and (4) 11. The major product of the following reaction is :

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[Main Jan. 12, 2019 (II)] (a) CH3CH = C = CH2 (b) (c) CH3CH = CHCH2NH2 (d) CH3CH2 C CH 12. Which hydrogen in compound (E) is easily replaceable during bromination reaction in presence of light? [Main Jan. 10, 2019 (I)] CH3 – CH2 – CH = CH2 

(a) (b) (c) (d) 13.



 

- hydrogen - hydrogen - hydrogen - hydrogen The major product of the following reaction is : [Main 2018]

(a)

(b)

(c) 14.

(d) Which of the following will most readily give the dehydrohalogenation product?

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[Main Online April 15, 2018 (I)] (a)

(b)

(c) 15.

(a) (b) (c) (d) 16.

(d) Which one of the following reagents is not suitable for the elimination reaction ? [Main Online April 10, 2016] NaI NaOEt/ EtOH NaOH / H2O NaOH / H2O – EtOH The synthesis of alkyl fluorides is best accomplished by : [Main 2015]

(a) (b) (c) (d) 17.

Finkelstein reaction Swarts reaction Free radical fluorination Sandmeyer's reaction In SN2 reactions, the correct order of reactivity for the following compounds: [Main 2014] CH3Cl, CH3CH2Cl, (CH3)2CHCl and (CH3)3CCl is: (a)

(b) (c) (d)

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18. For the compounds CH3Cl, CH3Br, CH3I and CH3F, the correct order of increasing C-halogen bond length is: [Main Online April 9, 2014] (a) CH3F < CH3Cl < CH3Br < CH3I (b) CH3F < CH3Br < CH3Cl < CH3I (c) CH3F < CH3I < CH3Br < CH3Cl (d) CH3Cl < CH3Br < CH3F < CH3I 19. The order of reactivity of the given haloalkanes towards nucleophile is : [Main Online April 23, 2013] (a) RI > RBr > KCl (b) RCl > RBr > RI (c) RBr > RCl > RI (d) RBr > RI > RCl 20. KI in acetone, undergoes SN2 reaction with each of P, Q, R and S. The rates of the reaction vary as [Adv. 2013]

(a) (b) (c) (d) 21.

P>Q>R>S S>P>R>Q P>R>Q>S R>P>S>Q The reagent(s) for the following conversion,

is/are [2007] (a) alcoholic KOH (b) alcoholic KOH followed by NaNH2 (c) aqueous KOH followed by NaNH2

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(d) Zn/CH3OH 22. What would be the product formed when 1−bromo−3−chlorocyclobutane reacts with two equivalents of metallic sodium in ether? [2005S] (a) (b) (c) (d) 23.

Identify the set of reagent / reaction conditions 'X' and 'Y' in the following set of transformations [2002S]

(a) (b) (c) (d) 24.

X = dilute aqueous NaOH, 20°C; Y = HBr/acetic acid, 20°C X = concentrated alcoholic NaOH, 80°C; Y = HBr/acetic acid, 20°C X = dilute aqueous NaOH, 20°C; Y = Br2/CHCl3, 0°C X = concentrated alcoholic NaOH, 80°C; Y = Br2/CHCl3, 0°C A solution of (+) –2–chloro–2–phenylethane in toluene racemises slowly in the presence of small amount of SbCl5, due to the formation of [1999 - 2 Marks] carbanion carbene free-radical carbocation (CH3)3CMgCl on reaction with D2O produces : [1997 - 1 Mark] (CH3)3 CD

(a) (b) (c) (d) 25. (a)

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(b) (CH3)3OD (c) (CD3)3CD (d) (CD3)3OD 26. Isobutyl magnesium bromide with dry ether and ethyl alcohol gives : [1995S] (a) (b) (c) (d) 27.

The chief reaction product of reaction between n-butane and bromine at 130ºC is : [1995S] (a) CH3CH2CH2CH2Br (b) (c) (d) 28. (a) (b) (c) (d) 29.

(a) (b)

CH3CH2CH(Br)CH3 CH3 – CH2CH(Br)CH2Br CH3CH2C(Br2)CH3 1-Chlorobutane on reaction with alcoholic potash gives [1991 - 1 Mark] 1-butene 1-butanol 2-butene 2-butanol The number of structural and configurational isomers of a bromo compound, C5H9Br, formed by the addition of HBr to 2-pentyne respectively are [1988 - 1 Mark] 1 and 2 2 and 4

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(c) 4 and 2 (d) 2 and 1 30. n-Propyl bromide on treatment with ethanolic potassium hydroxide produces [1987 - 1 Mark] (a) Propane (b) Propene (c) Propyne (d) Propanol 31. The reaction conditions leading to the best yields of C2H5Cl are : [1986 - 1 Mark] (a) C2H6 (excess) + Cl2 (b) C2H6 + Cl2 (c) C2H6 + Cl2 (excess) (d) C2H6 + Cl2 32.

Among the following compounds, the most acidic is [2011]

(a) (b) (c) (d) 33. (a) (b) (c) (d)

p-nitrophenol p-hydroxybenzoic acid o-hydroxybenzoic acid p-toluic acid The number of stereoisomers obtained by bromination of trans-2butene is [2007] 1 2 3 4

34.

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compound on hydrolysis in aqueous acetone will give [2005S] (i)

(ii)

(iii) (a) Mixture of (i) and (ii) (b) Mixture of (i) and (iii) (c) Only (iii) (d) Only (i) 35. Which of the following compounds exhibits stereoisomerism? [2002S] (a) (b) (c) (d) 36. (a) (b) (c) (d) 37.

(a)

2-methylbutene-1 3-methylbutyne-1 3-methylbutanoic acid 2-methylbutanoic acid An SN2 reaction at an asymmetric carbon of a compound always gives [2001S] an enantiomer of the substrate a product with opposite optical rotation a mixture of diastereomers a single stereoisomer The order of reactivity of the following alkyl halides for a SN2 reaction is [2000S] RF > RCl > RBr > RI

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(b) RF > RBr > RCl > RI (c) RCl > RBr > RF > RI (d) RI > RBr > RCl > RF 38. Which of the following has the highest nucleophilicity? [2000S] (a) F (b) OH– (c) –

(d) 39. (a) (b) (c) (d) 40.

How many optically active stereoisomers are possible for butane-2, 3diol? [1997] 1 2 3 4 The Cl—C—Cl angle in 1,1,2,2-tetrachloroethene tetrachloromethane respectively will be about

and [1988]

(a) (b) (c) (d)

120º and 109.5º 90º and 109.5º 109.5º and 90º 109.5º and 120º

41.

For the given compound X, the total number of optically active stereoisomers is ______. [Adv. 2018]

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42. In the following monobromination reaction, the number of possible chiral products is [Adv. 2016]

43. The total number of alkenes possible by dehydrobromination of 3bromo-3-cyclopentylhexane using alcoholic KOH is [2011] 44.

The total number of monohalogenated organic products in the following (including stereoisomers) reaction is __________.

[Main Sep. 03, 2020 (I)] 45.

Vinyl chloride on reaction with dimethyl copper gives ............... . [1997 - 1 Mark]

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46.

The interaction of elemental sulphur with Grignard reagent gives .............. [1991 - 1 Mark] 47. Isomers which are ............... mirror images are known as ............... . [1988] (superimposable, non-superimposable, enantiomers, diastereomers, epimers) 48. The compound prepared by the action of magnesium on dry ethyl bromide in ether is known as ............. reagent. [1982 - 1 Mark] 49. The halogen which is most reactive in the halogenation of alkanes under sunlight is ................. . (chlorine, bromine, iodine) [1981 - 1 Mark]

50.

2, 3, 4-Trichloropentane has three asymmetric carbon atoms.

[1990] 51. During SN1 reaction, the leaving group leaves the molecule before the incoming group is attached to the molecule. [1990] 52. The reaction of vinyl chloride with hydrogen iodide to give 1-chloro1-iodoethane is an example of anti-Markovnikov’s rule. [1989 - 2 Marks] 53. Iodide is a better nucleophile than bromide. [1985 - ½ Mark] 54.

For the following compounds, the correct statement(s) with respect to nucleophilic substitution reaction is(are) [Adv. 2017]

(I)

(II)

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(III)

(IV)

(a) I and III follow SN1 mechanism (b) I and II follow SN2 mechanism (c) Compound IV undergoes inversion of configuration (d) The order of reactivity for I, III and IV is : IV > I > III 55. The correct statement(s) about the compound given below is (are) [2008]

(a) (b) (c) (d) 56.

The compound is optically active The compound possesses centre of symmetry The compound possesses plane of symmetry The compound possesses axis of symmetry Which of the following have asymmetric carbon atom?

(a)

(b)

[1989]

(c)

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(d)

57.

The Fischer projection of D-erythrose is shown below.

D-Erythrose and its isomers are listed as P, Q, R, and S in Column-I. Choose the correct relationship of P, Q, R, and S with D-erythrose from Column II. [Adv. 2020] Column-I Column-II P.

1. Diastereomer

Q.

2. Identical

R.

3. Enantiomer

S. (a) P → 2, Q → 3, R → 2, S → 2 (b) P → 3, Q → 1, R → 1, S → 2 (c) P → 2, Q → 1, R → 1, S → 3

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(d) P → 2, Q → 3, R → 3, S → 1 58.

(a) (b) (c) (d) 59.

(a) (b) (c) (d) 60.

An ‘Assertion’ and a ‘Reason’ are given below. Choose the correct answer from the following options : [Main April 12, 2019 (II)] Assertion (A) : Vinyl halides do not undergo nucleophilic substitution easily. Reason (R) : Even though the intermediate carbocation is stabilized by loosely held p-electrons, the cleavage is difficult because of strong bonding. Both (A) and (R) are wrong statements. Both (A) and (R) are correct statements and (R) is the correct explanation of (A). Both (A) and (R) are correct statements but (R) is not the correct explanation of (A). (A) is a correct statement but (R) is a wrong statement. Statement -1: Molecules that are not superimpossable on their mirror images are chiral. [2007] Statement -2: All chiral molecules have chiral centres. If both Statement-1 and Statement-2 are correct, and Statement-2 is the correct explanation of the Statement-2. If both Statement-1 and Statement-2 are correct, but Statement -2 is not the correct explanation of the Statement-1. If Statement-1 is correct but Statement-2 is incorrect. If Statement-1 is incorrect but Statement-2 is correct. Cyclobutyl bromide on treatment with magnesium in dry ether forms an organometallic (A). The organometallic reacts with ethanal to give an alcohol (B) after mild acidification. Prolonged treatment of alcohol (B) with an equivalent amount of HBr gives 1–bromo–1– methylcyclopentane (C). Write the structures of (A), (B) and explain how (C) is obtained from (B). [2001 - 5 Marks]

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61.

Identify the pairs of enantiomers and diastereomers from the following compounds I, II and III [2000]

62.

Write the structural formula of the major product in each of the following cases :

(i) [2000 - 1 Mark] (ii) C6H5CH2CHClC6H5

2 Products [1998 - 2 Marks]

(iii) [1993 - 1 Mark] (iv) [1992 - 1 Mark] (v) [1992 - 1 Mark] (vi) bromoethane reacts with one-half of the molar quantity of silver carbonate. [1981 - ½ Mark] [1992 - 1 Mark] (vii) chloroform reacts with aniline in the presence of excess alkali

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[1981 - ½ Mark] 63. An alkyl halide, X, of formula C6H13Cl on treatment with potassium tertiary butoxide gives two isomeric alkenesY and Z (C6H12). Both alkenes on hydrogenation give2, 3-dimethylbutane. Predict the structures of X, Y and Z. [1996 - 3 Marks] 64. Draw the stereochemical structures of the products in the following reaction : [1994 - 4 Marks]

65. Arrange the following in : Increasing reactivity in nucleophilic substitution reactions CH3F, CH3I, CH3Br, CH3Cl [1992] 66. 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 give the structures of Y and Z. [1989 - 4 Marks] 67. What effect should the following resonance of vinyl chloride have on its dipole moment? [1987 - 1 Mark] – CH = Cl+ CH2 = CH – Cl 68. Write the structure of all the possible isomers of dichloroethene. Which of them will have zero dipole moment? [1985 - 2 Marks]

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69.

State the conditions under which the following preparation are carried out. Give the necessary equations which need not be balanced : (i) Lead tetraethyl from sodium-lead alloy [1983 - 1 Mark] (ii) Methyl chloride from aluminium carbide [1983 - 1 Mark] 70. (a) Show by chemical equations only, how you would prepare the following from the indicated starting materials. Specify the reagents in each step of the synthesis. (i) Hexachlorethane, C2Cl6, from calcium carbide. (ii) Chloroform from carbon disulphide. (b) Give one chemical test which would distinguish between C2H5OH from CHCl3. [1979]

1.

The decreasing order of reactivity of the following compounds towards nucleophilic substitution (SN2) is : [Main Sep. 03, 2020 (II)]

(a) (II) > (III) > (I) > (IV) (b) (II) > (III) > (IV) > (I)

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(c) (III) > (II) > (IV) > (I) (d) (IV) > (II) > (III) > (I) 2. The major product of the following reaction is: [Main Jan. 10, 2019 (II)]

(a) (b) (c) (d) 3.

The major product obtained in the following reaction is : [Main 2017]

(a) (±)C6H5CH(OtBu)CH2C6H5 (b) C6H5CH = CHC6H5 (c) (+)C6H5CH(OtBu)CH2C6H5 (d) (–)C6H5CH(OtBu)CH2C6H5 4. In a nucleophilic substitution reaction: , which one of the following undergoes complete inversion of configuration? [Main Online April 9, 2014] (a) C6H5CHC6H5Br (b) C6H5CH2Br (c) C6H5CHCH3Br (d) C6H5CCH3C6H5Br

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5.

Compound (A), C8H9Br, gives a yellow precipitate when warmed with alcoholic AgNO3. Oxidation of (A) gives an acid (B), C8H6O4. (B) easily forms anhydride on heating. Identify the compound (A). [Main 2013]

(a)

(b)

(c)

(d) 6. (a) (b) (c) (d) 7.

The Wurtz-Fittig reaction involves condensation of : [Main Online April 22, 2013] two molecules of aryl halides one molecule of each of aryl-halide and alkyl-halide. one molecule of each of aryl-halide and phenol. two molecules of aralkyl-halides. Aryl fluoride may be prepared from arene diazonium chloride using : [Main Online April 9, 2013]

(a) (b) (c) (d)

HBF4/∆ HBF4/NaNO2,Cu, ∆ CuF/HF Cu/HF

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8.

The major product of the following reaction is – [2008]

(a)

(b)

(c)

(d)

9. (a) (b) (c) (d)

When phenyl magnesium bromide reacts with tert −butanol, the product would be [2005S] Benzene Phenol ter−butylbenzene ter−butyl phenyl ether

10.

.

How many structures for F are possible? [2003S] (a) 2 (b) 5

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(c) 6 (d) 3 11. The reaction of toluene with chlorine in presence of ferric chloride gives predominantly : [1986 - 1 Mark] (a) benzoyl chloride (b) m-chlorotoluene (c) benzyl chloride (d) o- and p-chlorotoluene 12. Chlorobenzene can be prepared by reacting aniline with : [1984 - 1 Mark] (a) hydrochloric acid (b) cuprous chloride (c) chlorine in presence of anhydrous aluminium chloride (d) nitrous acid followed by heating with cuprous chloride 13.

m-Chlorobromobenzene is an isomer ofm-bromochlorobenzene. [1985 - ½ Mark]

14.

Choose the correct option(s) for the following set of reactions [Adv. 2019]

(a)

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(b)

(c)

(d) 15.

(a) (b) (c) (d) 16. (a) (b) (c) (d) (e) 17.

Benzyl chloride (C6H5CH2Cl) can be prepared from toluene by chlorination with [1998 - 2 Marks] SO2Cl2 SOCl2 Cl2 NaOCl Aryl halides are less reactive towards nucleophilic substitution reaction as compared to alkyl halides due to: [1990 - 1 Mark] The formation of less stable carbonium ion Resonance stabilization Longer carbon-halogen bond The inductive effect sp2 hybridized carbon attached to the halogen. Given below are certain matching type questions, where two columns (each having 4 items) are given. Immediately after the columns the matching grid is given, where each item of Column I has to be matched with the items of Column II, by encircling the correct match(es). Note that an item of Column I can match with more than one item of Column II. All the items of Column II must be matched. Match the following :

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[2006 - 6M] Column I (A) C6H5CH2CD2Br with

on

Column II reaction(p) E1 reaction

C2H5O– gives C6H5– CH=CD2 (B) PhCHBrCH3 and PhCHBrCD3, (q) E2 reaction both react with the same rate (C) C6H5CH2CH2Br on treatment (r) E1cB with C2H5O– and C2H5OD gives reaction C6H5CD=CH2 (D) C6H5CH2CH2Br reacts faster (s) First order than C6H5CD2CH2Br on reaction reaction with C2H5O– in ethanol

18.

(a) (b) (c) (d)

Read the following Statement-1 (Assertion) and Statement -2 (Reason) and answer as per the options given below : Statement-1 : Bromobenzene upon reaction with Br2/Fe gives 1,4dibromobenzene as the major product. [2008S] Statement-2 : In bromobenzene, the inductive effect of the bromo group is more dominant than the mesomeric effect in directing the incoming electrophile. Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 Statement-1 is True, Statement-2 is False Statement-1 is False, Statement-2 is True

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19.

Give reasons for the following : Acidic solution

(i)

Neutral

solution.

Explain. [2005 - 1 Mark] (ii) 7-Bromo-1, 3, 5-cycloheptatriene exists as ionic compound, while 5bromo-1, 3-cyclopentadiene does not ionise even in presence of Ag+ ion. Explain. [2004 - 2 Marks] 20. Write the structural formula of the major product in the following case : Me

I + Cu + heat

[1997 - 1 Mark] 21. How will you prepare m-bromoiodobenzene from benzene (in not more than 5-7 steps)? [1996 - 2 Marks] 22. Give reasons for the following : Aryl halides are less reactive than alkyl halides towards nucleophilic reagents [1994] 23. What happens when excess chlorine is passed through boiling toluene in the presence of sunlight? [1987 - 1 Mark]

1.

The major organic compound formed by the reaction of 1, 1, 1trichloroethane with silver powder is:

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[Main 2014] (a) (b) (c) (d) 2. (a) (b) (c) (d) 3.

Acetylene Ethene 2 - Butyne 2 - Butene The major product formed when 1, 1, 1-trichloropropane is treated with aqueous potassium hydroxide, is: [Main Online April 19, 2014] propyne 1-propanol 2-propanol propionic acid Chlorobenzne reacts with trichloro acetaldehyde in the presence of H2SO4.

The major product formed is: [Main Online April 11, 2014] (a)

(b)

(c)

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(d)

Topic-1 : Preparation and Properties of Haloalkanes 1. (d) 2. (d) 3. (d) 4. (c) 5. (c) 6. (d) 7. (b) 8. (b) 9. (a) 10. (c) 11. (d) 12. (b) 13. (b) 14. (a) 15. (a) 16. (b) 17. (b) 18. (a) 19. (a) 20. (b) 21. (b) 22. (d) 23. (b) 24. (d) 25. (a) 26. (b) 27. (b) 28. (a) 29. (b)

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30. (b) 31. (a) 32. (c) 33. (a) 34. (a) 35. (d) 36. (d) 37. (d) 38. (c) 39. (b) 40. (a) 41. (7) 42. (5) 43. (5) 44. (8) 50. (False) 51. (True) 52. (False) 53. (False) 54. (a, b, c) 55. (a, d) 56. (c, d) 57. (c) 58. (d) 59. (c) Topic-2 : Preparation and Properties of Haloarenes 1. (b) 2. (d) 3. (b) 4. (c) 5. (d) 6. (b) 7. (a) 8. (a) 9. (a)

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10. (d) 11. (d) 12. (d) 13. (False) 14. (a, b) 15. (c) 16. (b, e) 17. (A)-q; (B) -p, s; (C) -r, s; (D) - q 18. (c) Topic-3 : Polygen Compounds 1. (c) 2. (d) 3. (c)

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1.

The major product of the following reaction is : [Main Sep. 05, 2020 (II)]

(a)

(b)

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(c)

(d)

2.

When neopentyl alcohol is heated with an acid, it slowly converted into an 85 : 15 mixture of alkenes A and B, respectively. What are these alkenes? [Main Sep. 04, 2020 (I)]

(a)

(b)

(c)

(d) 3.

The gas evolved on heating CH3MgBr in methanol is : [Main Online April 9, 2016] (a) Methane(b)Ethane (c) Propane (d) HBr

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4.

The

most

suitable

reagent is:

for

the

conversion

of

[Main 2014] (a) (b) (c) (d) 5.

KMnO4 K2Cr2O7 CrO3 PCC (Pyridinium chlorochromate) Which one of the following statements is not correct? [Main Online April 11, 2014] (a) Alcohols are weaker acids than water (b) Acid strength of alcohols decreases in the following RCH2OH > R2CHOH > R3COH (c) Carbon-oxygen bond length in methanol, CH3OH is shorter than that of C – O bond length in phenol. (d) The bond angle

in methanol is 108.9°.

6.

An unknown alochol is treated with the “Lucas reagent” to determine whether the alcohol is primary, secondary or tertiary. Which alcohol reacts fastest and by what mechanism : [Main 2013] (a) secondary alcohol by SN1 (b) tertiary alcohol by SN1 (c) secondary alcohol by SN2 (d) tertiary alcohol by SN2 7. Rate of dehydration of alcohols follows the order: [Main Online April 9, 2013] (a) 2° > 1° > CH3OH > 3° (b) 3° > 2° > 1° > CH3OH (c) 2° > 3° > 1° > CH3OH (d) CH3OH > 1° > 2° > 3° 8. The best method to prepare cyclohexene from cyclohexanol is by using [2005S] (a) Conc. HCl + ZnCl2

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(b) Conc. H3PO4 (c) HBr (d) Conc. HCl 9. The product of acid catalyzed hydration of 2-phenylpropene is [2004S] (a) (b) (c) (d) 10.

3-phenyl-2-propanol 1-phenyl-2-propanol 2-phenyl-2-propanol 2-phenyl-1-propanol 1−propanol and 2−propanol can be best distinguished by

[2001S] (a) oxidation with alkaline KMnO4 followed by reaction with Fehling solution (b) oxidation with acidic dichromate followed by reaction with Fehling solution (c) oxidation by heating with copper followed by reaction with Fehling solution (d) oxidation with concentrated H2SO4 followed by reaction with Fehling solution 11. The compound that will react most readily with NaOH to form methanol is [2001S] (a) (b) CH3OCH3 (c) (d) (CH3)3CCl 12. Which of the following compounds is oxidised to prepare methyl ethyl ketone? [1987 - 1 Mark] (a) 2-Propanol (b) l-Butanol (c) 2-Butanol (d) t-Butyl alcohol 13. HBr reacts fastest with : [1986 - 1 Mark]

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(a) (b) (c) (d) 14. (a) (b) (c) (d) 15. (a) (b) (c) (d) 16. (a) (b) (c)

2-methylpropan-2-ol propan-1-ol propan-2-ol 2-methylpropan-1-ol An industrial method of preparation of methanol is : [1984 - 1 Mark] catalytic reduction of carbon monoxide in presence of ZnO–Cr2O3 by reacting methane with steam at 900ºC with a nickel catalyst by reducing formaldehyde with lithium aluminium hydride by reacting formaldehyde with aqueous sodium hydroxide solution The compound which reacts fastest with Lucas reagent at room temperature is [1981 - 1 Mark] butan-1-ol butan-2-ol 2-methylpropan-1-ol 2-methylpropan-2-ol Which of the following is basic [1980] CH3 – CH2 – OH OH – CH2 – CH2 – OH H–O–O–H

(d) 17.

Ethyl alcohol is heated with conc H2SO4 the product formed is [1980]

(a) (b) C H 2

6

(c) C2H4 (d) C2H2

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18.

Total number of hydroxyl groups present in a molecule of major product P is ___ [Adv. 2019]

19.

The number of hydroxyl group(s) in Q is [Adv. 2015]

20.

The number of chiral centres present in [B] is ________. [Main Sep. 04, 2020 (I)]

21.

Glycerine contains one .................. hydroxy group.

22.

23.

[1997 - 1 Mark] Ethanol vapour is passed over heated copper and the product is treated with aqueous NaOH. The final product is ............... . [1983 - 1 Mark] Sodium ethoxide is prepared by reacting ethanol with aqueous sodium hydroxide.

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[1986 - 1 Mark] 24.

(a) (b) (c) (d) 25.

The correct combination of names for isomeric alcohols with molecular formula C4H10O is/are [Adv. 2014] tert-butanol and 2-methylpropan-2-ol tert-butanol and 1, 1-dimethylethan-1-ol n-butanol and butan-1-ol isobutyl alcohol and 2-methylpropan-1-ol with HBr gives

The reaction of

[1998 - 2 Marks] (a) (b) CH3CH2CHBr

OH

(c) CH3CHBrCH2

Br

(d) CH3CH2CHBr

Br

26.

(a) (b) (c) (d)

Read the following statement and explanation and answer as per the options given below : [1988 - 2 Marks] Statement (S) : Solubility of n-alcohols in water decreases with increase in molecular weight. Explanation (E) : The relative proportion of the hydrocarbon part in alcohols increases with increasing molecular weight which permits enhanced hydrogen bonding with water. Both (S) and (E) are correct and (E) is the correct explanation of (S). Both (S) and (E) are correct but (E) is not the correct explanation of (S). (S) is correct but (E) is wrong. (S) is wrong but (E) is correct.

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27.

Identify (X) and (Y) in the following reaction sequence. [2005 - 2 Marks] (Y)

28.

A biologically active compound, bombykol (C16H30O) is obtained from a natural source. The structure of the compound is determined by the following reactions. [2002 - 5 Marks] (a) On hydrogenation, bombykol gives a compound A, C16H34O, which reacts with acetic anhydride to give an ester; (b) Bombykol also reacts with acetic anhydride to give another ester, which on oxidative ozonolysis (O3/H2O2) gives a mixture of butanoic acid, oxalic acid and 10-acetoxydecanoic acid. Determine the number of double bonds in bombykol. Write the structures of compound A and bombykol. How many geometrical isomers are possible for bombykol? 29. Complete the following reaction with appropriate reagents : [1999 - 3 Marks]

30.

Explain briefly the formation of the products giving the structures of the intermediates. [1999 - 3 Marks]

(i)

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(ii) 31. Give reasons for the following : Acid catalysed dehydration of t-butanol is faster than that of n-butanol. [1998 - 2 Marks] 32. An optically active alcohol A (C6H10O) absorbs two moles of hydrogen per mole of A upon catalytic hydrogenation and gives a product B. The compound B is resistant to oxidation by CrO3 and does not show any optical activity. Deduce the structures of A and B. [1996 - 2 Marks] 33. 3,3-Dimethylbutan-2-ol loses a molecule of water in the presence of concentrated sulphuric acid to give tetramethylethylene as a major product. Suggest a suitable mechanism. [1996 - 2 Marks] 34. When t-butanol and n-butanol are separately treated with a few drops of dilute KMnO4, in one case only the purple colour disappears and a brown precipitate is formed. Which of the two alcohols gives the above reaction and what is the brown precipitate? [1994 - 2 Marks] 35. Compound ‘X’ (molecular formula, C5H8O) does not react appreciably with Lucas reagent at room temperature but gives a precipitate with ammonical silver nitrate. With excess of MeMgBr, 0.42 g of ‘X’ gives 224 mL of CH4 at STP. Treatment of ‘X’ with H2 in presence of Pt catalyst followed by boiling with excess HI, gives n-pentane. Suggest structure for ‘X’ and write the equation involved. [1992 - 5 Marks] 36. State with balanced equations what happens when : Ethylene glycol is obtained by the reaction of ethylene with potassium permanganate. [1991 - 1 Mark]

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37.

38.

39. (ii) (i) 40. 41.

42.

A ketone ‘A’ which undergoes haloform reaction gives compound B on reduction. B on heating with sulphuric acid gives compound C, which forms monoozonide D, D on hydrolysis in presence of zinc dust gives only acetaldehyde. Identify A, B and C. Write down the reactions involved. [1989 - 4 Marks] A hydrocarbon A (molecular formula C5H10) yields 2-methylbutane on catalytic hydrogenation. A adds HBr (in accordance with Markownikoff’s rule) to form a compound B which on reaction with silver hydroxide forms an alcohol C, C5H12O. Alcohol C on oxidation gives a ketone D. Deduce the structures of A, B, C and D and show the reactions involved. [1988 - 5 Marks] Outline the reaction sequence for the conversion of 1-propanol from 2-propanol (in three steps) [1982 - 1 Mark] ethyl alcohol to vinyl acetate. (in not more than 6 steps) [1986 - 3 Marks] Give a chemical test/suggest a reagent to distinguish between methanol and ethanol. [1985 - 1 Mark] An alcohol A, when heated with conc. H2SO4 gives an alkene B. When B is bubbled through bromine water and the product obtained is dehydrohalogenated with excess of sodamide, a new compound C is obtained. The compound C gives D when treated with warm dilute H2SO4 in presence of HgSO4. D can also be obtained either by oxidizing A with KMnO4 or from acetic acid through its calcium salt. Identify A, B, C and D. [1983 - 4 Marks] An organic liquid (A), containing C, H and O with boiling point : 78°C, and possessing a rather pleasant odour, on heating with concentrated sulphuric acid gives a gaseous product (B) – with the empirical formula, CH2. ‘B’ decolourises bromine water as well as alkaline KMnO4 solution and takes up one mole of H2 (per mole of ‘B’) in the presence of finely divided nickel at high temperature. Identify the substances ‘A’ and ‘B’.

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[1979]

1.

(a) (b) (c) (d) 2.

The increasing order of boiling points of the following compounds is : [Main Sep. 05, 2020 (II)]

I < III < IV < II I < IV < III < II IV < I < II < III III < I < II < IV Consider the following reaction : [Main Sep. 03, 2020 (II)]

The product 'P' gives positive ceric ammonium nitrate test. This is because of the presence of which of these –OH group(s)? (a) (ii) only (b) (iii) and (iv) (c) (iv) only (d) (ii) and (iv) 3. Arrange the following compounds in increasing order of C – OH bond length: methanol, phenol, p-ethoxyphenol [Main Jan. 08, 2020 (I)]

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(a) (b) (c) (d) 4.

methanol < p-ethoxyphenol < phenol phenol < methanol < p-ethoxyphenol phenol < p-ethoxyphenol < methanol methanol < phenol < p-ethoxyphenol What will be the major product when m-cresol is reacted with propargyl bromide (HC º C — CH2Br) in presence of K2CO3 in acetone ? [Main April 12, 2019 (II)]

(a)

(b)

(c)

(d) 5.

The major product of the following reaction is : [Main Jan. 11, 2019 (I)]

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(a)

(b)

(c)

(d) 6.

The increasing order of the pKa values of the following compounds is: [Main Jan. 10, 2019 (I)]

(a) (b) (c) (d) 7.

C II The reaction of phenol with benzoyl chloride to give phenyl benzoate is known as : [Main Online April 23, 2013] Claisen reaction Schotten-Baumann reaction Reimer-Tiemann reaction Gatterman-Koch reaction The increasing order of boiling points of the below mentioned alcohols is [2006 - 3M, –1] 1,2-dihydroxybenzene 1,3-dihydroxybenzene

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(III)1,4-dihydroxybenzene (IV)Hydroxybenzene (a) I < II < IV < III (b) I < II < III < IV (c) IV < II < I < III (d) IV < I < II < III 15. Which of the following acids has the smallest dissociation constant ? [2002S] (a) CH3CHFCOOH (b) FCH2CH2COOH (c) BrCH2CH2COOH (d) CH3CHBrCOOH 16. In the following compounds, [1996]

The order of acidity is : (a) III > IV > I > II (b) I > IV > III > II (c) II > I > III > IV (d) IV > III > I > II 17. When phenol is reacted with CHCl3 and NaOH followed by acidification, salicyladehyde is obtained. Which of the following species are involved in the above mentioned reaction as intermediate? [1995S] (a)

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(b)

(c)

(d) 18. (a) (b) (c) (d) 19. (a) (b) (c) (d) 20. (a) (b) (c) (d)

Chlorination of toluene in the presence of light and heat followed by treatment with aqueous NaOH gives [1990 - 1 Mark] o-Cresol p-Cresol 2, 4-Dihydroxytoluene Benzoic acid Phenol reacts with bromine in carbon disulphide at low temperature to give [1988 - 1 Mark] m-bromophenol o- and p-bromophenol p-bromophenol 2, 4, 6-tribromophenol When phenol is treated with excess bromine water, it gives: [1984 - 1 Mark] m-Bromophenol o- and p-Bromophenol 2, 4-Dibromophenol 2, 4, 6-Tribromophenol

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21. An organic compound (C8H10O2) rotates plane-polarized light. It produces pink color with neutral FeCl3 solution. What is the total number of all the possible isomers for this compound? [Adv. 2020] 22. The number of resonance structures for N is [Adv. 2015]

23.

A solution of phenol in chloroform when treated with aqueous NaOH gives compound P as a major product. The mass percentage of carbon in P is ______ . (to the nearest integer) (Atomic mass: C = 12; H = 1; O = 16) [Main Sep. 06, 2020 (II)]

24.

Phenol is acidic because of resonance stabilization of its conjugate base, namely ............... . [1990 - 1 Mark] 25. Formation of phenol from chlorobenzene is an example of ............... aromatic substitution. [1989 - 1 Mark] 26. The acidity of phenol is due to the ............... of its anion. [1984 - 1 Mark

27.

The reactivity of compound Z with different halogens under appropriate conditions is given below: [Adv. 2014]

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The observed pattern of electrophilic substitution can be explained by (a) The steric effect of the halogen (b) The steric effect of the tert-butyl group (c) The electronic effect of the phenolic group (d) The electronic effect of the tert-butyl group 28. The major product(s) of the following reaction is(are) [Adv. 2013]

(a) (b) (c) (d) 29.

P Q R S In the reaction

the intermediate (s) is (are) [2010]

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(a)

(b)

(c)

(d)

30.

The reaction of

with HBr gives [1998 - 2 Marks]

(a) (b) CH3CH2CHBr

OH

(c) CH3CHBrCH2

Br

(d) CH3CH2CHBr

Br

and NaOH followed by 31. When phenol is reacted with acidification, salicylaldehyde is obtained. Which of the following

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species are involved in the above mentioned reaction as intermediates? [1995 - 2 Marks] (a)

(b)

(c)

(d) 32.

Phenol is less acidic than : [1986]

(a) (b) (c) (d)

acetic acid p-methoxyphenol p-nitrophenol ethanol

33.

Match each item of the right hand column with an appropriate item in the left hand column for each of the following sections: [Multiple Concepts 1985 - 2½ × 4 = 10 Marks] A. (i) spinel (a) MgAl2O4 (ii) feldspar (b) PbCO (iii) cerussite (c) KAlSi3O8

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(v) (e) B.

(ix) (i) (x) (j) C.

D.

34.

(i) (a) (ii) (iii) (iv) (v) (e) (vi)

(iv) malachite (d) MgSO4.H2O kisserite Cu(OH)2.CuCO3 (vi) liquid air (f) Deacon process (vii) Na2CO3 (g) Parke process (viii) nitric oxide (h) Claude process silver Ostwald process chlorine Solvay process (xi) phenol (k) coloured glass (xii) Na2S2O3 (l) antichlor (xiii) salicylic acid (m) refractory material (xiv) quick lime (n) antiseptic (xv) CuO (o) analgesic (xvi) Aston (p) radium (xvii) Priestley (q) radioactivity (xviii) Ramsay (r) oxygen (xix) Marie Curie (s) inert gas (xx) Bacquerel (t) mass spectrum Match the following, choosing one item from column X and one from column Y. [Multiple Concepts 1982 - 3 Marks] X Y pyrolysis of alkanes elimination reaction benzene+chloroethane (b) saponification (+anhydrous AlCl3) CH3COOC2H5 (c) Wurtz reaction + NaOH preparation of alkanes (d) Friedel-Crafts reaction phenol Reimer-Tiemann + CHCl3 (NaOH) reaction C2H5Br + alc. KOH (f) cracking

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PASSAGE -I Riemer-Tiemann reaction introduces an aldehyde group, on to the aromatic ring of phenol, ortho to the hydroxyl group. This reaction involves electrophilic aromatic substitution. This is a general method for the synthesis of substituted salicylaldehyde as depicted below.

35. (a) (b) (c) (d) 36. (a) (b) (c) (d) 37.

Which one of the following reagents is used in the above reaction ? [2007] aq.NaOH + CH3Cl aq.NaOH + CH2Cl2 aq.NaOH + CHCl3 aq.NaOH + CCl4 The electrophile in the reaction is [2007] : CHCl + CHCl2 : CCl2 CCl3 The structure of the intermediate I is [2007]

(a)

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(b)

(c)

(d)

38.

(c) (d)

Statement -1: Phenol is more reactive than benzene towards electrophilic substitution reactions. Statement-2:In the case of phenol, the intermediate carbocation is more resonance stabilized. [2000S] If both Statement-1 and Statement-2 are correct, and Statement -2 is the correct explanation of the Statement-2. If both Statement-1 and Statement-2 are correct, but Statement -2 is not the correct explanation of the Statement-1. If Statement-1 is correct but Statement-2 is incorrect. If Statement-1 is incorrect but Statement-2 is correct.

39.

Outline the reaction sequence for the conversion of

(a) (b)

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(i) [2003 - 2 Marks] (ii) phenol to acetophenone [1989 - 1½ Marks] 40. A compound D (C8H10O) upon treatment with alkaline solution of iodine gives a yellow precipitate. The filtrate on acidification gives a white solid E (C7H6O2). Write the structures of D and E and explain the formation of E. [1996 - 2 Marks] 41. Complete the following with appropriate structures : (i)

SO3H

..... ..... [1992 - 1 Mark]

(ii) [1986 - 1 Mark] 42. Give reasons for the following : Phenol is an acid but it does not react with sodium bicarbonate. [1987 - 1 Mark] 43. A compound of molecular formula C7H8O is insoluble in water and dilute sodium bicarbonate but dissolves in dilute aqueous sodium hydroxide. On treatment with bromine water, it readily gives a precipitate of C7H5OBr3. Write down the structure of the compound. [1985 - 2 Marks] 44. What happens when p-xylene is reacted with concentrated sulphuric acid and the resultant product is fused with KOH. [1984 - 2 Marks]

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45. State with balanced equations what happens when : acetic anhydride reacts with phenol in presence of a base. [1982 - 1 Mark]

1.

The major product [B] in the following reactions is : [Main Sep. 04, 2020 (II)]

(a) CH2 = CH2 (b) (c) (d) CH3 – CH2 – CH = CH – CH3 2. The major aromatic product C in the following reaction sequence will be : [Main Sep. 02, 2020 (I)]

(a)

(b)

(c)

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(d) 3.

An organic compound ‘A’ (C9H10O) when treated with conc. HI undergoes cleavage to yield compounds 'B' and ‘C’. ‘B’ gives yellow precipitate with AgNO3 where as ‘C’ tautomerizes to ‘D’. ‘D’ gives positive iodoform test. ‘A’ could be : [Main Sep. 02, 2020 (II)]

(a) (b) (c) (d) 4.

1-methyl ethylene oxide when treated with an excess of HBr produces: [Main Jan. 07, 2020 (I)]

(a) (b) (c) (d) 5.

The major product of the following reaction is: [Main April 10, 2019 (I)]

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(a)

(b)

(c)

(d)

6.

The major product of the following reaction is :

[Main April 8, 2019 (I)] (a)

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(b)

(c)

(d) 7.

The major product formed in the following reaction is :

[Main 2018] (a)

(b)

(c)

(d) 8.

Consider the reaction sequence below : [Main Online April 10, 2016]

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, is

(a)

(b)

(c)

(d) 9. (a) (b) (c) (d) 10.

(a)

Allyl phenyl ether can be prepared by heating: [Main Online April 9, 2014] C6H5Br + CH2 = CH – CH2 – ONa CH2 = CH – CH2 – Br + C6H5ONa C6H5 – CH = CH – Br + CH3 – ONa CH2 = CH – Br + C6H5 – CH2 – ONa An ether (A), C5H12O, when heated with excess of hot concentrated HI produced two alkyl halides which when treated with NaOH yielded compounds (B) and (C). Oxidation of (B) and (C) gave a propanone and an ethanoic acid respectively. The IUPAC name of the ether (A) is : [Main Online April 9, 2013] 2-ethoxypropane

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(b) ethoxypropane (c) methoxybutane (d) 2-methoxybutane 11. The reaction products of C6H5OCH3 + HI

is : [1995S]

(a) (b) (c) (d)

C6H5OH + CH3I C6H5I + CH3OH C6H5CH3 + HOI C6H6 + CH3OH

12.

Total number of isomers, considering both structural and stereoisomers of cyclic ethers with the molecular formula C4H8O is [Adv. 2019]

13.

Aliphatic ethers are purified by shaking with a solution of ferrous salt to remove ............... which are formed on prolonged standing in contact with air. [1992 - 1 Mark]

14.

The ether

when treated with HI

produces [1999 - 3 Marks] (a) (b) (c)

I

(d)

OH

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15.

P

Match the chemical conversions in List I with the appropriate reagents in List II and select the correct answer using the code given below the lists : [Adv. 2013] List I List II 1. (i) Hg(OAc)2; (ii) NaBH4

Q.

2. NaOEt

R.

3. Et-Br

S.

4. (i) BH3; (ii) H2O2/NaOH

(a) (b) (c) (d) 16.

17.

Codes : P Q R S 2 3 1 4 3 2 1 4 2 3 4 1 3 2 4 1 Match the following, choosing one item from column X and the appropriate item from column Y. [Multiple Concepts, 1983 - 2 Marks] X Y (i) Decarboxylation (a) Addition reaction (ii) Ozonolysis (b) Soda lime (iii) Williamson’s synthesis (c) Structure of alkene (iv) Dichloroethylene (d) Ether An organic compound (P) of molecular formula C5H10O is treated with dil. H2SO4 to give two compounds (Q) and (R) both of which

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18.

19.

20. 21. (i) (ii) 22.

23.

respond iodoform test. The rate of reaction of (P) with dil. H2SO4 is 1010 faster than the reaction of ethylene with dil. H2SO4. Identify the organic compounds, (P), (Q) and (R) and explain the extra reactivity of (P). [2004 - 4 Marks] Write the structural formula of the main organic product formed when : 2 Products. [1998 - 2 Marks] Write the intermediate steps for the following reaction.

[1998 - 1 Mark] 2, 2-Dimethyloxirane can be cleaved by acid (H ). Write mechanism. [1997 - 2 Marks] Which of the following is the correct method for synthesising methylt-butyl ether and why? (CH3)3CBr + NaOMe CH3Br + NaO-t-Bu [1997 - 2 Marks] An organic compound containing C, H and O exists in two isomeric forms A and B. An amount of 0.108 g of one of the isomers gives on combustion 0.308 g of CO2 and 0.072 g of H2O. A is insoluble in NaOH and NaHCO3 while B is soluble in NaOH. A reacts with conc. HI to give compounds C and D. C can be separated from D by ethanolic AgNO3 solution and D is soluble in NaOH. B reacts readily with bromine water to give compound E of molecular formula, C7H5OBr3. Identify, A, B, C, D and E with justification and give their structures. [1991 - 6 Marks] Give reasons for the following : +

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Sodium metal can be used for drying diethyl ether but not ethanol. [1982 - 1 Mark] 24. A compound (X) containing C, H and O is unreactive towards sodium. It does not add bromine. It also does not react with Schiff’s reagent. On refluxing with an excess of hydriodic acid, (X) yields only one organic product (Y). On hydrolysis, (Y) yields a new compound (Z) which can be converted into (Y) by reaction with red phosphorus and iodine. The compound (Z) on oxidation with potassium permanganate gives a carboxylic acid. The equivalent weight of this acid is 60. What are the compounds (X), (Y) and (Z)? Write chemical equations leading to the conversion of (X) to (Y). [1981 - 3 Marks]

Topic-1 : Preparation and Properties of Alcohols 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17.

(a) (b) (a) (d) (c) (b) (b) (b) (c) (c) (a) (c) (a) (a) (d) (a) (c)

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18. 19. 20. 23. 24. 25. 26.

(6) (4) (4) (False) (a, c, d) (b) (c) Topic-2 : Preparation and Properties of Phenols

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 27.

(b) (a) (c) (a) (b) (d) (a) (a) (a) (d) (d) (c) (b) (d) (c) (d) (d) (d) (b) (d) (6) (9) (69) (a,b,c)

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28. 29. 30. 31. 32. 35. 36. 37. 38.

(b) (a, c) (b) (a, d) (a, c) (c) (c) (b) (a) Topic-3 : Preparation and Properties of Ether

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 14. 15. 16.

(b) (a) (c) (b) (c) (b) (d) (d) (b) (a) (a) (10) (a, d) (a) (i)-(b); (ii)-(c); (iii)-(d); (iv)-(a)

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1.

The major products of the following reaction are: [Main Sep. 06, 2020 (I)]

(a)

(b)

(c)

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(d) 2.

The major product [R] in the following sequence of reactions is : [Main Sep. 04, 2020 (II)]

(a)

(b)

(c)

(d) 3.

The major product [C] of the following reaction sequence will be : [Main Sep. 04, 2020 (II)]

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(a)

(b) (c)

(d) 4.

The increasing order of the reactivity of the following compounds in nucleophilic addition reaction is : Propanal, Benzaldehyde, Propanone, Butanone [Main Sep. 03, 2020 (II)] (a) Benzaldehyde < Butanone < Propanone < Propanal (b) Butanone < Propanone < Benzaldehyde < Propanal (c) Propanal < Propanone < Butanone < Benzaldehyde (d) Benzaldehyde < Propanal < Propanone < Butanone 5. The compound A in the following reactions is : [Main Sep. 03, 2020 (II)]

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(a) (b) (c) (d) 6.

The increasing order of the following compounds towards HCN addition is : [Main Sep. 02, 2020 (I)]

(a) (b) (c) (d) 7.

(i) < (iii) < (iv) < (ii) (iii) < (iv) < (i) < (ii) (iii) < (i) < (iv) < (ii) (iii) < (iv) < (ii) < (i) Identify (A) in the following reaction sequence: [Main Jan. 09, 2020 (I)]

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(a)

(b)

(c)

(d)

8.

In the following reaction A is: [Main Jan. 09, 2020 (II)]

(a)

(b)

(c)

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(d) 9.

The major product in the following reaction is: [Main Jan. 08, 2020 (II)]

(a)

(b)

(c)

(d) 10.

What is the product of following reaction?

Hex-3-ynal [Main Jan. 07, 2020 (I)] (a) (b) (c)

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(d) 11.

The major product(s) obtained in the following reaction is/are : [Main April 12, 2019 (I)]

(a) (b) (c) (d) 12.

Major products of the following reaction are : [Main April 10, 2019 (I)]

(a)

(b)

(c) (d) CH3OH and HCO2H 13.

The major product of the following reaction is: [Main April 9, 2019 (I)]

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(a) (b) (c)

(d) 14.

In the following reaction [Main April 9, 2019 (II)]

Carbonyl compound + MeOH

acetal

Rate of the reaction is the highest for: (a) Acetone as substrate and methanol in excess. (b) Propanal as substrate and methanol in stoichiometric amount. (c) Propanal as substrate and methanol in excess. (d) Acetone as substrate and methanol in stoichiometric amount. 15. An organic compound neither reacts with neutral ferric chloride solution nor with Fehling solution. It however, reacts with Grignard reagent and gives positive iodoform test. The compound is : [Main April 8, 2019 (I)] (a)

(b)

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(c)

(d)

16.

In the following reaction [Main Jan. 12, 2019 (I)]

Aldehyde Alcohol t HCHO BuOH CH3CHO MeOH The best combination is: (a) CH3CHO and tBuOH (b) HCHO and MeOH (c) CH3CHO and MeOH (d) HCHO and tBuOH 17. The total number of optically active compounds formed in the following reaction is: [Main Online April 15, 2018 (II)]

(a) (b) (c) (d) 18.

Zero Six Four Two The correct sequence of reagents for the following conversion will be : [2017]

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(a) [Ag(NH3)2]+ OH–, H+/CH3OH, CH3MgBr (b) CH3MgBr, H+/CH3OH, [Ag(NH3)2]+ OH– (c) CH3MgBr, [Ag(NH3)2]+ OH–, H+/CH3OH (d) [Ag(NH3)2]+ OH–, CH3MgBr, H+/CH3OH 19.

A compound of molecular formula C8H8O2 reacts with acetophenone to form a single cross-aldol product in the presence of base. The same compound on reaction with conc. NaOH forms benzyl alcohol as one of the products. The structure of the compound is: [Main Online April 9, 2017]

(a)

(b)

(c)

(d)

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20.

(a) (b) (c) (d) 21.

The correct statement about the synthesis of erythritol (C(CH2OH)4) used in the preparation of PETN is : [Main Online April 10, 2016] Thy synthesis requires three aldol condensations and one Cannizzaro reaction. Alpha hydrogens of ethanol and methanol are involved in this reaction. The synthesis requires two aldol condensations and two Cannizzaro reactions. The synthesis requires four aldol condensations between methanol and ethanol. The major product of the following reaction sequence is [Adv. 2016]

(a)

(b)

(c)

(d)

22.

Which compound would give 5 - keto - 2 - methylhexanal upon ozonolysis ? [2015]

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(a)

(b)

(c)

(d)

23.

In the reaction sequence [Main Online April 11, 2015] B ; the product B is :

(a) CH3– CH2 – CH2– CH2– OH (b) CH3–CH = CH – CHO (c) CH3–CH2 –CH2–CH3 (d) 24.

(a) (b) (c) (d)

Which one of the following reactions will not result in the formation of carbon-carbon bond? [Main Online April 9, 2014] Reimer-Tieman reaction Friedel Craft’s acylation Wurtz reaction Cannizzaro reaction

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25. Tischenko reaction is a modification of: (a) Aldol condensation [Main Online April 11, 2014] (b) Claisen condensation (c) Cannizzaro reaction (d) Pinacol-pinacolon reaction 26. The major product in the following reaction is

[Adv. 2014] (a)

(b)

(c) (d) 27. (a) (b) (c) (d) 28. (a)

Formaldehyde can be distinguished from acetaldehyde by the use of : [Main Online April 9, 2013] Schiff's reagent Tollen's reagent I2/Alkali Fehling’s solution Clemmensen reduction of a ketone is carried out in the presence of : [Main Online April 22, 2013] LiAlH4

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(b) Zn-Hg with HCl (c) Glycol with KOH (d) H2 with Pt as catalyst 29. The major product H of the given reaction sequence is CH3 — CH2 — CO — CH3

G

H [2012 - II]

(a)

(b)

(c)

(d) 30.

In the following reaction,

the structure of the major product 'X' is [2007] (a)

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(b)

(c)

(d)

31.

The compound (X) is [2005S]

32.

(a) CH3COOH (b) BrCH2 − COOH (c) (CH3CO)2O (d) CHO − COOH How will you convert butan−2−one to propanoic acid? [2005S]

33.

(a) Tollen’s reagent (b) Fehling’s solution (c) NaOH/I2/H+ (d) NaOH/NaI/H+ The correct order of reactivity of PhMgBr with

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[2004S] is (a) (b) (c) (d)

(I) > (II) > (III) (III) > (II) > (I) (II) > (III) > (I) (I) > (III) > (II)

34.

Major product is : [2003S]

(a)

(b)

(c)

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(d)

35.

The product of acid hydrolysis of P and Q can be distinguished by [2003S] ,

(a) (b) (c) (d) 36.

(a) (b) (c) (d) 37.

Lucas Reagent 2,4–DNP Fehling’s Solution NaHSO3 A mixture of benzaldehyde and formaldehyde on heating with aqueous NaOH solution gives [2001S] benzyl alcohol and sodium formate sodium benzoate and methyl alcohol sodium benzoate and sodium formate benzyl alcohol and methyl alcohol The appropriate reagent for the following transformation is [2000S]

(a) Zn(Hg), HCl (b) NH2NH2/OH– (c) H2/Ni (d) NaBH4

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38.

Which one of the following will most readily be dehydrated in acidic condition ? [2000S]

(a) (b) (c)

(d) 39.

The enol form of acetone, after treatment with D2O, gives. [1999 - 2 Marks]

(a) (b) (c) (d) 40.

In the Cannizzaro reaction given below, [1996 - 1 Mark] ,

the slowest step is (a) the attack of –OH at the carbonyl group, (b) the transfer of hydride to the carbonyl group,

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(c) the abstraction of proton from the carboxylic acid, (d) the deprotonation of PhCH2OH. 41. m-Chlorobenzaldehyde on reaction with conc. KOH at room temperature gives [1991 - 1 Mark] (a) potassium m-chlorobenzoate and m-hydroxybenzaldehyde (b) m-hydroxybenzaldehyde and m-chlorobenzyl alcohol (c) m-chlorobenzyl alcohol and m-hydroxybenzyl alcohol (d) potassium m-chlorobenzoate and m-chlorobenzyl alcohol. 42. The enolic form of acetone contains [1990 - 1 Mark] (a) 9 sigma bonds, 1 pi-bond and 2 lone pairs (b) 8 sigma bonds, 2 pi-bonds and 2 lone pairs (c) 10 sigma bonds, 1 pi-bond and 1 lone pair (d) 9 sigma bonds, 2 pi-bonds and 1 lone pair 43. Polarisation of electrons in acrolein may be written as [1988 - 1 Mark] (a)

= CH –

(b)

= CH – CH =

(c)

=

(d)

= CH – CH =

44.

(a) (b) (c) (d) 45.

=O

– CH = O

The compound that will not give iodoform on treatment with alkali and iodine is : [1985 - 1 Mark] acetone ethanol diethyl ketone isopropyl alcohol The Cannizzaro reaction is not given by

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[1983 - 1 Mark] (a) (b) (c) (d) 46.

(a) (b) (c) (d) 47. (a) (b) (c) (d) 48. (a) (b) (c) (d) 49.

trimethylacetaldehye acetaldehyde benzaldehyde formaldehyde When acetaldehyde is heated with Fehling’s solution it gives a precipitate of [1983 - 1 Mark] Cu CuO Cu2O Cu + Cu2O + CuO A compound that gives a positive iodoform test is [1982 - 1 Mark] 1-pentanol 2-pentanone 3-pentanone pentanal The reagent with which both acetaldehyde and acetone react easily is [1981 - 1 Mark] Fehling’s reagent Grignard reagent Schiff’s reagent Tollen’s reagent Among the following, the number of reaction(s) that produce(s) benzaldehyde is [Adv. 2015]

I.

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II.

III.

IV. 50.

Consider all possible isomeric ketones, including stereoisomers of MW = 100. All these isomers are independently reacted with NaBH4 (NOTE: stereoisomers are also reacted separately). The total number of ketones that give a racemic product(s) is/are [Adv. 2014] 51. In the scheme given below, the total number of intramolecular aldol condensation products formed from ‘Y’ is

[2010] 52.

Consider the following reactions 2-methyl-2-butene

The mass percentage of carbon in A is ______. [Main Jan. 09, 2020 (II)] 53.

The structure of the enol form of CH3–CO–CH2–CO–CH3 with intramolecular hydrogen bonding is .................. [1993 - 1 Mark] 54. The structure of the intermediate product, formed by the oxidation of toluene with CrO3 and acetic anhydride, whose hydrolysis gives

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benzaldehyde is ............... . [1992 - 1 Mark] 55. Fehling’s solution ‘A’ consists of an aqueous solution of copper sulphate, while Fehling’s solution ‘B’ consists of an alkaline solution of ............... . [1990 - 1 Mark] 56. Each item from (i) to (x) given below indicates a reaction type, a process or a homologue. Match each of these items with the related phrase by writing the correct phrase in the corresponding vacant space given under each. The correct phrase must be picked only from those given below within brackets : (Bayer’s process, Nucleophilic addition, Free radical substitution, Ostwald’s process, Homologous pair, Cyanamide process, Electrophilic substitution, Homolytic addition, Thermite process, Nucleophilic substitution) [Multiple Concepts, 1981 - 1 × 10 = 10 Marks] (i) Cyclopropane, chlorine and light ................................................... (ii) Welding ................................................... (iii) Propanone and sodium bisulphite ................................................... (iv) Production of ammonia ................................................... (v) Chloromethane and methanol ................................................... (vi) Ore purification ................................................... (vii) Ethanal and methanal ................................................... (viii)Benzene, nitric acid and sulphuric acid ................................................... (ix) Production of nitric acid ................................................... (x) Propene, hydrogen bromide and a peroxide catalyst

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................................................... 57.

The reaction of methyl magnesium iodide with acetone followed by hydrolysis gives secondary butanol. [1987 - 1 Mark] 58. The yield of ketone when a secondary alcohol is oxidized is more than the yield of aldehyde when a primary alcohol is oxidized. [1983 - 1 Mark] 59. Benzaldehyde undergoes aldol condensation in an alkaline medium. [1982 - 1 Mark]

60.

In the reaction scheme shown below, Q, R, and S are the major products. [Adv. 2020]

The correct structure of

(a) S is

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(b) Q is

(c) R is

(d) S is

61.

Compounds P and R upon ozonolysis produce Q and S, respectively. The molecular formula of Q and S is C8H8O. Q undergoes Cannizzaro reaction but not haloform reaction, whereas S undergoes haloform reaction but not Cannizzaro reaction

(i)

P

(ii) R The option(s) with suitable combination of P and R, respectively, is (are) [Adv. 2017] (a)

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(b)

(c)

(d) 62.

The correct statement(s) about the following reaction sequence is (are) [Adv. 2016]

Q (major) + R (minor) Q (a) (b) (c) (d) 63.

R is steam volatile Q gives dark violet coloration with 1% aqueous FeCl3 solution S gives yellow precipitate with 2, 4–dinitro-phenylhydrazine S gives dark violet coloration with 1% aqueous FeCl3 solution Positive Tollen’s test is observed for [Adv. 2016]

(a)

(b)

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(c)

(d) 64.

The major product of the following reaction is [Adv. 2015]

(a)

(b)

(c)

(d) 65.

After completion of the reactions (I and II), the organic compound(s) in the reaction mixtures is (are)[Adv. 2013] Reaction I :

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Reaction II :

(a) (b) (c) (d) 66.

Reaction I : P and Reaction II : P Reaction I : U, acetone and Reaction II : Q, acetone Reaction I : T, U, acetone and Reaction II : P Reaction I : R, acetone and Reaction II : S, acetone In the following reaction, the product(s) formed is (are) [Adv. 2013]

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(a) (b) (c) (d) 67.

(a) (b) (c) (d)

P (major) Q (minor) R (minor) S (major) The smallest ketone and its next homologue are reacted with NH2OH to form oxime [2006 - 5M, –1] Two different oximes are formed Three different oximes are formed Two oximes formed are optically active All oximes formed are optically active

68.

The major products P and Q are [2006 - 5M, –1] (a)

(b)

(c)

(d) 69.

Which of the following will undergo aldol condensation? [1998 - 2 Marks]

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(a) (b) (c) (d) 70.

acetaldehyde propanaldehyde benzaldehyde trideuteroacetaldehyde Which of the following will react with water? [1998 - 2 Marks]

(a) CHCl3 (b) Cl3CCHO (c) CCl4 (d) CICH2CH2Cl 71. A new carbon-carbon bond formation is possible in (a) Cannizzaro reaction [1998 - 2 Marks] (b) Friedel–Craft alkylation (c) Clemmensen reduction (d) Reimer–Tiemann reaction 72. Which of the following are examples of aldol condensation? [1989 - 1 Mark] CH3CHOHCH2CHO (a) 2CH3CHO (b) 2CH3COCH3 CH3COH(CH3)CH2COCH3 (c) 2HCHO (d) C6H5CHO + HCHO

CH3OH C6H5CH2OH

73. Keto-enol tautomerism is observed in [1988 - 1 Mark] (a) (b) (c)

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(d) 74.

Base catalysed aldol condensation occurs with :

(a) propionaldehyde [1984 - 1 Mark] (b) benzaldehyde (c) 2-methylpropionaldehyde (d) 2, 2-dimethylpropionaldehyde 75. The correct match between Item - I (starting material) and Item - II (reagent) for the preparation of benzaldehyde is : [Main Sep. 06, 2020 (II)] Item - I (I) Benzene

(a) (b) (c) (d) 76.

(A)

Item - II (P) HCl and SnCl2, H3O+ (II) Benzonitrile (Q)H2, Pd-BaSO4, quinoline S and (III)Benzoyl (R) CO, HCl and Chloride AlCl3 (I) - (Q), (II) - (R) and (III) - (P) (I) - (P), (II) - (Q) and (III) - (R) (I) - (R), (II) - (P) and (III) - (Q) (I) - (R), (II) - (Q) and (III) - (P) Match the entries in Column I with the correctly related quantum number(s) in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS [2008] Column I Column II (p) sodium fusion extract of the compound gives Prussian blue colour with FeSO4

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(B)

(q) gives positive FeCl3 test

(C)

(r) gives white precipitate with AgNO3 (s) reacts with aldehydes to form

(D)

the corresponding hydrazone derivative

Passage-1 In the following reactions [Adv. 2015]

77.

Compound X is

(a)

(b)

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(c)

(d) 78.

The major compound Y is

(a)

(b)

(c)

(d)

Passage-2 Two aliphatic aldehydes P and Q react in the presence of aqueous K2CO3 to give compound R, which upon treatment with HCN provides compound S. On acidification and heating, S gives the product shown below. [2010]

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79.

The compounds P and Q respectively are :

(a)

(b)

(c)

(d) 80.

The compound R is :

(a)

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(b)

(c)

(d)

81.

The compound S is :

(a)

(b)

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(c)

(d)

Passage -3 In the following reaction sequence, products I, J and L are formed. K represents a reagent. [2008]

82.

The structure of the product I is –

(a) (b) (c)

83. (a)

(d)

The structures of compounds J and K, respectively, are and SOCl2

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(b)

and SOCl2 (d)

(c) and CH3SO2Cl 84.

The structure of product L is

(a) (b) (c)

(d)

Passage -4 A tertiary alcohol H upon acid catalysed dehydration gives a product I. Ozonolysis of I leads to compounds J and K. Compound J upon reaction with KOH gives benzyl alcohol and compound L, whereas K on reaction with KOH gives only M. [2008]

85.

Compound H is formed by the reaction of + PhMgBr

(a)

(b)

+

PhCH2MgBr (c)

+ PhCH2MgBr

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(d) 86.

+ The structure of compound I is

(a)

(b)

(c)

(d)

87. The structure of compounds J, K and L respecitvely, are – (a) PhCOCH3, PhCH2COCH3 and PhCH2COO–K+ (b) PhCHO, PhCH2CHO and PhCOO–K+ (c) PhCOCH3, PhCH2CHO and CH3COO–K+ (d) PhCHO, PhCOCH3 and PhCOO–K+ Each of this question contains STATEMENT-1 (Assertion/Statement) and Statement-2 (Reason/Explanation) and has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct. (a) Statement-1 is True, Statement-2 isTrue; Statement-2 is a correct explanation for Statement-1 (b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (c) Statement-1 is True, Statement-2 is False (d) Statement-1 is False, Statement-2 is True. 88. Statement-1 : p-Hydroxybenzoic acid has a lower boiling point than o-hydroxybenzoic acid. Statement-2 : o-Hydroxybenzoic acid has intramolecular hydrogen bonding. [2007] 89. Statement-1 : Dimethyl sulphide is commonly used for the reduction of an ozonide of an alkene to get the carbonyl compounds.

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Statement-2 : It reduces the ozonide giving water soluble dimethyl sulphoxide and excess of it evaporates. [2001S] 90.

A monomer of a polymer on ozonolysis gives two moles of CH2O and one mole of CH3COCHO. Write the structure of monomer and write all – ‘cis’ configuration of polymer chain. [2005 - 2 Marks]

91. [2003 - 4 Marks] (it gives negative test with Fehling solution but responds to iodoform test). (both gives positive Tollen’s test but do not give iodoform test). primary alochol. Identify from A to G. 92. Compound ‘A’ of molecular formula C9H7O2Cl exists in keto form and predominantly in enolic form ‘B’. On oxidation with KMnO4, ‘A’ gives m–chlorobenzoic acid. Identify ‘A’ and ‘B’. [2003 - 2 Marks] 93. Identify (A), (B), (C), (D) and (E) in the following schemes and write their structures : [2001 - 5 Marks]

94.

An organic compound A, C6H10O on reaction with CH3MgBr followed by acid treatment gives compound B. The compound B on ozonolysis gives compound C, which in presence of a base gives 1-

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acetylcyclopentene D. The compound B on reaction with HBr gives compound E. Write the structures of A, B, C and E. Show how D is formed from C ? [2000 - 5 Marks] 95. (i)

Write the structural formula of the main organic product formed when : Identify A, B, C and give their structures.

[2000 - 3 Marks] (ii) [2000 - 1 Mark] (iii) [1997 - 1 Mark] (iv) [1997 - 1 Mark] (v) [1997 - 1 Mark] (vi)

+

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[1997 - 1 Mark] (vii) Complete the following reaction with appropriate structure. [1996 - 1 Mark]

(viii) H3CO

CHO + HCHO [1992 - 1 Mark]

(ix)

propanal [1985 - 1 Mark]

(x) 96.

97.

methanal reacts with ammonia [1981 - ½ Mark] Complete the following reaction with appropriate structures of products/reagents : [1998 - 2 + 2 Marks]

Write the intermediate steps for the following reaction.

[1998 - 2 Marks] 98. An aldehyde A (C11H8O), which does not undergo self aldol condensation, gives benzaldehyde and two moles of B on ozonolysis. Compound B, on oxidation with silver ion gives oxalic acid. Identify the compounds A and B. [1998 - 2 Marks] 99. Suggest appopriate structures for the missing compounds. (The number of carbon atoms remains the same throughout the reactions.) [1996 - 3 Marks]

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100. An organic compound A, C8H6, on treatment with dilute sulphuric acid containing mercuric sulphate gives a compound B, which can also be obtained from a reaction of benzene with an acid chloride in the presence of anhydrous aluminium chloride. The compound B, when treated with iodine in aqueous KOH, yields C and a yellow compound D. Identify A, B, C and D with justification. Show how B is formed from A. [1994 - 3 Marks] 101. Arrange the following in increasing order of expected enol content [1992 - 1 Mark] CH3COCH2CHO, CH3COCH3, CH3CHO, CH3COCH2COCH3 102. Complete the following reactions : (i)

H3C – CHO

? [1988 - 1 Mark]

(ii)

?

? [1988 - 1 Mark]

(iii)

?

[1988 - 1 Mark] 103. An unknown compound of carbon, hydrogen and oxygen contains 69.77% carbon and 11.63% hydrogen and has a molecular weight of 86. It does not reduce Fehling solution, but forms a bisulphite addition compound and gives a positive iodoform test. What are the possible structures for the unknown compound?

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[1987 - 5 Marks] 104. A white precipitate was formed slowly when silver nitrate was added to a compound (A) with molecular formula C6H13Cl. Compound (A) on treatment with hot alcoholic potassium hydroxide gave a mixture of two isomeric alkenes (B) and (C), having formula C6H12. The mixture of (B) and (C), on ozonolysis, furnished four compounds: [1986 - 4 Marks] (i) CH3CHO; (ii) C2H5CHO; (iii) CH3COCH3 and (iv) What are the structures of (A), (B) and (C)? 105. Give a chemical test/suggest a reagent to distinguish between acetaldehyde from acetone. [1987 - 1 Mark] 106. Arrange the following in : Increasing reactivity towards HCN [1985] CH3CHO, CH3COCH3, HCHO, C2H5COCH3 107. Give reasons for the following : (i) Explain why o-hydroxybenzaldehyde is a liquid at room temperature while p-hydroxybenzaldehyde is a high melting solid. [1999 - 2 Marks] + (ii) In acylium ion, the structure R – C ≡ O is more stable than R – C+ = O. [1994 - 1 Mark] (iii) Iodoform is obtained by the reaction of acetone with hypoiodite but not with iodide ion. [1991 - 1 Mark] (iv) Hydrazones of aldehydes and ketones are not prepared in highly acidic medium. [1986 - 1 Mark] (v) Suggest a reason for the large difference between the boiling points of butanol and butanal, although they have almost the same

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solubility in water. [1985 - 2 Marks] 108. Outline the reaction sequence for the conversion of (i) Carry out the following transformation in not more than three steps. [1999 - 3 Marks]

(ii) acetylene to acetone [1985 - 1 Mark] (iii) methanal to ethanal (the number of steps should not be more than three). [1981 - 2 Marks]

1.

Which of the following derivatives of alcohols is unstable in an aqueous base? [Main Sep. 05, 2020 (I)] (a)

(b)

(c) (d) RO – CMe3 2.

The increasing order of the acidity of the α-hydrogen of the following compounds is : [Main Sep. 05, 2020 (I)]

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(a) (D) < (C) < (A) < (B) (b) (B) < (C) < (A) < (D) (c) (A) < (C) < (D) < (B) (d) (C) < (A) < (B) < (D) 3. An organic compound (A) (molecular formula C6H12O2) was hydrolysed with dil. H2SO4 to give a carboxylic acid (B) and an alochol (C). ‘C’ gives white turbidity immediately when treated with anhydrous ZnCl2 and conc. HCl. The organic compound (A) is : [Main Sep. 04, 2020 (I)] (a)

(b)

(c) (d) 4.

[P] on treatment with Br2/FeBr3 in CCl4 produced a single isomer C8H7O2Br while heating [P] with sodalime gave toluene. The compound [P] is : [Main Sep. 04, 2020 (I)]

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(a)

(b)

(c)

(d) 5.

An organic compound [A], molecular formula C10H20O2 was hydrolyzed with dilute sulphuric acid to give a carboxylic acid [B] and an alcohol [C]. Oxidation of [C] with CrO3 – H2SO4 produced [B]. Which of the following structures are not possible for [A]? [Main Sep. 03, 2020 (I)]

(a)

(b)

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(c) CH3CH2CH2COOCH2CH2CH2CH3 (d) (CH3)3 – C – COOCH2C(CH3)3

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6.

Consider the following molecules and statements related to them : [Main Sep. 03, 2020 (II)]

(A)

(B) (1) (B) is more likely to be crystalline than (A) (2) (B) has higher boiling point than (A) (3) (B) dissolves more readily than (A) in water Identify the correct option from below : [Main Sep. 03, 2020 (II)] (a) (b) (c) (d) 7.

(1) and (2) are true (1) and (3) are true only (1) is true (2) and (3) are true Arrange the following labelled hydrogens in decreasing order of acidity : [Main Sep. 02, 2020 (II)]

(a) (b) (c) (d) 8.

(ii) > (i) > (iii) > (iv) (iii) > (ii) > (iv) > (i) (ii) > (iii) > (iv) > (i) (iii) > (ii) > (i) > (iv) The most suitable reagent for the given conversion is: [Main Jan. 08, 2020 (I)]

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(a) (b) (c) (d) 9.

B2 H6 NaBH4 LiAlH4 H2/Pd An unsaturated hydrocarbon X absorbs two hydrogen molecules on catalytic hydrogenation, and also gives following reaction:

B(3-oxo-hexanedicarboxylic acid) X will be : [Main Jan. 08, 2020 (II)] (a)

(b)

(c) (d) 10. The major products of the following reaction are :

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[Main April 12, 2019 (I)]

(a)

(b)

(c)

(d)

11.

The major product of the following reaction is:

CH3CH = CHCO2CH3 [Main April 9, 2019 (I)]

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(a) (b) (c) (d) 12.

CH3CH2CH2CO2CH3 CH3CH = CHCH2OH CH3CH2CH2CH2OH CH3CH2CH2CHO The major product of the following reaction is: [Main April 8, 2019 (II)]

(a)

(b)

(c)

(d)

13. The major product of the following reaction is :

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[Main Jan. 11, 2019 (I)] (a)

(b)

(c)

(d) 14. The decreasing order of ease of alkaline hydrolysis for the following esters is [Main Jan. 10, 2019 (I)]

(a) (b) (c) (d) 15.

III > II > IV > I III > II > I > IV IV > II > III > I II > III > I > IV The major product obtained in the following reaction is:

[Main Jan. 10, 2019 (II)]

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(a)

(b)

(c)

(d) 16. The correct decreasing order for acid strength is: [Main Jan. 9, 2019 (I)] (a) NO2CH2COOH > FCH2COOH> CNCH2COOH > ClCH2COOH (b) FCH2COOH > CNCH2COOH> NO2CH2COOH > ClCH2COOH (c) CNCH2COOH > NO2CH2COOH> FCH2COOH > ClCH2COOH (d) NO2CH2COOH > CNCH2COOH> FCH2COOH > ClCH2COOH 17. The correct order of acid strength of the following carboxylic acid is [Adv. 2019]

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(a) (b) (c) (d) 18.

I > II > III > IV II > I > IV > III I > III > II > IV III > II > I > IV The major product of the given reaction is: [Main Online April 16, 2018]

(a)

(b)

(c)

(d)

19. The increasing order of the acidity of the following carboxylic acids is: [Main Online April 15, 2018 (II)]

(a) (b) (c) (d) 20.

III < II < IV < I I < III < II < IV IV < II < III < I II < IV < III < I Sodium salt of an organic acid 'X' produces effervescences 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] (a) C6H5COONa (b) HCOONa

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(c) CH3COONa (d) Na2C2O4 21. The major product obtained in the following reaction is : [2017]

(a)

(b)

(c)

(d) 22. Bouveault–Blanc reduction reaction involves : [Main Online April 9, 2016] (a) (b) (c) (d) 23.

Reduction of an acyl halide with H2/Pd Reduction of an anhydride with LiAlH4 Reduction of an ester with Na/C2H5OH Reduction of a carbonyl compound with Na/Hg and HCl. The correct order of acidity for the following compounds is

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[Adv. 2016] (a) (b) (c) (d) 24.

(a) (b) (c) (d) 25.

I > II > III > IV III> I > II > IV III > IV > II > I I > III > IV > II In the presence of a small amount of phosphorous, aliphatic carboxylic acids react with chlorine or bromine to yield a compound in which α-hydrogen has been replaced by halogen. This reaction is known as : [Main Online April 10, 2015] Wolff - Kishner reaction Rosenmund reaction Etard reaction Hell - Volhard - Zelinsky reaction In the reaction,

the product C is: [2014] (a) (b) (c) (d) 26.

Acetaldehyde Acetylene Ethylene Acetyl chloride An organic compound A upon reacting with NH3 gives B. On heating B gives C. C in presence of KOH reacts with Br2 to give CH3CH2NH2. A is : [2013] (a) CH3COOH (b) CH3CH2CH2COOH (c)

(d) CH3CH2COOH

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27. Monocarboxylic acids are functional isomers of: [Main Online April 23, 2013] (a) Ethers (b) Amines (c) Esters (d) Alcohols 28. The compound that does NOT liberate CO2, on treatment with aqueous sodium bicarbonate solution, is [Adv. 2013] (a) Benzoic acid (b) Benzenesulphonic acid (c) Salicylic acid (d) Carbolic acid (Phenol) 29. The compound that undergoes decarboxylation most readily under mild condition is [2012] (a)

(b)

(c)

(d) 30. The carboxyl functional group (– COOH) is present in [2012] (a) picric acid (b) barbituric acid (c) ascorbic acid

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(d) aspirin 31. The major product of the following reaction is [2011]

(a) (b) (c) (d)

a hemiacetal an acetal an ether an ester

32. In the reaction the structure of the product T is : [2010]

(a)

(b)

(c)

(d)

33. The compounds P, Q and S

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were separately subjected to nitration using HNO3/H2SO4 mixture. The major product formed in each case respectively, is : [2010] (a)

(b)

(c)

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(d)

34. The correct acidity order of the following is [2009S]

(I)

(a) (b) (c) (d) 35.

(II)

(III)

(IV)

(III) > (IV) > (II) > (I) (IV) > (III) > (I) > (II) (III) >(II) >(I) > (IV) (II) > (III) > (IV) > (I) In the following reaction sequence, the correct structures of E, F and G are [* implies 13C

labelled carbon) [2008] (a)

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(b)

(c)

(d) 36. 4−Methylbenzenesulphonic acid reacts with sodium acetate to give [2005S]

(a)

(b)

(c)

(d)

37. An enantiomerically pure acid is treated with a racemic mixture of an alcohol having one chiral carbon. The ester formed will be [2003S] (a) Optically active mixture (b) Pure enantiomer

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(c) Meso compound (d) Racemic mixture 38. Ethyl ester

. The product P will be [2003S]

(a)

(b)

(c)

(d)

39. Benzoyl chloride is prepared from benzoic acid by [2000S] (a) (b) (c) (d)

Cl2, SO2Cl2 SOCl2 Cl2, H2O

40. When propionic acid is treated with aqueous sodium bicarbonate, CO2 is liberated. The ‘C’ of CO2 comes from [1999 - 2 Marks] (a) methyl group (b) carboxylic acid group (c) methylene group (d) bicarbonate 41. The organic product formed in the reaction

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[1995S] C6H5COOH (a) (b) (c) (d)

C6H5CH2OH C6H5COOH & CH4 C6H5CH3 & CH3OH C6H5CH3 & CH4

42. The total number of carboxylic acid groups in the product P is [Adv. 2013]

43. Amongst the following, the total number of compounds soluble in aqueous NaOH is

[2010] 44. In the following reaction, compound Q is obtained from compound P via an ionic intermediate. [Adv. 2020]

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What is the degree of unsaturation of Q?

45. Formic acid when heated with conc. H2SO4 produces ............... . [1983 - 1 Mark] 46. The boiling point of propionic acid is less than that ofn-butyl alcohol, an alcohol of comparable molecular weight. [1991 - 1 Mark] 47. Hydrolysis of an ester in presence of a dilute acid is known as saponification. [1983 - 1 Mark]

48. Choose the correct option(s) for the following reaction sequence [Adv. 2019]

Consider Q, R and S as major products. (a)

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(b)

(c)

(d)

49. The reaction(s) leading to the formation of 1,3,5-trimethylbenzene is (are) [Adv. 2018] (a) (b)

(c)

(d)

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50. Reagent(s) which can be used to bring about the following transformation is (are) [Adv. 2016]

(a) (b) (c) (d) 51.

LiAlH4 in (C2H5)2O BH3 in THF NaBH4 in C2H5OH Raney Ni/H2 in THF With reference to the scheme given below, which of the given statement(s) about T, U, V and W is (are) correct ?

[2012] (a) (b) (c) (d) 52.

T is soluble in hot aqueous NaOH U is optically active Molecular formula of W is C10H18O4 V gives effervescences on treatment with aqueous NaHCO3. Identify the binary mixture(s) that can be separated into individual compounds, by differential extraction as shown in the given scheme. [2012]

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(a) C6H5OH and C6H5COOH (b) C6H5COOH and C6H5CH2OH (c) C6H5CH2OH and C6H5OH (d) C6H5CH2OH and C6H5CH2COOH 53. Which of the following reactants on reaction with conc. NaOH followed by acidification gives following lactone as the product? [2006 - 5M, –1]

(a)

(b)

(c)

(d) 54. Which of the following compounds will react with ethanolic KCN? [1984 - 1 Mark] (a) ethyl chloride (b) acetyl chloride (c) chlorobenzene (d) benzaldehyde 55. Which of the following compounds will give a yellow precipitate with iodine and alkali?

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[1984 - 1 Mark] (a) (b) (c) (d)

2-Hydroxypropane acetophenone methyl acetate acetamide

Directions [Qs. 56-57] : Answer the following by appropriately matching the lists based on the information given in the paragraph. List–I includes starting materials and reagents of selected chemical reactions. List–II gives structures of compounds that may be formed as intermediate products and/ or final products from the reactions of List–I. [Adv. 2019] List–I

List–II

(I)

(P)

(II)

(Q)

(III)

(R)

(IV)

(S)

(T)

(U)

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56. Which of the following options has correct combination considering List–I and List–II? (a) (I), (S), (Q), (R) (b) (II), (P), (S), (U) (c) (I), (Q), (T), (U) (d) (II), (P), (S), (T) 57. Which of the following options has correct combination considering List – I and List –II? (a) (IV), (Q), (R) (b) (IV), (Q), (U) (c) (III), (S), (R) (d) (III), (T), (U) Directions [Qs. 58-60] : By appropriately matching the information given in the three columns of the following table. Columns 1, 2 and 3 contain starting materials, reaction conditions, and type of reactions, respectively. [Adv. 2017] Column 1 Column 2 Column 3 (I) Toluene (i) NaOH/Br2 (P) Condensation (II) Acetophenone(ii) Br2/hv (Q) Carboxylation (III)Benzaldehyde (iii) (CH3CO)2O/ (R) Substitution CH3COOK (IV)Phenol (iv) NaOH/CO2 (S) Haloform 58. For the synthesis of benzoic acid, the only CORRECT combination is (a) (II) (i) (S) (b) (IV) (ii) (P) (c) (I) (iv) (Q) (d) (III) (iv) (R) 59. The only CORRECT combination that gives two different carboxylic acids is (a) (II) (iv) (R) (b) (IV) (iii) (Q) (c) (III) (iii) (P) (d) (I) (i) (S) 60. The only CORRECT combination in which the reaction proceeds through radical mechanism is (a) (III) (ii) (P) (b) (IV) (i) (Q) (c) (II) (iii) (R) (d) (I) (ii) (R) 61. Different possible thermal decomposition pathways for peroxy esters are shown below. Match each pathway from List-I with an appropriate structure

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from List-II and select the correct answer using the code given below the lists. [Adv. 2014]

List-I List-II P. Pathway 1. P Q. Pathway 2. Q R. Pathway 3. R

S. Pathway 4. S

Code : P Q R S (a) 1 3 4 2 (c) 4 1 2 3 (b) 2 4 3 1 (d) 3 2 1 4 62. Match the reactions in Column I with appropriate types of steps/reactive intermediate involved in these reactions as given in Column II. [2011]

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Column I

Column II (p)Nucleophilic substitution

(A)

(B)

(q)Electrophilic substitution

(C)

(r) Dehydration

(D)

(s) Nucleophilic addition

(t) Carbanion 63. Match each of the compounds given in Column-I with the reaction(s), that they can undergo, given in Column-II. [2009] Column-I

Column-II

(A)

(p) Nucleophilic substitution

(B)

(q) Elimination

(C)

(r) Nucleophilic addition

(D)

(s) Esterification acetic anhydride (t) Dehydrogenation

with

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64. Match the compounds/ions in Column I with their properties/reactions in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS. [2007] Column Column II I (A)C6H5CHO(p) gives precipitate with 2, 4dinitrophenylhydrazine (B)CH3C ≡ (q) gives precipitate with AgNO3 CH (C)CN– (D)I–

(r) is a nucleophile (s) is involved in cyanohydrin formation

Passage-1 An organic acid P (C11H12O2) can easily be oxidized to a dibasic acid which reacts with ethylene glycol to produce a polymer dacron. Upon ozonolysis, P gives an aliphatic ketone as one of the products. P undergoes the following reaction sequences to furnish R via Q. The compound P also undergoes another set of reactions to produce S. [Adv. 2018]

65. The compound R is (a)

(b)

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(c)

(d)

66. The compound S is [Adv. 2018]

(a)

(b)

(c)

(d)

Passage-2 Treatment of benzene with CO/HCl in the presence of anhydrous AlCl3/CuCl followed by reaction with Ac2O/NaOAc gives compound X as the major product. Compound X upon reaction with Br2/Na2CO3, followed by heating at 473 K with

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alc. KOH furnishes Y as the major product. Reaction of X with H2/Pd-C, followed by H3PO4 treatment gives Z as the major product. [Adv. 2018] 67. The compound Y is

(a)

(b)

(c)

(d) 68. The compound Z is (a)

(b)

(c)

(d) Passage-3 The reaction of compound P with CH3MgBr (excess) in (C2H5)2O followed by addition of H2O gives Q. The compound Q on treatment with H2SO4 at 0ºC gives R. The reaction of R with CH3COCl in the presence of anhydrous AlCl3 in CH2Cl2

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followed by treatment with H2O produces compound S. [Et in compound P is ethyl group]

[Adv. 2017] 69.

The product S is

(a)

(b)

(c)

(d)

70. (a) (b) (c) (d)

The reactions, Q to R and R to S, are Dehydration and Friedel–Craft’s acylation Aromatic sulfonation and Friedel–Craft’s acylation Friedel–Craft’s alkylation, dehydration and Friedel–Craft’s acylation Friedel–Craft’s alkylation and Friedel–Craft’s acylation Passage-4

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P and Q are isomers of dicarboxylic acid C4H4O4. Both decolorize Br2/H2O. On heating, P forms the cyclic anhydride. Upon treatment with dilute alkaline KMnO4, P as well as Q could produce one or more than one from S, T and U.

[Adv. 2013] 71. (a) (b) (c) (d) 72.

Compounds formed from P and Q are, respectively Optically active S and optically active pair [T, U] Optically inactive S and optically inactive pair [T, U] Optically active pair [T, U] and optically active S Optically inactive pair [T, U] and optically inactive S In the following reaction sequences V and W are respectively

(a)

(b)

(c)

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(d)

Passage-5 In the following reaction sequence, the compound J is an intermediate.

J (C9H8O2) gives effervescences on treatment with NaHCO3 and also give positive Baeyer’s test. [2012] 73. The compound I is

(a)

(b)

(c)

(d)

74. The compound K is

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(a)

(c)

(b)

(d)

Each of this question contains STATEMENT-1 (Assertion/Statement ) and STATEMENT-2 (Reason/Explanation) and has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct. (a) Statement-1 is True, Statement-2 isTrue; Statement-2 is a correct explanation for Statement-1 (b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (c) Statement-1 is True, Statement-2 is False (d) Statement-1 is False, Statement-2 is True. 75. Statement-1 : Acetic acid does not undergo haloform reaction. Statement-2 : Acetic acid has no alpha hydrogens. [1998 - 2 Marks] 76. Statement-1 : Acetate ion is more basic than the methoxide ion. Statement-2 : The acetate ion is resonance stabilized [1994 - 2 Marks] 77. Match the Ka values [2003] Ka (a) Benzoic acid

6.4 × 10

–5

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(b)

30.6 × 10–5

(c)

10.2 × 10–5

(d)

3.3 × 10–5

(e)

4.2 × 10–5

78. A racemic mixture of (±) 2–phenylpropanoic acid on esterification with (+) 2–butanol gives two esters. Mention the stereochemistry of the two esters produced. [2003 - 2 Marks] 79. Identify (X), (Y) and (Z) in the following synthetic scheme and write their structures. Ba*CO3 + H2SO4 → (X) gas [C* denotes C14] [2001 - 5 Marks] Explain the formation of labelled formaldehyde (H2C*O) as one of the products when compound (Z) is treated with HBr and subsequently ozonolysed. Mark the C* carbon in the entire scheme. 80. An organic compound A, C8H4O3, in dry benzene in the presence of anhydrous AlCl3 gives compound B. The compound B on treatment with PCl5, followed by reaction with H2/Pd (BaSO4) gives compound C, which on reaction with hydrazine gives a cyclic compound D (C14H10N2). Identify A, B, C and D. Explain the formation of D from C. [2000 - 5 Marks] 81. Write the structural formula of the main organic product formed when : (i)

Write the structures of the products A and B.

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[2000 - 2 Marks] (ii)

[1997 - 1 Mark] (iii)

. [1995 - 2 Marks] (iv)

[1995 - 1 Mark] (v)

[1995 - 2 Marks] (vi) [1994 - 1 Mark] (vii) [1993 - 2 Marks] (viii) benzene [1985 - 1 Mark] (ix)

ethyl acetate is treated with double the molar quantity of ethyl magnesium bromide and the reaction mixture poured into water.

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[1981 - ½ Mark] 82. Explain briefly the formation of the products giving the structures of the intermediates.

83.

84. (i)

(ii) (iii)

85.

(i)

[1999 - 5 Marks] An ester A (C4H8O2), on treatment with excess methyl magnesium chloride followed by acidification, gives an alcohol B as the sole organic product. Alcohol B, on oxidation with NaOCl followed by acidification, gives acetic acid. Deduce the structures of A and B. Show the reactions involved. [1998 - 6 Marks] Give reasons for the following : Although phenoxide ion has more number of resonating structures than benzoate ion, benzoic acid is a stronger acid than phenol. Why? [1997 - 2 Marks] Formic acid is a stronger acid than acetic acid; [1985 - 1 Mark] Acetic acid can be halogenated in the presence of red P and Cl2 but formic acid cannot be halogenated in the same way. [1983 - 1 Mark] Which of the following carboxylic acids undergoes decarboxylation easily? Explain briefly. [1995 - 2 Marks] C6H5–CO–CH2–COOH

(ii) C6H5–CO–COOH

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(iii) (iv) 86. An organic compound A, C8H6, on treatment with dilute sulphuric acid containing mercuric sulphate gives a compound B, which can also be obtained from a reaction of benzene with an acid chloride in the presence of anhydrous aluminium chloride. The compound B, when treated with iodine in aqueous KOH, yields C and a yellow compound D. Identify A, B, C and D with justification. Show how B is formed from A. [1994 - 3 Marks] 87. In the following reactions identify the compounds A, B, C and D. [1994 - 1 × 4 = 4 Marks] PCl5 + SO2 → A + B A + CH3COOH → C + SO2 + HCl 2C+ (CH3)2Cd→

2D

+

CdCl2

88.

An organic compound ‘A’ on treatment with ethyl alcohol gives a carboxylic acid ‘B’ and compound ‘C’. Hydrolysis of ‘C’ under acidic conditions gives ‘B’ and ‘D’. Oxidation of ‘D’ with KMnO4 also gives ‘B’. ‘B’ on heating with Ca(OH)2 gives ‘E’ (molecular formula, C3H6O). ‘E’ does not give Tollen’s test and does not reduce Fehling’s solution but forms a 2,4dinitrophenylhydrazone. Identify ‘A’, ‘B’, ‘C’, ‘D’ and ‘E’. [1992 - 3 Marks] 89. Compound ‘X’, containing chlorine on treatment with strong ammonia gives a solid ‘Y’ which is free from chlorine. ‘Y’ analysed as C = 49.31%, H = 9.59% and N = 19.18% and reacts with Br2 and caustic soda to give a basic compound ‘Z’. ‘Z’ reacts with HNO2 to give ethanol. Suggest structures for ‘X’, ‘Y’ and ‘Z’. [1992 - 1 Mark] 90. Arrange the following in : Increasing order of acid strength : [1991] ClCH2COOH (I), CH3CH2COOH (II), ClCH2CH2COOH (III), (CH3)2CHCOOH (IV),

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CH3COOH (V) 91. (i)

C6H5COOH

C D

C6H5CN

E;

Identify C, D and E. [1991 - 2 Marks] (ii) H3C – CH = CH–CHO

F G

H;

Identify F, G and H. 92.

93.

94. (i) (ii) (iii) 95.

[1991 - 2 Marks] Compound A (C6H12O2) on reduction with LiAlH4 yielded two compounds B and C. The compound B on oxidation gave D, which on treatment with aqueous alkali and subsequent heating furnished E. The latter on catalytic hydrogenation gave C. The compound D was oxidized further to give F which was found to be a monobasic acid (molecular weight = 60.0). Deduce the structures of A, B, C, D and E. [1990 - 4 Marks] The sodium salt of a carboxylic acid, A, was produced by passing a gas, B, into an aqueous solution of caustic alkali at an elevated temperature and pressure. A, on heating in presence of sodium hydroxide followed by treatment with sulphuric acid gave a dibasic acid, C. A sample of 0.4 g of acid C, on combustion gave 0.08 g of water and 0.39 g of carbon dioxide. The silver salt of the acid C weighing 1.0 g on ignition yielded 0.71 g of silver as residue. Identify A, B and C. [1990 - 5 Marks] Outline the reaction sequence for the conversion of Ethanoic acid to a mixture of methanoic acid and diphenyl ketone. [1990 - 2Marks] Ethanal to 2-hydroxy-3-butenoic acid. [1990 - 2 Marks] acetic acid to tertiary-butyl alcohol. [1989 - 1½ Marks] Complete the following reactions :

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CH3COOH

ClCH2COOH

?

[1988 - 1 Mark] 96. An organic compound (A) on treatment with acetic acid in the presence of sulphuric acid produces an ester (B), (A) on mild oxidation gives (C), (C) with 50% potassium hydroxide followed by acidification with dilute hydrochloric acid generates (A) and (D), (D) with phosphorus pentachloride followed by reaction with ammonia gives (E), (E) on dehydration produces hydrocyanic acid. Identify the compounds A, B, C, D and E. [1987 - 5 Marks] 97. Complete the following with appropriate structures : (i) (CH3CO)2O CH3COOH + ? [1986 - 1 Mark] (ii) ?

– CH = CH – CHO

[1986 - 1 Mark] 98. A liquid (X), having a molecular formula C6H12O2 is hydrolysed with water in the presence of an acid to give a carboxylic acid (Y) and an alcohol (Z). Oxidation of (Z) with chromic acid gives (Y). What are the structures of (X), (Y) and (Z)? [1986 - 3 Marks] 99. Arrange the following in increasing ease of hydrolysis CH3COOC2H5, CH3COCl, (CH3CO)2O, CH3CONH2. [1986 - 1 Mark] 100. Write down the reactions involved in the preparation of the following using the reagents indicated against it in parenthesis : Propionic anhydride from propionaldehyde [AgNO3/NH4OH, P2O5]. [1984 - 2 Marks] 101. State the conditions under which the following preparations are carried out. Give the necessary equations which need not be balanced. (i) Ethanol from acetylene [1983 - 1 Mark] (ii) Acetic acid from methyl iodide [1983 - 1 Mark]

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102. An alkene (A) on ozonolysis yields acetone and an aldehyde. The aldehyde is easily oxidized to an acid (B). When B is treated with bromine in presence of phosphorus, it yields a compound (C) which on hydrolysis gives a hydroxy acid (D). This acid can also be obtained from acetone by reaction with hydrogen cyanide followed by hydrolysis. Identify the compounds A, B, C and D. [1982 - 2 Marks] 103. Outline the accepted mechanism of the following reaction. Show the various steps including the charged intermediates. [1981 - 3 Marks]

104. Write the chemical equation to show what happens when ethyl acetate is treated with sodium ethoxide in ethanol and the reaction mixture is acidified. [1981 - 2 Marks]

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1.

Which of these will produce the highest yield in Friedel Crafts reaction? [Main Jan. 09, 2020 (I)]

(a)

(b)

(c)

(d) 2.

A compound ‘X’ on treatment with Br2/NaOH, provided C3H9N, which gives positive carbylamine test. Compound ‘X’ is :

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[Main Jan. 11, 2019 (II)] (a) CH3COCH2NHCH3 (c) CH3CH2CH2CONH2 3. The compounds A :

(a) (b) (c) (d) 4.

(b) CH3CH2COCH2NH2 (d) CH3CON(CH3)2 and B in the following reaction are, respectively [Main Jan. 9, 2019 (I)]

A = Benzyl alcohol, B = Benzyl cyanide A =Benzyl chloride, B = Benzyl cyanide A = Benzyl alcohol, B = Benzyl isocyanide A = Benzyl chloride, B = Benzyl isocyanide The increasing order of nitration of the following compounds is : [Main Online April 15, 2018 (I)]

(a) (A) < (B) < (D) < (C) (b) (A) < (B) < (C) < (D) (c) (B) < (A) < (C) < (D) (d) (B) < (A) < (D) < (C) 5. Which of the following compounds will form significant amount of meta product during mono-nitration reaction ? [Main 2017] (a)

(b)

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(c)

(d) 6.

The correct stability order of the following resonance structures is [2009]

(a) (I) > (II) > (IV) > (III) (b) (I) > (III) > (II) > (IV) (c) (II) > (I) > (III) > (IV) (d) (III) > (I) > (IV) > (II) 7. In the following reaction, [2006] → Nitrogen containing compound + KCl + H2O. The nitrogen containing compound is (a)

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(b) (c) (d) 8.

(a) (b) (c) (d) 9. (a) (b) (c) (d)

When benzenesulfonic acid and p-nitrophenol are treated with NaHCO3, the gases released respectively are [2006] SO2, NO SO2, NO2 CO2, CO2 SO2, CO2 Benzamide on reaction with POCl3 gives [2004S] aniline chlorobenzene benzylamine benzonitrile

10.

The product A will be [2003S]

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(a)

(b)

(c)

(d)

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11.

The most unlikely representation of resonance structures of pnitrophenoxide ion is [1999 - 2 Marks]

(a)

(b)

(c)

(d)

12.

Butanenitrile may be prepared by heating : [1992 - 1 Mark]

(a) Propyl alcohol with KCN (b) Butyl alcohol with KCN (c) Butyl chloride with KCN

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(d) Propyl chloride with KCN 13. The formation of cyanohydrin from a ketone is an example of : [1990 - 1 Mark] (a) Electrophilic addition (b) Nucleophilic addition (c) Nucleophilic substitution (d) Electrophilic substiution 14. The compound that is most reactive towards electrophilic nitration is : [1985 - 1 Mark] (a) toluene (b) benzene (c) benzoic acid (d) nitrobenzene

15.

Amongst the three isomers of nitrophenol, the one that is least soluble in water is ............... . [1992 - 1 Mark]

16.

When nitrobenzene is treated with Br2 in presence of FeBr3, the major product formed is m-bromonitrobenzene. Statements which are related to obtain the m-isomer are [1992 - 1 Mark] The electron density on meta carbon is more than that on ortho and para positions The intermediate carbonium ion formed after initial attack of Br+ at the meta position is least destabilised Loss of aromaticity when Br+ attacks at the ortho and para positions and not at meta position Easier loss of H+ to regain aromaticity from the meta position than from ortho and para positions. The products of reaction of alcoholic silver nitrite with ethyl bromide are [1991 - 1 Mark]

(a) (b) (c) (d) 17.

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(a) (b) (c) (d) (e)

ethane ethene nitroethane ethyl alcohol ethyl nitrite

18.

Match each of the compounds in Column I with its characteristic reaction(s) in Column II. [2009] Column I Column II (A) CH3CH2 CH2CN (p) Reduction with Pd–C/H2 (B) CH3 CH2 OCOCH3 (q) Reduction with SnCl2/HCl (C) CH3–CH= CH–CH2OH (r) Development of foul smell on treatment with chloroform and alcoholic KOH (D) CH3CH2CH2CH2NH2 (s) Reduction with diisobutylaluminium hydride(DIBAL-H) (t) Alkaline hydrolysis

Read the following Statement-1(Asseration) and Statement -2 (Reason) and answer as per the options given below : (a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (c) Statement-1 is True, Statement-2 is False (d) Statement-1 is False, Statement-2 is True 19. Statement - 1: Benzonitrile is prepared by the reaction of chlorobenzene with potassium cyanide.

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Statement - 2 : Cyanide (CN–) is a strong nucleophile. [1998 - 2 Marks] 20. Statement - 1: p-Nitrophenol is a stronger acid thano-nitrophenol. Statement - 2 : Intramolecular hydrogen bonding makes the o-isomer weaker than the p-isomer. [1989 - 2 Marks] 21.

Give reasons for the following :

(i)

but

[2005 - 1 Mark] (ii)

(a)

but

(b) No release of (iii)

[2005 - 1 Mark] Nitrobenzene does not undergo Friedel-Craft’s alkylation [1998 - 2 Marks]

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(iv)

o-Nitrophenol is steam volatile whereas p-nitrophenol is not; [1985 - 1 Mark] 22. Identify (A) to (D) in the following series of reactions. [2004 - 4 Marks]

23.

Write structures of the products A, B, C, D and E in the following scheme. [2002 - 5 Marks]

24.

Identify the major product in the following reactions :

(i)

[2000 - 1 Mark]

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(ii) [2000 - 1 Mark] (iii) [1993 - 1 Mark] 25.

Complete the following with appropriate structures :

(i)

+ (COOEt)2 + EtONa [1997 - 1 Mark]

(ii) [1992 - 1 Mark] (iii)

CONH2

[1992 - 1 Mark] 26. Show with equations how the following compounds are prepared (equations need not be balanced) : (i) 4-nitrobenzaldehyde from benzene. [1994 - 2 Marks] (ii) p-bromonitrobenzene from benzene in two steps. [1993 - 2 Marks] (iii) toluene to m-nitrobenzoic acid? [1987 - 1 Mark]

1.

The increasing order of pKb values of the following compounds is : [Main Sep. 06, 2020 (I)]

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(a) (b) (c) (d) 2.

II < IV < III < I I < II < IV < III II < I < III < IV I < II < III < IV Which of the following compounds can be prepared in good yield by Gabriel phthalimide synthesis? [Main Sep. 06, 2020 (II)]

(a) (b) (c)

(d) 3.

The most appropriate reagent for conversion of C2H5CN into CH3CH2CH2NH2 is : [Main Sep. 05, 2020 (I)] (a) NaBH4 (b) CaH2

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(c) LiAlH4 (d) Na(CN)BH3 4. The increasing order of basicity of the following compounds is: [Main Sep. 05, 2020 (I)]

(a) (b) (c) (d) 5.

(A) < (B) < (C) < (D) (B) < (A) < (D) < (C) (D) < (A) < (B) < (C) (B) < (A) < (C) < (D) In the following reaction sequence, [C] is : [Main Sep. 04, 2020 (II)]

(a) (b) (c) (d)

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6.

Three isomers A, B and C (mol. formula C8H11N) give the following results :

A and C R (product of A) + S (product of C) R has lower boiling point than S alkali-insoluble product B A, B and C, respectively are : [Main Sep. 03, 2020 (II)] (a)

(b)

(c)

(d) 7.

The major product Z obtained in the following reaction scheme is: [Main Jan. 09, 2020 (I)]

(a)

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(b)

(c)

(d) 8.

Consider the following reactions,

The compound [P] is: [Main Jan. 09, 2020 (II)]

(a)

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(b)

(c)

(d) 9.

The decreasing order of basicity of the following amines is: [Main Jan. 09, 2020 (II)]

(a) (b) (c) (d) 10.

(A) > (C) > (D) > (B) (C) > (A) > (B) > (D) (B) > (C) > (D) > (A) (C) > (B) > (A) > (D) The increasing order of pKb for the following compounds will be: [Main Jan. 07, 2020 (I)] NH2 – CH = NH,

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(a) (b) (c) (d) 11.

(B) < (C) < (A) (A) < (B) < (C) (C) < (A) < (B) (B) < (A) < (C) In the following reaction sequence,

the major product B is: [Main Jan. 07, 2020 (II)]

(a)

(b)

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(c)

(d)

12.

Hinsberg’s reagent is: [Main April 9, 2019 (II)]

(a) (b) (c) (d) 13. (a) (b) (c) (d) 14.

C6H5COCl SOCl2 C6H5SO2Cl (COCl)2 Which of the following amines can be prepared by Gabriel phthalimide reaction ? [Main April 8, 2019 (I)] n-butylamine triethylamine t-butylamine neo-pentylamine In the following compound,

the favourable site/s for protonation is/are : [Main Jan. 11, 2019 (II)] (a) (a) and (e) (b) (b), (c) and (d)

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(c) (a) and (d) (d) (a) 15. Arrange the following amines in the decreasing order of basicity : [Main Jan. 9, 2019 (I)]

(a) (b) (c) (d) 16.

I > II > III III > I > II III > II > I I > III > II The increasing basicity order of the following compounds is: [Main Jan. 9, 2019 (II)] (A) CH3CH2NH2 (B)

(C) (D) (a) (b) (c) (d) 17.

(D) < (C) < (B) < (A) (D) < (C) < (A) < (B) (A) < (B) < (C) < (D) (A) < (B) < (D) < (C) The increasing order of diazotisation of the following compounds is: [Main Online April 15, 2018 (II)]

(A)

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(B) (C) (D) (a) (b) (c) (d) 18.

(D) < (C) < (B) < (A) (A) < (D) < (B) < (C) (A) < (B) < (C) < (D) (A) < (D) < (C) < (B) Among the following compounds, the increasing order of their basic strength is : [Main Online April 9, 2017]

(I)

(II)

(III)

(IV) (a) (b) (c) (d) 19.

(I) < (II) < (IV) < (III) (I) < (II) < (III) < (IV) (II) < (I) < (IV) < (III) (II) < (I) < (III) < (IV) The order of basicity among the following compounds is

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[Adv. 2017]

(a) II > I > IV > III (b) IV > II > III > I (c) IV > I > II > III (d) I > IV > III > II 20. In the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br2 used per mole of amine produced are : [Main 2016] (a) Two moles of NaOH and two moles of Br2. (b) Four moles of NaOH and one mole of Br2. (c) One mole of NaOH and one mole of Br2. (d) Four moles of NaOH and two moles of Br2. 21. The test to distinguish primary, secondary and tertiary amines is : [Main Online April 9, 2016] (a) Sandmeyer's reaction (b) Carbylamine reaction (c) Mustard oil test (d) C6H5SO2Cl 22. In the reaction

the product E is : [Main 2015]

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(a)

(b)

(c)

(d) 23.

Arrange the following amines in the order of increasing basicity. [Main Online April 10, 2015]

(a)

(b)

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(c)

(d)

24. (a) (b) (c) (d) 25.

(a) (b) (c) (d) 26.

(a) (b) (c) (d) 27.

On heating an aliphatic primary amine with chloroform and ethanolic potassium hydroxide, the organic compound formed is: [Main 2014] an alkanol an alkanediol an alkyl cyanide an alkyl isocyanide The final product formed when methyl amine is treated with NaNO2 and HCl is: [Main Online April 19, 2014] Diazomethane Methylalcohol Methylcyanide Nitromethane A compound with molecular mass 180 is acylated with CH3COCl to get a compound with molecular mass 390. The number of amino groups present per molecule of the former compound is : [Main 2013] 2 5 4 6 The order of basicity of amines in gaseous state is : [Main Online April 23, 2013]

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(a) (b) (c) (d) 28.

1° > 2° > 3° > NH3 3° > 2° > NH3 > 1° 3° > 2° > 1° > NH3 NH3 > 1° > 2° > 3° Amongst the compounds given, the one that would form a brilliant colored dye on treatment with NaNO2 in dil. HCl followed by addition to an alkaline solution of β-naphthol is [2011 - II]

(a) (b)

(c)

(d) 29.

The major product of the following reaction is [2011 - I]

(a)

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(b)

(c)

(d)

30.

In the compound given below, the correct order of the acidity of the positions X, Y and Z is [2004S]

(a) (b) (c) (d) 31.

Z>X>Y X>Y>Z X>Z>Y Y>X>Z The major product obtained when Br2/Fe is treated with

[2004S]

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(a)

(b)

(c)

(d)

32.

[2003S] (a)

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(b) (c) (d) 33.

Compound 'A' (molecular formula C3H8O) is treated with acidified potassium dichromate to form a product 'B' (molecular formula C3H6O). 'B' forms a shining silver mirror on warming with ammonical silver nitrate. 'B' when treated with an aqueous solution of H2NCONHNH2.HCl and sodium acetate gives a product 'C'. Identify the structure of 'C'. (a) CH3CH2CH = NNHCONH2 [2002S] (b) (c) (d) CH3CH2CH = NCONHNH2 34. The correct order of basicities of the following compounds is [2001S] 1. 2. 3.

CH3 − CH2 − NH2 (CH3)2NH

4. (a) 2 > 1 > 3 > 4 (b) 1 > 3 > 2 > 4

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(c) 3 > 1 > 2 > 4 (d) 1 > 2 > 3 > 4 35. Among the following, the strongest base is [2000S] (a) (b) (c) (d) 36.

(a) (b) (c) (d) 37. (a) (b) (c) (d) 38. (a) (b) (c) (d) 39. (a) (b) (c)

C6H5NH2 p-NO2.C6H4NH2 m-NO2.C6H4.NH2 C6H5CH2NH2 In the reaction p-chlorotoluene with KNH2 in liq. NH3,the major product is: [1997 - 1 Mark] o-toluidine m-toluidine p-toluidine p-chloroaniline. Amongst the following, the most basic compound is : [1990 - 1 Mark] Benzylamine Aniline Acetanilide p-Nitroaniline Carbylamine test is performed in alcoholic KOH by heating a mixture of : [1984 - 1 Mark] chloroform and silver powder trihalogenatedmethane and a primary amine an alkyl halide and a primary amine an alkyl cyanide and a primary amine Acetamide is treated separately with the following reagents. Which one of these would give methylamine? [1983 - 1 Mark] PCl5 NaOH + Br2 soda lime

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(d) hot conc. H2SO4 40. The compound which on reaction with aqueous nitrous acid at low temperature produces an oily nitrosoamine is [1981 - 1 Mark] (a) methylamine (b) ethylamine (c) diethylamine (d) triethylamine 41. Schemes 1 and 2 describes the conversion of P and Q and R to S, respectively, scheme 3 describes the synthesis of T from Q and S. The total number of Br atoms in a molecule of T is ____ [Adv. 2019] Scheme 1 :

Scheme 2 :

Scheme 3 :

42.

Consider the reaction sequence from P to Q shown below. The overall yield of the major product Q from P is 75%. What is the amount in grams of Q obtained from 9.3 mL of P? (Use density of P = 1.00 g mL–1; Molar mass of C = 12.0, H = 1.0, O = 16.0 and N = 14.0 g mol– 1 ) [Adv. 2020]

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43.

In the following reaction sequence, the amount of D (in g) 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) [Adv. 2018]

.

44.

The high melting point and insolubility in organic solvents of sulphanilic acid are due to its........structure. [1994 - 1 Mark] 45. In an acidic medium, ................. behaves as the strongest base. (nitrobenzene, aniline, phenol) [1981 - 1 Mark]

46.

Consider the following four compounds I, II, III, and IV.

Choose the correct statement(s).

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[Adv. 2020 (a) The order of basicity is II > I > III > IV. (b) The magnitude of pKb difference between I and II is more than that between III and IV. (c) Resonance effect is more in III than in IV. (d) Steric effect makes compound IV more basic than III. 47. In the following reactions, the product S is [Adv. 2015]

(a) (b)

(c)

(d) 48.

In the reaction shown below, the major product(s) formed is/are [Adv. 2014]

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(a)

(b)

(c)

(d)

49.

Hydrogen bonding plays a central role in the following phenomena [Adv. 2014] (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 50. In the reaction 2X + B2H6 → [BH2(X)2] + [BH4]– the amine(s) X is (are) [2009] (a) NH3

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(b) CH3NH2 (c) (CH3)2NH (d) (CH3)3N 51. A positive carbylamine test is given by [1999 - 2 Marks] (a) (b) (c) (d) 52.

N, N—dimethylaniline 2, 4–dimethylaniline N–methyl-o-methylaniline p-methylbenzylamine Among the following compounds, which will react with acetone to give a product containing > C = N–bond ? [1998 - 2 Marks] (a) C6H5NH2 (b) (CH3)3N (c) C6H5NHC6H5 (d) C6H5NHNH2. 53. p-Chloroaniline and anilinium hydrochloride can be distinguished by [1998 - 2 Marks] (a) Sandmeyer reaction (b) NaHCO3 (c) AgNO3 (d) Carbylamine test 54. Examine the following two structures for the anilinium ion and choose the correct statement from the ones given below: [1993 - 1 Mark]

(a) II is not an acceptable canonical structure because carbonium ions are less stable than ammonium ions. (b) II is not an acceptable canonical structure because it is non-aromatic.

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(c) II is not an acceptable canonical structure because its nitrogen has 10 valence electrons. (d) II is an acceptable canonical structure. 55.

Reaction of

with a mixture of Br2 and KOH gives R-NH2

as the main product. The intermediates involved in this reaction are : [1992 - 1 Mark] (a) (b) R – NHBr (c) R – N = C = O (d)

56. [Adv. 2014] List - I

List - II

P.

1. Scheme I

Q.

2. Scheme II

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R.

3. Scheme III

S.

4. Scheme IV

(a) (b) (c) (d)

Code: P 1 3 3 4

Q 4 1 4 1

R 2 4 2 3

S 3 2 1 2

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PASSAGE - 1 Treatment of compound (O) with KMnO4/H+ gave (P), which on heating with ammonia gave (Q). The compound (Q) on treatment with Br2/NaOH produced (R). On strong heating, (Q) gave (S), which on further treatment with ethyl 2-bromopropanoate in the presence of KOH followed by acidification, gave a compound (T). [Adv. 2016]

57. The compound (R) is (a)

(b)

(c)

(d)

58. The compound (T) is

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(a) (b) (c) (d)

glycine alanine valine serine

PASSAGE - 2 The conversion of an amide to an amine with one carbon atom less by the action of alkaline hypohalite is known as Hofmann bromamide degradation.

In this reaction, RCONHBr is formed from which the reaction has derived its name. Hofmann reaction is accelerated if the migrating group is more electron-releasing. Hofmann degradation reaction is an intramolecular reaction. 59. How can the conversion of (i) to (ii) be brought about? [2006 - 5M, –2] (a) KBr (b) KBr + CH3ONa (c) KBr + KOH (d) Br2 + KOH 60. Which is the rate determining step in Hofmann bromamide degradation? [2006 - 5M, –2] (a) Formation of (i) (b) Formation of (ii)

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(c) Formation of (iii) (d) Formation of (iv) 61. What are the constituent amines formed when the mixture of (i) and (ii) undergoes Hofmann bromamide degradation? [2006 - 5M, –2]

(a)

,

(b)

,

and

(c)

and

(d)

and

62.

and

Statement-1 : In strongly acidic solutions, aniline becomes more reactive towards electrophilic reagents. Statement-2 : The amino group being completely protonated in strongly acidic solution, the lone pair of electrons on the nitrogen is

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no longer available for resonance. (a) (b) (c) (d)

[2001S] Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 Statement-1 is True, Statement-2 is False Statement-1 is False, Statement-2 is True

63.

[2005 - 4 Marks] (i) Identify (X) and (Y) (ii) Is (Y) optically active? (iii) Give structure(s) of intermediate(s), if any, in the formation of (Y) from (X). 64. There is a solution of p–hydroxybenzoic acid and p-aminobenzoic acid. Discuss one method by which we can separate them and also write down the confirmatory tests of the functional groups present. [2003 - 4 Marks] 65. Explain briefly the formation of the products giving the structures of the intermediates.

[1999 - 2 Marks] 66. Compound A (C8H8O) on treatment with NH2OH. HCl gives B and C. B and C rearrange to give D and E, respectively, on treatment with acid. B, C, D and E are all isomers of molecular formula (C8H9NO). When D is boiled with alcoholic KOH an oil F (C6H7N) separates out. F reacts rapidly with CH3COCI to give back D. On the other hand, E

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67.

on boiling with alkali followed by acidification gives a white solid G (C7H6O2). Identify A-G. [1999 - 7 Marks] Complete the following with appropriate structures : 2 products

(i)

[1998 - 2 Marks] 2 products

(ii)

[1998 - 2 Marks] (iii)

NH2 +

COCl

? [1986 - 1 Mark]

68. Give reasons for the following : (i) Dimethylamine is a stronger base than trimethylamine. [1998 - 2 Marks] (ii) Cyclohexylamine is a stronger base than aniline. [1982 - 1 Mark] 69. Acetophenone on reaction with hydroxylamine hydrochloride can produce two isomeric oximes. Write structures of the oximes. [1997 - 2 Marks] 70. Write the structure of the foul-smelling compound obtained when aniline is treated with chloroform in the presence of KOH. [1996 - 1 Mark] 71. Identify, A (C3H9N) which reacts with benzensulphonyl chloride to give a solid, insoluble in alkali. [1993 - 1 Mark] 72. A basic, volatile nitrogen compound gave a foul smelling gas when treated with chloroform and alcoholic potash. A 0.295 g sample of the substance. dissolved in aq. HCl and treated with NaNO2 solution at 0°C, liberated a colorless, odourless gas whose volume corresponded to 112 mL at STP. After the evolution of the gas was complete, the aqueous solution was distilled to give an organic liquid which did not contain nitrogen and which on warming with alkali and iodine gave a yellow precipitate. Identify the original substance. Assume that it contains one N atom per molecule. [1993 - 4 Marks]

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73.

A mixture of two aromatic compounds A and B was separated by dissolving a chloroform followed by extraction with aqueous KOH solution. The organic layer containing compound A, when heated with alcoholic solution of KOH produced a compound C(C7H5N) associated with an unpleasant odour. The alkaline aqueous layer on the other hand, when heated with chloroform and then acidified gave a mixture of two isomeric compounds D and E of molecular formula C7H6O2. Identify the compounds A, B, C, D, E and write their structures. [1990 - 4 Marks] 74. Give a chemical test and the reagents used to distinguish between the following pair of compounds : Ethylamine and diethylamine. [1988 - 1 Mark] 75. An organic compound A, containing C, H, N and O, on analysis gives 49.32% carbon, 9.59% hydrogen and 19.18% nitrogen. A on boiling with NaOH gives off NH3 and a salt which on acidification gives a monobasic nitrogen free acid B. The silver salt of B contains 59.67% silver. Deduce the structures of A and B. [1988 - 3 Marks] 76. Arrange the following : (i) methylamine, dimethylamine, aniline, N-methylaniline in incerasing order of base strength. [1988 - 1 Mark] (ii) p-toluidine, N, N-dimethyl-p-toluidine, p-nitroaniline, aniline in increasing basicity [1986 - 1 Mark] 77. Write balanced equations for the following reaction : Acetamide is reacted with bromine in the presence of potassium hydroxide. [1987 - 1 Mark] 78. Show with equations how the following compounds are prepared (equations need not be balanced) : (i) aniline to chlorobenzene [1985 - 1 Mark] (ii) Acetoxime from acetaldehyde using the reagents, [K2Cr2O7/H+, Ca(OH)2 and NH2OH.HCl].

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[1984 - 2 Marks] (iii)

Aniline from benzene [1983 - 1 Mark]

(iv)

chlorobenzene from aniline (in two steps). [1982 - 1 Mark]

(v)

n-propyl amine from ethyl chloride (in two steps) [1982 - 1 Mark]

1.

Aniline dissolved in dilute HCl is reacted with sodium nitrate at 0 °C. This solution was added dropwise to a solution containing equimolar mixture of aniline and phenol in dil. HCl. The structure of the major product is: [Main April 9, 2019 (I)]

(a) (b) (c) (d) 2.

The major product of the following reaction is

[Adv. 2017]

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(a)

(b)

(c)

(d) 3.

(a) (b) (c) (d) 4.

(a) (b) (c) (d) 5.

Conversion of benzene diazonium chloride to chlorobenzene is an example of which of the following reactions? [Main Online April 12, 2014] Claisen Friedel-craft Sandmeyer Wurtz Complete reduction of benzene-diazonium chloride with Zn/HCl gives: [Main Online April 11, 2014] Aniline Phenylhydrazine Azobenzene Hydrazobenzene Fluorination of an aromatic ring is easily accomplished by treating a diazonium salt with HBF4. Which of the following conditions is correct about this reaction?

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[Main Online April 10, 2016] (a) (b) (c) (d)

NaF/Cu Cu2O/H2O Only heat NaNO2/Cu

6.

[2003S] (a) (b) (c) (d)

7.

Aniline reacts with mixed acid (conc. HNO3 and conc. H2SO4) at 288 K to give P (51 %),Q (47%) and R (2%). The major product(s) of the following reaction sequence is (are) [Adv. 2018]

major product(s)

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(a)

(b)

(c)

(d)

8.

The product(s) of the following reaction sequence is (are) [Adv. 2016]

(a)

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(b)

(c)

(d)

9.

In the following reactions, the major product W is

[Adv. 2015]

(a)

(b)

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(c)

(d)

10.

The major product of the reaction is [Adv. 2015]

(a)

(b)

(c)

(d) 11.

Benzenediazonium chloride on reaction with phenol in weakly basic medium gives [1998 - 2 Marks] (a) diphenyl ether

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(b) p-hydroxyazobenzene (c) chlorobenzene (d) benzene 12.

Match the reaction in Column I with appropriate options in Column II. [2010] Column-I Column-II (p) Racemic (A) mixture (B)

(q)

Addition

reaction

(C)

(r)

Substitution

reaction (D)

(s)

Coupling reaction

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(t)

13.

(i)

Carbocation intermediate

Show with equations how the following compounds are prepared (equations need not be balanced) : Convert

to

in not more than four steps. [2004 - 4 Marks]

(ii) [2003 - 2 Marks] (iii)

Aniline → Benzylamine (in 3 steps) [2000 - 3 Marks]

(iv)

benzamide from nitrobenzene [1994 - 2 Marks]

(v)

4-nitroaniline to 1, 2, 3-tribromobenzene.

[1990 - 2 Marks] (vi) benzaldehyde to cyanobenzene. (in not more than 6 steps) [1986 - 2 Marks] 14. How would you synthesise 4–methoxyphenol from bromobenzene in NOT more than five steps? State clearly the reagents used in each step and show the structures of the intermediate compounds in your synthetic scheme. [2001 - 5 Marks] 15. Complete the following reaction with appropriate reagents:

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[1999 - 4 Marks] 16. Complete the following with appropriate structures : 2, 4-Dinitroaniline [1995 - 1 Mark]

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1. (a) (b) (c) (d) 2. (a) (b) (c) (d) 3.

Which one of the following statements is not true? [Main Sep. 06, 2020 (II)] Lactose contains α-glycosidic linkage between C1 of galactose and C4 of glucose. Lactose is a reducing sugar and it gives Fehling’s test. Lactose (C11H22O11) is a disaccharide and it contains 8 hydroxyl groups. On acid hydrolysis, lactose gives one molecule of D(+)-glucose and one molecule of D(+)-galactose What are the functional groups present in the structure of maltose? [Main Sep. 04, 2020 (I)] One ketal and hemiketal Two acetals One acetal and one hemiacetal One acetal and one ketal Consider the following reactions : [Main Sep. 02, 2020 (I)] Acetal

(i) Glucose + ROH (ii) Glucose

A

acetyl derivative acetyl derivative

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(iii) Glucose

acetyl derivative

'x', 'y' and 'z' in these reactions are respectively. (a) 5, 4 & 5 (b) 4, 6 & 5 (c) 4, 5 & 5 (d) 5, 6 & 5 4. The correct observation in the following reactions is :

[Main Sep. 02, 2020 (II)] (a) (b) (c) (d) 5. (a) (b) (c) (d) 6. (a) (b) (c) (d) 7. (a) (b) (c) (d) 8.

Formation of blue colour Gives no colour Formation of red colour Formation of violet colour Which of the following statement is not true for glucose? [Main Jan. 08, 2020 (I)] Glucose exists in two crystalline forms and Glucose gives Schiff’s test for aldehyde Glucose reacts with hydroxylamine to form oxime The pentaacetate of glucose does not react with hydroxylamine to give oxime Two monomers in maltose are: [Main Jan. 08, 2020 (II)] -D-glucose and -D-glucose -D-glucose and -D-galactose -D-glucose and -D-fructose -D-glucose and -D-glucose Which of the following statements is correct ? [Main Jan. 07, 2020 (II)] Gluconic acid can form cyclic (acetal/hemiacetal) structure Gluconic acid is a dicarboxylic acid Gluconic acid is a partial oxidation product of glucose Gluconic acid is obtained by oxidation of glucose with HNO3 Which of the given statements is INCORRECT about glycogen ?

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(a) (b) (c) (d) 9. (a) (b) (c) (d) 10. (a) (c) (d) 11. (a) (b) (c) (d) 12. (a) (b) (c) (d) 13. (a) (b)

[Main April 12, 2019 (II)] It is a straight chain polymer similar to amylose. Only a-linkages are present in the molecule. It is present in animal cells. It is present in some yeast and fungi. Amylopectin is composed of : [Main April 10, 2019 (I)] -D-glucose, C1 – C4 and C1– C6 linkages β-D-glucose, C1 – C4 and C2 – C6 linkages β-D-glucose, C1 – C4 and C1 – C6 linkages -D-glucose, C1 – C4 and C2 – C6 linkages Number of stereo centers present in linear and cyclic structures of glucose are respectively: [Main April 10, 2019 (II)] 5&4 (b) 4 & 4 5&5 4&5 Which of the following statements is not true about sucrose? [Main April 9, 2019 (I)] It is a non reducing sugar The glycosidic linkage is present between C1 of α-glucose and C1 of βfructose It is also named as invert sugar On hydrolysis, it produces glucose and fructose Glucose on prolonged heating with HI gives : [Main 2018] n-Hexane 1- Hexene Hexanoic acid 6-iodohexanal Among the following, the incorrect statement is: [Main Online April 16, 2018] Cellulose and amylose have 1,4-glycosidic linkage Lactose contains β-D-galactose and β-D-glucose

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(c) Maltose and lactose have 1,4-glycosidic linkage (d) Sucrose and amylose have 1,2-glycosidic linkage 14. Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution? [Main 2017]

(a)

(b)

(c)

(d)

15. (a) (b) (c) (d) 16.

The incorrect statement among the following is : [Main Online April 9, 2017] α-D-glucose and β-D-glucose are anomers. α-D-glucose and β-D-glucose are enantiomers. Cellulose is a straight chain polysaccharide made up of only β-Dglucose units. The penta acetate of glucose does not react with hydroxyl amine. Accumulation of which of the following molecules in the muscles occurs as a result of vigorous exercise ?

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[Main Online April 11, 2015] (a) (b) (c) (d) 17. (a) (b) (c) (d) 18. (a) (b) (c) (d) 19. (a) (b) (c) (d) 20. (a) (b) (c) (d) 21. (a) (b) (c) (d) 22.

Glycogen Glucose Pyruvic acid L-lactic acid Complete hydrolysis of starch gives : glucose only [Main Online April 10, 2015] galactose and fructose in equimolar amounts glucose and galactose in equimolar amounts glucose and fructose in equimolar amounts Which of the following will not show mutarotation? [Main Online April 12, 2014] Maltose Lactose Glucose Sucrose Synthesis of each molecule of glucose in photosynthesis involves : [Main 2013] 18 molecules of ATP 10 molecules of ATP 8 molecules of ATP 6 molecules of ATP Which of the following statement is not correct? [Main Online April 25, 2013] Amylopectin is a branched polymer of α - glucose. Cellulose is a linear polymer of β-glucose Glycogen is the food reserve of plants All proteins are polymers of α - amino acids. Natural glucose is termed D-glucose because : [Main Online April 23, 2013] – OH on the second carbon is on the right side in Fischer projection – OH on the sixth carbon is on the right side in Fischer projection. – OH on the fifth carbon is on the right side in Fischer projection. It is dextrorotatory. The following carbohydrate is

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[2011 - II]

(a) a ketohexose (b) an aldohexose (c) an α-furanose (d) an α-pyranose 23. The correct statement about the following disaccharide is

[2010] (a) (b) (c) (d) 24.

Ring (A) is pyranose with α - glycosidic link Ring (A) is furanose with α - glycosidic link Ring (B) is furanose with α - glycosidic link Ring (B) is pyranose with β - glycosidic link Cellulose upon acetylation with excess acetic anhydride/H2SO4 (catalytic) gives cellulose triacetate whose structure is [2008S]

(a)

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(b)

(c)

(d)

25. (a) (b) (c) 26. (a) (b) (c) (d)

The two forms of D−glucopyranose obtained from the solution of D−glucose are called [2005S] Isomers Anomers Epimers (d) Enantiomers The pair of compounds in which both the compounds give positive test with Tollen’s reagent is [2004S] Glucose and Sucrose Fructose and Sucrose Acetophenone and Hexanal Glucose and Fructose

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27.

When the following aldohexose exists in its D-configuration, the total number of stereoisomers in its pyranose form is : [2012] CHO — CH2 — CHOH — CHOH — CHOH — CH2OH 28. The number of chiral carbons present in sucrose is ____________. [Main Sep. 05, 2020 (II)]

29. (a) (b) (c) (d) 30.

Which of the following statement(s) is (are) true? [Adv. 2019] Oxidation of glucose with bromine water gives glutamic acid. Hydrolysis of sucrose gives dextrorotatory glucose and laevorotatory fructose. The two six-membered cyclic hemiacetal forms of D-(+)- glucose are called anomers. Monosaccharides cannot be hydrolysed to give poly-hydroxy aldehydes and ketones The Fischer presentation of D-glucose is given below. [Adv. 2018]

The correct structure(s) of β-L-glucopyranose is (are)

(a)

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(b)

(c)

(d)

31. For ‘invert sugar’, the correct statement(s) is(are) (Given : specific rotations of (+) -sucrose, (+)-maltose, L-(-)-glucose and L-(+) fructose in aqueous solution are+ 66°, +140°, –52° and +92°, respectively) [Adv. 2016] (a) ‘invert sugar’ is prepared by acid catalyzed hydrolysis of maltose (b) ‘invert sugar’ is an equimolar mixture of D-(+)-glucose and D-(-)fructose (c) specific rotation of ‘invert sugar’ is –20° (d) on reaction with Br2 water, ‘invert sugar’ forms saccharic acid as one of the products 32. The structure of D-(+)-glucose is [Adv. 2015]

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The structure of L-(–)-glucose is (a)

(b)

(c)

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(d)

33.

The correct statement(s) about the following sugars X and Y is (are) [2009S]

(a) (b) (c) (d)

X is a reducing sugar and Y is a non-reducing sugar X is a non-reducing sugar and Y is a reducing sugar The glucosidic linkages in X and Y are α and β, respectively The glucosidic linkages in X and Y are β and α, respectively

34.

Match the following, choosing one item from column X and the appropriate item from column Y. [1983 - 2 Marks] (i) Lucas test (a) Phenol (ii) Neutral FeCl3 test (b) Glucose (iii) Dye test (c) Tertiary alcohol

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(iv) Tollen’s test

(d) Aniline

35.

This question contains Statement-1 (Assertion) and Statement-2 (Reason) and has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct. Statement-1 : Glucose gives a reddish-brown precipitate with Fehling's solution. because Statement-2 : Reaction of glucose with Fehling's solution give CuO and gluconic acid. [2007] (a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 (b) Statement-1 is True, Statement-2 is True; Statement-2 is not a correct explanation for Statement-1 (c) Statement-1 is True, Statement-2 is False (d) Statement-1 is False, Statement-2 is True. 36.

Which of the following will reduce Tollen’s reagent? Explain. [2005 - 2 Marks]

A

B

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37.

The Fisher projection of D-glucose is drawn below. [2004 - 2 Marks]

(i) Draw the Fisher projection of L-glucose. (ii) Give the reaction of L-glucose with Tollen’s reagent

1. (a) (b) (c) (d) 2. (a) (b) (c) (d) 3. (a) (b) (c) (d) 4.

Which of the following is not an essential amino acid? [Main Sep. 05, 2020 (I)] Tyrosine Leucine Valine Lysine Which of the following will react with CHCl3 + alc. KOH? [Main Sep. 04, 2020 (I)] Adenine and proline Thymine and proline Adenine and lysine Adenine and thymine The peptide that gives positive ceric ammonium nitrate and carbylamine tests is: [Main April 9, 2019 (II)] Ser – Lys Gln – Asp Lys – Asp Asp – Gln Among the following compounds most basic amino acid is: [Main Jan. 12, 2019 (I)]

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(a) (b) (c) (d) 5.

Asparagine Lysine Serine Histidine The correct structure of histidine in a strongly acidic solution (pH = 2) is : [Main Jan. 12, 2019 (II)]

(a)

(b)

(c)

(d)

6.

The increasing order of pKa of the following amino acids in aqueous solution is: [Main Jan. 9, 2019 (I)]

Gly Asp Lys Arg

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(a) Asp < Gly < Arg < Lys (b) Gly < Asp < Arg < Lys (c) Asp < Gly < Lys < Arg (d) Arg < Lys < Gly < Asp 7. The correct sequence of amino acids present in the tripeptide given below is: [Main Jan. 9, 2019 (II)]

(a) (b) (c) (d) 8.

Val - Ser - Thr Thr - Ser - Val Leu - Ser - Val Thr - Ser - Leu Which of the following will not exist in zwitter ionic form at pH = 7 ? [Main Online April 15, 2018 (I)]

(a)

(b)

(c)

(d) 9.

Among the following, the essential amino acid is : [Main Online April 8, 2017] (a) Alanine (b) Valine (c) Aspartic acid

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(d) Serine 10. Observation of "Ruhemann's purple" is a confirmatory test for the presence of [Main Online April 10, 2016] (a) Starch (b) Reducing sugar (c) Protein (d) Cupric ion 11. Glycosidic linkage is actually an : [Main Online April 23, 2013] (a) Carbonyl bond (b) Ether bond (c) Ester bond (d) Amide bond 12. The number of chiral carbons (s) present in peptide, Ile-Age-Pro, is ______. [Main Sep. 05, 2020 (I)] 13. The number of chiral centres present in threonine is ____________. [Main Sep. 04, 2020 (II)] groups present in a tripeptide Asp - Glu 14. The number of Lys is __________. [Main Sep. 03, 2020 (II)]

15.

The structure of a peptide is given below.

[Adv. 2020]

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If the absolute values of the net charge of the peptide at pH = 2, pH = 6, and pH = 11 are |z1|, |z2|, and |z3|, respectively, then what is |z1| + |z2| + |z3|? 16.

The total number of distinct naturally occurring amino acids obtained by complete acidic hydrolysis of the peptide shown below is [Adv. 2014] 17. A tetrapeptide has —COOH group on alanine. This produces glycine (Gly), valine (Val), phenyl alanine (Phe) and alanine (Ala), on complete hydrolysis. For this tetrapeptide, the number of possible sequences (primary structures) with —NH2 group attached to a chiral center is [Adv. 2013]

18.

The substituents R1 and R2 for nine peptides are listed in the table given below. How many of these peptides are positively charged at pH = 7.0 ? [2012]

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19.

A decapeptide (Mol. wt. 796) on complete hydrolysis gives glycine (Mol. wt. 75), alanine and phenylalanine. Glycine contributes 47.0% to the total weight of the hydrolysed products. The number of glycine units present in the decapeptide is [2011] 20. The total number of basic groups in the following form of lysine is [2010]

21.

The mass percentage of nitrogen in histamine is ______. [Main Jan. 09, 2020 (I)]

22.

Following two amino acids lysine and glutamine form dipeptide linkage. What are two possible dipeptides? [2003 - 2 Marks]

23.

Aspartame, an artificial sweetener, is a peptide and has the following structure :

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[2001 - 5 Marks]

(i) Identify the four functional groups. (ii) Write the zwitterionic structure. (iii) Write the structures of the amino acids obtained from the hydrolysis of aspartame. (iv) Which of the two amino acids is more hydrophobic? 24. Write the structures of alanine at pH = 2 and pH = 10. [2000 - 2 Marks]

1. (a) (b) (c) (d) 2.

Which of the following statements is not true about RNA? [Main April 12, 2019 (I)] It controls the synthesis of protein. It has always double stranded helix structure. It usually does not replicate. It is present in the nucleus of the cell. Among the following compounds, which one is found in RNA ? [Main Jan. 11, 2019 (I)]

(a)

(b)

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(c)

(d)

3.

Which of the following is the correct structure of adenosine ? [Main Online April 15, 2018 (I)]

(a)

(b)

(c)

(d) 4.

Thiol group is present in :

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[Main 2016] (a) (b) (c) (d) 5.

Cysteine Methionine Cytosine Cystine Which of the vitamins given below is water soluble ? [Main 2015]

(a) (b) (c) (d) 6. (a) (b) (c) (d) 7. (a) (b) (c) (d) 8. (a) (b) (c) (d) 9.

Vitamin E Vitamin K Vitamin C Vitamin D Which one of the following bases is not present in DNA? Quinoline Adenine [Main 2014] Cytosine Thymine The reason for double helical structure of DNA is the operation of: [Main Online April 19, 2014] Electrostatic attractions van der Waals forces Dipole - Dipole interactions Hydrogen bonding Among the following vitamins the one whose deficiency causes rickets (bone deficiency) is [Main Online April 25, 2013] Vitamin A Vitamin B Vitamin D Vitamin C Which of the following structures represents thymine ? [Main Online April 22, 2013]

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(a)

(b)

(c)

(d)

10. (a) (b) (c) (d) 11.

Which of the following enzyme converts starch into maltose? [Main Online April 9, 2013] Diastase Maltase Zymase Invertase Match the following: [Main Jan. 07, 2020 (I)]

(i) Riboflavin (a) Beriberi (ii) Thiamine (iii) Pyridoxine

(b) Scurvy (c) Cheilosis

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(iv) Ascorbic acid (d) Convulsions (a) (i) – (a), (ii) – (d), (iii) – (c), (iv) – (b) (b) (i) – (c), (ii) – (d), (iii) – (a), (iv) – (b) (c) (i) – (c), (ii) – (a), (iii) – (d), (iv) – (b) (d) (i) – (d), (ii) – (b), (iii) – (a), (iv) – (c)

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1.

(a) (b) (c) (d) 2.

Which one of the following polymers is not obtained by condensation polymerisation? [Main Sep. 05, 2020 (II)] Nylon 6,6 Buna - N Bakelite Nylon 6 Which of the following is a thermosetting polymer ? [Main April 12, 2019 (I)]

(a) Bakelite (b) Buns-N (c) Nylon 6 (d) PVC 3.

The correct name of the following polymer is: [Main April 12, 2019 (II)]

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(a) Polyisobutane (b) Polytert-butylene (c) Polyisoprene (d) Polyisobutylene 4.

Which of the following is a condensation polymer? [Main April 10, 2019 (I)]

(a) Buna - S (b) Neoprene (c) Teflon (d) Nylon 6, 6 5.

The structure of Nylon-6 is:

[Main April 8, 2019 (II)]

(a) (b) (c) (d) 6.

(a) (b) (c) (d) 7. (a)

Which of the following polymers is synthesized using a free radical polymerisation technique ? [Main Online April 10, 2016] Terylene Melamine polymer Nylon 6, 6 Teflon Which one is classified as a condensation polymer? [Main 2014] Dacron

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(b) Neoprene (c) Teflon (d) Acrylonitrile 8. Which one of the following is an example of thermosetting polymers? [Main Online April 19, 2014] (a) Neoprene (b) Buna-N (c) Nylon 6, 6 (d) Bakelite 9. Which of the following polymer is a polyamide ? [Main Online April 23, 2013] (a) Terylene (b) Nylon (b) Rubber (d) Vulcanised rubber 10. Which of the following is a polyamide? [Main Online May 19, 2012, April 23, 2013] (a) Teflon (b) Orlon (c) Nylon (d) Terylene 11.

Write the matched set (of three) for each entry incolumn A: [Multiple Concepts, 1984 - 1 × 5 = Mark] (i)

A Asbestos

(ii) Fluorocarbons (iii) Lithium metal (iv) Nitric oxide

B (a) molecular(1) sieve (b) paramagnetic (2) (c) refrigeration (3) (d) agent

C air pollutant

carcinogen fluorescent paint reducing(4) electron donor

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(v) Zeolites (vi) Zinc oxide

1.

(e) semi-(5) ion exchanger conductor (f) silicates of (6) propellant (Ca + Mg)

Which polymer has ‘chiral’ monomer(s)? [Main Jan. 09, 2020 (II)]

(a) (b) (c) (d) 2. (a) (b) (c) (d) 3.

Neoprene Buna-N Nylon 6, 6 PHBV Preparation of Bakelite proceeds via reactions: [Main Jan. 08, 2020 (II)] Electrophilic addition and dehydration Condensation and elimination Electrophilic substitution and dehydration Nucleophilic addition and dehydration The major product of the following reaction is:

[April 9, 2019 (I)]

(a)

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(b)

(c)

4.

(d)

Which of the following compounds is a constituent of the polymer ? [April 9, 2019 (II)]

(a) (b) (c) (d) 5. (a) (b) (c) (d) 6. (a) (b) (c)

N-Methyl urea Formaldehyde Methylamine Ammonia Poly- -hydroxybutyrate-co- -hydroxy copolymer of __________.

valerate

(PHBV)

is

a

[Jan. 12, 2019 (I)] 3-Hydroxybutanoic acid and 4-Hydroxypentanoic acid 2-Hydroxybutanoic acid and 3-Hydroxypentanoic acid 3-Hydroxybutanoic acid and 2-Hydroxypentanoic acid 3-Hydroxybutanoic acid and 3-Hydroxypentanoic acid The two monomers for the synthesis of nylon 6, 6 are : [Main Jan. 12, 2019 (II)] HOOC(CH2)4COOH, H2N (CH2)6NH2 HOOC(CH2)6COOH, H2N(CH2)6NH2 HOOC(CH2)4COOH, H2N(CH2)4NH2

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(d) HOOC(CH2)6COOH, H2N(CH2)4NH2 7. The polymer obtained from the following reactions is [Jan. 11, 2019 (I)]

(a)

(b)

(c)

(d) 8.

The homopolymer formed from 4-hydroxy-butanoic acids is : [Jan. 11, 2019 (II)]

(a)

(b)

(c) 9.

(d) Major product of the following reaction is:

[Main Jan. 9, 2019 (I)]

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(a)

(b)

(c)

(d)

10. Which of the following statements is not true? [Main Online April 15, 2018 (II)] (a) Chain growth polymerisation involves homopoly-merisation only (b) Chain growth polymerisation includes both homo-polymerisation and copolymerisation (c) Nylon 6 is an example of step-growth polymerisation (d) Step growth polymerisation requires a bifunctional monomer 11. The formation of which of the following polymers involves hydrolysis reaction? [Main 2017] (a) Nylon 6 (b) Bakelite (c) Nylon 6, 6 (d) Terylene

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12.

Which of the following is a biodegradable polymer ? [Main Online April 9, 2017]

(a) (b) (c) (d) 13. (a) (b) (c) (d) 14. (a) (b) (c) (d) 15. (a) (b) (c) (d) 16.

Which of the following statements about low density polythene is FALSE? [Main 2016] Its synthesis requires dioxygen or a peroxide initiator as a catalyst. It is used in the manufacture of buckets, dust-bins etc. Its synthesis requires high pressure. It is a poor conductor of electricity. On complete hydrogenation, natural rubber produces [Adv. 2016] ethylene–propylene copolymer vulcanised rubber polypropylene polybutylene Which polymer is used in the manufacture of paints and lacquers ? [Main 2015] Polypropene Polyvinyl chloride Bakelite Glyptal Which one of the following structures represents the neoprene polymer ? [Main Online April 11, 2015]

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(a) (b) (c) (d) 17.

Structure of some important polymers are given. Which one represents Buna-S? [Main Online April 9, 2014]

(a) (b) (c) (d) 18. (a) (b) (c) (d) 19. (a) (b) (c) (d)

Bakelite is obtained from phenol by reacting with: [Main Online April 25, 2013] Acetal CH3CHO HCHO Chlorobenzene The polymer used for optical lenses is : [Main Online April 22, 2013] polypropylene polyvinyl chloride polythene polymethyl methacrylate

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20.

(a) (b) (c) (d) 21. (a) (b) (c) (d) 22.

If a polythene sample contains two monodisperse fractions in the ratio 2 : 3 with degree of polymerization 100 and 200, respectively, then its weight average molecular weight will be : [Main Online April 9, 2013] 4900 4600 4300 5200 Among cellulose, poly (vinyl chloride), nylon and natural rubber, the polymer in which the intermolecular force of attraction is weakest is [2009S] Nylon Poly (vinyl chloride) Cellulose Natural rubber The total number of lone-pairs of electrons in melamine is [2013]

23.

Sulphur acts as ............... agent in vulcanization of rubber. [1989 - 1 Mark]

24.

Choose the correct option(s) from the following. [Adv. 2019]

(a) Nylon-6 has amide linkages (b) Cellulose has only -D-glucose units that are joined by glycosidic linkages (c) Teflon is prepared by heating tetrafluoroethene in presence of a persulphate catalyst at high pressure (d) Natural rubber is polyisoprene containing trans alkene units 25. The correct functional group X and the reagent/reaction conditions Y in the following scheme are [2011 - II]

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X – (CH2)4 – X (a) (b) (c) (d)

X = COOCH3, Y = H2/Ni/heat X = CONH2, Y = H2/Ni/heat X = CONH2, Y = Br2/NaOH X = CN, Y = H2/Ni/heat

26. The correct match between Item-I and Item - II is: [Main Sep. 06, 2020 (II)] Item - I Item - II (a) Natural rubber (I) 1, 3-butadiene + styrene (b) Neoprene (II) 1, 3-butadiene + acrylonitrile (c) Buna-N (III) Chloroprene (d) Buna-S (IV) Isoprene (a) (A) - (III), (B) - (IV), (C) - (I), (D) - (II) (b) (A) - (III), (B) - (IV), (C) - (II), (D) - (I) (c) (A) - (IV), (B) - (III), (C) - (II), (D) - (I) (d) (A) - (IV), (B) - (III), (C) - (I), (D) - (II) 27. The correct match between Item – I and Item- II is : [April 10, 2019 (II)] Item-I Item-II (a) High density (I) Peroxide catalyst polythene (b) Polyacrylonitrile (II) Condensation at high temperature & pressure (c) Novolac (III) Ziegler-Natta catalyst (d) Nylon 6 (IV) Acid or base catalyst (a) (a) → (IV), (b) → (II), (c) → (I), (d) → (III) (b) (a) → (II), (b) → (IV), (c) → (I), (d) → (III) (c) (a) → (III), (b) → (I), (c) → (II), (d) → (IV) (d) (a) → (III), (b) → (I), (c) → (IV), (d) → (II)

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28.

(a) (b) (c) (d) 29.

Match the polymers in column - A with their main uses in column - B and choose the correct answer: [Main Online April 10, 2015] Column - A Column - B (A) Polystrene (i) Paints and lacquers (B) Glyptal (ii) Rain coats (C) Polyvinyl chloride (iii) Manufacture of toys (D) Bakelite (iv) Computer discs (A) - (ii), (B) - (i), (C) - (iii), (D) - (iv) (A) - (ii), (B) - (iv), (C) - (iii), (D) - (i) (A) - (iii), (B) - (iv), (C) - (ii), (D) - (i) (A) - (iii), (B) - (i), (C) - (ii), (D) - (iv) Match the chemical substances in Column I with type of polymers/type of bonds in Column II. [2007] Column I Column II (A) Cellulose (p) Natural polymer (B) Nylon-6, 6 (q) Synthetic polymer (C) Protein (r) Amide linkage (D) Sucrose (s) Glycoside linkage

30.

Consider the Assertion and Reason given below. Assertion (A) : Ethene polymerized in the presence of Ziegler Natta Catalyst at high temperature and pressure is used to make buckets and dustbins. Reason (R) : High density polymers are closely packed and are chemically inert. [Main Sep. 06, 2020 (I)] Choose the correct answer from the following : (a) (A) is correct but (R) is wrong. (b) Both (A) and (R) are correct but (R) is not the correct explanation of (A). (c) Both (A) and (R) are correct but (R) is the correct explanation of (A). (d) (A) and (R) both are wrong.

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31.

(a) (b) (c) (d) 32.

Assertion : Rayon is a semisynthetic polymer whose properties are better than natural cotton. Reason : Mechanical and aesthetic properties of cellulose can be improved by acetylation. [Main Online April 9, 2016] Both assertion and reason are correct, but the reason is not the correct explantion for the assertion Both assertion and reason are correct, and the reason is the correct explantion for the assertion Assertion is incorrect statement, but the reason is correct. Both assertion and reason are incorrect. Give the structures of the products in each of the following reactions. [2000 - 2 Marks]

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1. (a) (b) (c) (d) 2.

If a person i s sufferi ng from t he defi ci ency of noradrenaline, what kind of drug can be suggested? [Main Sep. 05, 2020 (I)] Anti-inflammatory Antidepressant Antihistamine Analgesic The following molecule acts as an : [Main Sep. 05, 2020 (II)]

(a) Antiseptic (b) Anti-depressant

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(c) Anti-bacterial (d) Anti-histamine 3. The mechanism of action of "Terfenadine" (Seldane) is : [Main Sep. 04, 2020 (II)] (a) Activates the histamine receptor (b) Inhibits the secretion of histamine (c) Helps in the secretion of histamine (d) Inhibits the action of histamine receptor 4. The antifertility drug “Novestrol” can react with : [Main Sep. 03, 2020 (I)] (a) ZnCl2/HCl; FeCl3; Alcoholic HCN (b) Br2/ water; ZnCl2/HCl; FeCl3 (c) Alcoholic HCN; NaOCl; ZnCl2/ HCl (d) Br2/ water; ZnCl2/ HCl; NaOCl 5. Noradrenaline is a/an: [Main April 9, 2019 (II)] (a) Antacid (b) Neurotransmitter (c) Antidepressant (d) Antihistamine 6. The predominant form of histamine present in human blood is (pKa, Histidine – 6.0) [Main 2018] (a)

(b)

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(c)

(d) 7. (a) (b) (c) (d) 8. (a) (b) (c) (d) 9. (a) (b) (c) (d) 10. (a) (b) (c) (d) 11. (a) (b)

The reason for “drug induced poisoning” is : [Main Online April 8, 2017] Binding reversibly at the active site of the enzyme Bringing conformational change in the binding site of enzyme Binding irreversibly to the active site of the enzyme Binding at the allosteric sites of the enzyme Which of the following is a bactericidal antibiotic ? [Main Online April 10, 2016] Ofloxacin Tetracycline Chloramphenicol Erythromycin The artificial sweetener that has the highest sweetness value in comparison to cane sugar is : [Main Online April 9, 2016] Sucralose Aspartane Saccharin Alitame Which of the following compounds is not an antacid ? [Main 2015] Phenelzine Ranitidine Aluminium hydroxide Cimetidine Which artificial sweetener contains chlorine ? [Main Online April 11, 2015] Sucralose Alitame

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(c) Aspartame (d) Saccharin 12. Which one of the following is used as Antihistamine? [Main Online April 11, 2014] (a) Omeprazole (b) Chloranphenicol (c) Diphenhydramine (d) Norethindrone 13. Aminoglycosides are usually used as: [Main Online April 12, 2014] (a) antibiotic (b) analgesic (c) hypnotic (d) antifertility 14. Which of the following statements about aspirin is not true? [Main Online April 22, 2013] (a) It is effective in relieving pain. (b) It is a neurologically active drug. (c) It has antiblood clotting action. (d) It belongs to narcotic analgesics. 15. H1 – Receptor antagonists is a term associated with : [Main Online April 23, 2013] (a) Antiseptics (b) Antihistamins (c) Antacids (d) Analgesics 16.

The number of chiral carbons present in the molecule given below is _________. [Main Sep. 02, 2020 (I)]

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17.

The number of chiral centres in penicillin is _______. [Main Jan. 08, 2020 (I)]

18. 19.

The number of chiral carbons in chloramphenicol is ______. [Main Jan. 07, 2020 (I)] The number of sp2 hybridised carbons present in “Aspartame” is ________. [Main Jan. 07, 2020 (II)]

20.

Match the following drugs with their therapeutic actions : [Main Sep. 03, 2020 (II)] (i)

Ranitidine

(A) Antidepressant

(ii) Nardil (Phenelzine)

(B) Antibiotic

(iii) Chloramphenicol

(C) Antihistamine

(iv) Dimetane

(D) Antacid

(Brompheniramine) (E) Analgesic (a) (i)-(A); (ii)-(C); (iii)-(B); (iv)-(E) (b) (i)-(D); (ii)-(A); (iii)-(B); (iv)-(C) (c) (i)-(E); (ii)-(A); (iii)-(C); (iv)-(D) (d) (i)-(D); (ii)-(C); (iii)-(A); (iv)-(E) 21. A chemist has 4 samples of artificial sweetener A, B, C and D. To identify these samples, he performed certain experiments and noted

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the following observations: [Main Jan. 09, 2020 (I)] (I) A and D both form blue-violet colour with ninhydrin. (II) Lassaigne extract of C gives positive AgNO3 test and negative Fe4[Fe(CN)6]3 test. (III)Lassaigne extract of B and D gives positive sodium nitroprusside test. Based on these observations which option is correct? (a) A : Aspartame; B : Saccharin; C : Sucralose; D : Alitame (b) A : Alitame; B : Saccharin; C : Aspartame; D : Sucralose (c) A : Saccharin; B : Alitame; C : Sucralose; D : Aspartame (d) A : Aspartame; B : Alitame; C : Saccharin; D : Sucralose 22.

(A) (B) (C) (a)

The correct match between item (I) and item (II) is : [Main Jan. 11, 2019 (I)] Item - I Item - II Norethindrone (P) Antibiotic Ofloxacin (Q) Antifertility Equanil (R) Hypertension (S) Analgesics (A) (Q); (B) (R); (C) (S)

(b) (A) (c) (A)

( Q); (B) (P); (C) (R) (R); (B) ( P); (C) ( S)

(d) (A) (R); (B) (P); (C) (R) 23. The correct match between Item I and Item II is: [Main Jan. 11, 2019 (II)] Item I

Item II

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(A) Allosteric effect

(a) (b) (c) (d) 24.

(A) (B) (C) (D) (a) (b) (c) (d)

1.

(P)

Molecule binding to the active site of enzyme (B) Competitive (Q) Molecule crucial for inhibitor communication in the body (C) Receptor (R) Molecule binding to a site other than the active site of enzyme (D) Poison (S) Molecule binding to the enzyme covalently (A) (R); (B) (P); (C) (Q); (D) (S) (A) (P); (B) (R); (C) (Q); (D) (S) (A) (R); (B) (P); (C) (S); (D) (Q) (A) (P); (B) (R); (C) (S); (D) (Q) The correct match between items of List-I and List-II is: [Main Online April 16, 2018] List-I List-II Phenelzine (p) Pyrimidine Chloroxylenol (q) Furan Uracil (r) Hydrazine Ranitidine (s) Phenol (A)-(s); (B)-(r); (C)-(q); (D)-(p) (A)-(r); (B)-(s); (C)-(p); (D)-(q) (A)-(r); (B)-(s); (C)-(q); (D)-(p) (A)-(s); (B)-(r); (C)-(p); (D)-(q)

Which of the following is an anionic detergent? [Main 2016]

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(a) (b) (c) (d)

Cetyltrimethyl ammonium bromide. Glyceryl oleate. Sodium stearate. Sodium lauryl sulphate.

2.

Each entry in column X is in some way related to the entries in column Y and Z. Match the appropriate entries. [Multiple concepts, 1988]

(i) X Animal charcoal

kJ deg–1

Z watch spring

Invar

cm–1

1.3805 × 10–26

Nichrome

Co, Ni

sugar refining

Rydberg

Fe, Ni

cutlery

Stainless steel Boltzmann

Fe, Cr, Ni, C C, Ca3(PO4)2

109677 heating element

(ii)

Y

X

Y

Z

Friedel-Craft’s

Oil

Alkenes

Fermentation

Lewis acid

Soap

Dehydrohalogenation Cuprous

Anhydrous

chloride

AlCl3

Sandmeyer

Yeast

Chlorobenzene

Saponification

Alcoholic alkali Ethanol

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1. (a) (b) (c) (d) 2. (a) (b) (c) (d) 3. (a) (b) (c)

While titrating dilute HCl solution with aqueous NaOH, which of the following will not be required? [Main Sep. 02, 2020 (I)] Burette and porcelain tile Pipette and distilled water Clamp and phenolphthalein Bunsen burner and measuring cylinder If you spill a chemical toilet cleaning liquid on your hand, your first aid would be : [Main Sep. 02, 2020 (II)] vinegar aqueous NaOH aqueous NaHCO3 aqueous NH3 The strength of an aqueous NaOH solution is most accurately determined by titrating: (Note: consider that an appropriate indicator is used)[Main Jan. 08, 2020 (I)] Aq. NaOH in a pipette and aqueous oxalic acid in a burette Aq. NaOH in a burette and aqueous oxalic acid in a conical flask Aq. NaOH in a burette and concentrated H2SO4 in a conical flask

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(d) Aq. NaOH in a volumetric flask and concentrated H2SO4 in a conical flask 4. Sodium metal on dissolution in liquid ammonia gives a deep blue solution due to the formation of: (a) sodium-ammonia complex [Main Jan. 10, 2019 (II)] (b) sodamide (c) sodium ion-ammonia complex (d) ammoniated electrons 5. The green colour produced in the borax bead test of a chromium (III) salt is due to [Adv. 2019] (a) Cr2(B4O7)3 (b) Cr2O3 (c) Cr(BO2)3 (d) CrB 6. An alkali is titrated against an acid with methyl orange as indicator, which of the following in a correct combination? [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 7. For standardizing NaOH solution, which of the following is used as a primary standard? [Main Online April 16, 2018] (a) Sodium tetraborate (b) Ferrous ammonium sulfate (c) Oxalic acid (d) dil. HCl 8. A solution containing a group–IV cation gives a precipitate on passing H2S. A solution of this precipitate in dil.HCl produces a white

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(a) (b) (c) (d) 9.

(a) (b) (c) (d) 10.

(a) (b) (c) (d) 11.

(a)

precipitate with NaOH solution and bluish–white precipitate with basic potassium ferrocyanide. The cation is : [Main Online April 8, 2017] Co2+ Ni2+ Mn2+ Zn2+ The hottest region of Bunsen flame shown in the figure below is : [Main 2016]

region 3 region 4 region 1 region 2 The cation that will not be precipitated by H2S in the presence of dil. HCl is : [Main Online April 10, 2015] 2+ Pb Cu2+ Co2+ As3+ Sodium Carbonate cannot be used in place of (NH4)2CO3 for the identification of Ca2+, Ba2+ and Sr2+ ions (in group V) during mixture analysis because : [Main Online April 9, 2013] 2+ Mg ions will also be precipitated.

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(b) Concentration of CO32– ions is very low. (c) Sodium ions will react with acid radicals. (d) Na+ ions will interfere with the detection of Ca2+, Ba2+, Sr2+ ions. 12. Upon treatment with ammoniacal H2S, the metal ion that precipitates as a sulfide is [Adv. 2013] (a) Fe(III) (b) Al(III) (c) Mg(II) (d) Zn(II) 13. Passing H2S gas into a mixture of Mn2+, Ni2+, Cu2+ and Hg2+ ions in an acidified aqueous solution precipitates [2011] (a) CuS and HgS (b) MnS and CuS (c) MnS and NiS (d) NiS and HgS 14. 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] 2+ (a) Pb (b) Hg2+ (c) Cu2+ (d) Co2+

15. A solution when diluted with H2O and boiled, gives a white precipitate. On addition of excess NH4Cl/NH4OH, the volume of precipitate decreases leaving behind a white gelatinous

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precipitate. Identify the precipitate which disolves in NH4OH/NH4Cl [2006 - 3M, –1] (a) (b) (c) (d)

Al (OH)3 Zn(OH)2 Ca(OH)2 Mg(OH)2

16. A metal nitrate reacts with KI to give a black precipitate which on addition of excess of KI is converted into orange colour solution. The cation of the metal nitrate is [2005S] (a) Hg2+ (b) Bi3+ (c) Pb2+ (d) Cu+ 17. A solution which is 10-3 M each in Mn2+, Fe2+, Zn2+ and Hg2+ is treated with 10-16 M sulphide ion. If Ksp of MnS, FeS, ZnS and HgS are 10-15, 10-23, 10-20 and 10-54 respectively, which one will precipitate first? [2003S] (a) FeS (b) MgS (c) HgS (d) ZnS [Y] a colourless gas with irritating smell, [Y] + 18. [X] + H2SO4 green solution. [X] and [Y] are: K2Cr2O7 + H2SO4 [2003S] 2– (a) SO3 , SO2 (b) Cl–,HCl (c) S2–, H2S (d) CO32–, CO2

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19. A gas 'X' is passed through water to form a saturated solution. 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'. [2002S] (a) X = CO2, Y = Cl2 (b) X = Cl2, Y = CO2 (c) X = Cl2, Y = H2 (d) X = H2, Y = Cl2 20. An aqueous solution of a substance gives a white precipitate on treatment with dilute hydrochloric acid, which dissolves on heating. When hydrogen sulfide is passed through the hot acidic solution, a black precipitate is obtained. The substance is a [2002S] salt (a) (b) (c) (d) 21.

(a) (b) (c) (d) 22.

(a) (b) (c) (d)

Cu2+ salt Ag+ salt Pb2+ salt Identify the correct order of solubility of Na2S, CuS and ZnS in aqueous medium [2002S] CuS > ZnS > Na2S ZnS > Na2S > CuS Na2S > CuS > ZnS Na2S > ZnS > CuS An aqueous solution contains Hg2+, Hg22+, Pb2+ and Cd2+. The addition of HCl (6N) will precipitate : [1995S] Hg2Cl2 only PbCl2 only PbCl2 and Hg2Cl2 PbCl2 and HgCl2

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23. Which one among the following pairs of ions cannot be separated by H2S in dilute hydrochloric acid? [1986 - 1 Mark] 3+ 4+ (a) Bi , Sn (b) Al3+, Hg2+ (c) Zn2+, Cu2+ (d) Ni2+, Cu2+ 24. The ion that cannot be precipitated by both HCl and H2S is [1982 - 1 Mark] 2+ (a) Pb (b) Cu+ (c) Ag+ (d) Sn2+

25. Among PbS, CuS, HgS, MnS, Ag2S, NiS, CoS, Bi2S3 and SnS2, the total number of BLACK coloured sulphides is [Adv. 2014] 26.

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? [Adv. 2020]

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27.

To measure the quantity of MnCl2 dissolved in an aqueous solution, it was completely converted to KMnO4 using the reaction, MnCl2 + K2S2O8 + H2O→ KMnO4 + H2SO4 + HCl (equation not balanced). Few drops of concentrated HCl were added to this solution and gently warmed. Further, oxalic acid (225 mg) was added in portions till the colour of the permanganate ion disappeared. The quantity of MnCl2 (in mg) present in the initial solution is ______. (Atomic weights in g mol–1: Mn = 55, Cl = 35.5) [Adv. 2018] 28.

The formula of the deep red liquid formed on warming dichromate with KCl in concentrated sulphuric acid is ............. . [1993 - 1 Mark] 29. If metal ions of group III are precipitated by NH4Cl and NH4OH without prior oxidation by conc. HNO3 ............. is not completely precipitated. [1984 - 1 Mark]

30.

From the acidic solution containing copper (+2) and zinc (+2) ions, copper can be selectively precipitated using sodium sulphide. [1987 - 1 Mark]

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31.

Addition of ammonium chloride to a solution containing ferric and magnesium ions is essential for selective precipitation of ferric hydroxide by aqueous ammonia. [1985 - ½ Mark]

32.

With reference to aqua regia, choose the correct option(s)

(a) (b) (c) (d) 33.

(a) (b) (c) (d) 34.

(a) (c) 35.

(a)

[Adv. 2019] Reaction of gold with aqua regia produces NO2 in the absence of air Reaction of gold with aqua regia produces an anion having Au in +3 oxidation state Aqua regia is prepared by mixing conc. HCl and conc. HNO3 in 3 : 1 (v/v) ratio The yellow colour of aqua regia is due to the presence of NOCl and Cl2 The correct option(s) to distinguish nitrate salts of Mn2+ and Cu2+ taken separately is (are) [Adv. 2018] 2+ Mn shows the characteristic green colour in the flame test Only Cu2+ shows the formation of precipitate by passing H2S in acidic medium Only Mn2+ shows the formation of precipitate by passing H2S in faintly basic medium 2+ Cu /Cu has higher reduction potential than Mn2+/Mn (measured under similar conditions) The reagent(s) that can selectively precipitate S2– from a mixture of S2– and SO42– in aqueous solution is (are) [Adv. 2016] CuCl2 (b) BaCl2 Pb(OOCCH3)2 (d) Na2[Fe(CN)5NO] The pair(s) of ions where BOTH the ions are precipitated upon passing H2S gas in presence of dilute HCl, is (are) [Adv. 2015] Ba2+, Zn2+

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(b) (c) (d) 36.

(a) (b) (c) (d) 37.

(a) (b) (c) (d) 38. (a) (b) (c) (d) (e)

Bi3+, Fe3+ Cu2+, Pb2+ Hg2+, Bi3+ Which of the following statement(s) is (are) correct with reference to the ferrous and ferric ions? [1998 - 2 Marks] 3+ Fe gives brown colour with potassium ferricyanide. Fe2+ gives blue precipitate with potassium ferricyanide. Fe3+ gives red colour with potassium thiocyanate. Fe2+ gives brown colour with ammonium thiocyanate. Which of the following statement(s) is (are) correct when a mixture of NaCl and K2Cr2O7 is gently warmed with conc. H2SO4? [1998 - 2 Marks] A deep red vapour is evolved The vapours when passed into NaOH solution gives a yellow solution of Na2CrO4 Chlorine gas is evolved Chromyl chloride is formed The reagents, NH4Cl and aqueous NH3 will precipitate [1991 - 1 Mark] 2+ Ca Al3+ Bi3+ Mg2+ Zn2+

PASSAGE-1 An aqueous solution of metal ion M1 reacts separately with reagents Q and R in excess to give tetrahedral and square planar complexes, respectively. An aqueous solution of another metal ion M2 always forms tetrahedral complexes with these reagents. Aqueous solution of M2 on reaction with reagent S gives white precipitate which dissolves in excess of S. The reactions are summarized in the scheme given below:

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Scheme:

39. Reagent S is [Adv. 2014] (a) K4[Fe(CN)6] (b) Na2HPO4 (c) K2CrO4 (d) KOH 40. M1, Q and R, respectively are [Adv. 2014] (a)

Zn , KCN and HCl

(c) (d)

(b) Ni2+, HCl and KCN Cd2+, KCN and HCl Co2+, HCl and KCN

2+

PASSAGE-2 An aqueous solution of a mixture of two inorganic salts, when treated with dilute HCl, gave a precipitate (P) and a filtrate (Q). The precipitate (P) was found to dissolve in hot water. The filtrate (Q) remained unchanged, when treated with H2S in a dilute mineral acid medium. However, it gave a precipitate (R) with H2S in an ammoniacal medium. The precipitate (R) gave a coloured solution (S), when treated with H2O2 in an aqueous NaOH medium. [Adv. 2013-II] 41. The coloured solution (S) contains (a) Fe2(SO4)3

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(b) CuSO4 (c) ZnSO4 (d) 42. (a) (b)

Na2CrO4 The precipitate (P) contains Pb2+ Hg22+

(c) Ag+ (d) Hg2+ PASSAGE-3 p-Amino-N, N-dimethylaniline is added to a strongly acidic solution of X. The resulting solution is treated with a few drops of aqueous solution of Y to yield blue coloration due to the formation of methylene blue. Treatment of the aqueous solution of Y with the reagent potassium hexacyanoferrate(II) leads to the formation of an intense blue precipitate. The precipitate dissolves on excess addition of the reagent. Similarly, treatment of the solution of Y with the solution of potassium hexacyanoferrate (III) leads to a brown coloration due to the formation of Z. [2009] 43. The compound X is (a) NaNO3 (b) NaCl (c) Na2SO4 (d) Na2S 44. The compound Y is (a) MgCl2 (b) FeCl2 (c) FeCl3 (d) ZnCl2 45. The compound Z is (a) Mg2[Fe(CN)6]

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(b) Fe[Fe(CN)6] (c) Fe4[Fe(CN)6]3 (d) K2Zn3[Fe(CN)6]2 46.

(a) (b) (c) (d) 47.

(a) (b) (c) (d)

48.

Read the following statement and explanation and answer as per the options given below : [1998 - 2 Marks] Assertion : Sulphate is estimated as BaSO4 and not as MgSO4. Reason : Ionic radius of Mg2+ is smaller than that of Ba2+. If both assertion and reason are correct, and reason is the correct explanation of the assertion. If both assertion and reason are correct, but reason is not the correct explanation of the assertion. If assertion is correct but reason is incorrect. If assertion is incorrect but reason is correct. Read the following statement and explanation and answer as per the options given below : [1989 - 2 Marks] 2+ Assertion : A very dilute acidic solution of Cd and Ni2+ gives yellow precipitate of CdS on passing hydrogen sulphide. Statement : Solubility product of CdS is more than that of NiS. If both assertion and statement are correct and statement is an explanation of assertion. If assertion is correct and statement is wrong, statement is not an explanation of assertion. If assertion is wrong and statement is correct, statement is not an explanation of assertion. If both assertion and statement are wrong and statement is not explanation of assertion. A mixture consists of A (yellow solid) and B (colourless solid) which gives lilac colour in flame.

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(a) Mixture gives black precipitate C on passing H2S(g) through its aqueous solution. (b) C is soluble in aqua-regia and on evaporation of aqua-regia and adding SnCl2, gives greyish black precipitate D. The salt solution with NH4OH gives a brown precipitate. (i) The sodium carbonate extract of the salt withCCl4/FeCl3 gives a violet layer. (ii) The sodium carbonate extract gives yellow precipitate with AgNO3 solution which is insoluble in NH3. Identify A and B, and the precipitates C and D [2003 - 4 Marks] 49. When a white crystalline compound X is heated with K2Cr2O7 and concentrated H2SO4, a reddish brown gas A is evolved. On passing A into caustic soda solution, a yellow coloured solution of B is obtained. Neutralizing the solution B with acetic acid and on subsequent addition of lead acetate, a yellow precipitate C is obtained. When X is heated with NaOH solution, a colourless gas is evolved and on passing this gas into K2HgI4 solution, a reddish brown precipitate D is formed. Identify A, B, C, D and X. Write the equations of reactions involved. [2002 - 5 Marks] 50. A white substance (A) reacts with dilute H2SO4 to produce a colourless gas (B) and a colourless solution (C). The reaction between (B) and acidified K2Cr2O7 solution produces a green solution and a slightly coloured precipitate (D). The substance (D) burns in air to produce a gas (E) which reacts with (B) to yield (D) and a colourless liquid. Anhydrous copper sulphate is turned blue on addition of this colourless liquid. Addition of aqueous NH3 or NaOH to (C) produces first a precipitate, which dissolves in the excess of the respective reagent to produce a clear solution in each case. Identify (A), (B), (C), (D) and (E). Write the equations of the reactions involved. [2001 - 10 Marks] 51. Write the chemical reactions associated with the ‘borax bead test’ of cobalt (II) oxide. [2000 - 3 Marks]

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52.

53.

54.

(i) (ii) 55.

56.

57.

An aqueous blue coloured solution of a transition metal sulphate reacts with H2S in acidic medium to give a black precipitate A, which is insoluble in warm aqueous solution of KOH. The blue solution on treatment with KI in weakly acidic medium, turns yellow and produces a white precipitate B. Identify the transition metal ion. Write the chemical reactions involved in the formation of A and B. [2000 - 4 Marks] An aqueous solution containing one mole of HgI2 and two moles of NaI is orange in colour. On addition of excess NaI, the solution becomes colouress. The orange colour reappears on subsequent addition of NaOCl. Explain with equations. [1999 - 3 Marks] A white solid is either Na2O or Na2O2. A piece of red litmus paper turns white when it is dipped into a freshly made aqueous solution of the white solid. [1999 - 4 Marks] Identify the substance and explain with balanced equation. Explain what would happen to the red litmus if the white solid were the other compound. During the qualitative analysis of a mixture containing Cu2+ and Zn2+ ions, H2S gas is passed through an acidified solution containing these ions in order to test Cu2+ alone. Explain briefly. [1998 - 2 Marks] A colourless inorganic salt (A) decomposes completely at about 250°C to give only two products, (B) and (C), leaving no residue. The oxide (C) is a liquid at room temperature and neutral to moist litmus paper while the gas (B) is a neutral oxide. White phosphorus burns in excess of (B) to produce a strong white dehydrating agent. Write balanced equations for the reactions involved in the above process. [1996 - 3 Marks] Calcium burns in nitrogen to produce a white powder which 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 - 2 Marks]

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58.

59.

60.

61. (i) (ii) (iii)

62.

A scarlet compound A is treated with conc. HNO3 to give a chocolate brown precipitate B. The precipitate is filtered and the filtrate is neutralised with NaOH. Addition of KI to the resulting solution gives a yellow precipitate C. The precipitate B on warming with conc. HNO3 in the precence of Mn(NO3)2 produces a pink-coloured solution due to the formation of D. Identify A, B, C and D. Write the reaction sequence. [1995 - 4 Marks] 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 Al2(SO4)3. Identify A, B and C [1994 - 5 Marks] An orange solid (A) on heating gave a green residuce (B), a colourless gas (C) and water vapour. The dry gas (C) on passing over heated Mg gave a white solid (D). (D) on reaction with water gave a gas (E) which formed dense white fumes with HCl. Identify (A) to (E) and give reactions involved. [1993 - 3 Marks] A light bluish green crystalline compound responds to the following tests : Its aqueous solution gives a brown precipitate or colour with alkaline K2[HgI4] solution. Its aqueous solution gives a blue colour with K3[Fe(CN)6] solution. Its solution in hydrochloric acid gives a white precipitate with BaCl2 solution. Identify the ions present and suggest the formula of the compound. [1992 - 4 Marks] The gas liberated on heating a mixture of two salts with NaOH, gives a reddish brown precipitate with an alkaline solution of K2[HgI4]. The aqueous solution of the mixture on treatment with BaCl2 gives a white precipitate which is sparingly soluble in conc. HCl. On heating the mixture with K2Cr2O7 and conc. H2SO4, red vapours of A are produced. The aqueous solution of the mixture gives a deep blue colouration of B with potassium ferricyanide solution. Identify the

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63.

64.

65.

(ii) (iii) (iv)

radicals in the given mixture and write the balanced equations for the formation of A and B. [1991 - 4 Marks] When 20.02 g of a white solid X is heated 4.4 g of an acid gas A and 1.8 g of a neutral gas B are evolved, leaving behind a solid residue Y of weight 13.8 g. A turns lime water milky and B condenses into a liquid which changes anhydrous copper sulphate blue. The aqueous solution of Y is alkaline to litmus and gives 19.7 g of white precipitate Z with barium chloride solution. Z gives carbon dioxide with an acid. Identify A, B, X, Y and Z [1989 - 5 Marks] A hydrated metallic salt A, light green in colour, on careful heating gives a white anhydrous residue B. B is soluble in water and its aqueous solution reacts with NO to give a dark brown compound C. B on strong heating gives a brown residue D and a mixture of two gases E and F. The gaseous mixture when passed through acidified permanganate, discharges the pink colour and when passed through acidified BaCl2 solution, gave a white precipitate. Identify A, B, C, D, E and F. [1988 - 3 Marks] A mixture of two salts was treated as follows : [1987 - 5 Marks] (i) The mixture was heated with manganese dioxide and concentrated sulphuric acid when yellowish green gas was liberated. The mixture on heating with sodium hydroxide solution gave a gas which turned red litmus blue. Its solution in water gave blue precipitate with potassium ferricyanide and red colouration with ammonium thiocyanate. The mixture was boiled with potassium hydroxide and the liberated gas was bubbled through an alkaline solution of K2HgI4 to give brown precipitate. Identify the two salts. Give ionic equations for reactions involved in the tests (i), (ii) and (iii).

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66.

67.

(i) (ii) (iii) (iv) (v) (vi)

68. The

1.

When 16.8 g of white solid X were heated, 4.4 g of acid gas A that turned lime water milky was driven off together with 1.8 g of a gas B which condensed to a colourless liquid. The solid that remained, Y, dissolved in water to give an alkaline solution, which with excess barium chloride solution gave a white precipitate Z. The precipitate effervesced with acid, giving off carbon dioxide. Identify A, B and Y and write down the equation for the thermal decomposition of X. [1984 - 4 Marks] Compound A is a light green crystalline solid. It gives the following tests: [1980] It dissolves in dilute sulphuric acid. No gas is produced. A drop of KMnO4 is added to the above solution. The pink colour disappears. Compound A is heated strongly. Gases B and C, with pungent smell, come out. A brown residue D is left behind. The gas mixture (B) and (C) is passed into a dichromate solution. The solution turns green. The green solution from step (iv) gives a white precipitate E with a solution of barium nitrate. Residue D from step (iii) is heated on charcoal in a reducing flame. It gives a magnetic substance. Name the compounds A, B, C, D and E Account for the following. Limit your answer to two sentences: precipitation of second group sulphides in qualitative analysis is carried out with hydrogen sulphide in presence of hydrochloric acid and not nitric acid [1979]

Consider the following reactions:

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‘B’ + ‘C’

'A' is : [Main Sep. 06, 2020 (I)] (a) (b) (c) (d) 2.

The Kjeldahl method of Nitrogen estimation fails for which of the following reaction products? [Main Sep. 03, 2020 (I)]

(1)

(2)

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(3)

(4) (a) (b) (c) (d) 3.

(3) and (4) (1) and (4) (1), (3) and (4) (2) and (3) In Carius method of estimation of halogen, 0.172 g of an organic compound showed presence of 0.08 g of bromine. Which of these is the correct structure of the compound? [Main Sep. 02, 2020 (I)] (a) H3C – CH2 – Br

(b)

(c)

(d) H3C – Br 4. Two compounds A and B with same molecular formula (C3H6O) undergo Grignard’s reaction with methyl-magnesium bromide to give products C and D. Products C and D show following chemical tests.

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C and D respectively are : [Main Sep. 02, 2020 (II)] (a)

(b)

(c)

(d)

5.

A, B and C are three biomolecules. The results of the tests performed on them are given below: Molisch’s Test Barfoed Test Biuret Test A Positive Negative Negative

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B Positive C Negative

Positive Negative

Negative Positive

A, B and C are respectively: [Main Jan. 09, 2020 (II)] (a) (b) (c) (d) 6.

A = Glucose, B = Fructose, C = Albumin A = Lactose, B = Glucose, C = Albumin A = Lactose, B = Glucose, C = Alanine A = Lactose, B = Fructose, C = Alanine Kjeldahl’s method cannot be used to estimate nitrogen for which of the following compounds? [Main Jan. 08, 2020 (II)] (a) C6H5NH2 (b) CH3CH2 – C  N (c) C6H5NO2 (d) 7.

(a) (b) (c) (d) 8.

A solution of m-chloroaniline, m-chlorophenol and m-chlorobenzoic acid in ethyl acetate was extracted initially with a saturated solution of NaHCO3 to give fraction A. The left over organic phase was extracted with dilute NaOH solution to give fraction B. The final organic layer was labelled as fraction C. Fractions A, B and C, contain respectively: [Main Jan. 07, 2020 (I)] m-chlorobenzoic acid, m-chloroaniline andm-chlorophenol m-chlorobenzoic acid, m-chlorophenol andm-chloroaniline m-chlorophenol, m-chlorobenzoic acid andm-chloroaniline m-chloroaniline, m-chlorobenzoic acid andm-chlorophenol Consider the following reaction: [Main Jan. 07, 2020 (I)]

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The product ‘X’ is used: (a) (b) (c) (d) 9.

(a) (b) (c) (d) 10.

(a) (b) (c) (d) 11. (a) (b) (c) (d) 12. (a)

in protein estimation as an alternative to ninhydrin in acid base titration as an indicator as food grade colourant in laboratory test for phenols A chromatography column, packed with silica gel as stationary phase, was used to separate a mixture of compounds consisting of (A) benzanilide (B) aniline and (C) acetophenone. When the column is eluted with a mixture of solvents, hexane: ethyl acetate (20:80), the sequence of obtained compounds is: [Main Jan. 07, 2020 (II)] (B), (C) and (A) (B), (A) and (C) (C), (A) and (B) (A), (B) and (C) An organic compound ‘A’ is oxidized with Na2O2 followed by boiling with HNO3. The resultant solution is then treated with ammonium molybdate to yield a yellow precipitate. Based on above observation, the element present in the given compound is: [Main April 12, 2019 (I)] Nitrogen Phosphorus Fluorine Sulphur Which one of the following is likely to give a precipitate with AgNO3 solution ? [Main April 12, 2019 (II)] CH2 = CH — Cl CCl4 CHCl3 (CH3)3CCl The principle of column chromatography is : [Main April 10, 2019 (I)] Gravitational force.

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(b) Capillary action. (c) Differential absorption of the substances on the solid phase. (d) Differential adsorption of the substances on the solid phase. 13. The organic compound that gives following qualitative analysis is: [Main April 9, 2019 (I)] Test Inference (a) Dil. HCl Insoluble (b) NaOH solution soluble (c) Br2/water Decolourization (a) (b) (c) (d) 14. An organic compound is estimated through Dumas method and was found to evolve 6 moles of CO2, 4 moles of H2O and 1 mole of nitrogen gas. The formula of the compound is: [Main Jan. 11, 2019 (I)] (a) C12H8N (b) C12H8N2 (c) C6H8N2 (d) C6H8N 15. If dichloromethane (DCM) and water (H2O) are used for differential extraction, which one of the following state-ments is correct? [Main Jan. 10, 2019 (I)] (a) DCM and H2O would stay as lower and upper layer respectively in the S.F. (b) DCM and H2O will make tur bid/colloidal mixture

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(c) DCM and H2O would stay as upper and lower layer respectively in the separating funnel (S.F.) (d) DCM and H2O will be miscible clearly 16. Which of the following tests cannot be used for identifying amino acids? [Main Jan. 10, 2019 (II)] (a) Biuret test (b) Barfoed test (c) Ninhydrin test (d) Xanthoproteic test 17. Which of the following compounds will be suitable for Kjeldahl’s method for nitrogen estimation? [Main 2018] (a)

(b)

(c)

(d)

18.

For the estimation of nitrogen, 1.4 g of an organic compound was digested by Kjeldahl method and the evolved ammonia was absorbed in 60 mL of

sulphuric acid. The unreacted acid required 20 mL of

sodium hydroxide for complete neutralization. The percentage of nitrogen in the compound is: [Main Online April 10, 2015, 2014] (a) (b) (c) (d)

6% 10% 3% 5%

19. Match the following : [Main Sep. 06, 2020 (II)]

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(I)

Column-I Test / Method Lucas Test

Column-II Reagent (A) C6H5SO2Cl / aq. KOH

(II)

Dumans method

(B) HNO3 / AgNO3

(III) Kjeldahl’s method (C) CuO / CO2 (IV) Hinsberg Test

(D) Conc. HCl and ZnCl2 (E) H2SO4

(a) (b) (c) (d) 20.

(a) (b) (c) (d) 21.

(I)-(D), (II)-(C), (III)-(B), (IV)-(A) (I)-(B), (II)-(D), (III)-(E), (IV)-(A) (I)-(D), (II)-(C), (III)-(E), (IV)-(A) (I)-(B), (II)-(A), (III)-(C), (IV)-(D) The correct match between Item-I and Item-II is: [Main Jan. 9, 2019 (I)] Item-I Item-II (drug) (test) A Chloroxylenol P Carbylamine test B Norethindrone Q Sodium hydrogen carbonate test C Sulphapyridine R Ferric chloride test D Penicillin S Bayer’s test A→R;B→P;C→S;D→Q A→Q;B→S;C→P;D→R A→R;B→S;C→P;D→Q A→Q;B→P;C→S;D→R The correct match between item ‘I’ and item ‘II’ is: [Main Jan. 10, 2019 (II)] Item ‘I’ Item ‘II’ (compound) (reagent) (A) Lysine (P) 1-Naphthol (B) Furfural (Q) Ninhydrin (C) Benzyl alcohol (R) KMnO4

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(D) Styrene

(a) (b) (c) (d) 22.

(a) (b) (c) (d) 23.

(S) Ceric ammonium nitrate (A) → (Q); (B) → (P); (C) → (S); (D) → (R) (A) → (Q); (B) → (P); (C) → (R); (D) → (S) (A) → (R); (B) → (P); (C) → (Q) : (D) → (S) (A) → (Q); (B) → (R); (C) → (S); (D) → (P) The correct match between Item I and Item II is: [Main Jan. 9, 2019 (II)] Item I Item II (A) Benzaldehyde (P) Mobile phase (B) Alumina (Q) Adsorbent (C) Acetonitrile (R) Adsorbate (A) → (Q) ; (B) → (P) ; (C) → (R) (A) → (R) ; (B) → (Q) ; (C) → (P) (A) → (Q) ; (B) → (R) ; (C) → (P) (A) → (P) ; (B) → (R) ; (C) → (Q) The desired product X can be prepared by reacting the major product of the reactions in LIST-I with one or more appropriate reagents in LIST-II. (given, order of migratory aptitude: aryl > alkyl > hydrogen) [Adv. 2018]

List - I

List- II 1. I2, NaOH

P. + H2SO4

2. [Ag(NH3)2]OH

Q. + HNO2

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3. Fehling solution

R. + H2SO4

4. HCHO, NaOH

S. + AgNO3

5. NaOBr The correct option is (a) (b) (c) (d)

P-1; Q-2, 3 ; R-1, 4; S-2, 4 P-1, 5 ; Q-3, 4 ; R-4, 5; S-3 P-1, 5; Q-3, 4 ; R-5; S-2, 4 P-1, 5; Q-2, 3 ; R-1, 5; S-2, 3

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1. (c)

1st component n1 M1 n1M1

mole m.w mass [ mass = mole × m.w.] Mass of solution = n1M1 + n2M2

2nd component n2 M2 n2 M 2

Mole fraction of the 2nd component

Mass of solution = n1M1 + n2M2

Volume of solution

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2.

(b)

35

Molar ratio

x

Cl

37

Cl

(1 – x)

Mavg. = 35 × x + 37 (1 – x) = 35.5 = 35x + 37 (1 – x) = 35.5

So, ratio of 35Cl : 37Cl 3.

(c) Except (3) all postulates was given by the Dalton.

4.

(c) No. of moles of H2O (n1) =

=1

No. of moles of NaOH (n2)

Mole fraction of NaOH =

= 0.167

Molality = = 5.

= 11.11 m

(c) As we know,

Molarity = ⇒ 0.1 = So, no. of moles of sugar = 0.2 mole ∴ Mass of sugar = No. of moles of sugar ×

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Molar mass of sugar = 0.2 × 342 = 68.4 g 6.

(d)

So, x = 1, y = 2 Equation for combustion of CxHy

Oxygen atoms required As mentioned,

Now putting the values of x and y ⇒

⇒ z = 1.5

∴ Molecule (CxHyOz) can be written as C1H2O3/2 ⇒ C2H4O3 7. (d) Given percentage of chlorine in an hydrocarbon = 3.55% i.e., 100 g of chlorohydrocarbon has 3.55 g of chlorine. 1 g of chlorohydrocarbon will have

g of chlorine.

Atomic wt. of Cl = 35.5 g/mol Number of moles of Cl Number of atoms of Cl = 0.001 mol × 6.023 × 1023 mol–1

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= 6.023 × 1020 8. (c) Percentage (by mass) of elements given in the body of a healthy human adult is :Oxygen = 61.4%, Carbon = 22.9%, Hydrogen = 10.0% and Nitrogen = 2.6% Total weight of person = 75 kg ∴ Mass due to 1H is If 1H atoms are replaced by 2H atoms. Mass gain by person would be = 7.5 kg 9.

(c)

10.

(N)

Volume of O2 used Volume of air = 375 – 75 = 300 mL Total volume of gases after combustion = vol. of CO2 + vol. of air = 330 mL Volume of CO2 = 330 – 300 = 30 mL 15 mL CxHy gives = 30 mL CO2 1 mL CxHy gives At constant T and P; Volume ∝ mole ∴ 1 mol CxHy = 2 mol CO2 x=2

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4 x + y = 20 y = 20 – 4 × 2 = 12 Hence, formula of the hydrocarbon is C2H12. 11. (d) Since the compound undergoing combustion is an alkane. Hence the combustion reaction can be written as

Since volumes are measured at constant T & P, hence according to Avogadro's law Volume ∝ mole 1 L alkane requires

L of O2

5 L alkane requires 25 L of O2

Hence alkane is propane (C3H8). 12. (b) 2 mole of water softner require 1 mole of Ca2+ ion So, 1 mole of water softner require

will be maximum uptake.

Thus, 13.

mole of Ca2+ ion

(b)

Since Cl2– does not exist. So, X2Cl3 is incorrect. The correct formula should be XCl3.

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14. (b) Given mass of solute (w) = 120 g mass of solvent (w) = 1000 g Mol. mass of solute = 60 g density of solution = 1.12 g/ mL From the given data, Mass of solution = 1000 + 120 = 1120 g or Volume of solution

or = 1 litre =

Now molarity (M) = 15.

(d) 18 g, H2O contains = 2 g H ∴ 0.72 g H2O contains

= 44 g CO2 contains = 12 g C ∴ 3.08 g CO2 contains = ∴C:H=

; 0.07 : 0.08 = 7 : 8

∴ Empirical formula = C7H8 16.

(a) From molarity equation M1V1 + M2V2 = MV(total)

M= 17.

(c) The relation between molarity (M) and molality (m) is d=

On putting value

, M2 = Mol. mass of solute

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1.252 = On solving m = 2.79 18. (a) TIPS/Formulae : Atomic weight in gms = 6.023 × 1023 atoms = 1 Mole atoms (i) Number of atoms in 24 g of C =

× 6.023 × 1023 = 2 × 6.023 × 1023 atom

= 2 mole atoms (ii) Number of atoms in 56 g of Fe =

× 6.023 × 1023 = 6.023 × 1023 atom

= 1 mole atoms (iii) Number of atoms in 27 g of Al =

× 6.023 × 1023 = 6.023 × 1023 atom

= 1 mole atoms (iv) Number of atoms in 108 g of Ag =

× 6.023 × 1023 = 6.023 × 1023 atom

= 1 mole atoms 24 g of C has maximum number of atoms. 19. (d) TIPS/Formulae : (i) Mass of one electron = 9.108 × 10–31 kg (ii) 1 mole of electron = 6.023 × 1023 electrons Weight of 1 mole of electron = Mass of one electron × Avogadro Number = 9.108 × 10–31 × 6.023 × 1023 kg No. of moles of electrons in 1 kg = 20.

= (b) TIPS/Formulae : Empirical Formula = n × Molecular formula

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Solution : Since the molecular formula is n times the empirical formula, therefore, different compounds having the same empirical formula must have different molecular weights. 21.

(a) (a) 18 g of H2O = 6.02 × 1023 molecules of H2O ∴ 36 g of H2O = 2 × 6.02 × 1023 molecules of H2O = 12.04 × 1023 molecules of H2O

(b) 28 g of CO = 6.02 × 1023 molecules of CO (c) 46 g of C2H5OH = 6.02 × 1023 molecules of C2H5OH (d) 108 g of N2O5 = 6.02 × 1023 molecules of N2O5 ∴ 54 g of N2O5 =

× 6.02 × 1023 molecules of N2O5

= 3.01 × 1023 molecules of N2O5 ∴ 36 g of water has highest number of molecules. 22. (a) No. of e– in C = 6 and in O = 8 ∴ Total no. of e– in CO2 = 6 + 8 × 2 = 22 23. (c) Let mass of oxygen = 1g, Then mass of nitrogen = 4g Mol. wt. of N2 = 28g, Mol. wt. of O2 = 32g 28 g of N2 has = 6.02 × 1023 molecules of nitrogen molecules of nitrogen

4 g of N2 has = =

molecules of nitrogen

32 g of O2 has = 6.02 × 1023 molecules of oxygen ∴

1g of O2 has = molecules of oxygen

Thus, ratio of molecules of oxygen : nitrogen

24.

(d) 4 Al + 3O2

2 Al2O3

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At. wt. of Al = 27 Thus 4 × 27 g of Al reacts with oxygen = 3 × 32 g ∴ 27 g of Al reacts with oxygen =

× 27 g = 24 g

25.

=

(c) No. of nitrogen atoms = =

No. of oxygen atoms =

=2

=5

∴ Formula of compound is N2O5. 26. (5) and Mass ratio of C : H is C : O is So, mass mole mole ratio C 12 1 1 H 3 3 3 O 16 1 1 Empirical formula ⇒ CH3O As compound is saturated acyclic, so molecular formula is C2H6O2.

∴ Number of moles of O2 required to oxidise 2 moles of (X) = 5. 27. (14.00) Mass percent of HNO3 = 63 Thus, 100 g of nitric acid solution contains 63 g of nitric acid by mass. No. of moles =

=1

Volume of 100 g of nitric acid solution =

=

= 71.4 mL

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Molarity = =

= 14 M

28.

(4) R = NA × k = 6.023 × 1023 × 1.380 × 10–23 = 8.312 which has 4 significant figures 29. (55.55) 1 litre water = 1 kg i.e. 1000 g water (

d = 1000 kg/m3)

moles of water So, molarity of water = 55.55M 30. (15.05) In pure iron oxide (FeO), iron and oxygen are present in the ratio 1 : 1. However, here number of Fe2+ present = 0.93 or No. of Fe2+ ions missing = 0.07 Since each Fe2+ ion has 2 positive charge, the total number of charge due to missing (0.07) Fe2+ ions = 0.07 × 2 = 0.14 To maintain electrical neutrality, 0.14 positive charge is compensated by the presence of Fe3+ ions. Now since, replacement of one Fe2+ ion by one Fe3+ ion increases one positive charge, 0.14 positive charge must be compensated by the presence of 0.14 Fe3+ ions. In short, 0.93 Fe2+ ions have 0.14 Fe3+ ions × 100 = 15.05%

100 Fe2+ ions have = 31.

(24) According to problem, three atoms of M combine with 2 atoms of N ∴ Formula of compound is M3N2 (Where M is the metal)

Equivalent wt of N =

(

valency of N in compound is 3)

28 g N combines with = 72g metal 14/3 N combines with =

= 12

Eq. wt. of metal = 12

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At

wt

of

metal

=

Eq.

wt

×

valency = 12 × 2 = [Valency of metal = 2]

24

32. TIPS/Formulae : 1 Mole = 6.023 × 1023 molecules = Molecular weight in gms. Weight of 6.023 × 1023 (Avogadro’s number) molecules of CuSO4.5H2O = Molecular wt. of CuSO4.5H2O = 249 g. ∴ Weight of 1 × 1022 molecules of CuSO4.5H2O =

= 4.14 g

33. Carbon (C – 12) 34. 6.02 × 1024 18 mL H2O = 18 g H2O ( density of water = 1 g/cc) = 1 mole of H2O. 1 Mole of H2O = 10 × 6.02 × 1023 electrons ( Number of electrons present in one molecule of water = 2 + 8 = 10) = 6.02 × 1024 electrons 35. TIPS/Formulae : (i) Volume of virus = r2 (Volume of cylinder) (ii) Mass of single virus = (iii) Molecular mass of virus = Mass of single virus × 6.02 × 1023 Volume of virus = r2

= 0.884 10–16 cm3 Weight of one virus = 1.178 10–16 g Mol. wt. of virus = 1.178 10–16 = 7.09 107

6.02

1023

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36.

(a)

From this it is evident, that 111 mg CaCl2 will give CaCO3 = 100mg 1 mg CaCl2 will give CaCO3 =

= 0.90 mg

95 mg MgCl2 gives CaCO3 = 100 mg 1 mg MgCl2 gives CaCO3 =

= 1.05 mg

Total CaCO3 formed by 1 mg CaCl2 and 1 mg MgCl2 = 0.90 + 1.05 = 1.95 mg Amount of CaCO3 present per litre of water = 1.95mg wt of 1 ml of water = 1g = 103 mg wt of 1000 ml of water = 103 × 103 = 106mg Total hardness of water in terms of parts of CaCO3 per 106 parts of water by weight = 1.95 parts. = 20

(b) Eq wt of Ca++ = Ca2+ + Na2CO3

CaCO3 + 2Na +

1 milliequivalent of Ca2+ = 20 mg 1 milliequivalent of Na2CO3 is required to soften 1 litre of hard water. 2MgO (c) 2Mg + O2 2 × 24 =48g 32g 2(24 + 16) = 80g 32g of O2 reacts with = 48g Mg 0.5g of O2 reacts with =

= 0.75g

Weight of unreacted Mg = 1.00 – 0.75 = 0.25g Thus, Mg is left in excess. Weight of MgO formed = MgO + H2SO4 (40g)

= 1.25g

MgSO4 + H2O

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According to reaction 40 g MgO is dissolved in 1000 mL of 1 N. H2SO4 40 g MgO is dissolved in 2000 mL 0.5 N H2SO4 1.25 MgO is dissolved in mL of 0.5 N H2SO4

=

= 62.5mL of 0.5N H2SO4 37.

Given P = 1 atm V = 1L, T = 127°C = 127 + 273 = 400 K PV = nRT (Ideal gas equation) = 0.0304

or Mol.

wt

=

=

92.10

Emperical formula = C7H8 Emperical formula, wt = 12 × 7 + 1 × 8 = 92 =1 Molecular formula = n × empirical formula = 1(C7H8) = C7H8 38. (i) No. of C atoms in 14g of 14C = 6.02 × 1023 No. of C atom in 7 mg (7/1000g) of 14C =

= 3.01 × 1020

No. of neutrons in 1 carbon atom = 7 Total no. of neutrons in 7 mg of 14C = 3.01× 1020 ×7 = 21.07 × 1020 Wt of 1 neutron = wt of 1 hydrogen atom

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= Wt of 3.01 × 1020 × 7 neutrons = 3.5 × 10–3g

=

1. (c) 1 mol of urea = 2 mol of NH3 60 g of urea = 2 mol of NH3 0.6 g of urea =

× 0.6 mol = 0.02 mol of NH3

mol of NH3 = mol of HCl ∴ mol of HCl = 0.02 mol ⇒ Normality of HCl = 0.2 N Volume of HCl = 100 mL 2. (a) According to the stoichiometry of balanced equation, 28 g N2 reacts with 6 g H2

3.

∴ For 56 g of N2, 12 g of H2 is required. (d) Moles of CO2 evolved = ∴ ∴

Moles of NaHCO3 = 10–5 Mass of NaHCO3 = 84 × 10–5 g = 0.84 × 10–3 g = 0.84 mg

4.

∴ % by weight = × 100 = 8.4% (d) No. of moles of oxygen in 0.16 g of oxygen molecule

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=

mol 2 NaCl + 3O2

2NaClO3

According to the reaction, 3 moles of O2 = 2 moles of NaCl = 2 moles of AgCl Molar mass of AgCl = 143.5 g/mol moles AgCl

0.005 moles of O2 will ppt. =

= 0.0033 moles of AgCl ∴ Mass of AgCl (in g) obtained will be = 143.5 g/mol × 0.0033 moles = 0.48 g. 5. (b) Given chemical eqn M2CO3 + 2HCl → 2MCl + H2O + CO2 1g

0.01186 mol

From the above chemical eq . nM2CO3 = nCO2 n

∴ Molar mass of Molar mass = 84.3 g/mol 6.

(d) The molar mass of H3AsO4 is 3 × 1 + 79 + 4 × 16 = 142g/mol ∴ Number of moles of H3AsO4 =

mol.

∴ Number of moles of As2S5 = 7.

(c) Let the weight of acetic acid initially be w1 in 50 mL of 0.060 N solution.

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N=

(Normality = 0.06 N)

0.06 = ⇒

= 0.18 g = 180 mg.

After an hour, the strength of acetic acid = 0.042 N so, let the weight of acetic acid be w2 N= 0.042 = ⇒ w2 = 0.126 g = 126 mg So amount of acetic acid adsorbed per 3g = 180 – 126 mg = 54 mg ∴ amount of acetic acid absorbed per g = 54/3 = 18 mg 8. (d) Weight of hydrated BaCl2 = 61 g Weight of anhydrous BaCl2 = 52 g Loss in mass = 9 g Assuming BaCl2.xH2O as hydrate Mass of H2O = 9 g Moles of H2O =

= 0.5 mol

Grass molecular wt. of BaCl2 = 208 g/mol. % of H2O in this hydrated BaCl2 =

× 100 = 14.75%

% of H2O in BaCl2.xH2O = Thus, On solving x = 2 Hence, the formula of hydrated salt is BaCl2.2H2O 9. (a) TIPS/Formulae :

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Write the reaction for chemical change during reaction and equate moles of products formed. [Co(NH3)5SO4] Br has ionisable Br– ions & [Co(NH3)5 Br] SO4 has ionisable SO4– – ion. Mixture X has 0.02 mol. of [Co(NH3)5SO4] Br and 0.02 mol of [Co(NH3)5Br] SO4 in 2 L of solution Conc. of [Co(NH3)5SO4] Br and [Co(NH3)5Br]SO4 = 0.01 mol/L for each of them. (i) 1 L mixture of X + excess AgNO3 → Y

No. of moles of Y = 0.01 (ii) Also 1 L mixture of X + excess BaCl2 → Z

∴ moles of Z = 0.01. 10. (a) TIPS/Formulae : Equivalents of H2C2O4.2H2O = Equivalents of NaOH

(At equivalence point)

Strength of H2C2O4 . 2H2O (in g/L) = = 25.2 g/L

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Normality of H2C2O4 . 2H2O = =

= 0.4N

Using normality equation : N1V1 = N2V2 (H2C2O4.2H2O)

(NaOH)

0.4 × 10 = 0.1 × V2 or V2 =

= 40 mL.

11. (b) TIPS/Formulae : (i) Find change in oxidation number of Cr atom. (ii) Eq. wt. = In iodometry, K2Cr2O7 liberates I2 from iodides (NaI or KI). Thus, it is titrated with Na2S2O3 solution. 2Na2S2O3+ I2 → 2NaI + Na2S4O6 O.N. of Cr changes from + 6 (in K2Cr2O7) to +3. i.e. +3 change for each Cr atom Thus, one mole of K2Cr2O7 accepts 6 mole of electrons. ∴ Equivalent weight = 12. (d) TIPS/Formulae : (i) H3PO3 is dibasic acid as it contains two –OH groups.

(ii) Normality = Molarity × basicity of acid. (iii) Basicity of H3PO3 = 2

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∴ Normality = 0.3 × 2 = 0.6 13. (b) For equivalent weight of MnSO4 to be half of its molecular weight, change in oxidation state must be equal to 2. It is possible only when oxidation state of Mn in product is + 4. Since oxidation state of Mn in MnSO4 is + 2. So, MnO2 is correct answer. In MnO2, O.S. of Mn = +4 ∴ Change in O.S. of Mn = +4 – (+2) = +2 14. (d) TIPS/Formulae : (i) Volume of substance changes with temperature and mass is not effected by change in temperature. (ii) Find expression which does not have volume term in it. (a) Molarity – Moles of solute/volume of solution in L. (b) Normality – g equivalents of solute/volume of solution in L. (c) Formality – g formula wt./volume of solution in L. (d) Molality – Moles of solute/mass of solvent in kg Molality does not involve volume term. It is independent of temperature. 15. (a) TIPS/Formulae : Molality A molal solution is one which contains one mole of solute per 1000 g of solvent. 16. (i) (ii) (iii)

(d) TIPS/Formulae : Write balanced chemical equation for chemical change. Find limiting reagent. Amount of product formed will be determined by amount of limiting reagent. The balanced equation is : → Limiting reagent is Na3PO4 (0.2 mol), BaCl2 is in excess. From the above equation :

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2.0 moles of Na3PO4 yields Ba3(PO4)2 = 1 mole ∴ 0.2 moles of Na3PO4 will yield Ba3(PO4)2 = 0.1 mol. 17.

(a) The change involved is i.e. it involves only one electron Eq.wt = [

18.

Mol. wt. = M]

(a)

NOTE : Ag2O is thermally unstable and decompose on heating liberating oxygen] Mol. wt. of Ag2CO3 = 108 × 2 + 12 + 16 × 3 = 276 g ∴ 276 g of Ag2CO3 on heating gives residue = 2 × 108 = 216 g of Ag ∴

2.76 g of Ag2CO3 on heating gives =

× 2.76 = 2.16g

19.

of Ag (18) Complete combustion of hydrocarbons can be represented by the following reaction.

For propane combustion reaction is

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Similarly, for butane is

For 1 mol of C4H10 required O2 = ∴For 2 mol of C4H10 required O2 =

mol × 2 = 13 mol

20. (50) M. eq. of K2Cr2O7 = M. eq. of FeC2O4

V = 50 mL 21. (85)

Moles of KMnO4 = Equivalents of H2O2 = Equivalents of KMnO4 Equivalents of KMnO4 = 2 × 10–3 × 5 = 0.01 Moles of H2O2 Mass of pere H2O2 = 0.005 × 34 = 0.170 g Percentage purity 22.

(3400) and Mole of H2

Mole of N2

N2(g) + 3H2(g) Initial mole Final mole 0

100

2NH3 (g) 500

300

200

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Limiting reagent

Mass of NH3 formed = 200 × 17 = 3400 g 23. (10) Normality

24.

(47) Let total mole of solution = 1 So, mole of glucose = 0.1 Mole of H2O = 0.9 % (w/w) of H2O = 47.368 = 47.37. 25.

(25)

Number of mole of x So molar mass of x = 100 g Molarity

.

26. (10) Phosphinic acid is hypophosphorous acid (H3PO2). For neutrization, (N1V1)acid = (N2V2)base 0.1 × 10 = 0.1 × (VmL)NaOH VNaOH = 10 mL 27. (10.00)

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ppm =

= 10

28. (4.96) 10 ppm = ⇒ Mass of Fe = 1 g Molar mass of FeSO4.7H2O = 278 56 g of iron present in 1 mole of FeSO4.7H2O in of salt = 4.96 g

1 g of Fe present 29. (2130)

NaClO3(s) + Fe(s) → NaCl(s) + FeO(s) + O2(g) Moles of NaClO3 = Moles of O2 Moles of O2 =

=

= 20 mol

Mass of NaClO3 = 20 × 106.5 = 2130 g 30. (6) KMnO4 + KI → MnO2 + I2 Eq of KMnO4 = Eq of I2 4×3=n×2 n=6 31. (2992)

Total mass = 12 × 172 + 4 × 232 = 2992 g 32. (6.47) PbS + O2 → Pb + SO2 No. of moles of O2 =

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Moles of Pb formed = ∴ Mass of Pb formed =

× 207 = 6468.75 g

= 6.46875 kg = 6.47 kg 33. (8.09) TIPS/Formulae : Use molarity equation to find volume of H2SO4 solution. CuSO4 + H2O + CO2 For 123.5 gms of Cu(II) carbonate 98 g of H2SO4 are required. For 0.5 gms of Cu(II) carbonate weight of H2SO4 reqd.=

g=

0.39676 g H2SO4 Weight of required H2SO4 = 0.39676 g Weight of solute in grams

or Volume of H2SO4 solution = 8.097 mL 34. (85) Reaction involved titration is

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+ 6HCl → 3ICl + 3KCl + 3H2O 20 mL. of stock KI solution ≡ 30 mL. of Molarity of KI solution =

KIO3 solution

=

Millimoles in 50 mL. of KI solution

= 50 ×

= 15

Millimoles of KI left unreacted with AgNO3 solution = 2 × 50 ×

= 10

∴ Millimoles of KI reacted with AgNO3 = 15 – 10 = 5 Millimoles of AgNO3 present in AgNO3 solution = 5 Molecular weight of AgNO3 = 170 ∴ Wt. of AgNO3 in the solution = 5 × 10–3 × 170 = 0.850 g % AgNO3 in the sample = 35.

× 100 = 85%

(6.0) Mass of Fe2O3 in the sample =

Number of moles of Fe2O3 =

× 1 = 0.552 g

= 3.454 × 10–3

Number of moles of Fe3+ ions= 2 × 3.454 × 10–3 = 6.9 × 10–3 mol = 6.90 mmol Since, only 1 electron is exchanged in the conversion of Fe3+ to Fe2+, the molecular mass is the same as equivalent mass. ∴ Amount of Fe2+ ion in 100 mL. of sol. = 6.90 meq Volume of oxidant used for 100 mL of Fe2+ sol. = 17 × 4 = 68 mL. Amount of oxidant used = 68 × 0.0167 mmol = 1.1356 mmol Let the number of electrons taken by the oxidant = n

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∴ No. of meq. of oxidant used = 1.1356 × n Thus 1.1356 × n = 6.90 ⇒ n = 36.

=6

(10.43) TIPS/Formulae :

Molality = Mass of H2SO4 in 100 ml of 93% H2SO4 solution = 93 g ∴ Mass of H2SO4 in 1000 ml of the H2SO4 solution = 930 g Mass of 1000 ml H2SO4 solution = 1000 × 1.84 = 1840 g Mass of water in 1000 ml of solution = 1840 – 930 g = 910 g = 0.910 kg Moles of H2SO4 =

=

∴ Moles of H2SO4 in 1 kg of water =

×

= 10.43 mol

∴ Molality of solution = 10.43m 37. (6.5) TIPS/Formulae : No. of equivalents of KMnO4 = No. of equivatents of hydrazine sulphate. ——→ Change in oxidation state for each N2H4 = 2 × 2 = 4 Equivalent weight of N2H6SO4 = Normality of KMnO4 = 5 × 450

= 32.5 (

valence factor = 5)

Number of equivalents of KMnO4 = 20 × =

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and if weight of hydrazin sulphate be x g then equivalents of hydrazine sulphate = ∴

=

or x =

= 0.065 g

Hence, wt. of N2H6SO4 in 10 mL solution = 0.065 g ∴ Wt. of N2H6SO4 in 1000 mL solution = 6.5 g 38. (39.6) Given .....(i) and

..(ii)

[On multiplying (i) by 5 and (ii) by 4 and then adding the resulting equations] Molecular weight of NH2OH = 33 Thus 4000 mL of 1M MnO4– would react with NH2OH = 330g ∴ 12 mL of 0.02 M KMnO4 would react with NH2OH

∴ Amount of NH2OH present in 1000 mL of diluted solution

Since, 10 mL of sample of hydroxylamine is diluted to one litre ∴ Amount of hydroxyl amine in one litre of original solution = 39.6 g 39.

(0.58) The complete oxidation under acidic conditions can be represented as follows:

Since 34 g of H2O2 = 2000 mL of 1N . H2O2

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∴ 34 g of H2O2 = 2000 mL of 1N KMnO4 or

of 1N KMnO4

Therefore the unknown normality or 0.588 N 40. (12.15) TIPS/Formulae : (i) Find the volume of CO2 at NTP (ii) Find molecular wt. of metal carbonate (iii) Find the wt. of metal (iv) Calculate equivalent weight of metal Given P1 = 700 mm, P2 = 760 mm, V1 = 1336 mL, V2 = ? T1 = 300 K, T2 = 273 K , or V2 =

=

= 1119.78 mL = 1.12 L at NTP 1.12 L of CO2 is given by carbonate = 4.215 g Molecular weight of metal carbonate =

× 22.4

= 84.3 Metal carbonate is MCO3 = M + 12 + 48 = M + 60 Atomic weight of M = 84.3 – 60 = 24.3 Eq. wt. of metal = 41.

× M. wt.

(4.87) TIPS/Formulae : Write the balance chemical equation and use mole concept for limiting reagent. AgNO3 + NaCl NaNO3 + AgCl 170 g 58.5 g 143.5 g

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From the given data, we find AgNO3 is limiting reagent as NaCl is in excess. 170.0 g of AgNO3 precipitates AgCl = 143.5 g ∴ 5.77 g of AgNO3 precipitates AgCl =

× 5.77 = 4.87 g

42. (59.33) 3 MnO2 → Mn3O4 + O2 3 (54.9 + 32) (3 × 54.9 + 64) = 260.7 g = 228.7 g Let the amount of pyrolusite ignited = 100.00 g ∴ Wt. of MnO2 = 80 g (80% of 100 g = 80 g) Wt. of SiO2 and other inert substances = 15 g Wt. of water = 100 – (80 + 15) = 5 g According to equation, 260.7 g of MnO2 gives = 228.7 g of Mn3O4 ∴ 80 g of MnO2 gives =

× 80 = 70.2 g of Mn3O4

NOTE : During ignition, H2O present in pyrolusite is removed while silica and other inert substances remain as such. ∴ Total wt. of the residue = 70.2 + 15 = 85.2 g Calculation of % of Mn in ignited Mn3O4 3 Mn = Mn3O4 3 × 54.9 = 164.7 g 3 × 54.9 + 64 = 228.7g Since, 228.7 g of Mn3O4 contains 164.7 g of Mn 70.2 g of Mn3O4 contains =

× 70.2 = 50.55 g of Mn

Weight of residue = 85.2 g Hence, percentage of Mn is the ignited sample

= 59.33%

43. TIPS/Formulae : Molality = =

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Molality =

= 0.4m

44. (b) TIPS/Formulae : Write reaction for titration between Na2CO3 and HCl. Method : (Half neutralisation) (Complete neutralisation)

From these reaction it is clear that (i) 2 moles of HCl are required for complete neutralization of Na2CO3. (ii) Titre value using phenolphthalein corresponds only to neutralisation of Na2CO3 to NaHCO3, i.e. half of value required by Na2CO3 solution. (iii) Titre value using methyl orange corresponds to complete neutralisation of Na2CO3 ∴ Both S and E are correct but S is not correct explanation of E. 45. TIPS/Formulae : Write the balanced chemical reaction for change and apply mole concept. The given reactions are

∴ Meq. of ∴ mM of MnO2

Meq of Na2C2O4 = 10 × 0.2 × 2 = 4

=

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Now Since, eq. wt. of MnO2 is derived from KMnO4 and MnSO4 both, thus it is better to proceed by mole concept mM of mM of MnO2 × (2/5) = 4/5 Also

∴ mM of H2O2 = mM of KMnO4 × ∴

or

2KMnO4 + 5H2O2 + 3H2SO4 K2SO4 + 2MnSO4 + 8H2O + 5O2 2KMnO4 + 3MnSO4 + 2H2O 5MnO2 + 2H2SO4 + K2SO4 MnO2 + Na2C2O4 + 2H2SO4 MnSO4 + 2CO2 + Na2SO4 + 2H2O 46. TIPS/Formulae : Write the reactions taking place, balance them and equate moles of I2 and Na2S2O3. KIO3 + 5KI 3K2O + 3I2 i.e., 2 I5+ + 10e– 2I– →

+ 2e–

Now liberated I2 reacts with Na2S2O3 I2 + 2e– 2I– 2S2O32S4O62- + 2e– ∴ millimole ratio of I2 : S2O3 = 1 : 2 Thus, m mole of I2 liberated = m mole of Na2S4O6 used [M is molarity of thiosulphate] Also m mole of KIO3 =

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Now m mole ratio of KIO3 : I2 = 1 : 3 Thus,

47. TIPS/Formulae : Find the milliequivalents and equate them as per data given in question. For Fe3O4 3FeO 2e + Fe3(8/3)+ 3Fe2+ Thus, valence factor for Fe3O4 is 2 and for FeO is 2/3. For, Fe2O3 2FeO; 2e + Fe23+ 2Fe2+ ...(1) Thus, valence factor for Fe2O3 is 2 and for FeO is 1. Let Meq.of Fe3O4 and Fe2O3 be a and b respectively. Meq. of Fe3O4 + Meq. Fe2O3 = Meq. of I2 liberated = Meq. of hypo used a+b= Now, the Fe2+ ions are again oxidised to Fe3+ by KMnO4. Note that in the change Fe2+ Fe3+ + e–; valence factor of Fe2+ is l. Thus, Meq. of Fe2+ (from Fe3O4) + Meq. of Fe2+ (from Fe2O3) = Meq. of KMnO4 used .... (2) 2+ If valence factor for Fe is 2/3 from Eq. (1), then Meq. of Fe2+ (from Fe3O4) = a If valence factor for Fe2+ is 1 then Meq. of Fe2+ (from Fe3O4) = 3a/2 … (3) 2+ Similarly, from Eq. (2), Meq. of Fe from (Fe2O3) = b. 3a/2 + b = 0.25 5 12.8 100/50 = 32 or 3a + 2b = 64 ....(4) From Eqs. (3) and (4) Meq. of Fe3O4 = a = 9 & Meq. of Fe2O3 = b = 18.5

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and % of Fe3O4 = and % of Fe2O3 = 48. The formula of Glauber’s salt is Na2SO4.10H2O Molecular mass of Na2SO4.10H2O = [2 × 23 + 32.1 + 4 × 16] + 10 (1.01 × 2 + 16) = 322.3 g mol–1 Weight of the Glauber’s salt taken = 80.575 g Out of 80.575 g of salt, weight of anhydrous Na2SO4 =

× 80.575 = 35.525 g

Number of moles of Na2SO4 per dm3 of the solution =

= 0.25

Molarity of the solution = 0.25 M Density of solution = 1077.2 kgm–3 =

gm cm–3 = 1.0772 g cm–3

Total weight of sol = V × d = 1 dm3 × d = 1000 cm3 × 1.0772 gcm–3 = 1077.2 g Weight of water = 1077.2 – 35.525 = 1041.67 g Molality of sol. =

× 1000 g = 0.2399 = 0.24 m

Number of moles of water in the solution = = 57.87

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Mole fraction of Na2SO4 =

=

= 0.0043 = 4.3 × 10–3 49. Calculation of number of moles in 45 mL. of 0.025 M Pb(NO3)2 Moles of Pb(NO3)2 = 0.25 ×

= 0.01125

∴ Initial moles of Pb2+ = 0.01125 Moles of = 0.01125 × 2 = 0.02250 [1 mole Pb(NO3)2 2 moles of NO3] Calculation of number of moles in 25 mL. of 0.1 M chromic sulphate Moles of chromic sulphate Cr2(SO4)3 = 0.1 × Moles of

= 0.0025 moles

= 0.0025 × 3 = 0.0075 [1 Mole of chromic sulphate

3

moles of SO42–] Moles of PbSO4 formed = 0.0075 [SO42– is totally consumed] Moles of Pb2+ left = 0.01125 – 0.0075 = 0.00375 Moles of left = 0.02250 [NO3– remain unreacted] Moles of chromium ions = 0.0025 × 2 = 0.005 Total volume of the solution = 45 + 25 = 70 mL. ∴ Molar concentration of the species left (i) (ii)

Pb2+ =

× 1000 = 0.05357 M =

(iii) Cr3+ = 50.

× 1000 = 0.3214 M × 1000 = 0.0714 M

1.5 g of sample require = 150 mL. of

HCl

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∴ 2 g of sample require =

mL. of

= 200 mL. of

HCl HCl

On heating, the sample, only NaHCO3 undergoes decomposition as given below. 2NaHCO3 → Na2CO3 + H2O + CO2 ↑ 2 moles 1 mole 1 mole 2 equ. Neutralisation of the sample with HCl takes place as given below. NaHCO3 + HCl → NaCl + H2O + CO2 1 eq. 1 eq. Na2CO3 + 2HCl → 2NaCl + H2O + CO2 1 mole 2 mole 2 eq. 2 eq. Hence, 2 g sample ≡ 200 ml. of M/10 HCl = 200 ml. of N/10 HCl = 20 meq = 0.020 eq Number of moles of CO2 formed, i.e. n=

=

= 0.005

Moles of NaHCO3 in the sample (2 g) = 2× 0.005 = 0.01 Equivalent of NaHCO3 = 0.01 Wt. of NaHCO3 = 0.01 × 84 = 0.84 g % of NaHCO3 =

= 42%

Equivalent of Na2CO3 = 0.02 – 0.01 = 0.01 Wt. of Na2CO3 = 0.01 × 53 = 0.53 g ∴ % of Na2CO3 =

= 26.5%

∴ % of Na2SO4 in the mixture = 100 – (42 + 26.5) = 31.5% 51. In the given problem, a solution containing Cu2+ and

is

titrated first with KMnO4 and then with Na2S2O3 in presence of KI. In

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titration with KMnO4, it is the

ions that react with the

ions. The concerned balanced equation may be written as given below. +5

+ 16H+ → 2Mn2+ + 10CO2 + 8H2O

Thus, according to the above reaction 2 mmol ≡ 5 mmol However, No. of mmol of

used in titration = Vol. in ml × M

= 22.6 × 0.02 = 0.452 mmol Since 2 mmol

≡ 5 mmol

0.452 mmol

≡

× 0.452 = 1.130 mmol

Titration with Na2S2O3 in the presence of KI. Here Cu2+ react and the reactions involved during titration are 2Cu2+ + 2I– → 2Cu+ + I2 , 2 + I2 → 2I– + Thus 2Cu2+ ≡ I2 ≡ 2 used in titration

No. of m mol of

= 0.05 × 11.3 = 0.565 mmol Now since 2 mmol 0.565 mmol

≡ 2 mmol Cu2+ [From above equation] =

× 0.565 mmol Cu2+ = 0.565 mmol Cu2+

∴ Molar ratio of Cu2+ to Balanced equations in two cases Case I. Mn+7 + 5e– → Mn+2 C2+3 → 2C+4 + 2e– Case II. 2Cu+2 + 2e– → Cu2+

=

=1:2

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2I– → I2 + 2e– and I2 + 2e– → 2I– 2S2+2 → S4+3/2 + 2e– 52. TIPS/Formulae : No. of equivalents of KMnO4 = No. of equivatents of reducing agents. Case I. Reaction of NaOH with H2C2O4 and NaHC2O4. (i) H2C2O4 + 2NaOH → Na2C2O4 + 2H2O (ii) NaHC2O4 + NaOH → Na2C2O4 + H2O Number of milliequivalents of NaOH = N × V = 3.0 × 0.1 = 0.3 ∴

Combined normality of the mixture titrated with NaOH =

= 0.03

Case II. Reaction of C2O4– ion and KMnO4 (iii) 5C2O4– + MnO4– + 16H+ → 2Mn2+ + 10CO2 + 8H2O KMnO4 will react in same manner with both NaHC2O4 and H2C2O4 as it can be seen from the above reaction. Number of milliequivalents of KMnO4 = 4.0 × 0.1 = 0.4 ∴ Combined normality of the mixture titrated with KMnO4 =

= 0.04

The difference (0.04 N – 0.03 N = 0.01 N) is due to NaHC2O4 The total normality of NaHC2O4 will be = 0.01 + 0.01 = 0.02 N From equation (ii) in case I. Eq. wt. of NaHC2O4 = 112 Amount of NaHC2O4 in one litre of solution formed = 0.01 × 112 = 1.12 g and amount of H2C2O4 = 2.02 – Wt. of NaHC2O4 = 2.02 – 1.12 = 0.90 g 53. TIPS/Formulae : Let the amount of NaNO3 in the mixture = x g ∴ The amount of Pb(NO3)2 in the mixture = (5 – x) g Heating effect of sodium nitrate and lead nitrate

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2NaNO2

+

2PbO2+ Now since, 170 g of NaNO3 gives = 32 g of O2 ∴ x g of NaNO3 gives =

× x g of O2

Similarly, 662 g of Pb(NO3)2 gives = 216 g of gases (5 – x) g of Pb(NO3)2 gives =

× (5 – x) g of gases

Actual loss, on heating, is 28% of 5 g of mixture = ∴

(NO2 + O2)

(NO2 + O2)

= 1.4 g × (5 – x) = 1.4

32 x × 662 + 216(5 – x) × 170 = 1.4 × 170 × 662 21184 x + 183600 – 36720 x = 157556 – 15536 x = – 26044, x = 1.676 g Wt. of NaNO3 = 1.676 g and Wt. of Pb(NO3)2 = 5 – 1.676 g = 3.324 g 54. TIPS/Formulae : No. of equivalents of KMnO4 in neutral medium = No. of equivalents of reducing agent. Assuming that KMnO4 shows the following changes during its oxidising nature. Acidic medium Mn7+ + n1e– → Mna+ ∴ n1 = 7 – a 7+ – b+ Neutral medium Mn + n2e → Mn ∴ n2 = 7 – b 7+ – c+ Alkaline medium Mn + n3e → Mn ∴ n3 = 7 – c

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Let V ml. of reducing agent be used for KMnO4 in different medium. ∴ Meq. of reducing agent = Meq. of KMnO4 in acid medium Meq. of KMnO4 in neutral medium = Meq. of KMnO4 in alkaline medium ⇒ 1 × n1 × 20 = 1 × n2 × 33.4 = 1 × n3 × 100 ⇒ n1 = 1.667 n2 = 5 n3 Since n1, n2 and n3 are integers and n1 is not greater than7 ∴ n3 = 1 Hence, n1 = 5 and n2 = 3 ∴ Different oxidation states of Mn in Acidic medium Mn7+ + 5e– → Mna+ or a = + 2 Neutral medium Mn7+ + 3e– → Mnb+ or b = + 4 7+ – c+ Alkaline medium Mn + 1e → Mn or c = + 6 Further, same volume of reducing agent is treated with K2Cr2O7, and therefore Meq. of reducing agent = Meq. of K2Cr2O7 1 × 5 × 20= 1 × 6 × V [ Cr+6 + 6e– → 2Cr+3] V = 16.66 mL ∴ 1M = 6 × 1N 55. (i) Weight of sugar syrup = 214.2 g Weight of sugar in the syrup = 34.2 g ∴ Weight of water in the syrup = 214.2 – 34.2 = 180.0 g Mol. wt. of sugar, C12H22O11 = 342 ∴ Molal concentration =

= 0.56

(ii) Mol. wt. of water, H2O = 18 ∴ Mole fraction of sugar = = 0.0099 56.

(i)

From the given half-cell reaction,

Here, Eq. wt. of NaBrO3 =

=

= 25.17

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[

number of electron involved = 6]

Now we know that Meq. = Normality × Vol. in ml. = 85.5 × 0.672 = 57.456 AlsoMeq. =

× 1000

=

× 1000 × 1000 = 57.456 g

∴

= 1.446 g

Molarity of NaBrO3 = =

= 0.112 M

(ii) From the given half-cell reaction, Eq. wt. of NaBrO3 =

=

= 30.2

[Number of electron involved per BrO3–

]

Thus, the amount of NaBrO3 required for preparing 1000 mL. of 1 N NaBrO3 = 30.2 g ∴ The amount of NaBrO3 required for preparing 85.5 mL of 0.672 N NaBrO3. = Hence, Molarity =

= 1.7532 g = 0.1344 M

57. TIPS/Formulae : (i) Find normality of acid mixture and Na2CO3 . 10H2O. Equate them to find volume of H2SO4.

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(ii) Meq. of H2SO4 = V × N =

eq.

(iii) Equivalent of SO42– = equivalents of H2SO4 × Eq. wt. of SO4– – N × V (mL.) = meq. Acid mixture contains 5 mL of 8N, HNO3, 4.8 mL of 5N, HCl and say, ‘V’ mL of 17 M ≡ 34 N, H2SO4. [1M H2SO4 = 2NH2SO4] N of the acid mixture = =

[Total volume = 2 L = 2000 mL]

or, Nmixture = Eq. of wt. of Na2CO3.10H2O = =

= 143

N of Na2CO3=

=

=

= 0.069N

N1V1 = N2V2 or 30 × Nmixture = 42.9 × 0.069 (acid) (sod. carbonate) ∴ Nmixture = Hence

= 0.0986 N = 0.0986

64 + 34 V = 0.0986 × 2000, 64 + 34 V = 197.2

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34 V= 197.2 – 64.0 = 133.2 ∴ or V =

= 3.9 mL.

Hence, meq. of H2SO4 = V × N of H2SO4 = 3.9 × 34 = 132.6 meq. = 0.1326 eq. of H2SO4 = 0.1326 eq. of = 0.1326 × 48 g of

= 6.3648 g of

are in 3.9 mL of 17M H2SO4

58. TIPS/Formulae : Equivalents of A oxidised = Equivalents of A reduced. Since in acidic medium, An+ is oxidised to AO3–, the change in oxidation state from (+5) to (+n) = 5 – n ∴ Total number of electrons that have been given out during oxidation of 2.68 × 10–3 moles of An+ = 2.68 × 10–3 × (5 – n) Thus the number of electrons added to reduce 1.61 × 10–3 moles of to Mn2+, i.e. (+7) to (+2) =1.61 × 10–3 × 5 [Number of electrons involved = + 7 – (+2) = 5] ∴ 1.61 × 10–3 × 5 = 2.68 × 10–3 × (5 – n) 5–n =

or

59.

TIPS/Formulae :

(i)

Mole fraction =

n=5–

≈2

(ii) 1 mole of Na2S2O3 gives 2 moles of Na+ and 1 mole of S2O32– Molecular wt. of sodium thiosulphate solution (Na2S2O3) = 23 × 2 + 32 × 2 + 16 × 3= 158

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(i)

The percentage by weight of Na2S2O3 =

= 37.92

[Wt. of Na2S2O3 = Molarity × Mol wt] (ii) Mass of 1 litre solution = 1.25 × 1000 g = 1250 g [ Mole fraction of Na2S2O3

density = 1.25g/l]

= Moles of water Mole fraction of Na2S2O3 (iii) 1 mole of sodium thiosulphate (Na2S2O3) yields 2 moles of Na+ and 1 mole of

.

Molality of Na2S2O3 =

= 3.87

Molality of Na+ = 3.87 × 2 = 7.74m Molality of = 3.87m 60. Weight of MCO3 and BaO = 4.08 g (given) Weight of residue = 3.64 g (given) ∴ Weight of CO2 evolved on heating = (4.08 – 3.64) g = 0.44 g = 0.01 mole Number of moles of MCO3

0.01 mole

Volume of 1N HCl in which residue is dissolved = 100 ml Volume of 1N HCl used for dissolution = (100 – 2.5 × 16) ml = 60 ml

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= 0.06 equivalents The

chemical

equation

for

dissolution

can

be

written

as

[Number of moles of BaO and MO = 1 + 1 = 2] Number of moles of BaO + Number of moles of MO = 0.03 Number of moles of BaO = (0.03 – 0.01) = 0.02 moles Molecular weight of BaO = 138 + 16 = 154 ∴ Weight of BaO = (0.02 × 154) g = 3.08 g Weight of MCO3 = (4.08 – 3.08) = 1.0 g Since weight of 0.01 mole of MCO3 = 1.0 g ∴ Mol. wt. of MCO3 Hence atomic weight of unknown M = (100 – 60) = 40 The atomic weight of metal is 40 so the metal M is Ca. 61. Following reactions take place3MnSO4.4H2O

Mn3O4 + 2FeSO4 + 4H2SO4 → Fe2(SO4)3 + 3MnSO4 Milliequivalents of FeSO4 in 30 mL of 0.1N FeSO4 = 30 × 0.1 = 3 m. eq. According to problem step (iv) 25 mL of KMnO4 reacts with = 3 m eq of FeSO4 Thus in step (iii) of the problem, 50 mL of KMnO4 reacts with =

+ 4H2O

m.eq. of FeSO4

= 6 meq of FeSO4 Milli eq. of 100 mL of 0.1N FeSO4 = 100 × 0.1 = 10 m eq.

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FeSO4 which reacted with Mn3O4 = (10–6) = 4 m eq. Milli eq of FeSO4 = Milli eq. of Mn3O4 ( Milli eq of oxidising agent and reducing agent are equal) 1 Meq of Mn3O4 = 3 Meq of MnSO4 . 4H2O 4 Meq of Mn3O4 = 12 Meq of MnSO4 . 4H2O Eq. wt of MnSO4.4H2O =

= 111.5

Wt of MnSO4.4H2O in sample = 12 × 111.5 = 1338 mg = 1.338g. 62. Weight of AgCl formed = 2.567 g Amount of AgCl formed due to MCl = 1.341 g ( NaCl does not decompose on heating to 300°C) Weight of AgCl formed due to NaCl = 2.567 – 1.341= 1.226g

143.5g of AgCl is obtained from NaCl = 58.5g 1.226 g of AgCl is obtained from NaCl =

= 0.4997 g Wt of MCl in 1 g of mixture = 1.000 – 0.4997 = 0.5003g 1.341 g of AgCl is obtained from MCl = 0.5003g 143.5g of AgCl is obtained from MCl

=

= 53.53 g

Molecular weight of MCl = 53.53 63. Volume of oxygen taken = 30 mL, Volume of unused oxygen = 15 mL Volume of O2 used = Volume of O2 added – Volume of O2 left = 30 – 15 = 15 mL

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Volume of CO2 produced = Volume of gaseous mixture after explosion – Volume of unused oxygen or Volume of CO2 produced = 25 – 15 = 10 mL Volume of hydrocarbon = 5 mL General equation for combustion of a hydrocarbon is as follows -

∴ Volume of CO2 produced = 5x, Since Volume of CO2 = 10 mL ∴ 5x = 10 ⇒ x = 2, Volume of O2 used = 15 mL ∴ 5 ⇒ 2+

= 15 ⇒ x + =3(

x = 2)

=3 ⇒ 8 + y = 12 ∴ y = 4

Hence, Molecular formula of hydrocarbon is C2H4. 64. TIPS/Formulae : (i) Find volume of H2 at N.T.P. (ii) Total amount of H2 liberated = H2 liberated by Mg & HCl + H2 liberated by Al & HCl. Conversion of volume of H2 to N.T.P Given conditions N.T.P conditions P1 = 0.92 atm. P2 = 1 atm. V1 = 1.20 litres V2 = ? T1 = 0 + 273 = 273 K T2 = 273 K Applying ideal gas equation, , V2 = = 1.104 litres = 1104 mL The relevant chemical equations are (i) 2 Al + 6HCl 2AlCl3 +

litres

3H2

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2 × 27 3 × 22400 = 54 g at NTP (ii) Mg + 2HCl MgCl2 + H2 24 g 22400 mL at NTP Wt. of alloy = 1 g Let the wt. of aluminium in alloy = x g ∴ Wt. of magnesium in alloy = (1 – x) g According to equation (i) 54 g of Al = 67200 mL of H2 at N.T.P ∴ x g of Al =

= 67200 mL

× x = 1244.4 x mL of H2 at N.T.P

Similarly, from equation (ii) 24 g of Mg = 22400 mL of H2 at N.T.P (1 – x) g of Mg =

× (1 – x) = 933.3 (1 – x) mL of H2

Hence total vol. of H2 collected at N.T.P = 1244.4 x + 933.3 (1 – x) mL But total vol. of H2 as calculated above = 1104 mL ∴ 1244.4 x + 933.3 (1 – x) = 1104 mL 1244.4 x – 933.3 x = 1104 – 933.3 311.1 x = 170.7, x = 0.5487 Hence 1 g of alloy contains Al = 0.5487 g ∴

Percentage of Al in alloy =

= 54.87%

% of Mg in alloy = 100 – 54.87 = 45.13%

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1.

(c)

For Li2+,

For He+,

2.

(a) In hydrogen spectrum maximum lines of Balmer series lies in visible region. 3. (c) Shortest wavelength → Max. energy (∞ → 1) For Lyman series of H atom,

For Balmer series of He+,

4.

(b) In the Balmer series of H–atom the transition takes place from the higher oribtal to n = 2. Therefore the longest wave length

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corresponds to n1 = 2 and n2 = 3. As the wave length decreases, the lines in the series converges. Hence, statement I, II, III are the correct statements among the given options. 5.

(c) r =

For Li2+, r = 6.

7.

=

(a) As the value of Z (atomic number) increases, energy of orbitals decreases (becomes more –ve value) ∴ Order of energy of 2s orbital is H > Li > Na > K. (a) For determined shortest wavelength, n2 = ∞

Lyman series Paschen series

8. (b) For H-atom = For Lyman series, ; ∴ For Balmer series, ;

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∴ So, 9.

(c) E = hν = E= =

KE = = 1.62 × 10–19 J [1J = kg.m2s–2] = 1 eV According to photoelectric effect, K.E. = hν – hv0 hv0 = hν – K.E. Work function (W0) = E – K.E. = 3.1 – 1 = 2.1 eV 10.

(a) n1 = 3, n2 = ∞ = 900 nm

11.

(c) According to Bohr’s model energy in nth state =

eV

For second excited state, of He+, n = 3

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∴ E3(He+) = 12. 13.

eV = –6.04 eV

(b) When temperature is increased, black body emitshigh energy radiation from higher wavelength to lower wavelength. (c) λ = 250 nm E=

KE = stopping potential = 0.5 eV E = W0 + K.E. 4.96 = W0 + 0.5 W0 = 4.46 4.5 eV 14. (d) Radius of nth Bohr orbit in H-atom = 0.53 n2Å Radius of II Bohr orbit = 0.53 × (2)2 = 2.12 Å 15. (d) For Lyman series (shortest wavelength) n1 = 1, n2 =

Longest wavelength = 1st line n1 = 3, n2 = 4

16.

(b) As electron of charge ‘e’ is passed through ‘V’ volt, kinetic energy of electron will be eV

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Wavelength of electron wave (λ) = ∴

λ= 17.

(a) Total energy =

where n = 2, 3, 4 .... Putting n = 2 ET = 18.

(d) Total energy of a revolving electron is the sum of its kinetic and potential energy. Total energy = K.E. + P. E. ; 19.

(c) The kinetic energy of the ejected electron is given by the equation

or

∴ or 20.

(a)

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= 1.214 × 10–7m 21.

(a) =

22.

=

(c) As per Bohr’s postulate, mvr =

KE =

So, So, KE =

Since, So, for 2nd Bohr orbit

KE =

=

23.

(b) Average atomic mass of Fe

24.

(d) TIPS/Formulae :

For hydrogen, n = 1 and Z = 1 ; ∴ rH = 0.529 For Be3+, n = 2 and Z = 4; ∴

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25.

26.

27. 28. 29.

(d) Rutherford's experiment was actually -particle scattering experiment. -Particle is doubly positively charged helium ion i.e., He - nucleus. (c) X-rays can ionise gases and cannot get deflected by electric and magnetic fields, wavelength of these rays is 150 to 0.1Å. Thus, the wavelength of X-rays is shorter than that of U.V. rays. (c) Difference in the energy of the energy levels involved in the transition. (a) Isotones have same number of neutrons. All atoms in triad (a) have same number of neutrons (= A – Z = 8). (d) E = so

30. 31.

32. 33. 34.

=

; λ1 = 2000 Å; λ2 = 4000 Å; =

=2

(b) Electrons in an atom occupy the extra nuclear region. (b) TIPS/Formulae : The following is the increasing order of wavelength or decreasing order of energy of electromagnetic radiations :

Among given choices, radiowaves have maximum wavelength. (b) The radius of nucleus is of the order of 1.5 × 10–13 to 6.5 × 10–13 cm or 1.5 to 6.5 Fermi (1 Fermi = 10–13 cm) (b) Bohr model can explain spectrum of atoms/ions containing one electron only. (d) NOTE : Energy is emitted when electron falls from higher energy level to lower energy level and energy is absorbed when electron moves from lower level to higher level. 1s is the lowest energy level of electron in an atom.

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An electron in 1s level of hydrogen can absorb energy but cannot emit energy. 35.

(d)

for neutron =

= 1; electron =

proton = 36.

37. 38. 39.

= 0; α-particle =

= 0.5;

= 1837

(a) According to Rutherford's experiment. "The central part consisting of whole of the positive charge and most of the mass, called nucleus, is extremely small in size compared to the size of the atom." (c) Rutherford’s scattering experiment led to the discovery of nucleus. (d) No. of neutrons = Mass number – Atomic number = 70 – 30 = 40. (4) Energy associated with incident photon = E=

= Photoelectric effect can take place only when Ephoton > Thus, number of metals showing photoelectric effect will be 4 (i.e. Li, Na, K and Mg). 40. (22.8) For maximum energy, n1 = 1 and n2 = ∞

Since RH is a constant and transition remains the same ;

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Hence, 41.

(27419) The shortest wavelength transition in the Balmer series corresponds to the transition n = . Hence, n1 = 2, n2 = ∞ n=2

= 27419.25 cm–1 42.

J

(2055) En of H =

∴ En of He+ =

× Z2 J

∴ E3 of He+ =

J

Hence, energy equivalent to E3 must be supplied to remove the electron from 3rd orbit of He+. Wavelength corresponding to this energy can be determined by applying the relation. E=

or λ =

=

= 2055 × 10–10 m = 2055 Å 43.

(220) TIPS/Formulae : (i) Energy of nth orbit = En =

(ii) Difference in energy = E1 – E2 = hν = or

λ= Given E1 = 2.17 × 10–11 Energy of second orbit = E2 = = 0.5425 × 10–11 erg

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∆E = E1 – E2 = 2.17 × 10–11 – 0.5425 × 10–11 = 1.6275 × 10–11 erg = 12.20 × 10–6 cm = 1220 Å

λ= 44.

(660) TIPS/Formulae : ∆E = E3 – E2 = hν =

or

λ= Given E2 = – 5.42 × 10–12 erg, E3 = – 2.41 × 10–12 erg λ= = 45. 46. 47. 48. =

= 6.604 × 10–5 cm = 660 nm

photons isobars; neutrons; 1.66 × 10–27 kg Mass of hydrogen atom =

= 0.166 × 10–23 g = 1.66 × 10–27 kg 49. True : β-particles are deflected more than α-particles because they have very-very large e/m value as compared to α-particles due to the fact that electrons are much lighter than He2+ species. 50. False : Gamma rays are electromagnetic radiations of wavelengths 10–9 cm to 10–10 cm. 51. (a, d) The energy of an electron on Bohr orbits of hydrogen atoms is given by the expression

Where n takes only integral values. For the first Bohr orbit, n = 1 and it is given that E1 = –13.6 eV

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Hence

of the given values of energy, only – 3.4 eV and –

1.5 eV can be obtained by substituting n =2 and 3 respectively in the above expression. 52. (b, d) In tritium (the isotope of hydrogen) nucleus there is one proton and 2 neutrons. ∴ n + p = 3. In deuterium nucleus there is one proton and one neutron∴ n + p = 2. 53. (a, c) α-particles pass through because most part of the atom is empty. 54. (a, c) Because they have isotopes with different masses. The average atomic mass is the weighted, mean of their presence in nature; eg. Cl35 and Cl37 are present in ratio 3 : 1 in nature. = 35.5

So A = 55.

and

(b, d)

have same number of neutrons (= A – Z) as

. 56.

(d)

57.

(c)

; ;

58.

Determination of number of moles of hydrogen gas, n= The concerned reaction is H2 → 2H; ∆H = 436 kJ mol-1

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Energy required to bring 0.0409 moles of hydrogen gas to atomic state = 436 × 0.0409 = 17.83 kJ Calculation of total number of hydrogen atoms in 0.0409 mole of H2 gas 1 mole of H2 gas has 6.02 × 1023 molecules molecules

0.0409 mole of H2 gas =

Since, 1 molecule of H2 gas has 2 hydrogen atoms 6.02 × 1023 × 0.0409 molecules of H2 gas = 2 × 6.02 × 1023 × 0.0409 = 4.92 × 1022 atoms of hydrogen Since, energy required to excite an electron from the ground state to the next excited state is given by E= =

= 1.632 × 10-21 kJ

Therefore, energy required to excite 4.92 × 1022 electrons = 1.632 × 10-21 × 4.92 × 1022 kJ = 8.03 × 10 = 80.3 kJ Therefore, total energy required = 17.83 + 80.3 = 98.17 kJ 59. Work done while bringing an electron infinitely slowly from infinity to proton of radius a0 is given as follows

NOTE : This work done is equal to the total energy of an electron in its ground state in the hydrogen atom. At this stage, the electron is not moving and do not possess any K.E., so this total energy is equal to the potential energy. T.E. = P. E + K. E. = P. E. =

...(1)

In order the electron to be captured by proton to form a ground state hydrogen atom it should also attain

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(It is given that magnitude of K.E. is half the magnitude of P.E. Note that P.E. is –ve and K.E is +ve) ∴ T.E = P. E. + K. E.

+

or P.E. = 2 × T.E. 60.

or P.E. =

Bond energy of I2 = 240 kJ mol–1 = 240 × 103 J mol–1 molecule–1 = 3.984 × 10–19 J molecule–1

Energy absorbed = 4.417 ×10–19 J Kinetic energy = Absorbed energy – Bond energy ∴ Kinetic energy = 4.417 × 10–19 – 3.984 ×10–19 J = 4.33 × 10–20 J ∴ Kinetic energy of each atom of iodine

61.

TIPS/Formulae : Number of waves =

where n = Principal quantum number or number of orbit Number of waves =

=

=3

62. For He+ ion, we have = Z2RH

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= (2)2RH Now for hydrogen atom

= RH

...(i)

= RH

...(ii)

Equating equations (i) and (ii), we get = Obviously, n1 = 1 and n2 = 2 Hence, the transition n = 2 to n = 1 in hydrogen atom will have the same wavelength as the transition, n = 4 to n = 2 in He+ species. 63.

TIPS/Formulae : ∆E = RhcZ2 Here, R = 1.0967 × 107 m–1 h = 6.626 × 10–34 J sec, c = 3 × 108 m/sec n1 = 1, n2 = 2 and for H-atom, Z = 1 E2 – E1 = 1.0967 × 107 × 6.626 × 10–34 × 3 × 108 ∆E = 1.0967 × 6.626 × 3 ×

× 10–19 J

= 16.3512 × 10–19 J eV = 10.22 eV

= ∆E =

= RhcZ2

= RZ2

= RZ2 ×

Given, λ = 3 × 10–8 m

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∴

= 1.0967 × Z2 ×

∴ Z2 =

× 107 ≈4 ∴Z=2

=

So, it corresponds to He+ which has 1 electron like hydrogen. 64. TIPS/Formulae : To calculate the energy required to remove electron from atom, n = ∞ is to be taken. Energy of an electron in the nth orbit of hydrogen is given by ergs

E = –21.7 × 10–12 × ∴ ∆E = –21.7 × 10–12 = –21.7 × 10–12

= –21.7 × 10–12 ×

= – 5.42 × 10–12 ergs Now we know that ∆E = hv ∴

∆E =

or

λ =

Substituting the values, λ = 65.

= 3.67 × 10–5 cm Let the % of isotope with At. wt. 10.01 = x ∴ % of isotope with At. wt. 11.01 = (100 – x)

At. wt. of boron = ⇒ 10.81 =

∴ x = 20

Hence, % of isotope with At. wt. 10.01 = 20% ∴ % of isotope with At. wt. 11.01 = 100 – 20 = 80%.

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1. 2.

(d) (a), (b) and (c) are according to quantum theory but (d) is statement of kinetic theory of gases. (d) 2πr = nλ r=

2π ×

= 4λ λ = 2π ×

3.

λ = 8πa0 (d) In photoelectric effect, = w + KE of electron

Given that KE of ejected photoelectron is very high in comparison to work function w. = KE

= New wavelength

4.

(d) Given λ = 1.5πa0 nλ = 2πr ...(i) Radii of stationary states (r) is expressed as: r = a0

...(ii)

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From eqn (i) and (ii) ;λ=

nλ = 1.5πa0 = 2πa0 =

= 0.75

5. (b) First Bohr orbit of H atom has radius r = 0.529 Å Also, the angular momentum is quantised. mvr =

∴ 6.

(d) Kinetic energy of any particle =

Also K.E. =

de-broglie wavelength = λ =

=

λ= Mass of electron < mass of neutron λ (electron) > λ (neutron) 7.

(a) de Broglie wavelength (λ) =

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= 10–33 m 8.

(b) h = 6.6 × 10–34 Js m = 1000 kg = 10 m/sec

v = 36 km/hr = ∴ λ= 9.

(a) TIPS/Formulae : According to de-Broglie’s equation λ=

=

Given, h = 6.6 × 10–34 Js, m = 200 × 10–3 kg v= λ=

= 2.38 × 10–10 m

10. (d) As packet of energy equal to hv; as wave having frequency v. 11. (5) Since,

⇒ For two gases, =5 12.

(222)

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So, (KE)max = 6.63 × 10–19 – 4.41 × 10–19 = 2.22 × 10–19 J = 222 × 10–21 J 13. 14. 15.

orbitals Heisenberg, de-Broglie; For hydrogen atom, Z = 1, n = 1 v = 2.18 ×

×

= 2. 18 × 106 ms–1

de Broglie wavelength, =

=

= 3.34 × 10–10 m = 3.3 Å For 2p, l = 1 ∴

Orbital angular momentum =

=

16. or λ = 6.627 × 10–35 m = 6.627 × 10–25Å probability of finding electron 17. within 2s sphere (at node) ( probability of finding an electron is zero at node) ∴ (Squaring the given value of

)

or

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1.

(a)

Thus, order of orbitals filled are 6s < 4f < 5d < 6p. 2. (c) Radial node = n – l – 1

Probability density (ψ2) can be zero for 3p orbital other than infinite distance. It has one radial node. Thus, statement (c) is correct. 3. (b) Under the given situation for n = 1, l = 0, 1, 2 n = 2, l = 0, 1, 2, 3 n = 3, l = 0, 1, 2, 3, 4 According to (n + l) rule of order of filling of subshells will be : 1s 1p 1d 2s 2p 3s 2d 3f Atomic number 6 1s2 1p4 Atomic number 9 1s2 1p6 1d1 Atomic number 8 1s2 1p6 Atomic number 13 1s2 1p6 1d5 Therefore option (b) is correct. Atomic number of first noble gas will be 18 (1s2 1p6 1d10). 4. (b) For n = 4 possible values of l = 0, 1, 2, 3; only l = 2 and l = 3 can have m = –2. So possible subshells are 2. 5. (b) The possible number of orbitals in a shell in term of ‘n’ is n2 ∴ n = 5 ; n2 = 25 6. (a) Probability of finding an electron will have maximum value at both ‘a’ and ‘c’. There is zero probability of finding an electron at ‘b’.

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7.

8.

9. (I) (II) (III) (IV) The

(b) The given probability density curve is for 2s orbital due to the presence of only one radial node. 1s and 2p orbital do not have any radial node and 3s orbital has two radial nodes. Hence, option (b) is correct. (a) Atomic numbers of N, O, F and Na are 7, 8, 9 and 11 respectively. Therefore, total number of electrons in each of N3–, O2–, F–, and Na+ are 10 and hence they are isoelectronic. (c) n+ n=4 = 2 4d 6 n=3 = 2 3d 5 n=4 = 1 4p 5 n=3 = 1 3p 4 energy of an atomic orbital increases with increasing n + . For identical values of n + , energy increases with increasing value of n. Therefore the correct order of energy is: 3p < 3d < 4p < 4d IV

II

III

I

10. (b) Number of orbitals in a shell = n2 = (5)2 = 25. 11. (d) Radial probability function curve for 1s is (D). Here P is 4πr 2R2.

12.

13.

(b) According to Aufbau principle, the sequence of filling electrons in sixth period is 6s – 4f – 5d – 6p i.e., (ns) → (n – 2) f → (n – 1)d → np (a) The electronic configuration of Rubidium (Rb = 37) is

Since last electron enters in 5s orbital

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Hence n = 5, l = 0, m = 0, 14.

(c) n = 3, l = 2 means 3d orbital

i.e. in an atom only one orbital can have the valueml = + 2 15. (a) TIPS/Formulae : Number of radial nodes = (n − l − 1) For 3s: n = 3, l = 0 (Number of radial node = 2) For 2p: n =2, l = 1 (Number of radial node = 0) 16. (c) 17. (d) NOTE : The quantum numbers +1/2 and –1/2 for the electron spin can represent any one among clockwise or anticlockwise spin direction. But if one value represents clockwise spin then the other value will represent anticlockwise spin. 18. (b) 3d 54s1 system is more stable than 3d44s2, hence former is the ground state configuration. 19. (a) px orbital being dumbell shaped, have number of nodal planes = 1, in yz plane. The electron density is only along x-axis (xy and zx planes), thus, in yz-plane, there will be zero electron density.

20.

(a) TIPS/Formulae : The two guiding rules to arrange the various orbitals in the increasing energy are: (i) Energy of an orbital increases with increase in the value of n + l. (ii) Of orbitals having the same value of n + l, the orbital with lower value of n has lower energy. Thus, for the given orbitals, we have (i) n + l = 4 + 1 = 5 (ii) n + l = 4 + 0 = 4

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(iii) n + l = 3 + 2 = 5 (iv) n + l = 3 + 1 = 4 Hence, the order of increasing energy is (iv) < (ii) < (iii) < (i) 21. (a) The expression for orbital angular momentum is Angular momentum For d orbital, l = 2. Hence, 22.

(b) TIPS/Formulae :

Orbital angular momentum (mvr) = For 2s orbital, (azimuthal quantum number) = 0 Orbital angular momentum = 0. 23. (c) TIPS/Formulae : Total nodes = n – l No. of radial nodes = n – l – 1 No. of angular nodes = l For 3p sub-shell, n = 3, l = 1 No. of radial nodes = n – l – 1 = 3 – 1 – 1 = 1 No. of angular nodes = l = 1 24. (c) Electronic configuration of chlorine is [Ne] 3s2, 3p5 Unpaired electron is found in 3p sub-shell. n = 3, l = 1, m = 1 25. (a) NOTE : Exactly half filled orbitals are more stable than nearly half filled orbitals. Cr (At. no. 24) has configuration [Ar] 3d5, 4s1. 26. (c) Configuration ns2, np5 means it requires only one electron to attain nearest noble gas configuration. So, it will be most electronegative element among given choices. 27. (b) According to Aufbau principle, the orbital of lower energy (2s) should be fully filled before the filling of orbital of higher energy

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starts. 28. (c) If l = 2, m ≠ –3, m will vary from –2 to +2. i.e. possible values of m are –2, –1, 0, + 1 and + 2. 29. (a) Rb has the configuration : 1s2 2s2p6 3s2p6d10 4s2p6 5s1; so n = 5, l = 0, m = 0 and s = +½ is correct set of quantum numbers for valence shell electron of Rb. 30. (d) One p-orbital can accommodate up to two electrons with opposite spin while p-subshell can accommodate upto six electrons. 31. (a) The principal quantum number (n) is related to the size of the orbital (n = 1, 2, 3.....) 32. (3) In one electron system, all orbitals of a shell are degenerate.

In case of many electron system, different orbitals of a shell are nondegenerate. Hence, in the second excited state, only three p-orbitals (2p) are degenerate. means ml can be +1 and –1. 33. (6) For n = 4, the total number of possible orbitals are :

Thus, total number of orbitals having | ml | = 1 is 6. The number of electrons with s = –1/2 is 6. 34. (9) Maximum number of orbitals when n = 3 is n2 = 32 = 9 ∴ Number of electrons with 35.

will be 9.

(10) For n = 3 and l = 2 (i.e., 3d orbital), the values of m varies from – 2 to +2, i.e. –2, –1, 0, +1, +2 and for each ‘m’ there are 2 values of ‘s’,

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i.e. +½ and –½. ∴ Maximum no. of electrons in all the five d-orbitals is 10. 36. 4s1, 3d5; The electronic configuration of Cr is : 1s2, 2s2, 2p6, 3s2, 3p6, 4s1, 3d5. ∴ Outermost electronic configuration is 3d5, 4s1. 37. orientation in space 38. antiparallel; or opposite 39.

False : The orbital

lie along X and Y axis where electron

density is maximum. 40. True : In case of hydrogen (single electron system), energy of electron depends only on the principal quantum number. Thus, 4s is in higher energy level than 3d. 41. False : The outer electronic configuration of the ground state chromium atom is 3d5 4s1, as half filled orbitals are more stable than nearly half filled orbitals. 42. (a, c) Given, azimuthal quantum no. (l) = 2 (d-subshell) Magnetic quantum no.(m) = 0(zero), which is for dz2orbital.

n2 = 42 ⇒ n = 4 Radial node = n – l – 1 = 4 – 2 – 1 = 1 Angular node = l = 2 Wave function corresponds to ψ4,2,0. It represents 4dz2-orbital which has only one radial node and two angular nodes. It experiences nuclear charge of 2e units. 43. (a, d) According to Hund’s rule pairing of electrons starts only when each of the orbital in a sub shell has one electron each of parallel spin. (a) and (d) are correct ground state electronic configurations of nitrogen atom in ground state. 44. (a, b, c) (a) 24Cr = 1s2, 2s22p6, 3s23p63d5, 4s1 = [Ar] 3d5,4s1

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(b) For magnetic quantum number (m), negative values are possible. For s - subshell, l = 0, hence m = 0 for p - subshell, l = 1, hence m = –1, 0, +1 (c) 47Ag = 1s2, 2s2 2p6, 3s23p63d10, 4s2 4p6 4d10, 5s1 Hence, 23 electrons have a spin of one type and 24 of the opposite type. (d) Oxidation state of N in HN3 is –1/3. 45. (c) No. of radial nodes = n – l – 1, For 2s orbital, n = 2 , l = 0 ∴ No. of radial nodes = 2 – 0 – 1 = 1 The plotted graph is correct for 2s-orbital, as wave function changes its sign at node.

46.

(a) E.C. of H : 1s1; for 1s orbital

For 1s orbital of hydrogen like species:

Then,

47. (c) (a) For 1s orbital :

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Probability density is maximum at nucleus for 1s orbital. (b) For 2s orbital:

At nucleus r = 0

Probability density

(c) In the wave function (ψ) expression for 1s-orbital of He+, there should be no angular part (θ). (d) Option (d) is correct from the previous question. 48. (A) - (q); (B) - (p), (q), (r, s); (C) - (p, q, r); (D) - (p, q, r) (a) Orbital angular momentum (b) (c) (d) 49.

i.e. L depends on

azimuthal quantum number only. To describe a hydrogen like one-electron wave function, three quantum numbers n, l and m are required. Further, to obey Pauli principle, fourth quantum number s is also required. To define size, shape and orientation of atomic orbitals, n, l and m are required respectively. Probability density (ψ2) of an electron can be determined from the value of n, l and m. (A) - (r); (B) - (q); (C) - (p); (D) - (s)

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(i)

where

∴ (i) – (c)

(ii)

(iii) Angular momentum of electron in lowest (1s) orbital = (iv) For 50-52 The spherically symmetric state S1 of Li2+ with one radial node is 2s. Upon absorbing light, the ion gets excited to state S2, which also has one radial node. The energy of electron in S2 is same as that of Hatom in its ground state.

∴

where E1 is the energy of H-atom in the ground state =

En =

for Li2+ En = E1 ⇒ n = 3

∴

State S2 of Li2+ having one radial node is 3p. Orbital angular momentum quantum number of 3p is 1. Energy of state 50. 51. 52.

(b) (c) (b)

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53.

Ground state electronic configuration of Si

3s 3px 3py 3pz is in accordance with Hund’s rule which states that electron pairing in any orbital (s, p, d or f) cannot take place until each orbital of the same sub-level contains 1 electron each of like spin.

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1.

(a) un = 1 nil = 0 enn = 9 So, atomic number = 109 2. (b) Phosphorus has atomic number 15. Its group number is 15, number of valence electrons is 5 and valency is 3. 3. (b) Elements with Z = 120 will belong to alkaline earth metals. Its electronic configuration may be represented as [Og] 8s2. 4. (a) If elements are arranged in order of their increasing atomic numbers, element coming at intervals of 2, 8, 8, 18, 18, 32 and 32 will have similar physical and chemical properties and thus grouped in one particular group. 5. (d) The electronic configuration of the given ions are as follows. 2+ = 1s2, (No unpaired electron) 12Mg 3+ = 1s2, 2s22p6, 3s23p63d1 (One unpaired electron) 22Ti 3+ = 1s2, 2s22p6, 3s23p63d2 (Two unpaired electrons) 23V 2+ = 1s2, 2s22p6,3s23p63d6 (Four unpaired electrons) 26Fe 6. (d) The electrons are not filled in d-subshell monotonically with increase in atomic number, among transition elements. It breaks at chromium and copper. 7. (101)

IUPAC symbol = Unu

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Atomic no. (Z) = 101 8. (a, c, d) Periodic table does not help to predict the stable valency states of the elements.

1.

(b) Elements with atomic number 21, 25, 42 and 72 belongs to transition metals. 2. (c) All are isoelectronic species, so more is the Zeff less will be the ionic size. ∴ Correct order of ionic radii is Al3+ < Mg2+ < Na+ < F– < O2– < N3– 3. (c) Belongs to actinoids series and they all belongs to 3rd group. So atomic no. 101 element is actinoids and atomic number 104 element belongs to 4th group. (endothermic) 4. (b) (exothermic) (endothermic) (endothermic) •

Electron gaining enthalpy (EGE) of H(g) is negative while that of Ar(g) is positive due to ns2np6 configuration. • Second EGE is always positive for an atom. • Ionization potential of an atom is positive. 5. (b) O2– F– Na+ Mg2+ Z 8 9 11 12 No. of e– 10 10 10 10 In isoelectronic species greater is Zeffe. smaller is radius so order is O2– > F– > Na+ > Mg2+. 6. (a) Charge/ radius ratio of Be and Al is same because of diagonal relationship. Remaining statements are correct.

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7.

(c) As difference in 3rd and 4th ionisation energies is high so atom contains 3 valence electrons. 8. (d) On moving left to right along a period in the periodic table atomic radius decreases while electronegativity, electron gain enthalpy and ionisation enthalpy increases, along a period. 9. (b) On moving left to right in a period, the acidic character of oxides increases. 3rd period element oxides.

Acidic character Atomic No. So, X have minimum atomic number while Z have maximum atomic number. Thus, the correct order of the atomic number is X Al; S > P; Se > Te; Ge > Ga. 20. (c) 71X = [Xe]6s24f145d1 ∴ Orbital occupied by last e– is 5d.

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21.

(c) Generally, electronegativity decreases down the group as the size increases. This can also be formulated as:

Electronegativity 22.

(d) On moving from left to right across a period, the electron affinity becomes more negative. On moving from top to bottom in a group, the electron affinity becomes less negative. Chlorine has exceptionally more negative electron affinity than fluorine, because adding an electron to fluorine (2p orbital) causes greater repulsion than adding an electron to chlorine (3p orbital) which is larger in size. 23. (a) Mg can form basic carbonate 3MgCO3 . Mg(OH)2 . 3H2O ↓ While Li can form only carbonate (Li2CO3), not basic carbonate. 24. (a) Isoelectronic species have same no. of electrons. Ions ⇒ O2– F– Na+ Mg2+ 8 + 2 9 + 1 11 – 1 12 – 2 – No. of e ⇒ 10 10 10 10 2– – + Therefore O , F, Na , Mg2+ are isoelectronic. 25. (c) Generally, the ionization enthalpies or energy increases from left to right in a period and decreases from top to bottom in a group. Several factor such as atomic radius, nuclear charge, shielding effect are responsible for change of ionization enthalpies. Here, Ist ionization enthalpy of A and B is greater than group I (Li 520 kJmol–1 to Cs374 kJmol–1), which means element A and B belong to group –2 and all three given ionization enthalpy values are less for element B means B will come below A. 26. (b) Alkali metals have the lowest ionization energy in each period, on the other hand Sc is a d - block element. Transition metals have smaller atomic radii and higher nuclear charge leading to high ionisation energy.

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27.

(b) Due to extra stable half-filled p-orbital electronic configurations of Group 15 elements, larger amount of energy is required to remove electrons compared to Group 16 elements. 28. (a) For isoelectronic species, size of anion increases as negative charge increases. Thus the correct order is (a). 29. (c) Tellurium (Te) has 5s25p4 valence shell configuration. It belongs to group 16 and present in period 5 of the periodic table. 30. (c) On moving down in a group atomic radii increases. 31. (a) For isoelectronic species ionic radii decreases as the charge on ion decreases. Further on moving down in a group ionic radii increases. Hence the correct order is O2– < N3– < S2– < P3– 32. (c) On moving down a group size increases, hence ionisation enthalpy decreases. Hence Se < S and Ba < Ca. Further, Ar being an inert gas has maximum IE. 33. (d) On moving down in a group atomic radii increases due to successive addition of extra shell hence, order is O < S < Se Further, As is in group 15 having one less electron in its p orbital compared to group 16, hence as has higher atomic radii than group 16 elements. i.e., O < S < Se < As 34. (a) Non-metallic oxides are acidic and metallic oxides are basic. Thus, the acidic order is CaO < CuO < H2O < CO2. 35. (b) Effective nuclear charge (i.e. Z/e ratio) decreases from F– to N3–, hence the radii follows the order: F– < O2– < N3–. Z/e for F– = 9/10=0.9, for O2– = 8/10 = .8, for N3 – = 7/10= 0.7 36. (a) Non-metallic oxides are acidic and acidic character decreases with increase in metallic character. 37. (a) TIPS/Formulae : (i) Hydrogen bonding increases the boiling point. (ii) Hydrogen bonds are formed in compounds having F or O or N with hydrogen. S, Se, Te cannot undergo hydrogen bond formation because of their larger size and lower electronegativity values.

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38. (i) (ii) Pb2+

(b) TIPS/Formulae : Ion having half filled or full filled orbital have extra stability. Larger the size of cation more will be its stability (5d10 6s2), has the most stable +2 oxidation state because here the dorbital is completely filled and is more stable than Fe2+ (3d6). Again Ag+ (4d10) is more stable as here again the d-orbital is completely filled and Ag2+ is not easily obtained. Pb2+ is more stable compared to Sn2+ (4d10 5s2) because of its large size. 39. (b) NOTE : Ionisation energy increases with increasing atomic number in a period, while it decreases on moving down a groups. IE of element with electronic configuration (d) is lowest because of its biggest size. Among the remaining three elements of the same period (3rd). IE of element with electronic configuration (b) is the highest due to greater stability of the exactly half-filled 3 p-subshell. 40. (b) Nitrogen, being smallest in size, can give up its lone pair of electrons most easily. 41. (d) TIPS/Formulae : For isoelectronic ions, ionic size Species

No. of e–

At. No.

N 10 7 –2 O 10 8 – F 10 9 + Na 10 11 + ∴ Na is smallest in size. 42. (a) NOTE : First ionisation potential increases from left to right in a period. IE1 of Mg is higher than that of Na because of increased nuclear charge and also than that of Al because in Mg a 3 s-electron has to be removed while in Al it is the 3 p-electron. The IE1 of Si is, however, higher than those of Mg and Al because of its increased nuclear charge. Thus, the overall order is Na < Mg > Al < Si. 43. (c) NOTE : Electronegativity increases on moving from left to right in a period and decreases on moving from top to bottom in a group. –3

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Si and P are placed in the 3rd period while C and N are placed in the 2nd period. Elements in 2nd period have higher electronegativities than those in the 3rd period. Since, N has smaller size and higher nuclear charge than C, its electronegativity is higher than that of C. Similarly, the electronegativity of P is higher than that of Si. Thus, the overall order is : Si, P, C, N. 44. (a) Ionisation potential of nitrogen is more than that of oxygen. This is because nitrogen has more stable half-filled p-orbitals. (N = 1s2, 2s2, 2p3, O = 1s2, 2s2, 2p4) 45. (a) TIPS/Formulae : (i) Noble gases do not have covalent radii. They have only van der Waal’s radii. (ii) Covalent radii is always smaller than corresponding van der Waal’s radii. Atomic radius of neon being van der Waal’s radius is larger than that of fluorine which is in fact is its covalent radius. 46. (c) Amongst B, C, N and O; N has the highest first ionization energy, because of its half filled 2p orbital which is more stable. 47. (c) NOTE : Ionization potential is amount of energy required to take out most loosely bonded electron from neutral atom. Its value depends on stability of atom (electronic configuration) C – 1s2 2s2 p2 C+ – 1s2 2s2 p1 N – 1s2 2s2 p3 N+ – 1s2 2s2 p2 O – 1s2 2s2 p4 O+ – 1s2 2s2 p3 F – 1s2 2s2 p5 F+ – 1s2 2s2 p4 (for second ionisation potential, IE2) As for IE2 the electron in all the cases is to be removed from 2p orbital so it must follow the order CC 48. (9) By observing the values of ionization enthalpy for atomic number (n + 2), it is observed that I2 >> I1. Which shows that number

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of valence shell electrons is 1 for atomic number (n + 2). Therefore element with atomic number (n + 2) should be an alkali metal. For atomic number (n + 3), I3 >> I2, which shows that it will be an alkaline earth metal. All the observations 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. 49. (2) Fluorine generally shows 0 and –1 oxidation states while sodium shows 0 and +1 oxidation state. 50. Electronegativity 51. Electron affinity 52. False : On moving down the group 13 (III) A the basic nature of hydroxides increases. The basic nature increases as the element becomes more electropositive or acquires more metallic character when we move down a group. 53. False : Halogens have high electron affinities which decrease as we move down the group. However, fluorine has lower value of E.A. than chlorine which is due to its small size and more repulsion between the electron added and electrons already present. Hence, the order Cl > Br > F. 54. False : Ionisation energy decreases on moving down in group IA from Li to Cs, the reducing property should increase in the same order, i.e., from Li to Cs which is found to be so, except an anomaly in lithium which is found to be the strongest reducing agent; because of its higher oxidationkj potential (Eº). 55. (a, d) NO ⇒ Neutral B2O3 ⇒ Acidic CrO ⇒ Basic All other oxides are amphoteric 56. (d) Higher the (+) charge, smaller will be radii. 57. (a, b) TIPS/Formulae : For dissolution, Hydration energy > Lattice energy. BaSO4 is sparingly soluble in water because its hydration energy is lesser than the lattice energy and thus, ions are not separated from each other. On the contrary in Na2SO4, the hydration energy is more than its

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lattice energy. Thus, ions are separated from each other and pass in solution state. 58. (c) The first ionisation energy of Be (1s22s2) is greater than that of B(1s22s2p1) because removal of electron from 2s orbital is tough compare to that from 2p orbital. This is because 2s-orbital is closer to nucleus the 2p-orbital. Reason is incorrect as 2p-orbital is higher in energy. 59. (i) Al3+ < Mg2+ < Li+ < K+ In these Al3+ & Mg2+ are isoelectronic species, so in these size decreases with increase in atomic number because increase in atomic number increases Zeff. Size Due to anomalous behaviour and diagonal relationship, the ionic radius of Li+ is slightly bigger than Mg 2+. In Li+ & K+, K+ is bigger in size than Li+ because on moving from top to bottom in a group, the group size increases. (ii) Increasing order of basic character : NiO < MgO < SrO < K2O < Cs2O The basic character of oxides increases when we move down the group. So, K2O < Cr2O and MgO < SrO. Further, higher the group number lesser is the basic character. Hence, NiO is the least basic. (iii) Increasing order of ionic size : Mg2+ < Na+ < F– < O2– < N3– NOTE : All the above ions are isoelectronic having 10 electrons each. In such a case, the greater the nuclear charge, the greater is the attraction for electrons and smaller is the ionic radius. Hence, N3– has the highest and Mg2+ has the least ionic size. (iv) Ca2+ < Ar < Cl– < S2–; All of these are isoelectronic. In such cases, the greater the nuclear charge, the greater is the attraction for electrons and smaller is ionic size.

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(v) The first ionization potential of the 3rd period elements follows the order : Na < Al < Mg < Si NOTE : Ionisation energy increases across a period but not regularly. Mg (1s2, 2s2p6, 3s2) is more stable because the electron is to be removed from 3s which is difficult as compared to Al (1s2, 2s2p6, 3s2p1) where electron is to be removed from 3p. (vi) Among oxides, the acidic strength increases with oxidation state. So Na2O2 is least acidic and P2O5 is most acidic. Further Na2O2 and MgO are basic, ZnO is amphoteric and P2O5 is acidic. (vii) O2– > F– > Na+ > Mg2+ NOTE : All the above ions are isoelectronic having 10 electron each. In such case, the greater the nuclear charge, the greater is the attraction for electrons and smaller is the ionic radius. Hence, O2– has the highest and Mg2+ has the least ionic size. 60. C (1s22s2p2) has high nuclear charge than B (1s22s2p1). Thus, carbon has higher IE, value than Boron. Further, for second ionization energy (IE2) in C+ (1s22s2p1) the electron is to be removed from 2p which is easy as compared to B+ (1s22s2), where it has to be removed from 2s.

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1.

(d)

CCl4

=

CH4

= 0 due to symmetrical structure but

CHCl3

0 . So

dipole moment order is : 2.

CHCl3 > CH4 = CCl4 (c) KCl is an ionic compound while others (PH3, O2, B2H6, and H2SO4) are covalent compounds.

3.

(c)

Dipole moment = (Distance between opposite charges) × (charge, q) µ=q×d So, greater the distance between the opposite charges higher the dipole. Due to the resonance the greater charge separation occurs between charges due to linearity.

4.

(N) (a) NO+ = 7 + 8 – 1 = 14 e–. O2 = 16 e– i.e not isoelectronic.

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(b) Boron forms only covalent compounds. This is due to its extremely high ionisation energy. (c) Compounds of Tl+ are much more stable than those of Tl3+. (d) LiAlH4 is a versatile reducing agent in organic synthesis. are 3, 2 and 1.5 respectively. 5. (c) The bond order of N2, O2, and Since higher bond order implies higher bond dissociation energy, hence the correct order will be 6.

(b) The geometry of IF5 is square pyramide with an unsymmetric charge distribution, therefore this molecule is polar.

7.

(a) Hybridisation of S in H2S =

8.

9.

(6 + 2 + 0 – 0) = 4

S has sp3 hybridisation and 2 lone pair of electrons in H2S. It has angular geometry and so it has non-zero value of dipole moment. (d) Critical temperature of water is higher than O2 because H2O molecule has dipole moment which is due to its V-shape. (c) (a)

[Cl]– – It has ionic and non-polar covalent

bond (b) H – C ≡ N - It has ionic and polar covalent bonds. (c) It has polar and non polar both type of covalent bonds.

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(d)

It has non polar covalent bonds only. 10. (c) TIPS/Formulae : (i) Non metallic oxides are more covalent (or less ionic) as compared to metallic oxides. (ii) Higher the polarising power of cation (higher for higher oxidation state of similar size cations) more will be covalent character. (i) P2O5 will be more covalent than other metallic oxides. (ii) Oxidation state of Mn is + 7 in Mn2O7, oxidation state of Cr in CrO3 is + 6 and oxidation state of Mn is + 2 in MnO. MnO is most ionic. NOTE : P2O5, being a non-metallic oxide will definitely be more covalent than the other metallic oxides. Further, we know that higher the polarising power of the cation (higher for higher oxidation state of the similar size cations) more will be the covalent character. Here Mn is in +7 O.S in Mn2O7, Cr in +6 in CrO3 and Mn in +2 in MnO. So, MnO is the most ionic and Mn2O7 is the most covalent. 11. (d) No. of e– in CH3+ = 6 + 3 – 1 = 8 No. of e– in H3O+ = 3 + 8 – 1 = 10 No. of e– in NH3 = 7 + 3 = 10 No. of e– in CH3– = 6 + 3 + 1 = 10 H3O+, NH3 and CH3– are isoelectronic. 12. (b) In N2, similar atoms are linked to each other and thus there is no polarity. 13. (b) TIPS/Formulae : Dipole moment of compound having regular geometry and same type of atoms is zero. It is vector quantity. The zero dipole moment of BF3 is due to its symmetrical (triangular planar) structure. The three fluorine atoms lie at the corners of an equilateral triangle with boron at the centre. NOTE : The vectorial addition of the dipole moments of the three bonds gives a net sum of zero because the resultant of any two dipole

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moments is equal and opposite to the third. The dipole moment of NH3 is 1.46 D indicating its unsymmetrical structure. The dipole moment of CH2Cl2 (the molecule uses sp3 hybridisation but is not symmetric) is 1.57D. 14. (d) In covalent bonds, between two identical non-metal, atoms share the pair of electrons equally between them, e.g. : F2, O2, N2. , covalent 15. (c) Ionic bond or electrovalent between Cu2+ and and coordinate in

;

ion.

16. (b) In regular tetrahedral structure, dipole moment of one bond is cancelled by opposite dipole moment of the other bonds. 17. (b) TIPS/Formulae : (i) Dipole moment is vector quantity. When vector sum of all dipoles in molecule will be zero, then molecule will not have net dipole moment. (ii) NOTE : For net dipole moment to be equal to zero, all the atoms attached to central atom must be identical and geometry must be regular.

Methyl chloride

Methylene chloride

18.

Carbon tetrachloride

Chloroform

Carbon tetrachloride having regular geometry and identical atoms attached to bonds has zero dipole moment. (a) NOTE : Isoelectronic species have same number of electrons. Electrons in CO = 6 + 8 = 14

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Electrons in CN– = 6 + 7 + 1 = 14 Electrons in O2– = 8 + 8 + 1 = 17 Electrons in O2+ = 8 + 8 – 1 = 15 CO and CN– are isoelectronic. 19. (c) NOTE : Dipole moment is vector quantity. In trigonal planar geometry (for sp2 hybridisation), the vector sum of two bond moments is equal and opposite to the dipole moment of third bond. It forms hydrogen bonds with water 20. (b) 21. (c) 22. (a) H2 H–H + – 23. (a) X Y Electropositive elements forms cation and electronegative elements forms anion. after forming the bonds, C has only 6 e– in its valence shell. 24. (d) 25. (c) In KCN, ionic bond is present between K+ and CN– and covalent bonds are present between carbon and nitrogen C ≡ N. 26. (6) Polar and paramagnetic molecules will be attracted/deflected near charged comb. Polar molecules : HF, H2O, NH3, H2O2, CHCl3, C6H5Cl(6-polar molecules) Nonpolar molecules : O2, CCl4, C6H6 27. (4)

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Total No. of molecules containing covalent bond between two atoms of the same kind are 4. 28. (80.09) Dipole moment, µ = e × d coulombs metre For KCl d = 2.6 × 10–10 m For complete separation of unit charge (electronic charge) (e) = 1.602 × 10–19 C Hence µ = 1.602 × 10–19 × 2.6 × 10–10 = 4.1652 × 10–29 Cm µKCl = 3.336 × 10–29 Coulomb meter (given) ∴ % Ionic character of KCl = 29.

× 100

= 80.09% Three centred two electron bonds or banana bond; NOTE : The formation of three centred two electron bond is due to one empty sp3 orbital of one of the B atom, 1s orbital of the bridge hydrogen atom and one of the sp3 (filled) orbital of the other B-atom. This forms a delocalized orbital covering the three nuclei giving the shape of a banana. Thus, also known as banana bonds.

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30.

2; N ≡ N (N2) has 1σ and 2π bonds. (A triple bond consists of 1σ and 2π bonds) 31. False : The C – F distance is less than the C – Cl, although the former involves more charge separation. However, here bond distance has more dominating effect causing dipole moment of CH3Cl to be more than that of CH3F. 32. False : The presence of polar bonds in a polyatomic molecule does not always lead to a definite dipole moment. This is because the dipole moment is a vector quantity and when the bond moment of one bond is cancelled by the equal but opposite bond moment due to other bond(s), the molecule has zero dipole moment, e.g. CO2, CH4, CCl4 etc. 33. False : Symmetrical molecules with polar bonds have zero dipole moment. 34. True : Sigma bond is formed by the overlapping of two s-orbitals or one s and one p or the two p orbitals of the two different atoms. Thus, linear overlap of two p-orbitals results in formation of a σ– bond.

35. 36.

(b, c) Dipole moment (µ) value of BF3, SF6, BeCl2, CO2, BCl3 is Zero. (b, c) Alkanes (a) and (d) don’t have dipole moment because of symmetry in them.

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trans 2-pentene

cis 3-hexene

These alkenes are not symmetrical and so, they have dipole moment. 37. (a) Both assertion and reason are correct. The reason explains the assertion as the central O-atom cannot have more than 8 electrons (octet). 38. (c) LiCl is a covalent compound due to the large size of the anion (Cl–) and small size of Li+ cation. This generates large amount of polarisation in bond. :: 39. O3; O = O → O; COCl2; O = 40.

;

Lewis dot structure

Neutral molecule

O22–;

F2;

CO32– ;

SO3;

CN–;

N2;

NCS–;

CO2;

1.

(d) H2O - 104.5° (sp3 with 2 lone pair at O) NH3 - 107° (sp3 with 1 lone pair at N) CH4 - 109.5° (sp3) H2S - 92° (sp3 with 2 lone pair at O) Lone pair-bond pair repulsion in H2S will increase because 'S' has lower electronegativity than 'O'. So there will be lesser electron density on 'S' and thus H–S–H bond angle will be smaller than H2O. 2. (a) (a) [Ni(CN)4]2– = dsp2 (b) BrF5 = sp3d 2 (c) XeF4 = sp3d2 (d) [CrF6]3– = d2sp2 3. (a) For AB4 compound possible geometry are

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No. of Bond pair No. of lone pair Hybridisation 4 0 sp3 4 1 sp3d 4 2 sp3d2 Structure with sp3d2 hybridisation is polar due to lone pair moment while in other possibilities molecules is non-polar. Square pyramidal can be polar due to lone pair moment as the bond pair moments will get cancelled out. 4. (a) St. No. = Bond pair + Lone Pair (i) = (5 + 2) = 7 So, hybridisation is = sp3d3 and structure is pentagonal planar. (ii) XeO3F2 St. No. = 5 So, hybridisation is = sp3d and structure is trigonal bipyramidal. 5. (b) SF4 Bond pair = 4 Lone pair = 1 Steric number = 5, So, hybridisation is sp3d.

Geometry is trigonal bipyramidal but shape is "See Saw". 6. (d) ICl5 is sp3d2 hybridised (5 bp, 1 lp)

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ICl4– is sp3d2 hybridised (4 bp, 2 lp)

7.

8.

(a) Species ICl2¯

Hybridisation sp3d

ICl¯4

sp3d 2

BrF¯2

sp3d

IF¯ 6

sp3d 3

(c)

∴ Total number of lone pair of electrons is 9. 9. (a) NF3 has trigonal pyramidal geometry. N atom has one lone pair and three bond pairs of electrons. The electron pair geometry is tetrahedral and molecular geometry is trigonal pyramidal. The bond angles are lower than tetrahedral bond angles due to lone pair - lone pair and lone pair - bond pair repulsions. N atom is sp3 hybridised. 10. (d) XeOF2 XeOF4

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11.

12.

(d) (a) BrF5

sp3d2

(b) SF6

sp3d2

(c) [CrF6]3–

sp3d2

(d) PF5

sp3d

Hybridisation (H) = [No. of valence electrons of central atom + No. of monovalent atoms attached to it + (–ve charge if any) – (+ve charge if any)] NO2+= i.e. sp hybridisation NO2–= i.e. sp2 hybridisation NO3–= i.e. sp2 hybridisation The Lewis structure of NO2 shows a bent molecular geometry with trigonal planar electron pair geometry hence the hybridization will be sp2. (c)

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13.

(d)

All molecules have sp3d hybridisation and 2 lone pairs. Hence all have identical (T-shape). 14. (d) The structure of CaC2 is Ca2+ i.e, one π and two σ bonds. 15. (a) The structure of IF6– is distorted octahedral. This is due to presence of a “weak” lone pair.

16.

(d) OSF2 :

. It has 1 lone pair.

(Shape is trigonal pyramidal)

The shapes of SO3, BrF3 and

are triangular planar respectively.

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17.

(d)

18.

(a) NOTE : Isoelectronic species have same number of electrons and isostructural species have same type of hybridisation at central atom. NO3– ; No. of e– = 7 + 8 × 3 + 1 = 32, hybridisation of N in NO3– is sp3 CO32– ; No. of e– = 6 + 8 × 3 + 2 = 32, hybridisation of C in CO32– is sp3 ClO3– ; No. of e– = 17 + 8 × 3 + 1 = 42, hybridisation of Cl in ClO3– is sp3 SO3; No. of e– = 16 + 8 × 3 = 40, hybridisation of S in SO3 is sp2 NO3– and CO32– are isostructural and isoelectronic. 19. (a) H3N → BF3 where both N, B are attaining tetrahedral geomerty. 20.

(b) TIPS/Formulae : H =

[V + M – C + A]

Hybridisation of N in NH3 =

[5 + 3 – 0 + 0] = 4

sp3

Hybridisation of Pt in [PtCl4]2–

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=

[2 + 4 – 0 + 2] = 4

dsp2

Hybridisation of P in PCl5 =

[5 + 5 – 0 + 0] = 5

sp3d

Hybridisation of B in BCl3 =

[3 + 3 – 0 + 0] = 3

21.

(b) For

sp2

:

;

∴ sp hybridisation For

:

;

∴ sp2 hybridisation For

:

;

∴ sp3 hybridisation 22. (d) The structure of species can be predicted on the basis of hybridisation which in turn can be known by knowing the number of hybrid orbitals (H) in that species

For SF4 : S is sp3d hybridised in SF4. Thus SF4 has 5 hybrid orbitals of which only four are used by F, leaving one lone pair of electrons on sulphur. For CF4 :

∴ sp3 hybridisaion

Since, all the four orbitals of carbon are involved in bond formation, no lone pair is present on C having four valence electrons For XeF4 :

, ∴ sp3d2 hybridization of the six

hybrid orbitals, four form bond with F, leaving behind two lone pairs

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of electrons on Xe. 23.

(b) ∴ Boron, in BF3 , is sp2 hybridised leading to trigonal planar shape.

24.

(b) TIPS/Formulae : H =

(i)

CO2; H =

(V + M – C + A)

(4 + 0 – 0 + 0) = 2

sp hybridisation. (ii) SO2; H =

(6 + 0 – 0 + 0) = 3

sp2 hybridisation. (iii) CO; H = 25.

(4 + 0 – 0 + 0) = 2

sp hybridisation. (c) Structure of a molecule can be ascertained by knowing the number of hybrid bonds in the molecule. Thus

In NF3 : Thus, N in NF3 is sp3 hybridized as 4 orbitals are involved in bonding. :

In

Thus, N in

is sp2 hybridized as 3 orbitals are involved in bonding

In BF3 : Thus, B in BF3 is sp2 hybridized and 3 orbitals are involved in bonding In H3O+ : Thus, O in

is sp3 hybridized as 4 orbitals are involved in bonding.

Thus, isostructural pairs are [NF3, H3O+] and [ 26.

(a) H =

, BF3].

(V + M – C + A)

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where H = No. of orbitals involved in hybridisation (viz. 2, 3, 4, 5, 6) and hence, nature of hybridisation (viz. sp2, sp3, sp3d, sp3d2) can be ascertained. V = No. of electrons in valence shell of the central atom, M = No. of monovalent atoms, C = Charge on cation, A = Charge on anion, For

, we have, ⇒H=

(7 + 1) = 4 or sp3 hybridisation as 4 orbitals are involved

27. (c) TIPS/Formulae : 4σ bonds – sp3 hybridisation 2σ and 2π bonds – sp2 hybridisation 1σ and 3π bonds – sp hybridisation [For hybridization only σ-bonds are considered]

(a)

(b)

(c)

(d)

(a) 3σ, 1π (b) 3σ, 1π (c) 4σ (d) 3σ, 1π (CH3)3COH has 4σ bonds and thus it has sp3 hybridisation. 28. (c) From amongst given species PH3, NH3 and SbH3 are all sp3 hybridised. Their central atom has both bond pair as well as lone pair of electrons. The lone pair occupy the fourth orbital. CH3+ has only three pairs of electrons so it is sp2 hybridised. 29.

(c) TIPS/Formulae : H = For SO2, H =

(V + M – C + A)

(6 + 0 + 0 – 0) = 3

sp2 hybridisation.

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30.

(6)

Sum of number of lone pairs = 1 + 2 + 0 + 3 = 6 31. (4) Cl – Be – Cl Hybridization sp Structure linear

Hybridisation sp Structure linear

Hybridisation sp Structure Linear

Hybridisation sp 2 Structure Trigonal planar

Hybridisation sp 3 Structure Angular

Hybridisation sp 3d Structure linear

Hybridisation sp 3d Structure Linear

Hybridisation sp 3d Structure Linear

Hybridisation sp Structure Linear

Only BeCl2, 32.

, N2O and NO2 are linear with sp-hybridisation.

(4)

XeF4 :

Square planar (sp3d2)

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SF4 :

See-saw (sp3d)

SiF4 :

Tetrahedral (sp3)

BF4– :

Tetrahedral (sp3)

BrF4– :

Square planar (sp3d2)

Square planar (dsp2)

[Cu(NH3)4]2+ :

[FeCl4]2– :

Tetrahedral (sp3)

[CoCl4]2– :

Tetrahedral (sp3)

[PtCl4]2– :

Square planar (dsp2)

NOTE : Shape of SF4 No. of sigma bonds = 4 No. of lone pair with central atom = 1

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Thus, the arrangement of electron pair is trigonal bipyramidal. If the electron pair takes an axial position then the shape will be trigonal pyramidal and if it takes an equatorial position then the shape of molecule will be see-saw.

Trigonal pyramidal

See-saw shape

To reduce the lp-bp repulsions, the molecule takes the shape of see-saw. 33. (0) According to VSEPR theory, number of electron pairs around central atom (Br) are 6. (Five are bond pairs and one is lone pair ) Its geometry is octahedral but due to lone pair –bond pair repulsion, the four fluorine atoms at corner are forced towards the upper fluorine atom thus reducing F–Br–F angle from 90° to 84.8°.

34.

Planar;

is a carbocation and such a species has a planar shape

due to sp2 hybridisation. 35. CO2; Bond angle in CH4 is 109°.28’, in H2O it is 104.5° and in CO2 it is 180°. So, it is maximum in case of CO2. 36. False : sp2 hybrid orbitals do not have equal s and p character. They have 33.3% s-character and 66.7% p-character. 37. False : Only two orbitals are used since C in benzene is in sp2 hybridised state. 38. True : SnCl2 has 2 bond pairs and one lone pair of electrons. It is sp2 hybridised and is trigonal planar in shape.

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39.

(b,

c)

Compound

Number of on central atom

lone

pairs

BrF5 → 1 ClF3 → 2 XeF4 → 2 SF4 → 1 + 40. (b, c, d) [O = N = O] ; [N ≡ C – O]–; S = C = S It can be seen from the structure shown above that CS2 being sp hybridized has a linear shape and other two molecules are isoelectronic to CS2, so they are also linear. SnCl2 and SO2 are sp2 hybridised and are not linear. 41. (a, c) CO2, HgCl2 and C2H2 have linear structure (sp hybridization), while SnCl2 is trigonal planar (sp2 hybridisation). NO2 has angular structure (V-shape). 42. First determine the total number of electron pairs around the central atom.

Thus, in XeF4, Xe is sp3d2 hybridised. The structure of the molecule is octahedral and shape is square planer with two lone pair of electrons.

For OSF4 : Thus, the central atom (S) is sp3d hybridised leading to trigonal bipyramidal structure with no lone pair of electrons.

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43.

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PCl5 : sp 3d Trigonal bipyramidal

BrF5 : sp 3d 2 Square pyramidal

NOTE : To decrease lp - bp repulsions, lone pair takes the axial position in BrF5. 44.

In H2S, no. of hybrid orbitals =

Hence, here sulphur is sp3 hybridised, so

or

NOTE : Due to repulsion betwen lp - lp; the geometry of H2S is distorted from tetrahedral to V-shape. In PCl3, no. of hybrid orbitals Hence, here P shows sp3 - hybridisation

or

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Thus, due to repulsion between lp - bp, geometry is distorted from tetrahedral to pyramidal. 45. The structure of OF2 is similar to H2O and involves sp3 hybridization on O atom. The bond angle in F – O – F is not exactly 109º28’, but distorted (103º) due to presence of lone pair of electrons on O as well as F leading to V shape or tetrahedral positions with two positions occupied by lone pair of electrons of the molecule.

Oxidation number of F = –1 ∴ Oxidation number of O = +2

1.

(b) When two H-atoms come closer then initially due to attraction P.E. is –ve, which decreases more as atoms come closer and after reacting to a minimum value as repulsion starts dominating so, P.E. increases then.

2.

(a)

3.

(d) Molecular orbital configuration for NO is

Species Bond order NO+ 3 NO2+ 2.5 NO– 2 NO 2.5 Bond strength is directly proportional to the bond order, so NO– has minimum bond strength. 4.

(a)

=

B.M.

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1.73 = n =1 O2+ = 1s2

*1s2

2s2

*2s2

2p2z

2p2x

*2s2

2p2z

2p2x

= 2p2y *2p1x = *2p0y O2– = 1s2

*1s2

2s2

= 2p2y *2p2x = *2p1y 5.

(b) Total number of electrons in CN– = 6 + 7 + 1 = 14 Molecular orbital distribution

Bond order =

=3

CN– is diamagnetic because all electrons are paired. 6. (a) Configuration of C2 Configuration of C2– Bond order

C2 has s-p mixing and the HOMO is 2px = 2py and LUMO is 2pz. So, the extra electron will occupy bonding molecular orbital and this will lead to increase in bond order. – So, C2 has more bond order than C2. 7. (b) The molecules with no unpaired electrons are diamagnetic. Molecule No. of unpaired electrons NO 1 CO Zero

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O2 2 B2 2 Since CO has no unpaired electron. Hence CO is diamagnetic. 8.

(d) and diamagnetic species has no unpaired electron in their molecular orbitals. No. of unpaired Bond Magnetic electrons order character 0 3 diamagnetic

O2 9.

2

2

paramagnetic

0

1

diamagnetic

1 2 paramagnetic has least bond length and is diamagnetic.

(d)

B.O. = B.O. =

= 2.5 = 2

bond + 0.5

bond

10.

(c) Electronic configuratios of Li2+ and Li2–: Li2+: σ1s2 σ*1s2 σ2s1 Li2–: σ1s2 σ*1s2 σ2s2 σ*2s1 Now, Bond order of Li2+ = Bond order of Li2– = Here, both Li2+ and Li2– have positive bond order, thus both are stable. 11. (d)

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Species No. of e–s

Elec. conf. Bond order

(4 – 1 = 3) (2 + 1 = 3) (2 + 2 = 4) (4 – 2 = 2) Molecule having zero bond order will not be a viable molecule. 12. (a) As in the resonance structure of hydrogen azide, it can be seen that number of N - N bond for bond (I) 2. Hence for bond (I), bond order will be < 2 whereas for bond (II) , number of bond 2. Thus its bond order will be > 2. 13. (d) An antibonding π orbital best describes the given diagram of a molecular orbital. Two orbitals laterally overlap to form π bond. Out of phase combination of these two p-orbitals give π* MO. 14. (d) Total electron Mag. Character NO+ 1 4 diamagnetic CO 1 4 diamagnetic 18 diamagnetic B2 10 paramagnetic 15. (b) CO (14) → No unpaired electron, hence diamagnetic. 16. (b) Assertion is correct but reason is incorrect. Bonding MO shows constructive interference of the combining electron waves. 17. (a) Nitric oxide is paramagnetic in the gaseous state because of the presence of one unpaired electron in its outermost shell. The electronic configuration of NO is σKKσ2s2 σ*2s2 σ2pz 2π2px 2= π2 py 2π*2px 1= π*2py 0

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18. (b) The molecular orbital configuration of the molecules given is Total no. of electrons in NO = 7(N) + 8(O) = 15 Hence E.C. of NO = Due to presence of one unpaired electron NO is paramagnetic. Except NO all are diamagnetic due to absence of unpaired electrons. 19. (c) C2 = σ1s2 σ*1s2 σ2s2 σ* 2s2 σ2p2z π2p1x π2p1y Thus, only C2 will be paramagnetic 20. (c) H2+2 = σ1s0 σ*1s0 Bond order for H2+2 = He2 = σ1s2σ*1s2 Bond order for He2 = So both H2+2 and He2 do not exist. 21. (a, b) The molecular orbital structures of C2 and N2 are

Both N2 and C2 have paired electrons, hence they are diamagnetic. 22. (N) All options are correct, (a) (b) (c) It is a pale blue gas. At – 249.7°, it forms violet black crystals. (d) It is diamagnetic in nature due to absence of unpaired electrons. 23. (b) Li2 = σ1s2 σ*1s2 σ2s2 ∴ Bond order = Li+2= σ1s2 σ*1s2 σ2s1

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B.O. Li2– = σ1s2 σ*1s2σ2s2σ*2s1 B.O. = The bond order of Li+2 and Li–2 is same but Li+2 is more stable than Li–2 because Li+2 is smaller in size and has 2 electrons in antibonding orbitals whereas Li–2has 3 electrons in antibonding orbitals. Hence Li+2 is more stable than Li–.2 24. (N) None of the given option is correct. The molecular orbital configuration of the given molecules is H2 = σ1s2 σ1s0 (no anti-bonding electron) Li2 = σ1s2 σ∗1s2 σ2s2(two anti-bonding electrons) B2 = σ1s2 σ∗1s2 σ2s2 σ∗2s2 (4 anti-bonding electrons) Though the bond order of all the species are same (B.O = 1) but stability is different. This is due to difference in the presence of no. of antibonding electron. Higher the no. of anti-bonding electron lower is the stability hence the correct order is H2 > Li2 > B2 25. (a) For NO Total no. of electrons = 15; B.O = 2.5 Mag. behaviour = paramagnetic For NO+ Total no. of electrons = 14; B.O = 3 Mag. behaviour = diamagnetic 26. (a) Molecular orbital configuration of B2(10) as per the condition will be

σ1s2, σ* 1s2, σ2s2, σ* 2s2, Bond order of B2 = 27.

, B2 will be diamagnetic.

(a) Molecular electronic configuration of

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Therefore, bond order

Bond order 28.

∴ NO– has different bond order from that in CO. (d) (a) In Na2O2, we have ion. Number of valence elctrons of the two oxygen in

ion = 8 × 2 + 2 =18 which are present as

follows σ1s2, σ*1s2, σ2s2, σ*2s2,

, {

=

, {

=

∴ Number of unpaired electrons = 0, hence,

is

diamagnetic. (b) No. of valence electrons of all atoms in O3 = 6 × 3 = 18. Thus, it also does not have any unpaired electron, hence, it is diamagnetic. (c) No. of valence electrons of all atom in N2O = 2 × 5 + 6 = 16. Hence, here also all electrons are paired. So, it is diamagnetic. (d) In KO2, we have No. of valence electrons of all atoms in =2 × 6 + 1 = 13, Thus, it has one unpaired electron, hence it is paramagnetic. 29. (b)

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Bond order (two unpaired electrons in antibonding molecular orbital)

Bond order (One unpaired electron in antibonding molecular orbital) Hence, O2 as well as is paramagnetic, and bond order of

is greater

than that of O2. 30. (c) O2– is the only species having unpaired electron. 31. (b) NOTE THIS STEP : Write configuration of all species. Half filled and full filled orbitals are more stable as compared to nearly half filled and nearly full filled orbitals. Li– = 1s2, 2s2 ; Be– = 1s2, 2s2, 2p1 B– = 1s2, 2s2, 2p2 ; C– = 1s2, 2s2, 2p3 Be– will be least stable. It has lowest I.E. 32. (a) Number of electrons in each species are , Each of the species has 14 electrons which are distributed in MOs as below

Bond order = 33.

(d) KEY CONCEPT

(i)

Bond length ∝

(ii) Bond order is calculated by either the help of molecular orbital theory or by resonance.

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(i)

Bond order of CO as calculated by molecular orbital theory = 3

(ii) Bond order of CO2 (by resonance method) =

=

=2

(iii) Bond order in CO32– (by resonance method) =

= 1.33

Order of bond length of C – O is CO < CO2 < CO32– 34. (b) Calcium carbide is an ionic compound (Ca2+ C2–) which produces acetylene on reacting with water. Thus, the structure of C2– is [C C]2–. It has one and two bonds. [ A triple bond consists of one σ and two π-bonds] 35. (d) O2 = Oxygen (Z = 8) has following molecular orbital configuration of O2. O2 (16e–) = σ 1s2, σ* 1s2, σ 2s2, σ* 2s2, σ2p2x, {π 2p2y = π 2p2z , {π* 2p1y = π* 2p1z i.e., 2 unpaired and 14 paired electrons. 36. (b) H2O molecule can form four hydrogen bonds per molecule, two via lone pairs and two via hydrogen atoms.

37.

(a)

(17e–) – K K σ 2s2 σ* 2s2 σ 2p2x, {π2p2y = π2p2z , {π*2p2y = π* 2p1z Thus, has one unpaired electron; hence it is paramagnetic. Other species have no unpaired electron. All of them have 14 electrons. 38. (a) TIPS/Formulae : Molecule having sp3 hybridisation and one lone pair of electron will have pyramidal structure. (i) CO32– and NO3– have tetrahedron structure.

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(ii) In PCl3, P is sp3 hybridised and has one lone pair of electrons, hence it is pyramidal in shape. 39. (a) TIPS/Formulae : Compound having sp hybridisation will have linear shape. CO2 or (O=C=O) which has C in sp hybrid state has linear shape. 40. (a) NOTE : Compounds having F or O or N attached to H form hydrogen bond. CH3CH2OH C2H5– O –C2H5 CH3CH2Cl ethanol

diethyl ether

ethyl chloride

Trimethyl amine

41.

Ethanol having H attached to O atom will form hydrogen bond. Rest of the compounds do not hydrogen bonds. (c) NOTE : Greater the difference between electro-negativities of two covalently bonded atoms, more will be strength of hydrogen bond. ∴ F – H ..........F bond is strongest due to largest difference in electronegativity of atoms and smallest size of F atom.

42. (a) NOTE THIS STEP : Write the electronic configuration of each species according to molecular orbital theory. NO (15e–) – 1 unpaired electron. CO (14e–) – no unpaired electron CN– (14e–) – O2 (16e–) – σ 1s2, σ* 1s2, σ 2s2, σ* 2s2, σ2p2x,{π 2p2y = π 2p2z , {π* 2p1y = π* 2p1z ; Two unpaired electrons.

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43.

(b) TIPS/Formulae : sp type of hybridization involves the intermixing of one s and one p (say px) orbitals to give two equivalent hybrid orbitals, known as sp hybrid orbitals. The two sp hybrid orbitals are directed diagonally, i.e., in a straight line with an angle of 180º (collinear orbitals). The other two p orbitals (say py and pz) remain pure.

+

44.

(d) TIPS/Formulae : Hydrogen bonding is formed in those compounds in which F or O or N atoms are attached to hydrogen atom. HCl does not have F or N or O It does not form hydrogen bond. 45. (a) CO2 having sp hybridation has linear shape. 46. (6) (H2, Cl2, Be2, C2, N2, F2) H2 : σ1s2 (Diamagnetic) : σ1s2, σ*1s1 (Paramagnetic) Li2 : σ1s2, σ*1s2, σ2s2 Be2 : σ1s2, σ*1s2, σ2s2, σ*2s2 B2

: σ1s2, σ*1s2, σ2s2, σ*2s2,

(Diamagnetic) (Diamagnetic) = (Paramagnetic)

C2

: σ1s2, σ*1s2, σ2s2, σ*2s2,

= (Diamagnetic)

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N2 : σ1s2, σ*1s2, σ2s2, σ*2s2,

= (Diamagnetic) ,

: σ1s2, σ*1s2, σ2s2, σ*2s2, = F2

(Paramagnetic)

: σ1s2, σ*1s2, σ2s2, σ*2s2,

,

=

(Diamagnetic)

47. Increases, decreases; ∴ Bond order in N2 = 3 and Bond order in Thus, conversion of N2 to

decreases bond order (from 3 to 2.5) and

hence, increases the N – N bond distance. [ Bond distance increases with decrease in B.O.) = 2.5 Bond order in O2 = 2 and Bond order in Thus, conversion of O2 to

increases bond order (from 2 to 2.5) hence,

decrease in O – O bond distance. 48.

sp3; Hybridisation (H)=

[No. of valence electron in central atom +

No. of monovalent atoms – Charge on cation + Charge on anion] For N in NH4+, hybridisation (H) =

(5 + 4 – 1 + 0) = 4

sp3 hybridisation. 49.

and

form strongest hydrogen bonds

because of largest difference in electronegativities of bonded atoms. 50. (a, c) (a) The molecular orbital energy configuration of C2–2 is In the MO of C2–2, there is no unpaired electron hence, it is diamagnetic.

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(b) Bond order of O22+ is 3 and O2 is 2 therefore bond length of O2 is greater than O22+ (c) The molecular orbital energy configuration of N2+ is Bond order of N+2= –

The molecular orbital energy configuration of N2 is Bond order of N–2= (d) He+2 has less energy in comparison to two isolated He atoms because some energy is released during the formation of He2+ from 2 He atoms. 51. (a, c) The outer most shells of C, N & O has 4, 5 and 6 electrons respectively. Thus, CN– and NO+ each has 10 electrons to accommodate in the molecular orbitals. So, their bond order is same. has 13 and CN+ has 12 electrons in outermost orbits. 52.

(c)

In the given diagrams, +ve phase is shown by darkening the lobes and –ve phase is by white lobes. When two same phase overlap, it forms bonding molecular orbital otherwise antibonding molecular orbital. Also axial overlap produces σ-bond and sideways overlap produces πbond. For example : σ-antibonding orbital π -bonding orbital

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P.

d – d (σ bonding)

Q.

p – d ( π bonding)

R.

p – d (π antibonding)

S. 53. 54.

d – d (σ antibonding)

(A) –p, r, t ; (B) – s,t ; (C) – p, q, r ; (D) – p, r, s MO configuration of O2 : ,

Bond order = Since, O2 molecule has two unpaired electrons, it is paramagnetic. 55. (i) Increasing order of bond dissociation energy. F2 < Cl2 < O2 < N2 NOTE : Fluorine-fluorine bond energy is less than the Cl–Cl because of larger repulsion between the non-bonded electrons of the two smaller fluorine atoms (chlorine atoms are larger in size; hence their lone pair of electrons exert less repulsion than fluorine). Oxygen having two pairs of lone pair of electrons on each atom exert less repulsion than that of chlorine or fluorine each having three lone pairs of electrons. Nitrogen having only one lone pair of electrons exert minimum repulsion, hence it is the most stable. (ii) H-bonding is an electrostatic attractive force between covalently bonded hydrogen atom of one molecule and an

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electronegative atom (F, O, N). Further, higher the electronegativity and smaller the size of the atom, the stronger is the hydrogen bond. NOTE : Although Cl has the same electronegativity as nitrogen, it does not form effective hydrogen bonds. This is because of its larger size than that of N with the result its electrostatic attractions are weak. Similarly, sulphur forms a very weak hydrogen bond due to its low electronegativity, although oxygen present in the same group forms a strong hydrogen bond. Hence, the order is S < Cl < N < O < F (iii) In KO2, O2 is present as O2–, while in , O2 is present as O2+. Write down the MO configuration of O2, O2– and O2+. O2 : σ 1s2 , σ∗ 1s2 ,σ 2s2 ,σ∗ 2s2 ,σ 2p2x, {π 2py2,= π 2p2z, {π∗ 2py1 = π∗ 2p1z. Thus, the bond order = 2 O2– : Same as above except π∗ 2p2y, π∗ 2p1z in place of

.

Thus, the bond order in O2– = 1.5 O2+ : Same as in O2 except π∗ 2py1 = π∗ 2p0z in place of

.

∴ Bond order in O2+ = 2.5 ∴ Bond order in the three species is O2+ > O2 > O2– or O2[AsF4] > O2 > KO2 Bond length order : KO2 > O2 > O2 [AsF4] 56. Benzene has 12σ and 3π bonds. 57. H2O molecules are held together by hydrogen bonding which is stronger force of attraction but H2S molecules are held together by van der Waals forces of attraction, which are weaker forces. As a result, water molecules come closer and exist in liquid state.

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1.

(a)

As n, T and V constant so So, 2.

(b) For ideal gas PV = nRT

; PM = dRT ; ⇒d∝

;d∝P

So, graph between d Vs T is not straight line. 3. (a) Ethyl acetate is polar molecule so dipole-dipole interaction and London dispersion will be present in its liquid state. 4. (d) Among given intermolecular forces, ionic interactions are stronger as compared to van der Waal interaction Thus, correct order is ion-ion > ion-dipole > dipole-dipole 5. (d) Ideal gas equation: PV = nRT After putting the values, we get 200 × 10 = (0.5 + x) × R × 1000 (total no. of moles = 0.5 + x) 2 – 0.5 R = xR;

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∴ x= 6.

(b) At 27°C or 300 K number of moles of an ideal gas = n1 At T2 K number of moles of the ideal gas = n2

Number of moles escaped = n2 = n1 –

=

PV = nRT (Ideal gas equation) At constant volume and pressure. n T2 =

; n1T1 = n2T2 T1; T2

7.

(c) From ideal gas equation PV = nRT where n = m/M So, PV = mRT/M P = mRT/MV P = dRT/M At constant temperature and pressure, d ∝ M ∴ d1/d2 = M1/M2 (Here d1 and M1 are density and molecular mass of ammonia whereas d2 and M2 are density and molecular mass of hydrogen chloride) d1/d2 = 17/36.5 : d1/d2 = 0.46 8.

(c) Density (ρ) =

(1 bar = 0.987 atm)

Let the molar mass of gas be x

Given ρgas =

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∴ x = 112 g/mol. 9. (a) For a given mass of an ideal gas, the volume and amount (moles) of the gas are directly proportional, if the temperature and pressure are constant. i.e V µ n Hence in the given case. Initial moles and final moles are equal (nT)i = (nT)f

; 10.

(b) Hydrogen bond is a type of strong electrostatic dipole-dipole interaction and dependent on the inverse cube of distance between the molecules. 11. (c) At high pressure and low temperature, gaseous atoms or molecules get closer to each other and van der Waal forces operates. So molecules or atoms start attracting each other. Hence a gas deviate the most from its ideal behaviour. While in ideal behaviour we consider that gases do not attract each, i.e., there is no intermolecular forces of attraction. 12. (b) According to Graham’s Law Diffusion: or Since rate of diffusion = ∴ or

=

=

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V2 = 14.1 13. (b) TIPS/Formulae : Use Grahams’ law of diffusion = 14.

(c) For positive deviation: PV = nRT + nPb

Thus, the factor nPb is responsible for increasing the PV value, above ideal value. b is actually the effective volume of molecule. So, it is the finite size of molecules that leads to the origin of b and hence, positive deviation at high pressure. 15. (c) TIPS/Formulae : Find the volume by either V = RT/P (PV = RT) or P1V1 = P2V2 and and match it with the values given in graph to find correct answer. Volume of 1 mole of an ideal gas at 273 K and 1 atm is 22.4 L and that at 373 K and 1 atm pressure is calculated as ;

16.

(c) Mass of 1 L of vapour = volume × density = 1000 × 0.0006 = 0.6 g V of liquid water =

=

= 0.6 cm3

17.

(c) For an ideal-gas behaviour, the molecules of a gas should be far apart. The factors favouring this condition are high temperature and low pressure.

18.

(b) Under identical conditions,

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As rate of diffusion is also inversely proportional to time, we will have, [M1 = 2 and t1 = 5] ;

(a) Thus, For He, (b) For O2, (c) For CO, (d)

For CO2,

19.

(b)

;

N2O4 (g) At start

At equilibrium

100/92 mol = 1.08 mol 80/92 mol = 0.86 mol

2 NO2(g) 0 20/46 mol =0.43 mol

According to ideal gas equation, at two conditions At 300 K; P0V = n0RT0 1 × V = 1.08 ×R × 300 …(i) At 600 K; P1V = n1RT1 P1× V = (0.86 +0.43) × R × 600 …(ii) Divide (ii) by (i), ; 20.

21.

(a) Due to increase in the temperature, the kinetic energy of the gas molecules increases resulting in an increase in average molecular speed. The molecules are bombarded to the walls of the container with a greater velocity resulting in an increase in pressure. (b) TIPS/Formulae : d=

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It means density of gas is directly proportional to pressure and inversely proportional to temperature. Density of neon will be maximum at highest pressure and lowest temperature. (b) is correct answer. =2=

=

22.

(a)

23.

(b) TIPS/Formulae :

,

or

Mx= 64

Rate of diffusion ∝ Molecular mass of HCl > molecular mass of NH3 HCl diffuses at slower rate and white ammonium chloride is first formed near HCl bottle. =

=

24.

(d)

25.

(b) Pressure exerted by hydrogen will be proportional to its mole fraction.

Mole fraction of H2 = 26.

=

(a) TIPS/Formulae : Mole fraction of O2 =

Partial pressure of O2 ∝ Mole fraction of O2 Mole fraction of O2 = 27. 28.

=

(b) The temperature at which a real gas behaves like an ideal gas is called Boyle’s temperature or Boyle’s point. (2.22) P1 = 5 bar P2 = 1 bar

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V1 = 1 m3 T1 = 400 K

V2 = 3 m3 T2 = 300 K

and n1 =

n2 =

Let the new volume of compartment A and B respectively be (1 + x) and (3 – x). = ⇒ ⇒

= =

⇒ 5(3 – x) = 4(1 + x) ⇒x= VA = 1 + x = 1 + 29.

=

= 2.22

(4) Diffusion coefficient ∝ λµ

Since ∴ Diffusion coefficient

Thus 30.

or (7) [Unknown compound is not following the ideal gas behaviour.] PHe = 1 – 0.68 = 0.32 atm, n = 0.1

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V=

=

=7L

31. (750) At constant temperature and number of moles

32.

(–5246.49) Potential energy of H-atom is taken as zero when electron and nucleus are at infinite distance. ∴ P.E. of a H-atom with electron in its ground state = –27.2 eV (from Bohr’s Model) At internuclear distance ‘d0’ electron-electron repulsion and nucluesnucleus repulsion are absent. P.E. of two H-atom = –2 × 27.2 eV = –54.4 eV kJ/mol

=

= –5242.42 kJ/mol PCl3 + Cl2 33. (4.53) PCl5 Initial moles 1 0 0 Moles at eq. 1–0.4 0.4 0.4 Total moles at equilibrium = 1 – 0.4 + 0.4 + 0.4 = 1.4 =1+

Also or

= 1.4

= 1.4

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Exp. mol. wt. of PCl5 or m. wt. of mixture = Now using, PV = =

d= 34.

RT for mixture =

= 4.53 g/litre +

(0.4)

Initial pressure 600 0 0 Final pressure 600-P 2P P/2 P moles when V and T are constant. (where moles equivalent to pressure P are decomposed) Total pressure = 600 – P + 2 P + P/2 = 960 mm of Hg P = 240 mm Hg Thus, moles of N2O5 decomposed 35.

(2.197) Since the pressures of gases are different, and the temperature is constant, the rate at which molecules of the two gases diffuse is directly proportional to the pressure. This rate of diffusion is also directly proportional to the distance travelled by the gas. Hence,

r1 (of HCl gas) at pressure P = 60 = and r2 (of NH3) at 1 atm. pressure P = 40 =

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

From (i) and (ii) = P= 36.

= ×

× = 2.197 atm

(85.2) Let NH3 diffuse through = x cm

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HCl diffuses through = y cm

According to Graham’s law of diffusion =

=

= 1.465

x = 1.465 y ... (1) x + y = 200 cm ... (2) From these equations; y = 85.2 cm Distance between P and X = y = 85.2 cm. 37. (3.42) TIPS/Formulae : RT ⇒ P =

PV =

=d

∴d=

Substituting the value, we get d= 38.

= 3.42 g/litre

(0.25)

Using ideal gas equation PV = nRT 39.

Inversely, time;

.

40.

R.

41.

False : An ideal gas cannot be liquefied as there exists no intermolecular attraction between molecules.

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42.

(b)

For gas A, a = 0, Z =

implies Z varies linearly with

pressure. . Hence, Z does not vary linearly with

For gas B, b = 0, Z =

pressure. Given the intersection data for gas C, it is possible to find the values of ‘a’ and ‘b’. All van der Waal gases, like gas C, give positive slope at high pressures. 43. (c) According to Graham’s law of diffusion for two gases undergoing diffusion at different pressures through same hole

44.

(d)

Pressure exerted by H2 is proportional to its mole fraction.

Mole fraction of H2 =

45.

=

=

(a, b)At constant temp., when gas expands, the K.E. of the molecules remains the same, but the pressure decreases.

46.

(b) No work is required to tear apart the molecules due to the absence of attractive forces in an ideal gas.

47.

(d) (I) Ion-ion interaction energy (II) Dipole-dipole interaction energy (III) London dispersion

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48.

(c) According to Graham’s law of diffusion, if all conditions are identical, r=

As in this question, all conditions are identical for X and Y, then

⇒ d = 48 – 2d ⇒ 3d = 48 ⇒ d = 16 cm 49. (d) The general formula of mean free path (λ) is λ= (d = diameter of molecule, p = pressure inside the vessel) Since d and p are same for both gases, ideally their λ are same. Hence, it must be the higher drift speed of X due to which it is facing more collisions per second with the inert gas in comparison to gas Y. Hence, X faces more resistance from inert gas than Y and hence, covers lesser distance than that predicted by Graham’s law. 50. (a) d = 0.36 kg m–3 = 0.36 g/L (i) From Graham’s Law of diffusion

where Mv = MW of the vapour (ii) Thus, occupies 1 L volume, so 1 mol occupies

=

50.25L

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Thus, molar volume of vapour = 50.25 L (iii) Assuming ideal behaviour, the volume of the vapour can be calculated by

Compressibility factor (Z) = (iv) Z is greater than unity, hence it is the short range repulsive force that would dominate. ( actual density is less than given density) (b)

( 51.

K, Boltzmann constant = R/N)

We know that or

or

M = 252 .

Thus, compound of xenon with fluorine is XeF6 . 52. (I) Given P = 1 atm, w = 12 g; T = (t + 273)K; V = V litre (II) If T = t + 10 + 273 = t + 283 K ; V = V litre,

Using gas equation, Case I.

... (1)

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Case II.

...(2)

From (1) and (2), t = –173ºC or t = 100 K Also from (1), on substituting t and m (120), V = 0.82 litre 53. Weight of liquid = 148 – 50 = 98 g Volume of liquid =

= 100 mL = volume of vessel

It means, vessel of 100 mL contains ideal gas at 760 mm Hg at 300 K. Weight of gas = 50.5 – 50 = 0.5g using, PV = nRT

Molecular weight of gas (m) = 123 54. Let the volume of ethane in mixture = x litre ∴ Volume of ethene = (40 – x) litre Combustion reactions of ethane and ethene are : (i) or (ii) Volume of O2 required for complete combustion of ethane

[For x

litres] Volume of O2 required for complete combustion of ethene = (40 – x) × 3 [For (40 – x) L] ∴ Total volume of O2 required Calculation of number of moles (n) P = 1 atm,

; R = 0.082 L atm K–1 mol–1; T = 400

K

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Since Mass of n moles of O2

= 130

or 8528 = 32 x + 240 × 32

32x = 848

or Hence,

mole

fraction

(%)

of

ethane

Mole fraction (%) of ethene = 33.75% 55. Mixture Krypton rmix = 1.16 rKr = 1 Mmix = ? MKr = 84 We know that or or Determination of the composition of the equilibrium mixture: Let the fraction of Cl2 molecules dissociated at equlibrium = x 2Cl Total Cl2 Initially 1 0 1 At equilibrium 1 – x 2x 1– x + 2x = 1 + x = 1+ x ∴ ∴

x = 0.137 = 13.7%.

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56. TIPS/Formulae : (i) He does not react with oxygen. (ii) KOH absorbs only CO2. NOTE : When the mixture of CO, CH4 and He gases(20 mL) are exploded by an electric discharge with excess of O2, He gas remains as such and the other reactions involved are : ...(i) ...(ii) Let the volumes of CO and CH4 to be ‘a’ mL and ‘b’ mL in the mixture then Volume of He gas = [20 – (a + b)] mL For the initial contraction of 13 mL, Volume of left hand side in the above reactions – 13 = Volume of right hand side.

= [20 – (a + b)] + a + b [neglect the volume of H2O (l)] (Since for gases, volume α no. of moles) ∴

a + 2 b = 13 or

a + 4b = 26

...(iv)

NOTE THIS STEP : The CO2 produced above in reactions (ii) & (iii), (a + b) mL, reacts with KOH sol for a further contraction of 14 mL. CO2(g) + 2KOH(l) → K2 CO3(l) + H2O(l) (a + b) ∴ a + b = 14 ...(v) Solving (iv) & (v) we get, a = 10 mL and b = 4 mL , and He = 100 – ( 20 + 50) = 30% 57. TIPS/Formulae : Partial pressure = Mole fraction × Total pressure

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∴

pHe

=

xHe

×

P

=

×

20

=

16

bar

∴

= 20 – 16 = 4 bar

Now applying the formula

∴ ∴ Composition of the mixture (He : CH4) effusing out= 8 : 1 58. Calculation of volume of gas : Weight of cylinder with gas = 29.0 kg Weight of empty cylinder = 14.8 kg ∴ Weight of gas in the cylinder = 14.2 kg Pressure in cylinder = 2.5 atm ∴ No. of moles (n) in 14.2 kg (14.2 × 103 g) of butane =

n=

= 244.83 mol

Applying gas equation, V=

=

= 2412 litres

[27°C = 273 + 27 = 300] Calculation of pressure in cylinder after use. Weight of cylinder after use = 23.2 kg Weight of empty cylinder = 14.8 kg ∴ Wt. of unused gas = 8.4 kg Thus P =

=

moles of butane = 1.478 atm [V = 2412 L]

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Calculation of volume of used gas at 2.5 atm and 27ºC. Weight of used gas = 14.2 – 8.4 = 5.8 kg Pressure under normal usage conditions = 1 atm =

V=

= 2463 litres = 2.463 m3 59. Using gas equation; PV = nRT Total no. of moles of gases in the mixture (n) =

=

= 0.7308 mol.

Thus, no. of moles of unknown gas = 0.7308 – 0.7 = 0.0308 mol. Now we know that = Also we know that ∴ M2 =

= = M1 or M2 =

× 2 = 1033

60. 2NO + O2 → 2NO2 → N2O4 Calculating the number of moles of NO and O2 by applying the formula, n = Moles of NO in the larger flask =

= 0.0107

[250 mL = 0.250 L] Moles of O2 in the smaller flask =

= 0.0032

[100 mL = 0.100 L] The reaction takes place as follows. N2O4 2NO + O2

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Mols before 0.0107 0.0032 0 reaction Mole after (0.0107 – 0 0.0032 reaction 2 × .0032) Moles of NO left = 0.0107 – 0.0064 = 0.0043 NOTE : Oxygen will be completely changed into NO2 which in turn is completely converted into N2O4 which solidifies at 262 K. Hence at 220 K, the dimer is in the solid state and only NO present in excess will remain in the gaseous state occupying volume equal to 350 mL. Hence, pressure (P) of NO gas left =

=

= 0.221 atm

[Total volume = 0.250 + 0.100 = 0.350 L] 61. Applying the general gas equation RT

PV = nRT =

Here, Mol. wt. of acetylene i.e., C2H2 (M) = 26, P =

atm,

T = 50°C = 50 + 273 = 323 K ∴ 62.

or V =

V=

= 5.23 L

Volume of balloon

Let no. of balloons to be filled = n ∴ Total volume occupied by n balloons = 4.851× n Volume of H2 present in cylinder = 2.82 L (given) ∴ Total volume of H2 at NTP = (4.851n + 2.82)L P1 = 1 atm P2 = 20 atm V1 = 4.85 × n + 2.82 L V2 = 2.82 L T1 = 273 K T2 = 300 K or

=

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∴n= NOTE : In equilibrium with balloon, cylinder will hold 2.82 L of gas (it will not empty completely). 63. NOTE THIS STEP : First we should calculate the number of moles of the gas under the given conditions by the relation PV = nRT Here P = 7.6 × 10–10 mm Hg atm. = 1 × 10–12 atm.

=

V = 1 litre, T = 273 + 0 = 273K, Putting the values in equation n=

=

R = 0.082 litre atm./K/mol

moles

Now since 1 mole = 6.023 × 1023 molecules moles =

molecules

= 2.7 × 1010 molecules 64. From ideal gas equation, PV = nRT ⇒ PV =

RT or M = m

Let the molecular wt. of A and B be MA and MB respectively. Then MA = 2 ∴

=

; MB = =

=

Therefore, the ratio MA : MB = 1 : 3 65. Following reaction takes places in tube 2CO C + CO2 Volume of mixture of CO and CO2 = 1L Let volume of CO2 in mixture = x Volume of CO in mixture = 2x Original volume of CO in mixture = 1 – x

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Total volume of gases after reaction = (1 – x) + 2x = 1 + x 1 + x = 1.6 ( It is given total volume after reaction = 1.6L) x = 0.6 L Volume of CO2 = 0.6 L Volume of CO = 0.4 L CO2 : CO = 3 : 2 66.

Given,

mass of gas = 3.7g, mass of hydrogen = 0.184g T1 = 298K, T2 = 17°C = 273 + 17 = 290K =

Moles of H2 = n1 =

= 0.092

=

Moles of gas = n2 =

For hydrogen P1V1 = n1RT1 .........(i) For gas P1V1 = n2RT2 .........(ii) ( Pressure and volume of gas are same) ∴ From equation (i) and equation (ii) or 1 = or n2 = or

1.

or = 0.0945 ∴ M =

= 39.15

(d) Vrms > Vaverage > Vmps >

2.

=

>

(b) Graph shows symmetrical distribution of speed therefore, the most probable and the average speed should be same in square of speeds. Higher speed will give more contribution, therefore the root mean square speed must be greater than the average speed.

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3.

(a)

From curve,

∴ 4.

(under given condition)

(a)

After dissociation,

5.

(c) At high pressure real gas particles are easily compressed.

6.

(b) Number of moles of

Number of moles of ∴ Ratio 7.

(d) Vrms (O2) = Vrms(He)

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or ∴ 8.

(c) Most probable speed (C*) =

Average Speed Root mean square velocity (C) =

9.

(d) r.m.s. velocity Vr m s =

i.e.,

∴ T2 = 4T1 10. (c) Average KE = E = ∴ or 11.

(d) URMS

=

Using

ideal gas equation, PV = nRT = ∴ URMS =

RT;

where d is the density of the gas

at constant pressure,

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12.

(c) TIPS/Formulae : ⇒

Urms =

; or

13.

(c) The expression of root mean square speed is

Hence,

14.

(a) The mean free path, λ =

or λ α

, where a = molecular diameter

∴ Smaller the molecular diameter, longer the mean free path. Hence, H2 is the answer. 15.

(d) Pressure exerted by the gas, P =

...(1)

Here, u = root mean square velocity m = mass of a molecule, n = No. of molecules of the gas Hence, (a) & (b) are clearly wrong. [explained from (1)]

Again u2 =

Here, M = Molecular wt. of the gas; Hence, (c) is wrong Further, Average K.E. = 16.

(a) Uav =

KT; Hence, (d) is true.

;

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= or

or

=

or

=

= 0.6 m/sec.

17.

(b) Average kinetic energy depends only on temperature and does not depend upon the nature of the gas.( K.E. = 3/2 KT)

18.

(a) Urms : Uav =

19.

(4)

:

or

:

= 1.086 : 1

;

Given vrms = vmp ⇒

=

⇒ 20.

(434) TIPS/Formulae :

= 1.085 × Cav = 1.085 × 400 = 434 ms–1 21.

1 : 16;

22.

900; Energy of one mole of an ideal monoatomic gas

=

RT R = 8.314 J K–1 mole–1 = 1.99 cal K–1 mole–1 T = 27°C = 273 + 27 = 300 K ⇒ E=

× 1.99 × 300 = 900 cal

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23.

False : K.E. =

24.

False : The pressure exerted by the gas is the result of collisions of the molecules on the walls of the container. (a, b, c) According to kinetic theory of gases, all gases at a given temperature have same average kinetic energy.

25.

;

KT, and cannot be zero at 0ºC or 273 K.

(absolute temp)

Root mean square velocity is directly proportional to square root of absolute temperature and inversely proportional to square root of molecular weight of the gas. (absolute temp) : 26. (a) (b) (c) (d) 27.

(a, b, c, d) According to kinetic theory of gases : Collision between the molecules as well as with the walls of the container is perfectly elastic in nature. Momentum is defined as m. . Hence, heavier molecules transfer more momentum to the walls of the container. According to Maxwell-Boltzmann distribution of molecular speed, very few molecules have either very high or very low speeds. A gas molecule moves in a straight line unless it collides with another molecule or walls of the container with constant velocities. (a) At constant volume. (from PV = nRT) ∴

...... (i)

. Collision frequency is directly proportional to Thus collision frequency Hence, on increasing the collosion pressure, increases frequency. 28. Given V= 1L = 10–3 m3, P = 7.57 × 10–3 Nm–2, R = 8.314J, n = 2 × 1021/6.023×1023moles

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PV = nRT or = Urms =

= 274.13 K =

m/s = 494.15 m/s

(given) 29.

∴ Ump = 0.82 × Urms = 0.82 × 494.15 = 405.2 m/sec TIPS/Formulae : Average velocity =

and Most probable velocity = Given -For CO2 Average velocity at T1 = Most probable velocity at T2 = 9 × 104 cm/sec =

m/sec.

= 9 × 102 m/sec. ∴ 9 × 102 =

...(A) [Average velocity at T1K]

and 9 × 102 =

...(B)

[Most probable velocity at T2K] On solving, T1 = 1682.5 K, T2 = 2143.4 K 30. Given T = 20°C = 20 + 273 = 293K R = 8.314 × 107 erg per degree per mol

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M (of O3) = 48 = 3.9 × 104 cm sec–1 31. TIPS/Formulae : Total kinetic energy = n (3/2 RT) where n = Number of moles of the gas R = Gas constant T = Absolute temperature Molecular weight of methane, CH4 = 12 + 4 × 1 = 16 ∴ Number of moles of methane in 8.0 g of methane = 0.5

=

R = 8.314 joules/K/mole, T = 27 + 273 = 300 K ∴ Total kinetic energy of the molecules in 8.0 g of methane at 27ºC = n × 3/2 RT = 0.5 × 3/2 × 8.314 × 300 = 1870.65 joules. ∴ Average kinetic energy = = 6.21 × 10–21 joules/molecule

1.

(a) If values of ‘b’ for two gases are same but values of ‘a’ are different, then the gas having a larger value of ‘a’ will occupy lesser volume. Since, it, will have larger force of attraction and, therefore, lesser distance between its moleucles. If values of ‘a’ for two gases are same but values of ‘b’, are different then the smaller value of ‘b’ will occupy lesser volume and, therefore, will be more compressible.

2.

(a) Critical temperature = Ar

= 0.4; Ne

= 0.12

Kr

= 5.1; Xe

= 0.82

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Value of

is highest for Kr. Therefore, Kr has highest value of critical

temperature. 3.

(a) P = PV – Pb = RT

⇒

Z=1+

[Where Z (compressibility factor) = PV/RT ] Slope of Z vs P curve (straight line) =

4.

∴ Higher the value of b, more steep will be the curve. Constant ‘b’ value depends on size of atoms or molecules. (a) According to van der Waals equation for one mole of gas

At very high pressure So,

is negligible.

P (V – b) = RT ; PV – Pb = RT on dividing RT on both sides. compressibility factor. 5. •

•

(d) A solution of CH3OH and water shows positive deviation from Raoult's law, it means by adding CH3OH intermolecular force of attraction decreases and hence surface tension decreases. By adding KCl in water, intermolecular force of attraction bit increases, so surface tension increases by small value.

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•

By adding surfactant like CH3(CH2)11OSO3–Na+, surface tension decreases rapidly and after forming micelle it slightly increases.

6.

(b) Compressibility factor

(For one mole of real gas) van der Waals equation At low pressure, volume is very large and hence correction term b can be neglected in comparison to very large volume of V. i.e. ; ; Hence, 7.

(a) Given

Which can also be written as

At low pressure and high temperature the effect of

and b is negligible

hence PV = nRT.

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8.

(b)

PV + a/V = RT; PV = RT – a(V) y = RT – a(x) So, slope = a =

=

= 1.5

9.

(b) Correction factor for attractive force for n moles of real gas is given by the term mentioned in (b). 10. (b) Upon increase of temperature, the internal energy of water or any system increases resulting in decrease in intermolecular force and hence, decrease in surface tension. Surface tension decreases with increase in mobility due to increase in temperature. 11. (b) (PV)Observed / (PV)Ideal < 1 Vobs < Videal, Vobs < 22.4 litre. 12. (b) The compressibility factor of a gas is defined as

For an ideal gas, pVm = RT. Hence Z = 1 13. (c) NOTE : The value of ‘a’ indicates the magnitude of attractive forces between gas molecules. Value of ‘a’ ∝ size of molecule. inert gas will have minimum value of ‘a’ followed by H2O, C6H6 and C6H5CH3

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14.

(c) ‘a’ is directly related to forces of attraction. Hence greater the value of ‘a’, more easily the gas is liquified.

15.

(c)

16.

intermolecular forces. (0.99) van der Waals’ equation for one mole of a gas is

(V – b) = RT; Here

represents the

...(1) Given that volume occupied by CO2 molecules, ‘b’ = 0 Hence, (1) becomes

or

Using R = 0.082, T = 273K, V = 22.4 L for 1 mole of an ideal gas at 1 atm pressure.

17. 18.

False : The constant ‘a’ reflects the intermolecular attraction between gaseous molecules. The constant ‘b’ reflects the actual volume of one mole of gaseous molecules. (c) P(V–b) = RT ⇒ PV – Pb = RT ⇒ ⇒

Hence Z > 1 at all pressures. This means , repulsive tendencies will be dominant when interatomic distance are small. This means, interatomic potential is never negative but becomes positive at small interatomic distances. 19. (a, c) Van der Waals equation is [For n moles of a gas)

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a, b are van der Waals constants. The ideal gas equation is PV = nRT [For n moles of a gas] where P is pressure excerted by ideal gas and V is volume occupied by ideal gas. In van der Waals equation the term

represents the pressure

exerted by the gas and (V– nb) the volume occupied by the gas. At low pressure, when the gas occoupies large volume the intermolecular distance between gaseous moleculas is quite large and in such case there is no significant role played by intermolecular forces and thus the gas behaves like an ideal gas thus (a) is correct NOTE : Under high pressure the intermolecular distance decreases and the intermolecular forces play a significant role and the gas shows a devation from ideal behaviour. Thus (b) is not correct. a, b i.e. the van der Waals coefficients defined on the nature of gas and are independent of temperature so (c) is correct. The pressure

is not lower than P so, (d) is not correct.

Hence, the correct answer is (a, c). 20. (A) : (p) and (s) Because 200 atm pressure is very large. For H2 gas, at very high pressure Z > 1. (B) : (r) Since P ~ 0, it means very low pressure, so ideal behaviour is observed. (C) : (p) and (q) Since P is 1 atm, Z for CO2 would be less than 1 and attractive forces will be dominant. (D) : (r) In real gas with very high molar volume, molecules will be very far apart from each other due to which van der Waal's forces as well as actual volume occupied by molecules will be negligible. NOTE : When Z > 1, the force of attraction between the molecules are very feeble. 21. (a) 'a' indicates the magnitude of the attractive forces among the gas molecules, which increases in NH3 due to H-bonding.

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22.

The van der Waal equation (for one mole) of a real gas is ; ...(i)

NOTE THIS STEP : To calculate the intercept P → 0, hence Vm → ∞ due to which the last two terms on the right side of the equation (i) can be neglected. ∴ PVm = RT + Pb, comparing this with y = mx + c intercept = RT 23.

We know that, Compressibility factor , ∴ NOTE : Further when volume of a gas molecule is negligible, van der Waal’s equation becomes

or

or

Substituting the values a = (0.082 × 0.1119 × 273) – (100 × 0.1119 × 0.1119) = 1.253 atm L2 mol–2 24. van der Waals equation for n moles of gas is

Given V = 4 litre; P = 11.0 atm, T = 300 K; b = 0.05 litre mole–1, n =2 Thus, a = 6.46 atm litre2 mol–2

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1.

(a) A system at higher temperature has (randomness). S and ∆S are related with T as:

greater

entropy

∆S = Thus both S and ∆S are function of temperature. 2. (d) In expansion against vacuum Pext = 0 W = – Pext ∆V = 0 W=0 3. (a) We know that heat and work are not state functions but q + w = ∆U is a state function. H – TS (i.e. G) is also a state function. 4. (b) From first law of thermodynamics, ∆U = q + w For adiabatic process, q = 0 ∴ ∆U = w For isothermal process, ∆U = 0 ⇒ q = – w For cyclic process, ∆U = 0 ⇒ q = – w For isochoric process, w = 0 ⇒ ∆U = q 5. (c) ∆U = n Cv ∆T = 5 × 28 × 100 = 14 kJ ∆(PV) = nR (T2 – T1) = 5 × 8 × 100 = 4 kJ 6. (a) For ideal gas, Cp and Cv are dependent on temperature only. Cp =

R (Independent of P)

Cv =

R (Independent of V)

Thus, Cp will not change with pressure. 7. (d) In the process of synthesis of ammonia from N2 and H2, number of moles decreases which implies that the change in entropy will be

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negative. 2NH3(g) N2(g) + 3H2 (g) 8. (a) From thermodynamics relation. ∆G° = –RT n K ∆H° – T∆S° = RT n K = nK or

nK=

For exothermic reaction, ∆H° = –ve slope = So from graph, lines should be A & B. 9. (b) (a) When diamond is converted into graphite (it is heated to 1500°C) entropy is increased, ∆S > 0 (b) When pressure increases then molecules of gas will come closer and intermolecular distance decreases, so entropy will also decrease, ∆S < 0 (c) When we increase the temperature of a gas then randomness is increased as the kinetic energy gained by molecules. So, ∆S > 0 (d) H2 molecule is converted into atoms, the no. of particles increases. Thus entropy will increase, ∆S > 0 10. (c) From 1st law of thermodynamics ∆U = q + w For adiabatic process : q = 0 ∴ ∆U = w 11.

(d) q = – 3, (work done on the system).

12.

(a) w = –Pext (∆V) =

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100 mole H2O2 on decomposition give 50 mole O2. ∴ w = –(50) (8.3) (300) = –124500 J = –124.5 kJ. 13. (d) ∆G = ∆H – T∆S For a spontaneous reaction ∆G = –ve which is possible when both ∆H and ∆S are positive at high temperature and hence the reaction becomes spontaneous. 14. (c) From 1st law of thermodynamics qsys = ∆U – w = 0 – [–Pext.∆V] = 3.0 atm × (2.0 L – 1.0 L) = 3.0 L-atm ∴

= –1.013 J/K 15.

(a) Process is isothermal reversible expansion, hence ∆U = 0, therefore q = – w. Since q = + 208 J, w = – 208 J 16. (c) For isothermal reversible expansion. w= 17.

(a) TV γ–1 = Constant

(

change is adiabatic)

= For monoatomic gas γ = ∴

=

⇒ T(1)2/3 = T2(2)2/3

T2 = 18.

(b) In general, the molar heat capacity for any process is given by , when PVn = constant

Here

, i.e. PV –1 = constant

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For monoatomic gas, . 19. 20.

(c) ∆H = nCp ∆T solution; since ∆T = 0 so, ∆H = 0 (c) TIPS/Formulae : ∆H = ∆U +P2V2 – P1V1 Given, ∆U = 30.0 L atm P1 = 2.0 atm, V1 = 3.0 L, T1 = 95 K P2 = 4.0 atm, V2 = 5.0 L, T2 = 245 K ∆H = ∆U + P2V2 – P1V1 = 30 + (4 × 5) – (2 × 3) = 30 + 20 – 6 = 44 L atm. 21. (a) Work is not a state function because it depends upon the path followed. 22. (c) In a reversible process, the driving and the opposite forces are nearly equal, hence the system and the surroundings always remain in equilibrium with each other. 23.

(2) wd = wd = –

atm

ws = – 2.303 RT log ws = – 4.606 × 1.04 = – 4.8 L atm = 24.

= 1.72

2.0

(9) Energy released by combustion of 3.5 g gas = 2.5 × (298.45 – 298) kJ

Energy released by 1 mole of gas 25. (– 13538) From = –22477.57 J

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= – 13538 J 26. (48.00) Work done is given by the area under the trapezium. ∴ |w| =

(6 + 10) × 6 = 48 J

27. (6.25) ∆U = nCv∆T 5000 = 4 × Cv (500 – 300) Cv = 6.25 JK–1 mol–1 28. (935.00) SnO2(S) + C(S) —→ Sn(S) + CO2(g) ∆rHo = [–394] – [–581] = 187 kJ/mole = 187 × 103 J/mol ∆rSo = [52 + 210] – [56 + 6] = 200 Jk–1 mol–1 =

T= 29.

= 935 K

(557)

or So, the magnitude is 557 kJ mol–1. 30. (115.41) TIPS/Formulae : For adiabatic expansion, we have and ∆H = nCp ∆T. Solving, we get, T2 = 188.5 K No. of moles of argon gas, n = Now we know that [ Cp= Cv+ R = 12.48 + 8.314 = 20.8] 31. (319.1) 100 g of glucose = 1560 kJ

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Energy utilised in body =

× 1560 = 780 kJ

Energy left unutilised in body = 1560 – 780 = 780 kJ Energy to be given out = 1560 – 780 = 780 kJ Enthalpy of evaporation of water = 44 kJ/mole = 44 kJ/18 g of water [1 mole H2O = 18g water] Hence amount water to be perspired to avoid storage of energy = 32. 33. 34.

35. 36. P-V

× 780

= 319.1 g extensive (because its value depends on quantity of substance) isolated True; Heat capacity Monoatomic Diatomic Cv 3 R/2 5 R/2 Cp 5 R/2 7 R/2 Thus, the heat capacity of diatomic gas is higher than that of a monoatomic gas. True; It only tells that if the heat gained by one end would be exactly equal to heat lost by the other. It does not predict the direction. (a, b, c) work done is applicable for reversible isobaric as well as isothermal and adiabatic process.

For vander waals equation

..... (i) equation (i) is not applicable to irreversible process. Therefore work done is calculated assuming pressure constant throughout the process. 37. (b, c) A– C ⇒ isochoric process A– B ⇒ isothermal process

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B– C ⇒ isobaric process (a) qAC = ∆UAC = nCV, m (T2 – T1) = ∆ UBC WAB = –nRT1

(pressure is not constant)

(b) WBC = –P2(V1–V2) = P2(V2–V1) qBC = ∆HBC = nCP, m (T2–T1) = ∆HAC (c) ∆HCA = nCP, m(T1–T2) (d) ∆UCA = nCV, m(T1–T2) ∆HCA < ∆UCA since both are negative (T1 < T2) and Cp, m > CV, m 38. (a, b, c)

(a)

During irreversible compression, maximum work is done on the gas (corresponding to shaded area) when P1 = P2 (d) When T1 = T2 ⇒ ∆U = nCV∆T = 0 In reversible adiabatic expansion, T2 < T1. ∴ ∆T = –ve and also ∆U = –ve (b) In free expansion, Pext = 0, ∴ W = 0 From Ist law of thermodynamics ∆U = q + W ∴ ∆U = q If expansion is carried out isothermally, ∆U = 0 Hence q = 0 ∴ It is adiabatic process. If carried out adiabatically (q = 0), ∴ ∆U = 0 ∴ It is an isothermal process. (c) During adiabatic expansion, the final temperature is less than the initial temperature. Therefore, the final volume in adiabatic

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expansion will also be less than the final volume in isothermal expansion. This can be graphically shown as

The magnitude of work done by the gas is equal to the area under the curve. As seen from the figure the area under curve in reversible isothermal is more. Hence, the magnitude of work done is lesser in adiabatic reversible expansion as compared to the corresponding work in isothermal expansion. 39. (a, b, c) Since the vessel is thermally insulated, q = 0 Further since, Pext = 0, so w = 0, hence ∆U = 0 Since ∆T = 0, T2 = T1, and P2V2 = P1V1 However, the process is adiabatic irreversible, so we can’t apply P2V2γ = P1V1γ. 40. (a, c, d) T1 = T2 because process is isothermal. Work done in adiabatic process is less than in isothermal process because area covered by isothermal curve is more than the area covered by the adiabatic curve. In adiabatic process expansion occurs by using internal energy, hence, it decreases while in isothermal process temperature remains constant, that's why no change in internal energy. [Entropy is a state function, 41. (a, c) hence additive] [Work done in Y → Z is zero because it is an isochoric process]. 42. (a, b) Mass independent properties (molar conductivity and electromotive force) are intensive properties. Resistance and heat capacity are mass dependent, hence extensive properties. 43. (a, d) Internal energy and molar enthalpy are state functions. Work (reversible or irreversible) is a path function.

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44.

(b, d) Properties independent of mass are intensive properties. Hence, (b) and (d) which are independent of mass are the obvious choices. 45. A- (r, t); B -(p, q, s); C -(p, q, s); D- (p, q, s, t) (A) → r, t H2O(s) H2O(l) It is at equilibrium at 273 K and 1 atm. So, ∆Ssys is negative As it is equilibrium process so ∆G = 0 (B) → p, q, s Expansion of 1 mole of an ideal gas in vacuum under isolated condition Hence, w = 0 and qp = CpdT ( dT = 0) ⇒ q=0 ∆U = CvdT ( dT = 0) ∆U = 0 (C) → p, q, s Mixing of two ideal gases at constant temperature Hence, ∆T = 0 ∴ q = 0; ∆U = 0 also w = 0 (∆U = q + w) (D) → p, q, s, t Reversible heating and cooling of gas follows same path; also initial and final position is same. Hence,

46. (c) K → L ⇒ V increasing at constant P Hence, T increases (Heating). L → M ⇒ P decreasing at constant V Hence, T decreases (Cooling), M → N ⇒ V decreasing at constant P Hence, T decreases (Cooling), N → K ⇒ P increasing at constant V Hence, T increases (Heating).

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47. (b) L to M and N to K, both are having constant volume therefore, these processes are isochoric. 48. (a) Statement 1 is true because it is not possible to convert whole of heat to work. For such a conversion, we need an efficiency of 100% but so far, we have not been able to get such a machine (carnot engine). Statement 2 is true because it is not possible to convert the whole of heat absorbed from a reservoir into work. Some of the heat is always given to the sink. Also, statement 2 is correct explanation for statement 1. Thus, the correct choice is option (a). 49. (b) Assertion : For isothermal expansion, ∆T = 0 ⇒ ∆U = 0 For an ideal gas, work done against vacuum is zero, i.e.W = 0 Now, ∆Q = ∆U + W ⇒ ∆Q = 0 Thus, assertion is correct. Reason : By kinetic theory of ideal gases, the volume occupied by the molecules of an ideal gas is zero. Thus, reason is correct, but it is not the correct explanation of the assertion. 50. TIPS/Formulae : For adiabatic process, W = P (V2 – V1) Here P1 = 1 bar, P2 = 100 bar, V1 = 100 mL, V2 = 99 mL; For adiabatic process, q = 0 ∴ ∆U = W ∆U = q + W = q – P(V2 – V1) since W = – P(V2 – V1) = 0 – 100 (99 – 100) = 100 bar mL ∆H = ∆U + ∆(PV) = ∆U + (P2V2 – P1V1) = 100 + [(100 × 99) – (1 × 100)] = 100 + ( 9900 – 100) = 9900 bar mL 51. Helium molecule is monoatomic so it has just three degrees of freedom corresponding to the three translational motion at all temperature and hence Cv value is always 3/2 R. Hydrogen molecule is diatomic which are not rigidly held so they vibrate about a well defined average separation. For hydrogen molecule we have rotational and vibrational motion both besides translational motion. These two

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additional contributions increase its total heat capacity. Contribution from vibrational motion is not appreciable at low temperature but increases from 0 to R on raising temperature.

52.

(i)

(ii) Total work (W) = W1 + W2 + W3

= – 20 + 13.87 = – 6.13 L atm = 121.95 K Since the system has returned to its initial state i.e. the process is cyclic, so so q = –W = – (– 6.13) L.atm = 620.7 J NOTE : In a cyclic process heat absorbed is completely converted into work. (iii) Entropy is a state function and since the system has returned to its initial state, so ∆S = 0 . Similarly ∆H = 0 and ∆U = 0 for the same reason i.e. U and H are also state functions having definite values in a given state of a system. 53. If heat is absorbed at constant pressure, then qp = ∆E – (– P∆V) or qp= E2 – E1 – [–P(V2 – V1)] or qp = (E2 + PV2) – (E1 + PV1) = H2 – H1 = ∆H

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1.

2. (1) (2) (3) (4) 3. 4. 5.

(c) ∆sol.H° = ∆latticeH° + ∆Hyd.H° 4 = 788 + ∆Hyd.H° ∆Hyd. H° = – 784 kJmol–1 (d) For ideal gas U and H are function of temperature. Compressibility factor for an ideal gas is 1. CP – CV = R for all processes. (d) A-B bond has highest intermolecular potential energy among the given molecules. Hence, it is strongest bond and has maximum bond enthalpy. (c) ∆Hatomisation = ∆Hvap + Bond energy Hence x > y (a) ∆H – ∆U = ∆ngRT ∆ng = no. of moles of product in gaseous state - no. of moles of reactant in gaseous state. ∴ ∆H – ∆U = – 4RT

6.

7.

8.

(d)

= – 3263.9 + (–3.71) = – 3267.6 kJ mol–1 (b) ∆H = ∆U + ∆ngRT 2HI (g) → H2(g) + I2(g) ; ∆ng = (1 + 1) – 2 = 0 ∴ ∆H = ∆U + 0 CO2(g); (c) C(graphite) + O2 (g)

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∆rH° = – 393.5 kJ/mol–1 ...(i) H2(g) +

O2(g)

H2O(l);

∆rH° = – 285.8 kJ/mol–1 ...(ii) CH4(g) + 2O2(g); CO2(g) + 2H2O(l) ∆H°r = + 890.3 kJ/mol–1 ...(iii) CH4(g); C(graphite) + 2H2 (g) ∆H = ? [Eq. (i) + Eq. (iii)] + [2 × Eq. (ii)] = Eq (iv) ∴ [∆H1 + ∆H3] + [2 × ∆H2] = ∆H4 [(– 393.5) + (890.3)] + [2(–285.8)] = –74.8 kJ/mol 9. (d)

...(iv)

∆H = ∆U + ∆ngRT Given, ∆H = –3RT Here ∆ng = np – nr = 0 – 1 = –1 ∆H = ∆U + ∆RT

10.

(a) C(graphite) → C(diamond) (Isothermally) ∆ rG° = ∆G°(diamond) – ∆G°(graphite) = 2.9 – 0 = 2.9 kJ mol–1 Gibbs free energy is the maximum useful work, then –∆G = wmax = ∆PV –2.9 × 103 = –∆P × 2 × 10–6

11.

=1.45× 104 bar = 14500 bar P = ∆P + P0 = 14500 + 1 = 14501 bar (b) Given C(s) + O2(g) → CO2(g); ∆H = –393.5 kJ mol–1 …(i)

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CO(g) +

O2(g) → CO2(g); ∆H = –283.5 kJ mol–1

…(ii)

∴ Heat of formation of CO = eqn (i) – eqn (ii) = –393.5 – (–283.5) = –110 kJ 12. (b) In CH4, 4 × BE(C – H) = 360 kJ/mol ∴ BE(C – H) = 90 kJ/mol In C2H6, BE(C – C) + 6 × BE(C – H) = 620 kJ/mol ∴ BE(C – C) = 80 kJ/mol ∴ BE(C – C) =

J/mol

Now, ∴ λ = 1.49 × 10–6 m ( 3 ∴ λ = 1.49 × 10 nm 13. (a) Bomb calorimeter gives ∆U of the reaction Given, ∆U = –1364.47 kJ mol–1 ∆ng = – 1 ∆H = ∆U + ∆ngRT

1 nm = 10–9 m)

= = – 1366.95 kJ mol–1 14.

(c) Given

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–3∆HN–H = – 1056 15.

(b) Given conditions are boiling conditions for water due to which system is in equilibrium.

∆Stotal = 0 ∆Ssystem + ∆Ssurroundings = 0 ∆Ssystem = – ∆Ssurroundings For process, ∆Ssystem > 0 ∆Ssurroundings < 0 16. (c) Given H2(g) +

→ H2O (l) ;

∆H° = – 285.9 kJ mol–1 H2(g) +

… (i)

→ H2O (g) ;

∆H° = – 241.8 kJ mol–1 … (ii) We have to calculate H2O(l) → H2O (g) ; ∆H° = ? On substracting eqn. (ii) from eqn. (i) we get H2O(l) → H2O(g) ; ∆H° = – 241.8 – (– 285.9) = 44.1 kJ mol–1 17. (c) The standard enthalpy of the combustion of glucose can be calculated by the eqn. C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) ∆HC = 6 × ∆Hf (CO2) + 6 × ∆Hf (H2O) – ∆Hf [C6H12O6] ∆H° = 6 (–400) + 6(–300) – (–1300) = –2900 kJ/mol For one gram of glucose, enthalpy of combustion ∆H° = – 18.

(b) The species in its elemental form has zero standard molar enthalpy of formation at 298 K. At 298K, Cl2 is gas while Br2 is liquid.

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19.

(a) Since, liquid is passing into gaseous phase so entropy will increase and at 373 K during the phase transformation it remains at equilibrium. So, ∆G = 0. 20. (b) ∆G° = ∆ H° – T∆S° , ∆G° = – 2.303 RTlog10 K – 2.303 RT log10 K = ∆H° – T∆S° 2.303 RT log10 K = T∆S° – ∆H° = ;

∴ T = 400 K

21.

(b)

22.

(b) TIPS/Formulae : is the enthalpy change when 1 mole of the substance is formed from its elements in their standard states.

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In (a) carbon is present in diamond however, standard state of carbon is graphite. Again, in (d) CO (g) is involved so it can’t be the right option. Further in (c) 2 moles of NH3 are generated. Hence, the correct option is (b). 23. (b) , ∆H = ? ∆H = Given, ∆Hf CO2(g) = –393.5 kJ/mol = – 110.5 kJ/mol = – 241.8 kJ/mol – = [– 110.5 + (– 241.8)] – (– 393.5 + 0) = 41.2 kJ mol–1 24. (d) TIPS/Formulae : ∆H = ∆E + ∆nRT For ∆H ∆E, ∆n 0 Where ∆n = no. of moles of gaseous products – no. of moles of gaseous reactants (a) ∆n = 2 – 2 = 0 (b) ∆n = 0 ( they are either in solid or liquid state) (c) ∆n = 1– 1= 0 ( C is in solid state) (d) ∆n = 2 – 4 = – 2 (d) is correct answer 25. (a) TIPS/Formulae : Heat capacity at constant volume (qv) = ∆E Heat capacity of constant pressure (qp) = ∆H or ∆H – ∆E = ∆nRT ∆n = no. of moles of gaseous products – no. of moles of gaseous reactants = 12 – 15 = –3 ∆H – ∆E = – 3 × 8.314 × 298 J = – 7.43 kJ.

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26.

(189494) mol

∆n = 1 41000 = ∆U + 1 × 8.314 × 373 J For 5 moles, 27. (– 326400)

28. (–2.70) ∆U = 2.1 kcal = 2.1 × 103 cal ∆ng = 2 ∆H = ∆U + ∆ngRT = 2.1 × 103 + 2 × 2 × 300 = 2100 + 1200 = 3300 cal ∆G = ∆H – T∆S = 3300 – 300 × 20 = 3300 – 6000 = –2700 cals = –2.7 kcal 29. (–192.5) C(s) + O2(g) → CO2(g) ; ∆CH0 [Cgraphite] = –286.0 kJ/mol

...(i)

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H2(g) +

O2(g) → H2O(g) ; ∆CH0[H2(g)] = –393.5 kJ/mol

C2H6 (g) +

...(ii)

O2(g) → 2CO2(g) + 3H2O(g);

∆CH0 [C2H6 (g)] = –1560 kJ/mol ...(iii) The reaction of formation of ethane is 2C(s) + 3H2 (g) →C2H6(g) ∴ ∆f H0 of C2H6(g) = 2 × ∆C0H[Cgraphite] + 3 × ∆CH0[H2(g)] – ∆CH0[C2H6(g)] = 2 × (–286.0) + 3(–393.5) – (–1560) = –192.5 kJ/mol 30. (– 14.6) (i) (ii) (i) – (ii) then ∆G° = –78 + 178 = 100 kJ/mol = 105 J/mol Now for the above reaction

To prevent the above reaction: ∆G > 0

105 + 8 × 1250 ln

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Now, > –10 – 2ln10 > –10 – 2×2.3

(Given ln10 = 2.3)

∴ Minimum 31.

(–2035) The chemical reaction for combustion of diborane is B2H6 (g) + 3O2 (g) B2O3 (s) + 3H2O (g), ∆H = ? For this the enthalpy change can be calculated in the following way. ; of O2 = 0) can be obtained by adding

and

, i.e.

– 286 + 44 = – 242 kJ mol–1 ∆H = [–1273 + 3 × (– 242)] – 36 kJ mol–1 = – 1273 – 726 – 36 = –2035 kJ mol–1 32. (309.16) .....(i) Given S(s) + 3F2(g) SF6(g) ; ∆H = –1100 kJ S(s) S(g); ∆H = 275 kJ .....(ii) 1/2 F2(g) F(g); ∆H = 80 kJ .....(iii) To get SF6(g) S(g) + 6F(g) we can proceed as (ii) + 6 (iii) - (i) SF6(g) S(g) + 6F(g); ∆H= 1855 kJ Thus, average bond energy for S-F bond = 33.

(–2091.32)

C(s) + O2(g)

(g)

CH3CH = CH2(g);

∆H = –33.0 kJ CO2(g); ∆H = – 393.5kJ

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

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H2(g) + 1/2O2(g) H2O(l); ∆H = – 285.8 kJ ...(iii) 3C(s) + 3H2(g) CH3–CH = CH2(g); ∆H = 20.42 kJ ...(iv) The required reaction is + To calculate the value of ∆H follow the following steps. (iv) – (i) yields 3C + 3H2

[3

...(v) (ii) + 3

;

(iii)] – (v) yields ] +

34.

35.

(–266)

∆H = 53.42 kJ

(9/2)O2

3CO2

+

3H2O;

∆H = –2091.32 kJ The required thermochemical equation is

= [715 + 2 × 436 +249] – [3 × 415 + 356 + 463 + 38 ] = – 266 kJ mol–1 (–152) Standard enthalpy of hydrogenation of cyclohexene (–119kJ –1 mol ) means the enthalpy of hydrogenation of one C = C double bond. Now benzene has three C = C double bonds, the enthalpy of the reaction would be = 3 × (–119) = – 357 kJ mol–1

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Actual enthalpy of the reaction can be evaluated as follows.

∴ Resonance energy = 36.

= –357 – (–205) = –152kJ mol–1 (–72) nCH2 = CH2 → (CH2 – CH2)n NOTE : During the polymerisation of ethylene, one mole of ethylene breaks i.e. one C = C double bond breaks and the two CH2 – groups are linked with C – C single bonds thus, forming three single bonds (two single bonds are formed when each CH2 – group of ethylene links with one CH2 – group of another ethylene molecule). But in the whole unit of polymer, number of single C–Cbonds formed/mole of ethylene is 2.

e.g. Number of single bonds formed by 4 moles of ethylene = 8 Energy released = Energy due to formation of 2 C–C single bonds = 2 × 331 = 662 kJ/mol of ethylene Energy absorbed = Energy due to dissociation of 1 C=C double bond = 590 kJ/mol of ethylene ∴ Enthalpy of polymerisation/mol of ethylene or ∆Hpolymerisation = 590 – 662 kJ/mol = –72 kJ/mole 37. (–55.7) From the given data, we can write : (i)

H2 +

O2 → H2O;

∆H1 = –285.8 kJ/mol

(ii) CH4 + 2O2 → CO2 + 2H2O; ∆H2 = –890 kJ mol (iii) C2H6 +

O2 → 2CO2 + 3H2O;∆H3 = –1560 kJ/mol

(iv) C(s) + O2 → CO2; ∆H4 = –393.5 kJ/mol (v) 3C(s) + 4H2 → C3H8(g); ∆H5 = –103.8 kJ/mol The required reaction is C3H8(g) + H2(g) →C2H6(g) +

CH4(g), ∆H = ?

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It can be obtained by the following calculations. 3 × (iv)–(v) + 5(i) – (iii) – (ii) In other words, ∆H = 3∆H4 – ∆H5 + 5∆H1 – ∆H2 – ∆H3 ∴ ∆H = 3(–393.5) – (–103.8) + 5(–285.8) + 890 + 1560 = –2609.5 + 2553.8 = –55.7 kJ/mol 38. (50.90) Combustion of C2H4 and CH4 takes place as follows: C2H4 + 3O2 → 2CO2 + 2H2O 1 vol.

2 vol.

CH4 + 2O2 → CO2 + 2H2O 1 vol.

1 vol.

Let the vol. of CH4 in mixture = x L ∴ Vol. of C2H4 in the mixture = (3.67 – x) L Vol. of CO2 produced by x L of CH4 = x L and Vol. of CO2 produced by (3.67 – x) L of C2H4 = 2 (3.67 – x) L ∴ Total vol. of CO2 produced = x + 2 (3.67 – x) or 6.11 = x + 2 (3.67 – x) or x = 1.23 L ∴ Vol. of CH4 in the mixture = 1.23 L and Vol. of C2H4 in the mixture = 3.67 – 1.23 = 2.44 L Vol. of CH4 per litre of the mixture =

= 0.335 L

Vol. of C2H4 per litre of the mixture =

= 0.665 L

Now, we know that volume of 1 mol. of any gas at 25ºC (298 K) =

= 24.45 L

[ Volume at NTP = 22.4L] Heat evolved due to combustion of 0.335 L of CH4 =–

= –12.20 kJ [given, heat evolved by combustion of 1L =

891 kJ] Similarly, heat evolved due to combustion of 0.665 L of C2H4 = – = –38.70 kJ ∴ Total heat evolved = 12.20 + 38.70 = 50.90 kJ 39. (–121) The required reaction is

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C6H10(g) + H2(g) → C6H12(g), Cyclohexene

Cyclohexane

∆H1 = ?

The given facts can be written as : H2(g) + ½O2(g)→ H2O(l), ∆H2 = –241 kJ/mol C6H10(g) +

...(1) ...(2)

O2(g) → 6CO2(g) + 5H2O; ∆H3 = –3800 kJ/mol ...(3)

C6H12(g) + 9O2(g) → 6CO2(g) + 6H2O, ∆H4 = –3920 kJ/mol

...(4) The required reaction (1) can be obtained by adding equations (2) and (3) and subtracing (4) from the sum of (2) and (3). C6H10(g) + H2(g) → C6H12(g) ∆H1 = (∆H2 + ∆H3) – ∆H4 = [– 241 + (–3800)] – (– 3920) = (– 241 – 3800) – (– 3920) = – 4041 + 3920 = – 121 kJ/mole 40. (3.94) Fe2O3 + 2Al → 2Fe + Al2O3 2 × 56 + 48 = 160

2 × 27 = 54

Heat of reaction = 399 – 199 = 200 kcal [Al & Fe are in their standard states] Total weight of reactants = 160 + 54 = 214 g ∴ Fuel value/gram = Volume of Al = Volume of Fe2O3 =

= 0.9346 kcal/g

= 20 cc = 30.77 cc

Total volume = 20 + 30.77 = 50.77 cc ∴ Fuel value per cc =

= 3.94 kcal/cc

41. (–372.0) The required chemical reaction. 2C2H6 + 7O2 4CO2 + 6H2O; ∆H = x Note that since 2 moles of ethane are reacting, the ∆H of the reaction will be ½ x.

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The thermochemical equations for the given data are written as below. (i) C(s) + O2(g) CO2(g); ∆H = –94.1 kcal (ii) H2(g) + ½O2(g) H2O(g); ∆H = –68.3 kcal (iii) 2C(s) + 3H2(g) C2H6(g); ∆H = –21.1 kcal We know that ∆H = HProducts – HReactants ∆H = + –( + ) ∆H = 4 × (– 94.1) + 6 × (– 68.3) – (2 × (– 21.1) + 0) = –376.4 – 409.8 + 42.2 = –744.0 kcal/2 mole of ethane = –372.0 kcal/mole of ethane 42. (–22) Bond H–H Cl – Cl H – Cl ∆H disso. 104 kcal 58 kcal 103 kcal Formation of hydrogen chloride can be represented as H – H + Cl – Cl → 2H – Cl Thus, the reaction involves Cleavage of one H – H bond, ∆H = 104 kcal Cleavage of one Cl – Cl bond, ∆H = 58 kcal Formation of two H – Cl bonds, ∆H = 2 × (– 103) kcal ∴ ∆H of the reaction = (104 + 58) – 2(103) = 162 – 206 = – 44 kcal Now, since the enthalpy of formation of a compound is the change in heat content accompanied in the formation of one mole of the compound, the enthalpy of formation of HCl gas = 43.

(41.104)

(i)

H2(g) +

(ii) C2H2(g) +

= – 22 kcal

The given data can be written as follows O2(g) → H2O(l); ∆H = –68.3 kcal O2(g) → H2O(l) + 2CO2(g);

∆H = –310.6 kcal (iii) C2H4(g) + 3O2(g) → 2H2O(l) + 2CO2(g); ∆H = –337.2 kcal The required thermochemical equation is C2H2(g) + H2(g) → C2H4(g)

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The required equation can be obtained by subtracting equation (iii) from the sum of equations (i) and (ii), thus ∆H of the required equation can be calculated as below. ∆H = [–68.3 + (–310.6)] – (–337.2) = [–68.3 – 310.6] + 337.2 = –378.9 + 337.2 = –41.7 kcal ∆E, the heat of reaction for the hydrogenation of acetylene at constant volume is given by : ∆E = ∆H – ∆nRT Here∆n = Moles of the gaseous products – Moles of the gaseous reactants = 1 – (1 + 1) = –1 Substituting the values of ∆H, ∆n, R and T in ∆E = ∆H – ∆nRT = –41.7 – (–1 × 2 × 10–3 × 298)

= –41.7 + 2 × 10–3 × 298 = –41.7 + 0.596 = 41.104 kcal 44. (54.20) The required equation is : 2C(s) + H2(g) → C2H2; ∆H = ? Write the thermochemical equations for the given data (i)

C2H2 (g) +

O2(g) → 2CO2(g) + H2O(l) ;

∆H = –310.62 kcal (ii) C(s) + O2(g) → CO2(g) ; ∆H = –94.05 kcal (iii) H2(g) +

O2(g) → H2O(l) ; ∆H = –68.32 kcal

For getting the above required reaction, we will have to NOTE : (a) Bring C2H2 in the product that can be done by reversing the equation (i) to give equation (iv). (b) Multiply equation (ii) by 2 to get 2C atoms in the reactants and thus equation (v) is obtained. (c) Keep equation (iii) as such. (d) Add equations (iv), (v) and (iii).

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(iv) 2CO2 + H2O → C2H2 + (v)

2C + 2O2 → 2CO2;

(iii) H2 +

O2 → H2O;

O2;

∆H = 310.62 kcal

∆H = –188.10 kcal ∆H = –68.32 kcal

On adding, 2C + H2 → C2H2; ∆H = 54.20 kcal Hence, the standard heat of formation of C2H2(g) = 54.20 kcal 45. (101.19) The required reaction in terms of dissociation energy is OH(g) → O(g) + H(g); ∆H = ? This equation can be achieved by (a) reversing the equation (i), (b) dividing equation (ii) and (iii) each by 2, and (c) adding the three resulting equations. OH(g) →

H2(g) +

O2(g); ∆H = + 10.06 kcal [Reversing eq (i)]

H2(g) → H(g)

∆H = – 52.09 kcal

O2(g) → O(g);

∆H = – 59.16 kcal

OH(g) → O(g) + H(g); ∆H = – 101.19 kcal (adding) Thus, one mole of OH(g) needs 101.19 kcal of energy to break into oxygen and hydrogen gaseous atoms. Hence the bond energy of O–H bond is 101.19 kcal. 46. zero; In a closed vessel, ∆V = 0 47. endothermic 48. (a, c) Enthalpy of formation is the enthalpy change for formation of 1 mole of substance from its element present in the most stable natural form. 49.

(b, d)

∆Ssurr =

For endothermic reaction, if Tsurr increases, unfavourable change in entropy of the surroundings decreases.

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For exothermic reaction, if Tsurr increases, favourable change in entropy of the surroundings decreases. 50. (b, c, d) All combustion reactions are exothermic in nature. (b) Decomposition reactions are endothermic in nature. (c) More stable compound is converting into less stable compound. Thus, reaction is endothermic. (d) More stable allotrope is converting into less stable allotrope. Thus, reaction is endothermic. 51. A – p, r, s ; B – r, s ; C – t ; D – p, q, t (A) CO2 (s) → CO2 (g) It is phase transition. The process is endothermic (sublimation). Gas is produced, so entropy increases. (B) On heating CaCO3 decomposes. So, process is endothermic. The entropy increases as gaseous product is formed. (C) 2H• → H2(g) Entropy decreases as number of gaseous particles decreases. (D) It is phase transition. White and red P are allotropes. Red P is more stable than white. So ∆H is –ve. 52. (i) N2O4(g) 2NO2(g) Initially Reaction quotient

∆G = ∆Gº – 2.303 RT log Kp = 0 – 2.303 × 298 log 10 = – 56.0304 L atm. (ii) The negative value of ∆G indicates that the reaction is spontaneous and will lie in the right direction, (forward). 53. TIPS/Formulae :

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Calculation of ∆G values : Thus for the equilibrium B

A

or Similarly for the equilibrium B

C

or Similarly for equilibrium, A

C

Hence, we have that B A, B C, A C, Thus, the correct order of stability, B > C > A 54. For following reaction

∆G° can be calculated as follows : ∆G° = ∆Gp° − ∆G°R = = – 394.4 – (–137.2 +

) = – 257.2 kJ mol–1

Since, ∆G° = ∆H° − Τ∆S° or – 257.2 = ∆H°– 300(0.094) ∆H°= – 285.4 kJ/mol NOTE : ∆H° is negative, so the reaction is exothermic and since ∆G° is negative so the reaction is spontaneous.

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55.

TIPS/Formulae :

For ionisation ∆HHydration > ∆Hionisation Total hydration energy of Al3+ & 3Cl– ions of AlCl3 (∆Hhydration) = (Hydration energyof Al3+ + 3 × Hydration energy of Cl–) = [–4665 + 3 (–381)]kJ mole–1 = –5808 kJ mole–1 NOTE : This amount of energy is more than that required for the ionisation of Al into Al3+ (Ionisation energy of Al to Al3+= 5137 kJ mol–1). Due to this reason, AlCl3 becomes ionic in aqueous solution. In aqueous solution it exists in ionic form as below AlCl3 + 6H2O [Al(H2O)6]3+ + 3Cl– AlCl3 + aq. AlCl3(aq.); ∆H = ? H = Energy released during hydration – Energy used during ionisation = –4665 – 3 381 + 5137 = – 671 kJ/mol Thus, formation of ions will take place. 56. Combustion of CH4 and C4H10 takes place as follows CH4 + 2O2 → CO2 + 2H2O, ∆H = –809 kJ mol–1 C4H10 + 13/2O2 → 4CO2 + 5H2O, ∆H = –2878 kJ mol–1 In order to get the same calorific output due to C4H10,

the rate of supply of butane =

= 0.281 x L/hr

Rate of supply of oxygen = 0.28 x ×

× 3 = 5.481 x L/hr

57. For C3H8 : 3C + 4H2 → C3H8; ∆H1 = ? For C2H6 : 2C + 3H2 → C2H6; ∆H2 = ? ∴ ∆H1 = –[2(C–C) + 8(C–H)] + [3Cs→g + 4(H–H)] ...(1) ∴ ∆H2 = –[1(C–C) + 6(C–H)] + [2Cs→g + 3(H–H)] ...(2) Let bond energy of C–C be x kcal and bond energy of C–H be y kcal ∴ By eq. (1) ∆H1 = –(2x + 8y) + [3 × 172 + 4 × 104] ...(3) ∆H2 = –(x + 6y) + [2 × 172 + 3 × 104] ...(4) Also given C + O2 → CO2; ∆H = –94.0 k cal ...(5) H2 + ½O2 → H2O; ∆H = –68.0 kcal ...(6) C2H6 + (7/2)O2 → 2CO2 + 3H2O; ∆H = –372 k cal ...(7)

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C3H8 + 5O2 → 3CO2 + 4H2O; ∆H = –530 k cal (8) By inspection method : 2 × (5) + 3 × (6) – (7) gives 2C + 3H2 → C2H6; ∆H2 = –20 k cal ...(9) and 3 × (5) + 4 × (6) – (8) gives 3C + 4H2 → C3H8; ∆H1 = –20 k cal (10) ∴ By eq. (3), (4), (9) and (10) x + 6y = 676 2x + 8y = 956 ∴ x = 82 k cal and y = 99 k cal Bond energy of C–C bond = 82 k cal and Bond energy of C–H bond = 99 k cal

...

...

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1.

(c)

2.

(c)

,

at at = –14.29 kJ/mol

kJ/mol

kJ/mol 3.

; Kc

(d) ;

For ;

4.

(c) (i) As reaction is endothermic (∆Hz = +ve) so on decrease in temperature equilibrium will shift towards reactant side. (ii) On increase in pressure by adding inert gas (N2) at same temperature, no shifting will take place. The equilibrium changes only if the added gas is a reactant or product involved in the reaction.

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5.

(b) At equilibrium, rate of forward reaction = Rate of backward reaction. ; ...(i) (d)

6.

;

...(ii)

On adding equations (i) and (ii), we get

7.

(d) Equilibrium constant

8.

(a) Equilibrium constant has no relation with catalyst. Catalyst only affects the rate of the reaction. Catalyst, V2O5 in the given reaction, is used to speed up the reaction. 9.

(a) We know that,

Now, ∆ng = + 1 Þ Kp = Kc (RT)1 Hence, Kp Kc 10. (c) G = G° + RT lnQ At equilibrium; G = 0 and Q = Keq G° = – 2.303 RT log Kw = –2.303 × 8.314 × 298 × log10–14 = 79.9 kJ/mol 80 kJ/mol 11.

(c) Given: A2 + B2 ⇒ 2AB

2AB

A2 +B2; K =

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6AB

3A2 + 3B2; K2 =

The relation between K1 and K2 is K2 = K1–3 12. (b) Volume ↑ P ↓, reaction proceeds in which direction where the number of moles of gases increases. ∆ng = (2 + 1) – 2 = 1 13. (d) Perturbation Shifts reaction towards Removal of CO Left Removal of CO2 Right Addition of CO2 Left Addition of Fe2O3 No change (This is a solid compound. Its concentration has no effect on the equilibrium.) 14.

(a) Given,

No. of moles initially At eqm.

Kc =

C + D

A + B 1

1

1

1

1–a

= 100;

1–a

1+a

1+a

= 10

On solving; a = 0.81 [D]At eq = 1 + a = 1 + 0.81 = 1.81 15.

(a)

At eqm., Total pressure = 2p = 10 bar ∴ p = 5; Now Kp = ( px) ( py) = p2 = 25. 16.

(b)

2∆G°f (NO2) =

= 2∆G°f (NO2) – 2∆G°f (NO) – ∆G°f (O2) + 2∆G°f (NO) + ∆G°f (O2)

∆G = ∆G° + RTlnKp At equilibrium,

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∆G = 0, Q = Kp ; ∆G° = –RTlnKp ∆G°f (O2) = 0 ∴ ∆G°f (NO2) =

[2 × 86600 – R(298)ln (1.6 × 1012)]

17. (d) Volume of ice is greater than that of water. The direction in which the reaction will proceed can be predicted by applying Le-Chatelier's principle Pressure ∝ So equilibrium, will shift forward. 18.

(b)

where

number of gaseous moles in product – number of gaseous moles in reactants

= 19. (d) On adding inert gas at constant volume the total pressure of the system is increased, but the partial pressure of each reactant and product remains the same. Hence no effect on the state of equilibrium. 20. (b)

Hence, Kc = Given, at equilibrium ∴ (a – x) = (1.5a – 2x) ∴ a = 2x On solving Kc = 4

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21.

(d)

Then

(Where s is the solubility) or

Similarly for or and for or From the given values of Ksp for MX, MX2 andM3X, we can find the solubilities of those salts at temperature, T. Solubility of Solubility of

or

=

or

Solubility of =

or 10–4

Thus the solubilities are in the order MX > M3 X> MX2 i.e. the correct answer is (d). 22. (a) In a reversible reaction, catalyst speeds up both the forward and backward reactions to the same extent, so (c) is wrong. At equilibrium,

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or On adding N2, reaction will move to forward direction because of LeChatelier's principle. 23. (d) TIPS/Formulae : At constant temperature Kp or Kc remains constant. For the equilibria : 2NO2(g) N2O4(g) ∆n = 1 Since, temperature is constant so Kc or Kp will remain constant. Further, since volume is halved, the pressure will be doubled so will decrease so as to maintain the constancy of Kc or Kp. From Le-Chatelier's principle reaction will move backwards. 24. (d) At constant temperature, Kp remains constant. With change of pressure, x will change in such a way that Kp remains a constant. pA+q + 25. (a) ApBq(s) pS

qS

26.

LS = (pS) .(qS) = p .q .S( p + q) (d) At initial stage of reaction, concentration of each product will increase and hence, Q will increase.

27.

(d)

p

q

p

q

∆n = 2 – 4 = –2 (R in L.atm.K–1 mole–1). 28. (a) The given reaction will be exothermic in nature due to the formation of three X - Y bonds from the gaseous atoms. The reaction is also accompanied with the decrease in the gaseous species (i.e. ∆n is negative). Hence, the reaction will be affected by both temperature

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and pressure. The use of catalyst does not affect the equilibrium concentrations of the species in the chemical reaction. 29. (d) NOTE : In case of alkaline earth hydroxides solubility increases on moving down the group. Be(OH)2 has lowest solubility and hence, lowest solubility product. 30. (a) For a precipitation to occur Solubility product < Ionic product Given Ksp = 1.8 × 10–10 Calculating ionic products in each Ionic product = [Ag+] [Cl–] = = 2.5 × 10–9 which is greater than Ksp (1.8 × 10–10).

31.

(d) As all the reactants and products are present in aqueous form in (d), so it is a reversible reaction. In others, either solid or gas is generated, which is insoluble or volatile and hence, makes the reaction unidirectional. 32. (a) Statement (a) is correct and the rest statements are wrong. Kp depends only on temperature, hence at constant temp. Kp will not change. at equilibrium 33. (d) (a) Inter-molecular forces are more in liquid phase than that in gas phase. (b) Due to difference in inter-molecular forces, the potential energy is different. (c) Due to difference in potential energy, the total energy of the molecules are different. (d) Kinetic energy depends on temperature only. Hence, both the molecules of liquid and solid have same kinetic energy. 34. (b) TIPS/Formulae : For precipitation to occur, ionic product > solubility products

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Given, Ionic product of

=

Calculate I.P. in each case (a) I.P. of CaF2 = (10–4) × (10–4)2 = 10–12 (b) I.P. of CaF2 = (10–2) × (10–3)2 = 10–8 (c) I.P. of CaF2 = (10–5) × (10–3)2 = 10–11 (d) I.P. of CaF2 = (10–3) × (10–5)2 = 10–13 I.P > solubility in choice (b) only. ppt of CaF2 is obtained in case of choice (b) only. 35. (d) Only temperature affects the equilibrium constant. 36. (b) TIPS/Formulae : (i) According is Le-Chateliers principle, exothermic, reactions are favoured at low temperature. (ii) According to Le-Chateliers principle, the reaction in which n < 0, are favoured at high pressure. Given It is exothermic reaction ∴ Yield of SO3 is maximum at low temperature n = 2 – 3= – 1 or n < 0 ∴ Yield of SO3 is maximum at high pressure. 37. (b) In molten state, the cations and anions become free and flow of current is due to migration of these ions in opposite directions in the electric field. 38. (7) Let the solubility of AgCl is x mol litre-1 and that of CuCl is y mol litre-1

∴ Ksp of AgCl = [Ag+] [Cl–] 1.6 × 10–10 = x (x + y) ....(i) Similarly, Ksp of CuCl = [Cu+][Cl–]

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1.6 × 10–6 = y(x + y) On solving, (i) and (ii) [Ag+] = 1.6 × 10–7 ∴ x = 7 39. (16)

t = teq

1– a

1.5 – a

… (ii)

0.5 + 2a

0.5 + 2a = 1; a = 0.25

40. (0.25)

Now, = 0.25 41. (8.93) t=0 At equilibrium

0.03M 0.1 M (0.03 – x) (0.1 – x)

Since, Kc >> 103 ; 0.03 – x = 0 ∴ x = 0.03

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and 0.1 – x = 0.07 (conc. of solid is taken as 1)

= 8.93 × 10–17 comparing (i) with given value in question we get

(i)

Y = 8.93. 42. (0.983) ∴ K1 =

H+ + HS–

H2S

Further, HS–

H+ + S2–

∴ K2 =

Dissociation constant of H2S, K = K1 × K2 i.e. K = 1 × 10–7 × 1.3 × 10–13 = 1.3 × 10–20 Now we know that Ksp = [M 2+] [S 2–] ⇒ 6 × 10–21 = 0.05 × [S2–] [S 2–] =

= 1.2 × 10–19

Substituting the various values in the following relation K= 1.3 × 10–20 =

[H2S] = 0.1 M

[H+]2 = [H+] =

= 1.04 × 10–1

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pH = –log [H+]; pH = –log (1.04 × 10–1) = 1.0 – log 1.04 = 1.0 – 0.017 = 0.983 43. (1.886) 2AB(g) A2(g) + B2(g)

At start At equ.

∴ [A2] =

1 1–x

2 2–x

, [B2] =

∴K=

0 2x

, [AB] = = 50

On solving we get, 23x2 – 75x + 50 = 0; x = 2.317 or 0.943 The value 2.317 is inadmissable because initial concentration of reactants is 2 moles and so x = 0.943 ∴ Moles of AB formed = 2 × 0.943 = 1.886 44. no change; (Kp of an equilibrium reaction is independent of the pressure of the system.) 45. Kp = Kc (RT)∆n; Here ∆n = No. of moles of gaseous products – no. of moles of gaseous reactants R = gas constant, and T = absolute temperature. 46. TIPS/Formulae : Dissolution of NaOH is exothermic. False : When a solute like NaOH is added to a solvent exothermic dissolution takes place. An increase in temperature always favour endothermic process. So, solutes having exothermic dissolution shows a decrease in their solubility with temperature. 47. True : Lower the pressure, lower will be boiling point. More liquid will vapourise and temperature decreases. 48.

False : K for A2 + B2 K´ for AB

2AB is

½ A2 + ½ B2 is

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or

(K´)2=

=

∴

K´ =

49. (b) Initially, on increasing temperature, rate of reaction will increase, so % yield will also increase with time. But at equilibrium, % yield at high temperature (T2) would be less than at T1 as reaction is exothermic so the graph is

50. (a, b, d) (A) ∆H = CP(rxn) ∆T Hence, enthalpy depends on temperature. CaO(s) + CO2(g) Κp = (B) CaCO3(s) (C) Keq depends only on temperature and not on Pressure. (D) Enthalpy of reaction is independent of the catalyst. Catalyst generaly changes activation energy. 51. (b) Ksp = 1.1 × 10–12 = [Ag+]2 1.1 × 10–12 = [0.1]2 [s] ; s = 1.1 × 10–10 CO2(g) + H2(g) 52. (d) CO(g) + H2O(g)

A catalyst simply helps in attaining the equilibrium earlier. Addition of inert gas has no effect on a reaction because in it, n = 0. This equilibrium is not based upon volume because in it, n = 0. On increasing the amount of CO, KC will decrease but it is constant at constant temperature, so for maintaining the constant value of KC, the amount of CO2 increases.

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53. (c, d, e) (c) Introducing an inert gas at constant pressure will increase the volume. As a result, concentration of reactants and products will decrease. Thus, the reaction will move in forward direction to increase the concentration. (d) Increasing the volume of container, will move the reaction in forward direction. (e) Introducing PCl5 at constant volume will favour the forward direction to decrease the concentration of PCl5. 54. (a) At constant volume, concentrations do not change. 2NaNO2 (s) + O2 (g) 55. (c, d) 2NaNO3 (s) According to Le-chatelier principle an increase in pressure always favours the reaction, where volume or moles decrease (i.e. reverse direction). As heat is added, i.e. reaction is endothermic and is supported in forward direction with increase in temperature. NaNO3 and NaNO2 both are solid. Thus, they will not effect the position of equilibrium. 56. (a, b, c, d) The reaction is exothermic, hence increasing temperature will favour backward reaction (i.e. conc. of C2H4 increases) removing H2 and adding C2H6 favours backward reaction. By decreasing the pressure, reaction will move towards preparing more moles i.e. backward reaction. 57. (d) We know that for every chemical reaction at equilibrium, change in Gibb's free energy ( G = 0) is zero. However, charge in standard Gibb's free energy ( G°) may or may not be zero. Thus, statement 1 is False. For a spontaneous reaction, at constant temperature and pressure, the reaction proceeds in the direction in which G is < 0 i.e. in the direction of decreasing Gibb's energy (G) so, statement 2 is True. 58. (d) The statement-1 is clearly wrong in context to Le-Chateliers principle, which states that “increase in temperature shifts the equilibrium in the forward direction of those reactions which proceed with absorption of heat (endothermic reactions), and in the backward direction of those reactions which proceed with the evolution of heat (exothermic reactions).” Statement -2 is clearly true again according to Le-chatelier principle.

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59.

NH4HS(s)

NH3(g) 0

Initial moles

+ H2S(g) 0

Moles at eq.

Given V = 2 litre, T = 300K,

n = 2 – 0= 2

Kc = [NH3][H2S] = 8.1 10–5 mol2 litre-2 Also Kp = Kc(RT) = 8.1 10–5 (0.082 300)2 = 4.90 10–2 atm2 NOTE : Addition of more NH4HS on this equilibrium will cause no effect because concentration of NH4HS is not involved in formula of Kp or Kc. Ag+ + Cl–

60. For AgCl; AgCl ∴ Ksp = [Ag+] [Cl–]

.....(i)

Again it is given that Ag+ + 2NH3 ; Kc = 6.2 × 10–8 [Ag(NH3)2]+ or Ag+ + 2NH3

[Ag(NH3)2]+ ;

Kf = ∴

or

NOTE THIS STEP : Since the formation constant of the complex is very high, most of the [Ag+] which dissolves must be converted into complex and each Ag+ dissolved also requires dissolution of Cl–. ∴ [Cl–] = [Ag (NH3)2]+ and let it be c M Equation (i) becomes ⇒

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= 0.2903 × 10–2 or c = 0.538 × 10–1 = 0.0538 M 61. The concerned chemical reaction is Calculation of [Ag+] left in the solution : Ksp(Ag2CO3) = [Ag+]2 [CO32–]

Concentration of Cl– left = 0.0026 g/L = 7.33 ×10–5 M

=

∴ Ksp(AgCl) = [Ag+] [Cl–] = (2.34 × 10–6 ) (7.33 × 10–5) = 1.71 × 10–10 62. TIPS/Formulae : Consider common ion effect Conc. of Ag+ ions = Conc. of AgNO3 = 0.03 M Most of these Ag+ ions will be present in the form of [Ag(CN)2]–. 0.03 M AgNO3 requires 2 × 0.03 M = 0.06 M CN– to form [Ag(CN)2]– ∴Conc. of free CN– at equilibrium will be 0.1 – 0.06 = 0.04 M Ag+ + 2CN– [Ag(CN)2]– ∴ K =

; 4.0 × 10–9 =

[Ag+] = 63. Initial mole

= 7.5 × 10–18 M 2AB2(g) 1

2AB(g) + B2(g) 0

0

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Moles at equb

1–x

x

Total moles at equb. = 1 – x + x + =

=1+

P= P=

pAB = =

P P

P=

∴ Kp =

=

P

=

= Kp ≈ 64.

P

(i)

or

x= CO(g)

Moles at start Moles at equb. or

0.15 (0.15 – x) (0.15 – 0.08)

+ 2H2(g) a (a – 2x) (a – 0.16)

CH3OH (g) 0 0.08 0.08

∴ Total moles at equb. = 0.15 – 0.08 + a – 0.16 + 0.08 = a – 0.01 Total moles at equilibrium can also be calculated from the following relation n=

=

= 0.345

∴ 0.345 = a – 0.01 [Comparing (i) and (ii)] or a = 0.355 Thus, Moles of CO at equilibrium =

0.15 – 0.08 = 0.07

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Moles of H2 at equilibrium = 0.355 – 0.16 = 0.195 Moles of CH3OH at equilibrium = 0.08 Substituting the values in the relation, =

Kc =

= 187.85 mole–2 litre2 [ V = 2.5 L] Calculation of Kp Kp = Kc (RT)∆n = 187.85 × (0.0821 × 750)–2 = 0.05 atm– 2 [ ∆n = –2] (ii) Calculation of final pressure when there is no reaction Moles of CO = 0.15; Moles of H2 = 0.355 ∴ Total moles = 0.15 + 0.355 = 0.505 PV = nRT P × 2.5 = 0.505 × 0.0821 × 750 ⇒ P = 12.438 atm. 65. Let the solubility of Ca(OH)2 in pure water = S moles/litre Ca2+ + 2OH– Ca(OH)2 S

2S

Then Ksp = [Ca ] [OH ] 4.42 × 10–5 = S × (2S)2; 4.42 × 10–5 = 4S3 S = 2.224 × 10–2 = 0.0223 moles litre–1 2+

– 2

∴ No. of moles of Ca2+ ions in 500 mL. of solution = λ

=

0.01115 NOTE THIS STEP : Now when 500 mL. of saturated solution is mixed with 500 mL of 0.4M NaOH, the resultant volume is 1000 mL. The molarity of OH– ions in the resultant solution would therefore be 0.2 M. ∴ [Ca2+] =

=

= 0.001105 M

Thus, No. of moles of Ca2+ or Ca(OH)2 precipitated = 0.01115 – 0.001105 = 0.010045 Mass of Ca(OH)2 precipitated

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= 0.010045 × 74 = 0.7433 g = 743.3 mg [mole wt. of Ca(OH)2 = 74] 66. Ag2CO3 + K2C2O4 → Ag2C2O4 + K2CO3 Moles at start Moles after reaction

Excess

0.1520 0.1520–0.0358 = 0.1162

0 0.0358

0 0.0358

left unreacted

Molar concentration of K2C2O4 or = 0.2324 moles L–1

=

] [K2CO3] = [

[

500 mL = 0.5 L

] at equilibrium = 0.07156 moles L–1

=

Given that Ksp for Ag2C2O4 = 1.29 × 10–11 mol3 L–3 at 25ºC ] = 1.29 × 10–11 So, [Ag+]2 [ or [Ag+]2 × 0.2324 = 1.29 × 10–11 × 10–11

Hence [Ag+]2 = Then Ksp for Ag2CO3 ] =

= [Ag+]2 [

× 0.0716

= 3.974 × 10–12 mol3 L–3 67. Let the total number of moles of all gases at equilibrium point = n P = 4.92 atm. V = 5L –1 –1 R = 0.0821 atm. L mol K T = 273 + 327 = 600K By applying the formula PV = nRT n=

=

= 0.5 moles

(i) Calculation of the number of moles of the individual gases at equilibrium point. No. of moles of CH3OH formed = 0.1 (Given) ∴ No. of moles of CO (also)= 0.1

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[ moles of CO = moles of CH3OH formed] Hence, No. of moles of H2 = 0.5 – (0.1 + 0.1) = 0.3 ∴ Molar concentration of various species will be = 0.02; [H2] =

[CH3OH] = [CO] = ∴ Kc =

=

= 0.06

= 277.78 mol–2 L2

(ii) Calculation of Kp. We know that Kp = Kc × (RT) ∆n = 277.78 2

= 68.

× (0.0821 × 600)– (∆n = 1 – 3 = –2)

= 0.1144 atm–2

TIPS/Formulae : p(OH) for basic buffer

We know that pOH= pKb + log

or –log 1.8 × 10–5 + log

pOH = 5 – log 1.8 + log 5 = 5.6989 – 0.2552 – log [OH–] = 5.4437; log [OH–] = –5.4437 [OH–] = 3.5999 × 10–6 [Taking antilog] Ksp for Mg(OH)2 = [Mg2+] [OH–]2 6 × 10–10 = [Mg2+] [3.5999 × 10–6]2 [Mg2+] =

= 0.4629 × 102

= 46.29 mole ion/L Ksp for Al(OH)3 = [Al3+] [OH–]3 6 × 10–32 = [Al3+] (3.5999 × 10–6)3 [Al3+] =

= 1.286 × 10–15 mol ion/L

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69.

NOTE : Since the reaction is carried out at constant volume, change in partial pressure of a species will be directly proportional to the change in its amount. Hence, we can write 2SO3(g) 2SO2(g) + O2(g)

Initial pressure Equb. pressure

0 2x

2 atm 2 atm + x

1 atm 1 atm – 2x

Where 2x is the change in partial pressure of SO3 at equilibrium. Substituting the expression of partial pressure in the expression. For Kp, we get Kp = or 900 atm–1 = Assuming x is very small as compared to 1 900 atm–1= On usual calculations, x = 0.0118 atm Thus,

= 2x = 2 × 0.0118 atm = 0.0236 atm = 2 atm + x = 2 + 0.0118 = 2.0118 atm = 1 atm – 2x = 1 – 0.0236 = 0.9764 atm

70.

(i) Before dissociation After dissociation

2NO2

N2O4

1

0

1–α

2α

∴ Total moles = 1 – α + 2 α = 1 + α

∴ Kp =

=

where P is total pressure

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Kp =

= 0.266 atm

[

α =

0.25] =

(ii) Kp = ⇒ α = 0.64

0.266 =

∴ Percentage dissociation = 63 % 71. Initial concentration of each gas = 1 mole Let the No. of moles of NO2 reacted at equilibrium = x SO3(g) + NO2(g) Then, SO2(g) + NO2(g) At equilibrium

(1 – x)

(1 – x)

(1 + x)

= Kc

Now we know that, or

or

(1 + x)

= 16 (

V = 1L)

= 4 or 1 + x = 4 – 4x or 5x = 3

or x=

= 0.6

∴ Thus, the concentration of NO at equilibrium = 1 + x = 1 + 0.6 = 1.6 moles Concentration of NO2 at equilibrium = 1 – x = 1 – 0.6 = 0.4 moles 72. Solubility of Mg(OH)2 in water S = 9.57 × 10–3 g/litre =

= 1.65 × 10–4 mole/litre

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[ Molar mass for Mg(OH)2 = 58] Mg2+ + 2OH– Mg(OH)2 Ksp = (S) (2S)2 = 4S3 = 4 (1.65 × 10–4)3 = 1.8 × 10–11 approx. Calculation of solubility of Mg(OH)2, say, x, in Mg(NO3)2 or [Mg2+] = x + 0.02; [OH–] = x Ksp = [Mg2+] [2OH–]2 or 1.8 × 10–11 = (x + 0.02) (2x)2 Neglecting x in comparison to 0.02 (common ion effect) ⇒ 4x2 =

= 9 × 10–10 or 2x = 3 × 10–5

x = 1.5 × 10–5 moles/litres = 1.5 × 58 × 10–5 = 8.7 × 10–4 g/litre. 73. Let x be the degree of dissociation of PCl5(g), then PCl3(g) + Cl2(g) PCl5(g)

Initial At equilibrium

3 3(1 – x)

0 3x

1 1 + 3x

∴ Total number of moles at equilibrium = 3 (1 – x) + 3x + 1 + 3x = 3 (1 + x) + 1 Using the gas equation : PV = nRT ∴n = Here, P = 2.05 atm., V = 100 litres, R = 0.082 atm/deg., T = 273 + 227 = 500 K =5∴

∴

n=

or

3 + 3x + 1 = 5 or 3x = 5 – 4 or

3(1 + x) + 1 = 5 x=

= 0.333

Hence perecentage dissociation of PCl5 = 0.333 × 100= 33.3% Calculation of Kp for the reaction : Kp =

=

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= =

=

Substituting, x = 1/3 and P = 2.05 atm., we get =

Kp =

= 0.41

74.

TIPS/Formulae : For precipitation to occur ionic product > Ksp. Mixture solution contains 0.1 M Ag+ and 0.1 M Hg22+. Ksp of Hg2I2 = 2.5 × 10–26 is much smaller than Ksp of AgI which is 8.5 × 10–17. [I–] concentration needed to precipitate Hg2I2 is calculated as : Hg22+ + 2I– Hg2I2 =

[I–] =

= 5.0 × 10–13 M

Similarly, [I–] concentration needed to precipitate AgI is :

[I–] =

=

= 8.5 × 10–16 M

NOTE : Since [I–] concentration needed to ppt. AgI is smaller than that needed to ppt. Hg2I2, AgI is completely precipitated first. AgI starts precipitation with [I–] = 8.5 × 10–16 M. However, Hg2I2 starts precipitating with AgI only when molar concentration of I– reaches 5.0 × 10–13 M. [Ag+] left when Hg2I2 begins to ppt. is given by

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=

= 1.7 × 10–4 M × 100

Thus % [Ag+] left unprecipitated = = 0.17% Hence % Ag precipitated = 99.83% 75. Initial moles 1 3 N2(g) + 3H2(g) +

Eq. moles

1 – 0.0025

3 – 0.0075

0

2NH3(g)

2 × 0.0025

Eq. conc.

Now we know that Kc = Since 0.0025 and 0.0075 are very small, 1–0.0025 and 3–0.0075 may be taken as 1 and 3 respectively. Substitute the various values

=

Kc =

×

= 1.48 × 10–5 litre2 mol–2 For the equilibrium, N2(g) +

H2(g)

Kc’ =

NH3(g) =

= = 3.82 × 10–3 litre mol–1

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1. (b) (A) 0.01 M HCl [H+] = 10–2, pH = –log 10–2 = 2 pOH = 14 – 2 = 12 (B) 0.01 M NaOH [OH–] = 10–2, pOH = – log[OH] = 2 (C) 0.01 M CH3COONa pH = pH > 7 pOH < 7 (D) 0.01 M NaCl, pH = 7, pOH = 7 Decreasing order of pOH value is, (A) > (D) > (C) > (B). 2. (d) Millimoles at start Millimoles after reaction

10

20

0

0

0

10

10

10

Buffer solution contains CH3COONa (10 millimole) and CH3COOH (10 millimole) which is a acidic buffer. 3. (c) At equivalence point pH is 7 and pH increases with addition of NaOH so correct graph is (c). Pb2+(aq) + 2Cl–(aq) 4. (c) PbCl2 Given; Ksp = 1.6 × 10–5 = 0.1005

[Pb2+] = [Cl–] =

= 0.1

Q = [Pb2+] [Cl–]2 = 0.1005 × (0.1)2 = 1.005 × 10–3 Q > Ksp

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5.

(b) Cr(OH)3



Ksp = s.(3s)3 6 × 10–31 = 27.s4 ; s = [OH–] = 3s = 3 ×

6.

= (18 × 10–31)1/4 M (b) From the given curve,

if [X] = 1 mM then [Y] = 2 mM Salt is XY2 Ksp = [X][Y]2 10–3) (2 × 10–3)2 = 4 × 10–9 M3 7. (b) Temperature plays a significant role on pH measurements. As the temperature rises, molecular vibrations increase which results in greater ability of water to ionise and form more hydrogen ions. As a result, the pH will drop. So assertion is incorrect. The dissociation of water molecules into ions is bond breaking and is therefore an endothermic process (energy must be absorbed to break the bonds). So reason is also incorrect. 8. (c) Let the solubility of Al(OH)3 in 0.2M NaOH solution be s. Then, and [Al3+] = s and [OH–] = 3s + 0.2 » 0.2 Ksp = 2.4 × 10–24 = [Al3+] [OH]3 2.4 × 10–24 = s(0.2)3

9.

(d)

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= 5.35 pH = 5.35 10. (b) Graph A and B, both represents the titration curve between strong acid and strong base, i.e., HCl and NaOH but, the pH of NaOH is more than 7 and during the titration it decreases, so graph (A) is correct. 11. (b) Zr3(PO4)4 Ksp =[Zr4+]3 [PO43–]4 = (3S)3(4S)4 Ksp = 6912 S7 S= 12.

(b) CH3COOK is a salt of weak acid (CH3COOH) and strong base (KOH). FeCl3 is a salt of weak base [Fe(OH)3] and strong acid (HCl). Pb(CH3COO)2 is a salt of weak base Pb(OH)2 and weak acid (CH3COOH) Al(CN)3 is a salt of weak base [Al (OH)3] and weak acid (HCN). 13.

(b)

∴

[S2–] = 3 × 10–20 14.

(d)

Ksp = [Pb2+][Cl–]2 Ksp = 4s3 = 32 × 10–9

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s3 = 8 × 10–9 s = 2 × 10–3 M

15.

(b) The salt (AB) given is a salt of weak acid and weak base. Hence the pH can be calculated by the following formula

= 6.9 16. (c) (Unionized, weak acid and common ion effect) (ionized)

Given, pH = 6, [H+] = 1 × 10–6

17. (b) i.e, conjugate base of hydrazoic acid is N3. 18. (d) Solutions which resist the change in the value of pH when small amount of acid or base is added to them are known as buffers.

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19. (d) pH = 1 ; H+ = 10–1 = 0.1 M pH = 2 ; H+ = 10–2 = 0.01 M ∴ M1 = 0.1 V1 = 1 M2 = 0.01, V2 = ? From M1V1 = M2V2 0.1 × 1 = 0.01 × V2 V2 = 10 L ∴ Volume of water added = 10 – 1 = 9 L 20. (d) Given [OH– ] = 5 × 10–2 ∴ pOH = – log 5 × 10–2 = – log 5 + 2 log 10 = 1.30 pH + pOH = 14 pH = 14 – pOH = 14 – 1.30 = 12.70 21. (d) Let the weak manoacidic base be BOH, then the reaction that occurs during titration is BOH + HCl → BCl + H2O Equilibrium : Using the normality equation,

=

Substituting various given values, we get = or

=

= 2.5 × 3 = 7.5 mL

Then the concentration of BCl in resulting solution is given by [BCl] =

=

or 0.1 M

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Since Thus

= 10–2 or

or 10–2 – 10–2 h = 0.1 h2 or 0.1 h2 + 10–2 h – 10–2 = 0

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(Solving this quadratic equation for h, we get)

= = =

=

[Neglecting the negative term]

= 0.27 ∴ [H+] = C. h = 0.1 × 0.27 = 2.7 × 10–2 M Thus, the correct answer is [d]. 22. (b) NOTE : For basic buffer pH is more than 7. CH3NH2 + HCl Initial moles Moles after mixing

0.1 0.02

→

0.08 0

0 0.08

As it is a basic buffer solution. pOH = pKb + log

= – log 5 × 10–4 + log 4

= 3.30 + 0.602 = 3.902 pH = 14 – 3.902 = 10.09; [H+] = 7.99 × 10–11 ≈ 8 ×10–11 M 23. (c) NOTE : For isotonic solutions, osmotic pressure is same.

Since both solutions are isotonic, therefore, 0.004 + 2x = 0.01 ; x = 3 × 10–3 ∴ % Dissociation

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Alternate eqn. Total no. of moles = 0.004 × x + 2x + x 24. (b) ;

Hence, % hydrolysis = 10–4 × 100 = 0.01 25. (a) TIPS/Formulae : For oxyacids containing similar central atom, the acid strength increases with the increase in the number of oxygen atom attached to the central atom and not attached to any other atom. TIPS/Formulae : Higher the oxidation number of the central atom, higher is the acidity of the species. Thus, acidity follows the order Oxi. No. of Cl HClO < HClO2 < HClO3 < HClO4 +1 +3 +5 +7 26. (b) The characteristics of the given solutions are: NaCl – neutral solution NH4Cl – slightly acidic due to the following reaction NaCN – slightly alkaline due to the following reaction HCl – highly acidic The pH of the solution will follow the order highly acidic < slightly acidic < neutral < slightly alkaline i.e. HCl < NH4Cl < NaCl < NaCN 27. (a) Among oxyacids of the same type formed by different elements, acidic nature increases with increasing electronegativity. In general, the strength of oxyacids decreases as we go down the family in the periodic table. HOCl (I) > HOBr (II) > HOI (III) 28. (c) Salts of weak base and strong acid get hydrolysed in aqueous solution forming an acidic solution. AlCl3 + 3H2O

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—→ Al(OH)3 + 3HCl (weak)

29.

(strong)

(acidic)

(b) TIPS/Formulae : H2O H+ (1 – α) c

+ OH–

αc

αc

α = 1.9 × 10–7 ; Density of water = ∴c=

× 1000 = 55.56 moles/l

∴ [H+] = 55.56 × 1.9 × 10–9 = 1.055 × 10–7 [ 1.9 × 10–% = 1.9 × 10–9] ∴ Kw = [H+] [OH–] = (1.055 × 10–7)2 = 1.0 × 10–14 30. (d) (a) It is not correct answer because 100 ml M/10 HCl will completely neutralise 100 ml M/10 NaOH and the solution will be neutral. (b) After neutralisation resultant solution will be acidic due to presence of excess of HCl. (c) After netralisation resultant solution will be basic due to presence of excess of NaOH. (d) M. eq. of HCl = 75 N/5 = 15 Meq M.eq. of NaOH =

= 5 Meq

M. eq. of HCl left = 10

[HCl] =

= M/10

pH = –log [H+] = 31.

(d) Since HCl is stronger than CH3COOH hence acts as acid. On the other hand Cl– is a stronger base than and is the conjugate base of HCl. HCl + CH3COOH acid 1

base2

Cl– base1

+ acid 2

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32.

(a) Due to hydrolysis of FeCl3, backward reaction will not take place.

33.

(a) TIPS/Formulae : (i) Higher the electronegatively of central atom, higher will be the acidic strength. (ii) In case of same atom, higher the value of oxidation state of the metal, higher will be its acidic strength. The electronegativity of Cl > S. Oxidation no. of Cl in ClO3(OH) = + 7 Oxidation no. of Cl in ClO2(OH) = + 5 Oxidation no. of S in SO(OH)2 = + 4 Oxidation no. of S in SO2(OH)2 = + 6 ClO3(OH) is the strongest acid. 34. (d) TIPS/Formulae : In acidic medium, weak acids are unionized due to common ion effect and they are completely ionised in alkaline medium. Aspirin (or acetyl salicylic acid) is unionised in stomach (where pH is 2-3 ) and is completely ionised in small intestine (when pH is 8). 35. (d) (a) is a neutral solution due to both cationic and anionic hydrolysis (Ka = Kb = 1.8×10–5); (b) is acidic solution due to cationic hydrolysis; (c) is acidic solution due to cationic hydrolysis; (d) is basic solution due to anionic hydrolysis. 36. (c) CaO, CaCO3 and Ca(OH)2 dissolve in CH3COOH due to formation of (CH3COO)2Ca. But CaC2O4 does not dissolve as CH3COO– is a stronger conjugate base than . 37.

(N) NOTE : Electron acceptors or elements having incomplete octet are Lewis acids. (i) BF3 (B has 6 e– in valance shell), AlCl3 (Al has 6 electrons in valance shell), BeCl2 (Be has 4 e– in valance shell) are electron defecient compounds and hence Lewis acids.

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(ii) SnCl4 has complete octet but due to empty d-orbitals, it can accept electron pair and can be considered as a Lewis acid. 38. (a) Base + H+ → (conjugate acid) (base) + H+ → NH3 (conjugate acid) 39.

(d) TIPS/Formulae : The pH of the solution at the equivalence point will be greater than 7 due to salt hydrolysis. So an indicator giving colour in basic medium will be suitable. Phenolphthalein is a good indicator if the base is strong because strong base immediately changes the pH at end point. 40. (c) TIPS/Formulae : The equilibrium constant for the nuetralization of a weak acid with a strong base is given by = 41.

= 1.0 × 1010

(a) For a basic buffer, pH = 14 – pKb –

pH = 14 – pKb – log

= 14 – (–log 10–10) – log 1

⇒ pH = 4 42. (b) For pure water, [H3O+] = [OH–] ⇒ Kw = 10–6 × 10–6 = 10–12 43. (a) TIPS/Formulae : (i) Lower the oxidation state of central atom, weaker will be oxy acid. (ii) Weaker the acid, stronger will be its conjugate base. Oxidation state of Cl in HClO is + 1, in HClO2 is + 3, in HClO3 is + 5, and in HClO4 is + 7 ∴ HClO is the weakest acid and so its conjugate base ClO– is the strongest Bronsted base. 44 (d) TIPS/Formulae : (i) pH of acid cannot be more than 7.

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(ii) While calculating pH in such case, consider contribution of [H+] from water also. Molar conc. of HCl = 10–8. (given) ∴ pH = 8. But this cannot be possible as pH of an acidic solution can not be more than 7. So, we have to consider [H+] coming from H2O. Total [H+] = [H+]HCl + Ionisation of H2O : H2O H+ + OH– Kw = 10–14 = [H+] [OH–] Let x be the conc. of [H+] from H2O or [H+] = x = [ [H+] = [OH]– in water] ∴ 10–14 = (x + 10–8) (x) or x = 9.5 × 10–8 M [For quadratic equation

]

∴ Total [H+] = 10–8 + 9.5 × 10–8 = 10.5 × 10–8 or pH = – log (10.5 × 10–8) = 6.98 It is between 6 and 7. Trick : pH of acidic solution cannot be more than 7. Thus, only option (d) is possible. 45. (a) NOTE : Acidic buffer is mixture of weak acid and its salt with common anion. (a) CH3COOH + CH3COONH4 is acidic buffer. (b) NH4Cl + NH4OH is basic buffer. (c) H2SO4 + Na2SO4 is not buffer because both the compounds are strong electrolytes. (d) NaCl + NaOH is not buffer solution because both compounds are strong electrolytes. 46. (3) B + HA —→ BH+ + A– Volume of HA used till equivalence point = 6 mL At half of equivalence point, solution will be basic buffer with B and BH+ ∴ pOH = pKb + At half equivalence point [BH+] = [B] ∴ pOH = pKb = 14 – 11 = 3

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47.

(3) KCN, K2CO3 and LiCN are the salts of weak acid and strong base. So, their aqueous solutions turns red litmus paper blue.

48.

(6) Diprotic acids are H2SO4, H3PO3, H2CO3, H2S2O7, H2CrO4 and H2SO3.

49.

(8) pH of sodium salt of weak acid = 50. (9) At the end point, the solution contains only NaA whose concentration is

0.1/2 = 0.05 M Since, the salt NaA is formed by strong alkali (NaOH) and weak acid HA (indicated by its low Ka value), its pH can be evaluated by the following relation.

51.

(2) Given Ka = 1 × 10–5

∴ pKa = 5 The two conditions when colour indicator will be visible are derived by pH = pKa + log (i) pH = 5 + log 10 = 6 (ii) pH = 5 + log 0.1 = 4 Thus, minimum change in pH = 2 52.

(37)

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53. (10.60)

MH2SO4 =

MNaOH =

= 10–3 M = 10–3 M

After neutralisation [OH–] can be calculated as [OH–] = = [OH–] =

× 10–3

× 10–3

pOH = 3.397 pH = 14 – pOH = 14 – 3.397 = 10.603 54. (5.22) No. of moles 3g CH3 COOH =

0.5 mol = 50 m mol

No. of millimoles = Molarity × Volume in mL 250 mL of 0.1 MHCl = 250 × 0.1 = 25 m mol 500 mL solution = 50 m mol CH3COOH 20 mL solution

m mol CH3COOH

500 mL solution contains = 25 m mol HCl

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20 mL solution contains

= 1 m mol HCl

mL of 5M NaOH = HCl + NaOH 1

× 5 = 2.5 m mol NaOH

NaCl + H2O

2.5

1

Remaining NaOH = 2.5–1 = 1.5 m mol CH3COOH + NaOH (remaining) CH3COONa + Water 2 0.5

pH = pKa + log

1.5 0

0 1.5

0 –

= 4.74 + log3

= 4.74 + 0.48 = 5.22 55. (2.0)

56.

(0.20)

[S2–]max to prevent precipitation is given by [S2–]max = = 2.5 × 10–21 M H2S 2H+ + S–2 K1⋅K2 = KNet = 10–21 = [H+]2 =

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[H+] =

57.

M = 0.2 M

(4.47) S =

= = 4.47 × 10–3 M

Alternate solution : Let the solubility of salt AB be x. Ksp = 2 × 10–10 = x(x – y) ..... (i) Association constant of weak acid HA,

Let concentration of HB at equilibrium be y. It is given that pH of solution is 3 which means [H+] = 10–3 ..... (ii)

Putting value of y in equation (i) :

⇒

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⇒ ⇒ ⇒ ⇒ ⇒ Hence, Y = 4.47 58.

(4.86) Amount of SO2 in atmosphere =

= 10 × 10–6

Molar concentration of SO2 present in water = Amount of SO2 × Solubility of SO2 in water = 10 × 10–6 × 1.3653 mole L–1 = 1.3653 × 10–5 M Writing the concerned chemical equation + Initial conc.

1.3653 ×10 –5

Molar conc. at equb. 1.3653 × 10 -5 – x

0

0

x

x

Therefore

(pKa = 1.92, ∴ Ka = 10–1.92)

x2 = 1.2 × 10–2 (1.3653 × 10-5 – x) On solving, x = 1.364 × 10–5 Therefore, pH = – log (1.364 × 10–5) = 4.865 59. (11.30)

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pH of HCl = 2, pH of NaOH = 12 [HCl] = 10–2 M, [NaOH] = 10–2 M [OH–] =

or p[OH] = – log (2 × 10–1)

pOH = 2.6989; pH = 11.3010 p(OH)] 60. (11.5) pKb = 4.70, ∴ Kb = 10–4.7 Now we know that

[pH = 14 –

∴ Now we know that pOH = –log[OH–] pOH = –log 3.158 × 10–3 = 2.5 or, pH = 14 – 2.5 = 11.5 61. (6.5)For ammonium formate which is a salt of weak acid with weak base, we know that = 6.5 62. (78.36) Volume of blood = 10 mL (given) [H2CO3] in blood = 2 M (given) [NaHCO3] to be added = 5 M (given) Let volume of NaHCO3 added in 10 mL blood = V mL ∴ [H2CO3] in blood mixture = [NaHCO3] in blood mixture = ∴ pH = pKa + log or 7.4 = –log 7.8 × 10–7 + log

∴ V = 78.36 mL

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63. 64.

65.

I2. electron acceptors are Lewis acids. amphoteric; because amphoteric substances show properties of both acids and basic.

; Conjugate base = Acid – H+ ∴ Conjugate base of

is

66.

False : AlCl3 is a Lewis acid (although they do not have a proton, aprotic) because it accepts electrons (octet being incomplete). 67. (a) As ester hydrolysis is first order with respect [H+]. RHA = K[H+]HA [ester] RHX = K[H+]HX [ester] ;

= 0.01

68.

(c, d) Any solution of a weak acid and its salt with strong base acts as an acidic buffer solution. If volume of HNO3 solution added is less as compared to that of CH3COONa solution, it results in the formation of an acidic buffer solution. CH3COONa + HNO3 CH3COOH + NaNO3 Excess limiting reagent MV MV′ – – M (V – V′) 0 MV′ MV′ (V′ < V)

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69.

(a, c) A buffer solution is prepared by mixing a weak acid/base with salt of its conjugate base/acid. 70. (b, c) pH of 1 10-8 M is below 7 because it is an acid. H2PO4– + H2O HPO42– + H3O+ H2O + H2O OH– + H3O+ K (Auto protolysis constant of water i.e. ionic product of water) increases with temperature. For half neutralisation of a weak acid by a strong base, pH = pKa + [Salt] = [Acid], pH = pKa 71. (d) p – 1; q – 5; r – 4; s – 1 (p) NaOH + CH3COOH → CH3COONa + H2O m mole 10 × 0.1 20 × 0.1 = 1 m.mol = 2 m.mol ∴ Solution contains 1 m. mol CH3COOH & 1 m.mol CH3COONa in 30 mL solution. It is a Buffer solution. Hence, pH does not change with dilution. (q) NaOH + CH3COOH → CH3COONa + H2O m mole 20 × 0.1 20 × 0.1 = 2 m.mol = 2 m.mol ∴ Solution contains 2 m. mol of CH3COONa in 40 mL solution (salt of weak acid and strong base) For salts of weak acid and strong base : [H+]initial = On dilution upto 80 mL, new conc. will be = ∴ [H+]new = (r) m mole

= [H+]initial × HCl + 20 × 0.1 = 2m.mol

NH3 → 20 × 0.1 = 2 m.mol

NH4Cl

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∴ Solution contains 2 m. mol of NH4Cl in 40 mL solution (salt of strong acid and weak base) For salts of strong acid and weak base [H+]initial = On dilution upto 80 mL, new conc. will be = ∴ [H+]new =

.

=

(s) Ni(OH)2(s)

Ni2+ + 2OH–

it is sparingly soluble salt ∴ On dilution [OH–] conc. in saturated solution of Ni(OH)2 remains constant ∴ [H+]new = [H+]initial 72.

(b)

∴

K = p

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∴

Kp =

Since, P = 2 bar So, 73. As (b) At (c)

(c) (a) Correct statement. on decrease in pressure, reactions moves in direction where no. of gaseous molecules increase. Correct statement the start of reaction, QP < KP so dissociation of X2 take place spontaneousely. ∆G0° > 0 but at start ∆G 1, then ∆G° is negative. But it is given that, ∆G° is positive. (d) KP < 1 and KP = KC (RT)∆n ; ∆n = 1 ⇒ 74.

(a) Let the heat capacity of insulated beaker be C. Mass of aqueous content in expt. 1 = (100 + 100) × 1 = 200 g ⇒ ± Total heat capacity = (C + 200 × 4.2) J/K Moles of acid, base neutralised in expt. 1 = 0.1 × 1 = 0.1 ⇒ Heat released in expt. 1 = 0.1 × 57 = 5.7 KJ = 5.7 × 1000 J ⇒ 5.7 × 1000 = (C + 200 × 4.2) × ∆´T. 5.7 × 1000 = (C + 200 × 4.2) × 5.7

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⇒ (C + 200 × 4.2) = 1000 In second experiment, nCH3COOH = 0.2, nNaOH = 0.1 Total mass of aqueous content = 200 g ⇒ Total heat capacity = (C + 200 × 4.2) = 1000 ⇒ Heat released = 1000 × 5.6 = 5600 J. Overall, only 0.1 mol of CH3COOH undergo neutralization. ⇒ ∆Hneutralization of CH3COOH = = – 56000 J/mol = – 56 KJ/mol. ⇒ ∆Hdissociation of CH3COOH = 57 – 56 = 1 KJ/mol 75. (b) Final solution contain 0.1 mole of CH3COOH and CH3COONa each. Hence, it is a buffer solution. pH = pKa + = 5–log 2 + 76.

(b) Cu2+ ions get precipitated every quickely due to low Ksp value even at very low concentration of S2– ion. Ksp = [Cu2+] [S2–]

Due to high value of Keq, CuS precipitated easily. 77. (c) TIPS/Formulae : Among oxyacids, the acidic character increases with increase in oxidation state of the central atom. O.S. of N in HNO3 = + 5

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O.S.of N in HNO2 = + 3 Thus, HNO3 is stronger acid than HNO2. Hence, assertion is correct. Structure of HNO2 : ; Structure of HNO3 : The assertion is true but the reason is wrong as can be clearly seen from the above structures. 78. BaO > B2O3 > CO2 > SO3 > Cl2O7 Oxide basicity decreases in a period but increases in a group. 79. (i) The volume being doubled by mixing the two solutions, the molarity of each component will be halved i.e. [CH3COOH] = 0.1 M, [HCl] = 0.1 M. NOTE : HCl being a strong acid will remain completely ionised and hence H+ ion concentration furnished by it will be 0.1 M. This would exert common ion effect on the dissociation of acetic acid, (a weak acid). CH3COOH At start C At equilibrium C (1 – α)

0 Cα

0 Cα + 0.1

Since α is very very small, Cα2 can be neglected and (1 – α) can be taken as unity ∴ or [H+]Total = 0.1 + Cα, Cα is negligible as compared to 0.1. ∴ [H+]Total = 0.1 ∴ pH = 1 (ii)

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0.1 mole of NaOH will be consumed by 0.1 mole of HCl. Thus, 0.05 mole of NaOH will react with acetic acid according to the equation. CH3COOH + NaOH CH3COONa + H2O Initial moles 0.1 mol 0.05 mol 0 0 At equilibrium 0.05 mol 0 mol 0.05 mol 0.05mol

Thus, solution of acetic acid and sodium acetate will become acidic buffer. So, pH of the buffer will be

80. (given)

+ 6.7 × 10 –6M

6.7 × 10 –6M

2 × 6.7 × 10 –6M

= (6.7 × 10– 6) (2 × 6.7 × 10– 6)2 = 1.203 ×10–15 The buffer solution pH = 8 (given) pOH = 6 or [OH–] = 10–6 Thus, in this buffer we have, [Pb2+][OH–]2 = 1.203 or [Pb2+] × [10–6]2 = 1.203 10–15 [Pb2+] = 1.203 10–3 mol litre–1 81.

NH3 + H2O

NH4+ + H2O

NH4+ + OH–; Kb = 3.4

NH4OH + H+ ; Ka = 5.6

Kbase = Kf = 6.07 82.

H2O < CH3 –

10–15

1010

10-10

or 105 < –OH < –OCH3

Weaker the base, stronger is its conjugate acids H3O+ > CH3 H2 > H2O > CH3OH (Decreasing acidic order of the conjugate bases.) 83. Case I. Write the concerned chemical reaction

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BOH Moles before reaction Moles after reaction

x

+

HCl

—→

0.1 × 5 = 0.5

(x – 0.5)

BCl

0

0

+ H2O

0 0.5

0.5

Since, the solution represents a basic buffer, following Hendersen equation can be applied. pOH= –log Kb + log 14 – 10.04 = –log Kb + log

...(i)

Case II. BOH Moles at start Moles after adding 20 ml.

x

+

HCl

—→

0.1 × 20 = 2

(x – 2)

0

0

BCl + H2O 0

2

2 of 0.1N HCl

0

Again the solution is acting as basic buffer ∴ pOH = –log Kb + log 14 – 9.14 = –log Kb + log

...(ii)

Divide (i) by (ii), =

⇒ x = 0.088 mol L–1

Substituting x in (i) and solving for Kb 3.96 = –log Kb + log 84.

Kb = 1.828 × 10–5 Case I. CH3COOH

CH3COO– + H+

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At start At equib.

1 1–α

[H+] = cα = c ∴ [H+] =

0 α

0 α

= = 4.24 × 10–3 M

Thus pH = – log [H+] = –log 4.24 × 10–3 = 2.3724 Case II. pH after dilution = 2 × original pH = 2 × 2.3724 = 4.7448 Let conc. after dilution = c1 and degree of dissociation = α1 Since pH = – log [H+] 4.7448 = – log [H+] [H+] = 1.8 × 10–5 = c1α1 ∴ c1α1 = 1.8 × 10–5 Dissociation constant Since Ka = = 1.8 × 10–5 =

= ∴ α1 = 0.5

Substituting the value of α1 in the following relation c1α1 = [H+]; c1 × 0.5 = 1.8 × 10–5 c1 =

= 3.6 × 10–5 M

Since the number of moles of CH3COOH before and after dilution will be same ∴ Mole of CH3COOH before dilution = Mole of CH3COOH after dilute ( Mole = M × Vin litre) 1 × 1 = 3.6 × 10–5 × V ⇒ V = 2.78 × 104 litres

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85.

TIPS/Formulae :

pH = pKa + log If x moles of HCl are added then they will combine with NaCN to form x moles of very weak acid HCN. NaCN + HCl → NaCl + HCN At equilibrium :

(0.01–x)

x

x

x

For an acidic buffer, pH = – log Ka + log

or log

8.5 = (10 – log 4.1) + log = –0.8872

[log 4.1 = 0.6128]

= 0.1296 x = 8.85 × 10–3 M = 8.85 × 10–3 moles of HCl 86. (i) Amount of HCl added = 0.20 mole [H+] = 0.2 g litre–1 NOTE : Added H+ ions will combine with the acetate ions forming acetic acid with the result, concentration of acetate ions will decrease while that of acetic acid will increase. CH3COOH + Cl– CH3COO – + HCl before reaction after reaction

1 0.8

0.2 0

1 1.2

0 0.2

∴ Concentration of acetate ions after adding 0.20 mole of HCl. [CH3COO–] = 1.0 – 0.2 = 0.8 mole Similarly, concentration of acetic acid, [CH3COOH] = 1.0 + 0.2 = 1.2 mole

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Now, pH = –log Ka + log = –log 1.8 × 10–5 + log pH = 4.7447 + 0.3010 – 0.4771 = 4.5686 (ii) Amount of HCl added = 0.20 mole Out of 0.2 mole of [H+] added, 0.1 mole will combine with 0.1 mole of CH3COO– forming 0.1 mole of CH3COOH. CH3COO– + H+ → CH3COOH + Cl– before reaction after reaction

0.1 0

0.2 0.1

0.1 0.2

0 0.1

∴ Total concentration of acetic acid [CH3COOH] = 0.1 + 0.1 = 0.2 mole In presence of [H+], CH3COOH will not ionize. Therefore, pH of the solution will be due to the presence of H+ of HCl, i.e. 0.2 – 0.1 = 0.1 mole HCl pH = –log [H+] = –log [0.1] = 1 87. TIPS/Formulae : For acidic buffer Calculation of concentration of HCOOH. Here, c = 0.2 M; [H+] = 6.4 × 10–3 [H+] = cα or α =

⇒α=

= 3.2 × 10–2

NOTE : Thus the degree of dissociation of HCOOH is very low which on addition of sodium formate is further suppressed due to common ion effect. Since the degree of dissociation is very low (3.2 × 10–2), it can be neglected and hence [HCOOH] can be taken as0.2 M. Calculation of concentration of HCOO–, [HCOO–] It can be obtained in the following manner : HCOONa HCOO– + Na+ At start

1

0

0

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At equb.

0.25

0.75

0.75

∴ [HCOO ] = 0.75 –

For acidic buffer pH = – log Ka + log = – log 2.4 × 10–4 + log 88.

= 4.19

Among oxyacids of the same element, acidic nature increases with its oxidation number, e.g., HOCl < HOClO < HOClO2 < HOClO3 O.N. of Cl

+1

+3

+5

+7

89.

N2 < O2 < F2 < Cl2 i.e., NOTE : As the number of bonds increases the bond length decreases. So N2 < O2 < halogens. Among F2 and Cl2, bond length of Cl2will be higher because of higher atomic radii. 90. (i) From the dissociation of weak acid HA, HA H+ + A– It α is the degree of ionization of the acid HA, then [H+] = 0.1 α [ the acid is decimolar] [A–]= 0.1 α; [HA] = 0.1 (1 – α) Therefore, Ka = =

= (since acid is weak, 1 – α = 1)

Ka = 0.1 α2 or 4.9 × 10–8 = 0.1 α2 or α2 =

or α = 7 × 10–4

∴ Percentage ionization = 100 × 7 × 10–4 = 7 × 10–2% (ii) Calculation of pH [H+] = 0.1α = 0.1 × 7 × 10–4 mole/litre = 7 × 10–5 mole/litre

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Now since pH = – log [H+] = – log [7 × 10–5] = 5 – log 7 = 5 – 0.8451 = 4.1549 (iii) Concentration of OH– in decimolar solution [H+] = 7 × 10–5 mole per litre Now, Kw = [H+] [OH–] or 1.0 × 10–14 = 7 × 10–5 × [OH–] ∴ [OH–] =

= 1.43 × 10–10 mole per litre

91. TIPS/Formulae : (i) Find the moles of each species after reaction. (ii) pH = – logKa + Given, NaOH 0.2 M, 20 mL; CH3COOH 0.2 M, 50 mL Ka = 1.8 × 10–5 V of 0.2M NaOH required to make pH = 4.74 =? From the chemical reaction

It is evident that 70 mL of the product will contain (i) 30 mL of 0.2 M unused CH3COOH [unused CH3COOH = 50 – 20 = 30 mL] (ii) 20 mL of CH3COONa. ∴ No. of moles of CH3COOH in solution =

× 30 = 0.006 mole

Similarly, No. of moles of CH3COONa solution =

× 20 = 0.004 moles

pH = –log Ka + log Substituting the values of the various values pH = –log 1.8 × 10–5 + log

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= 4.7447 – 0.1761 = 4.5686 Calculation of the additional volume of 0.2 M NaOH required to make pH of solution 4.74. pH = –log Ka + log or 4.74 = –log 1.8 × 10–5 + log or 4.74 = 4.7447 + log ∴ log

= 0.0047 or

=

NOTE THIS STEP : Let x mL. be the volume of additional 0.2 M NaOH added to make the pH of the solution 4.74. This will further neutralise x mL of 0.2 M CH3COOH and produce x mL of 0.2 M sodium acetate. The resulting solution (70 + x) mL will now contain (i) (30 – x) mL of 0.2 M acetic acid. (ii) (20 + x) mL of 0.2 M sodium acetate. Number of moles of acetic acid in (70 + x) mL. solution =

× (30 – x) = 2 × 10–4 (30 – x)

Number of moles of CH3COONa in (70 + x) mL. solution =

× (20 + x) = 2 × 10–4 (20 + x)

Therefore,

= =

= or 20.22 + 1.011x = 30 – x

or 1.001 x + x = 30 – 20.22; 2.011x = 9.78 or x = 4.86 Therefore, the additional volume of 0.2 M NaOH required to make the pH of the solution 4.74 is 4.86 mL. 92. Suppose the number of moles of sodium propionate = x

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Then pH = pKa + log 4.75 = –log (1.34 × 10–5) + log or 4.75 = 5 – 0.1271 + log or 4.75 = 4.8729 + log log

= –0.1229

or

= Antilog [–0.1229] or

= 0.7536

x = 0.7536 × 0.02 = 1.5072 × 10–2 mol HCl H+ + Cl– 0.01 mole

0.01 mole

When 0.01 mole of HCl is added, there is (0.01 + 0.02) M of propionic acid and (0.015 – 0.010) M of propionate. Therefore, pH = –log (1.34 × 10–5) + log

= 4.09

The pH of a 0.010 molar HCl solution = –log 10–2 = 2 93. Phenolphthalein indicates half neutralization. Na2CO3 + H+

NaHCO3 + Na+

...(i)

Methyl orange indicates complete neutralisation NaHCO3 + H+

Na+ + H2O + CO2

...(ii)

∴ Volume of 0.1M H2SO4 required for complete neutralisation = 2 × 2.5 = 5.0 mL 0.1 M H2SO4 ≡ 0.2 N H2SO4 [For H2SO4 molarity = 2 × normality] ( Mol. wt. of H2SO4 = 98, and eq. wt. of H2SO4 = 49) ∴ 0.2 M H2SO4 ≡ 0.4 NH2SO4 N1 = normality of Na2CO3, V1 = volume of Na2CO3= 10 mL,

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N2 = normality of H2SO4 = 0.2, V2 = volume of H2SO4= 5.0 mL ∴ N1V1 = N2V2 ⇒ N1 × 10 = 0.2 × 5 ∴ N1 =

= 0.1 N

∴ Eq. wt. of Na2CO3 =

× molecular weight =

= 53

Strength of Na2CO3 = 53 × 0.1 = 5.3 g/L [ strength = normality × Eq. wt] For neutralization with methyl orange, volume of 0.2 M H2SO4 used = 2.5 mL = 2.5 mL of 0.4 N H2SO4 = 5 mL of 0.2 N H2SO4 [ N1V1 = N2V2] From 5 mL of 0.2 N H2SO4, 2.5 mL is used for neutralising NaHCO3 formed during first half neutralization Na2CO3. ∴ Volume of 0.2N H2SO4 used for neutralisation of NaHCO3 present in original solution = 5.0 – 2.5 = 2.5 mL ∴ N1V1 = N2V2 where N1 = Normality of NaHCO3, N2 = Normality of H2SO4= 0.2, V1 = Volume of NaHCO3 = 10 mL, V2 = Volume of H2SO4= 2.5 mL N1V1 = N2V2 ⇒ N1 × 10 = 0.2 × 2.5 N1 =

= 0.05 N

Eq. wt. of NaHCO3 = 84 ∴ Strength of NaHCO3 = 84 × 0.05 = 4.2 g/L

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1.

(a) In H3PO4 oxidation state of P is +5, which cannot be oxidised further to a higher oxidation state. Hence, it cannot act as reducing agent. 2. 3O2

(d) N2 + O2 2O3

2NaOH + H2SO4

2NO (Redox reaction) (Photochemical reaction)

Na2SO4 + 2H2O (Neutralisation reaction)

[Co(H2O)6]Cl3 + 3AgNO3

[CO(H2O)6](NO3)3 + 3AgCl (Neutralisation reaction)

3.

(a) In the reaction

4.

(b) Copper when exposed to moist air having CO2. It gets superficially coated with a green layer of basic carbonate CuCO3. Cu (OH)2. I– ( I2 is weakest oxidising agent) True : Cu+ is the intermediate oxidation state between Cu++ and Cu. If the reduction potential from the intermediate oxidation state to the lower one is more positive than from the higher to the intermediate, then the intermediate state will undergo disproportionation. Cu+ Cu Cu++

5. 6.

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7.

False : Copper metal does not reduces Fe2+ in an acidic medium, because the E° value for Cu is more. 8. (A - p, s); (B - r); (C - p, q); (D - p). is – A→ p, s; The reaction is redox reaction because the O.N. of O in 0.5 and that in O2 is zero. In

is –1.0. It involves reduction

oxidation reaction. Since here a part of molecule is oxidised and a part is reduced so it is disproportionation. B → r; The structure of Cr2O72– is given below

[NOTE : In any solution dichromate ions and chromate ions exist in equilibrium. In alkali solution, dichromate ions are converted into chromate ions and on acidification chromate ions are converted back into dichromate ion.] C → p, q; The reaction is In involves change in O.N of Mn from + 7 (in MnO4– ) to+ 2(in Mn2+) So, Mn is reduced and NO2– is oxidised to NO3–; it is a redox reaction. The structure of NO3– (one of the products) is trigonal planar. D → p, It is a redox reaction.

1.

(b) K2O : 2x – 2 = 0 x = +1 K2O2 : 2x – 2 = 0 x = +1 KO2 : x – 1 = 0 x = +1 Thus, potassium shows +1 state in all its oxides, superoxides and peroxides.

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2.

(d)

It is an example of disproportionation reaction, as Cu undergoes both oxidation and reduction. 3.

(a) MnO–4 + 5e–

(i)

FeC2O4

Mn2+

Fe3+ + 2CO2 + 3e– mole of acidified KMnO4

1 mole of FeC2O4 reacts with (ii) Fe2(C2O4)3

Fe3+ + CO2 + 6e–

1 mole of Fe2(C2O4)3 reacts with (iii) FeSO4

moles of KMnO4

Fe3+ + e–

1 mole of FeSO4 react with

moles of KMnO4

(iv) Fe2 (SO4)3 does not oxidise Total moles required =

=2

4.

(c) [Cr (H2O)6] Cl3 ⇒ x + 6 × 0 + (– 1) × 3 = 0 ⇒ x = +3 [Cr (C6H6)2] x+2×0=0;x=0 K2 [Cr (CN)2 (O)2 (O2) (NH3)] [Here (O)2 is OXO, (2x – 2) and (O2) is per OXO, (1x – 2)] 2 × 1 + x + 2 × (– 1) + 2 × (– 2) + (– 2) + 0 = 0 x=+6 5. (d) CrO2 Cl2 Let O. No. of Cr = x ∴ x + 2 (–2) + 2 (–1) = 0 x–4–2=0

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∴ x=+6 6. (a) The reaction given is Cr3+ + Fe3+ + CO2 Cr2O72– + Fe2+ + C2O42– 2Cr3+ Cr2O72– On balancing 2Cr3+ + 7H2O ......(i) 14H+ + Cr2O72– + 6e– Fe3+ + e– ......(ii) Fe2+ 2CO2 + 2e– .......(iii) C2O42– On adding all three equations, we get Cr2O72– + Fe2+ + C2O2–4 + 14H+ + 3e– 2Cr3+ + Fe3+ + 2CO2 + 7H2O Hence the total no. of electrons involved in the reaction = 3 7.

(a)

First half reaction

.... (i)

On balancing .... (ii) Second half reaction .... (iii) On balancing .... (iv) On multiplying eqn. (ii) by 5 and (iv) by 2 and then adding we get

8.

(c) In SO23– x + 3(– 2) = – 2; x = + 4 In S2O24– 2x + 4(– 2) = – 2

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2x – 8 = – 2 2x = 6; x = + 3 In 2x + 6(– 2) = – 2 2x = 10; x = + 5 hence the correct order is S2O24– < SO2–3 < S2O26– 9. (d) The following reaction occurs: + 7H2O From the above equation, we find that Mohr's salt (FeSO4.(NH4)2SO4.6H2O) and dichromate reacts in 6 : 1 molar ratio. 10. (b) TIPS/Formulae : The highest O.S. of an element is equal to the number of its valence electrons (i) [Fe(CN)6]3–, O.N. of Fe = + 3, [Co(CN)6]3–, O.N. of Co = + 3 (ii) CrO2Cl2, O.N. of Cr = +6, (Highest O.S. of Cr) [MnO4]– O.N of Mn = + 7 (Highest O.S. of Mn) (iii) TiO3, O.N. of Ti = + 6, MnO2 O.N. of Mn = + 4 (iv) [Co(CN)6]3–, O.N. of Co = + 3, MnO3, O.N. of Mn = + 6 11. (c) TIPS/Formulae : (i) In a disproportionation reaction same element undergoes oxidation as well as reduction during the reaction. (ii) In decomposition reaction a molecule breaks down to more than one atoms or molecules It is disproportionation reaction because Cl is both oxidised (+ 1 to + 5) and reduced ( + 1 to – 1) during reaction. 12. (d) TIPS/Formulae : (i) In an ion sum of oxidation states of all atoms is equal to charge on ion and in a compound sum of oxidation states of all atoms is always

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zero. Oxidation state of Mn in MnO4– = + 7 Oxidation state of Cr in Cr(CN)63– = + 3 Oxidation state of Ni in NiF62– = + 4 Oxidation state of Cr in CrO2Cl2 = + 6 13. (a) TIPS/Formulae : (i) Oxidation state of element in its free state is zero. (ii) Sum of oxidation states of all atoms in compound is zero. O.N. of S in S8 = 0; O.N. of S in S2F2 = + 1; O.N. of S in H2S = –2; 14. (a) TIPS/Formulae : Balance the reaction by ion electron method. → 2CO2 + 2e–] × 5 Oxidation reaction : Reduction reaction : + 8H+ + 5e– → Mn2+ + 4H2O] × 2 2

Net reaction : + 16H+ + 5

→ 2Mn2+ + 10CO2 + 8H2O

15. (b) TIPS/Formulae : (i) Write balance chemical equation for given change. (ii) Identify most electronegative element and find its oxidation state. BaO2 + H2SO4 → BaSO4 + H2O2 Oxygen is the most electronegative element in the reaction and has the oxidation states of –1 (in H2O2) and –2 (in BaSO4). In H2O2, peroxo ion is present. 16. (c) 2 + 2 (2 + x – 4) = 0 [ Ba(H2PO2)2 is neutral molecule] or 2x – 2 = 0 ⇒ x = + 1 17. (b) TIPS/Formulae : Sum of oxidation state of all atoms in neutral compound is zero. Let the oxidation state of iron in the complex ion be x; then [Fe(H2O)5(NO)]2+ x + 5 × 0 + 0= + 2. ∴ 18. (c) NOTE :

x=+2

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The sum of oxidation states of all atoms in compound is zero. Calculation of O.S. of C in CH2O. x + 2 + (–2) = 0 ⇒ x = 0 19. (c) TIPS/Formulae : (i) Find oxidation state of N in N2H4. (ii) Find change in oxidation number with the help of number of electrons given out during formation of compound Y. N2H4 → Y + 10 e–, Calculation of O.S. of N in N2H4 : 2x + 4 = 0 ⇒ x = –2 The two nitrogen atoms will balance the charge of 10 e. Hence, oxidation state of N will increase by +5, i.e. from –2 to +3. 20. (6) The oxidation states of iron in these compounds will be In A, x + 5(–1) + (–1) = –4 ⇒ x = +2 In B, y + 4(–2) = –4 ⇒ y = +4 In C, z = 0 The sum of oxidation states will be = 4 + 2 + 0 = 6.

21.

(5)

Difference in oxidation number = 5 – 0 = 5 22. (19)

23.

(19) Compound

Oxidation state of transition element

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(i) K2Cr2O7 x=+6 (ii) KMnO4 y=+7 (iii) K2FeO4 z=+6 So, (x + y + z) = 6 + 7 + 6 = 19. 24. NOTE : Sum of oxidation states of all atoms (elements) in a neutral compound is zero. TIPS/Formulae : As YBa2Cu3O7 is neutral. (+ 3) + 2 (+ 2) + 3 (x) + 7 (–2) = 0 or 3 + 4 + 3x – 14 = 0 ⇒ 3x + 7 – 14 = 0

or

x=

25. (a, b, d) Balanced chemical equation is

26. HI < I2 < ICl < HIO4; O.N. of I in I2 = 0, HI = –1, ICl = +1, HIO4 = +7. 27. TIPS/Formulae : Balance the atoms as well as charges by ion electron/oxidation number method. While balancing the equations, both the charges and atoms must balance. (i) 6Ag++ AsH3 + 3H2O → 6Ag + H3AsO3 + 6H+ (ii)

+ 6I– + 6H2SO4 → Cl– + 6

+ 3I2 + 3H2O

+ 3H2O

(iii) 4S + 6OH– → 2S2– +

+ 2H2O + 5Pb2+

(iv) 2Mn2+ + 5PbO2 + 4H+ → 2

Cl– + ClO– + H2O

(v) Cl2 + 2OH– (vi) 2Ce3+ +

2

(vii) 2HNO3 + 6HCl

2NO + 3Cl2 + 4H2O

(viii)

+ 2Ce4+

+ 3C2H4O + 8H+ 3C2H4O2 + 2Cr3+ + 4H2O

(ix) 4Zn +

+ 10H+

4Zn2+ +

+ 3H2O

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28.

Balance the reactions by ion electron method. (i) Cu2O + 2H+ → 2Cu2+ + H2O + 2e–] × 3 + 4H+ + 3e– → NO + 2H2O] × 2 3Cu2O + 14H+ + 2

......(1) .....(2)

→ 6Cu2+ + 2NO + 7H2O

(ii) K4[Fe(CN)6] + 6H2SO4 + 6H2O → 2K2SO4 + FeSO4 + 3(NH4)2SO4 + 6CO (iii) C2H5OH + 4I2 + 8OH– →CHI3 +

1.

+ 5I– + 6H2O

(b) Higher the reduction potential, higher will be oxidising power. So,

Bi3+ < Ce4+ < Pb4+ < Co3+ 2. (c) More positive is the reduction potential stronger is the oxidising agent. Reduction potential is maximum for S2O82–, therefore, it is the strongest oxidising agent amongst the given species. 3. (b) Higher the oxidation potential, higher will be the reducing power. So, the order of reducing behaviour is: Ca > Mg > Zn > Ni 4. (a) Reaction involved:

The number of electrons involved in producing one mole of CO2 is 5.

1. (d)

(a)

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(b)

(c)

(d) Notice that the oxidation state of oxygen goes from –1 on the H2O2 to –2 on the H2O means H2O2 is being reduced. On the other hand the oxidation state of sulfur is going from –2 on the PbS to +6 on the PbSO4. i.e Sulfur is being oxidised. 6.

(c)

Hence H2SO3 is the reducing agent because it undergoes oxidation. 7. (a) If an electronegative element is in its lowest possible oxidation state in a compound or in free state. It can function as a powerful reducing agent. e.g. I –

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1.

(d) Dihydrogen of high purity (> 99.95%) is obtained by the electrolysis of Ba(OH)2 using Ni electrodes.

2.

(d)

This reaction is called water gas shift reaction. 3. (b) Zn + 2NaOH Na2ZnO2 + H2 Zn + 2HCl ZnCl2 + H2 NaOH and HCl reacts with a certain amount of zinc to produce equal number of moles of H2. 1 2 3 4. (a) 1H 1H(D) 1H(T) Number of neutron 0 1 2 (x) (y) (z) Total number of neutrons in three isotopes of hydrogen = 0 + 1 + 2 = 3 5. (c) Zn + 2NaOH ¾®Na2ZnO2 + H2 Zn + H2SO4 ¾® ZnSO4 + H2 Zn is an amphoteric element. 6. (a) (i) Saline hydrides with water produces H2 gas. (ii) (iii) PH3 is electron rich whereas CH4 is electron precise hydride. (iv) HF and CH4 are molecular hydrides as they are covalent molecules. 7. (c) NaH is an ionic hydride which is also known as saline hydride. 8. (a) There are three isotopes of H out of which only tritium is radioactive, which emits low energy – particles. Its half life is 12.33 years. 9. (c) Hydrogen has three isotopes: Protium (1H1), deuterium (1H2) and tritium (1H3). 10. (d) H2 is a highly inflammable gas.

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11. (a) Both S & E are true and E is the correct explanation of S.

1. 2. 3. 4.

5.

(a) Clark's method is used to remove temporary hardness, using lime water (or) Ca(OH)2 from water. (a) In distilled water, there are only neutral water molecules therefore, it does not conduct electricity. (b) Synthetic resin method is more efficient than zeolite process as it can exchange both cations as well as anions. (a) Temporary hardness is caused by bicarbonates of calcium and magnesium. On boiling following changes occurs,

(d) When steam is passed over red hot coke, an equimolar mixture of CO and H2 is obtained.

The gaseous mixture thus obtained is called water gas or syn gas (synthesis gas). 6. (c) In CuSO4·5H2O, four H2O molecules are directly coordinated to the central metal ion while one H2O molecule is hydrogen bonded. 7.

(d) Moles of Ca(HCO3)2 =

Moles of Mg(HCO3)2 =

= 0.005

= 0.005

Hardness in terms of CaCO3 in ppm =

= 104 ppm

8.

(d) 10–3 M CaSO4 ≡ 10–3 M CaCO3 ⇒ 10–3 M CaCO3 means 10–3 moles of CaCO3 are present in 1 L Molar mass of CaCO3 = 40 + 12 + 48 = 100 g/mol

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10–3 mol = W = 10–3 × 100 g = 100 mg i.e. 100 mg of CaCO3 is present in 1 L solution. Hardness of water = Number of milligram of CaCO3 per litre of water. ∴ Hardness of water = 100 ppm 9. (c) Only bicarbonates cause temporary hardness, whereas chlorides and sulphates cause permanent hardness. 10. (a) There is extensive intermolecular hydrogen bonding in the condensed phase instead of intramolecular H-bonding. 11. (b) Heavy water acts as moderator. This is used in nuclear reactors to slow down the speed of fast moving neutrons. (bicarbonate) 12. (b) Only temporary hardness which is due to ions is removed by boiling. 13. (b) Heavy water is D2O hence number of electrons = 2 + 8 = 10 number of protons = 10 Atomic mass of D2O = 4 + 16 = 20, hence number of neutron = atomic mass – number of protons = 20 – 10 = 10 14. (c)

15. 16.

(c) Heavy water is D2O, deuterium oxide. (b) Temporary hardness of water is due to presence of bicarbonates of Ca and Mg and it is removed by adding Ca(OH)2 to hard water and precipitating these soluble bicarbonates in the form of insoluble salts. CaCO3↓ + 2H2O Ca(HCO3)2 + Ca(OH)2 Mg(HCO3)2 + 2Ca(OH)2 2CaCO3↓ + Mg(OH)2↓ + 2H2O 17. (100.00) 10–3 molar MgSO4 10–3 moles of MgSO4 present in 1 L solutions.

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10–3 M MgSO4 10–3 M CaCO3 10–3 M CaCO3 = 10–3 × 100g CaCO3 in 1L water ppm(in term of CaCO3) = ppm(in term of CaCO3) = 100 ppm 18.

(b, c, d) Temporary hardness is due to bicarbonates of calcium and magnesium. Temporary hardness can be removed by Clark’s process, which involves the addition of slaked lime, Ca(OH)2. Washing soda (Na2CO3) removes both the temporary and permanent hardness by converting soluble calcium and magnesium compounds into insoluble carbonates. 2CaCO3 ↓ + 2H2O Ca(HCO3) + Ca (OH)2 CaCO3 ↓ + 2NaHCO3 Ca(HCO3) + Na2CO3 NaOCl can remove the hardness as : 2HOCl + 2OH– 2OCl– + 2H2O CaCO3 ↓ + + 2H2O Ca (HCO3)2 + 2OH– H2CO3

19.

(a)

20.

(b, d) Na2 Al2Si2O8 . xH2O + Ca2+

→ CaAl2Si2O8 . xH2O + 2Na+ Na2 Al2Si2O8 . xH2O + Mg2+ → MgAl2Si2O8 . xH2O + 2Na+

1. 2.

(a) H2O2 has open book like structure, which is non-planar. It is a colourless viscous liquid but in large quantity appears blue in colour. (a) For H2O2

Molarity Molarity

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3. (a) All the statements are correct. 4. (b) 11.2 V strength of H2O2 means, 11.2 L of O2 is liberated at STP.

11.2 L of O2 at STP = 0.5 mol  No. of moles of H2O2 = 1 mol i.e., 1 L of given H2O2 solution has 1 mole of H2O2 (i.e., 34 g) Strength =

× 100 = 3.4%

5. 6.

(c) Volume strength = 11.35 × molarity = 11.35 (d) H2O2 acts as oxidising agent as well as reducing agent in both acidic and basic medium. H2O2 acts as oxidant: H2O2 + 2H+ + 2e– → 2H2O (acidic medium) H2O2 + 2e– → 2OH– (basic medium) H2O2 acts as reductant:H2O2 → 2H+ + O2 + 2e– (acidic medium) H2O2 + 2OH– → 2H2O + O2 + 2e– (basic medium) 7.

(c) [Fe(CN)6]4– +

[Fe(CN)6]3– + 8. 9.

H2O2 + H+ → [Fe(CN)6]3– + H2O

H2O2 + OH– → [Fe(CN)6]4– + H2O +

O2

(c) H2O2 has oxidizing and reducing properties both. (b) The reducing agent loses electron during redox reaction i.e. oxidised itself.

(i)

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(ii) (iii) (iv) 10.

(a) H2O2 acts as a reducing agent only in presence of strong oxidising agents (i.e., MnO4–) in acidic as well as alkaline medium.

11. (a) KIO4 + H2O2 → KIO3 + H2O + O2 Thus, H2O2 is acting as a reducing agent 2NH2OH + H2O2 → N2 + 4H2O Here H2O2 is acting as an oxidising agent 12. (b) Volume strength = Normality × 5.6 = 1.5 × 5.6 = 8.4 L 13.

(100) Molarity of H2O2 solution

Volume strength = 8.9 × 11.2 = 99.68 V 100 V 14. (1.344) Meq. of H2O2 = Meq. of Na2S2O3 = Meq. of I2 = Meq.of KI w = 0.102 g (equating Meq. in 25 mL solution)

15.

(4.48)

i.e. 254 g of I2 is released by 34 g H2O2

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∴ 0.508 g of I2 will be released by

5 mL of H2O2 sol. contains 0.068 g of H2O2. ∴ 1 mL of H2O2 sol contains NOTE : The strength of H2O2 is generally calculated in terms of volume strength. According to which, 10 volume of H2O2 means that 1 mL of H2O2 sol gives 10 mL of O2 at STP.

or or

i.e., 68 g of H2O2 gives 22,400 mL of O2 at STP 1 mL of H2O2 sol of H2O2 gives

or 1 mL of H2O2 sol gives 4.48 mL of O2 i.e. strength of H2O2 sol is 4.48 volumes. 16. When H2O2 acts as oxidising agent, following reaction takes place: 2OH– H2O2 + 2e– While regarding its action as reducing agent, the following reaction takes place: O2 + 2H2O + 2e– H2O2 + 2OH– Example of oxidising character of H2O2 in alkaline medium Here, Cr3+ (Cr is a first row transition metal) is oxidised to Cr6+. Example of reducing character of H2O2 in alkaline medium : Here, Fe3+ (Fe is a first row transition metal) is reduced to Fe2+. K2SO4 + 2MnSO4 + 3H2O + 5 [O] 17. 2KMnO4 + 3H2SO4

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18.

(i) The oxidation state of oxygen in H2O2 (i.e. –1) can be changed to 0 or –2 i.e. oxygen in H2O2 exists in an intermediate oxidation state with respect to O2 and O2–. Hence, it acts both as an oxidising and reducing agent. (ii) H2O2 is a better oxidising agent than H2O because oxidation number of oxygen in H2O2 is –1 and that in water it is –2. So, H2O2 easily reduces to –2 oxidation number. 19. The mixture of N2H4 and H2O2 (in presence of Cu (II) catalyst) is used as a rocket propellant because the reaction is highly exothermic and large volume of gases are evolved, which can propel a rocket. N2(g)↑ + 4H2O(g)↑ N2H4 + 2H2O2

Hydrazine Hydrogen peroxide

20.

Ferricyanide is oxidised to ferrocyanide on treatment with alkali 2K3[Fe(CN)6] + H2O2 + 2KOH → 2K4[Fe(CN)6] + 2H2O + O2

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1. 2.

3.

(b) On heating in excess air Li form oxide sodium form peroxide while K, Rb, Cs form superoxide. (d) Cesium has lowest ionisation enthalpy and hence it shows photoelectric effect to the maximum extent. So, it is used in photo electric cell. (d) Li and Mg do not form solid bicarbonate, but react with N2 to give nitrides.

4. (b) (a) Pb3O4 is insoluble in water or do not react with water. (b) 2KO2 + 2H2O → 2KOH + H2O2 + O2 (c) Na2O2 + 2H2O → 2NaOH + H2O2 (d) Li2O2 + 2H2O → 2LiOH + H2O2 5.

(c) Lithium nitrate decomposes into its oxide on heating.

6.

(a) Na2CO3.10H2O → Na2CO3.H2O + 9H2O (X)

(Y)

Na2CO3 .H2O → Na2CO3 + H2O (Y)

(Z)

X = Washing soda Z = Soda ash

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7.

(d) Hydration energy is inversely proportional to the size of ion.

Li+ < Na+ < K+ < Rb+ < Cs+

Size

Li+ > Na+ > K+ > Rb+ > Cs+

Hydration energy

8.

(b)

9.

(c) K+ ions are the most abundant cations within cell fluids, where they activate many enzymes, participate in the oxidation of glucose to form ATP and, along with sodium ions, they are responsible for the transmission of nerve signals. 10. (b) Amongst the given alkali metals, only lithium can react with N2 in air to form lithium nitride. LiCl + AlCl3 + SiH4 11. (b) SiCl4 + LiAlH4 12. (c) Compounds Nature KO2 Superoxide BaO2 Peroxide SiO2 Oxide CsO2 Superoxide 13. (b) 4 Li + O2 → 2Li2O Lithium monoxide 2Na + O2 → Na2O2 Sodium peroxide K + O2→ KO2 Potassium superoxide 14. (d) Covalent character increases, when the cation has small size and high charge density. Among all these, Mg2+ size is smallest, so MgCl2 tends to be more covalent in nature. 15. (d) Na2O2 is the peroxide of sodium not super oxide. The formula of sodium superoxide is NaO2. 16.

(b)

IE1 = 5.1 eV

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(because the reaction is reverse) 17. (d) Higher the lattice enthalpy, lower will be solubility i.e., lattice enthalpy Since the lattice enthalpy of alkali metals follow the order Li > Na > K > Rb Hence the correct order of solubility is LiF < NaF < KF < RbF 18. (b) Oscillation of loose electrons. 19. (a) NOTE : Acidic and basic salts cannot exist together. Since NaHCO3 is an acid salt of H2CO3, it reacts with NaOH to form Na2CO3 and H2O. NaHCO3 + NaOH → Na2CO3 + H2O 20. (b) Na2O2 + H2SO4 (20% ice cold) → Na2SO4 + H2O2 21. (d) The free ammoniated electrons make the solution of Na in liquid NH3 a very powerful reducing agent. NOTE : The ammonical solution of an alkali metal is rather favoured as a reducing agent than its aqueous solution because in aqueous solution the alkali metal being highly electropositive evolves hydrogen from water (thus H2O acts as an oxidisng agent) while its solution in ammonia is quite stable, provided no catalyst (transition metal) is present. 22. Hydrogen, anode.

23. 24.

of solvated electrons. False : Sodium when burnt in excess of oxygen gives monoxide and sodium peroxide (Na2O2) and not sodium oxide. 4Na + O2 → 2Na2O; 2Na + O2 → Na2O2 25. True : The metallic bonding decreases with increase in atomic size and thus the tendency to show metallic bonding among alkali metals

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26. 27.

decreases from Li to Cs and thus close packing of atoms in crystal lattice decreases from Li to Cs resulting in an increase in softness. (a, b) (a, b) Blue colour is due to the presence of solvated (ammoniated) electrons, while electrical conductance is due to the presence of ions.

NOTE : Sodium in liquid ammonia forms NaNH2 only in presence of a catalyst like Pt black, iron oxide etc. . 28. 29. 30. (iii) 31.

1.

2.

(b) Statement-1 is correct. Statement-2 is also correct but not the correct explanation becuase blue colour of the solution is due to the solvated electrons. LiF has more ionic character while LiI has more covalent character. The latter is due to the greater polarizability of larger iodide ion than the fluoride ion. (iv) 2K3[Fe(CN)6] + H2O2 + 2KOH → 2K4[Fe(CN)6] + 2H2O + O2 NaCl + NH4OH + CO2 → NH4Cl + NaHCO3 (i) Potassium carbonate cannot be manufactured by Solvay process, since, unlike sodium hydrogen carbonate, potassium hydrogen carbonate is rather too soluble in water to be precipitated like NaHCO3. (c) BeSO4 and MgSO4 are readily soluble in water due to greater hydration enthalpies of Be2+ and Mg2+ ions, dominate over their lattice enthalpies and therefore their sulphates are highly soluble. (d) Solubility of BeSO4 is highest among the given metal sulphates

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BeSO4 > MgSO4 > CaSO4 > SrSO4 > BaSO4 Solubility order for hydroxide is Be(OH)2 < Mg(OH)2 < Ca(OH)2 < Sr(OH)2 < Ba(OH)2 Thus BeSO4 is soluble and Be(OH)2 is insoluble. BeO does not form rock salt like structure. In the solid state, it adopts the hexagonal neurtzite structure form while in the vapour phase, it is present as discrete diatomic covalent molecules. 3. (b) LiCl is soluble in pyridine. 4. (c) Gypsum on heating to 393 K forms plaster of Paris.

5.

(c) 3Mg(OH)2 + 2NH3

4NH3 + CuSO4 → 6.

(b) Thermal stability of alkaline earth metal carbonates increases down the group. MgCO3 < CaCO3 < SrCO3 < BaCO3 7. (a) In manufacturing of aircraft an alloy of Mg and Al called magnalium is used due to its light weight and high strength. 8. (b) Mg burns in air and produces a mixture of nitride and oxide. 9. (c) BeCl2 in vapour phase exists as dimer (below 1200 K temperature) whereas, in solid state BeCl2 has chain structure. 10. (c) According to Fajan’s rule, greater the polarising power of cation greater would be the covalent character. Since, Be2+ has maximum polarising power among given cations. Therefore, BeX2 would be most covalent alkaline earth metal halides among the given halides. 11. (a) Be(OH)2 is amphoteric in nature. 12. (c) Be is transparent to X-rays, so it is used in makingX-ray tube windows.

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13.

14. 15.

16. 17.

18.

(d) The chances of formation of hydrate decreases with the decrease in the charge density down the group. This is why, Ba(NO3)2 does not crystallise with water molecules. (a) Quick lime is commercial name of CaO. (a) The solubility of sulphates of alkaline earth metals decreases as we move down the group from Be to Ba due to the reason that ionic size increases down the group. The lattice energy remains constant because sulphate ion is so large, so that small change in cationic sizes do not make any difference. However the hydration energy decreases from Be to Ba apprciably as the size of the cation increase down the group. Hence, the solubility of sulphates of alkaline earthmetal decrease down the group mainly due to decreasing hydration energy from Be to Ba. Thus the order will be Mg > Ca > Sr > Ba (d) BeSO4 has its hydration enthalpy greater than its lattice enthalpy. (c) The amount of heating required depends on the extent of polarisation. More polarization required less energy. The smaller the positive ion is, the higher the charge density, and the greater effect it will have on the hydroxide ion. As the positive ions get larger down the group, they affect on the hydroxide ions. Therefore, the hydroxides become more thermally stable down the group. (a)

19.

(b) In going from top to bottom in a group, the first ionization potential decreases, thus Be > Mg > Ca 20. (b) The increasing thermal stability is BeCO3 < MgCO3 < CaCO3 < K2CO3 (IV) (II) (III) (I) NOTE : Increasing size of cation decreases its polarization ability towards carbonate, making the compound more stable. 21. (d) Glauber’s salt is Na2SO4.10 H2O. 22. (a) Ca is obtained by electrolysis of molten mixture of CaCl2 mixed with CaF2. + H2O2 + 8H2O 23. (c)

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24. (b) CaO + H2O Ca(OH)2 + Heat CaCO3 CaO + CO2 25. higher effective nuclear charge. 26. Anhydrous HCl All the water of crystallisation cannot be removed by heating hydrated MgCl2. HCl checks the hydrolysis of MgCl2 by its own water of crystalization. 27. False : Although 4 molecules of water of crystallisation are removed by heating, the remaining two react with MgCl2 as per the equation given below : MgCl2 + 2H2O → MgO + 2HCl + H2O NOTE : In order to avoid this to happen, MgCl2.2H2O is dehydrated in presence of HCl gas, which checks, (being in excess) the hydrolysis of MgCl2 by its own water of crystallisation. 28. (a) 29. (a) H2O2 (hydrogen peroxide) SrO2 and BaO2 (Barium peroxide) contain peroxide ions. 30. (b) (I) Ca(OH)2 is used in white wash. (II) NaCl is used in preparation of washing soda (Na2CO3). is used for making casts of statues.

(III)

(IV) CaCO3 is used as an antacid. 31. (d) (i) Na2CO3.10H2O Solvay process (ii) Mg(HCO3)2 Temporary hardness (iii) NaOH Castner-Kellner process (iv) Ca3Al2O6 Portland cement 32. BeCl2 is hydrolysed due to high polarising power and presence of vacant p-orbitals in Be-atom. (Be = 33.

)

3Ca(OH)2 + 2Cl2

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34.

SrSO4 > CaSO4 > MgSO4 > BeSO4 (Based upon size of cation or ionic character) 35. NOTE : Smaller the size of cation, higher will be hydration tendency because hydration energy of cation is inversely proportional to size of cation. The size of alkaline earth metal ions are smaller than the size of alkali metal ions. So, in crystalline form, the salts of alkaline earth metals have more water molecules than those of alkali metals. 36.

M may be either Ca or Ba. NOTE : It is not magnesium because Mg(OH)2 has very low solubility in water. If we consider Ba as M then A is Ba, B is Ba3N2, C is Ba(OH)2, D is BaCO3. 37. N3– being smaller in size and high charge present on it make it more susceptible to hydrolysis : Cl–being a weak conjugate base does not undergo hydrolysis. Mg2+ get hydrolysed as :

38.

Ca5(PO4)3F + 5H2SO4 + 10H2O 3H3PO4 + 5CaSO4.2H2O + HF

39. In sea water Mg exists as MgCl2. On treating sea water with slaked lime Mg(OH)2 is obtained. +

→ Mg(OH)2↓ + CaCl2

On reacting Mg(OH)2 with HCl, MgCl2 is obtained. Mg(OH)2 + 2HCl → MgCl2 + H2O From MgCl2, Mg is obtained by reduction of MgCl2 with CaC2.

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MgCl2 + CaC2 → Mg + CaCl2 + 2C 40.

CaCO3 + CO2 + H2O →

NOTE : Suspension of lime stone is CaCO3. 41. Bleaching powder, Ca(OCl)2, can be prepared by passing chlorine through Ca(OH)2 solution. Ca(OCl)2.CaCl2.Ca(OH)2.2H2O 3Ca(OH)2 + 2Cl2 Slaked Lime

Bleaching Powder (a mixture of Ca(OCl) and basic chloride)

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1.

(d) H3N3B3Cl3 + LiBH4

B3N3H6 + LiCl + BCl3

(A)

H3N3B3Cl3 + 3MeMgBr

(B)

H3N3B3(Me)3+ 3MgBrC

(A)

(C)

2.

(d) Fullerene (C60) contains 20 hexagons (six membered) rings and 12 pentagons (five membered rings): B2O3 + 3H2O 3. (a) B2H6 + 3O2 2H3BO3 + 6H2 B2H6 + 6H2O 4. (a) SiH4 : Electron precise hydride B2H6, GaH3 and Al H3 are electron deficient 5. (b) Be and Al show diagonal relationship due to which these two elements have similar electronegativity. 6.

(c) White gelatinous ppt.

Al2O3 is used as adsorbent in chromatography. Thus, metal ‘M’ is Al. 7.

(a)

BCl3 is electron deficient but it does not form dimer like Al, Ga or In because its electron deficiency is complemented by the formation of co-

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8.

9. 10.

11.

ordinate bond between lone pair of electron of chlorine and empty unhybridised p-orbital of boron forming pπ-pπ bonding. (a) Lithium hydride react with diborane to produce lithiumborohydride. (b) Atomic radii increases on moving down a group. However, due to poor shielding effect of d orbit, atomic radius of Ga is smaller than Al (anomaly). Thus, the correct order is Ga < Al < In < Tl. (b) cis-1,2-diol forms chelated complex ion with the product, [B(OH)4]– causing the reaction to proceed in forward direction.

(a) The central boron atom in boric acid, H3BO3 is electrondeficient. NOTE : Boric acid is a Lewis acid with onep–orbital vacant. There is no d-orbital of suitable energy in boron atom. So, it can accommodate only one additional electron pair in its outermost shell.

12.

(c) AlCl3 exists as a dimer (Al2Cl6). It is a strong Lewis acid as it has an incomplete octet and has a tendency to gain electrons. AlCl3 undergoes hydrolysis easily and forms an acidic solution. AlCl3 + 3H2O → Al(OH)3 + 3HCl Option (c) is true that AlCl3 sublimes at 100ºC under vacuum.

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AlCl3 is a Lewis acid. 13. (6) 3B2H6 + 18CH3OH → 6B(OCH3)3 + 18H2 14. (6) Coordination number of Al is 6. It exists in ccp lattice with 6 coordinate layer structure. 15. sodium hydroxide;

16.

False :

Bond distance between aluminium-chlorine bond forming bridge is greater (2.21 Å) than the distance between aluminium-chlorine bond present in the end (2.06 Å). 17. (a, b, d) (a) Structure of Al2(CH3)6

(b)

Structure of B2H6

(c)

Structure of Al2Cl6

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(d)

18.

BCl3 is stronger lewis acid as the bond formed with the base will involve 2p orbital overlap which is stronger than 3p orbital overlap in the case of AlCl3. (a, c, d) Structure of borax

Correct formula of borax is Na2[B4O5(OH)4].8H2O (A) Borax has tetranuclear. [B4O5(OH)4]2– unit (B) Only two 'B' atom lie in same plane (C) two Boron are sp2 and two are sp3 hybridised. (D) one terminal hydroxide per boran atom. 19. (b, d)H3BO3 does not undergo self ionization. However, it acts as a weak acid in water (hence it is a weak electrolyte in water). Addition of cis-diols (e.g., ethylene glycol) to aqueous solution of orthoboric acid leads to complex formation, thus acidity of aqueous solution of orthoboric acid is increased.

It arranges to planar sheets due to H-bonding. 20. (A)-(q), (B)-(s), (C)-(p), (D)-(r) (A) Hydrolysis of BiCl3

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(B)

Upon dilution by water, we will again get the white ppt of Al(OH)3. (C) Pyrosilicates are formed from 2 moles of orthosilicates as :

(D) 21.

(a) Both statements are true. Boron forms only covalent compounds (bonds) because small sized B ion polarizes the corresponding anion largely. 22. (c) Statement-1 is correct but statement-2 incorrect. Orthoboric acid (H3BO3) is soluble in water and behaves as weak monobasic acid. It does not donate protons like most acids, but rather it accepts OH– ions. It is, therefore, Lewis acid, and is better written as B(OH)3. H3O+ + [B(OH)4]–; pKa = 9.25 B(OH)3 + 2H2O 23. (c) Assertion is correct because Al(OH)3 can react with acid and bases. + 3H2O

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Reason is incorrect because the electronegativity difference between Al–O and O–H are different. 24. HF is weakly dissociated, while KF is highly dissociated giving a high concentration of F– which leads to the formation of soluble AlF63–. AlF3 + 3 KF → K3[AlF6] Since BF3 is more acidic than AlF3, it pulls out F– from AlF63– reprecipitating AlF3. K3[AlF6] + 3BF3 → 3KBF4 + AlF3 ↓ 25. NOTE : When hot concentrated HCl is added to borax (Na2B4O7.10H2O) the sparingly soluble H3BO3 is formed which on subsequent heating gives B2O3 which is reduced to boron on heating with Mg, Na or K.

↓

[NOTE : Normally this reaction takes place in the presence of Lewis acid (AlCl3)]

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26.

Since B2O3 is formed by reaction of (Y) with air, (Y) therefore should be B2H6 in which % of hydrogen is 21.72. The compound (X) on reduction with LiAlH4 gives B2H6. Thus it is boron trihalide. The reactions are shown as: 2B2H6 + 3LiX + 3AlX3 4BX3 + 3LiAlH4 (X)

(Y)

B2H6 + 3O2 (Y)

(X =Cl or Br)

B2O3 + 3H2O + heat

Structure of B2H6 is as follows:

Thus, the diborane molecule has four two-centre- two -electron bonds (2c – 2e bonds) also called usual bonds and two three-centre-two -electron bonds (3c – 2e) also called banana bonds. Hydrogen attached to usual and banana bonds are called Ht (terminal H) and Hb (bridged H) respectively. NaAlO2 27. Al + NaOH Al(OH)3 Al2O3 (Alumina) 28.

(i) Foul odour, on damping of Al2S3 is due to formation of H2S gas, which smells like rotten eggs.

(ii)

2Al + 2NaOH

+ H2

29. In excess of NaOH, the hydroxide of Al becomes soluble due to the formation of meta-aluminate. + Al3+ → Al(OH)3 + 30. (i) (ii) AlBr3 + K2Cr2O7 + H3PO4 → K3PO4 + AlPO4 + H2O + Br2 + Cr3+

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1.

(a)

Due to backbonding of lone pair electrons of nitrogen into vacant d-orbitals of Si, trisilylamine (SiH3)3N is planar. In trimethylamine (CH3)3N, there is no backbonding and hence it is more basic. 2. (d) Carbon-carbon bond length is maximum in diamond because diamond has all single bonds while graphite, C70 and C60 have single and double bonds. Carbon allotrope C – C bond length Diamond 154 pm Graphite 141.5 pm C60 138.3 pm and 143.5 pm C70 eight type of bond lengths from 0.137 pm to 0.146 pm. 3. (b) The catenation property among 14th group elements is based on bond enthalpy value of bond between the same element. The decreasing order of bond enthalpy values is Bond enthalpy in (kJ/mol) ∴ Decreasing order of catenation is C > Si > Ge ≈ Sn 4. (d) Catenation power of the elements decreases as we move down in the group. Therefore, Pb does not show catenation property. 5. (b) CCl4 cannot be hydrolysed due to absence of d orbitals at carbon atom. 6. (b, d) BCl3 and AlCl3, both have vacant p-orbital and incomplete octet, thus they behave as Lewis acids.

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SiCl 4 can accept lone pair of electron in d-orbital of silicon, hence it can act as Lewis acid. 7.

(d)

8.

(b) Two dimensional sheet structures are formed when three oxygen atoms of each [SiO4]4– tetrahedral are shared. 9. (c) TIPS/FORMULAE : It appears at the first sight that Me2SiCl2 on hydrolysis will produce Me2Si(OH)2 which ultimately upon loss of water, will form Me2Si = O. But silicon atom, because of its very large size in comparison to oxygen, is unable to form π-bond. Thus, the product of hydrolysis is polymeric in nature.

10. (a) CO is an example of neutral oxide. 11. (d) Due to inert pair effect, Pb4+ is not a stable oxidation state. Both Pb4+ and I– are large in size. It is difficult for the atoms to come close together to form the compound. 12. (b) Graphite shows moderate electrical conductivity due to the presence of unpaired or free fourth valence electron on each carbon atom. 13. (c) Earlier the graphite present in pencil was thought to be lead, as it resembles to graphite in appearance.

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14. (3) Total cationic charge = Total anionic charge 2n + 6 + 24 = 36 ⇒ n = 3 15. inert-pair effect ; When ns2 electrons of outermost shell do not participate in bonding it is called inert pair and the effect is called inert pair effect. 16. glass 17. Fullerene 18. Trialkylchlorosilanol; The hydrolysis of R3SiCl, yields R3Si(OH) which condenses to give R3 Si – O – SiR3 ——→ R3 – Si – O – Si – R3 19. Silicones; 20. True : The property of catenation in carbon is due to the fact that in carbon atom, the number of valence electrons (4) is equal to the number of valence orbitals (one 2s + three 2p). Hence, carbon in the tetravalent state is fully saturated, i.e., it has neither any vacant orbital nor any lone pair of electrons on its atom due to which the C – C bond is extremely stable. NOTE : The reason for greater tendency of carbon for catenation than that in silicon may further be explained by the fact that the C – C bond energy is approximately of the same magnitude as the energies of the bond between C and other elements. On the other hand, the Si – Si bond is weaker than the bonds between silicon and other elements. 21. True : In diamond, each carbon atom is in sp3 hybridised state and is linked to four other neighbouring carbon atoms held at the corners of a regular tetrahedron by covalent bonds. Owing to very strong covalent bonds by which the atoms are held together, diamond is the hardest substance known. Graphite has a two dimensional sheet like structure and carbon in sp2 hybridised state is attached to three other carbon atoms by three σ bonds forming a hexagonal planar structure. Due to wide separation and weak interlayer bonds, the two adjacent layers can easily slide over each other; hence graphite is soft. 22. True : Graphite is better lubricant on the moon than on the earth because of lack of gravitation pull on the moon, where friction is already less than earth.

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23.

False : CCl4 is most stable as compared to other tetrachlorides of the group because of its high thermal stability. 24. False : CCl4 gives phosgene with superheated steam CCl4 + H2O → COCl2 + 2HCl 25. False : PbO2 is a dioxide and it does not give hydrogen peroxide when it reacts with a dilute acid. PbO2 + 4HCl → PbCl2 + Cl2 + 2H2O 26. (a, b) (a) SnCl2. 2H2O is a reducing agent since Sn2+ tends to convert into Sn4+. (b) SnO2 + 2KOH + 2H2O → K2[Sn(OH)6] (c) In presence of conc. HCl, PbCl2 exists as H2[PbCl4] PbCl2 + 2HCl → H2[PbCl4] Chloroplumbous acid

(d) Pb3O4 + 4HNO3 → PbO2 + 2Pb(NO3)2 + 2H2O (2PbO + PbO2) (mixture of oxides)

It is not a redox reaction. 27.

(c, d)

There is a coordinate bond between NMe3 and SnCl2 due to sharing of lone pairs of NMe3 with SnCl2. Structure of X, Y, Z are respectively :

28.

(b) (CH3)2SiCl2 form linear polymer on hydrolysis and (CH3)3SiCl is a chain terminator.

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29. (b, d) (a) Diamond is harder than graphite. (b) Graphite is good conductor of electricity as each carbon is attached to three C-atoms leaving one valency free, which is responsible for electrical conduction, while in diamond, all the four valencies of carbon are satisfied, hence insulator. (c) Diamond is better thermal conductor than graphite. Whereas electrical conduction is due to availability of free electrons; thermal conduction is due to transfer of thermal vibrations from atom to atom. A compact and precisely aligned crystal like diamond thus facilitates fast movement of heat. (d) In graphite, C – C bond acquires double bond character, hence higher bond order than in diamond. 30. (b) Silicon is used in solar cells. 31. (a) A - Silica gel packets are used to absorb moisture and keep things dry i.e. as drying agent. B - Silicon is a semiconductor and is used in transistors. C - Silicone is used as sealant. D - Silicates are widely used in ion-exchange beds in domestic and commercial water purification, softening, and other applications. 32. (A)-(t), Pb(N3)2 is an explosive (B)-(s), Al2O3 is used to prepare artificial gem. (C)-(v), Extraction of copper involves self-reduction process. (D)-(q), Fe3O4 is a magnetic material. 33. A-(c)-(ii); B-(e)-(iii); C-(a)-(iv); D-(b)-(v); E-(d)-(i). 34. (b) In diamond, each C-atom is covalently bonded to four other Catoms to give a tetrahedral unit, so it shows sp3 hybridisation. Therefore, each C-atom forms four sigma bonds with neighbouring Catoms. In diamond eachC-atom utilizes its four unpaired electrons in bond formation. These bonding electrons are localized. Due to this reason diamond is a bad conductor of electricity. In graphite each Catom is covalently bonded to threeC-atoms to give trigonal geometry. Each C-atom in graphite is sp2-hybridized. Three out of four valence electrons of each C-atom are used in bond formation while the fourth electron is free to move in the structure of graphite. Due to this reason graphite is a good conductor of electricity.

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35.

(c) NOTE : In group 14 elements, the lower (and not higher) oxidation states are more stable for heavier members of the group due to inert pair effect. Thus, Pb4+ is less stable as compared to Sn4+ (lead is heavier than Tin). Therefore, Pb4+ acts as a strong oxidising agent than Sn4+. Hence, statement 1 is false and statement 2 is true. 36. (c) SiCl4 undergoes hydrolysis due to the presence of empty d orbitals in the valence shell of Si, while C has no vacant d orbitals to accommodate electron pairs donated by water molecules during hydrolysis. SiCl4 is a covalent compound because the difference electronegativity of Si and C is not high. Si + 2MgCl2 (or ZnCl2) 37. (i) SiCl4 + 2Mg(or Zn) (CH3)2SiCl2 + 2MgCl2 (ii) SiCl4 + 2CH3MgCl

Polymerisation continues on both ends to give linear silicone. Si + 2MgCl2 (iii) SiCl4 + 2Mg Na2SiO3 + C Si + Na2CO3 OR

38.

In cyclic

, three tetrahedra of

are joined together

sharing two oxygen atoms per tetrahedron.

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Structure of Dark circle represents Si and open circle represents oxygen atom/ion

39. Sn + 2KOH + 4H2O → K2Sn(OH)6 + 2H2 40. (i) SnCl4 + 2C2H5Cl + Na C4H10 + Na2[SnCl6] 4AlCl3 ↑ + 3Si (ii) 3SiCl4 + 4Al (molten)

41.

(i) In (SiH3)3N, lone pair of electrons on nitrogen is involved in pπ – dπ back bonding, while in (CH3)3N no such pπ – d π back bonding is possible because of absence of d orbitals in carbon so (CH3)3N is more basic than (SiH3)3N. (ii) In MgCl2, Mg is sp hybridised while in SnCl2, Sn is sp2 hybridised (hence the molecule is angular). (iii) In graphite, out of four valence electrons, only three form covalent bonds (sp2 hybridisation) with three other carbon atoms. This forms hexagonal rings as sheets of one atom thickness. These sheets are held together by weak attractive forces. One electron of each carbon atom is free and this enables these thin sheets to slide over one another. For this reason, graphite is a soft material with lubricating properties. (iv) Solid CO2 is technically known as dry ice because it sumblimes without leaving any stain on surface. (v) Carbon exists in various allotropic forms like diamond, graphite, coal, etc. Diamond consists of a three-dimensional structure of sp3 hybridised carbon atoms bonded through very strong covalent bonds. It makes it hard and useful as an abrasive. Graphite, on the other hand, is made up of a two dimensional sheet like structure made of sp2 hybridised carbon atoms. These layers of carbon atoms are held together by relatively weak van der Waal’s forces and

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42.

can, therefore, slip over one another imparting lubricating properties to graphite. H2SnO3 + 4NO2 + H2O (i) Sn + 4HNO3 Metastanic acid

(ii) 2Al + 2NaOH + 2H2O

2NaAlO2 + 3H2

Sod. aluminate

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1.

(c)

5-Amino-4-hydroxymethyl-2-nitrobenzaldehyde

2.

(b)

4-Bromo-2-methylcyclopentane carboxylic acid

3.

(c)

(2, 5-dimethyl 1-6-oxo-hex-3-enoic acid)

4.

(d)

3-(ethynyl-2-hydroxy-4-methyl-hex-3-en-5-ynoic acid)

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5.

(d)

3, 5-dimethyl-4-propylhept-1-en-6-yne

6.

(b)

2-Chloro-1-methyl-4-nitrobenzene

7.

(d) IUPAC name: 4-Bromo-3-methylpent-2-ene

8.

(d)

9.

(b)

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10.

(b)

4, 4 - Dimethylpentene

11.

(b)

3-Ethyl-4-methylhexane

12.

(b)

13. 14.

(b) – CN has highest priority. Further, the sum of locants is 7 in (b) and 9 in (d). (b) Carboxylic acids are named as oyl chlorides.

15.

(d)

16.

(b)

17.

(c)

18. 19. 20. 21.

Butane-1, 4-dioic acid. . Hyperconjugation. vicinal, adjacent or stable, different. (a, b)

(3-Methyl-1-butene)

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IUPAC name (a)

(b)

1-Chloro-4-methylbenzence

22.

(i)

4-Chlorotoluene

IUPAC name is

3–(N, N-dimethylamino)–3–methylpentane.

(ii)

(iii) Pent-2-en-1-oic acid

23.

1.

(b) Enolic form of acetylacetone (b) is quite stable due to H-bonding which leads to stable 6-membered ring.

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2.

(b)

3.

(d)

At (2),

‘S’configuration

At (3),

‘R’configuration

4.

(c)

In 1-phenyl-2-butene, the two groups around the doubly bonded carbons are different. This compound can show cis-and trans-isomerism.

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5.

(d)

The possible structural isomers = 5 6. (a) Pentan-2-one and pentan-3-one are position isomers. (b), (c), (d) contain different compounds aldehyde and ketones. These exhibit functional group is omerism. 7. (c) Optically active compounds contain an asymmetric (chiral) carbon atom (a carbon atom attached to four different atoms or groups). Therefore, all acids except maleic acid exhibit optical isomerism. 8.

(b)

9.

(d) Number of isomers (six) can be derived by keeping the position of any one halogen (say Br) fixed and changing the position of the other halogen one by one.

10.

(b) NOTE : A compound which consists of at least one asymmetric carbon atom is capable of showing the phenomenon of optical isomerism. The structure cannot show geometrical isomerism as one of the carbons along the double bond has identical group (methyl). Tautomerism is

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11.

not possible because of the absence of –CO group. It shows optical isomerism because it has chiral C atom with four different groups, H, CH3, COOH and (CH3)2C = CH. (a) Stereoisomers which are mirror image of each other are enantiomers and the one which are not mirror images are diasteromers. Conformation of the molecule is the spatial arrangement of the atoms of a given molecular structure that are obtained merely by rotation about a sigma bond in the molecule. is an isomer of

12.

(d)

13.

(d) The first three are isomers of diethyl ether, C2H5OC2H5 (C4H10O).

,

,

,

14.

(b, c)

(a)

All carbons are sp2 hybridised. Thus, they will form trigonal planar structure. But the conformer about C2–C3 bond can be in another plane. Hence, the molecule is non-planar in some of its conformations. (b)

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All carbons are either in sp or sp2 bybridisation. Due to the presence of C2– C3 single bond, we can rotate the molecule about this bond. If we rotate the groups attached to C3 carbon, the molecule still remains planar as theC1–C2 part of the molecule is linear. The alkyne group has infinite rotational symmetry along its axis. After rotation the molecular plane will be different but the molecule will always remain in a plane. (c)

C1 – carbon is sp2 hybridised. Thus, it forms 3σ-bonds in x-y plane. It also forms a π-bond which must be perpendicular to this plane. Therefore, it will use the remaining pz-orbital to form π-bond with C2-carbon. C2 – carbon is sp-bybridised. Thus,it forms two linearsp-orbitals to form two σ-bonds, i.e. one with carbon and the other with oxygen. The remaining py orbital will form a π-bond with oxygen. Oxygen atom will use its pxorbital to form σ-bond with carbon, i.e. (sp – px) σ-bond. Its py-orbital is used in making π-bond and the pz-orbital is already full. Hence, the C2 – O part of the molecule is linear in xdirection. Hence, we can say that the whole moleucle is planar. (d) The given compound is allene, which is non-planar as explained in page no.

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15. 16.

(b,c,d) E and F ; and also E and G differ in position of atom (H), so these are tautomers (not resonating structures). Geometrical isomers are also diastereomers. (a,c,d) TIPS/Formulae : For a carbonyl compound to show tautomerism, it must have at least one H at the α – carbon atom. (a), (c) and (d) show tautomerism. CH = CH - OH

CH2 – CHO

(enol form)

Tautomerism is not possible

O

17.

(keto form)

(a, d) Option (a) In n-butane, Cl can add at either the first or second carbon giving two isomers.

Option (b) :

will give three isomers with

Cl group at either of the CH3 groups, second C-atom and 3rd C-atom. Option (c) Benzene forms only one single derivative. mers will again give two iso with Cl at either Option (d) : one of the CH3 groups or at the central C-atom.

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18.

1.

(d) Acidic strength –I, –M effect. Due to strong – I, and –M effect of three – COOCH3 groups, it has most acidic Hydrogen. 2. (d) Glycerol can be separated from spent-lye in soap industry by using reduce pressure distillation technique. 3. (c) Basicity order can be determined by the cummulative effect of the factors on the electron density of concerned atom. 4. (a) Liquid having lower boiling point comes out first in fractional distillation. Simple distillation can’t be used as boiling point difference is very small. 3-Methylpantane will show greater boiling point (63°C) comparative to isohexane due to symmetrical structure. Therefore isohexane distilled out first. 5. (b) Electron withdrawing group like (NO2) increase stability of alkoxide ion by dispersal of negative charge. In (B) and (C) structures negative charge is in conjugation with double bond and also stabilised by electron withdrawing effect of nitro group. 6. (d) Higher the basicity of a base, stronger will be its nucleophilic character. Further a stronger base has a weaker conjugate acid. The acidic order of the conjugate acid of the given bases is H3O+ > CH3SO3H > CH3COOH > H2O Thus the increasing order of the basic nature or nucleophilicity of the given species is

7.

(c)

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(i)

(ii) (iii)

(iv) 8.

9.

10.

Hence, correct order of basicity will be : (ii) < (i) < (iv) < (iii). (c) Among the substituents attached to the given compounds, fluorine has maximum electronegativity. so it will push electron pair towards itself. In option (b), the two F groups are attached opposite to each other, thus net dipole moment will cancel out and reduce its polarity. In option (d), the F groups are attached in slightly opposite direction, thus this also decreases its polarity. But in option (c), the compound has the two F groups along same direction, thus net dipole moment will increase in this direction and therefore it will exhibit maximum polarity. Hence the compound in option (c) has maximum polarity. (b) Since, adsorption of I > II, I is firmly attached to column (stationary phase). Hence, it moves slowly and will cover little distance, while II is loosely attached to column (stationary phase). Hence, it moves faster and will cover large distance. (d)

is non-aromatic and hence least reasonance stabilized,

whereas other three are aromatic.

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11.

(b) Pent-l-ye and glycerol are separated by distillation under reduced pressure. Under the reduced pressure, the liquid boil at low temperature and the temperature of decomposition will not reach. e.g. glycerol boils at 290 °C with decomposition but at reduced pressure it boils at 180° C without decomposition. 12.

(d)

In both the molecules the bond moments are not cancelling with each other and hence the molecules has a resultant dipole. is more stable than (CH3)3 because resonance 13. (d) stabilisation effect in Ph3 is more pronounced as compared to hyperconjugation stabilisation effect in (CH3)3 , overall stability order among free radical is : Triphenylmethyl > benzyl > allyl > tertiary alkyl > secondary > primary > methyl > vinyl 14. (c) Carbocations are planar, hence can be attacked on either side to form racemic mixture.

15.

(d) Higher stability of allyl and benzyl carbocations is due to dispersal of positive charge by resonance

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whereas in alkyl carbocations dispersal of positive charge on different hydrogen atoms is due to inductive effect. Hence the correct order of stability will be

16. 17.

18.

(a) 3° carbocations are most stable. (d) NOTE : Heterolytic fission occurs when the two atoms differ considerably in their electronegativities. O – H bond undergoes cleavage most readily because O and H differ markedly in their electronegativity and further oxygen being highly electronegative can accommodate the negative charge more effectively developed after the cleavage. (a) TIPS/Formulae : The bond angle in sp3, sp2 and sp hybridisad carbon atoms is respectively 109.28', 120º and 180º.

19.

(c) Carbon bonded with a triple bond (i.e. C1) is sp hybridised. Carbon bonded with a double bond (C2) is sp2 hybridised.

20.

(c) (a)

(b)

(c)

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[Note : dipole moment is a vector quantity]. 21.

(d)

Hybridisation in C1 = sp2, C2 = sp, C3 = sp2, C4 = sp3. 22. (a) TIPS/Formulae : The bond length decreases in the order.

On the basis of the size of the hybrid orbitals, sp orbital should form the shortest and sp3 orbital the longest bond with other atom. 23. (6)

a = 3 Hyperconjugative H's b = 2 Hyperconjugative H's c = 1 Hyperconjugative H 24. (50) Mass of organic compound = 1.6 g Mass of AgBr = 1.88 g Moles of Br = Moles of AgBr = Mass of Br = 0.01 × 80 = 0.80 g % of Br Alternate Method : % of Br

25.

sp ; Ag – C

C – Ag

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26.

cyclopropane, because it has maximum deviation, from the normal bond angle of 109°28' present in alkanes. In cyclopropane, bond angle is 60°, therefore, deviation d, .

27.

propadiene;

28.

tert-Butyl carbonium ion is more stable due to hyperconjugation and +I effect of the three methyl groups. (a, b, c)

29. (I) (II)

(III)

pka = 33.3 H – CH3

pka = 43

(IV) (V) pka = 50

pka = 16

(a) I

(Resonance stabilised anion of I)

+ H+

(b) IV

Conjugate base of IV

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(c) (d)

30.

Aromatic –NO2 is –I group (electron withdrawing group), which increases acid strength of III. Acid strength order : IV > V > I > II > III

(a)

In tert- butyl cation, carbon bearing positive charge has one vacant porbital hence it is σ–p (empty) conjugation or hyperconjugation.

In 2-butene, hyperconjugation is between σ→π∗ bond. 31. (b, c) (b) and (c) being antiaromatic, are unstable at room temperature. 32. (d) Order of acidic strength CH3OH > CH ≡ CH > C6H6 > C2H6 ; CH3OH is most acidic because O is more electronegative than C and capable of accommodating negative charge in CH3O– Although alcohols are neutral towards the litmus paper. 33. (a) TIPS/Formulae : Conjugate base of strong acid is weak while conjugate base of a weak acid is strong. Acidic strength of acids, of the given conjugate bases HOH > CH ≡ CH > NH3 > CH3.CH3 Hence, the order of strength of bases, > > CH ≡ C – > OH– CH3 34. (b, d)

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35.

36.

1, 4-Dichlorobenzene (p-dichlorobenzene) and trans-1, 2dichloroethene have zero dipole moment because of their symmetrical structures. (a,b,c) Resonating structures differ in bonding pattern.

(i)

Anti conformer

Gauche conformer

Given, mole fraction of anti conformer = 0.82 ∴ mole fraction of gauche conformer = 0.18 1 = µ(anti) × 0.82 + µ(gauche) × 0.18 1 = 0 × 0.82 + µ(gauche) × 0.18 ∴ 1 = µ(gauche) × 0.18 µ(gauche) =

[

µ(anti) = 0]

= 5.55 D

(ii)

37.

Presence of an electron-attracting group increases acidity of the compound. Thus, acidity order:

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38.

39.

TIPS/Formulae : Diethyl ether acts as a Lewis base and anhydrous AlCl3 as a Lewis acid. Anyhydrous AlCl3 is more soluble in diethyl ether because the oxygen atom of ether donates its pair of electrons to the vacant orbital of electron deficient aluminium of AlCl3 through the formation of coordinate bond. In case of hydrated AlCl3 aluminium is not electron deficient as oxygen atom of water molecule has already donated its pair of electrons to meet the electron deficiency of aluminium.

40.

(i)

CH C –, C– is sp hybridised and more electronegative than the of CH2 = which is sp2 hybridised. Thus, the former can better accommodate electron pair and hence, less basic. (ii) In biphenyl, one of the phenyl groups acts as electron donor and the other electron acceptor due to mesomeric effect. This makes it more reactive than benzene.

,etc.

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(iii) In formic acid, resonance is not possible, with the result, there are two types of C – O bonds. In sodium formate, resonance is possible, so both of the C – O bonds have same bond length.

41.

(ii) Isobutane Butanol


0; ∆Hmixing = 0 Since, there is no heat exchange between system and surrounding; ∆Ssurroundings = 0 × ∆G = ∆H – T∆Ssystem < 0 32. TIPS/Formulae : Molality,

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Mole fraction,

,

p1 = x1 p°1 ∴ x1 =

=

= 0.9868

x2 (solute) = 1 – 0.9868 = 0.0132 Molality, m =

× 1000 = = 0.7503 mol kg–1

NOTE: Molality =

33. Molecular weight of CH3OH = 12 + 3 + 16 + 1 = 32 Molecular weight of C2H5OH = 24 + 5 + 16 + 1 = 46 According to Raoult’s law Ptotal = p1 + p2 where Ptotal = Total vapour pressure of the solution p1 = Partial vapour pressure of one component p2 = Partial vapour pressure of other component Again, p1 = Vapour pressure (p°1) × mole fraction Similarly, p2 = Vapour pressure (p°2) × mole fraction Mole fraction of CH3OH =

= 0.49

Mole fraction of ethanol =

= 0.51

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NOTE THIS STEP : Thus now let us first calculate the partial vapour pressures, i.e., p1 and p2 of the two components. Partial vapour pressure of CH3OH(p1) = 88.7 × 0.49 = 43.48 mm Partial vapour pressure of C2H5OH(p2) = 44.5 × 0.51 = 22.69 mm ∴ Total vapour pressure of the solution = 43.48 + 22.69 mm = 66.17 mm Mole fraction of CH3OH in vapour =

= 0.65

34.

If they form an ideal solution which obeys Raoult’s Law and for which ∆Hmixing = 0 and ∆Vmixing = 0 Thus, we can separate two volatile and miscible liquids by fractional distillation if they should not form azeotropic solutions. 35. The chemical equation for the combustion of organic compound CxH2yOy can be represented as : CxH2yOy + 2xO2 = x CO2 + y H2O + x O2 The gases obtained after cooling = x + x = 2x ∴ 2x = 2.24 litres or x =

= 1.12 litres

Number of moles of CO2 =

⇒ x=

mole = 0.05 mole

Moles of H2O formed (y) =

= 0.05;

x : y = 0.05 : 0.05 = 1 : 1 The empirical formula of the organic compound is CH2O ...(1) The mole fraction of the solute (A)

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= relative decrease in vapour pressure of the solvent (B) =

or

=

[MA = mol.wt. of A]

or

=

or

=

or MA = 150.6

Molecular wt. of the organic compound (CH2O)n = 150 Molecular wt. of CH2O = 12 + 2 + 16 = 30 ∴ 30 × n = 150 or n =

=5

∴ Molecular formula of the given organic compound is (CH2O)5 or C5H10O5. 36.

M=

× 1000 =

× 1000 = 15.81 M

M1V1 = M2V2 M1 = 15.81, V1 = ? M2 = 0.2, V2 = 1 L = 1000 mL ∴ 15.81 × V1 = 0.2 × 1000 or V1 = 37.

= 12.65 mL

∴ Amount of acid to be used to make 1 L of 0.2 M H2SO4 = 12.65. TIPS/Formulae :

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(i)

Molarity =

(ii) N1V1 = N2V2 A 13% solution (by weight) contains 13g of solute (i.e. H2SO4) per 100 g of solution. =

Moles of solute =

= 0.1326

Volume of solution in L =

= ∴

= 0.0980 Litre = 1.35 M

Molarity of solution =

Again, Molality = Mass of solute in 100 mL of solution = 13 g [ 13 % solution] Mass of solvent = Mass of solution – Mass of solvent = 100 – 13 = 87 g ∴ Molality =

= 1.57 m

Normality = Molarity ×

or 1.35 ×

= 2.70 N

N1 = 2.70, V1 = 100 mL, N2 = 1.5, V2 = ? N1V1 = N2V2; 2.70 × 100 = 1.5 × V2 or V2 =

= 180 mL.

∴ 100 mL of this acid should be diluted to 180 mL to prepare 1.5 N solution.

1. (c) Osmotic pressure (p) = CRT Since, there are two solutes i.e. urea and glucose. ∴ π = (C1 + C1) RT

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= 0.2 × 0.0821 × 300 = 4.926 atm 2. (c) Relative lowering of vapour pressure, is given by,

Given,

3.

(d) Dissociation of Potassium Sulphate (K2SO4), K2SO4

i (Van’t Hoff factor) = 3 We know that, DTf = iKf m where, Kf is molal depression constant and m is molality. = 3 × 4 × 0.03 = 0.36 K 4.

(b) ;

5.

i = 1.8 (d) According to the question we can write Tb = Kbm Kb(1) = 2 Kb = 2 Km–1 Tf = Kfm Kf(2) = 2 Kf = 1 Km–1

So, 6. (a) (b) (c) (d)

Kb = 2Kf (d) Number of particles (i) [Co(H2O)6]Cl3 4 [Co(H2O)5Cl]Cl2⋅H2O 3 [Co(H2O)4Cl2]Cl⋅2H2O 2 [Co(H2O)3Cl3]⋅3H2O 1

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z ; where Remember, greater the no. of particles, lower will be the freezing point. Compound (d) will have the highest freezing point due to least number of particles. 7. (d) In benzene, (CH3COOH)2 2CH3COOH 1–α

α/2

i = 1 – α + α/2 = 1 – α/2 Here α is degree of association ∆Tf = i × Kf × m

α = 0.945 % degree of association = 94.6% 8. (c) Molality (experimental) solvent Molality (theortical) =

Moles before dissociation Moles after dissociation

1

0

1–α

2α

0 α

Von't Hoff factor (i)

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Na2SO4 is ionised 81.5% means α = 0.815 = 2.63.

9.

x = 45.07 g (c) As T increase, V.P. increases ∆Tf = Kf × m (m = Molality of solute) 273 – Tf′ =

∴ T′f = 270 K Thus, freezing point of solution = 270K. Further, as T increases, vapour pressure increases. Hence, these facts coincide with the curve given in (c). 10. (a) According to Raoult's Law …(i) Here P° = Vapour pressure of pure solvent, Ps = Vapour pressure of solution WB = Mass of solute, WA = Mass of solvent MB = Molar mass of solute, MA = Molar Mass of solvent Vapour pressure of pure water at 100 °C (by assumption = 760 torr) By substituting values in equation (i) we get, …(ii) On solving (ii) we get Ps = 752.4 torr 11. (d) Using relation,

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where w1, M1 = mass in g and mol. mass of solvent w2, M2 = mass in g and mol. mass of solute P° = 185 torr, Ps = 183 torr

(Mol. mass of acetone = 58) M2 = 64.68 ≈ 64 ∴ Molar mass of substance = 64 12. (d) Acetic acid contain carboxylic group – COOH which can form Hbonding so acetic acid dimerises.

13.

(a) π = i CRT

Since the osmotic pressure of all the given solutions is equal. Hence all are isotonic solutions. 14. (a) Given, πob = 10.8 atm πnor = CRT = 0.10 × 0.0821 × 298 = 2.446 Now experimental value of (i) =

= 15.

(c)

0.1 =

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mB = 16.

(a) From Raoult's law,

When the concentration of solute is much lower than the concentration of solvent,

...(i) From elevation in boiling point, ∆Tb = Kb × m 2 = 0.76 × m ...(ii)

m=

From (i) and (ii), p = 724 mm 17. (a) ∆Tf = i × Kf × m Where m = Molality of the solution (i.e. number of moles of solute per 1000 g of the solvent) Here, m = Thus, ∆Tf = 4 × 1.86 ×

= 2.3 × 10–2

∴ Tf = 0 – 2.3 × 10–2 = – 2.3 × 10–2 ºC 18. (a) Molecular weight of naphthoic acid C11H8O2 = 172 gmol–1. The theoretical value of depression in freezing point = Kf × molality = van't Hoff factor,

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= 19.

.

(a) TIPS/Formulae :

(i) (ii) ∆Tb = i × Kb × m

i = 1 + 2α Assuming 100% ionization So, i = 1 + 2 = 3 ∆Tb = 3 × 0.52 × 0.1 = 0.156 ≈ 0.16 20.

21.

(a) NOTE : At the freezing point liquid and solid remain in equilibrium. If a solution of a non-volatile solute is cooled to a temperature below the freezing point of solution, some of liquid solvent will separate as a solid solvent and thus the concentration of solution will increase. (b) Benzoic acid exists as dimer in benzene.

[Normal molecular mass = 122 amu and observed molecular mass = 244 amu, in case of complete association] 22. (a) Depression in freezing point, ∆Tf = i × Kf × m , where n = no. of ions produced by

van’t Hoff factor, i =

complete dissociation of 1 mole of HX. H+ + X– ⇒ n = 2 HX ∴i=

= 1.2

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[For 20 % ionisation,

]

∴ ∆Tf = 1.2 × 1.86 × 0.2 = 0.45 [ m = 0.2] Hence, freezing point of solution is 0 – 0.45 = –0.45°C 23. (d) TIPS/Formulae : The salt that ionises to least extent will have highest freezing point. [ i.e., minimum ∆ Tf ] Glucose, being non electrolyte, gives minimum no. of particles and hence, minimum ∆T f or maximum F. pt. 24. (a) NOTE : The salt producing highest number of ions will have lowest freezing point. ; K2SO4 gives highest number of particles (2 + 1 = K2SO4 → 2K+ + 3). Glucose, being non-electrolyte gives minimum no. of particles and hence minimum ∆Tf or maximum F. pt. 25. (a) Added HgI2 forms a complex with KI in the solution as follows 2KI + HgI2 → K2[HgI4] As a result, number of particles decreases and so ∆Tf increases. NOTE : Depression in freezing point is a colligative property which is not observed here. 26. (5.0) Molality of CaCl2 solution = 0.05 m Molality of CrCl3.xNH3 = 0.10 m

Since, co-ordination number of Cr is 6. ∴ The complex is [Cr(NH3)5Cl]Cl2. ∴ x = 5. 27. (1) Given ∆Tf = 0.0558°C as we know, ∆Tf = i × Kf × m ⇒ 0.0558 = i × 1.86 × 0.01 Therefore, the complex will be [Co(NH3)5Cl]Cl2. Hence, number of chloride in co-ordination sphere is 1.

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28.

(2) i=

= 29.

= 1 + 2 × 0.5 = 2 (167)

For NaCl : For Glucose : When 1 L of NaCl solution and 2 L glucose solution are mixed. and

30.

(177)

For isotonic solution, (For protein i = 1)

31. (1.75) Tf = iKf m 0.2 = 2 × 2 ×

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w=

=

= 1.75 g

32. (2.18)

Kf = 2.0 K kg mol–1; m = 0.5 Tf = Kfm = 0.5 × 2 Tinitial = 272 K n = 0.1 mol V = 1 dm3 = 2.176 atm

Pgas =

After releasing piston P1V1 = P2V2 2.176 × 1 = 1 × V2 V2 = 2.18 dm3 33. (1.03) We know that,

m2 = 64 g

34. (0.05) From graph For solvent ‘X’ ∆Tb(x) = 362 – 360 = 2 ∆Tb(x) = mNaCl × Kb(x)

...(1)

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For solvent ‘Y’ ∆Tb(y) = 368 – 367 = 1 ∆Tb(y) = mNaCl × Kb(y) Dividing equation (1) by (2) ⇒

...(2)

=2

For solute ‘S’ Given solute S dimerizes in solvent. Hence, 2(S) → S2 1 (1 – α) α/2

............ initial ............ after dimerisation

For solvent 'Y', αY = 0.7

For solute 'S', it is given that

35.

(75)

Total number of moles at equilibrium =

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So, the percentage of phenol that dimerises = 75%. 36. (0.228) TIPS/Formulae : Weight of water = 500 × 0.997 = 498.5 g (weight = volume × density) No. of moles of acetic acid = Since 498.5 g of water has 0.05 moles of CH3COOH 1000 g of water has = Therefore, molality of the solution = 0.1 Determination of van’t Hoff factor, i No. of moles at start No. of moles at equb.

1 1 – 0.23

0

0 0.23

0.23

Therefore, van’t Hoff factor = =

= 1.23

Now, we know that, ∆Tf = i × Kf× m = 1.23 × 1.86 × 0.1 = 0.228 K 37. (23.44) Depression in freezing point, ∆Tf = 0 – (–0.30) = 0.30 Now we know that ∆Tf = Kf m ∴ According to Raoult’s law

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On usual calculations,

= 23.51 – 0.068 p = 23.44 mm Hg 38. (156.056) TIPS/Formulae : Given Wt. of benzene (solvent), W = Volume × density = 50 × 0.879 = 43.95 g Wt. of compound (solute), w = 0.643 g Mol. wt. of benzene, M = 78; Mol. wt. of solute, m = ? Depression in freezing point, ∆Tf = 5.51 – 5.03 = 0.48 Molal freezing constant, Kf = 5.12 Now we know that, =

m= 39.

= 156.056

(746.3) TIPS/Formulae : First find moles of Ca(NO3)2 and water. to find vapour pressure of

Then use the expression solution. Let initially 1 mole of Ca(NO3)2 is taken Degree of dissociation of Ca(NO3)2 =

= 0.7

Ionisation of Ca(NO3)2 can be represented as Ca2+ + Ca(NO3)2 At start At equilibrium

1 1 – 0.7

0 0.7

0 2 × 0.7

∴ Total number of moles in the solution at equilibrium

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= (1 – 0.7) + 0.7 + 2 × 0.7 = 2.4 No. of moles when the solution contains 1 g of calcium nitrate instead of 1 mole of the salt (164 is the mol. wt. of Cal. nitrate)

=

∴ No. of moles of the solute in the solution containing 7 g of salt, i.e., × 7 = 0.102

n=

No. of moles of water (N) = =

= 5.55

Applying Raoult’s law,

⇒

= 40.

= = 0.0180

⇒ p = 760 – (760 × 0.0180) = 746.3 mm Hg (65.25) TIPS/Formulae : According to Raoult’s law, =

Here, pº = 640 mm p = 600 mm w = 2.175 g W = 39.0 m=? M = 78 Substituting the various values in the above equation for Raoult's law : = =

⇒ m = 65.25

41.

Kf ; Depression in freezing point, ∆Tf = Kf . m, where Kf is the molar depression constant or cryoscopic constant of the solvent. 42. (a, d) The freezing point of a solvent depresses as a non-volatile solute is added to a solvent. According to Raoult's law, when a non-

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volatile solute is added to a solvent the vapour pressure of the solvent decreases. At the freezing point, it will be only the solvent molecules which will solidify. 43. (d) The solution M is a mixture of ethanol and water. In it the mole fraction of ethanol is 0.9 and mole fraction of water is 0.1 (1.0 – 0.9 = 0.1). Also given are:Standard freezing point of ethanol = 155.7K. Freezing point depression constant (Kf) for ethanol = 2.0 kg mol–1 Molecular weight of ethanol (C2H5 OH) = 46 now molality (m) of solution = Using the formula :) = Kf × m, Depression in freezing point ( Substituting various values, we get =

= 4.83 K.

∴ Freezing point of solution 'M' = (155.7 – 4.83) K = 150.9 K i.e. (d) is the correct answer. NOTE: Here, solvent is ethanol and solute is water. 44. (b) Given : vapour pressure of pure ethanol ( ) = 40 mm Hg Mole fraction (XA) of ethanol in solution = 0.9 Using the formula : Total pressure (P) = Substituting the given values, we get P = 40 × 0.9 = 36.0 mm Hg i.e. (b) is the correct answer. NOTE: It is given that the solute is non-volatile, in the question. 45. (b) Given: Standard boiling point of water = 373K Boiling point elevation constant of water (Kb) = 0.52 Kg mol–1 Molecular weight of water (H2O) = 18 Mole fraction of ethanol in solution = 0.1 (1.0 – 0.9 = 0.1) molality (m) =

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using the relation ∆Tb = Kb × m Substituting the given values, we get =

= 3.20 K

∴ Boiling point of solution = (373 + 3.20) K = 376.2 K i.e. option (b) is correct answer. 46. (i) TIPS/Formulae : ∆Tb = Kb × M In first case,

or

or M = 122

Thus, benzoic acid exists as a monomer in acetone (ii) In second case,

⇒ M' = 224

or

NOTE : Double the expected molecular weight of benzoic acid (244) in benzene solution indicates that benzoic acid exists as a dimer in benzene. 47. TIPS/Formulae : Higher the value of dipole-dipole interaction higher is b.p. Higher value of Kb of a solvent suggests larger polarity of solvent molecules which in turn leads to higher dipole – dipole interaction implies higher boiling point due to dipole – dipole interaction. Therefore, the correct order of Kb values of the three given solvents is Mathematically, or 48.

Element

%

Relative no. of atoms

Simplest ratio

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C

42.86

H

2.40

N

16.67

O

38.07

∴ Empirical formula of the minor product is C3H2NO2 Molar empirical formula mass of the minor product = 3 × 12 + 2 × 1 + 1 × 14 + 2 × 16 = 84 g mol–1 Let M be the molar mass of the minor product. For 5.5 g of the minor product dissolved in 45 g benzene, the molality (m) of the solution

Substituting this in the expression of elevation of boiling point, ∆Tb = Kbm ⇒ 1.84 K = (2.53 K kg mol–1) or M = 168 g mol–1 No. of unit of empirical formula in molecular formula = Hence, the molecular formula of the minor product is 2 (C3H2NO2), i.e., C6H4(NO2)2. The product is m-dinitrobenzene 49.

.

TIPS/Formulae :

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= Here, w and m are wt. and molecular wt. of solute, W and M are wt. and molecular weight of solvent. p = Pressure of solution; pº = Normal vapour pressure Let the initial (normal) pressure (pº) = p ∴ Pressure of solution =

×p=

p

m = 60, M = 18, W = 100 g ∴

⇒

=

+ 5.55 ⇒

=

= =

= 5.55 or w = 111 g × 1000

Molality = =

= 18.52 m

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1.

(d)

Among given electrolytes, CH3COOH is weak electrolyte. 2. (N) Ionic mobility increases with increase in temperature, which increases the conductance of the solution, while conductance of NaCl solution is independant of temperature above 400°C. Because the temperature is not given, so none of the option is correct. 3. (d) (Λm0) NaBr – (Λ0m)NaI + Λ0mBr– – (Λ0m +Λ0mI–) = Λm0 + Λ0mBr– – Λ0m – Λ0mI = Λ0m = Λm0Br– + Λ0mI– (Λ0m)KBr – (Λ0m)NaBr = Λ0mK+ + Λ0mBr– – (Λ0mNa+ + Λ0mBr– ) – Λ0mBr– = Λ0mK+ + Λ0mBr– – Λ0m = Λ0mK+ + Λ0m (Λ0m)NaBr – (Λ0m)NaI (Λ0m)KBr – (Λ0m)NaBr 4. (a) Order of acidic strength is HCOOH > C6H5COOH > CH3COOH More the acidic strength more will be the dissociation of acid into ions and more will be the conductivity. Thus, order of conductivity will be,

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HCOOH > C6H5COOH > CH3COOH (A)

>

(C)

>

(B)

5.

(b) Conductivity of an electrolyte is the conductance of 1 cm3 of the given electrolyte. It increases with the increase in concentration of electrolyte due to increase in the number of ions per unit volume. Molar conductivity (lm) is the conductance of a solution containing 1 mole of the electrolyte. It increases with the decrease of concentration due to increase in the total volume having one mole of electrolyte. Thus, interionic attraction increses and degree of ionisation decreases. Therefore, (S1) is wrong and (S2) is correct. 6. (a) According to the Faraday’s law of electrolysis, nF of current is required for the deposition of 1 mol According to the reaction, 2 F of current deposits = 1 mol ∴ 0.1 F of current deposits =

= 0.05 mol

7. (c) The reaction involved in the process is given below PbSO4 Pb2+ + Pb2+ + H2O PbO2 + 4H+ + 2e– So, half cell reaction is PbSO4 Pb4+ + 2e– According to the reaction: PbSO4 → Pb4+ + 2e– We require 2F for the electrolysis of 1 mol or 303 g of PbSO4 ∴ Amount of PbSO4 electrolysed by 0.05F = 8.

9.

× 0.05 = 7.575 g ≈ 7.6 g (d) Sodium stearate at low concentration (i.e., below CMC) behaves as normal strong electrolyte, but at higher concetration (i.e. above CMC) exhibits colloidal behaviour due to the formation of micelles. for Thus, plot (d) correctly represents relation between Λm and sodium strearate. (c)

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27.66 g of B2H6 (1 mole) requires 3 moles of oxygen (O2) for complete burning. Now the oxygen is produced by the electrolysis of H2O. On electrolysis : 1 mole O2 is produced by 4F charge ∴ 3 mole O2 will be produced by 12F charge. Q = It 12 × 96500 C = I × t 12 × 96500 C = 100 × t

= 3.2 hours 10. (d) Reduction at cathode: 2e– + 2H2O → H2 + 2OH– (valence factor) H2 = 2 At NTP 22400 mL of H2 = 1 mole of H2 112 mL of H2 = Moles of H2 produced = 0.005 = I = 1.0 A 11.

(a)

Since Cr3+ is having least reduction potential, so Cr is the best reducing agent. 12. (c) No reaction will occur as the Zn is placed above Cu in electrochemical series. Hence there will be no2 displacement reaction.

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13.

(d) 2F i.e. 2 × 96500 C deposit Cu = 1 mol = 63.5g 14. (c) EMF of galvanic cell = 1.1 volt If Eext < EMF then electrons flow steadily from anode to cathode, while if Eext > EMF then electrons flow from cathode to anode as polarity is changed. 15. (a) Given for 0.2 M solution R = 50 Ω κ = 1.4 S m–1 = 1.4 × 10–2 S cm–1 Now, R = ⇒

cm–1

For 0.5 M solution R = 280 Ω cm–1 =

= 2.5 × 10–3 S cm–1 Now, = 5 S cm2 mol–1 = 5 × 10–4 S m2 mol–1 16. (c) According to Debye Huckle onsager equation, 17.

(c) As the value of reduction potential decreases the reducing power increases i.e. C < B < D < A

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(0.85). (0.6). (–0.76) (– 1.2) 18. (d) Reducing character decreases down the series. Hence the correct order is Al < Fe2+ < Br– Cu2+ + 2e– 19. (c) Cu i.e, to deposit 1 mole of Cu at cathode from Cu2+ SO42– solution = 2 moles of electrons are required i.e, To deposit Thus total no. of electrons required = 20. (b) W = Zit where Z = Electrochemical equivalent Eq. wt. of copper = Z= W = Zit = 21. (d) AgNO3 (aq) + KCl (aq) → AgCl (s) + KNO3 (aq) Conductivity of the solution is almost compensated due to formation of KNO3(aq). However, after end point, conductivity increases more rapidly due to addition of excess AgNO3 solution. 22. (b) Give :I = 10 milliamperes ; F = 96500 C mol–1 t = ? ; Moles of H2 produces = 0.01 mol From the law of electrolysis, we have Equivalents of H2 produces = Substituting given values, we get

or

= 19.3 × 104 sec.

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23.

24. 25.

(b) As we go down the group 1 (i.e. from Li+ to K+), the ionic radius increases, degree of solvation decreases and hence effective size decreases resulting in increase in ionic mobility. Hence, equivalent conductance at infinite dilution increases in the same order. (d) Charge of one mole of electrons = 96500 C ∴ 1 gram equivalent of substance will be deposited by one mole of electrons. ∴

(c)

Here E1 & E2 are equivalent weights of the ions. 26.

(6)

The formula for conductance is

κ = 6 × 10–5S cm–1

pH = 4 ∴ [H+] = 10–4 = cα = 0.0015 α

Also,

Hence, Z 27. (3) 1 → HX

2 → HY

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= 0.01

= 0.1

∴ = 0.1

= 10–3

pKa(HX) – pKa(HY) = –log 28.

= –log10–3 = 3

(11)

For synthesis of 1 mole of

, 6F of charge is required.

Current efficiency = 60% ∴ To synthesis 1 mole of To synthesis

, 10F of charge is required.

moles of KClO3, charge

Q = I. t

29.

∴ t = 11 h. (60) Charge (Q) = It = 2 × 8 × 60 = 960 C

Theoritical mass of Cr3+

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So, efficiency 30. (13.32)

E0cell = 1.23 – 0.00 = 1.23 V G0cell = –nF E0cell = –2 × 96500 × 1.23 J Work derived from this fuel cell using 70% efficiency

= 0.7 × 2 × 96500 × 1.23 × 10–3 = 166.17 J For insulated vessel, q = 0 Therefore for monoatomic gas, w= U 166.17 = nCv,m T where

T = 13.32K 31.

(0.154) i =

ampere

No. of equivalents of Zn2+ which are lost =

=

=

Milli equivalents of Zn2+ which are lost 3.646 Initial value of Zn2+ = 300 × 0.160 × 2 = 96 Mili equivalents of Zn2+ left in solution = 96 – 3.646 = 92.354 [ZnSO4] = 32.

M

Molarity of Zn2+ = 0.154 M (19.06) TIPS / FORMULAE : Watt = Volt × Current ⇒ 100 = 110 × Current

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or

Current =

=

amp.

Now we know that, × 10 × 3600 ×

Q=i×t=

= 0.339 F

Molecular mass of cadmium = 112.4 = 19.06 g

Wt. of cadmium deposited = 33.

(125.09) Volume of the surface = area × thickness cm =

= 80 cm2 ×

cm3

Mass of Ag deposited = Volume × Density =

× 10.5 g/cm3 =

Cell reaction : Ag+ + e– → =

We know that,

g

Ag

=

E = Eq. wt. of Ag = 108 ∴

= =

34.

∴ t = 125.09 sec (27171.96) Wt. of Cu deposited = Zit

Electrochemical equivalent of Cu =

= 31.75

Volume of surface = area × thickness = 10 × 10 × 10–2 = 1 cc Weight of Cu = density × volume = 8.94 × 1 = 8.94 g According to Faraday’s laws of electrolysis 31.75 g of Cu is deposited by = 96500 coulombs of electricity ∴ 8.94 g of Cu is deposited by =

× 8.94

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35.

= 27171.96 coulombs increased; Formed salt will be a strong electrolyte.

36. (a) (P)

Initially conductivity increases because on neutralisation ions are created. After that it becomes practically constant because X alone cannot form ions. (Q)

(R)

(S) 37 38.

Number of ions in the solution remains constant as only AgNO3 precipitated as AgI. Thereafter, conductance increases due to increase in number of ions. Initially conductance decreases due to the decrease in the number of ions as OH– is getting replaced by CH3COO– which has poorer conductivity. Thereafter, it slowly increases due to the increase in number of H+ ions. Initially it decreases due to decrease in H+ ions and then increases due to the increase in OH– ions. (i)-(s); (ii)-(r); (iii)-(q); (iv)-(p) Cl2 + 2e– (b) Reaction at anode: 2Cl– moles of Cl– = 4 × 500 × 10–3 = 2 moles Cl2 =

39. (d) 500 mL of 4.0 molar NaCl has 2 mole of NaCl. By electrolysis, we can get a maximum of 2 moles of sodium which can combine with exactly 2 moles of mercury to give amalgam. ∴ The maximum weight of amalgam which can be formed from this solution = weight of 2 mole of sodium + weight of 2 mole of mercury = 2 × 23 + 2 × 200 = 446g Na 40. (d) Na+ + e– Total number of moles of Na+ discharged at cathode = 2mole ∴ The number of electron required for this purpose

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= 2 mole ∴ Total charge required = 2 faraday = 2 × 96500 = 193000 coulombs. 41. Given : NOTE THIS STEP : To find the specific conductivity (κ) of the final solution of AgBr in which AgNO3 (10–7 M) is mixed, we must find the individual κ of the ions. or Again, Calculation of molar concentration of ions : Concentration, Let x be the molar concentration of Ag+ from AgBr

or, and

Similarly,

and

So, the correct answer is 55. 42. w = Zit

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Z for Cu =

; t = 16 × 60 sec

=

g

Wt. of Cu at 50% electrolysis of CuSO4 g

=

Wt. of Cu at 100% electrolysis of CuSO4 g = 0.198 × 63.5 ×10–4g

= CuSO4

Cu

= 0.198 × 10–4 mol.

Conc. of CuSO4 = 0.198 ×10–4 = 7.95 × 10–5 mol/L 43. Volume of Ag

cm3

Surface area = 44. CrO3 + 6H+ + 6e– ——→ Cr + 3H2O Eq. wt. of Cr = = (i)

96500 coulomb deposit =

g Cr

∴ 24000 coulomb deposit =

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= 2.1554 g of Cr (ii) Also given, wCr = 1.5 g, i = 12.5 ampere, t = ?, ECr = ∴ w=

or 1.5 =

∴ t = 1336.15 second 45. C6H5NO2 + 6H+ + 6e–

C6H5NH2 + 2H2O

Eq. wt of C6H5NO2 = current efficiency = 50%

12.3 = i × t = Q = 115800 Coulomb Energy used = 115800 × 3 = 347.4 kJ. 46. The chemical reactions taking place at the two electrodes are At cathode : Cu2+ + 2e– → Cu H+ + OH– H2O NOTE : Only Cu2+ ions will be discharged so as these are present in solution and H+ ions will be discharged only when all the Cu2+ ions have been deposited. At anode : 2OH– → H2O + O + 2e– O + O → O2 Thus in first case, Cu2+ ion will be discharged at the cathode and O2 gas at the anode. Let us calculate the volume of gas (O2) discharged during electrolysis. According to Faraday’s second law 31.75 g Cu ≡ 8 g of oxygen ≡ 5.6 litres of O2 at NTP 0.4 g Cu =

× 0.4 litres of O2 at NTP

= 0.07055 litres = 70.55 mL

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As explained earlier, when all the Cu2+ ion will be deposited at cathode, H+ ions will start going to cathode liberating hydrogen (H2) gas, i.e. H H + H → H2 H+ + e– NOTE THIS STEP : However, the anode reaction remains same as previous. Thus in the second (latter) case, amount of H2 collected at cathode should be calculated. 8 g of O2 ≡ 1 g of H2 5.6 litres of O2 at NTP = 11.2 litres of hydrogen Quantity of electricity passed after 1st electrolysis, i.e. Q = i × t = 1.2 × 7 × 60 = 504 coulombs = 29.24 mL of O2.

504 coulombs will liberate = Similarly, = 11.2 ×

H2 liberated by 504 coulombs = 58.48 mL

(Twice the volume of O2 liberated in latter phase = 2 × 29.24 = 58.48 mL) Total volume of O2 liberated = 70.55 + 29.24 = 99.79 mL Vol. of H2 liberated = 58.48 mL 47. For the given reactions, it is obvious that 22.4 litres of H2 gas require 2 Faraday electricity. ∴ 67.2 litres of H2 will produce = 6 Faraday electricity Q = C × t; 6 × 96500 = C × 15 × 60 C=

= 643.3 ampere

Calculation of amount of Cu deposited by 6 F Since 1 F deposits =

= 31.75 g of Cu

6 F will deposit = 31.75 × 6 g = 190.50 g 48. In lead storage battery, the anodic and cathodic reactions during discharge (or operation or working) are as : (i) Anodic reaction : Pb(s) + SO2–4 (aq) → PbSO4(s) + 2e– (ii) Cathodic reaction :

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PbO2(s) + SO2–4 (aq) + 4H+ (aq) + 2e–

→ PbSO4(s) + 2H2O(l) In both the half cell reactions, H2SO4 is consumed and hence, conc. of H2SO4 decreases during the working (discharging, of the battery. For the withdrawl of 2F = 2 × 96500 C of electric charge, 2 mol of H2SO4 are consumed. Density of H2SO4 solution (used as electrolyte) falls during working of the cell. Both reactions get reversed on charging the battery, leading to regeneration of H2SO4 as : Formerly anode but now cathode (recharging) PbSO4 (s) + 2e– → Pb(s) + SO2–4(aq) Formerly cathode but now anode : PbSO4(s) + 2H2O(l) → PbO2(s) + SO2–4(aq) + 4H+(aq) + 2e– Molarity of H2SO4 before electrolysis

= 5.15 Moles of H2SO4 before electrolysis = 3.15 × 3.5 = 18.025 Molarity of H2SO4 after electrolysis

= 2.32 Moles of H2SO4 after electrolysis = 2.32 × 3.5 = 8.12 The overall discharging reaction is : Here, 2SO2–4 requires 2e– hence n-factor = 1 i.e. Equivalent mass of H2SO4 = 98/1 = 98 Moles or equivalents of H2SO4 used = 18.025 – 8.12 = 9.905 Number of coulomb required = 9.905 × 96500

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i × t = 955350 A-s = 265.375 A-h 49. Gold deposited in the first cell = 9.85 g At. wt. of Gold = 197, Oxidation number of gold = +3 Eq. Wt. of Gold = W = Zit Charge required to deposit 1 g eq. of gold = 1F = 96,500 C ∴ Charge required to deposit 9.85 g of gold or g eq. of gold =

C

= 965 × 5 × 3 C = 14475 C According to Faraday’s second law, = ⇒ Wt. of Cu deposited = Current =

1.

=

A=

= 4.7625 g A = 0.8042 A

(d) For the concentration cell,

Anode : Cathode : Overall : As If

, then Ecell is +ve.

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So, C2 > C1. Thus, 2. So, if

relation is correct.

(b) = 1.1 V no electron will flow

At,

> 1.1 V cell act as electrolytic cell and electron will flow from Cu

At,

to Zn. < 1.1 V cell act as electrochemical cells so Zn dissolve and Cu

deposit. 3. (b) Cu2+ + 2e– Cu, G1° = – 2F(0.34) ...(i) Cu+ + e– Cu, G2° = – F(0.522) ...(ii) Subtract (ii) from (i) Cu2+ + e– Cu+, G3° = –F(E0) G1° – G2° = G3° –FE° = –2F(0.34) + F(0.522) E° = 0.68 – 0.522 = 0.158 V 4. (a) G° = –nFE°cell = –2 × (96000) × 2 V = – 384000 J/mol = –384 kJ/mol 5. (c)

= 3z – 2y

=x

(given)

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∴

= x – 3z + 2y (n = 2e–)

6. (a) ° ° ° Ecell = (ER.P. )cathode – (ER.P. )anode = 0.80 – (–0.76) = 1.56 V for 2e– = 0.78 V

°  Ecell for 1e– =

7.

(d) We know that,

ln K = G = – nF E°cell G = – RT lnk Now, After putting the given values, we get = 160

ln K = K = e160 8.

(a)

E0 for 9.

(b) Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s)

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∆G = ∆G° + 2.303 RT log10Q ; Q = [∆G° = –nFE°] = –2 × F × 1.1 Given [Zn2+] = 10[Cu2+] ∴ ∆G = –2F (1.1) + 2.303 RT log1010 = 2.303 RT – 2.2F 10. (b) Galvanization is the process by which zinc is coated over corrosive (easily rusted) metals to prevent them from corrosion. 11. (c) Corrosion of iron can be minimized by forming an impermeable barrier at its surface. 12.

(d) At anode : H2(g)

At cathode : M4+ (aq) + 2e–

2H+ (aq) + 2e– M2+ (aq)

Net cell reaction : H2(g) + M4+ (aq)

2H+ (aq) + M2+ (aq)

–

Now, Ecell =

or, 0.092 = (0.151 – 0) – ∴

= 102 ⇒ x = 2

...(i) (a) (a) ... (ii) (b) Now multiplying equation (ii) by two and subtracting from equation (i) 13.

= – 1.18 + (– 1.51) = – 2.69 V (–ve value of EMF (i.e. ∆G = +ve) shows that the reaction is nonspontaneous)

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14.

(d) Higher the value of standard reduction potential, stronger is the oxidising agent, hence MnO4– is the strongest oxidising agent. 15. (d) Here n = 4, and [H+] = 10–pH = 10– 3 Applying Nernst equation E = Eº –

= 1.67 – 0.105 = 1.565 V 16.

(a) Fe(s) → Fe2+ +2e– ; E° = 0.44 V 2H+ + 2e– + ½O2 → H2O(l) ;E° = +1.23 V

Fe(s) + 2H+ + ½O2 → Fe2+ + H2O ; E°cell = 0.44 + 1.23 = 1.67 V ∴ ∆G° = –nFE°cell = –2 × 96500 × 1.67 = –322 kJ 17. (b) Cell reaction : Using Nernst equation

At 298 K, E = 0.2905

or 18.

or

.

(c) NOTE : In an electrolytic cell, electrons do not flow themselves. It is the migration of ions towards oppositely charged electrodes that indirectly constitutes the flow of electrons from cathode to anode through internal supply.

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19.

(a) MnO4–will oxidise Cl– ion according to the following equation:

The cell corresponding to this reaction is as follows : Pt, Cl2 (1 atm) | Cl– || MnO4–, Mn2+, H+ | Pt E°cell = 1.51 – 1.40 = 0.11 V E°cell being +ve, ∆G° will be –ve and hence, the above reaction is will not only oxidise Fe2+ ion but also Cl– ion feasible. simultaneously. So, the quantitative estimation of aq Fe(NO3)2 cannot be done by this. 20. (c) The salt used to make ‘salt−bridge’ must be such that the ionic mobility of cation and anion are of comparable order so that they can keep the anode and cathode half cells neutral at all times. KNO3 is used becasue velocities of K+ and NO3– ions are nearly same. – 21. (b) For M + + X → M + X, E°cell = 0.44 – 0.33 = 0.11V is positive, hence reaction is spontaneous. 22. (a) The given order of reduction potentials is Z > Y > X. A spontaneous reaction will have the following characteristics: (i) Z reduced and Y oxidised, (ii) Z reduced and X oxidised and (iii) Y reduced and X oxidised Hence, Y will oxidise X and not Z. 23. (c) We have Half -cell Half-cell reaction Cu2+ | Cu Cu2+ + 2e– = Cu Cu2+ | Cu+ Cu2+ + e– = Cu+ Cu+ | Cu

Cu+ + e– = Cu

From the half-cell reactions, it follows that i.e., or

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24.

= 2(0.337 V) – 0.153 V = 0.521 V (a) H2O is more readily reduced at cathode than Na+. It is also more readily oxidized at anode than SO2–4. Hence, the electrode reactions are

25. (b) TIPS/FORMULAE : (i) In a galvanic cell oxidation occurs at anode and reduction occurs at cathode. (ii) Oxidation occurs at electrode having higher oxidation potential and it behaves as anode and other electrode acts as cathode. (iii) ECell = EC – EA (substitute reduction potential at both places). and

26.

∴ Zn is anode and Fe is cathode. Ecell = EC – EA = –0.41 – (–0.76) = 0.35V. (a) Water is reduced at the cathode and oxidized at the anode instead . of Na+ and Cathode : 2H2O + 2e– → H2 + 2OH– ; E° = –0.83 V

Anode : H2O → 2H+ +

O2 + 2e– ; E° = –1.23 V

Note : The standard electrode, reduction potential of Na+ is less than that of water.

The standard electrode, oxidation potential of SO2–4 is less than that of water.

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27.

(c) NOTE : Oxidation is loss of electron and in a galvanic cell it occurs at anode. Reduction is gain of electron and in a galvanic cell it occurs at cathode. Cell representation : Anode / Anodic electrolyte || Cathodic electrolyte / Cathode Reaction at Anode : H2 → 2H+ + 2e– Reaction at Cathode : AgCl + e– → Ag + Cl– 28. (c) The reduction potentials (as given) of the ions are in the order : Ag+ > Hg22+ > Cu2+ > Mg2+ Mg2+ (aq.) will not be reduced as its reduction potential is much lower than that of water (–0.83 V). Hence, the sequence of deposition of the metals will be Ag, Hg, Cu. 29. (a) More negative is the value of reduction potential, higher will be the reducing property, i.e., the power to give up electrons. 30.

(10)

2.3 = lnx; x = 10 31. (4) X Y; M+

M3+ + 2e–

∆G° = – 193 kJ mol–1 E° = – 0.25V

Hence ∆G° for oxidation will be ∆G° = – nFE° = –2 × 96500 × (–0.25) = 48250 J = 48.25 kJ 48.25 kJ energy oxidises one mole M+

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∴

193 kJ energy oxidises

32.

(– 6)

33.

(96500)

mole M+ = 4 mole M+

= –0.16 – 0.34 = –0.50 V = –2 × 96500 × (–0.5) = 96500 J 34. (144)

35.

(1.52) E = 1.23 –

log[H+]4

= 1.23 + 0.0591 × pH = 1.23 + 0.0591 × 5 = 1.23 + 0.2955 = 1.52 V 36. (2.15) At equilibrium state Ecell = 0; E0cell = 0.01 V Sn + Pb2+ → Sn2+ + Pb E=

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0= – 0.01 = = = 101/3 = 2.15 37.

(– 11.62) A (s) | An+ (aq, 2M) || B2n+ (aq, 1M) | B (s) Reactions

Overall reaction :

; Now, ∆G° = ∆H° – T∆S° = 2∆G° – T∆S° (Given ∆H° = 2∆G°) Τ∆S° = ∆G° = – R × 2 ln2 = –8.3 × 2 × 0.7 = –11.62 JK–1 mol–1 38. (0.05) Daniel cell is : Zn | Zn2+ | | Cu2+ | Cu

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Let there be two Daniel cells with their Ecell as given below: Zn | Zn2+ (C1) | | Cu2+ (C = ?) | Cu, Ecell = E1 Zn | Zn2+ (C2) | | Cu2+ (C = 0.5 M) | Cu Ecell = E2 where E2 > E1 According to question, E2 – E1 = 0.03 and C2 = C1 The cell reaction is . So, Thus,

;

and

;

Since, same ZnSO4 is used in both cells C1 = C2 So,

or C = 0.05 M 39. (–0.22) At pH = 14; [H+] = 1 10-14 M; [OH–] = 100 = 1M ( [H+] [OH–] = 1 10–14) Cu (OH)2 ionises as follows: Cu2+ + 2OH– Cu(OH)2 Ksp of Cu(OH)2 = [Cu2+][OH–]2 1.0 10–19 = [Cu2+][1]2 ; [Cu2+] = 1.0 10–19 M The standard reduction potential of Cu2+/Cu is represented in the form of following equation: On applying Nernst equation

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= 0.34 – 0.56 = – 0.22V 40.

(0.792)

Initial conc. Eqilb.

conc.

2Hg +2Fe3+

Hg22+

1.0 × 10 –3 0.05

0 ×

10 –3

+ 2Fe2+ 0 0.95

×

10 –3

NOTE : At equilibrium, E = 0

On usual calculations, 41.

negative, greater; Among the various metals, since sodium has the minimum reduction potential, it must be strongest reducing agent. In general, more the reduction potential lesser is its reducing action. 42. False : When the temperature is 273, the value of the factor will come out as 0.0541 instead of 0.0591. The value 0.0591 comes out at 298 K and not at 273 K. 43. (a) Salt bridge is introduced to keep the solutions of two electrodes separate, so that the ions in electrodes do not mix freely with each other. Salt bridge maintains the diffusion of ions from one electrode to another. 44. (a, b, d) The species having less reduction potential with respect to NO–3(E° = + 0.96 V) will be oxidised by NO–3. These species are V,

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45.

Fe and Hg. (a) NOTE : More negative or lower is the reduction potential, more is the reducing property. Thus, the reducing power of the corresponding metal will follow the reverse order, i.e. Y > Z > X.

46. (d) (P)

⇒

⇒

3 × x = 1 × 0.77 + 2 × (–0.44)

⇒

(Q)

(R)

(S) x × 1 + 2 × (– 0.91) = 3 × (– 0.74) x – 1.82 = – 2.22 ⇒ x = – 0.4V 47. (d) At anode : M(s) + 2X– (aq) At cathode : M+2 (aq) + 2e– Thus, here n = 2

MX2(aq) + 2e– M(s)

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48.

∆G = – nFEcell = – 2 × 96500 × 0.059 × 10–3 kJ/mole = – 11.4 kJ/ mole (b) M|M 2+ (aq) || M 2+ (aq) |M 0.001 M

Anode : Cathode : __________________________

⇒

⇒ 10 – 2 × 10– 3 = M2+ (aq)a = solubility = s MX2 M 2++ 2X – S

2S

Ksp = S.(2S) ⇒ Ksp = 4s3 = 4 × (10–5)3 = 4 × 10–15 49. (b) According to Nernst equation, 2

= + ve Hence, |Ecell| = Ecell = 0.070 V and ∆G < 0 for the feasibility of the reaction. Note: For concentration cell, 50.

(c) From above equation

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⇒ So, Ecell = 0.13988V 140 mV 51.

(c) E° =

; E° = 0.82V

E° is positive hence, iodide ion is oxidized by chlorine. 52.

(d) = 1.50 + (– 1.23) = 0.27 V

Reaction is feasible. [ ∴Eo is positive] 53. (a) The precipitate formed in this reaction is of Fe4 [ Fe(CN)6]3. 54. (d) In the given reaction, Ag+ ions are reduce to Ag and Glucose is oxidised to gluconic acid as per the given reactions, and

Hence,

55. (a)

For the reaction,

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So, Eoxidation increases over

by 0.65 V.

56. (b) During Tollen’s test, oxidation of silver ion requires an alkaline medium. Under these conditions it forms insoluble silver oxide, hence to dissolve this oxide a complexing agent, ammonia is added, which brings silver ion as diamminosilver (I) ion, [Ag(NH3)2]+. It is a soluble complex. 57. (a) From the given details, the reactions can be written as:

Hence, cell representation is Ag (s) | AgCl (s) | (aq) | | =

(aq) | Ag (s)

(AgCl) –

= – 109 – (– 129 + 77) = – 57 kJ/mol = – 57000 J/mol We know that, =–nF – 57000 = – 1 × 96500 × ( = Again or

n = electron transferred = 1)

= 0.59 volts

= = = (

log

[AgCl (s) = 1 and

=

)

or 0.59 = – 0.059log or log

= – 10

= 10–10

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(b) When Zn is added to 100 mL of saturated AgCl solution. 2 +

Ag; Eº = 0.80 V

+2

Zn; Eº = – 0.76 V

= = 0.80 – ( – 0.76) = 1.56 V = 1.56 = = 52.9 NOTE : As the value of equilbrium constant is very high so the reaction moves in forward direction completely. from (a) = [(

= =

Ag+ in 100 mL of solution = 58.

=

] = 10–6 mol.

The required reaction can be obtained in the following way.

On adding, Now we know that – n F Eº = – 0.59 F or or

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;

; 59.

(i)

Kc = 1010

The half cell reactions are

At anode At cathode The cell reaction

(ii) TIPS/FORMULAE : We know that ∆So = nF n F dEo dT

No. of transferred electrons = 1 faraday number = 96500 coulombs Difference of electrode potential temperatures = (0.21 – 0.23) = –0.02V Difference of two temperatures = (35°C – 15°C) = 20°C

at

two

different

; ∴

;

so

= –22195 Jmole = –49987 J/mole.

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(iii)

of cell = The corresponding cell is represented as : In form of oxidised electrode potential E°cell = E°anode – E°Cathode = E°Ag/Ag + – E°Ag/AgCl/Cl– = – 0.80 – (0.22) = 0.58V E°cell = AgCl(s)

Ag+ + Cl–

E°cell = Therefore or log10Ksp = –9.8139 =

; Ksp = 1.54 × 10–10

Ksp of AgCl = 1.54 × 10–10 (mole Litre–1)2 Solubility of AgCl = 60.

Given,

Thus for

; to be positive, following reaction should occur

Hence, Ce4+/Ce3+ electrode will act as cathode andFe3+ / Fe2+ electrode will act as anode. Therefore, current will flow from Ce electrode to Fe electrode. Current will decrease with time. 61. Note that the given cell will not work as electrochemical cell since The equation for electro-chemical cells will be:

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Cu Cu2+ + 2e– 2Ag+ + 2e– 2Ag Thus, e.m.f. of cell Cu | Cu2+ || Ag+ |Ag will be

[Ag+] = 1M and [Cu2+] = 1M ⇒ After the passage of 9.65 ampere for 1 hr i.e. 9.65 60 60 Coulomb charge,during which the cell reactions are reversed, the Ag metal passes in solution state and Cu2+ ions are discharged. The reactions during the passage of current are: 2Ag 2Ag+ + 2e– and Cu2+ + 2e– Cu Thus, Ag+ formed =

= 0.36 eq. = 0.36 mole

Cu2+ discharged =

= 0.36 eq. = 0.18 mole

Thus, [Ag+] left = 1 + 0.36 = 1.36 mole [Cu2+] left = 1–0.18 = 0.82 mole. Now e.m.f. can be given as: =

+ 0.010 V

Thus, increases by 0.010 V. 62. The cell reaction can be written as: E = 0.164 V At cathode :

+ e–

At anode : Ag

Ag+anode + e–

Net reaction : Ag+cathode

Ag Ag+anode ; E = 0.164 V

Thus here, n = 1, E = 0.164 V, [Ag+] cathode = 0.1 M Let the solubility of Ag2CrO4 be S M

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Since, Ag2CrO4 gives 2 Ag+. ∴ Here, concentration of [Ag+]anode = 2 S M

(for a concentration cell, Eo = 0) ∴ 0.164 = 0.164 = ∴ 2S = 1.697 × 10–4

or 0.164 =

Hence, S = 0.8485 × 10–4 M For

63.

2 Ag+ + CrO42–

∴ Ksp = 4 × (0.8485 × 10–4)3 = 2.44 × 10–12 For the change 2Fe3+ + 3I– 2Fe2+ + I3–, = 0.77 – 0.54 = 0.23 V

At equilibrium, Ecell = 0 Thus, 0.23 =

( Using Nernst equation)

log Kc ∴ KC = 6.26

64.

107

; E° = 0.68V

E°cell = 1.44 – 0.68 = + 0.76 V at equilibriums, Ecell = 0 ; 0.76 =

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or log10 Kc =

Kc = 7.6 ×1012

65.

The thin protective layer of oxides of aluminium is formed which protects the metal from further attack of water and air and make it stable.

66.

(i)

Given

= +0.40 V; = –0.40 V;

Since

for Fe/FeO >

= – 0.87 V

= +0.87V for NiO/Ni2O3.

Redox changes can be written as At anode : Fe(s) + 2OH– → FeO(s) + H2O(l) + 2e– At cathode : Ni2O3(s) + H2O(l) + 2e– → 2NiO(s) + 2OH– Cell reaction : Fe(s) + Ni2O3(s) → FeO(s) + 2NiO(s) (ii) Ecell =

+

= 0.87 + 0.40 = 1.27 V It is independent of conc. of KOH. (iii) Electrical energy = nFEcell = 2 × 96500 J V–1 × 1.27 V = 2.45 × 105 J 67. Eº = Standard reduction potential of the Ag+/Ag electrode = 0.799 V AgI (s) Ag+ + I– Ksp = [Ag+] [I–] = 8.7 × 10–17 (given) If ‘S’ is the solubility of AgI, then Ksp = S2 ∴S=

=

= 9.327 × 10–9 mol L–1

∴ [Ag+] = [I–] = 9.327 × 10–9 M Reaction : Ag+ + e– ——→ Ag ∴E = Eº –

log

= 0.799 V –

log

V

[ Activity of the electrode material in pure solid state is taken as one]

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= 0.799 – 0.059 log 0.1072 × 109 V = 0.799 – 0.474 = 0.325 V Again, L.H.S. electrode reaction : Ag → Ag+ + e– R.H.S. electrode reaction : AgI(s) + e– → Ag + I– Cell reaction : AgI(s) → Ag+ + I– K = Equilibrium constant = [Ag+] [I–] = 8.7 × 10–17 The standard cell emf Eº and the equilibrium constant K are related by the expression: =

log K at 298 K, Here, n = 1, K = 8.7 × 10–17

= 0.059 log 8.7 × 10–17 = 0.059 [0.9395 – 17] = –0.948 V But

=

∴

=

– +

= –0.948 + 0.799 = –0.149 V

68. For the half-cell reaction NO3–(aq) + 2H+(aq) + e– ——→ NO2(g) + H2O(l) The Nernst equation is E = Eº –

log

Substituting the values in case of (i) E = 0.78 –

log

= 0.78 + 0.059 log 64 = 0.887 V

Substituting the value in the Nernst equation in case (ii) E = 0.78 –

log

= 0.78 – 0.059 log 10–14

= 0.78 – (0.059) × (14 )= –0.046 V 69. 2Cl–(aq) + 2H2O = 2OH–(aq) + H2(g) + Cl2(g) Reaction at anode : 2Cl– → Cl2 + 2e– Reaction at cathode : 2H2O + 2e– → H2 + 2OH– i=

× 25 = 15.4 amperes

Weight of Cl2 deposited = 1 kg or 1000 g

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We know that

;

=

t = 175300 sec. or 48.69 hours No. of moles of Cl2 thus produced =

= 14.08

Amount of OH– released in the electrolysis = 2 × 14.08 moles = 28.16 moles ∴ Molarity with respect to OH– =

= 1.408 M

70.

Ag | AgCl(s), KCl(0.2M) || KBr(0.001M), AgBr(s) | Ag Anode Cathode –10 Ksp(AgCl) = 2.8 × 10 Ksp(AgBr) = 3.3 × 10–13 At anode, 1Ag → 1Ag+ + e– At cathode, 2Ag+ + e– → 2Ag ∴ Cell reaction 1Ag + 2Ag+ → 2Ag + 1Ag+ NOTE : The subscripts 1 and 2 on Ag denote the species concerned with anode and cathode respectively. Applying Nernst equation E = Eº – =0–

log log

[1Ag] = [2Ag] = 1 ( these are in solid state) –10 Ksp (AgCl) = 2.8 × 10 or [1Ag+] [Cl–] = 2.8 × 10–10 [1Ag+]

=

= 14 × 10–10

Ksp (AgBr) = 3.3 × 10–13 or [2Ag+] = ∴E=–

(

[Cl–] = 0.2)

(

[Br–] = 0.001)

[2Ag+] [Br–] = 3.3 × 10–13

= 3.3 ×10–10 log

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= –0.059 log

= –0.059 × 0.6276 = –0.037 V

Since, emf is negative this shows that the reaction is non-spontaneous. NOTE : For the reaction to be spontaneous, its emf should be positive i.e. E = 0.037 V and its polarities should be reversed i.e. anode should be made cathode and vice-versa. So, the galvanic cell is : Ag | AgBr(s), KBr || AgCl(s), KCl | Ag In other words, Ag | AgBr acts as anode and AgCl | Ag acts as cathode. 71. The following chemical cell sets up : Zn | Zn2+ || Ni2+ | Ni The net cell reaction is : Zn + Ni2+ Zn2+ + Ni The e.m.f. is given by Ecell =

–

–

log

= –0.24 – (–0.75) – 0.0295 log = 0.51 – 0.0295 log At equilibrium Ecell = 0 Let x mol L–1 be the concentration of Ni2+ at equilibrium. Then [Zn2+] = 1 – x [ 1 mole of Ni2+ gives 1 mole of Zn2+] ∴ 0.0295 log or log

= 0.51

=

= 17.29 or

or x = 72.

= 1.95 × 1017

= 5.128 × 10–18 mol l–1 = 0.337 and +

= 0.799 V = 0.799 – 0.337 = 0.462 V

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∴ Cu + 2Ag+ → Cu2+ + 2Ag;

= 0.462 V

Hence, the galvanic cell in question will consist of anode of copper and cathode of silver. Calculation of concentration : Ecell=

– =

0.462 =

log log

[

Ecell = 0]

log

[ n = 2]

= log (10–2) – log [Ag+]2 = – 2 – 2log [Ag+] ⇒ [Ag+]= 1.48 × 10–9 M 73.

TIPS/FORMULAE : or

Let us split the desired reaction into two half cell reactions: Oxidation half reaction : H2O +

H2(g)

H3O+ + e–

Eº = 0.00 V

Reduction half reaction : H2O + e–

H2 + OH–

Eº = –08.277 V

Net reaction : 2H2O H3O+ + OH–

= –0.8277 V

So, the number of electrons involved in redox reaction, (n) = 1 We know that

=

log Kc

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log Kc = 74.

=

= –14.005

Kc = Antilog [–14.005] = 9.88 × 10–15 TIPS/FORMULAE : For a concentration cell Ecell =

log

NOTE : It is a concentration cell as both the electrodes are made of same element. Negative electrode acts as anode in a galvanic cell. [H+] = 10–6 M At anode; At cothode; 2H + + 2e–

H2

or 0.118 =

Ecell = log

=

[H+] = ?

=2 ⇒

log

= 10–4 M

75. Half cell reactions will be Zn2+ + 2e– Zn H+ + e–

......(i)

H2 or 2H+ + 2e–

We know that

=

H2 –

......(ii)

ln

Here, R = 8.314 Jmol–1 deg–1, T = 298 K, F = 96,500 coul/equi, n = 2, = 0.76 V. Substituting the values in the above equation = 0.76 – Similarly,

ln =

= 0.79 V –

ln

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=0–

ln

= 0.05915 log10 [H+] = –0.05915pH ( –log10 [H+] = pH) Now since E =

+

0.28 = 0.79 – 0.05915 pH ⇒ pH = 76.

= 8.62

(i)

The two half cell reactions can be written as below : Oxidation half reaction : Zn → Zn2+ + 2e– Reduction half reaction : Cu2+ + 2e– → Cu Thus the cell reaction will be : Zn + Cu2+ → Zn2+ + Cu (ii) EMF of cell, Eºcell = Eºcathode – Eºanode Eºcell = 0.350 – (– 0.763) = 0.350 + 0.763 volts = 1.113 volts (iii) Since emf of the cell is positive, the reaction as written is spontaneous.

77.

(a) 119g Sn deposits from = 380g SnCl2 ∴ 0.119g Sn deposits from =

= 0.380g SnCl2

380g SnCl2 gives = 261g SnCl4 ∴

0.380 g SnCl2 gives =

∴

Wt of SnCl2 left after decomposition

= 0.261g SnCl4

= 19.00 – 0.380 = 18.620 g. Ratio SnCl2 : SnCl4 71.34 : 1 ⇒ 18.620 : 0.261 (b) At Cathode; Na+ + e– 2Na + H2O

Na

2NaOH + H2

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At anode; Cl– Cl + Cl

Cl + e– Cl2

2OH– + Cl2 OCl– + 2HOCl Na+ + ClO3–

Cl– + OCl– + H2O ClO3– + 2Cl– + 2H+ NaClO3 Sod.Chlorate

On prolonged electrolysis Na+ + ClO4–

NaCl O4 Sod. perchlorate

(c) Charge on N = 3 No. of ions in 14 g of N3– = 6.02 × 1023 3–

No. of ions in 1g of N3– = No. of electronic charges on 1 g N3– = Charge on 1 g of N3– =

Coulombs

( Charge on one electron is 1.6 × 10–19 Coulombs) = 2.06 × 104 Coulombs

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1. (d) Rate = k [A]n log[Rate] = log k + n log [A] Slope = n [n is order of the reaction] ∴ Correct sequence for the order of the reaction is (iv) > (ii) > (i) > (iii) 2.

(b)

3. 4.

(b) Zero order reaction is always multi step reaction. (c) For a given reaction,

rate

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rate 5. (c) Rate (R) = k[A]a [B]b Exp I Exp II Exp III Exp IV Exp V From Exp I & II,

Similarly, from exp I & III we get

From Exp. II & IV,

From Exp. I & V,

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6.

(b) Among the given bases (A) and (B), t-butoxide being bulky base favours elimination reaction and ethoxide favours substitution reaction. when Z = CH3CH2O–, (substitution reaction favoured)

and when Z =

,(elimination reaction favoured)

Hence, A > B and ke(B) > ke(A) 7. (d) Rate of forward reaction = kf [NO]2[H2]2 Observed rate = kf [NO]2[H2] Observed rate = Rate of backward reaction = 8.

(b)

rate of reaction = According to the question

= 1.67 × 10–2 M min–1 9. (a) In graph (i), ln [Reactant] vs time is linear with positive intercept and negative slope. Hence it is 1st order. In graph (ii), [Reactant] vs time is linear with positive intercept and negative slope. Hence, it is zero order.

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10.

(d) = k1[A] – k2[B] = 0

[B] = 11. (a) Rate constant of decomposition of X = 0.05 µg/year. Unit of rate constant confirms that the decomposition of X is a zero order reaction. For zero order kinetics, [X] = [X]0 – kt kt = [X]0 – [X] t= t=

=

= 50 years

12. (c) For the reaction 2X B, follow zeroth order Rate equation is Kt = [A]0 – [A] For the half-life; t = t1/2 and [A] = 0.1 K t1/2 = 0.2 – 0.1

Time required to reach from 0.5 M to 0.2 M Kt = [A]0 – [A] t = (0.5 – 0.2); t = 18 hour 13. Now,

(d) Given: = k–1[A]2 – k1[A2] = –2k–1[A]2 + 2k1[A2] = 2k1[A2] – 2k–1[A]2

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14.

(a)

Generally m = order of reaction a – x = unreacted r1 = 1 torr s–1, when 5% reacted r2 = 0.5 torr s–1, when 33 % reacted

2 = (1.41)m ⇒ 2 = ⇒m=2

15.

(d)

First order reaction as half life is constant. 16.

(d) H2O2 (aq) → H2O (aq) +

O2 (g)

For a first order reaction k= Given a = 0.5, (a – x) = 0.125, t = 50 min ∴k=

= 2.78 × 10–2 min–1

r = k[H2O2] = 2.78 × 10–2 × 0.05 = 1.386 × 10–3 mol min–1

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Now,

= ∴

=

∴

=

=

=

= 6.93 × 10–4 mol min–1

17. 18.

(d) Rate constant is independent of concentration. (c) Reactions of higher order (>3) are very rare due to very less chances of many molecules to undergo effective collisions. 19. (c) Rate = k [A][B] = R R' = k [A][2B]

⇒ 2R = R' i.e., rate become doubles. 20.

(d) Let rate of reaction =

Now from the given data 1.2 × 10 – 3 = k [0.1]x[0.1]y 1.2 × 10 – 3 = k [0.1]x[0.2]y 2.4 × 10 – 3 = k [0.2]x[0.1]y Dividing equation (i) by (ii)

= k[A]x [B]y .....(i) .....(ii) .....(iii)

⇒ We find, y = 0 Now dividing equation (i) by (iii) ⇒ We find, x = 1 Hence,

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21. (c) Given Total time (T ) = 1 hr = 60 min From T = n × n= Now from the formula Where N0 = initial amout N = amount left after time t Hence the amount of substance left after 1 hour will be 22.

(b) M → N r = k [M]x when concentration = 2M; rate = 8r, thus 8r = k[2M]x 8 = (2)x x=3 23. (d) For a zero order reaction Rate constant = k = 2 × 10–2 = a – 0.5 = 0.5 a = 1.0 M 24. (d) For P, if t50% = x then t75% = 2x This is true only for first order reaction. So, order with respect to P is 1. Further, the graph shows that concentration of Q decreases with time. So, rate with respect to Q, remains constant. Hence, it is zero order wrt Q. So, overall order is 1+ 0 = 1 25. (a) The values of rate constants k0, k1 for zero order and first order reaction, respectively, are given by the following equation: [where Ars = initial concentration,

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and t1/2 = half-life period] and substituting various given values, we get ... (i) ... (ii)

and Dividing (ii) by (i), we get

=

mol–1 litre = 0.5 mol–1 litre

= 0.5 mol–1 dm3 [1 litre = 1dm3] Thus, the correct answer is (a). 26. (d) TIPS/Formulae : Overall order = sum of orders w.r.t each reactant. Let the order be x and y for G and H respectively

Applying r = k [G]x [H]y we get, x = 1, y = 2 For (1) and (3), the rate is doubled when conc. of G is doubled ∴x=1 keeping that of H constant i.e., From (2) and (3), y = 2 ∴ Overall order is 3. 27. (d) Order of a reaction can be fractional. Rest of all are true. [NOTE : Order of a reaction can be determined experimentally] 28. (c) TIPS/Formulae : For first order reaction,

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Rate = k [conc. of reactant] Since 0.1 M of A changes to 0.025 M in 40 minutes, t1/2 of reaction = 40/2 = 20 minutes Rate of reaction of A = 3.47 ×10–4 M min–1 29. (c) TIPS/Formulae :

30.

(a) NOTE : Individual rates of reactants and products become equal when each of these is divided by their respective stoichiometric coefficient. With time, concentration of reactants decreases and is represented by negative sign whereas concentration of products increases and is represented by positive sign. The given reaction is ∴

Correct relationship amongst the rate expression is shown in (a). 31. (b) NOTE : The rate of photochemical process varies with the intensity of absorption. Since, greater the intensity of absorbed light, more photons will fall at a point, and further, each photon causes one molecule to undergo reaction. 32. (d) TIPS/Formulae : Find the order of reaction and then use appropriate equation. As unit of k is sec–1 , reaction is of first order, mol/L 33.

(d) It is a characteristic constant of a particular reaction at a given temperature. It does not depend upon initial concentration of the reactants, time of reaction and extent of reaction.

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34.

(a) It is a constant of a particular reaction at a given temperature. It does not depend upon initial concentration of the reactants, time of reaction and extent of reaction. → Mn2+ + 35. (8) 8H+ + 5[Fe(H2O)2(Ox)2]2– + + 5 [Fe(H2O)2(Ox)2]– + 4H2O Rate = Hence, 36.

(9)

× 10 = 9

37.

(0)

Since, rate = ∆x/∆t is constant for different concentration values, order of reaction is zero. 38.

(60)

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39.

(23.03) t1/2 = 6.93 years, a = 10–6 g t1/2 = K=

= 0.1

For Ist order reaction, K= t= =

= = 23.03 years

40. (6.75) Sol. Rate of the reaction = K[A]x [B]y [C]z Compairing experiment 1 with 2 we get that, y=0 Compairing experiment 1 with 3 we get that, z=1 Compairing experiment 1 with 4 we get that, x=1 putting values of x, y, z in rate equation for experiment 1. 6 × 10–5 = K × (0.2)1 × (0.1)1 × (0.1)0 K = 3 × 10–3 Now, for the given concentration of A, B and C, rate of reaction will be, Rate = 3 × 10–3× 0.15 × 1 × 0.15 = 6.75 × 10–5 Therefore, value of Y = 6.75. 41. (2.30) Given rate constant of the reaction = 5 × 10–4sec–1 Thus, it is a first order reaction.

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1–x x 3 t = Y × 10 sec.

x/2

x = 0.90 atm For a first order reaction,

42.

(24.14) r1 = kc1 and r2 = kc2 Since, rate of first order reaction is directly proportional to the concentration of its reactant, ∴ According to first order reaction

On substituting the various values k = 0.0287 min–1

43. (0.749) CH3 – O – CH3(g) → CH4(g) + CO(g) + H2(g)

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Given t½ = 14.5 min, initial pressure = 0.40 atm, t = 12 min. Now, k =

= 4.78 × 10–2 min–1

Writing first order equation and substituting the given values, we get log

4.78 × 10–2 =

which gives x = 0.175 atm Since, volume and temp. are constant, final pressure : CH3 – O – CH3(g) → CH4(g) + CO(g) + H2(g) 0.4 0.4 – 0.175

0.175

0.175

0.175

Hence, total pressure = 0.4 – 0.175 + 3 × 0.175 = 0.749 atm 44. 1.765 × 10–4 kg/hr; N2(g) + 3H2(g) ——→ 2NH3(g) Rate of reaction =

[Rate of disappearance of H2] =

[Rate of

appearance of NH3] ⇒

=

or

= 0.01 kg/hr = ∴ 45. 46. 47. 48.

=

=

= × 1000 =

mole/hr =

mole/hr kg/hr

= 1.765 × 10–4 kg/hr. acidic, Pseudo first Product of active masses of reactants at that time True : The rate of reaction of first order is directly proportional to the concentration of reacting substance. (a)

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⇒

rate =

49.

(a, d)

t=0

P0

t = t1/3 t=t P0 – x 2x x So, Pt = P0 – x + 2x + x = P0 + 2x or 2x = Pt – P0

or

or or ln(3P0–Pt) = – kt + ln 2P0 Graph between ln(3P0 – Pt) vs ‘t’ is a straight line with negative slope. Since, rate constant is a constant quantity and independent of initial concentration. So, graph (a) and (d) are correct. 50. (a, b, d) For first order reaction [A] = [A]0e–kt Hence, concentration of [NO2] decreases exponentially.

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Also, t1/2 =

. Which is independent of concentration and t1/2 decreases

with the increase of temperature.

51.

(a, b, d)

The relevant expressions are as follows.

Choice (a) log Kp = Choice (b) log [X] = log [X]0 + k t Choice (c) P/T = constant (V constant) Choice (d) PV = constant (T constant). 52. (a, d) In first order reaction, if is the degree of dissociation then or

The Arrhenius equation is, Plot of reciprocal concentration of the reactant vs time is linear. Dimensions of pre-exponential factor ‘A’ are equivalent to dimensions of k, which is T –1 for a first order reaction. 53. (b, c) As rate = k [RCl], so it is first order reaction. On decreasing the concentration of RCl to half, the rate will also be halved. Rate will also increase with temperature. (half-life period) 54. (i) From the given data, it is evident that the for the decomposition of X (g) is constant (100 minutes) therefore, the order of reaction is one. =

(ii) Rate constant, k =

= 6.93 × 10–3 min–1

(iii) Time taken for 75% completion of reaction = 2 × 100 = 200 minutes =2 (iv)

3Y

2X Initial pressure After time t

800 (800 – 2P)

0 3P

+

2Z

0 2P

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when the pressure of X is 700 mm of Hg then, 800 – 2 P = 700 2 P = 100 ; P = 50 mm of Hg Total pressure = 800 – 2P + 3 P + 2P = 800 + 150 = 950 mm of Hg 55. (a) From the rate law expression, R0 = k [A0]a [B0]b and from the table it is clear that : (i) when the concentration of [A0] is doubled, keeping [B0] constant (see readings 1 and 2), the rate also doubles i.e. rate is directly proportional to [A0] or a = 1. (ii) when the concentration of [B0] is reduced, keeping [A0] constant (see readings 1 and 3), the rate remains constant. i.e. rate is independent of [B0] or b = 0 Thus, rate equation becomes R0 = k [A0] (b) 56. Let the number of moles of A left after 100 min = x Total number of moles after 100 min = x + 12 + 0.525 Pmix = pA + pB

According to Raoult’s law

On solving we get, x = 9.9 Now according to first order kinetics,

57.

A

B; k = 4.5

10-3 min–1;

For first order reaction,

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Find (a – x) at t = 60 min. (a – x) = 0.7634 Thus, rate after 60 minute = k (a – x) = 4.5 × 10–3 × 0.7634 = 3.4354 × 10–3 M min–1 58. (i) According to Fig. in the given time of 4 hours (1 to 5) concentration of A falls from 0.5 to 0.3 M, while in the same time concentration of B increases from 0.2 M to 0.6 M. Decrease in concentration of A in 4 hours = 0.5 – 0.3 = 0.2 M Increase in concentration of B in 4 hours = 0.6 – 0.2 = 0.4 M Thus increase in concentration of B in a given time is twice the decrease in concentration of A. Thus n = 2. (ii) K =

=

= 1.2 M

(iii) Initial rate of conversion of A = Change in conc. of A during 1 hour =

= 0.1 mole litre–1 hour–1

59. 2a + a/2 = 584.5 mm Hg a = 584.5 ×

mm Hg

a = 233.8 mm Hg Therefore, initial pressure of N2O5 = 233.8 mm Hg Total pressure after 30 min = 284.5 mm Hg = 233.8 – x + 2x +

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or 233.8 +

= 284.5 or x = 33.8 mm Hg

Hence pressure of N2O5 after 30 min. = 233.8 – 33.8 = 200 mm Hg k=

log

=

log

k=

× 0.0679 = 5.2 × 10–3 min–1

60. TIPS/Formulae : For a first order reaction we know that k=

log

Here, t = 10 × 60 × 60 sec. and let a = 1, then substituting the values, we get log

1.5 × 10–6 =

= log 0.0234 = log Taking antilog, 1.055 = or 1.055 – 1.055 x = 1 ⇒ x =

= 0.052

Thus, 5.2% of the initial concentration has changed into product. Again we know that t1/2 =

=

= 462000 second = 128.33 hours

61. Assuming that the decomposition of N2O5 is a first order reaction, then

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k=

log

or log P =

=

log

+ log P0

Thus log P vs time graph is linear with slope =

if the given reaction

is of first order which is in accordance with the given statement. Thus, the reaction obeys first order reaction. 62. From data (i) and (ii), it is obvious that when the concentration of B is kept constant (0.01 mol litre–1) and the concentration of A is doubled (0.01 to 0.02 mol litre–1), the rate of reaction is also doubled (0.005 to 0.010 mol litre–1 min–1). Hence, the order of reaction with respect to A is 1. Similarly, from data (i) and (iii) it is obvious that when the concentration of A is kept constant (0.01 mol litre–1) and the concentration of B doubled (0.01 to 0.02 mol litre–1), the rate of reaction remains constant (0.005 mol litre–1 min–1). This shows that the order of reaction with respect to B is zero. Now we know that the rate of reaction, A + B → Products, is given by Rate r = k [A]1 [B]0 ⇒ r = k [A] ∴ k=

=

= 0.5 min–1

[ r = 0.005 mol/l/m and [A] = 0.01 mol/l] For a first order reaction : We know that t1/2 =

=

[

k = 0.5 min–1]

= 1.386 minutes

1.

(d) Arrhenius equation :

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Slope of graph

2.

(a) A

Product Product

A Given:k1 = k2 =

3.

(a) The rate constant of a reaction is given by

The rate constant in tpresence of catalyst is given by

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4.

(c) Arrhenius equation, k = Ae–Ea/RT

log k = log A – slope = – More negative the slope greater will be the Ea. So correct order is Ec > Ea > Ed > Eb 5.

(a)

So,

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6.

7.

Ea = 165.4 kJ/mol H” 166 kJ/mol (a) As we can see from the graph that activation enthalpy to form D from A + B is 15 – 5 = 10 kJ mol–1, whereas, to form C from A + B is 20 – 5 = 15 kJ mol–1. Therefore, activation enthalpy to form C is 5 kJ more than that to form D. (b) From Arrhenius equation,

slope = –y (given) –y = –Ea Ea = y 8. (d) From Arrhenius equation,

so,

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

On dividing equation (ii) by (i)

9.

(a)

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10.

(b) According to Arrhenius equation

11.

Ea = 60 kJ / mol (a) Activation energy can be calculated from the equation

Given

; T2 = 310 K ; T1 = 300 K

Ea = 53598.6 J/mol = 53.6 kJ/mol. 12. (b) X → Y ; ∆H = –135 kJ/mol, Ea = 150 kJ/mol For an exothermic reaction Ea(F.R.) = ∆H + E′a(B.R.) 150 = – 135 + E′a(B.R.) E′a(B.R.) = 285 kJ/mol 13. (a) As per Arrhenius equation

, the rate constant

increases exponentially with temperature. 14.

(d)

Also given On comparing equations, (1) and (2)

....(1) ....(2)

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log A = 6.0 ⇒ A = 106 s–1 and

;

⇒ Ea = 2000 × 2.303 × 8.314 = 38.29 kJ mol–1 15. (b) The Arrhenius equation is : k = A exp (–Ea/RT) As T ∞, exp (–Ea /RT) 1. Hence, k = A where A, the Arrhenius parameter, is 6.0 1014 s–1 [NOTE : 'A' is also known as frequency factor] 16. (c) A catalyst decreases the activation energy of the reactants and thus shortens time of reaction. So (c) is the correct option. 17. (100) The Arr henices equation is

Assuming A and Ea to be independent of temperature

18.

(84297.48)

19.

(– 3.98) For a first order reaction, kt = ln

At 300 K, k1 × 60 = ln

...(1)

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At 400 K, k2 × 40 = ln

...(2)

From equation (1) and (2), = ln

=

ln

=

ln

× 8.3 × 1200 = Ea

Given : Then,

20.

n n

= 0.4 = – 0.4

Ea = – 0.4 × 8.3 × 1200 Ea = – 3984 J/mol. Ea = – 3.984 kJ/mol. (– 8500)

Now, Rate constant of forward reaction Rate constant of reverse reaction Equilibrium constant

Now, ∆G° = –RT ln Keq = –2500 ln (4e2)

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= –2500 (ln 4 + ln e2) = –2500 (1.4 + 2) = –2500 × 3.4 = –8500 J/mol. 21. (100) According to Arrhenius equation k = Let Ea of the reaction in absence of catalyst = x kJ mol–1 Therefore Ea of the reaction in presence of catalyst = x – 20 kJ mol–1 The Arrhenius equations in the two conditions can thus be written as ...(i) ...(ii) Dividing equation (i) by (ii), we get or x = 100 kJ mol–1 22.

(20.34)

Calculation of k723 K Ea = 200 kJ mol–1 = 200 ×103 J mol–1 T2= 723 K, T1 = 653 K k653 K = 1.925 ×10–3min–1 We know that,

or On usual calculations, k723 K = 6.81 ×10–2 min–1 Calculation of time for 75% decomposition at 723 K Let the initial amount of H2O2, a = 1 ∴ Amount at the required time, (a – x) = 0.25 Substituting the values in the given relation, = 20.34 min

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23. (311.35) TIPS/Formulae : According to Arrhenius equation log k = log A – We know that k =

=

(t½ = 10 × 60 sec.)

= 1.555 × 10–3 Substituting the various values in the above equation, we get log 1.155 × 10–3 = log 4 × 1013 – On usual calculations, T = 311.35 K 24. very high temperature or zero activation energy. 25. True : The rate of a reaction increases with increase in temperature because at higher temperature more number of molecules attain the activation energy. Note : For a reversible exothermic reaction, the rate of both forward and backward reactions will increase with increase in temperature. But according to Le Chatelier's principle, increased temperature favour the backward reaction. 26. False : Catalyst lowers the energy of activation and therefore influences the rate as well as rate constant of the reaction. 27. False : Catalyst does not make a reaction more exothermic, but decreases the activation energy and hence, increase the rate of reaction. 28. (a, c) On increasing temperature, concentration of product decreases. Hence, reaction is exothermic ⇒ ∆H°< 0 ⇒

>

so,

Also, or

⇒

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or

( ∆G = –RT lnK)

or

∆H°–T1∆S° C2. Hence, age of fossil in nearby areas, ....... (i) And age in far off places,

..... (ii)

From (i) and (ii), Since, C1 > C2, R.H.S. is positive i.e., T1 > T2.

27.

(a)

A look at the above curve shows that for stable nuclei it shows a curvature towards x-axis from the line of 45° slope (dotted line) as the atomic number (i.e. number of protons) increases. So statement 1 is true. The proton - proton repulsion would overcome the attractive force of proton and neutron. Thus, statement 2 in True. Also this statement 2 is a correct explanation for statement 1. Therefore the correct answer is option (a).

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28.

(c) Nuclides having both even number of protons and neutrons have maximum stability. So the reason is incorrect. But the assertion is correct as 40Ca20 has even number of neutrons and protons as compared to 30Al13 , which has odd neutrons and protons. 29. TIPS/Formulae : (i) Sum of atomic masses of reactant = sum of atomic masses of products (ii) Sum of atomic numbers of reactant = sum of atomic numbers of products (i) The atomic number of the final stable product = 90 – 7 × 2 + 1 × 6 = 82 and the mass number of the final stable product = 234 – 7 × 4 + 0 = 206. Thus, the element should be 82Pb206. (ii) (iii)

30.

Let the rate constants of the above emission processes be k1, k2 and k3 respectively and the overall rate constant be k. Then, k = k1 + k2 + k3 =

.

Also, k1 = 0.38 k

Similarly, ,

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where t1, t2 and t3 are the partial half-lives for

emission,

emission

and electron capture processes, respectively. 31. Let the number of α-particles emitted = a and number of β-particles emitted = b Thus 4a + 206 = 238; Therefore a = 8 Further 2a – b + 82 = 92; Therefore b = 6 Composition of the ore indicates that it has 1 g of U and 0.1 gm of Pb; thus here Nt = 1 g Determination of N0 206 g Pb is obtained from 238 g of U 0.1 g Pb is obtained from = Therefore, initial amount of U (N0) = 1 + 0.1155 = 1.1155 Now we know that t= By usual calculations, t = 7.097 × 108 years. 32. 33.

year–1

Since, the decay involves two parallel paths Th227 ←––– Ac227 –––→ Fr223 Thus, Fractional yield of or

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Similarly, Fractional yield of

34.

The ratio of H3 : H1 : : 8 × 10–18 : 1 No. of H atoms in 18 g H2O = 2NA ∴ No. of H3 atoms in 18 g of H2O = 2NA × 8 × 10–18 = 2 × 6.023 × 1023 × 8 × 10–18 atoms ∴ No. of H3 atoms in 10 g H2O atoms

=

= 5.354 × 106 atoms No. of atoms left after 40 years are derived as follows using the relation t=

log

40 =

∴ N = 5.624 × 105 atoms

log

35. Minimum number of β-particles required in one minute = 346 No. of β-particles required for carrying out the experiment for 6.909 × 60 minutes = 346 × 6.909 × 60 = 143431 ∴ Amount of β-particles required =

= 2.3814 × 10–19 mol

Now we know that, λ = Further we know that, λ =

=

= 0.0104 hr–1 log

where a = Initial concentration of β-particles x = Consumed concentration of β-particles log

=

=

= – 0.0312

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or or

= 0.931

[Taking antilog] = 0.931

[

x = 2.3814 × 10–19]

On usual calculations, a = 3.451 × 10–18 mol 36. → NOTE : For emission of one α-particle, atomic mass decreases by 4 and atomic number by 2. Further, for the emission of one β-particle, the atomic mass does not change but the atomic number increases by 1. So, we first find the α-particles : Decrease in atomic mass = 234 – 206 = 28 No. of α-particles emitted =

=7

Hence, atomic number should have decrease to 90 – (7 × 2) = 76 Now, atomic number of Pb = 82, which is more by (82 – 76) = 6 This increase is due to emission of β-particles. Therefore, β-particles emitted =6 37. TIPS/Formulae :

Half life, t1/2 = 5770 years Let the original sample be 1 gram. ∴ After every 5770 years one-half of radioactive carbon would decay or disintegrate. Thus, 1 g sample becomes ½ g after 5770 years and

left after

11,540 years. ∴ 25% of radioactive carbon remains after 11540 years. Rate constant, k for first order reaction,

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k=

=

= 1.2 × 10–4 year–1.

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1.

(a) As adsorption is exothermic process, hence

decreases with

increase of temperature. 2. (c) (i) When gas is adsorbed on metal surface, ∆H becomes less negative with progress of adsorption. (ii) The gas having greater value of critical temperature (TC) is adsorbed more compared to lower TC containing gases. As TC(NH3) > TC(N2), so NH3 is adsorbed more than N2 gas. 3. (b) Initially, adsorption of gases at the surface of charcoal occurs rapidly which results in a sudden decrease in pressure. As the number of vacant sites at the surface of adsorbent decreases with the passage of time, rate of adsorption decreases. Therefore, pressure tends to be constant. 4. (a) According to Freundlich adsorption isotherm

Slope = = Slope =

5.

(d) Smaller the value of critical temperature of gas, lesser is the extent of adsorption.

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6.

∴ Least adsorption is shown by H2 (least critical temperature). (b) According to Freundlich adsorption isotherm.

This is the equation of straight line of type y = mx + c. Hence slope is 1/n (m) and intercept is log10k.

7.

(c)

According to Freundlich adsorption isotherm

8. log

(a)

According to Freundlich adsorption isotherm

= log k +

log p

Thus if a graph is plotted between log(x/m) and log p, a straight line will be obtained

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The slope of the line is equal to 1/n and the intercept on log x/m axis will correspond to slog k. 9. (d) The characteristics given suggests that this should be physical adsorption. Physical adsorption usually takes place at low temperature and decreases with increase in temperature. The force of attraction holding the adsorbate are van der Waal's forces. Heat of adsorption is low. It is reversible and forms multimolecular layer. It does not acquire any activation energy. 10. (c) Adsorption is a phenomena of attracting and retaining the molecules of a substance on the surface of a liquid or a solid resulting into a higher concentration of the molecules on the surface. After adsorption there is a decrease in the residual forces due to bond formation ∆G, ∆H & ∆S all are negative in the case of adsorption. 11. (b) The adsorption of methylene blue on activated charcoal is an example of physisorption which is exothermic, multilayer and does not have energy barrier. 12. (b) When a gas is adsorbed on the surface, the freedom of movement of its molecules becomes restricted. This causes decrease in the entropy of the gas after adsorbtion, i.e. ∆S becomes negative. Thus, adsorption to be sontaneous, ∆H has to be negative from the equation ∆G = ∆H – T∆S. 13. (a) Rate of physisorption increases with decrease of temperature

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14.

(6)

Slope

, so

Intercept

So k = Antilog (0.477) = 3

So, 15.

(2)

= 0.001 atm, T = 300 K, V = 2.46 Number of

=

molecules

× NA =

× 6.023 × 1023

= 6.016 × Now, total number of surface sites = Density × Total surface area × 1000 = 6.023 × = 6.023 × Sites occupied by

molecules =

× 6.023 ×

= 12.04 × No. of sites occupied by each N2 molecule

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= 16.

2 (48) Freundlich adsorption isotherm :

Slope Intercept

so k = Antilog (0.4771) = 3

So,

= 3 × (0.04)2 = 48 × 10–4 17. (b, c, d) Reaction on metal surface

This is an example of chemisorption. (c) There are 2 unpaired electrons in π*2p orbital of O2 molecule : In O–2the extra electron will go to π*2p orbital. (d) This will decrease the bond order and increase the bond length. 18. (a, c) Graph (I) and (III) represent physisorption because, in physisorption, the amount of adsorption decreases with the increase of temperature and increases with the increase of pressure. Graph (II) represent chemisorption, because in chemisorption amount of adsorption increases with the increase of temperature. Graph (IV) is showing the formation of a chemical bond, hence chemisorption. 19. (a, b, d) (a) ∆H is negative for adsorption (b) Fact based

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(d) Chemical bonds are stronger than van der Waal’s forces, so chemical adsorption is more exothermic. 20. Number of moles of acetic acid in 100 mL before adding charcoal = 0.05 Number of moles of acetic acid in 100 mL after adding charcoal = 0.049 Number of moles of acetic acid adsorbed on the surface of charcoal = 0.001 Number of molecules of acetic acid adsorbed on the surface of charcoal = 0.001 × 6.02 × 1023 = 6.02×1020 Surface area of charcoal = 3.01 × 102 m2(given) Area occupied by single acetic acid molecule on the surface of charcoal

1. (b) C + O2 → CO2 (No catalyst is required) In Ostwald process, Haber’s process and hydrogenation of vegetable oils, catalyst (solid) such as Pt/Rh, Fe and Ni are used. 2. (d) Enzymes are very specific biological catalysts possessing well defined active sites 3. (d) (A) - (ii), (B) - (i), (C) - (iv), (D) - (iii) 1. 2. 3. 4. (i)

(a) Above Kraft temperature the formation of micelles takes place and the conc. above which micelle formation become appreciable is called critical micelles conc. (a) In micelle formation, above "CMC" hydrocarbon chains are pointing towards the centre of sphere with COO– part remaining outward on the surface. (a) Egg white is used to stabilise the prepared colloidal solution of red ink. (d) Tyndall effect is observed only when the following two conditions are satisfied. The diameter of the dispersed particles is not much smaller than the wavelength of the light used.

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(ii) The refractive indices of the dispersed phase and the dispersion medium differ greatly in magnitude. 5. (c) Excess water of liquid

Due to presence of hydrophobic chain it forms micelle. 6. (c) Bredig's Arc method is used for preparation of colloidal sol's of less reactive metal like Au, Ag, Pt. 7. (a) It is an example of osmosis. Osmosis is the movement of solvent across a semipermeable membrane towards a higher concentration of solute (concentrated solution). 8. (a) According to Hardy-Schulze, Coagulation value or fluocculation value

order of coagulation power: K3[Fe(CN)6] > K2CrO4 > KBr = KNO3 = AlCl3

9.

10. 11. 12.

order of flocculation value : K3[Fe(CN)6] < K2CrO4 < KBr = KNO3= AlCl3 (b) Peptisation is the process of converting a precipitate into a colloidal sol by shaking it with dispersion medium in the presence of small amount of electrolyte. It is also known as deflocculation. (c) The process of electrophoresis is used to precipitate colloidal particles in lyophobic sols. (a) In aerosol, the dispersion medium is gas while the dispersed phase can be both solid or liquid. (b) Colloids phase Cheese (C) Milk (M) Smoke (S)

Dispersed

Dispersion

medium liquid liquid solid

solid liquid gas

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13 14. 15.

(b)

(c) (d)

16. 17.

18.

19.

(a) Haemoglobin and gold sol (metal) are examples of positive and negative sols, respectively. (b) For coagulation of a negatively charged arsenious sulphide sol, the cation which is in higher oxidation state will be most effective. (b) (a) When silver nitrate solution is added to potassium iodide solution, a negatively charged colloidal solution is formed due to selective adsorption of I– ion from the dispersion medium. However, if the order of addition is reversed, i.e, potassium iodide solution is added to silver nitrate solution, due to selective adsorption of Ag+ion from the dispersion medium, a positively charged colloidal solution is obtained. Freezing point of colloidal solution is same as that of true solution at same concentration of a solute. The depression in freezing point is a colloidal property and depends on the number of solute particles and independent of size or shape of solute particles. Colloidal particles are so small that they can pass through ordinary filter paper. Also, they cannot be seen with ordinary microscope. When excess of electrolyte is added to colloidal solution, colloidal particle will be precipitated. Although electrolytes in minute quantities are necessary for the stability of colloids, they cause coagulation of disperse phase if present in large quantities. (b) (a) Brownian movement is the random motion of particles suspended in a fluid (a liquid or gas) resulting from their collision with the fast moving atom or molecules in the liquid or gaseous state of the matter. That means smaller particles are responsible for the Brownian movement than for bigger particles. (a) Gold number i.e., The smaller the value of gold number of lyophilic sol, the greater is the protective action. Hence, gelatin will be better protective colloid. (d) Greater the surface area, greater the van der waal forces of attraction and therefore at lesser concentration micelle formation will

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take place. In case of

(CH3)3 BrO due to greater chain

length, greater will be van der waal forces. 20. (c) According to Hardy Schulze rule, greater the charge on cation, greater is its coagulating power for negatively charged sol (As2S3), hence the correct order of coagulating power : Na+ < Ba2+ < Al3+ 21. (b) The motion of a liquid through a membrane under the influence of an applied electric field is known as electroosmosis. 22. (c) Smoke is an example of solid particles dispersed in gas. 23. (c) As Sb2S3 is a negative sol, so Al2(SO4)3 will be the most effective coagulant due to higher positive charge on Al (Al3+) – Hardy-Schulze rule. 24. (b) We know that surface acting agents (i.e. surfactants) such as soaps and detergents belong to the class of micelles. A miceller system when dissolved in water, dissociates to give ions. The anion ion is consists of two parts. The polar groups such as COO– or water loving (i.e. hydrophilic) in nature. It is called head of the species. The hydrocarbon chain which is quite big in size is water repelling (i.e. hydrophobic) in nature. It is called tail of the species. The hydrocarbon chain aggregates into the micelle above the critical concentration. NOTE : It may also be noted that the critical concentration for micelle formation decreases with increase in the molecular weight of the hydrocarbon chain of surfactant. Here, the two anions that are formed is in case of Option (b) is CH3 (CH2)11 OSO3–and option (d) CH3 (CH2)6 COO–. The molecular weight of hydrocarbon chain is more in case of "B" so, it has lower value of critical concentration for micelle formation in aqueous solution. 25. (d) Lyophilic sols are self stabilizing because these sols are reversible and are highly hydrated in the solution. NOTE: Lyophilic sols can be coagulated by adding a suitable solvent like alcohol and an electrolyte both. 26. (0.37)

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For 1 L sol 30 m mol of HCl is required For 1 L sol 15 m mol H2SO4 is required For 250 mL of sol, H2SO4 required =

× 250 m mol H2SO4

= 3.75 m mol of H2SO4 1 mol H2SO4 = 98 g 3.75 m mol of H2SO4 = 3.75 × 98 ×10–3 = 0.3675 g H2SO4 27. (a, b) (a) Since adsorption is an exothermic process, hence, enthalpy decreases during this process. On adsorption, the randomness of the adsorbate molecules decreases, thus, entropy decreases. (b) Higher the critical temperature of a gas, greater the amount of gas adsorbed thus, the extent of adsorption ethane will be more than nitrogen. (c) Cloud is an aerosol in which liquid is dispersed phase and gas is dispersion medium, Whereas, emulsion is liquid in liquid colloidal system. (d) Brownian motion of colloidal particles depends on the size of the particles as well as on viscosity of the solution. 28. (a, d) (a) When two or more ions are present in the dispersion medium, preferential adsorption of the ion common to the colloidal particle takes place. (d) This is called zeta potential which helps in increasing the stability of the lyophobic sol. 29. (b) Statement-1 is correct because the surfactant molecules aggregate to form micelles only at or above the critical micellar concentration (CMC). Although, statement-2 is also correct, i.e., the conductivity of the solution having surfactant molecules decreases

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sharply at CMC. However, statement-2 is not the explanation for statement-1. Each micelle contains at least 100 molecules and thus, with the formation of micelles, the number of ions in solution decreases and mobility of the bulkier micelle particles decreases. This, finally leads to decrease in conductivity of the solution.

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1. (a) (b) (c) (d)

2.

(d) Bronze contains 88-96% Cu and 4-12% Sn. Cast iron is used to make wraught iron. German silver contains 50% Cu, 30% Zn & 20% Ni. Brass contains 70% Cu and 30% Zn.

(b)

From graph it is evident that the temperature below which the oxide is stable, is the point at which free energy change shows a change from negative to positive. 3. (b) Ellingham diagram is the graph of ∆G0 vs T of any metal/ element oxide. Since For most metal oxide formation. metal oxide (s) metal (s) + oxygen (g) So, graph will be a straight line with –ve intercept and +ve slope. 4. (a) All statements are correct.

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(1)

CaO + CO2

(2) In extraction of Ag, it is extracted as an anionic complex [Ag(CN)2]–. (3) Ni is purified by Mond's process Ni + 4CO Ni(CO)4

Ni(CO)4 Ni + 4CO

(4) Van Arkel method is used to purify Zr and Ti. B + AO2 5. (b) A + BO2 G = –ve when the temperature is above 1400 °C, A reduces BO2. 6. (c) During the reduction of Fe2O3 to FeO, Fe2O3 is first reduced to Fe3O4 and then to FeO, thus follows reaction (b). In order to remove impurity of the ore limestone is added along with ore which is decomposed to CaO. The resulted CaO reacts with SiO2 to form CaSiO3 i.e. follows reaction (a) and is removed as a slag. 7. (a) Liquation method is used for purification of metals having low melting point. 8. (b) Froth floatation was discovered by washer women. It is a method of concentration of ores. 9. (a) Ti is refined by van Arkel method. Silver is leached by dilute solution of NaCN. Zincite is oxide ore of zinc i.e. ZnO. Aniline is a froth stabilizer. 10. (a) Magnetite Fe3O4 Sphalerite ZnS Cryolite Na3AlF6 Malachite CuCO3.Cu(OH)2 11. (d) Bauxite → Al(OH)3 Calamine → ZnCO3 Siderite → FeCO3 Malachite → CuCO . Cu(OH) 3

2

12. (a) Mond’s process is used for purification of Ni

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13. (b) In Hall-Heroult process, steel vessel with lining of carbon acts as cathode and graphite is used as anode. 14. (c) Hall Heroult’s process is an electrochemical process used in extraction of Al from alumina. 15. (a) Amongst the given ores; copper pyrite (CuFeS2), dolomite (MgCO3.CaCO3), malachite [CuCO3.Cu(OH)2], azurite [2CuCO3.Cu (OH)2] only copper pyrite contains both copper and iron. 16. (a) Calamine → ZnCO3 Malachite → CuCO3.Cu (OH)2 Magnetite → Fe3O4 Cryolite → Na3AlF6 17. (d)

18. (1) (2) (3) (4) 19.

(a) Froth floatation method is mainly applicable for sulphide ores. Malachite ore : Cu(OH)2 . CuCO3 Magnetite ore : Fe3O4 Siderite ore : FeCO3 Galena ore : PbS (Sulphide ore) (c) FeO is gangue and SiO2 is flux to form slag FeSiO3.

20. (b) In the metallurgy of aluminium, purified Al2O3 is mixed with Na3AlF6 or CaF2 which lowers the melting point of the mix and brings conductivity.

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21. (a) Calamine (ZnCO3) is an ore of zinc. 22. (c) The iron obtained from blast furnace contains about 4% C and impurities (eg. S, P, Si, Mn). This is known as pig iron. 23. (c) Reduction by powdered aluminium is known as Gold-Schmidt aluminothermic process. This process is employed in cases where metals have very high m.p. and are to be extracted from their oxides. 24. (a) Silver, copper and lead are commonly found in earth's crust as Ag2S (silver glance), CuFeS2 (copper pyrites) and PbS (galena) 25. (b) The reactions involved in cyanide extraction process are : + 4NaCN → 2Na [Ag(CN)2] + Na2S 4Na2S +

+ 2H2O → 2Na2SO4 + 4NaOH + 2S

2Na[Ag(CN)2] +

→ Na2 [Zn(CN)4] + 2 Ag ↓

26.

(d) (i) Haematite is Fe2O3 in which Fe is present in III oxidation state. (ii) Magnetite (Fe3O4) is an equimolar mixture of FeO and Fe2O3. Oxidation state of Fe in FeO is II. Oxidation state of Fe in Fe2O3 is III. 27. (b) NOTE : Extraction of Zn from ZnS (Zinc blende) is achieved by roasting followed by reduction with carbon. 2ZnO + 2SO2 2ZnS + 3O2 Zn + CO ZnO + C 28. (c) Cuprite : Cu2O; Chalcocite : Cu2S; Chalcopyrite : CuFeS2; Malachite: Cu(OH)2.CuCO3. We see that CuFeS2 contains both Cu and Fe. 29.

(a)

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30.

(a)

→ Mg+2 + 2Cl–

31.

(c) Acidic strength increases as the oxidation number of central atom increases. Hence acidic strength order is (+7) (+5) (+3) (+1) HClO4 > HClO3 > HClO2 > HClO 32. (c) Haematite ore (Fe2O3) is first reduced to cast iron which is then oxidised for removing carbon (impurity) as CO2. 33. (b) NOTE : During the extraction of copper, iron is present in the ore as impurity (FeS). The ore together with a little coke and silica is smelted; FeS present as impurity in the ore is oxidized to iron oxide, which then reacts with silica to form fusible ferrous silicate which is removed as slag. 2FeO + 2SO2↑ ; 2FeS + 3O2 FeO + SiO2 34.

(c) Al2O3 is electrolyte, while Na3AlF6 is used to decrease the melting point of Al2O3 and to increase the conductivity. 35. (c) Al reduces Fe2O3 or Cr2O3 to respective metals and acts as a reducing agent. Fe2O3 + 2Al → Al2O3 + 2 F CaO + CO2; 36. (a) CaCO3 + 37. 38. 39.

40.

——→ CaSiO3 (slag).

(c) Malachite is CuCO3 . Cu (OH)2 which is ore of copper. sintering, smelting. Magnesia and lime; calcium silicate NOTE : The lining of corverter is made of magnesia & lime. Slag formed consists of CaSiO3. Aluminium

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[Thermite reaction] 41. 42.

Tin. It is SnO2. (a, b, c, d) (a) 2Na[Al(OH)4] + CO2 —→ NaCO3 + H2O + 2Al(OH)3(↓) or Al2O3.2H2O (b) Function of Na3AlF6 is to lower the melting point of electrolyte. (c) During electrolysis of Al2O3, the reactions at anode are: C(graphite) + O2 —→ CO(g) + CO2(g) (d) The steel vessel with a lining of carbon acts as cathode. 43. (a, b, c) Gold extraction:

44. (a, b, c) Copper pyrite [CuFeS2]

Concentrated by froth floatation process

Roasting take place in reverberatory furnace

Copper matte (Cu2S + FeS)

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(Blister copper)

Refining of blister copper is done by poling followed by electrorefining but not by carbon reduction method. 45. (b, c, d) (a) (b) (c)

From above, Cu2O gives copper metal.

(d) From above, CuO also gives copper metal. 46. (c, d)Al from Al2O3 and Mg from MgCO3.CaCO3 are separately extracted by electrolytic reduction. 47. (a, c, d) Cassiterite (SnO2) contains impurity of S, As, Fe, Cu etc. Roasting : Concentrated ore is heated in a current of air impurities of iron and copper changed to their oxides and sulphates. SnO2 is treated with coke to reduce it to Sn. 48. 49.

50.

(b, c) To make the fused mixture very conducting and to reduce the temperature of the melt. (b, d) Both Mg and Al have their reduction potentials less than that of water [Eº = –0.83 V]. Hence, their ions in the aqueous solution can not be reduced. Instead water will be reduced : 2H2O + 2e– → H2 + 2OH– (a, b) Because of high melting point (2050ºC), pure alumina cannot be electrolysed. Hence a mixture of alumina, cryolite (m.p. 1000ºC)

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and calcium fluoride (to lower the temperature of the melt) is electrolysed at about 900ºC. NOTE : The function of cryolite is to increase the electrical conductivity of the electrolyte, and to lower the temperature of the melt. 51. A-(p, q, s), B-(t), C-(q, r), D-(r)

52. 53.

(A) - p; (B) - q; (C) - p, r; (D) - p, s (A)- p, r; (B) - p, r; (C) - q; (D) - s NOTE : The oxides and sulphides of less active metals like Hg, Cu & Pb are unstable to heat and hence no reducing agent is required. They undergo self reduction. ; (A)

; Hence, (A) → (p), (r) NOTE : The oxides of less electropositive metals like Pb, Zn, Fe, Sn, Cu, etc. are reduced by strongly heating them with coke or coal. ; (B) Hence, (B) → (p), (r) (C) Extraction from argentite (Ag2S) Na2S + 2AgCN; Ag2S + 2NaCN

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Zn, being more electropositive than Ag, displaces Ag from the complex. Hence, (C) → (q) (D) Among the halides of boron, BI3 is unstable because of the large size of Iodine and small size of boron atom. Hence, it decomposes to give boron. Thus, (D) → (s). 54. (i) (g) (ii) (h) (iii) (f) (iv)(c) (v) (e)(vi) (d) (vii) (b) (viii) (a) 55. (c) (i) (b) (ii) (c) (iii) (d) (iv) (a) Cu2S + SO2 ↑ + 2FeS 56. (b) 2CuFeS2 + O2 2FeO + 2SO2 ↑ 2FeS + 3O2 2Cu2O + 2SO2 ↑ 2Cu2S + 3O2 FeSiO3 57. (d) FeO + SiO2 6Cu + SO2 58. (c) Cu2S + 2Cu2O The reducing species is the one which gets oxidised. So, it is S2– ion getting oxidised to S4+. 59. Calcination of the ore A1 to form CO2 indicates that A1 should be a carbonate. Further, reaction of A1 with HCl and KI to evolve I2 indicates that A1 would also be hydroxide. So the possible formula for the ore, should be CuCO3.Cu(OH)2 which explains all the given reactions

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Roasting of A2 gives gas G whose nature is identified as SO2 as it gives green colour with acidified K2Cr2O7. So A2 should be sulphide of copper.

60.

Oxidation number of Pb in litharge (PbO) is +2 61. Haematite (Fe2O3) on burning with coke and lime at 2000 C results in the following reactions. CO2 C + O2 2CO CO2 + O 2Fe + 3CO2 (Reduction of Fe2O3 to form steel) 3CO + Fe2O3 (Steel)

SiO2 + CaO (Lime)

62.

CaSiO3

(Slag)

(Slag, CaSiO3 is used as building material)

(i)

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(xii) K2Cr2O7 + 4NaCl + 6H2SO4 → 2CrO2Cl2 ↑ + 4NaHSO4 + 2KHSO4 + 3H2O chromyl chloride (orange)

+ 4NaCN

(ii)

2 Na[Ag(CN)2] + Na2S

4Na2S + SO2 + 2H2O → 2 Na2SO4 + 4 NaOH + 2S [NOTE : Na2S is converted into Na2SO4 to avoid reversibility of first reaction] 2NaAg(CN)2 + Zn → Na2 [Zn(CN)4] + 2Ag Sod. Zincocyanide

63. Recovery of Pb from galena : 2PbS + 3O2 → 2PbO + 2SO2 ↑ PbS + 2PbO → 3Pb + SO2 64. Carbon monoxide is the actual reducing agent of haematite in blast furnace.

65.

(i) Excess of Air (used during roasting) is necessary for converting chalcocite (a sulphide ore) to oxide. Calcination does not convert it to oxide. (ii) Metals can be recovered by chemical methods because they occur as oxides, carbonates, sulphides which have to be calcined or roasted. 66. Equations for extraction of silver from its sulphide ore. Cyanide Process : + 2NaCN→

Na2S + 2AgCN

AgCN + NaCN →

Na[Ag(CN)2] Sod. argentocyanide (soluble) 2Na[Ag(CN)2] + Zn →Na2[Zn(CN)4] + 2Ag ↓ [NOTE : Zn is more electropositive than Ag.] 67. Extraction of aluminium from bauxite : At cathode : At anode :

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Net reaction :

68. Bleaching agent Chlorine Ammonium carbonate Smelling salt Aluminium Cryolite Tin Bell metal Calcium Fluorspar Ammonium phosphate Fertilizer Carbon Anthracite 69. (a) Galena is roasted in excess of air in a reverberatory furnace 2PbO + 2SO2 2PbS + 3O2 (air)

PbSO4 PbS + 2O2 It is followed by self reduction 2Pb + 2SO2 PbS + PbSO4 3Pb + SO2 PbS + 2PbO (D) Amalgamation (b) (i) Silver (A) Fused salt electrolysis (ii) Calcium (B) Carbon reduction (iii) Zinc (C) Carbon monoxide reduction (iv) Iron (E) Self reduction (v) Copper 1. 2. 3. 4. 5.

(b) Zn, Cd and Hg are purified by fractional distillation process. (b) Germanium (Ge), silicon (Si), boron (B), gallium (Ga) and indium (In) can be obtained in pure state by zone refining process. (d) Cast iron is made from pig iron which is used for production of wroght iron and steel. (b) Wrought iron is purest form of commercial iron. (b) On electrolysis of aqueous solution ofs-block elements, H2 gas is discharged at cathode.

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At cathode : 6.

(b, c, d) In electrolytic refining of blister Cu, acidified CuSO4 is used as electrolyte, pure Cu deposits at cathode and impurities settle as anode mud. Impure metal act as an anode. 7. (c) Liquation is used for Sn, zone refining is used for Ga, mond’s process is used for refining of Ni and van Arkel method is used for Zr, So, correct match is (I) - (c); (II)-(d); (III)-(b); (IV)-(a) 8. MgO is used for the lining of steel making furnace because it acts as basic flux and facilitates the removal of acidic impurities of Si, P and S from steel through slag formation. 9. NOTE : Zone refining is based on the difference in solubility of impurities in molten and solid state of the metal. This method is used for obtaining metals of very high purity. Ge, Si and Ga used as semi-conductors are refined in this manner. These metals can be easily melted and can easily crystallise out from the melt form.

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1. (d) (a) N2 is diamagnetic in nature. (b) (c) Liquid N2 is used in cryosurgery. (d) Because of its inertness, it is used where an inert atmosphere is required. 2. (c) 3.

(d) If NH3 is used in excess then NH4Cl is formed instead of NCl3.

4.

(c)

5.

(a) Aqua regia is HNO3 : HCl 1 : 3 + – AuCl–4 + NO + 2H2O Au + 4H + NO3 + 4Cl– 3Pt + 16H+ + 4NO–3 + 18Cl– 3PtCl62– + 4NO + 8H2O

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6.

(d)

No. of P = O bond = 2. P – OH bond = 4 P – O – P bond = 1. 7. (b) Pb(NO3)2 does not produce nitrogen gas on heating. (a) (b) (c) (d) 8.

(b)

Basicity of H3PO2 = 1 9. (b) Number of pentagons in C60 (Buckminsterfullerene) = 12 Number of triangles in P4 (White phosphorous) = 4 10. (d) Oxide oxidation state N2O +1 NO +2 N2O3 +3 NO2 +4 So, N2O < NO < N2O3 < NO2 11. (b)

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12.

(c) Structure of H3PO2:

Greater the number of P—H bonds present in the acid, greater will be its reducing property. 13. (d) (a) (b) (c) (d) 14.

NH3 is evolved in reaction (d). (c) The number of P–O bonds in P4O6 = 12

15.

(a)

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16.

(c) Let oxidation states of phosphorus in H3PO2, H3PO4, H3PO3 and H4P2O6 be w, x, y and z respectively. Thus, in H3PO2: 3 × (+ 1) + w + 2×(–2) = 0 ∴ w = +1 In H3PO4: 3 × (+ 1) + x + 4×(–2) = 0 ∴ x = +5 In H3PO3: 3 × (+ 1) + y + 3×(–2) = 0 ∴ y = +3 In H4P2O6: 4 × (+ 1) + 2z + 6×(–2) = 0 ∴ z = +4 Thus, the order of oxidation state is : H3PO4 > H4P2O6 > H3PO3 > H3PO2 17. (c) Phosphorous acids contain P in +3 oxidation state. Acid Formula Oxidation state of phosphorus Pyrophosphorous acid H4P2O5 +3 Pyrophosphoric acid H4P2O7 +5 Orthophosphorous acid H3PO3 +3 Hypophosphoric acid H4P2O6 +4 18.

(c) H4P2O6 has P–P linkage

19.

(b) In case of

, N atom is sp2 hybridised, the lone pair is

present in 2p orbital and it is transferred to empty d orbital of Si forming dπ – pπ bond. Hence nitrogen with sp2 hybridisation has trigonal planar shape. 20.

(a)

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21. (b) The slow decomposition of HNO3 is represented by the eqn. 4HNO3 → 4NO2 + 2H2O + O2 (yellow-brown)

22.

(c)

23.

(b)

24.

(d)

25.

(b)

Here N2 acts as a diluent and thus retards further oxidation. Reaction of P4 under other three conditions. (a) (c) In moist air, P4O6 is hydrolysed to form H3PO3 (d) In presence of NaOH, 26. 27.

(d) In P4, the P–P linkage is formed by sp3–sp3 hybridised orbital overlapping. So, the percentage of π-character will be 75%. (b) PbO2 is a powerful oxidizing agent and liberate O2 when treated with acids.

28.

(b) NO (g) + NO2 (g)

29.

(c) The ignition temperature of black phosphorus is highest among all its allotropes, hence is most stable. (a) The structure of H3PO3 is as follows: There are only two –OH groups and hence dibasic. The oxidation number of P in this acid is

30.

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+3. Whereas P can have +5 oxidation state also. Therefore, H3PO3 can be oxidised which means H3PO3 is a reducing agent.

31.

(c) NH3 does not react with CaO while other reacts with NH3

32.

(c) In cyclic metaphosporic acid number of P–O–P bonds is three.

33.

(b) In BCl3,

34. 35. 36.

37. 38.

; sp2 hybridization (bond angle =

120°). Similarly PCl3, AsCl3 and BiCl3 are found to have sp3hybridized central atom with one lone pair of electrons on the central atom. The bond angle ≤ 109°28′, since the central atoms belong to the same group. The bond angle of the chlorides decreases as we go down the group. Thus, the order of bond angle is, BCl3 > PCl3 > AsCl3 > BiCl3. N2 + Cr2O3 + 4H2O (d) (NH4)2Cr2O7 3Ca(OH)2 + 2PH3 ; i.e. 2 moles of phosphine (c) Ca3P2 + 6H2O are produced. (c) In P4O10, each P atom is linked to 4 oxygen atoms as can be confirmed by its structure. It is linked to three oxygen atoms by single bond and with one oxygen atom by double bond. [For structure, refer to Q.11 of Section A] (a) Least basic trihalogen of nitrogen is NF3 because of the highest electronegativity of fluorine. (d) NO2 is reddish brown coloured gas. Rest of the oxides are colourless.

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39.

(b)

The structure clearly shows the presence of covalent and co-ordinate bonds. 40. (a) Only nitrates of heavy metals and lithium decompose on heating to produce NO2. 41. (d) None; it reacts with all given compounds. It forms addition compounds with them because it is basic in nature and given compounds are acidic in nature. It can be dried over any metal oxide. 42.

(c)

It is redox reaction. (i) Oxidation number increases during oxidation reaction and decreases during reduction reaction. (ii) In a neutralisation reaction, acids and bases reacts together to form salt and water. 43.

(c)

44. (8) Number of lone pairs = 8 POCl3 + SOCl2 45. (4) PCl5 + SO2 POCl3 + 2HCl PCl5 + H2O POCl3 + SO2Cl2 + 2HCl PCl5 + H2SO4 10POCl3 6PCl5 + P4O10

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46.

four.

Each P atom is linked to 4 oxygen atoms

47. two;

[It contains two replaceable hydrogens.]

48.

white/ yellow; NOTE : In white phosphorus, each phosphorus atom is linked to the other three atoms by covelent bonds. PPP bond angle is 60°, due to which the molecule remains under strain and hence is active in nature. 49. –3; Nitrogen has 5 electrons in its valence shell. Thus, it can accept maximum 3 electrons. 50. True : The molecule of NO has eleven valence electrons (5 due to N and 6 due to O). It is impossible for all of them to be paired, hence the nitric oxide molecule contains an odd electron which makes gaseous nitric oxide paramagnetic. NOTE : In the liquid and solid states, nitric oxide is polymerised to a dimer which is diamagnetic. 51. True : NOTE : The central element in the metal hydrides of group 15 elements is although in sp3 hybrid state, the H – M – H bond angle is less than the normal tetrahedral bond angle of 109º 28’; e.g. the bond angle, H – N – H in NH3 is 106º 45’. This is due to greater repulsion between a lone pair and a bond pair of electrons than between the two bond pairs of electrons. The decrease in bond angle from 106º 45’ in ammonia to about 90º in AsH3 can be explained by the fact that in the latter case sp3 hybridisation

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52.

53. (a) (b)

(c) (d) 54.

becomes less and less distinct with the increasing size of their electron clouds, i.e., pure p orbitals (instaed of sp3 hybrid orbitals) are used forM – H bonding and the lone pair of electrons is present in spherical s-orbital. False : Red phosphorus is polymeric substance. It exist as chains of P4 tetrahedra linked together. Therefore, it is less volatile than white phosphorus. (a, b, c) Basic character of oxides increases on moving down the group therefore, Bi2O5 is more basic than N2O5. Covalent nature depends on electronegativity difference between bonded atoms. In NF3 , N and F are non-metals but in BiF3, Bi is metal while F is non metal therefore, NF3 is more covalent than BiF3. In PH3, hydrogen bonding is absent but in NH3, hydrogen bonding is present, therefore PH3 boils at lower temperature than NH3. Due to small size in N–N single bond, l.p. – l.p. repulsion is more than P–P single bond therefore, N–N single bond is weaker than the P–P single bond. (b, c) (a)

(b) (c) (d) Hence, only (NH4)2Cr2O7 and Ba(N3)2 can provide N2 gas on heating below 300°C 55. (b, d)

(a)

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(b) N2O5 is diamagnetic in nature (c) N2O5 → N2O5 contains one N–O–N bond but not N–N bond. (d) Na + N2O5 → NaNO3 + NO2↑ (Brown gas)

56.

(a, b, c)

57.

(a, b) When ammonium salt NH4NO3 or NH4NO2 (ammonium salts are colourless) is boiled with excess of NaOH, ammonia (NH3) gas is evolved as follows: NaNO2 + NH3 + H2O NH4NO2 + NaOH NaNO3 + NH3 + H2O NH4NO3 + NaOH The NH3 gas evolved is non-flammable gas. When the gas evolution ceases we are left with NaNO2 or NaNO3 in solution. These salts get reduced when Zn is added to this solution containing salt (NaNO2 or NaNO3). Again NH3 gas evolves.

Thus, the colourless salt [H] is either NH4NO2 or NH4NO3. Thus, (a) and (d) are correct answers. NH2.NH2 + H2O + Cl– 58. (c) 2NH3 + OCl– 59. (a, c, d) The four atoms in a P4 molecule are situated at the corners of a tetrahedron. There are six P - P single bonds with PPP bond angle equal to 60o. Each phosphorus has a lone pair of electrons.

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60.

(a, b, d) Sodium nitrate on decomposition upto 500oC gives NaNO2 and oxygen. 2NaNO2 + O2 2NaNO3 While at higher temperature (i.e. above 800oC), NaNO2 further decomposes into Na2O, N2 and O2. Na2O + 3/2O2 + N2 2NaNO2 61. (a, d) NH4NO3 ——→ N2O + 2H2O NH2OH + HNO2 ——→ N2O + 2H2O 62. (i) (d) (ii) (b) (iii) (a) (iv) (c) 63. (c) We know that phosphates have a biological significance in human, therefore statement (a) is not correct. Since nitrates are more soluble in water so they are less abundant in earth's crust where as phosphates are less soluble in water and so they are more abundant in earth's crust. Thus, statement (b) is False and statement (c) is correct. NOTE : In nitrates

nitrogen is in + 5 oxidation state which is

the highest oxidation state exhibited by nitrogen. Because of this nitrates can not be oxidized (oxidation means increase in oxidation state). Hence, statement (d) is not correct. The correct answer is (c). 64. (c) In case of group 15 (nitrogen group), on moving down the group, there occurs a decrease in bond angle of metal hydrides. This decrease in bond angle of metal hydrides of this group may be attributed to the increased p- character in the bond pair which results in more scharacter in lone pair orbital. NOTE : The directional character is more for sp3 hybrid orbital than on s- orbital. Thus, the correct answer is (c). 65. (b) The reaction between NaOH and white phosphorus (P4) can be represented as follows:

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NOTE : In this reaction, Phosphorus is oxidised as well as reduced, so it is a disproportionation reaction. ∴ The correct answer is (b). 66. (b) Nitrogen cannot form pentahalides because it cannot expand its octet due to non-availability of d-orbitals. So E is not correct explanation of S. 67. (CH3)3N and (Me3Si)3N are not isostructural, the former is pyramidal while the latter is trigonal planar. Silicon has vacant d orbitals which can accommodate lone pair of electrons from N (back bonding) leading to planar shape.

68.

6 CaO +

Moles of

=

=3

Moles of CaO = 3 × 6 = 18; wt. of CaO = 18 × 56 = 1008 g For structure of P4O10 : See question 20 of this section. 69. Elemental nitrogen exists as a diatomic molecule because nitrogen can form pπ - pπ multiple bonds which is not possible in case of phosphorus due to repulsion between non-bonded electrons of the inner core. There is no such repulsion in case of smaller nitrogen atoms as they have only 1s2 electrons in their inner core. 70. In such a case : A = Ca(OH)2, B = NH4HCO3, C = Na2CO3, D = NH4Cl and E = CaCl2

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71.

7H3PO4 + Ca5(PO4)3 F

5Ca(H2PO4)2

+ HF

Triple superphosphate

72. The poisonous element M may be As. So, on the basis of the given facts,

Hence, M = As; N = As H3

73.

Number of P – O single bonds = 12Number of P – O double bonds = 4 74. (i) 2P + 3I2 + 6H2O → 2H3PO4 + 6HI (ii) 2KNO3 + 10K → 6K2O + N2 75. Since carbon has no d-orbital, it cannot extend its coordination number beyond four, its halides are not attacked (hydrolysed) by water. On the other hand, silicon have vacant d-orbitals to which water molecules can coordinate and hence their halides are hydrolysed by water. As the charge on central atom increases, the tendency of attack of a nucleophile (OH–) increases. NOTE : Increasing order of extent of hydrolysis CCl4 < MgCl2 < AlCl3 < SiCl4 < PCl5 76. (i) KClO3 + 2H2C2O4 + H2SO4

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→ KHSO4 + HCl + 6CO2 + 3H2O (ii) (NH4)2SO4 + NO + NO2 → 2N2 + 3H2O + H2SO4 77. N2O has two principal resonance structures : :N≡ 78. (i) (ii)

(iii) (iv)

P4O10 + 6 PCl5 10 POCl3 [2HNO3 H2O + 2NO2 + [O]] 10

P4 + 20HNO3

4H3PO4 + 20NO2 + 4H2O

2KMnO4 + 2NH4OH → 2MnO2 + N2 + 2KOH + 4H2O (v) Na2CO3 + NO + NO2 → 2NaNO2 + CO2

NaOH + NH3 + H2O (vi) NaNO2 + 6H (vii) CaSO4 + 2NH3 + CO2 + H2O → CaCO3 ↓ + (NH4)2SO4 5Ca3P2 + 3P2O5↑ (viii) 15CaO + 4P4 Ca3P2 + 6H2O → 3Ca(OH)2 + 2PH3 ↑ ] × 5 (ix)

2H3PO2

15CaO + 4P4 + 30H2O → 15Ca(OH)2 + 3P2O5↑ + 10PH3 ↑ PH3 + H3PO4 Phosphine

(x)

4P + 10HNO3 + H2O → 5NO + 5NO2 + 4H3PO4

(xi)

4Sn +

→ 4Sn(NO3)2 + NH4 NO3 + 3H2O

(i) Phosphine gas (PH3) is evolved when white phosphorous is boiled with aqueous NaOH or alcoholic solution of potassium hydroxide. P4 + 3NaOH + 3H2O → NaH2PO2 + PH3 ↑ Sod. Hypophosphite

79.

(i) Nitrogen and fluorine both are small and have high electron density, they repel the bonded pair of electrons leading to larger bond length than expected. (ii) It is due to self ionization of NH3, the reaction is

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2NH3 →

+

Thus, on addition of NH4Cl the concentration of

radical increases

and therefore, NH4Cl acts as an acid in liquid NH3. (iii) NOTE : As compared to P, N atom has higher electronegativity and small size and shows H-bonding. Thus, ammonia molecule show association where as phosphine does not. (iv) NOTE : H3PO3 is a dibasic acid because it contains two OH groups in its molecule. In the two P–OH bonds, the hydrogen is ionisable. (v) Orthophosphorus acid is a dibasic acid as it has2–OH groups in its formula :

(vi) Liquor ammonia possesses high vapour pressure at room temperature and thus before opening a bottle of liquor ammonia, it should be cooled to lower the pressure of NH3 inside the bottle, otherwise NH3 will bump out of the bottle. (vii) In H3PO4 and H3PO3, the P atom is attached to 3 and 2 OH groups respectively. The H atom of theseP – OH bonds are ionisable. This clearly shows that H3PO4 is tribasic and H3PO3 is dibasic. 80.

(i)

(ii)

1.

(c) X = Na2SO3

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2.

(a) At room temperature, water is liquid and has boiling point 373 K due to hydrogen bonding. Whereas H2S is a gas and it has no hydrogen bonding. Hence boiling point of H2S is less than 300 K (boiling point of H2S is –60°C).

3.

(b)

4.

(c)

5.

(b) In KO2, the nature of oxygen species and the oxidation state of oxygen atom are superoxide (superoxide ion is O2–) and –1/2 respectively. Let x be oxidation state of oxygen. The oxidation state of K is +1. Hence + 1 + 2 (x) = 0 ⇒ x =

6.

(d)

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7.

(a) The S6 molecule has a chair-form hexagon ring with the approx same bond length as that in S8, but with some what smaller bond angles i.e. bond lengths are approx same but bond angles are different.

8.

(a)

9.

(b) The following reaction occurs 2 NaHSO4 + 8HCl. Na2S2O3 + 4Cl2 + 5H2O 10. (c) In KMnO4, manganese is already present in its highest possible oxidation state i.e. +7. So, no further oxidation is possible.

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11.

(c) Among oxyacids of sulphur, only Caro’s acid (H2SO5) and Marshall’s acid (H2S2O8) have the O – O linkage. 12. (d) In sulphur trioxide trimer S3O9 (also called γ−sulphur trioxide), two sulphur atoms are linked to each other via O atoms, hence there is no S –S bond. 13. (b) Na2SO3 + S 14.

Na2S2O3

(c)

(2 moles of sulphuric acid and 1 mole of H2O2)

But this is not one of the options.

This matches to option (c). 15. (b) HBr is not prepared by heating NaBr with conc. H2SO4 because HBr is a strong reducing agent and reduce H2SO4 to SO2 and is itself

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16. 17. 18. 19.

oxidised to bromine. NaBr + H2SO4 —→ NaHSO4 + HBr H2SO4 + 2HBr —→ SO2 + Br2 + 2H2O 2Hg + O2 (b) 2HgO (c) SO2 is highly soluble in water and therefore, cannot be collected over water. (d) All are colourless gases. (6)

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20.

(288) Concentrated nitric acid oxidises sulphur to sulphuric acid (+6 oxidation state). S8 + 48HNO3 → 8H2SO4 + 48NO2 + 16H2O Rhombic sulphur

1 mole S8 produces 16 × 18 g of H2O i.e. 288 g of H2O. 21. Nitric oxide. [NO] The mixture containing SO2, air and nitric oxide, when treated with steam, sulphuric acid is formed. → 2H2SO4 + [NO]

2SO2 + 22. (a, c, d)

Ozone is diamagnetic in nature (due to presence of paired electron) and both the O – O bond length are equal. It has a bent structure. 23. (a, c, d)

For 24-25

24.

(c)

26.

(a)

&

25. (b)

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(P) and (Q) are salts of HOCl and HClO3 respectively. 27.

(a)

28.

(c)

29.

Nitrogen and oxgen in air do not react to form oxides of nitrogen in atmosphere because the reaction between nitrogen and oxygen requires high temperature.s + 2 NaBr + 2 [A]

+

+

+2

[B] Brown and pungent smell

+

+

fumes

+

[C]

+

30.

+

;

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Oxidation states of ‘S’ are : + 4 in (A) , (+ 6) in B and + 2 in (C), + 2.5 in (D) 31. Sulphur trioxide produced in the contact process is absorbed by sulphuric acid forming H2S2O7. It is not dissolved in water as it gives a dense fog of sulphuric acid particles. The catalyst used in the contact process is vanadium pentoxide. 32. The reaction is NH2OH + 2H2SO4 HNO2 + 2H2SO3 + H2O (A)

(B)

(C) (D)

The structures of A, B, C and D are as follows.

(A)

33.

(B)

SO2 + PCl5

(C)

(D)

SOCl2

+ POCl3

Thionyl chloride

FeCl3 + 12HCl + 6SO2 FeCl3 . 6H2O + 6SOCl2 FeCl3 . 6H2O + 6CH3 – C(OCH3)2 – CH3 FeCl3 + 12CH3OH + 6CH3COCH3 34.

3PbS

+ PbS 3Pb + SO2

35.

The two resonating structures of ozone are :

36.

SO2 + H2O +

O2 → H2SO4

H2SO4 + 2NaCl → Na2SO4 + 2HCl 37. SiO2 < CO2 < N2O5 < SO3. Among oxides of the non-metals, the acidic strength increases with oxidation state. Hence SO3 (O.S. of S = +6) is most acidic followed by N2O5 (O.S. of N = +5) and CO2 and SiO2 (O.S. of C and Si = +4). Further CO2 is more acidic than SiO2 because of small size of C-atom. 38. (i) H2S oxidises into S,

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SO2 + 2H2S → 3S + 2H2O (ii) Sn +

→

(iii) Pb3O4 + 4HNO3 → 2Pb(NO3)2 + 2H2O + PbO2 ↓ 39. By boiling Na2SO3 solution with powder of sulphur in absence of air, sodium thiosulphate is prepared. Unreacted S is removed, filtrate is evaporated to give crystals of sod. thiosulphate. Na2SO3 + S → Na2S2O3 40. 2H2S + NaHSO3 + H+ → 3S ↓ + 3H2O + Na+ 41. (i) The reducing nature of SO2 is represented as + 2H+ + 2e– SO2 + 2OH– → Hence, with the increase of OH– (alkalinity) the forward reaction is favoured. (ii) NOTE : Oxygen is the 2nd most electronegative element after the fluorine and thus invariably show negative oxidation state. Further more, it has 2s22p4 configuration and thus requires only two electrons to complete its octet to show –2 oxidation state. Although sulphur also possess ns2np4 configuration but due to availability of dorbitals in their outer most shell –2, +2, +4, +6 oxidation state are also shown. Oxygen, however, shows only –2 oxidation state due to non-availability of d-orbitals in its outermost shell. (iii) Sulphur consists of S8 rings held together by weak van der Waal’s forces. As sulphur melts at 119ºC, these van der Waal’s forces are overcome and S8 rings slip and roll over one another giving rise to a clear mobile liquid. Above 160ºC, the S8 rings begin to open up and form long chains which gets tangled with each other, thereby gradually increasing the viscosity.

1.

(a) Generally, bond energy ∝

So bond energy order is C–F > C–Cl > C–Br > C–I 2. (a) 6NaOH + 3Cl2 → 5NaCl + NaClO3 + 3H2O (A, Sodium chlorate)

2Ca(OH)2 + Cl2 → Ca(OCl)2 + CaCl2 + H2O

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(B, Calcium hypochlorite)

3.

(a)

4.

(c) Due to strong H-bonding between HF molecules. HF has highest boiling point among the hydrogen halides. (a)

5. 6.

X : Ag , Y : Pb

(c) Cl2

+ NaOH

NaCl + NaClO + H2O

[cold and dilute]

7.

(a) NaF is composed of Na+ and F–.

Hence configuration of Na+ and F– do not match with the configuration given in the question. 8. (b) Order of reactivity of halogens But the interhalogen compounds are generally more reactive than halogens (except F2), since the bond between two dissimilar electronegative elements is weaker than the bond between two similar atoms i.e, X – X 9. (c) Chlorinated water is yellow in colour on standing following reaction occurs Cl2 + H2O → HCl + HOCl Thus, HCl and HOCl are formed. 10. (b) The molecular geometry of BrF5 is square pyramidal with asymmetric charge distribution on the central atom.

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11.

(c) The electron gain enthalpy order for halogens is Cl > F > Br > I Due to small size of fluorine the extra electron to be added feels more electron-electron repulsion. Therefore fluorine has less value for electron affinity than chlorine. 12. (b) The species called as pseudohalide ions are monovalent and made by electronegative atoms. They possess properties similar to halide ion. The corresponding dimers of these pseudohalide ions are called pseudohalogens. RCOO– is not a pseudohalide. KHF2 K+ + (HF2)– 13. (c) KF + HF 14. (c) CsBr3 may be represented as Cs+Br3– 15. (b) Chlorine is stronger oxidising agent than bromine therefore, chlorine water will liberate bromine from KBr solution. 2 KBr + Cl2 → 2 KCl + Br2 16. (b) Bleaching action of chlorine is only in presence of moisture where nascent oxygen is displaced from H2O. Cl2 + H2O → HCl + HClO (unstable) HClO → HCl + | O | 17. (d) HI and HBr (in that order) are the strongest reducing hydracids and hence they reduce H2SO4. HCl is quite stable and hence is oxidised by strong oxidising agent like KMnO4. HF is not a reducing agent. In the smallest F– ion, the electron which is to be removed during oxidation is closest to the nucleus and therefore most difficult to be removed. Therefore, HF is a poor reducing agent. Cl2 is more reactive than bromine. 18. (a) 19. (a) Due to highest bond dissociation energy. 20. (3) The structure of perchloric acid is

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The number Cl = O bond in HClO4 is 3. 21. (5) 22. (1.67)

Average bond order between Cl and O atom in NaClO3 =

= 1.67

complex ion; I2 + I– ——→ Hypobromous; bromite. KI3; complexes are more soluble in water as compared to normal salts. 26. HF; HF is the weakest of the three, because the ionisation (i.e. acidic character) of HX is a multistep process and when its ∆H, heat of ionisation, is calculated it comes out to be the minimum. This is due to the strong H – F bond, large heat of hydration (because of Hbonding) and low value of electron affinity of F-atom. 27. NaIO3

23. 24. 25.

28.

False : None amongst HBr and HI, exhibit hydrogen bonding. HI is a stronger acid than HBr because of its higher dissociation constant, Ka. HI has a stronger tendency to release protons to water molecules and hence, is a stronger acid. 29. False : Since halogens have high electron affinities, they easily pick up electrons from other substances. Hence, halogens are oxidising agents. The oxidising power decreases from fluorine to iodine. Since fluorine is the strongest oxidising agent it will oxidise any of the other halide ions in solution or when dry. Similarly, Cl2 will displace Br– and I– ions from their solutions and Br2 will displace I– ions. NOTE : In general, a halogen of low atomic number will oxidise the halide ion of higher atomic number. 30. True : Fe + 2HCl → FeCl2 + H2 [In FeCl2, Fe is in +2 state.]

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31. (a, b, d) (a)

Acidic order : Conjugate base order :

(b)

Hypochlorite ion (ClO–) :

Linear Trigonal pyramidal

Chlorate ion :

Perchlorate ion :

Tetrahedral

In chlorate ion bond angle changes due to presence of lone pair on chlorine atom. While there is no effect of lone pair on hypochlorite ion and perchlorate ion. (c) Disproportionation reaction of Hypochlorite ion : Chlorate ion : (d) 32.

(c, d)Energy, E

On moving down the group, the colour of the X2 molecule of group 17 elements changes gradually from yellow to violet. This happens because the amount of energy required for the excitation of the halogen atom decreases down the group. HOMO (π*)- LUMO (σ*) gap decreases down the group that makes π* to σ* excitation easier. Lesser the energy gap, more is the wavelength of light absorbed and hence, lesser is the wavelength of light emitted. 33. (a, b, d) (a) In both the acids central atom is sp3 hybridized.

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(b)

HClO4 is more acidic than HClO because ClO4– is more stable than ClO– due to resonance. (c) Cl2 + H2O → HCl + HClO (d) HClO4 is a stronger acid than H3O+, hence conjugate base of HClO4, i.e. ClO4– is weaker base than H2O. 34. (a, d) NH3 and CF2Cl2 (freon-12) are used as refrigerants. Ca2+ + –OCl + Cl– 35. (a) Ca(OCl) Cl – OCl (Hypochlorite ion) is anion of the acid HOCl which on dehydration gives Cl2O. H2O + Cl2O 2HOCl I2 + Products 36. (c) Bleach + 2KI Na2S4O6 + 2NaI I2 + 2Na2S2O3 Number of millimole of hypo = 0.25 × 48 = 2 × millimole of I2 ∴ Number of millimole of I2 =

=6

millimole of I2 = millimole of bleach Molarity of bleaching solution = 37.

38.

=

= 0.24

(c) F has slightly less electron affinity than chlorine because F has very small atomic size (only two shells). Hence, there is a tendency of electron-electron repulsion, which results in less evolution of energy in the formation of F– ion. Assertion is correct but reason is incorrect. More electronegative halogen displaces lesser electronegative halogen from its halide. Thus,

39. (a) 2KI + Cl2 2KCl + I2 Since, Cl2 is more powerful oxidising agent than I2, Cl2 is able to displace I– to form I2.

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2I–

I2 + 2e–,

E° = +0.54 V

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

On subtracting eq. (i) from eq. (ii), we get (b)

. is more powerful oxidising agent than

Here,

, so Cl is displaced

by I. ...(i) ..(ii) On subtracting eq. (i) from eq. (ii), we get 40.

At first, Bi(NO3)3 hydrolyses to give nitric acid which being an oxidising agent, oxidises potassium iodide liberating free iodine responsible for dark brown precipitate. Iodine dissolves in excess of potassium iodide forming soluble KI3 imparting yellow colour to solution. [Bi (OH)(NO3)2] + HNO3 Bi(NO3)3 + H2O – + – NO2 + 2H2O ] × 2 NO3 + 4 H + 3e – I2 + 2e– ] × 3 2I 2NO3– + 8H+ + 6I–

2NO + 4H2O +

KI + I2 41.

2NH3 + NaOCl → H2N.NH2 + NaCl + H2O Hydrazine

42. (i) 2FeBr2 + 3Cl2 → 2FeCl3 + 2Br2 (ii) 2SnCl2 + 2I2 → SnCl4 + SnI4 43. (i) HOCl < HOClO < HOClO2 < HOClO3 As the number of oxygen atoms increase, the -ve charge dispersal becomes more and more from Cl atom due to more electronegativity of oxygen atom. Due to more double bond

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(ii) 44. (ii) (iii)

character between Cl and O atoms, the bond length decreases and thus, bond strength increases. Hence, thermal stability increases. HI < HBr < HCl < HF The strength of H–X bond decreases from HF to HI. The larger is H–X bond length, lower is the bond energy, lesser is the bond strength. (i) 2NaIO3 + 5NaHSO3 → 2Na2SO4 + 3NaHSO4 + I2 + H2O NaClO3 + SO2 + 10H+ → NaCl + S + 5H2O NaBrO3 + F2 + 2NaOH → NaBrO4 + 2NaF + H2O + → I2 + H2O + + (iv)

(v)

This is a method used to prepare I2. 5NaHSO3 + 2NaIO3 → 3NaHSO4 + 2Na2SO4 + I2 + H2O 45. (i) The repulsive forces between fluorine atoms are high due to its small size and high electronegativity. It makes dissociation of F – F bond easy. So, bond dissociation energy of F2 is less than Cl2 (ii) The standard reduction potential of fluorine is highest and thus, it cannot be oxidized by any reagent. F2 + e–

F– ;

= maximum

(iii) Anhydrous HCl, being a covalent compound, is a bad conductor however, an aqueous solution of HCl is ionised to give H+ and Cl– ions and is a good conductor. (iii) NOTE : : HI cannot be prepared by heating hydrogen iodide with conc. H2SO4 because it is a strong oxidising agent and oxidises HI to I 2. H2SO4 + 2HI → SO2 + I2 + 2H2O Hence, HI is prepared by heating iodides with conc. phosphoric acid. 3KI + H3PO4 → K3PO4 + 3HI H3PO4 is not a strong oxidising agent. 2AlCl3 + 3 CO 46. (i) Al2O3 + 3C + 3Cl2 Alumina

Aluminium chloride

(ii)

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(iii)

SnO2 + 2C

Cassiterite

Sn + 2CO Tin

(iv) Na2SO4 + MnSO4 + 2H2O + Cl2 47. (i) HBr is a reducing agent and it reduces H2SO4 to SO2. (ii) Acids turn blue litmus red, so HClO also turns blue litmus red. The colour of litmus is decolourised because HClO is also a strong oxidising agent. 1.

(d)

(a) (b) (c) (d) 2.

(d) Radon is radioactive element and not present in atmosphere.

3.

(a)

4.

(a) Structure of XeO3F2

So, bond pairs = 5, π bonds = 3 lone pairs = 0

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5. 6.

7.

(a) Instantaneous dipole-induced dipole forces are most responsible in allowing xenon gas to liquify. (b) Xe. As we move down the group, the melting and boiling points show a regular increase due to corresponding increase in the magnitude of their van der waal forces of attraction as the size of the atom increases. (c) In XeOF4, Xe is sp3d2, hybridised having 6 bond pairs and 1 lone pair respectively.

8.

(d)

9.

(c)

10.

(d) XeO2F2 has trigonal bipyramidal geometry, but due to presence of lone pair of electrons on equitorial position, its shape is seesaw.

11. (b) In XeOF4, Xenon is sp3d2 hybridised and has one lone pair. 12. (19)

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Shape of XeF6 is distorted octahedral contains one lone pair e– s on central atom 3 lone pair e– s on each F atom surrounded by Xe. Total no. of lone pairs: 1 + 18 = 19 13. (d) (P) (Q) (R) (S) 14.

(A) – (p), (s)

(CH3)2SiCl2 + 2H2O

[(CH3)2Si–O]n. (B) – (p), (q), (r), (t) XeF4 + H2O Xe + XeO3 + H2F2 + O2 SiF4 + 2H2O SiO2 + 4HF glass H2[SiF6] SiF4 + 2HF Soluble hexafluorosilicic(IV) acid HCl + HOCl (C) – (p), (q) Cl2 + H2O [V(H2O)6]3+ (D) – (p), (q) VCl5 + 7H2O + 3Cl– + HCl + HOCl 15. (a) Argon, being a noble gas, will not react with the metals, thus, can be used in arc welding. 16. (c) In XeO3 there are total of 4 electron pairs around central atom. Out of which 3 are bonding electron pair and one is non-bonding electron pair. This combination provides sp3-hybridization and pyramidal shape.

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17.

(a) All xenon fluorides are strongly oxidizing, XeF4 can act as reducing agent (with F2) as well as oxidizing agent but XeF6 can only function as an oxidizing agent. 4Xe + 2XeO3 + 24HF + 3O2 6XeF4 + 12H2O XeF6 + 3H2O

XeO3 + 6HF

18. (i) (ii) Ammonia formed dissolves in water to form NH4OH (iii) (iv) (v) 19.

TIPS/Formulae : Use the formula where

H (hydridisation),

V = number of electron in valence shell of central atom M = number of monovalent atoms surrounding the central atom C = Charge on cation A = Charge on anion XeF2 :

Hence, hybridisation is sp3d, and thus its

shape is linear. XeF4 :

, Hence, hybridisation is sp3d2. and

thus its shape is square planar.

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XeO2F2 :

, Hence, hybridisation is sp3d. and

shape is see saw.

XeF2

XeF4

XeO2F2

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1.

(c) (i) W(VI) is more stable than Cr(VI) due to smaller size of atoms and also due to lanthanide contraction. (ii) Permanganate titrations in presence of HCl are unsatisfactory as HCl is oxidised to Cl2. (iii) Lanthanoid oxides are used as phosphors.

2.

(c)

Manganate ion is paramagnetic while permanganate ion is diamagnetic. 3. 4. 5.

(b) 26Fe = [Ar] 3d64s2. Third ionisation results into stable d5 configuration. (a) Atomic size of elements of 4d and 5d transition series are nearly same due to lanthanide contraction. (a)

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6.

(b) Electronic configuration of the given transition metal ions are : Sc3+ (Z = 21) 1s22s22p63s23p6 Ti2+ (Z = 22) 1s22s22p63s23p63d2

Ti3+ (Z = 22) 1s22s22p63s23p63d1 V2+ (Z = 23) 1s22s22p63s23p63d3 Since, magnetic moment is directly proportional to the number of unpaired electrons. The correct increasing order of magnetic moment is Sc3+ < Ti3+ < Ti2+ < V2+ because they have 0, 1, 2 and 3 unpaired electrons respectively. 7.

(a) Ruby is aluminium oxide (Al2O3) containing about 0.5 – 1% Cr3+ ions, which are randomly distributed in place of Al3+ ions.

8.

(a) Interstitial compounds are inert, i.e., they are chemically nonreactive.

9.

(c) Sc shows oxidation state of +3 only.

10.

(b)

11.

(d) As zinc has no unpaired of electrons to take part in the bond, it has least enthalpy of atomisation amongst the given transition elements.

12.

(a)

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13.

(d) (i)

ZnO + Na2O → Na2ZnO2 acid

(ii)

base

salt

ZnO + CO2 → ZnCO3 base

acid

salt

14.

(d) Ions E0 (V) Ti2+ –0.37 V2+ –0.26 Cr2+ –0.41 Mn2+ +1.57 Negative value of E0 means these metals librate hydrogen from dilute acid. 15.

(c) (Fe becomes passive on reaction with concentrated HNO3). However, cold relatively conc. HNO3 reacts with Fe as below. Fe + 6HNO3 → Fe(NO3)3 + 3NO2 + 3H2O Cu + 4HNO3 (conc.) → Cu(NO3) 2 + 2 NO2 + 2 H2O 4Au + 8NaCN + O2 + 2H2O → 4Na[Au (CN)2] + 4NaOH Zn + 2 NaOH → Na2ZnO2 + H2 16.

(c) Reaction of Zn with different concentration of HNO3 are as follows : Zn(NO3)2 + 2NO2↑ + 2H2O Zn + 4HNO3(60%) 3Zn(NO3)2 + 2NO↑ + 4H2O 3Zn + 8HNO3(30%) 4Zn(NO3)2 + N2O↑ + 5H2O 4Zn + 10HNO3(20%) 5Zn(NO3)2 + N2↑ + 6H2O 5Zn + 12HNO3(10%) 4Zn(NO3)2 + NH4NO3 + 3H2O 4Zn + 10HNO3(3%) Hence option (c) is correct. 17. (d) Out of all the four given metallic oxides, CrO2 is attracted by magnetic field very strongly. The effect persists even when the magnetic field is removed. Thus CrO2 is metallic and ferromagnetic in nature.

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18. (c) A ruby is a crystal of alumina, aluminum oxide (Al2O3) containing a trace of chromium (III) ions replacing some of the aluminum ions. In ruby, each Al3+ ion and Cr3+ ion is surrounded by six oxide ions in an octahedral arrangement. The origin of the color of emeralds is similar to that of the color of rubies. However, the bulk of an emerald crystal is composed of beryl, beryllium aluminum silicate (Be3Al2(SiO3)6) instead of the alumina which forms rubies. The color is produced by chromium (III) ions, which replace some of the aluminum ions in the crystal. In emeralds, the Cr3+ is surrounded by six silicate ions, rather than the six oxide ions in ruby. Therefore, the color (green) of emeralds is different from that of ruby. 19. (a) Aqueous solution of CoCl2 contains [Co(H2O)6]2+ which is pinkish in colour so option (d) is incorrect. Reduction potential of

is high so option(b) is

incorrect. Co2+ does not oxidises easily to Co3+. It is general case that symmetrical substituted octahedral complexes are less deeper in colour than tetrahedral complexes. So [CoCl4]2– is deep blue in colour. 20. (b) The equation in option (b) is correct since both charges as well as atoms are balanced. For the rest, (a) Given reaction in unfavourable in the forward direction (K2O is unstable, while Li2O is stable) (c) Given reaction is not balanced w.r.t. charge. (d) Given reaction will give K3[Cu(CN)4] as product instead of K2[Cu(CN)4]. 21. (d) In equation (a) Fe2(SO4)3, and in equation (b) FeSO4, on decomposition will form oxide instead of Fe. In equation (c) FeCl3 cannot be reduced when heated in air. Hence equation (d) is correct. 22. (d) Maganese exhibits the large number of oxidation states. The most common oxidation states of Mn are +2, +3, + 4, + 6 and +7. 23. (a) K2Cr2O7 + 4H2SO4 + 3H2S K2SO4 + Cr2 (SO4)3 + 7H2O + 3S 24. (a)

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(a) V = 3d 3 4s 2 ; V2+ = 3d 3 = 3 unpaired electrons Cr = 3d 5 4s 1 ; Cr2+ = 3d 4 = 4 unpaired electrons Mn = 3d 5 4s 2 ; Mn2+ = 3d 5 = 5 unpaired electrons Fe = 3d 6 4s 2 ; Fe2+ = 3d 6 = 4 unpaired electrons Hence the correct order of paramagnetic behaviour V2+ < Cr2+ = Fe2+ < Mn2+ (b) For the same oxidation state, the ionic radii generally decreases as the atomic number increases in a particular transition series, hence the order is Mn2+ > Fe2+ > Co2+ > Ni2+ (c) Larger size, least hydrated more stable in aqueous solution. As we move across the period (Sc3+ → Cr3+ → Fe3+ → Co3+ ), the ionic size usually decreases. Sc3+ with the large size as least hydrated and hence more stable. (d) Sc – (+ 2), (+ 3) Ti – (+ 2), (+ 3), (+ 4) Cr – (+ 2), (+ 3), (+ 4), (+ 5), (+ 6) Mn – (+ 2), (+ 3), (+ 4), (+ 5), (+ 6), (+ 7) i.e. Sc < Ti < Cr < Mn 25.

(d)

= – 0.41 V, = + 0.77 V = + 1.57 V,

= + 1.97 V

26.

(b) KMnO4 reacts with H2SO4 to form Mn2O7 which is highly explosive substance. K2SO4 + Mn2O7 + H2O 2KMnO4 + H2SO4 27. (a) The element with outer electron configuration 3d 54s2 is Mn which exhibits oxidation states from + 2 to + 7. 28. (a) CuSO4 will be absorbing orange-red colour and hence will be of blue colour. 29. (c) NOTE : Colour is due to d – d transitions. Coloured compounds contain partly filled d-orbital.

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The oxidation state of copper in various compounds is + 1 and + 2. In CuF2 it is in + 2 oxidation state. In + 2 state its configuration is

i.e. It has one unpaired electron due to this it is coloured. (NOTE : CuF2 possesses blue colour in crystalline form) 30. (b) In the presence of oxygen, Ag metal forms a water soluble complex Na [Ag (CN)2] with dilute solution of NaCN

31.

(a)

CuCN + 3 KCN 32.

(b) Colour of transition metal ion salt is due to d-d transition of unpaired electrons of d-orbital. Metal ion salt having similar number of unpaired electrons in d-orbitals shows similar colour in aqueous medium.

Number of unpaired electrons = 1 33. (a) (Same gas i.e., N2)

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34.

(a)

35.

(a) Stable oxidation state of Mn in alkaline medium is +6. So, MnO2 is oxidised to K2MnO4 (purple green) by atmospheric oxygen in KOH medium.

36.

(b)

37.

(b) The structure of

There are six normal Cr – O bonds and two bridged Cr – O bonds. The six normal Cr – O bonds are expected to be equivalent and different from those of the bridged Cr – O bonds. 38. (b) The electronic configurations of cations in the given salts are Only Cu2+ ion has one unpaired electron in 3d orbital and so, its salt is expected to be coloured. and in acidic medium is derived as 39. (a) The reaction of follows: + 8H+ + 5e-

Hence, 2 mole

Mn2+ + 4H2O]

2

5 mol

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i.e., 40.

mol

1 mol

(b)

Hence, green coloured powder blown in the air is Cr2O3. 41. (c) TIPS/FORMULAE : Chrome alum is K2SO4.Cr2(SO4)3.24H2O Sodium peroxide (Na2O2) will act as an oxidizing agent. It will oxidise Cr3+ to Cr6+ and Fe2+ to Fe3+.

Hence, the filtrate will be yellow in colour and the residue will be brown in colour. 42. (a) HgS does not dissolved in hot dil. HNO3. Cu (CN)2 + K2SO4 43. (d) CuSO4 + 2KCN Cu2(CN)2 + (CN)2 (Cyanogen) 2Cu(CN)2 2K3 [Cu(CN)4] Cu2(CN)2 + 6 KCN 2KCl + 2MnCl2 + 8H2O + 3Cl2 44. (a) 2KMnO4 + 16HCl 45. (c) Solder is an alloy containing Sn – 67% and Pb – 33%. 46. (b) In Zn–Cu couple, Zn is activated by Cu. It is used as a reducing agent in organic synthesis. The proportion of Zn is about 90% and it can be prepared by coating Zn with copper. 47. (d) Conc. HNO3 renders iron passive by forming a thin protective film of Fe3O4 on its surface. 48. (c) Hypo solution (Na2S2O3) is used in photography to remove the unaffected AgBr in the form of soluble complex.

49.

(b)

. It has 2 unpaired electrons.

3d orbital of Ni2+ ion. At No. of Ni = 28.

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50.

(b)

51. 52.

(b) German silver is alloy of Cu + Zn + Ni Basicity of hydroxides decreases on moving left to right in a (d) period. ZnSO4 + H2 53. (a) Zn + H2SO4 Na2[ZnO2] + H2 Zn + 2NaOH ∴ Ratio of H2 evolved is 1 : 1. 54.

(4)

Structure of CrO5 :

Number of oxygen atoms bonded to chromium through single bond is 4. 55. (6)

56.

∴ 8 moles of MnO–4form 6 moles of SO2–4. (7) K2Cr2O7 + KI + H2SO4 → K2SO4 + Cr2(SO4)3

2CuSO4 + KI → 2CuI + I2 + 2K2SO4

+ I2 + H2O

H2O2 + 2KI → 2KOH + I2 Cl2 + 2KI → 2KCl + I2 O3 + H2O + 2KI → 2KOH + O2 + I2 FeCl3 + 2KI → 2KCl + FeCl2 + I2 HNO3 + KI → KNO3 + I2 + NO 57. (6) Oxidation number of Mn in K2MnO4 is 6 K2MnO4 ; 2 + x – 8 = 0 x=6 58. (12.00)

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59.

Total number of Cr and O bonds is 12. (18.00) 4NaCl + K2Cr2O4 + 3H2SO4 (Conc.)

2CrO2Cl2 + K2SO4 (A)

+ 2NaSO4 + 3H2O CrO2Cl2 + NaOH (A)

Na2CrO4 + H2O + NaCl (B)

Na2CrO4 + H2SO4 + H2O2 (B)

60. 61. 62. 63. 64. 65. 66. 67.

68.

(dilute)

CrO5 + Na2SO4 + H2O (C)

The sum of total no. of atoms in one molecules each of A, B & C = 5 + 7 + 6 = 18.0 H2S; It is due to formation of sulphide of silver (Ag2S) which is black. FeSO4.7H2O, ZnSO4.7H2O; Hydration/solvation; [A substance dissolves when its ∆Hhydration > lattice energy]. Zinc; PbO2; Pb4+ can be easily reduced to Pb2+. K[Ag(CN)2] False : Dipositive zinc exhibits diamagnetism (and not paramagnetism) because it has no unpaired electron. True : Insolubility of AgCl in H2O is due to its high lattice energy. Further, AgCl forms a complex with conc. NaCl solution and is therefore soluble. True : Hydration energy of AgF is appreciably higher than its lattice energy because of smaller F– ion and thus AgF is soluble in water. In rest of the halides, lattice energy is more than hydration energy to make them insoluble.

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69.

(a, b, c) (W)

2K2MnO4 + 2H2O ¾→ 2KMnO4 + 2KOH (g) + H2 (W)

(X)

(mangnate ion)

K2MnO4 ¾→ 2K+ +

(Y)

(permangnate ion)

KMnO4 ¾→ K+ + (Z)

• sp3, Tetrahedral • Green colour

• •

•

•

• Paramagnatic

•

Disproportionation of

sp3, Tetrahedral Purple colour Diamagnetic undergoes in acidic medium but not in base,

concerned reaction is as under : 70.

(b, c, d)

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(a) (b) (c) (d) 71.

Oxidation state of Zn in Na2ZnO2 is +2 Bond order of Q is one for H2. ZnS is white in colour SO2 is angular in shape (a, b) (a) (b) The formed H2O2 will reduce Fe3+ to Fe2+.

72.

(a, b, c) Cr2+ is a reducing agent and Mn3+ is an oxidizing agent and both have electronic configuration d 4.

Above E° values explains reducing nature of Cr2+ and oxidizing behaviour of Mn3+. 73.

(a, b, c) Na + NH3 (excess) → Dilute solution of Na in liq. NH3 → Paramagnetic due to ammonioted e–. K + O2 (excess) → KO2 (O2– is paramagnetic) Cu + HNO3 (dil.) → Cu(NO3)2 + NO (NO is paramagnetic) 2-Ethylanthraquinol + O2 → 2-Ethylanthraquinone + H2O2 (H2O2 is diamagnetic) 74. (a, c, d) (a)

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(b)

(c) When the filterate containing KI3 add to start solution, the dissolved I2 will produce a blue colour solution. (d)

+ [Fe(CN)6]4– (aq) 75.

(b, c, d) Cu2+ ions will react with CN– and SCN– forming [Cu(CN)4]3– and [Cu(SCN)4]3– leading the reaction in the backward direction.

Cu2+ also combines with CuCl2 which reacts with Cu to produce CuCl pushing the reaction in the backward direction. 76. (a, d) Potassium permanganate will oxidize itself in this process. In acidic medium Change in oxidation state of Mn = 7 – 2 = 5 Thus electrons lost = 5

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In neutral medium Change in oxidation state of Mn = 7 – 4 = 3 ∴ Electrons lost = 3 77. (a, c) Mn makes steel harder and increases its elasticity and tensile strength. Further Mn acts as deoxidiser. MnO reacts with S present in cast iron, gets oxidised and then combine to form slag. 78. (b, c) Brass : Cu (60-80%), Zn (40-20%); Gun Metal : Cu (87%), Sn (10%), Zn (3%). 79. (c, d) Aqueous solution of Co(NO3)2 and CrCl3 in which Co2+ (d 7) and Cr3+ (d3) contains incompletely filled d-orbitals are coloured. 2K2MnO4 + 2H2O ; 80. (b, c) 2MnO2 + 4KOH + O2 2KOH + HCHO + 2 KMnO4 → K2MnO4 + H2O + HCOOH 81. (c) (A) V2O5 Preparation of H2SO4 in contact process (B) TiCl4+ Al(Me)3 Polyethylene (Ziegler-Natta catalyst) (C) PdCl2 Ethanol (Wacker’s process) (D) Iron oxide NH3 in (Haber’s process) 82. (A) –p, s ; (B) –q, s ; (C) –r, t ; (D) –q, t (A) (B) (C) (D) 83. (i) (g) (ii) (d) (iii) (e) (v) (f) (vi) (b) (vii) (c) 84. (b)

(iv) (a)

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85. 86.

(a) (c) (M)

(N)

Blue

(remaining)

(O)

Soluble

(remaining)

87.

88.

(Deep blue colour)

(a) Reactant should not be adsorbed strongly which might result into immobilisation that inhibit further adsorption on the catalyst’s surface. (b) 3d 4s Zn Zn2+

Zn2+ is diamagnetic because of absence of unpaired electrons. 89. (c) The statement-1 is correct + H2SO4 →

+ Na2SO4 + H2O

Oxidation state of Cr in K2CrO4 and K2Cr2O7 is +6, i.e. no change in O.S. So explanation is wrong. [B] = HCl 90. [A] =

+ (n + 2)

TiO2

+ 4 HCl

(moist air) white

= [Ar] 3d 0;

fumes [B]

= [Ar] 3d 1

TiCl4 is colourless since Ti4+ has no d electrons, henced-d transition is impossible. On the other hand, Ti3+ is coloured due to d-d transition. Ti3+ absorbs greenish yellow compound of white light, hence its

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91.

aqueous solution is purple which is complementary colour of greenish yellow in white light. Reaction involved in developing of a black and white photographic film.

+

Na2S2O3 + 2H+

+ 2Ag + 2HBr

2Na+ + H2SO3 +

92.

(i) Argentite is Ag2S. Silver is extracted from its ore argentite (silver glance, Ag2S) as follows : (1) Silver glance is concentrated by froth flotation. (2) Leaching : The concentrated ore is ground to fine powder and dissolved in dilute solution of sodium cyanide. Oxygen of air converts Na2S to Na2SO4 thereby preventing reaction to take place in the reversible direction. (3) Recovery of silver. Silver is precipitated out by adding electropositive metal, Zn. (ii) For development, activated grains are preferentially reduced by mild reducing agents like hydroquinone

( Reduction of activated AgBr to elemental silver.) The photographic film is permanently fixed by immediately washing out any non activated AgBr grains in hypo emulsion.

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hypo

soluble

93.

94.

2 Cu + H2O + CO2 + O2

CuCO3. Cu(OH)2

Green basic copper carbonate

95.

96.

Cu = 1s2, 2s2 2p6, 3s23p63d10, 4s1 2 2 6 2 6 10 2 30Zn = 1s , 2s 2p , 3s 3p 3d , 4s On the basis of configuration of Cu and Zn, first ionisation potential of Zn is greater than that of copper because in zinc the electron is removed from 4s2 configuration while in copper it is removed from 4s1 configuration. So more amount of energy is required for the removal of electron of 4s2 (completely filled orbital) than that of 4s1 while the second ionisation potential of Cu is higher than that of zinc because Cu+ has 3d10 (stable configuration) in comparison to Zn+ (4s1 configuration). + 4H+ → 3[Fe(H2O)6]2+ + 29

NO + 3[Fe(H2O)6]3+ + 2H2O [Fe(H2O)6]2+ + NO → [Fe(H2O)5NO]2+ + H2O 97. ZnO + 2NaOH → Na2ZnO2 + H2O Sod. Zincate

98. (i) CuSO4 + 4NH4OH → [Cu(NH3)4]SO4 + 4H2O or CuSO4 + 2NH4OH → Cu(OH)2 + (NH4)2SO4 Cu(OH)2 + 2(NH4)2SO4 + 2H2O + H2SO4

→ (ii) H2O2 → H2O + [O]

]×3

+ 3NaOH

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2Cr(OH)3 + 4NaOH + 3[O]

+ 6NaCl + 8H2O 99.

2Mn(OH)2 + 5NaBiO3 + 18H+ → 2MnO–4+ 5Bi3+ + 5Na+ + 11H2O 100. (i) CrO3 is acid anhydride of H2CrO4 (Chromic acid) [Anhydride are formed by loss of water from acid] CrO3 + H2O H2CrO4 In H2Cr2O4 , Cr is present in + 6 oxidation state. (ii) Cu2+ is reduced to Cu+ by I–, hence cupric iodide is converted into cuprous iodide so [CuI4]2– does not exist, Cl– cannot effect this change and thus [CuCl4]2– exists. (iii) Mercurous chloride changes from white to black when treated with ammonia due to the formation of finely divided mercury. (iv) Zinc is cheaper and stronger reducing agent than copper and zinc is volatile (v) The transition metals form coloured compounds and coloured complexes. They have vacant d-orbitals. Electrons take up energy from the visible region and move to higher energy levels. The visible colour of the substance is the complementary colour of the absorbed light. [NOTE : The colour is due to d-d transitions] (vi) It is because silver bromide being sensitive to light, reduces into metallic silver grains when light fall on it. Zn(NO3)2 + 2[H]] 4 101. (i) [Zn + 2HNO3 (dil)

(ii) (iii)

3[MnO4]2– + 4H+ → MnO2 + 2[MnO4]– + 2H2O

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(iv)

3SO2 +

(v)

(NH4)2S2O8 + 2H2O + MnSO4 → MnO2 + 2H2SO4 + (NH4)2SO4 AgBr + 2Na2S2O3 → Na3 [Ag(S2O3)2]

(vi)

+ 2H+ → 2Cr3+ +

+ H2O

+ NaBr

Sod. argento thiosulphate

(vii) K2Cr2O7 + 14HCl 2KCl + 2CrCl3 + 7H2O + 3Cl2 →

(viii) 2CuSO4 + 2Na2CO3 + H2O CuCO3.Cu(OH)2 + 2Na2SO4 + CO2 (ix) The individual reactions are 3Cu + 8HNO3 (dil.) → 2NO + 3Cu(NO3)2 + 4H2O Cu + 4HNO3 (conc.) → Cu(NO3)2 + 2NO2 + 2H2O For the molar ratio of 2 : 1 of NO and NO2, we will have 7Cu + 20HNO3 → 7Cu(NO3)2 + 4NO + 2NO2 + 10H2O (x) 2KMnO4 + 3MnSO4 + 2H2O → 5MnO2 + K2SO4 + 2H2SO4 This is known as Volhard method for estimation of manganese. (xi) 4Fe + 10HNO3 → 4Fe(NO3)2 + NH4NO3 + 3H2O (xii) K2Cr2O7 + 4NaCl + 6H2SO4 → 2CrO2Cl2 ↑ + 4NaHSO4 + 2KHSO4 + 3H2O chromyl chloride (orange)

(xiii)

(xiv) CoCl2 + 2KNO2 → Co(NO2)2 + 2KCl

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KNO2 + CH3COOH → CH3COOK + HNO2 Co(NO2)2+ 3KNO2 + 2HNO2 → K3[Co(NO2)6]↓ + NO + H2O Pot. cobaltinitrite (yellow ppt.)

NaCl + Na [Ag(CN)2] (xv) AgCl + 2NaCN 2Na[Ag(CN)2] + Zn → Na2[Zn(CN)4] + 2Ag↓ soluble

+ 4NaCN

(xvi)

2 Na[Ag(CN)2] + Na2S

4Na2S + SO2 + 2H2O → 2 Na2SO4 + 4 NaOH + 2S [NOTE : Na2S is converted into Na2SO4 to avoid reversibility of first reaction] 2NaAg(CN)2 + Zn → Na2 [Zn(CN)4] + 2Ag Sod. Zincocyanide

→ NOCl + 2H2O + 2(Cl)

(xvii)

AuCl3

Au + 2[Cl]

(xviii) 2 KMnO4 + 3H2SO4 → K2SO4 + 2MnSO4 + 3H2O + 5[O] 5 H2C2O4 + 5[O] → 10CO2 ↑ + 5H2O

(xix) K4[Fe(CN)6] + 6H2SO4 + 6H2O 2K2SO4 + FeSO4 + 3(NH4)2SO4 + 6CO (xx) (xxi) (yellow)

(xxii) 2

2KMnO4 + 4KOH + MnO2 → 3K2MnO4 + 2H2O 2K2CrO4 + H2SO4 → K2Cr2O7 + K2SO4 + H2O

(orange red)

+ 4H+ → MnO2 +

+ 2H2O

(xxiii) Fe2(SO4)3 + 2KI → 2FeSO4 + K2SO4 + I2 (xxiv) 2CuSO4 + SO2 + 2H2O + 2KCNS

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→ 102. SO2 +

+ K2SO4 + 2H2SO4

+ H+ →

+ Mn2+ + H2O

103. On standing FeCl3 is hydrolysed and produces colloidal solution of Fe(OH)3 which is in form of brown precipitate.

104. (i)

2Ag + 2H2SO4

+ SO2 + H2O + N2 + 4H2O

(ii) (NH4)2 Cr2O7 (iii) 2KMnO4 + 3H2SO4 + 5H2S

K2SO4 + 2MnSO4 + 105. (i)

ZnO + CO2

ZnCO3

(A)

(ii) ZnO + 2HCl

+ 8H2O

(C)

(B)

H2O + ZnCl2

(C)

(soluble)

4KCl + Zn2 [Fe (CN)6]↓

(iii) 2ZnCl2 + K4 [Fe(CN)6]

(white ppt)

(iv) ZnCO3 + HCl (A)

(v) CO2 + Ca(OH)2 (B)

CO2 + ZnCl2 (soluble)

CaCO3 + H2O (Milky)

(vi) CaCO3 + CO2 + H2O → Ca(HCO3)2 (soluble)

(vii) ZnCl2 + H2S

2HCl + ZnS ↓ (white) (D)

(viii) ZnCl2 + 2NaOH → 2NaCl + Zn(OH)2↓ (white) (E)

(ix) Zn(OH)2 + 2NaOH → Na2ZnO2 + H2 sod. zincate (soluble)

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106. (i) Since the compound (A) on strong heating gives two oxides of sulphur (C and D) which might be SO2 and SO3, it must be a sulphate. (ii) The reaction of compound (E) with thiocyanate to give blood red coloured compound (H) indicates that (E) must have Fe3+ ion. Thus the compound (A) must be ferrous sulphate, FeSO4.7H2O, which explains all given reactions as below (Fe2+ ion of FeSO4 is changed to Fe3+ during heating). FeSO4 + 7H2O↑ FeSO4 . 7 H2O Ferrous sulphate (A)

Ferrous sulphate

Fe2O3

2 FeSO4 Ferrous sulphate

Ferric oxide, (Blackish brown powder) (B)

(C)

(D)

+ 3H2O

Ferric chloride (Yellow solution), (E)

2FeCl3 + H2S (E)

FeCl3 + 3NH4CNS (E)

SO2 + SO3

2FeCl3

Fe2O3 + 6 HCl (B)

+

2FeCl2

+ 2HCl +

Apple green (G)

S

White turbidity (F)

Fe (CNS)3 + 3NH4Cl Ferric thiocyanate (blood red colour)(H)

1. (c) Europium (Eu) Atomic No. - 63 Electronic configuration - [Xe]4f 76s2 It shows only +2 and +3 oxidation state. 2. (a) Mischmetal is an alloy consisting mainly of lanthanoid metals. Alloy ⇒ Lan. metal Iron

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S, C, Ca, Al ⇒ traces 3. (b) Electronic configuration of

No. of unpaired electron = 7 4. 5.

(b) Eu2+ : [Xe]4f 7 ; Ce3+ : [Xe]4f 1 (d) Actinoids Oxidation state shown Th + 3, + 4 Ac + 3 Pu + 3, + 4, + 5, + 6, + 7 Np + 3, + 4, + 5, + 6, + 7 Bk + 3, + 4 Cm + 3, + 4, + 5 Lr + 3 ∴ Maximum oxidation state is shown by Np and Pu. 6. (b) Sm = 4f 6 6s2 Sm3+ = 4f 5 = Partially filled f orbital Sm3+ will be coloured Lu3+ = 4f 14 = colourless. 7. (d) Atomic radii follows the order Eu > Ce > Ho > N 185 pm 182 pm 177 pm 71 pm 8. (c) Due to lanthanoid contraction, size of atoms as well as ions of lanthanoid decreases.

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1.

(d) trans-[Co(en)2Cl2]+ (A) contains a plane of symmetry so (A) cannot be optically active, whereas, (B) cis [Co(en)2Cl2]+ does not contain any plane of symmetry, so (B) will be optically active. 2. (d) Only cis-[CrCl2(ox)2]3– shows optical isomerism while its trans form do not show optical isomerism due to presence of plane of symmetry. 3. (d) Conc. H2SO4 acts as dehydrating agent. Molar mass of given complex = 266.5 g/mol. On treating with conc. H2SO4 the mass lost by the complex

= 2 moles of H2O

Formula of the complex = [Cr(H2O)4Cl2]Cl . 2H2O 4. (c) (a) [Ni(NH3)4(H2O)2]2+ shows geometrical isomerism. (b) [Ni(en)3]2+ shows optical isomerism. (c) ⇒ sp3 hybridisation ⇒ tetrahedral So, [Ni(NH3)2Cl2] does not show isomerism. (d) [Pt(NH3)2Cl2] shows geometrical isomerism. B.M. = 3.83 B.M. 5. (d) = n = 3 (n = No. of unpaired e–) Therefore, oxidation number of Cr should be +3. Hence complex is Cr(H2O)6Cl3. Complex shows geometrical isomerism therefore formula of complex is [Cr(H2O)4Cl2]Cl.2H2O.

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IUPAC Name: Tetraaquadichlorido chromium(III) Chloride dihydrate 6. (d)

7.

(d) [Ma3b3] type complex shows facial and meridional isomerism. So, the complex [CO(NH3)3 (NO2)3] will show fac-and mer-isomers. 8. (c) [Pt(NH3)2Cl(NO2)] and [Pt(NH3)4ClBr]2+ are the Ma4bc and Ma2bc type of complexes. Each shows two geometrical isomers i.e., cis and trans isomers. 9. (a) [Pt (NH3)2Cl(NH2CH3)]Cl Diamine chlorido (methanamine) platinum (II) chloride 10. (c) [MA2B2] will not exhibit optical isomerism in both conditions. 11. (a) Co3+ with strong field ligand forms complex of low magnetic moment. t

=

0

t

=

× 18000 = 8000 cm–1

12. (d) [Co(en)2 Cl] Cl Cl– - monodentate ligand en - bidentate ligand ∴ Co-ordination Number of Co = (2 × 2 ) + 1 = 5 K3[Al(C2O4)3] C2O42–- bidentate ligand ∴ Co-ordination Number of Al = 2 × 3 = 6 13. (b) The maximum possible denticites of the given ligand towards transition metal ion is 6 and towards inner transition metal ion (due to greater ionic radii and more atomic orbitals) is 8.

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14. (d) K4[Th(C2O4)4(H2O)2] (oxalato) : bidentate ligand H2O (aqua): monodentate Co-ordination no. of Th = 2 × 4 + 2 = 10 15. (a) Reaction for the given condition can be written as: [CoCl6]3– + 2(en) [CoCl2(en)2]+ (1:2 mole ratio) (cis-trans-isomer) A = optically active (cis-isomer), violet B = optically inactive (trans-isomer), green

16. (b)

17. (b) The square planar complex of the type [Mabcd]n± , where all four ligands are different, has 3 geometrical isomers. But if one of the ligands is ambidentate, then 2 × 3 = 6 geometrical isomers are

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possible. But if two ligands are ambidentate, then 4 × 3 = 12 geometrical isomers are possible. In the given example, NO2– and SCN– are ambidentate ligands. 18. (d) Moles of complex = 0.01 mole Moles of ions precipitated with excess of AgNO3 ∴

= 0.02 moles Number of Cl– present in ionisation sphere

It means 2Cl– ions present in ionization sphere. Thus formula of the complex is [Co(H2O)5Cl]Cl2.H2O 19. (d) Optical isomerism occurs when a molecule is non-super imposable with its mirror image hence the complex cis[Co(en)2Cl2]Cl is optically active.

20. (a) Complexes having only one type of ligands are examples of homoleptic complex. 21. (d) Square planar complexes of type M[ABCD] form three isomers. Their position may be obtained by fixing the position of one ligand and placing at the trans position any one of the remaining three ligands one by one. 22. (d) It does not show optical and geometrical isomerism (B) and (C) shows only geometrical isomerism. 23. (b) The two possible isomers for the given octahedral complex are [M(NH3)5 SO4] Cl and [M(NH3)5 Cl] SO4. They respectively give

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chloride ion (indicated by precipitation with BaCl2) and SO4 ion (indicated by precipitation with AgNO3). Hence the type of isomerism exhibited by the complex is ionisation isomerism. 24. (c) Octahedral coordination entities of the type Ma3b3 exhibit geometrical isomerism. The compound exists both as facial and meridional isomers, both contain plane of symmetry

25. (a) The compound shows linkage isomerism because the ligand in the compound is an ambidenate ligand that can bond at more than one atomic site. and i.e., 26. (d) [Co(H2O)4(NH3)2]Cl3 = Diamminetetraaquacobalt (III) chloride. 27. (c) The correct structure of EDTA is

28. 29.

(b) Ionisation isomer of [Cr(H2O)4Cl(NO2)]Cl is [Cr(H2O)4Cl2]NO2. (c) The correct IUPAC name of the given compound is tetraamminenickel (II) - tetrachloronickelate (II) thus (c) is the correct answer. 30. (a) Co(NH3)4Br2Cl will show both geometrical and ionization isomerism. [Co(NH3)4Br2]Cl and [Co(NH3)4BrCl]Br are ionization isomers and geometrical isomers are

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cis

trans

cis trans – 31. (a) In [MnO4] , Mn is in +7 oxidation state. Electronic configuration of Mn (Z = 25) : [Ar] 3d 54s2 Electronic configuration of Mn7+ : [Ar] 3d 04s0 Central atom in other ions have definite number of d electrons. No. of electrons

32. (d) Organometallic compounds are those compounds in which metal atom is directly bonded with C-atom. H3C – Li. 33.

(6)

EDTA4– is a hexadentate ligand, since hexa means six and the ligand attaches six times.

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34.

(6)

Structure of cis-[Mn(en)2Cl2]

Number of cis N – Mn – Cl bonds, i.e. Mn – N and Mn – Cl bonds are in adjacent positions. 1 – Cl (a) – Mn – N (1) 2 – Cl (a) – Mn – N (2) 3 – Cl (a) – Mn – N (4) 4 – Cl (b) – Mn – N (4) 5 – Cl (b) – Mn – N (3) 6 – Cl (b) – Mn – N (1) Hence, the answer is 6. Note : All adjacent positions (at 90°) are cis to each other. 35. (5) [CoL2Cl2]– (L = H2NCH2CH2O–) L is unsymmetrical didentate ligand. So, the complete is equivalent to [M(AB)2a2] Possible G..I. are

36. (6) All the complexes given show cis-trans isomerism [Co (NH2 — CH2 — CH2 — NH2)2Cl2]+, [CrCl2(C2O4)2]3–

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37. (3)

38. (8)

Total no. of N – Co – O bonds are 8. 39. (6) m moles of [Cr (H2O)5Cl]Cl2 = 0.01 × 30 = 0.3 m moles of Cl– = 0.3 × 2 = 0.6 [1 mole of complex gives 2 Cl– ions] m moles of Ag+ = m moles of Cl–

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0.1 × V = 0.6 V = 6 mL 40. (3)

The number of geometrical isomers is 3. 41. (4) The number of water molecules directly bonded to the metal centre in CuSO4.5H2O is 4. SO42– . H2O

42. Hexammine cobalt (III) chloride 43. (b, d) The pair of complex ions [Co(NH3)4Cl2]+ and [Pt(NH3)2(H2O)Cl]+ show geometrical isomerism. The pair of complexes [Pt(NH3)3(NO3)]Cl and [Pt(NH3)3Cl]Br show ionisation isomerism. The other pairs given do not have same type of isomerism. The complex [CoBr2Cl2]2– will show sp3 hybridisation. The tetrahedral shape will not show any kind of isomerism. (c) The complex [PtBr2Cl2]2– will show dsp2 hybridisation. The square planar shape will show geometrical isomerism. Henc, (c) is not an answer. Note : Br and Cl are weak field ligand but in this case they will cause pairing of electrons. Most of the platinum complexes with four ligand are square planar in shape.

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44.

(c, d)

45.

(b) The geometrical isomers of [M(NH3)4Cl2] can be represented as follows:-

These isomers are optically inactive and they posses axis of symmetry. Both the statements are thus true. Out of two possible answers i.e. option (a) and (b) option (b) is correct as the statement 2 is not a correct explanation of statement 1. For a molecule to be optically active it should not possess alternate axis of symmetry. 46. The complex A does not react with concentrated H2SO4 implying that all water molecules are coordinated with Cr3+ ion. Hence, its structure would be [Cr(H2O)6]Cl3. The compound B loses 6.75% of its original mass when treated with concentrated H2SO4. This loss is due to the removal of water molecules which is/are not directly coordinated to Cr3+ ion. The mass of water molecules removed per mole of the complex

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=

molar mass of the complex =

266.5 g

= 17.98 g This corresponds to one mole of water. Hence, the structure of the compound B will be [Cr(H2O)5Cl](H2O)Cl2 NOTE : The compound C loses 13.5% of its mass when treated with concentrated H2SO4 which is twice of the mass lost by the compound B. Hence, the structure of the compound C will be [Cr(H2O)4Cl2] (H2O)2 Cl. 47. (i) [CoCl(NH3)5]2+ Formula of pentaamminechlorocobalt (III) (ii) LiAlH4 Formula of lithium tetrahydroaluminate (III) 48. (i) Pentaamminecarbonatochromium (III) chloride. (ii) Potassium hexacyanochromate (III) (iii) Pentaamminenitritocobalt (III) chloride

1.

(d)

[

Br is weak field ligand] d 5(Td) is high spin complex. B.M. So, 2.

(a) If

If

then fot

then for

3. (a) For d 6 configuration, high spin complex. (i) In case of octahedral field, CFSE = [– 0.4p + 0.6q]∆0 + n(P) = [– 0.4 × 4 + 0.6 × 2]∆0 + 0 = – 0.4∆0 (ii) In case of tetrahedral field, CFSE = [– 0.6p + 0.4q]∆t = [– 0.6 × 3 + 0.4 × 3]∆t = – 0.6∆t. 4.

(a)

No. of unpaired e– : 4

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No. of unpaired e– : 4 No. of unpaired e– : 4 No. of unpaired e– : 3 No. of unpaired e– : 3 No. of unpaired e– : 5 So [Cr(H2O)6]2+ and [Fe(H2O)6]2+ have same magnetic moment (spin only). 5.

(b)

CFSE = – [0.4p + 0.6q]∆0 + n(P) = [– 4 × 0.4 + 2 × 0.6]∆0 + 0 = – 0.4 ∆0. 6. (c) No. of unpaired electrons 2+ 2+ [Fe(en)(bpy)(NH3)2] : Fe – 3d 6 0 [Pd(gly)2] : Pd2+ – 3d8 0 [Co(OX)2(OH)2]– : Co5+ – 3d4 2 [Ti(NH3)6]3+ : Ti3+ – 3d1 1 Thus, [Co(OX)2(OH)2]– exhibits highest paramagnetic behaviour due to highest number of unpaired electrons. Ti3+ = 3d1 4s0 7. (d) [Ti(H2O)6]3+ ∴ Electronic configuration is CFSE = [– 0.4nt2g + 0.6neg] ∆0 + n(p) = [(–0.4) × 1 + 0] 20300 = – 8120 cm–1 kJ/mol = – 97 kJ/mol

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8.

(c) [Ru(en)3]Cl2 [Fe(H2O)6]2+

So, correct answer is (c). 9. (b) Spin only magnetic moment ∴ number of unpaired electrons = 4 Two possible arrangements are

So, option (b) is correct. 10. (c) With weak field ligands Mn(II) will be of high spin and with strong field ligands it will be of low spin. Ni(II) tetrahedral complexes will be generally of high spin due to sp3 hybridisation. Mn(II) is of light pink colour in aqueous solution. 11. (c) Number of Geometrical isomers (n) in square planar [Pd(F)(Cl)(Br)(I)]2– = 3

[Fe(CN)6]3–6 = [Fe(CN)6]3– Fe3+ = 3d5

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=

= 1.73 B.M.

CFSE = –0.4 0 × nt2g + 0.6 = –0.4 0 × 5 = –2.0 0 12. (c) As, s = Na4[Fe(CN)6] Fe2+ S.F.L. t2g6 eg0 [Cr(H2O)6] Br2 Cr2+ W.F.L. t2g3 eg1 (Et4N)2[CoCl4] Co2+ W.F.L. e4 t23 Na3[Fe(C2O4)3] Fe3+ S.F.L. t2g5 eg0 13. (d) (A) Ni(CO)4 ;

s

0

=0 s

s

s

× n2g

=

=

B.M. B.M.

B.M. = No. of unpaired e– 8

2

Ni = 3d 4s (SFL)

(B) [Ni(H2O)6] ; Cl2 Ni2+ = 3d 8(WFL)

0

0

2

(C) Na2[Ni(CN)4] ; Ni2+ = 3d 8 (SFL) 0 0 (D) PdCl2(PPh3)2 ; Pd2+ = 4d 8 0 0 Correct order of the calculated spin only magnetic moments of complexes A to D is (A) (C) (D) < B 14. (b) The covalant character of the bonding (M – C and M – C bonding) which exists between the metal and the carbon atom of the CO can only be explained by the molecular orbital theory. 15. (d) EDTA is used in the treatment of lead poisoning. Deferrioxime B is used in treatment of iron poisoning and D-penicillamine is used in

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treatment of heavy metal poisoning, while cis-platin is used for treating cancer. 16. (a) Cr3+ has d3 configuration and forms an octahedral inner orbitals complex, therefore the set of degenerate orbitals are (dxy, dyz and dxz) and (

and

).

17. (b) Complex

Metal ion No. of unpaired electrons 4– [V(CN)6] V2+ 3 [Ru(NH3)6]3+ Ru3+ 1 4– 2+ [Fe(CN)6] Fe 0 2+ 2+ [Cr(NH3)6] Cr 2 Spin magnetic moment ∝ no. of unpaired electrons. So the order of spin magnetic moment is: V2+ > Cr2+ > Ru3+ > Fe2+ 18. (a) Compounds having atleast one carbon metal (M – C) bond are known as organometallic compounds. It contains Mn-C bond. 19. (c) Electronic configuration of Mn2+ is, Mn2+ : 3d 5 Presence of 5 unpaired e– shows that the complex of Mn2+ has only weak field ligand (NCS–). 20. (a) The structure of Co2(CO)8 is represented as

It contains two bridging CO ligands and one metal – metal (Co – Co) bond. 21. (c) Wilkinson’s catalyst is [Rh(PPh3)3Cl] 22. (d) In case of similar metal atom or ion, the value of co-ordination number and the strength of the ligands determine the value of crystal field splitting energy.

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Greater the co-ordination number and strength of value of the ligand, greater will be the value of CFSE. Strength of ligands : CN– > NH3 > H2O > Cl– K3[Co(CN)6] has the highest crystal field splitting energy. 23. (c) Wilkinson catalyst is [RhCl(PPh3)3]. Rh belongs to 4d series, so it terms square planar complexes and contains dsp2 hybridisation.

24. (a)

25. (d) Metal

ion electron

Unpaired Magnetic moment B.M. (i) Cr2+ 4 B.M. (ii) Fe2+ 4 B.M. (iii) Co2+ 3 B.M. (iv) Mn2+ 5 Since (i) and (ii), each has 4 unpaired electron, so they will exhibit same magnetic moment. Thus option (d) is correct. 26. (a) Ni2+ with NH3 shows CN = 6 forming [Ni(NH3)6]2+ (sp3d2 hybridisation, Octahedral) Pt2+ with NH3 shows CN = 4 forming [Pt(NH3)4]2+ (d2sp hybridisation, square planner) Zn2+ with NH3 shows CN = 4 forming [Zn(NH3)4]2+ (sp3 hybridisation, tetrahedral) 27. (c) 28. (a) L → M charge transfer spectra. KMnO4 is colored because it absorbs light in the visible range of electromagnetic radiation. The permanganate ion is the source of color, as a ligand to metal (L → M) charge transfer takes place between oxygen's p orbitals and the empty d-orbitals on the metal. This charge transfer takes place when a

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photon of light is absorbed which leads to the purple color of the compound. 29. (a) The donor atoms, molecules or anions which donate a pair of electrons to the metal atom or ion and form a coordinate bond with it are called ligands. In methane there is no electrons for donation to central metal atom/ion, it is stable with complete octet configuration.

30. (b)

For a given metal ion, weak field ligands create a complex with smaller ∆, which will absorbs light of longer λ and thus lower frequency. Conservely, stronger field ligands create a larger ∆, absorb light of shorter λ and thus higher v i.e. higher energy. So order of ligand strength is L1 < L3 < L2 < L4 31. (c) [Co(NH3)6]3+

(d 2sp 3)

Octahedral and diamagnetic. 32. (c) 33. (b) [NiL4]2– ⇒ Diamagnetic in nature (given)

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d sp 2

So, no. of unpaired electron = 0 hybridisation – dsp2. 34. (a) [PdCl4]2– is dsp2 hybridised and square planar in shape. 35. (c) In [Cr (NO) (NH3) (CN)4]2– , Cr+(d5) has electronic configuration as : So, 1 unpaired electron is present. Thus, BM µ= 36. (b) The electronic configuration of central metal ion in complex ions P, Q and R are P = [FeF6]3–; Fe3+ : Q = [V(H2O)6]2+; V2+ R = [Fe(H2O)6]2+; Fe2+ Higher the no. of unpaired electron(s), higher will be magnetic moment. Thus, the correct order of spin only magnetic moment is Q < R < P 37. (c) In both states (paramagnetic and diamagnetic) of the given complex, Ni exists as Ni2+ whose electronic configuration is [Ar] 3d84s0. Ni2+ : In the above paramagnetic state the geometry of the complex is sp3 giving tetrahedral geometry. The diamagnetic state is achieved by pairing of electrons in 3d orbital.

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Thus, the geometry of the complex will be dsp2 giving square planar geometry. 38. (c)

Thus L, M, O and P are diamagnetic. 39.

(b) [NiCl4]2–. = 3d8 configuration with nickel in + 2 oxidation state, Cl– being weak field ligand does not compel for pairing of electrons.

So,

Hence, complex has tetrahedral geometry

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[Ni(CN)4]2– = 3d8 configuration with nickel in + 2 oxidation state, CN– being strong field ligand compels for pairing of electrons.

[NiCN4]–2

Hence, complex has square planar geometry.

[Ni(H2O)6] = 3d8 configuration with nickel in + 2 oxidation state. As with 3d8 configuration two d-orbitals are not available for d2sp3 hybridisation. So, hybridisation ofNi (II) is sp3d2 and Ni (II) with six co-ordination will have octahedral geometry.

40.

Note : With water as ligand, Ni (II) forms octahedral complexes. (b) [NiCl4]2–, O.S. of Ni = +2

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Ni(28) = 3d 8 4s2 Ni 2+

[NiCl4]2–

No. of unpaired electrons = 2 Magnetic moment, µ = 2.82 BM. 41. (a) Chromium in Cr(CO)6 is in zero oxidation state and has [Ar]18 3d54s1 as the electronic configuration. However, CO is a strong ligand, hence pairing up of electrons takes place leading to following configuration in Cr(CO)6.

Since, the complex has no unpaired electron, its magnetic moment is zero. 42. (b) NOTE : In carbonyls, O.S. of metal is zero. In [Ni(CO)4], the oxidation state of nickel is zero. Its configuration in Ni(CO)4 is

In [Ni(CN)4]2– the oxidation state of Ni is 2+ and its configuration is

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3

43. 44.

Thus, the hybridisations of nickel in these compounds are sp and 2 dsp respectively. (d) An anionic carbonyl complex can delocalise more electron density to antibonding pi-orbital of CO and hence, lowers the bond order. (b) The required reaction is

From the given equations, we have

45.

∴ The value of K is given by K = k1 × k2 = 6.8 × 10–3 × 1.6 × 10–3 = 1.08 × 10–5. (c) TIPS/FORMULAE :

Here Co is present as Co2+ ion which has 3 unpaired electrons. So the spin i.e. BM. magnetic moment will be 46.

(d) The configuration of Ni2+ is 3d 8. For the elements of the first transition series, Cl– behaves as a weak field/high spin ligand. Hence, Ni in [NiCl4]2– is sp3 hybridised leading to tetrahedral shape.

47. (c) NOTE : In metal carbonyl the metal is in zero oxidation state. In Ni(CO)4, O.N. of Ni = 0

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For Ni (Z = 28)

In presence of CO two 4s electrons pair up, thus

In Ni(PPh3)2Cl2, O.N. of Ni = +2 For Ni2+

PPh3 and Cl– can’t pair up d-electrons, leading to sp3 hybridization and tetrahedral geometry. 48. (c)

49.

(b) Mn2+ in MnSO4.4H2O has d 5 configuration (five unpaired electrons); Cu2+ in CuSO4.5H2O has d 9 configuration (one unpaired electron); Fe2+ in FeSO4.6H2O has d 6 configuration (four unpaired electron); and Ni2+ in NiSO4.6H2O has d8 configuration (two unpaired electron). Thus, CuSO4.5H2O has lowest degree of paramagnetism. 50. (1)

•

H-atom =1s

1

Paramagnetic

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• •

odd electron species

Paramagnetic

–

O2 (superoxide) = One unpaired electron in π* M.O.

Paramagnetic

•

S2 (in vapour phase) = same as O2, two unpaired e–s are present in π* M.O. Paramagnetic

•

Paramagnetic

• •

•

2+

6

(NH4)2[FeCl4] = Fe = 3d 4s 8

(NH4)2 [NiCl4] = Ni = 3d 4s

Paramagnetic

2

Ni2+ = 3d 8 4s0

,

+

K2MnO4 = 2K

0

Paramagnetic

6+

Mn = [Ar] 3d

1

Paramagnetic

• 0

K2CrO4

=

+

2K

3d ic [Ar]18 3d 64s2 51. (4) Fe(26) [Ar]18 3d 54s0 Fe3+ SCN– is weak field ligand hence pairing will not occur.

,

Cr

6+

=

[Ar]

Diamagnet

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∴

Fe3+

Unpaired electrons = 5 Magnetic moment =

B.M.

B.M. = 5.92 B.M. = CN is strong field ligand hence pairing will take place. –

∴ Fe3+ Unpaired electrons = 1 Magnetic moment =

B.M. = 1.732

Difference = 5.92 – 1.732 = 4.188 Hence, answer is (4). 52. (0.0) since No. of unpaired electrons = 0 ∴ Magnetic moment = 0 B.M. 53. (26.92) [ML6]Cl3 + 3AgNO3 → 3AgCl 0.3 g V mL, 0.125 M Number of moles of the complex = Number of moles of AgNO3 = 0.125 × V × 10–3 × 3 = 0.125 × V × 10–3 Or, V =

= 26.92 mL

54. (20.0)

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120° = 3; 90° = 6 ; 90° = 8; 180° = 2 180° = 1 Total = 10 Total = 10 Total number of 180°, 90° and 120° L–M–L bond angles = 10 + 10 = 20 55. Paramagnetism; [Mn(H2O)6]2+ shows paramagnetism because of presence of 5 unpaired electrons in the outer most orbital (3d5) of Mn2+. 56. False : Octahedral complexes of Fe(III) like [Fe(CN)6]3– are low spin (d2sp3 hybridization) with one unpaired electron and have magnetic moment of about 1.9 BM. On the other hand, complexes of Fe(II) like [Fe(CN)6]2– are low spin complex (d2sp3) with no unpaired electron and thus diamagnetic. 57. (a, c) (a) [FeCl4]–

(b) [Co(en)(NH3)2Cl2]+ has three geometrical isomers.

(c)

[FeCl4]–

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Number of unpaired electrons (n) = 5 Spin only magnetic moment = = 5.92 B.M. [Co(en)(NH3)2Cl2]+

Number of unpaired electrons (n) = 0 Spin only magnetic moment = =0 (d)

58.

[Co(en)(NH3)2Cl2]

+

d2sp3 hybridisation, octahedral geometry (a, b, d) (a) [Co(en)(NH3)3(H2O)]3+ has 2 geometrical isomers.

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(b) Compound [Co(CN)2(NH3)3(H2O)]+ will have three geometrical isomers.

(c) [Co(en)(NH3)3(H2O)]3+ is diamagnetic (d) [Co(en)(NH3)4]3+ has larger gap between eg and t2g than [Co(en) (NH3)3(H2O)]3+. So, [Co(en)(NH3)3(H2O)]3+absorbs light at longer wavelength as compared to [Co(en)(NH3)4]3+. 59. (b, c) (a) [(Fe(CO5)] & [Ni(CO)4] complexes have 18-electrons in their valence shell. (b) Due to strong ligand field, carbonyl complexes are predominantly low spin complexes. (c) As electron density increases on metals (with lowering oxidation state on metals), the extent of synergic bonding increases. Hence, MC bond strength increases. (d) While positive charge on metals increases and the extent of synergic bond decreases and hence, C–O bond becomes stronger. Note : For transition elements of 3d-series, the valence shells are 3dand 4s. 60. (b, c, d) Magnetic moment, µ = where, n = No. of unpaired electrons Given for X and Z, = 3.87 B.M. = 3.87 i.e. n2 + 2n – 15 = 0 ∴n=3 For complex Y (1 : 3 electrolyte) given

=0

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=0⇒n=0

i.e.

CoCl2.6H2O or [Co(H2O)6)] Cl2 (X) is pink coloured compound. On adding excess of HCl at room temperature, [Co(H2O)6]Cl2 (X) changes into [CoCl4]2– (Z) and on adding excess of NH3 and NH4Cl in the presence of air forms [Co (NH3)6] Cl3 (Y). [Co(H2O)6 ] Cl2

[Co(NH3)6 ] Cl3

(X) Pink

(Y)

(µ= 3.87 B.M) (µ = 0) 61. (a, c) The metal to ligand bonding creates a synergic effect which strengthens the bond between CO and the metal (Fe). As the bond strength between Fe – C increases, the bond order between C – O decreases and the effective bonding can be represented as Fe = C = O. This will increase the C – O bond length. Thus, the answers can be 1.13 Å or 1.15 Å. of a species is related to its number of 62. (c) The magnetic moment unpaired electrons (n) in form of following expressions. B.M The number of unpaired electrons in the given pairs are as follows:

or

Thus, here n = 4.

or

Thus, here n = 5 or

n = 0 ; NO or

n =1

The given combinations differ in the number of unpaired electrons. Hence, these can be differentiated by the measurement on the solid state magnetic moment of nitroprusside ion. 63. (b) Highest parmagnetic character will be shown by the ion having maximum number of unpaired electrons in their d-subshells.

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Cr3+ has 3 unpaired electrons ; 2+ has 4 unpaired electrons 26Fe 2+ has 1 unpaired electrons ; 29Cu 2+ has no unpaired electrons 30Zn So (a), (b) & (c) show paramagnetism. Out of which (b) has the highest paramagnetism. 64. (a) Wilkinson catalyst : [Rh(PPh)3Cl] Chlorophyll : C55H72O5N4Mg Vitamin B12 contains Co. Carbonic anhydrase contains Zn ion. 65. (c) (p) [FeF6]4–, Fe2+ = 3d6 and F– is weak field ligand ∴ Hybridization is sp3d2 (high spin complex) (q) [Ti(H2O)3Cl3], Ti3+ = 3d 1 (No effect of ligand field strength) ∴ Hybridization is d2sp3 (r) [Cr(NH3)6]3+, Cr3+ = 3d3 (No effect of ligand field strength) ∴ Hybridization is d 2sp3 (s) [FeCl4]2–, 3d6 and Cl– is weak field ligand ∴ Hybridization is sp3 (t) [Ni (CO)4], Ni = 3d10 and CO is strong field ligand ∴ Hybridization is sp3 (w) [Ni(CN)4]2–, Ni2+ = 3d 8 and CN is strong field ligand ∴ Hybridization is dsp2 66. (b) 25

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67.

(A) : (p), (q) and (s) In [Co(NH3)4(H2O)2]2+, Co is in + 2 state having 3d7 configuration, which makes it paramagentic due to odd electrons. Moreover, it is an octahedral complex showing cis-trans isomerism w.r.t., H2O.

(B) : (p), (r) and (s) In [Pt (NH3)Cl2], Pt is in + 2 state with configuration 5d8. Since NH3 is a strong field ligand, it will pair all the electrons making the complex diamagnetic. Moreover, it is a square planar complex showing cistrans isomerism.

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(C) : (q) and (s) In [Co(H2O)5Cl]Cl, Co is in + 2 state with 3d 7 configuration making it paramagnetic. (D) : (q) and (s) In [Ni(H2O)6]Cl2, Ni is in + 2 state with 3d 8 configuration. It is attached with weak field ligands, therefore it is paramagnetic. 68.

(a)

69.

(c) For [Ni(CN)4]2–; Ni = [Ar]3d 84s2; Ni2+ = [Ar]3d 8

Ni2+ However, CN– is a strong field ligand so it forces the 3d electrons to pair up and hence the effective configuration in this case will be: 2+ Ni in presence of CN–

Thus, [Ni (CN)4]2– exhibits dsp 2 hybridization and square planar shape. Since here, number of unpaired electrons is zero, the complex will be diamagnetic. In case of [NiCl4]2–, Cl– is a weak field ligand, so the effective configuration of Ni2+ in this complex will be as follows : Ni2+ in presence of Cl–

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So here, Ni2+ is sp3 hybridised and thus, tetrahedral in shape. Since, the complex has two unpaired electrons, it will be paramagnetic. 70. (a) Discussed above. 71. (a) In [Fe(H2O)5NO] SO4, Let the oxidation state of Fe be x. Then for [Fe(H2O)5 NO]2+, x + 1 = + 2 or x = + 2 – 1 = + 1 Hence, in this complex the oxidation state of Fe is + 1 Electronic configuration of Fe+ can be represented as Fe+ = 1s22s22p63s23p63d 7. This unexpected configuration is due to presence of NO+. Due to which 1 electron from 4s1 gets shifted to 3d– orbitals. The 3d 7 electrons in five 3d– orbitals can be shown as In it, we find 3 unpaired electrons. Because of the presence of unpaired electrons the complex is paramagnetic i.e. statement 1 is true. As is clear from above there are three unpaired electrons in this complex i.e. statement 2 is true. Since, paramagnetic behaviour is due to presence of unpaired electrons in it so statement 2 is correct explanation of statement 1. 72.

IUPAC name of (A) is pentaaquathiocyanatoferrate (III) ion IUPAC name of (B) is hexafluoroferrate (III) In [FeF6]3– coordination no. of Fe = 6 In [FeF6]3– oxidation state of Fe = + 3 It has 5 unpaired electrons, n = 5, Fe3+ is 3d 5 (B.M.) Magnetic moment (µ) = = 5.92 B.M.

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73.

(a)

(b) Charge on Ni in the complex is +2 and it is dsp2 hybridised (c) Since number of unpaired electrons in Ni2+ is zero, the complex is diamagnetic. 74. The spin magnetic moment, m of the complex is 1.73 BM. It means that nucleus of the complex, chromium ion has one unpaired electron. So the ligand NO is unit positively charged. 24Cr = 3d54s2, after pairing up of electrons, there shall be one unpaired electron left. Thus, chromimum can be in Cr+ state. IUPAC name : Potassium amminetetracyanonitrosochromate (I). (a) Electronic configuration of Cr+ :

(b) Electronic configuration of Cr+ under the influence of strong field ligand CN–

So, Hybridization : d2sp3; Shape : Octahedral

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75.

Cl– is a weak ligand which is unable to pair the electrons of Ni2+. Therefore, here hybridisation is sp3 and shape will be tetrahedral. Electronic configuration of Ni+2 (No. of electrons = 26) in presence of Cl– ion, a weak ligand.

Magnetic moment of [NiCl4]2– = On the other hand, CN– is a stong ligand which pairs up the electrons of Ni2+. Therefore, here hybridisation is dsp2 and shape will be square planar. Electronic configuration of Ni2+ in presence of CN– ion, a strong ligand.

For structure of [Ni(CN)4]2–, refer question 22 in Section (E).

Magnetic moment of [Ni(CN)4]2– = 76.

Compound (A) on treatment with AgNO3 gives white precipitate of AgCl, which is readily soluble in dil.aq. NH3.Therefore it has at least one Cl– ion in the ionization sphere. Furthermore, chromium has coordination number equal to 6. So, its formula is [Cr(NH3)4BrCl]Cl. Compound (B) on treatment with AgNO3 gives pale yellow precipitate of AgBr soluble in conc. NH3. Therefore it has Br– in the ionization sphere. So, its formula is [Cr(NH3)4Cl2]Br. Cr3+ (Z = 24)

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State of hybridization of chromium in both (A) and (B) is d2sp3. Spin magnetic moment of (A) or (B), = = 3.87 BM µspin = 77.

[Co(NH3)6]3+

Co3+

Octahedral complex, d 2sp 3 hybridisation

[Ni (CN)4]2– Ni2+ Ni2+

(after

rearrangement)

Square planar dsp 2 hybridisation

[Ni(CO)4] Ni = Ni (after rearrangement)

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Tetrahedral (sp 3 hybridisation)

78.

Magnetic moment ( ) =

BM

number of unpaired electrons = 1.73 BM for vanadium ion So, (1.73)2 = n(n + 2) 1.73 BM =

where n

3.0 = n2 + 2n or n2 + 2n – 3 = 0 n2 + 3n – n – 3 = 0 n(n + 3) – 1 (n + 3) = 0 (n – 1) (n + 3) = 0 Correct value of n = 1 Thus, no. of unpaired electrons in vanadium ion = 1 2 2 6 2 6 3 2 23V = 1s , 2s 2p , 3s 3p 3d , 4s It will have one unpaired electron if it will lose two electrons from 4s and two from 3d. Vanadium (IV) has one unpaired electron. 4+ 2 V = 1s , 2s2 2p6, 3s2 3p6 3d1 79. For the explanation of colouration of complexes, first of all find out the number of unpaired electrons present in outer available d-orbitals (i) [Ti(NO3)4] ; 22Ti4+ : [Ar] 3d 0 4s0 ; 29Cu+ : [Ar] 3d 10 4s 0 (ii) [Cu(NC CH3)4]+ (iii) [Cr(NH3)6]+3 3Cl–; 24Cr3+ : [Ar] 3d 3 4s 0 (iv) K3 [VF6]; 23V+3 : [Ar] 3d 2 4s 0 Due to the presence of unpaired electrons in d-orbitals, two complexes i.e., [Cr(NH3)6]3+ 3Cl– and K3 [VF6] are coloured. Others having all paired electrons are colourless.

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1. (d) Given reaction is SN1 reaction. In SN1 reaction Rate of reaction ∝ Stability of C+ (A)

(B)

(C)

(D) Stability of C+ : ii > i > iii > iv Reactivity order : B > A > C > D 2. (d) Ease of precipitation of AgBr depends upon the rate of formation of carbocation.

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Most stable carbocation due to +R effect of N. 3. (d) Due to conjugation of lonepair of Cl with π bond, partial double bond character decreases bond length that's why compound (d) has shortest C–Cl bond length. 4. (c) Above reaction is SN1 reaction as it proceeds via formation of carbocation. Polar protic solvent is more suitable for SN1 and so racemisation takes place. 5.

(c)

6.

(d)

7.

(b)

8.

(b) First reaction is SN1 in which rate does not depend on conc. of nucleophile but depends on reactant conc. Second reaction is E2 reaction in which rate depends on conc. of base as well as reactant conc.

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Therefore, changing in the concentration of base will have no effect on rate of reaction (1). 9. (a) E1 reaction proceeds via carbocation formation, therefore greater the stability of carbocation, faster will be the E1 reaction. Thus correct decreasing order of the given halides towards dehydrohalogenation by E1 is D>B>C>A 10. (c) (A) (CH3)2CCH(OH)CH3

(B) (CH3)2CHCH(Br)CH3

(C) (CH3)2CHCH(Br)CH3

Due to bulky nature of tertiary butoxide, the least hindered hydrogen is eliminated. Therefore, Hoffman product is formed. (D)

11. (d)

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12.

(b) Allylic H is easily replaced due to the greater stability of allylic free radical. CH3—CH2—CH==CH2 + Br2

13.

(b) CH3O– is a strong base and strong nucleophile, so favourable condition is SN2/E2. The given alkyl halide is 2° and β carbons are 4° and 2°, so sufficiently hindered, thus E2 dominates over SN2.

14.

(a) Here dehydrohalogenation goes by E1cB and most stable carbanion formation is favoured in (a). 15. (a) Alkyl chloride or bromide undergo substitution and get converted to an alkyl iodide on treatment with a solution of sodium iodide in acetone. e.g. CH3CH2CH2I + NaBr This reaction is also known as Finkelstein Reaction. 16. (b) Alkyl fluorides are more conveniently prepared by heating suitable chloro – or bromo-alkanes with organic fluorides such as AsF3, SbF3, CoF2, AgF, Hg2F2 etc. This reaction is called Swarts reaction. 17.

(b) Steric hindrance around the carbon atom having Cl will slow down the SN2 reaction, hence lesser the hindrance, faster will be the reaction. So, the order of reactivity is CH3Cl > (CH3)CH2 – Cl > (CH3)2CH – Cl > (CH3)3CCl 18. (a) The correct order of increasing bond length is CH3F < CH3Cl < CH3Br < CH3I

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19.

(a) For a given alkyl group, the order of reactivity is

decreasing halogen reactivity. This order depends on the carbon-halogen bond energy; the carbon-fluorine bond energy is maximum and thus fluorides are least reactive while carbon iodine bond energy is minimum hence iodides are most reactive. 20. (b)

Note: The electron with drawing groups (–I effect) stabilise the transition state formed in the rate determining step of SN2 reaction. Hence, the benzylic, allylic and

21.

22.

system increases the reaction rate to more

than 1° carbon system. But reactivity is higher for —CH3 than that for allylic (CH2=CH–) system. CH2 = CHBr (b) BrCH2 – CH2Br NOTE : Elimination of HBr from CH2= CHBr requires a stronger base because here, C – Br acquires partial double bond character due to resonance. (d) TIPS/Formulae : It is an example of intramolecular Wurtz reaction.

NOTE : Br– is a better leaving group than chloride. In this reaction alkali metal (Na) is electron donor.

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23. 24.

(b) Dehydrobromination by strong base (alc. NaOH) followed by Markownikoff addition of HBr. (d) Occurrence of racemization points out the formation of carbocation as intermediate, which being planar can be attacked from either side.

25.

(a)

26. (b) (CH3)2CHCH2MgBr (CH3)2 CHCH3 27. (b) TIPS/Formulae : The reaction proceeds via free radical mechanism. As 2º free radical is more stable than 1º, so CH3CH2CH(Br)CH3 would be formed. 28. (a) (Elimination reaction)

29.

NOTE : Alkyl halides give alcohols with aq. KOH, which is a substitution reaction. (b) Addition of HBr to 2-pentyne gives two structural isomers (I) and (II)

Each one of these will exist as a pair of geometrical isomers. Thus, there are two structural (I and II) and four configurational isomers cis-and trans-of I and II. 30.

(b)

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31.

(a) Chlorination beyond monochlorination during the preparation of alkyl halides in presence of UV light can be suppressed by taking alkane in excess. 32. (c) A mono-substituted benzoic acid is stronger than a monosubstituted phenol as former being a carboxylic acid. Among the given substituted benzoic acid, ortho-hydroxy acid is strongest acid although —OH causes electron donation by resonance effect which tends to decrease acid strength. It is due to a very high stabilisation of congugate base by intramolecular Hbond which outweigh the electron donating resonance effect of —OH.

The overall order of acid-strength of given four acids is orthohydroxybenzoic acid (pKa = 2.98). > Toluic acid (pKa = 4.37) > phydroxybenzoic acid (pKa = 4.58) >p-nitrophenol (pKa = 7.15) 33. (a) Anti addition of Br2 on trans alkene provides meso compound.

+ Br2

34.

(a) NOTE : This is an example of SN1 reaction involving carbocation as intermediate.

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This carbocation is especially stabilised through resonance in which group acts as a good electron donor. (A) (i) (B)

35.

36.

(ii) (d) 2-Methylbutanoic acid contains one asymmetric centre

(d) SN2 reaction at asymmetric carbon occur with inversion of configuration and a single steroisomer is formed bacuse the reactant and product are not enantiomer. Therefore the sign of optical rotation may or may not change.

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37.

(d) Rate of reaction will be R – I > R – Br > R – Cl > R – F. because I– is the best, while F– is the poorest leaving group among halide ions. 38. (c) –CH3 is the best nucleophile because carbon is least electronegative among the given options. The order is

39.

(b) The stereoisomers of butane -2,3-diol are

40.

(a) TIPS/Formulae : The bond angle in sp3, sp2 and sp hybridisad carbon atoms is respectively 109.28', 120º and 180º.

41.

(7) The given compound X is :

Trick : If the bottom double bond C = C is in trans position then there can be distinct possibilities with the C – C and C – OH. Similarly, there will be additional 4 distinct possibilities when the is in cis configuration. Now, out of these 8 double bond C = C distinct possibilities, only one configuration has a plane of symmetry. This makes it an optically inactive compound.

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We can also solve this problem by considering the two chiral centers at C and C . Number of stereoisomers : : = 22 = 4 Now, each of these 4 stereoisomers can be in cis or trans configuration about C = C bond. This makes the total number of stereoisomers =4×2=8 Total number of optically active stereoisomers = 8 – 1 = 7

42.

(5)

43. (5) Total no. of alkenes will be = 5

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44.

(8)

45.

(CH3)2Cu + CH2 = CHCl → CH3CH = CH2 Propene

46. 47.

Thioalcohol non-superimposable, enantiomers;

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48. 49.

Grignard (RMgX) Chlorine; because rate of formation of •CH3 (one of the propagating steps) is high when X• is Cl. • CH3 + H – X CH4 + X• 50. False : There are only two asymmetric (marked with *) carbon atoms.

Note that C-3 has two similar groups, i.e., CH3CHCl– 51. True : SN1 Reactions (unimolecular nucleophilic substitution reaction) take place in two steps. (i) Formation of carbocation which is slow and ratendeter mining step (ii) Attack of nucleophilic, a fast step. 52. False : CH2 = CHCl + HI → CH3CHCl(I) Vinyl chloride

1- Chloro-1-iodoethane

This is an example of Markownikoff’s rule as I– is added at the C with less number. of H-atoms. NOTE : anti-Markovnikov's rule is applicable only to HBr, but not to HI and HCl. 53. False : Iodide is bigger in size than bromide, hence its electrons are more dispersed than that of bromide, with the result, it is weaker nucleophile than bromide. 54. (a, b, c) (a) I and III both forms stable carbocations (I) (Resonance stabilised very stable carbocation intermediate)

(III)

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(Very stable tertiary carbocation intermediate)

Thus, they can follow SN1 mechanism. (b) I and II both are 1° alkyl halide and forms a stable transition state. (I)

Due to an electron withdrawing group, it forms a very stable transition state. (II) 1° alkyl halide always prefers SN2 mechanism. Thus, they can follow SN2 mechanism. (c) Compound (IV) can form a stable carbocation and also a stable transition state as :

(d) Order of reactivity for SN1 reaction depends on the stability of intermediate carbocation formed in the rate determining Slow step : Stability order of carbocations :

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>

(IV)

(III)

>

(I)

Order of reactivity for SN2 mechanism depends on the stability of T.S. and steric factor. Stability order of transition states :

>

>

Hence, option (d) is wrong. 55. (a, d)

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(a)

In both C2 and C3 carbons, the lowest priority group/atom (i.e. H) is in wedge position. Thus, we can directly assign the R–S configuration to the given compound. The R–S configuration of the compound is (2R, 3R). Hence, the compound is opticaly active. Option (b) and (c) can not be correct as the molecule is optically active. (d) If we make a conformer of the given compound by taking the C2 – C3 axis and rotating the groups on C3 carbon by 180°, then the compound can be represented as :

A Cn axis of symmetry is an axis about which the molecule can be rotated by 360°/n to produce a molecule indistinguishable from the original molecule. If we rotate the molecule by 180° along the axis which is perpendicular to C2 – C3 bond, then we will get the exactly same molecule. Hence, this molecule has C2 axis of symmetry. Note : Any optically active compound may or may not possess axis of symmetry but it cannot possess alternate axis of symmetry. 56. (c, d) An asymmetric carbon atom is one which is attached with 4 different groups. Hence, (c) & (d) are correct.

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57. (c)

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P-2, Q-1, R-1, S-3 58. (d) Due to partial double bond character of C–halogen bond, halogen leaves with great difficulty, if at all it does. Hence, vinyl halides do not undergo nucleophilic substitution easily. So, assertion is correct. Intermediate

59.

60.

carbocation is not stabilised by loosely held-π electrons because empty orbital, being at 90°, cannot overlap with p-orbitals of π bond. So, reason is wrong. (c) Statement-1 is correct. Statement-2 is incorrect because compound can be chiral even in the absence of chiral atoms. + Mg

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61.

In order to convert a molecule with two stereogenic centres to its enantiomer, the confuguration at both centres must be reversed. Reversing the configuration at only one stereogenic centre converts it to a distereomeric structure. Thus, structures I and III are enantiomers; while structures I and II as well as II and III are diastereomers.

62.

(i)

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(ii)

(iii)

(iv)

(v) (vi)

(vii) [NOTE : C6H5CH2 HCH3 and C6H5 HCH2CH3 carbocations are formed on addition of HBr on C6H5CH=CHCH3, the latter being benzylic carbocation, is stabilised due to resonance and hence Br– adds on it forming C6H5CHBr.CH2CH3 as the final product.] 63. Summary of the given facts

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The two isomeric precursors (Y and Z) of 2, 3-dimethylbutane are

Hence, the precursor of Y and Z should have following structure which explains all the given facts

64.

SN2 reaction leads to inversion in configuration.

65.

NOTE : A weaker base is a better leaving group. Thus, rate of reaction will be R – I > R – Br >R – Cl > R – F. because I– is the best, while F– is the poorest leaving groups among halide ions.

66.

% of Cl in X =

× 100 = 71.72%

Empirical formula of (X) Element % Relative no. of atoms Simplest ratio

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C 24.24 2.02 1 H 4.04 4.04 2 Cl 71.72 2.02 1 ∴ Empirical formula of (X) is CH2Cl Since, X has two isomers Y and Z; both react with KOH(aq). dihydroxy compound i.e. 2Cl atoms on adjacent Y carbon Z

CH3CHO i.e. Z should have 2Cl atoms on one C atom

Thus Z should be CH3CHCl2 (1, 1-dichlorethane) and Y should be CH2ClCH2Cl (1, 2-dichloroethane) Reactions : CH3CH(OH)2 CH3CHO CH3CHCl2 (Z)

Ethanal

CH2ClCH2Cl (Y)

CH2OHCH2OH

ethane-1, 2-diol

67.

TIPS/Formulae : Resonance decreases the dipole moment of vinyl chloride(CH2 = CHCl). The positive charge on Cl and a negative charge on C (developed by resonance) oppose each other and hence diminish the electronegativity of Cl and thus polarity (and dipole moment) of the bond. The dipole moments of vinyl chloride and chlorobenzene are 1.4D and 1.7D respectively, while the dipole moment of alkyl halides is 2–2.2D. 68. Dichloroethene exists in three isomeric forms.

trans-1, 2-Dichloroethene has zero dipole moment.

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69.

(i) 4C2H5Br + 4(Na–Pb) Pb(C2H5)4 + 4NaBr + 3Pb CH4

(ii) Al4C3 70. C2H2

(a) (i)

CH3Cl

CaC2 + 2H2O

Ca(OH)2 + C2H2

C2H6

CCl3 – CCl3

CCl4 + S2Cl2 (ii) CS2 + 3Cl2 CCl4 + 6S CS2 + 2S2Cl2 CHCl3 + HCl CCl4 + 2 [H] (b) Carbylamine test. CHCl3 + aq KOH + aniline (i.e. primary amine) Bad smelling isocyanide No reaction C2H5OH + aq KOH + aniline 1.

(b) SN2 reactions depend upon –I and –M effect on substrate. On increasing –I and –M effect, rate of SN2 reaction will increase. 2. (d) Sodium borohydride is a selective reducing agent. It reduces carbonyl group to alcoholic group, N-methylimino group (MeN = CH –) to 2° amines, but does not reduce an isolated carbon-carbon double bond. 3. (b) Elimination reaction is highly favoured if (a) Bulkier base is used (b) Higher temperature is used Hence in given reaction biomolecular elimination reaction provides major product.

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4.

(c) C6H5CHCH3Br being an optically active secondary alkyl bromide undergoes SN2 nucleophilic substitution reaction. Hence it undergoes complete inversion of configuration.

5.

(d)

6.

(b) Reaction between alkyl halides, aryl halides and sodium in presence of dry ether to give substituted aromatic compounds is known as Wurtz fitting reaction C6H5Cl + 2Na + ClCH3 →

7.

(a)

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8. (a) The product (a) will be formed. Nucleophilic substitution of an alkyl halide is easier as compared to that of an aryl halide. PhS– is a strong nucleophile and dimethyl formamide HCONMe2 is a highly polar aprotic solvent. These reagents favour SN2 reactions at 2° benzylic carbon. NOTE : In a SN2 reaction, the major product formed is inversion product.

9.

(a) Grignard reagents react with compounds containing active hydrogen to form hydrocarbons corresponding to alkyl (or aryl) part of the Grignard reagent. C6H5MgBr + Me3COH 6H6 + Me3COMgBr 10.

(d)

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So F can have three possible structures. 11. (d) TIPS/Formulae : The given reaction is an example of electrophilic substitution. Further, CH3 group in toluene iso, p-directing C6H5N2Cl

C6H5Cl

12.

(d) C6H5NH2

13.

False : m-Chlorobromobenzene and m-bromochloro-benzene is one and the same compound. (a, b)

14.

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15.

(c) Chlorination of toluene to form benzyl chloride is a free radical in presence of light. substitution reaction; only Cl2 can give 16. (b, e) TIPS/Formulae : Aryl halides are stable due to resonance stabilization. The resonating structures

stabilise the aryl halide. These structures include a double bond between C and Cl which is shorter and thus stronger than the usual C – Cl single bond. The sp2 hybridised carbon, being electronegative, makes the C – Cl bond shorter and stronger. 17. (A) - q; (B) - p, s; (C) - r, s; (D) - q E1 mechanism is encountered only with tertiary or secondary substrates and in presence of either a weak base or a base in low concentration. So, primary substrates will follow E2 mechanism, i.e. (A) → E2 and (D) → E2. Further, E1 mechanism (similar to SN1) proceeds by first order kinetics and is determined by the slower (first) step of the formation of carbocation. Hence, (B)→E1 and first order reaction. NOTE: Reaction of C6H5CH2CH2Br on treatment with C2H5O– in presence of C2H5OD gives C6H5CD=CH2. This reaction follows E1CB (Elimination unimolecular conjugate base) mechanism. 18. (c) When halogen is present directly on the benzene nucleus it produces two opposing effects namely + M (activating effect) and –I (deactivating)

The inductive effect (–I) of bromo group is more than the mesomeric effect (+M).

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This is the reason why bromo group deactivates the ring. But the directing influence with always governed by the mesomeric effect. Hence, due to +M effect of –Br, the incoming electrophile attacks at ortho and para positions. 19. (i) NOTE : The former halide is a 3° halide, hence it undergoes SN1 reaction forming HBr, as one of the products, which make solution acidic.

A 3° bromide

is an aryl halide so it does not undergo nucleophilic substitution reactions. Hence, the solution will remain neutral. (ii) TIPS/Formulae : 7-Bromo-1,3,5-cycloheptatriene is aromatic whereas 5-Bromo-1,3cycloheptadiene is non aromatic.

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20.

This is ullmann reaction 21.

22.

The low reactivity of halogen atom in aryl and vinyl halides towards nucleophiles is due to resonance.

Resonating structures of chlorobenzene

NOTE : Due to resonance, carbon-chlorine bond acquires partial double bond character, hence it becomes shorter and stronger and thus cannot be easily replaced by nucleophiles.

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23.

C6H5CH3

C6H5CH2Cl C6H5CHCl2

[NOTE : This follows free radical mechanism.]

1.

(c)

2.

(d)

+ 6AgCl

→

3.

(c) Chloral on reaction with chlorobenzene in the presence of a catalytic amount of sulphuric acid forms DDT (dichloro diphenyl trichloro ethane).

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1.

(a)

2.

(b)

3.

(a)

4.

(d) An excellent reagent for oxidation of 1° alcohols to aldehydes is PCC.

5.

(c) The C – O bond length in alcohols is 142 pm and in Phenol it is 136 pm. The C – O bond length in phenol is shorter than that in methanol due to the conjugation of unshared pair of electrons on oxygen with the ring, which imparts double bond character to the C – O bond.

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6.

(b) Reaction of alcohols with Lucas reagent proceeds through carbocation formation. Further 3° carbocations (from tertiary alcohols) are highly stable thus reaction proceeds through SN1 mechanism. 7. (b) The order of dehydration among three type of alcohols is 3° > 2° > 1° > CH3OH. This behaviour is related to the relative stabilities of carbocations (3° > 2° > 1°). 8. (b) TIPS/Formulae : Conc. HCl, HBr and conc. HCl + ZnCl2 all are nucleophiles, thus convert alcohols to alkyl halides. However, conc. H3PO4 is a good dehydrating agent which converts an alcohol to an alkene. 9. (c) NOTE : Addition of water to 2-phenylpropene follows Markovnikov’s rule.

2-Phenyl-2-propanol

10.

(c)

11.

(a) TIPS/Formulae : Compound (CH3)4N+I– is most reactive due to (i) better leaving group, I– and (ii) due to the fact that the methyl group, with positive N, is more electron deficient. Hence this group is more reactive towards nucleophile, OH–

+ (CH3)3N + I–

12.

(c) TIPS/Formulae : Secondary alcohols oxidise to produce kenone.

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CH3CHOHCH2CH3 2-Butanol

CH3COCH2CH3 Ethyl methyl ketone

13.

(a) Reactions involving cleavage of carbon-oxygen bond, (C – OH) follows the following order : Tertiary > Secondary > Primary 14.

(a)

15.

(d) Lucas test is based on the difference in the three types of alcohols (having 6 or less carbon) towads Lucas reagent (a mixture of conc. hydrochloric acid and anhydrous zinc chloride) at room temperature. ROH + HCl RCl + H2O

The tertiary alcohols produce turbidity immediately, the secondary alcohols give turbidity within 5 – 10 minutes, and the primary alcohols do not give turbidity at all, at room temperature. Thus, the order of reactivity of alcohol with Lucas reagent is tert. > sec. > pri. Hence, 2-methylpropan-2-ol (a 3° alcohol) reacts fastest. 16. (a) Among the given options, ethyl alcohol is most basic. 17.

(c)

18.

(6)

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19.

(4)

20.

(4)

Number of chiral centres, represented as (*) in the product (B) = 4

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21. 22.

Secondary. aldol (β-hydroxybutanal);

23.

False : Ethanol is not acidic enough to react with aq. NaOH. Thus, sod. ethoxide (C2H5ONa) is prepared by the reaction of Na metal with ethyl alcohol. 2 C2H5OH + 2Na → 2 C2H5ONa + H2 ↑ 24. (a, c, d) Isomeric alcohols with molecular formula C4H10O are

25.

(b) The mechanism of this reaction is represented as follows.

Benzylic carbonium ion (stable)

26.

(c) The solubility of alcohols in water can be explained due to the formation of hydrogen bond between the highly polarised –OH groups present both in alcohol and water. However, in higher alcohols the hydrocarbon character (alkyl chain) of the molecule increases and thus, alcohols tend to resemble hydrocarbon (which are insoluble in water) and hence, the solubility in water decreases. When the ratio of C to OH is more than 4, alcohols have little solubility in water. So, statement is correct but explanation is not.

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27.

28.

TIPS/Formulae : Let us summarise the given facts.

(i)

Hydrogenation of bombykol (C16H30O) to C16H34O (A) indicates the presence of two double bonds in bombykol. (ii) Reaction of A with acetic anhydride to form ester indicates the presence of an alcoholic group in A and hence, also in bombykol. (iii) Products of oxidative ozonolysis of bombykol ester suggests the structure of bombykol.

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The structure of the bombykol ester suggests that bombykol has the following structure : (Bombykol)

and the structure of A is CH3CH2CH2CH2CH2CH2CH2(CH2)8.CH2OH or C16H33OH. Four geometrical isomers are possible for the above bombykol structure (as it has two double bonds).

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29.

30. (i) NOTE : Since the large propenyl group is attached to the carbon atom bearing the hydroxyl group, so the reaction is likely to occur via SN1 mechanism.

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(ii)

NOTE : In the intermediate carbocation, I, carbon bearing positive charge has CH3 group which decreases the positive charge and hence prevents cyclisation of the compound. 31. Since, 3º carbocation (formed in case of t-butanol) is more stable than 1º (formed in n-butanol), dehydration in the former proceeds faster than in the latter. 32. TIPS/Formulae : (a) Since (B, C6H14O) is resistant to oxidation, it must be ter-alcohol. (b) Since (B) is optically inactive, it must have at least two similar alkyl groups .

Thus, the five carbon atoms can be adjusted into three alkyl groups (of which two are similar) either as –CH3, –CH3, and –C3H7, or as –C2H5, –C2H5 and – CH3, Thus the possible structure of alcohol (B) is either or

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Hence the corresponding compound (A) is either or (Option in active)

(Optically active)

However, the compound (A) is optically active, so (A) and hence also (B) should have right side structure.

(A)

(B)

33.

34. n-Butanol gives the following reaction in which the purple colour of KMnO4 changes to brown. tert-Alcohols are not oxidisable easily, hence purple colour of KMnO4 remains same.

The brown precipitate is of MnO2.

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35.

(i)

Since the compound X (C5H8O) does not react appreciably with Lucas reagent, it indicates that the compound has a primary alcoholic group (– CH2OH). (ii) Reaction of the compound X with ammonical silver nitrate to give a precipitate indicates that it has an acetylenic hydrogen atom, i.e., ≡C – H grouping is present. (iii) Treatment of X with H2/Pt followed by boiling with excess of HI gives npentane. It indicates that the compound does not have any branch. On the basis of the above points, compound X (C5H8O) may be assigned following structure. HC ≡ C – CH2 – CH2 – CH2OH (X) 4-Pentyn-1-ol (Mol. wt. 84, Eq. wt. = 42)

The above structure for the compound X is in accordance with its equivalent weight obtained from the given data. 224 ml. of CH4 at STP is obtained from 0.42 g 22400 ml. of CH4 at STP =

× 22400 = 42 g

∴ Eq. wt. of the compound X = 42 Reactions of the compound X : (i) AgC ≡ C.CH2CH2CH2OH ↓ (ii) MgBrC ≡ C.CH2CH2CH2OMgBr + 2CH4 (iii)

36. 37.

The compound A, a ketone, undergoes haloform reaction. Thus, it must contain CH3CO group.

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The compound C gives mono-ozonide D, which shows that the compound C contains a double bond. Since the hydrolysis of D gives only acetaldehyde, the compound C would be an alkene having four carbon atoms, i.e. CH3 – CH = CH – CH3 (butene-2). The compound B is obtained by the reduction of compound A (which contains CH3CO group). Hence, the compound B would be an alcohol, which on heating with H2SO4 gives (C). Hence B and A would be

The reactions involved : (A)

(B) 2CH3CHO

38.

TIPS/Formulae : For this type of problem, students are advised to summarise the whole problem in the form of reactions.

Let us draw some conclusions from the above set of reactions. (i) The molecular formula C5H10 (CnH2n) for A indicates that it is an alkene having one double bond. (ii) Since the alcohol C on oxidation gives a ketone D, C must be a secondary alcohol and hence B must be a secondary bromide. (iii) The structure of 2-methylbutane, the hydrogenated product of A, indicates that the secondary bromide must have following structure.

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(iv) Thus the corresponding olefin A must have structure A which on Markownikoff addition of HBr gives the bromide B, the other possible alkene A' will not give B when HBr is addd on it according to Markownikoff rule. ; Thus the reaction involved can be represented as below:

39.

(i)

CH2 = CH2 BrCH2.CH2Br

or

CH ≡ CH

CH3CHO

—→ CH3COOH (ii) CH3CH(OH)CH3

CH3.CH=CH2

2-Propanol

CH3.CH2.CH2Br

CH3.CH2.CH2OH 1-Propanol

40.

TIPS/Formulae : Iodoform test is used to distinguish methanol and ethanol. Ethanol gives iodoform test while methanol does not respond. C2H5OH + 4I2 + 6NaOH → CHI3 ↓ + 5NaI + HCOONa + 5H2O 41. The given problem can be sketched as below.

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NOTE THIS STEP : From the problem it appears that the compound C is an alkyne, hence D must be an aldehyde or ketone. Further, since D can be obtained from acetic acid through its calcium salt it may be either acetaldehyde or acetone. Hence going back, A must be ethyl alcohol, which explains the given set of reactions.

Hence, A is ethyl alcohol, CH3CH2OH B is ethylene, CH2 = CH2 C is acetylene, CH ≡ CH D is acetaldehyde, CH3 . CHO 42. ‘A’ is C2H5OH and ‘B’ is C2H4 C2H5OH

C2H4

(A) (Ethyl alcohol)

(B) (Ethene)

C2H4 + alk. KMnO4

; C2H4 + H2

C2H6

Eth Colourless

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1.

(b) (II) and (III) compounds almost have same boiling point. In the given options, option (b) will be the correct answer.

(I)

B.P.

202 °C

(II)

B.P.

279 °C

(III)

B.P.

284 °C

(IV)

B.P.

243 °C

2.

(a)

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Generally CAN test is done for alcohols which give pink or red colour. But for phenols and phenolic compounds it gives brown or black colour. So, this test helps to diffirentiate phenols from alcohols.

3.

(c)

4.

deciding factor to determine the order of bond length in given compounds. Phenol exhibits least C–OH bond length due to resonance, whereas methanol will show maximum bond length due to lack of resonance and pethoxyphenol will have some intermediate value of bond length. (a)

5.

(b)

6.

(d) Electron withdrawing substituents increase the acidic strength, while electron releasing groups decrease the acidic strength.

Resonance

Acidic strength

is

a

Ka 

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pKa : B < C < A < D

7.

(a)

8.

(a)

(a)

9.

(a)

+ CO2

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10.

(d)

Note : Gattermann Koch reaction is a variation of Gattermann reaction, and this reaction involves the use of carbon monoxide instead of HCN. 11. (d) In dye test, phenolic — OH group of p-naphthol is converted to — O– which activates the ring towards electrophilic aromatic substitution. 12. (c) Electron withdrawing substituents like –NO2, –Cl increase the acidity of phenol while electron releasing substituents like – CH3, – OCH3 decrease acidity. hence the correct order of acidity will be

III (– M, –I)

13.

I II (–I > +M) (+I, +HC) (+M)

IV

(b)

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14.

(d)

15. (c) TIPS/Formulae : (i) The acidity increases with the increase in electronegativity of the halogen present. (ii) The inductive effect decreases with increase in distance of halogen atom from the carboxylic group and hence, the strength of acid proportionally decreases. Smallest dissociation constant means weakest acid, which is BrCH2 CH2COOH because here Br (less electronegative than F) is two carbon atoms away from – COOH 16. (d) NOTE : –NO2 is an electron-attracting group where as –CH3 is an electron-releasing group. An electron - attracting substituent tends to disperse the negative charge of the phenoxide ion and thus, makes it more stable. This, in turn, increases the acid strength of phenol. The substituent in para position is more effective than in the meta position as the former involves a resonating structure bearing negative charge on the carbon attached to the electron withdrawing substituent. An electron - releasing substituent tends to intensify the negative charge of the phenoxide ion and thus makes it more unstable. This, in turn, decreases the acid strength of phenol. Hence, the order of acid strength is

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(IV) (I)

17.

(III) (II)

(d) TIPS/Formulae : Riemer-Tiemann reaction is an electrophilic substitution on the highly reactive phenoxide ring. HCCl3 + OH– → H2O + – :CCl3 : Cl3 → Cl– +

18.

(d) C6H5CH3 + Cl2(exc.)

C6H5CCl3 C6H5C(OH)3

19.

C6H5COOH

(b)

NOTE : Bromine inpresence of CS2 (non-polar solvent) at low temperature ionises easily. Further in absence of CS2, polyhalogenation in o- and ppositions takes place.

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20.

(d)

NOTE : Bromine in water (capalar) solvent ionises earily to give Br+ ions. Further the – OH group in phenol, being activating group, facilitates substitution in the o- and p-positions. 21. (6) C8H10O2 gives FeCl3 test means it has phenolic group. It rotates plane polarized light means it is optically active.

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22.

(9)

23.

(69)

Molecular formula of product 'P' = C7H6O2 So, mass % of C in 'P'

24. 25. 26. 27.

phenoxide ion nucleophilic resonance stabilization (a, b, c) —OH group is strongly activating and o, p-directing due to +M effect. Thus positions a, b and c are the sites for attack by an electrophile. However, sites b and c are not preferred bsy bulky electrophile due to steric crowding. Thus, more bulky electrophile (like I2) can attack only site a, which is least sterically hindered, a bit smaller electrophile (Br2) can attack at sites a and also b (relatively less sterically hindered site) and the smallest electrophile (Cl2) can attack all the three sites, viz., a, b and c (most sterically hindered site).

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28.

(b) Phenolic C—OH group is activating and o-p- directing group.

29.

(a, c)

Product of reaction of phenol with NaOH/Br2 is sodium salt of 2,4,6tribromophenol. Hence, species (I), (II), (III) are formed as intermediate. 30. (b) The mechanism of this reaction is represented as follows.

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Benzylic carbonium ion (stable)

31.

(a, d) TIPS/Formulae : The reaction involves electrophilic substitution on the highly reactive phenoxide ion. Here, the electrophile is dichlorocarbene which is formed by the action of strong alkali on chloroform.

32.

(a, c) Higher the stability of the corresponding anion, more will be the acidic character of the parent compound.

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Higher stability of acetate ions than phenoxide ion is due to equivalent resonating structures in the former 33. A. (i) (a)(ii) (c) (iii) (b) (iv) (e) (v) (d) B. (vi) (h) (vii) (j) (viii) (i) (ix) (g) (x) (f) C. (xi) (n) (xii) (l) (xiii) (o) (xiv) (m) (xv) (k) D.(xvi) (t) (xvii) (r) (xviii) (s) (xix) (p) (xx) (q) 34. (i) (f) (ii) (d) (iii) (b) (iv) (c) (v) (c) (vi) (a) 35. (c) Reagents for Reimer - Tiemann reaction are aq. NaOH + CHCl3. 36.

(c)

37.

(b)

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38.

(a) Due to +M effect of stable than the one in benzene.

39.

(i)

(ii)

40.

, its intermediate carbocation is more

C6H6

NOTE : The reaction of D (C8H10O ) with alkaline solution of iodine is an iodoform reaction. This reaction is possible if the compound D has

or

group. The high carbon content in D indicates that D is an aromatic compound containing a benzene ring. To account for the given formula, the compound D may be C6H5CH(OH)CH3. The given reactions are

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41.

(i)

(ii)

+ CHCl3 + NaOH

42.

Phenol (a weaker acid) reacts with NaHCO3 (a weaker base) to form phenoxide ion (a stronger base) and carbonic acid (a stronger acid). C6H5OH + NaHCO3 C6H5ONa + H2CO3

Weaker acid

Weaker base

Stronger base Stronger acid

Since, acid-base equilibria lies towards the weaker acid and weaker base, phenol does not decompose NaHCO3 (difference from carboxylic acids). RCOOH + NaHCO3 RCOONa + H2CO3 Stronger acid Stronger base

43.

Weaker base Weaker acid

(i)

The compound (C7H8O) is soluble in aq. NaOH but insoluble in NaHCO3, indicating it to have a phenolic group. (ii) The compound, on treatment with Br2 water, gives C7H5OBr3. Taking into account of molecular formulae of the two compounds, the parent compound seems to be cresol.

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(iii) Bromination of the compound reveals that it is m-cresol as it forms tribromo derivative. (iv) The reactions are

44.

45.

1.

(b)

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2.

(a)

3.

(c)

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4.

(b)

5.

(c)

6.

(b)

7.

(d)

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8.

(d)

9.

(b) C6H5ONa + Br – CH2 – CH = CH2 →

C6H5 – O – CH2 – CH = CH2 + NaBr

Allyl phenyl ether

10.

(a)

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Hence

the

IUPAC name of compound (A) is

11.

(a) C6H5OCH3

12.

(10)

C6H5OH + CH3I

The ether with 2 chiral carbons has a plane of symmetry in cis-configuration. Therefore, it will have 3 stereo-isomers. Total isomers = 1 + 2 + 1 + 2 + 3 + 1 = 10. 13. peroxides. On standing in contact with air, ethers are converted into unstable peroxides (R2O → O) which are highly explosive even in low concentrations. Hence, ether is always purified before distillation. Purification (removal of peroxides) can be done by washing ether with a solution of ferrous salt (which reduces peroxides to alcohols) or by distillation with conc. H2SO4 (which oxidises peroxides). 14. (a, d) TIPS/Formulae :

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The given ether is cleaved to give phenol as one of the products because benzyl (a stable carbocations) is formed as an intermediate.

15.

(a)

(P) (Q) (R)

(S)

16. (i) (b)(ii) (c)(iii) (d)(iv) (a) 17. (i) Molecular formula of P, C5H10O indicates 1º of unsaturation. So, it should have double bond. (ii) Acidic hydrolysis of P to Q and R, both of which responds iodoform test, indicates that Q and R should have following structure. CH3CH2OH, (CH3)2CHOH, CH3CHO or CH3COR The only possible linkage that can explain such hydrolysis is ether. Hence P should have following type of structure. C2 – Component – O – C3 – Component Further either the C2 – or the C3 – component should have double bond, thus the possible structure for P should be either of the following two structures which explains all the given reactions.

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or

Extra reactivity of P toward dil. H2SO4 than ethylene is due to formation of highly stable carbocation. or

3°carbocation

CH2 = CH2 1° carbocation

18.

(xvii)

19.

20.

TIPS/Formulae : The oxirane ring is cleaved via SN2 mechanism

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21.

The method given in (ii) is the correct method for the formation of ether because 3° alkyl halides in method (i) leads alkene as the main product. NOTE : 3° alkyl halides are easily dehydrohalogenated by base. (i) (ii) 22.

Empirical formula of A and B. Relative No. of atoms

% of C =

× 100 = 77.77

% of H =

× 100 = 7.40

∴ % of O = 100 – (77.77 + 7.40)

Simplest ratio

= 6.48 = 7.40 = 0.92

=7 =8 =1

= 14.83 ∴ Empirical formula of A and B = C7H8O Nature of (A) : Since A is insoluble in NaOH and NaHCO3, it can’t have –OH and –COOH groups. Further the reaction of A with conc. HI to give compounds C and D separable by means of ammonical AgNO3 and solubility of D in NaOH indicates that C and D are alkyl halide and phenol

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respectively. Thus A is an ether i.e. it is C6H5.O.CH3 which explains all the given reactions.

+

AgI

Nature of (B) : Solubility of B (C7H8O) in NaOH indicates that it is a phenol which is further confirmed by its reaction with bromine water to give compound E of molecular formula, C7H5OBr3. Further bromination of B to give tribromo product indicates that it is m-cresol.

23.

24.

Ethanol (due to the presence of active hydrogen atom, C2H5 – O – H) reacts with sodium metal, while ether has does no such hydrogen atom and hence, does not react with sodium and thus, can be dried by metallic sodium. TIPS/Formulae : The unreactivity of the compound (X) towards sodium indicates that it is neither an acid nor an alcohol, further its unreactivity towards Schiff’s base indicates that it is not an aldehyde. The reaction of compound (X) with excess of HI to form only one product indicates that it should be a symmetric ether. Hence, its other reactions are sketched as below. R–O–R

(X)

(Y)

2RI

2ROH (Z)

–COOH eq. wt. 60

Since, the carboxylic acid has equivalent weight of 60, it must be acetic acid (CH3COOH). Hence, Z must be ethyl alcohol, (Y) ethyl iodide and (X) diethyl ether.

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C2H5 – O – C2H5 + 2HI

2C2H5I

Diethyl ether (X)

Ethyl iodide (Y)

2C2H5OH Ethyl alcohol (Z)

CH3COOH

Acetic acid (Eq. wt. = 60)

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1.

(d)

2.

(c)

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3.

(a)

4.

(b) Rate of Nucleophillic addition reaction is directly proportional to the – I and –M effect of the substituents present in the substrate. Ketones are less susceptible to the nucleophillic addition, due to the presence of alkyl (R) group which has +I effect. Thus reactivity order is

(i) > (iv) > (ii) > (iii) 5.

(c)

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6.

(c) –I effect of NO2 increases reactivity towards nucleophilic addition reaction with HCN. – OCH3 group is electron donating due to resonance effect which decreases the reactivity towards nucleophillic addition.

7.

(b)

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8.

(a)

9.

(d)

10.

(d)

11.

(a)

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12.

(c)

13.

(c)

14.

(c) Generally, aldehydes are more reactive than ketones in nucleophilic addition reactions. \ Rate of reaction with alcohol to form acetal and ketal is

15.

(d)

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Test

Observation

Reaction with +ve hydrogen is present

Reason Electrophilic centre or

Fehling solution test

–ve

CHO group is absent

Neutral FeCl3 test

–ve

phenolic group is absent

Iodoform test

+ve

–COCH3 or –CH(OH)–

Grignard reagent

acidic

CH3 is present

16.

(b) Best combination is HCHO (more reactive aldehyde) and MeOH (less sterically hindered alcohol).

17.

(c) The total number of optically active compounds formed is four. The product has two chiral C atoms. Thus, it has 2n = 22 = 4 stereoisomers.

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18.

(a)

19.

(a)

20.

(a) The pentaerythritol is typically produced via a base-catalyzed reaction of acetaldehyde with excess formaldehyde. The aldol condensation of three moles of formaldehyde with one mole of acetaldehyde is followed by a crossed Cannizzaro reaction between pentaerythritol, the intermediate product, and formladehyde to give the final pentaerythritol product and sodium formate as a byproduct. These reactions are shown below

(Cannizaro's reaction)

21.

(a)

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22.

(d)

5-Keto-2-methylhexanal

23.

(b)

24.

(d) Cannizzaro’s reaction is a disproportionation reaction of aldehyde in which one molecule of aldehyde reduces to alcohol whereas other oxidises to salt of carboxylic acid

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C6H5CHO + KOH

C6H5CH2OH + C6H5COOK

25.

(c) Tishchenko reaction is a modification of Cannizzaro’s reaction. This reaction involves disproportionation of an aldehyde lacking a hydrogen atom in the alpha position in the presence of an alkoxide. The reaction product is an ester. Catalysts are aluminium alkoxide or sodium alkoxide. In Cannizzaro’s reaction the base is sodium hydroxide and the oxidation product is a carboxylic acid and the reduction product is an alcohol.

26. (d)

27.

(c) Only acetaldehyde and methyl ketones give iodoform test.

28.

(b)

29.

(a)

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30.

(b) The ring to which –NH group is attached is activated due to the lone pairs on N (+M and +E effects); while the ring to which –C = O is attached is deactivated. Hence, the electrophile NO2+would go to the para-position of the activated ring. 31. (c) TIPS/Formulae : This reaction is an example of “Perkin reaction”. The compound X should be (CH3CO)2O. In this step the carbanion is obtained by removal of an α–H atom from a molecule of an acid anhydride, the anion of the corresponding acid acting as a necessary base. 32.

(c)

33.

(c) TIPS/Formulae : Reaction of PhMgBr with carbonyl compounds is an example of nucleophilic addition on carbonyl group which increases with the increase in electron-deficiency of carbonyl carbon.

Thus the decreasing order should be

34.

(b)

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35.

(c) TIPS/Formulae : Ketone (non-reducing) and aldehyde (reducing) can be distinguished by Fehling solution.

Aldehyde

36.

(a)

TIPS/Formulae : Both compounds do not contain α-hydrogen hence undergo Crossed Cannizzaro reaction. Initially OH− attacks at the carbonyl carbon of HCHO than that of PhCHO because carbonyl carbon of HCHO is (i) more electrophilic than that of C6H5CHO (ii) less sterically hindered to give hydroxymethoxide which acts as hydride donor in next step to give sodium formate.

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:

etc.

Carboxyl carbon less electrons deficient

37.

(b) Zn(Hg), HCl cannot be used when acid sensitive group like –OH is present, but NH2NH2, OH– can be used.

38.

(a)

Due to π – σ – π conjugation, this dehydrated product will be most stable. 39. (b) This is a type of deuterium exchange reaction. Mechanism

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40.

Note : This reaction is fast in basic conditions where allα – H will replace by D. (b) The possible mechanism is

(i)

(ii) 2nd molecule

41.

NOTE : The slowest step is the transfer of hydride to the carbonyl group as shown in step (ii). (d) NOTE : m-Chlobenzaldehyde does not contains α-H atom and thus undergoes conc. KOH. Cannizzaro reaction with +

42.

(a)

No. of σ bonds in enolic form : 3 + 1 + 1 + 1 + 1 + 2 = 9 No. of π bonds in enolic form : 1 No. of lone pairs of electrons in enolic form = 2

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43.

(d) NOTE : –CHO withdraws electrons from the double bond or from a conjugated system towards itself. or

44.

(c) TIPS/Formulae : Iodoform test is given by compounds having –COCH3, CHOHCH3 and CH3CH2OH. Thus diethyl ketone does not give this test. 45. (b) TIPS/Formulae : The compound containing α-H atom (CH3CHO) does not undergo Cannizzaro's reaction. The other three reaction α-hydrogen. 46. (c) CH3CHO + 2Cu2+ + OH– → CH3COOH + Cu2O ↓ Fehling solution

47.

(red)

(b) TIPS/Formulae : Compounds having

groups, show positive iodoform test.

Hence, (pentanone-2) gives this test. 48. (b) Fehling solution, Schiff’s reagent & Tollen’s reagent react only with aldehydes, but Grignard reagents react both with aldehydes and ketones. 49. (4)

50.

(5) General molecular formula for ketones is CnH2nO ∴

CnH2nO = 100 or 12n + 2n + 16 = 100, n = 6

Possible isomeric ketones with 6 carbon atoms are

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Note that only isomer (III) has a chiral carbon so on reduction with NaBH4 it will give diastereomeric alcohols, while all other five isomers will give racemic mixture. For example :

Due to presence of one chiral carbon, the alcohol can be on both optically active forms and the compound will be racemic mixture.

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(III)

A pair of diastereomers from R-isomer of ketone

+

A pair of diastereomers from S-isomer of ketone

Note : All stereoisomers are reacted separately. 51. (1)

The number of intra molecular aldol condensation products (α, β –unsaturated carbonyl compound formed from Y is 1. 52.

(66.67)

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Compound A is Mass percentage of carbon =

= 66.67

53. 54.

C6H5CH(OCOCH3)2 benzylidene acetate;

55. sodium potassium tartarate. 56. (i)Free radical substitution (ii)Thermite process (iii)Nucleophilic addition (iv) Cyanamide (v)Nucleophilic subsitution (vi)Bayer’s process (vii)Homologous pair (viii) Electrophilic substitution (ix) Ostwald’s process

process;

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(x)Homolytic addition. 57. False. Grignard reagents react with ketones to form ter-alcohols; hence here, ter-butanol will be formed. 58. True : Aldehydes (from primary alchols) may further be oxidised easily to acids as compared to ketones (from secondary alcohols). 59. False : Benzaldehyde has no α-hydrogen atom, hence, it does not undergo aldol condensation but undergoes Cannizzaro reaction. 60. (b, d)

61.

(a, b)

(a)

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(b)

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62.

(b, c)

63.

Q is steam volatile, but not R. Q and R show positive test with 1% aqueous FeCl3 solution but not S. Q, R, S give yellow precipitate with 2, 4-dinitrophenyl hydrazine. (a, b, c) Aldehydes and α-Hydroxyketones show positive Tollen's test.

Silver mirror ↓ + 64. (a)

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65. (c) Reaction I :

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The reaction starts with one mole each of Br2 and acetone. In this reaction mechanism, we can see that consumption of bromine is more than that of acetone. Thus, acetone will be left after completion of reaction. The reaction will not stop after forming P or Q, but continue till the product I is obtained. Only option (c) matches with the result. Reaction II :

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Here, one mole of acetone reacts with one mole of Br2. This matches with the compound given in P. 66. (b, d)

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67. (b, c)

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68. (c)

69.

(a, b, d) TIPS/Formulae : Carbonyl compounds having α – H or α - D undergo aldol condensation.

(a)

(b)

(c) (d) 70.

(b) TIPS/Formulae : Three Cl of chloral makes its carbonyl carbon highly electron deficient, hence H2O, a nucleophile easily adds on it forming chloral hydrate, CCl3CH(OH)2, which is quite stable due to intramolecular H–bonding between two –OH groups.

71. (a)

(b, d) (No new C – C bond is formed)

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(b) (New C – C bond is formed)

(c) (No new C – C bond is formed )

(d) (New C – C bond is formed)

72.

(a, b) Aldehydes and ketones containing α-hydrogen atom undergo aldol condensation. 73. (b, d) Keto-enol tautomerism is shown in compounds having α-hydrogen on the C adjacent to the CO group. 74. (a, c) TIPS/Formulae : Aldehydes having at least one α-hydrogen atom undergo aldol condensation.

75.

(c) (I)

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(II)

(III)

76.

(A) - r, s; (B) - p, q; (C) - p, q, r;

(D) - p, s

(A) (p) does not have carbon thus, will not respond to (p). (r) because of presence of ion, it will give white precipitate of AgCl with AgNO3. (s) because of –NH2 group it will react with aldehydes to form hydrazone derivative. (B) (p) because of presence of C – N bond, it will form prussian blue colour with FeSO4. (q) presence of –OH group will give positive FeCl3 test. (r) It will give yellow precipitate of AgI. (s) because of absence of –NH2 group, it will not give hydrazone derivative. (C) In the similar way, it will respond to p, q and r. (D)

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(p) gives positive test with (p) (r) gives yellow precipitate of AgBr (s) it has a derivative of NH2 – NH2, thus it will react with aldehyde to form hydrazone derivative. For 77-78. (Passage-1)

77. (c)

78.

(d)

For 79-81. (Passage-2) Let us summarize the given facts of the problem. R

Structures of P, Q, R and S can be established on going backward from the known final product.

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Note : In the crossed aldol condensation product (R) from aldehydes (P) and (Q), dehydration will not occur readily because of formation of nonconjugated system. 79. (b) 80. (a) 81. (d) For 82-84 (Passage-3)

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82. (d) 83. (a) 84. (c) For 85-87. (Passage-4) Before answering these question let us complete the sequence of reactions given in data. The given compound (M) i.e. is the only product formed by the action of KOH on compound K. The compound K is

Compound K (i.e.

) is one of the products of ozonolysis of compound

I. Therefore the compound I may be

Thus J seems to be C6H5CHO and hence I is

Now we will try to answer the questions.

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85.

(b) As can be seen from above reaction sequence compound (I) is and it is formed by catalytic dehydration (acid catalysed) of a tertiary alcohol (compound H). Therefore, compound H is

(H) can be formed by the action of

with PhCH2MgBr as follows

Therefore, the correct answer is option (b) 86. (a) As can be seen form the above sequence of reactions the structure of compound (I) is Therefore, the correct answer is option (a).

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87.

(d) As can be seen from the above sequence of reactions the structures of compounds J, K and L respectively are

Thus, the correct answer is option (d), 88. (d) p-Hydroxybenzoic acid has higher boiling point than ohydroxybenzoic acid due to intermolecular hydrogen bonding. Thus, statement-1 is false. o-Hydroxybezoic acid shows intramolecular Hbonding thus, statement-2 is true.

89.

(a) TIPS/Formulae : Ozonide can be reduced by (CH3)2S to give carbonyl compounds and dimethyl sulphoxide.

90.

Thus, the possible polymer should be

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Structure of all cis configuration of the polymer.

All cis form

91.

(i) Formation of HCOONa and a primary alcohol due to Cannizzaro reaction of F and G indicate that either F or G should be HCHO. Thus, the alkene A should have CH2 = grouping. The remaining 5 C’s of A should have grouping = HCC4H9. (ii) Formation of only E by the ozonolysis of D indicates that D (isomer of A, C6H12) should have following structure

NOTE : Fehling's test is given by aldehydes and not ketones. (iii) Since A is isomer of D, former should have following structure.

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92. Hence, compound should have following part structure The enolic form of the compound is more stable than the keto form due to hydrogen bonding.

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93.

[C] 94.

TIPS/Formulae : The given reaction can be summarised as below :

Conclusions from the set of reactions (i) Carbon-hydrogen ratio in A indicates that it is a cyclic compound. (ii) Reaction of A with CH3MgBr indicates that it should have a ketonic group. (iii) As B undergoes ozonolysis to form C, It must have a double bond, and C must have two carbonyl groups. (iv) Reaction of C (a dicarbonyl compound) with a base gives a cyclic compound, it indicates that intramolecular condensation have occurred during this conversion. Thus, A is cyclohexanone which explains all the given reactions.

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95.

(i)

(ii)

(iii)

NOTE : The reaction is an example of benzil-benzilic acid type rearrangement.

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(iv)

ClCH2CH2CH2COPh

(v)

(vi) (vii)

(viii)

+ HCHO

+ HCOOK

(ix)

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(x)

Hexamethylenetetramine (Urotropine)

96.

LiAlH4 reduces only ketonic group to 2° alcoholic group without affecting double bond. This dehydration occurs in acidic conditions. 97.

98. Following informations are provided by the problem. (i) Since aldehyde A (C11H8O) gives C6H5CHO on ozonolysis, it must have a benzene nucleus and a side chain. The side chain should have five carbon(C11–C6=C5), three hydrogen (H8-H5=H3) and one oxygen atom, i.e., it should be C5H3O. Further the compound A has an aldehydic group, so the side chain can be written as C4H2CHO.

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(ii) Formation of two moles of B from one mole of A by ozonolysis indicates that the side chain must possess two unsaturated linkages, one of which must be alkyne type, suggested by very low number of hydrogen atoms. (iii) Further, since the aldehyde A does not undergo aldol condensation, hydrogen is absent and hence triple bond should be present between C2 and C3. (iv) Thus the side chain C4H2CHO of A can be written as – CH = CH – C C – CHO. (v) Thus compound A should possess following structure which explains all the given reactions.

[A]

Glyoxalic acid

[B]

Oxalic acid

99.

The last step is

intramolecular aldol condensation. 100. TIPS/Formulae :

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(i)

Formation of (B) from benzene and acid chloride in presence of anhydrous AlCl3 (Friedel-Craft reaction) indicates that it is a ketone, C6H5COR. (ii) Further the ketone (B) reacts with alkaline iodine forming yellow compound (D) (haloform reaction). This indicates that one of the alkyl groups in ketone (B) is –CH3. Hence, (B) should be C6H5.CO.CH3. (iii) Since ketone (B) is also formed from the hydrocarbon C8H6 (A) by reaction with dil. H2SO4 and HgSO4, the hydrocabon (A) must have an acetylenic hydrogen atom, i.e. ≡ C – H grouping. Hence, (A) must be C6H5C ≡ CH. Thus, compounds (A) to (D) are C6H5.C ≡ CH C6H5.CO.CH3 C6H5COOH CHI3 (A)

(B)

(C)

(D)

Formation of (B) from (A)

101.

The methyl group does stabilize the C = C in the enol form due to hyperconjugation. However, the methyl group also stabilizes the carbonyl double bond in the keto form. In fact, in simple carbonyl compounds, the methyl group stabilizes the carbonyl double bond much more than it stabilizes the enolic C = C bond. Thus, a carbonyl in a ketone is slightly stronger than that in an aldehyde.

As a result, simple aldehydes generally have a higher enol content than simple ketones.

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The two β-di carbonyl compounds have a much higher enol content than the two monocarbonyl compounds because of hydrogen bonding, conjugation and6 membered cyclic form. In fact, for these compounds, the enol form will be major form at equilibrium. Hence, the correct order of enol content is : CH3COCH3 < CH3CHO Ph–OH > R–C CH In between (iii) and (iv), (iii) is more acidic due to –M effect of –NO2. Thus, decreasing order of acidity is (ii) > (iii) > (iv) > (i). 8. (a) B2H6 is a very selective reducing agent and usually used to reduce acid to alcohol.

9.

(c)

10.

(d)

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11.

(b) LiAlH4 reduces esters to alcohols but does not reduce C = C.

CH3CH = CH – CO2Me

CH3CH = CHCH2OH

12. (a)

13.

(d)

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14.

(b) Rate of reaction Electrophillicity of carbonyl carbon, so E.W.G. increases the rate, while E.R.G. decreases the rate

15.

(d) Reaction involved:

16.

(d) The acidic strength of a compound or an acid depends on the inductive effect (– I). Higher the (–I) effect of a substituent higher will be acidic strength. Now, the decreasing order of (–I) effect of the given substituents is NO2 > CN > F > Cl. The correct decreasing order of acidic strength amongst the given carboxylic acids is: NO2CH2COOH > CNCH2COOH > FCH2COOH > ClCH2COOH 17. (a) Acidic strength depends upon the stability of anion formed higher the stability of the anion, higher will be acidic strength of the parent acid. Thus anion : HC º CCOO– > H2C = CHCOO– > p-C6H4 (OMe) COO– > sp sp2 H3CCH2COO– sp3 Therefore,

18.

(b) Acid chloride is more reactive than aldehyde. Hence, phenolic – OH will react with –COCl group first to form ester. This is followed by cyclisation in presence of conc. sulfuric acid.

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19.

(a) The increasing order of the acidity of the carboxylic acids is III < II < IV < I. In aromatic acids, electron withdrawing groups like –Cl, –CN, –NO2 increases the acidity, whereas electron releasing groups like –CH3, –OH, –OCH3, –NH2 decreases the acidity.

20.

(d) Na2C2O4 + H2SO4 → Na2SO4 + CO↑ + CO2↑ + H2O 'X'

(conc.)

Na2C2O4 + CaCl2 → CaC2O4↓ + 2NaCl 'X'

(white ppt.)

5CaC2O4 ↓ + 2KMnO4 + 8H2SO4 (purple)

K2SO4 + 5CaSO4 + 2MnSO4 + 10CO2 + 8H2O (colourless)

21.

(b) DIBAL-H is a reducing agent. It reduces both ester and carboxylic group into an aldehyde at low temperature.

22.

(c) The Bouveault reduction is a reaction in which an ester is reduced to primary alcohol using absolute ethanol and sodium.

Bouveault-Blanc reduction. 23. (a)

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24.

∴ Acidity order is I > II > III > IV (d) α-Substitution is occaured when a carboxylic acid having α-hydrogens is treated with chlorine or bromine in presence of small amount of red phosphorous. This reaction is commonly known as HVZ reaction.

25.

(c)

26.

(d)

CH3CH2NH2

Reaction (III) is a Hofmann bromamide reaction. Hence, C should be CH3CH2CONH2 which can be obtained from CH3CH2COO– NH4+(B).

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Thus (A) should be CH3CH2COOH

(A)

(B)

(C)

27.

(c) Mono-carboxylic acids are functional isomers of esters. e.g.,

28. (d) Carbolic acid (Phenol) is weaker acid than carbonic acid and hence does not liberate CO2 on treatment with aq. NaHCO3 solution. 29. (b) β-Ketoacids undergo decarboxylation easily. It is due to the stability of transition state formed after decarboxylation of β-keto acids. This is not possible in case of other acids.

30.

(d)

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31.

(b)

32.

(c)

33.

(c)

(OH group is strong activating and COOH group is weak deactivating group.)

(OCH3 group is more activating)

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(Benzene ring having –O– is activated, benzene ring having –CO is deactivated)

34.

(a) Carboxylic acid is stronger acid than phenol. The presence of electron withdrawing group (e.g. Cl) increases acidic strength, while presence of electron donating group (e.g. CH3) decreases acidic strength.

35.

(c)

36.

(a) This is simply a displacement reaction.

37.

(a) The optically active acid will react with d and l forms of alcohol present in the racemic mixture at different rates to form two diastereomers in unequal amounts leading to optical activity of the product. (a) Recall that, esters react with excess of Grignard reagents to form 3º alcohols having at least two identical alkyl groups corresponding to Grignard reagent.

38.

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Since, here Grignard reagent is CH3MgBr, the 3° alcohol should have at least two methyl groups. Hence, (a) is the correct choice. 39. (c) C6H5 COOH + SOCl2 C6H5COCl + SO2 + HCl 40.

41.

(d)

(a) TIPS/FORMULAE : LiAlH4 is a reducing agent, it reduces –COOH group to –CH2OH group. C6H5COOH C6H5CH2OH

42.

(2)

43.

(4) All carboxylic acids and phenols are soluble in aqueous NaOH. Thus, four compounds are soluble in aqueous NaOH. (18.00)

44.

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45.

CO;

46.

False : H-Bonding in propionic acid is stronger (carboxylic acids can form dimers) than that in butanol.

47.

False : Saponification is alkaline hydrolysis of esters.

48. (a, d)

49.

(a, b, d)

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(a)

(b)

(c)

(d) 50. (c, d) LiAlH4/(C2H5)2O : Reduces esters, carboxylic acids, epoxides and aldehydes and ketones. BH3 in T.H.F : Reduces –COOH and aldehydes to alcohols but does not reduce esters and epoxides. NaBH4 in C2H5OH : Reduces only aldehydes and ketones to alcohols. Raney Ni in T.H.F. : Does not reduce –COOH, –COOR and epoxides but it can reduce aldehydes to alcohols. 51. (a, c, d)

52.

(b, d)

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(a) Both are soluble in NaOH. Hence, inseparable. (b) Only benzoic acid (C6H5COOH) is soluble in NaOH and NaHCO3, while benzyl alcohol (C6H5CH2OH) is not. Hence, separable. (c) Although NaOH can enable separation between benzyl alcohol (C6H5CH2OH) and phenol (C6H5OH) as only the latter is soluble in NaOH. However, in NaHCO3, both are insoluble. Hence, inseparable. (d) α-Phenylacetic acid (C6H5CH2COOH) is soluble in NaOH and NaHCO3, while benzyl alcohol (C6H5CH2OH) is not. Hence, separable. 53. (d)

54. (a, b, d) NOTE : Ethyl chloride and acetyl chloride react with alc. KCN by nucleophilic substitution reaction while benzaldehyde undergoes benzoin condensation : C2H5Cl C2H5CN + KCl CH3COCl 2 C6H5CHO

CH3COCN + KCl C6H5CHOHCOC6H5

55.

(a, b) TIPS/Formulae : Iodoform reaction is given by the compounds containing –COCH3, –CH(OH)CH3 group and also CH3CH2OH and CH3CHO. 2-Hydroxypropane (CH3CHOHCH3) contains the grouping CH3CHOH— and acetophenone (C6H5COCH3) contains the grouping CH3CO–. 56. (b)

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57.

(a)

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58.

(a)

59.

(c) Perkin condensation of benzaldehyde with (CH3CO)2O/CH3COOK yields cis and trans form of cinnamic acid. C6H5CHO + (CH3CO)2O

Benzaldehyde

60.

Acetic anhydride

(d) Alkylbenzenes when treated with Br2 at high temperature, in the presence of sunlight and absence of halogen carrier undergo halogenation in the side chain via free radical formation.

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Free radical mechanism (Substitution reaction) 61. (a)

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62.

A – r, s, t ; B – p, s ;C - r, s; D – q, r

(A)

(B)

(C)

(D) 63.

(A) - p, q, t ; (B) - p, s, t ; (C) – r, s ; (D) – p

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(A)

(B)

(C)

(D) 64.

(A) - p, q, s; (B) - q; (C) - q, r, s; (D) - q, r (A) C6H5CHO forms ppt. of 2, 4 dibromophenylhydrazone (p), forms silver mirror with ammonical silver nitrate – Tollen's reagent (q), forms cyanohydrin with CN– (s). (B) CH3C ≡ CH gives ppt. with AgNO3 (q) (C) CN– reacts with AgNO3 to form ppt. of AgCN (q), it is a nucleophile (r) and forms cyanohydrin (s)

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(D) I– gives ppt. of AgI with AgNO3 (q), and it is a nucleophile (r) For 65-66 (Passage-1) (i) The oxidation product (a dibasic acid) of the compound P reacts with ethylene glycol and gives polymer dacron (a very well known reaction) points out that the oxidation product i.e. the dibasic acid is terephthalic acid. + n HOCH2.CH2OH Dacron (ii) Now since terephthalic acid is obtained by the oxidation of a carboxylic acid P(C11H12O2), P must have following partial structure (C11H12O2 – COOH – C6H4 = C4H7)

(iii) The compound P undergoes reduction by means of H2/Pd and also undergoes ozonolysis, it must have an unsaturated linkage. Further ozonolysis of P to form an aliphatic ketone indicates = C(R2) type of linkage for which P should have following

final structure,

which explains all the given reactions.

+ Other oxidation products The above structure of P also explains other given reactions.

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65. (a) 66. (b) For 67-68 (Passage-2)

67. 68.

(c) (a)

For 69-70 (Passage-3)

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69. (a) & 70. (a) For 71-72. (Passage-4)

71. (b)

P is in the cis-form, because it forms the cyclic anhydride.

72. (a)

For 73 - 74 (Passage-5) Reactions of compound J indicates that it has C = C linkage and – COOH group. Thus, J can be written as C6H5CH = CH COOH. Since, J is unsaturated carboxylic acid and it is formed by the reaction of compound I with (CH3CO)2O and CH3COONa,

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compound I should be an aldehyde (recall Perkin reaction). Thus, the whole series of reactions can be written as below.

73. 75.

(a) 74. (b) (c) Haloform reaction is shown by ketones havingR – CO – CH3 structure in which – CO – group is electron deficient, while

The – CO – in CH3COOH is not electron deficient due to supply of electrons by the H part-Acetic has three α–H atoms. 76.

(d) TIPS/Formulae : Acetate ion is reasonance stabilized while methoxide ion is not.

Hence, acetate ion is less basic than methoxide ion. 77. NOTE : Higher the Ka value, more stronger is the acid. Correct order of acidic strength of the given acids is

Hence, the Ka values of the five acids will be in the order.

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78.

The two enantiomers of 2-phenylpropanoic acid in the racemic mixture react with (+) – alcohol to form two diastereomers.

(A) and (B) are diasteromers. The bonds attached to the chiral carbon in both the molecules are not broken during the esterification reaction. (+) – Acid reacts with (+) –alcohol to give an (+)–(+)–ester, while (–) acid reacts with (+) alcohol to give (–)–(+)–ester. These two esters are diastereoisomers. (iv) Presence of electron withdrawing group increases the acidic character of the –COOH due to –I effect, Further the –I group (–Cl) from – COOH group, weaker is the acid, thus III is weaker than I while presence of electron-donating group (alkyl groups) decreases the acidic character due to +I effect. Thus, IV < II < V < III < I 79.

Formation of

from (Z)

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80.

81.

(i)

(Ester hydrolysis involves acyl-oxygen fission)

(ii)

(iii)

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(iv)

(Claisen condensation)

(v)

(vi)

NOTE : Esters react with excess of RMgX to form 3º alcohols having two alkyl groups corresponding to R of RMgX. Thus,

(vii) C6H5COOH + CH3MgI —→ CH4 + C6H5COO MgI

(viii)

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(ix)

82.

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83.

We know that esters on treatment with excess of methyl magnesium chloride either give secondary alcohols (from alkyl formates) or tertiary alcohols (from esters other than formates). However, tertiary alcohols are not easily oxidised, hence the alcohol should be secondary alcohol and thus ester is alkyl formate. Hence, ester A (C4H8O2) should be HCOOC3H7. Thus the various reactions and nature of compound B can be established as below.

84.

TIPS/Formulae : Both of the resonating structures of benzoate ion are equivalent, while it is not so in phenoxide ion.

Resonating structures of phenoxide ion

Resonating structures of benzoate ion

The benzoate ion is more stabilized because the negative charge on both structures is on the more electronegative oxygen atom, whereas in phenoxide ion, it is on the less

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electronegative carbon atoms. ;

Presence of CH3 gp in acetate ion shows +I effect and thereby intensifies charge on O– of acetate ion which is thus destabilized. Thus formate ion is more stable than acetate ion or HCOOH loses proton more easily than CH3COOH. TIPS/Formulae : Formic acid has no alkyl group i.e no α – H atom, hence it does not undergo halogenation, while acetic acid has a methyl group (i.e three α – H atoms) on which halogenation takes place. 85. (i) β-Keto acids are unstable and undergo decarboxylation most readily. This is due to formation of 6-membered cyclic transition state. 86. TIPS/Formulae :

(i)

Formation of (B) from benzene and acid chloride in presence of anhydrous AlCl3 (Friedel-Craft reaction) indicates that it is a ketone, C6H5COR. (ii) Further the ketone (B) reacts with alkaline iodine forming yellow compound (D) (haloform reaction). This indicates that one of the alkyl groups in ketone (B) is –CH3. Hence, (B) should be C6H5.CO.CH3. (iii) Since ketone (B) is also formed from the hydrocarbon C8H6 (A) by reaction with dil. H2SO4 and HgSO4, the hydrocabon (A) must have an acetylenic hydrogen atom, i.e. ≡ C – H grouping. Hence, (A) must be C6H5C ≡ CH. Thus, compounds (A) to (D) are C6H5.C ≡ CH C6H5.CO.CH3 C6H5COOH CHI3 (A)

(B)

(C)

(D)

Formation of (B) from (A)

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87.

88.

TIPS/Formulae :

(i)

Since, ‘E’ (C3H6O) forms a 2, 4-dinitrophenylhydrazone but does not reduce Tollen’s reagent and Fehling solution, it must be a ketone, CH3.CO.CH3. (ii) The compound ‘E’ (established as ketone) is obtained by heating compound ‘B’ with Ca(OH)2, ‘B’ must be CH3COOH. (iii) Compound ‘B’ is obtained by the oxidation of ‘D’, the latter must be ethyl alcohol, C2H5OH and hence ‘C’ must be ethyl acetate, CH3COOC2H5. (iv) Since, compound ‘A’ when treated with ethyl alcohol gives acetic acid ‘B’ and ethyl acetate ‘C’, it must be acetic anhydride.

89. For empirical formula of (Y) Element % Relative no. of atoms C 49.31 4.10 H 9.59 9.59 N 19.18 1.37 O 21.92 1.37

Simplest ratio 3 7 1 1

∴ Empirical formula of (Y) is C3H7NO. (Y) reacts with Br2 and NaOH to give (Z) and (Z) reacts with HNO2 to give ethanol and thus (Y) seems to have —CONH2 group.

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(Y) is formed from (X) having Cl on treatment with NH3 and So, (X) is CH3CH2COCl i.e., propanoyl chloride. CH3CH2COCl + NH3 → CH3CH2CONH2 (X)

(Y)

90.

Presence of electron withdrawing group increases the acidic character of the – COOH due to –I effect, Further the –I group (–Cl) from – COOH group, weaker is the acid, thus III is weaker than I while presence of electron-donating group (alkyl groups) decreases the acidic character due to +I effect. Thus, IV < II < V < III < I

91.

(i)

C6H5COOH C6H5CN

(ii) H3CCH = CHCHO

92.

TIPS/FORMULAE :

Molecular weight of the monobasic acid (RCOOH) indicates that the R– should be CH3– i.e., acid F should be acetic acid (CH3COOH, mol. wt. 15+45). Thus, compound D must be acetaldehyde CH3CHO, and compound B which on oxidation gives CH3CHO must be ethanol, CH3CH2OH. Acetaldehyde (D) on treating with aqueous alkali will undergo aldol condensation.

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Nature of A. Thus it is evident that reduction of A with LiAlH4 gives two alcohols; B (ethanol) andC (butanol). Hence, A must be an ester i.e., ethyl butanoate (CH3CH2CH2COOC2H5). CH3CH2CH2COOC2H5 Ethyl butanoate, A

CH3CH2CH2CH2OH + C2H5OH Butanol, C

93.

Ethanol, B

TIPS/Formulae : The given set of reactions can be represented as below :

Aq. NaOH

Sod. salt of acid A Dibasic acid, C

Calculation of molecular formula of C % of H =

× 100 = 2.22%

% of C =

× 100 = 27.30%

% of O = 100 – (2.22 + 27.30) = 71.48% `

By usual method, empirical formula of acid C = CHO2

Eq. wt. of acid C =

– 107 = 45

Mol. wt. of acid C = 45 × 2 = 90 ∴ Mol. formula of C = C2H2O4 Since it is dicarboxylic acid, it must have two –COOH groups. Hence C is Going back, compound C must be produced from sodium oxalate which in turn is produced from sodium formate. Hence A is formic acid and B is CO2. Thus the complete series of reactions can be written as below.

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94.

(i)

(ii)

(iii)

95.

P/Cl2, NH2CH2COONH4

96.

TIPS/Formulae :

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The above reactions lead to following conclusions. (i) Reaction of (A) with CH3COOH in presence of H2SO4 to form ester (B) indicates that (A) is an alcohol. (ii) Reaction of (C) with 50% KOH followed by acidification to give alcohol (A) and the compound (D) seems to be the Cannizzaro reaction. Hence, (C) must be an aldehyde and (D) must be an acid. The nature of (C) as aldehyde is again in consistent with the fact that it is obtained by the mild oxidation of (A) which has been established as an alcohol. (iii) Structure of acid (D) is established by its given facts. – COOH – COCl – CONH2 (D)

(E) HCN

Formation of HCN by the dehydration of (E) establishes that (E) is HCONH2 and hence (D) is HCOOH. (iv) Thus the alcohol (A) produced along with HCOOH during Cannizzaro reaciton of (C) must be CH3OH and hence, (C) must be HCHO. Thus, the various compounds are as below :

97.

(i)

98.

(X) is hydrolysed to give a carboxylic acid (Y) and an alcohol (Z) and thus (X) is an ester;

(ii)

–CHO + CH3CHO

.

Oxidation of alcohol (Z) gives acid (Y). R'OH RCOOH (Z)

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or RCH2OH ————→ RCOOH Hence (X), (Y) and (Z) are

(∴ R' is R—CH2)

,

99.

,

TIPS/Formulae : The weaker a base better is its leavability.

This is an example of nucleophilic substitution where the group .X (Cl, NH2, OC2H5, OCOCH3) is replaced by OH. The decreasing basic character of the four concerned groups is: > OR–> OCOR– > Cl– Hence, Cl– (the weakest base) will be lost most easily, while

(the strongest base)

will be lost with most difficulty. Thus, the order of hydrolysis becomes: CH3CONH2 < CH3COOC2H5 < (CH3CO)2O < CH3COCl. 100. CH3CH2CHO + [O]

101. (i)

C2H2

(ii) CH3I + KCN (alc)

CH3CH2COOH + 2Ag

CH3CHO

CH3CH2OH

CH3CN CH3COOH

Acetic acid

102. Ozonolysis of (A) to acetone and an aldehyde indicates the following part structure of alkene (A) :

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As per problem : RCHO

RCOOH [B]

Bromo compound [C]

Hydroxy acid [D]

Structure of (D) is determined by the reaction :

The compound (D) is obtained by hydrolysis of (C) with aqueous alkali. Since (C) is a bromo compound, therefore it has a bromo group whereas the compound (D) has a hydroxyl group. Therefore, structure of C is

The compound (C) is formed by bromination of compound (B), therefore, the compound (B) is

The compound (B) is formed by oxidation of an aldehyde therefore, the structure of the aldehyde is

The aldehyde and acetone are formed by ozonolysis of alkene. Therefore, the double bond in alkene should be between the carbon atoms of the two carbonyl compounds (the

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aldehyde and acetone). Therefore, the compounds and the reactions are identified as

103. CH3COCl + AlCl3 → CH3C+ = O +

104. First step is Claisen condensation.

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1. 2.

(c) Aniline and phenol form complex with lewis acid. Chlorobenzene produces highest yield in Friedel craft reaction among the given options. (c)

3.

(d)

4.

(a) In the given substituted benzene rings, the substitutents methoxy (– OCH3) and amino (–NH2) are strongly activating groups, while methyl (– CH3) is weakly activating and chloro (– Cl) is a deactivating group towards electrophilic aromatic substituation reaction. Since among methyl and methoxy group, methoxy group is more reactive than methyl group, (c) is more reactive than (d). Although amino group is strongly activating group, it gets protonated in presence of acid to form anilinium ion ( ) which is strongly deactivating. Hence, (a) is less reactive than (c) and (d).

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Chloro group is also deactivating group but less deactivating than (

).

Thus order is (a) < (b) < (d) < (c). Note: The activating groups increases the electron density on benzene ring and thus increases the rate of electrophilic aromatic substitution reaction. The deactivating groups decreases the electron density on benzene ring and thus decreases the rate of electrophilic aromatic substitution reaction. 5. (c) Nitration takes place in presence of conc. HNO3 + conc. H2SO4. In strongly acidic medium, amine is converted into anilinium ion (– NH+3); substitution is thus controlled not by –NH2 group but by –NH+3 group which, because of its positive charge, directs the entering group to the metaposition instead of ortho- and para.

6.

(b)

7.

(c) CHCl3 + RNH2 + KOH

8.

9.

R – N+ ≡ C– + 3KCl + 3H2O (N) None of the given options is correct. Benzene sulphonic acid, being stronger acid than carbonic acid, would liberate CO2 when treated with sodium bicarbonate, but p-nitrophenol, being less acidic than carbonic acid, will not liberate CO2. (d) NOTE : POCl3 is a dehydrating agent. Hence

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10.

(a) Two moles of NH–2 ions will abstract two moles of most acidic hydrogen out of the four moles of acidic hydrogen present per mole of the compound. The acidic strength is in the order: –COOH > –OH (phenolic having NO2 group) > –OH (phenolic) > alkynic H. 11. (c) N can't have more than 8 electrons in its valence shell as it does not have any d orbital. In (c), N has 10 electrons. 12. (d) TIPS/Formulae : In the formation of nitrile, number of carbon atoms in parent chain increases by one. CH3CH2CH2Cl + alc. KCN CH3CH2CH2CN Propyl chloride

13.

(b) TIPS/Formulae : The addition is initiated by the attack of CN– group which is a nucleophile.

14.

(a) TIPS/Formulae : Toluene has electron-donating methyl group. Hence, reacts fastest while others have either electron withdrawing groups (i.e. – COOH or – NO2 etc.) or no substituent.

15.

(b)

This is Hoffmann bromomide degradation reaction. 16. (a, b, d) (a) Nitro group decreases the electron density at the o- and p-positions in comparison to m-position due to –M effect. (b) The below intermediate is a resonance hybrid of three equivalent structures.

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Due to –NO2 group, electron density will decrease more from o- and p- positions. Hence, the attack of electrophile Br+ will be favourable at meta position. In another words, after attack of Br+ at meta position, intermediate will be least destabilised. (c) There will be loss of aromaticity when Br + attacks at any position because of formation of sp3 hybridised carbon. Thus, option (c) is incorrect. (d) Due to more electron density at meta position, heterolytic fission of C – H bond and removal of H+ will be easy from this position. Thus, option (d) is correct. 17. (c, e) TIPS/Formulae : Silver nitrite, being a salt of nitrous acid, occurs in two tautomeric forms.

ion from

may exist in two tautomeric forms, –O – N = O

(nitrite ion) forming alkyl nitrites, and

(nitro group) forming

nitroalkanes. 2 C2H5Br + 2AgNO2 → C2H5NO2 + C2H5ONO + 2 AgBr 18. (A)– p, q, s, t; (B) – s, t; (C) – p; (D) – r (A)

CH3CH2CH2CHO (B)

(C)

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(D)

19. (d) Chlorobenzene is resonance stabilized. Thus aryl halides (chlorobenzene) do not undergo nucleophilic substitution. Reason is correct. 20. (a) In o-nitrophenol, intramolecular H-bonding is possible because OH and NO2 groups are close to each other. This makes the ortho isomer less acidic as its capacity to donate a proton (H-atom) decreases. There is no such intramolecular H–bonding in the p-isomer. 21. (i) (a) group is electron releasing, hence o-, p-directing (b) –NO2 group is electron withdrawing, hence m-directing (ii) TIPS/Formulae : –NO2 group is electron withdrawing, hence mdirecting, whereas –CH2NO2 is not. (a) Given compound is an aryl fluoride having electron-withdrawing –NO2 group at para position of fluoride atom which activates the fluoride due to – M and – E effects for nucleophilic substitution (SNAr), hence reaction with NaOH will liberate F– as NaF. (b) The given compound is an aryl fluoride having –CH2NO2 group in the meta position which is not capable of activating aryl fluoride (absence of–M and –E effects because NO2 group is not conjugated to benzene ring) for nucleophilic substitution, hence aq. NaOH will not displace fluorine here, i.e. no F– will be formed. (iii) The nitro group in nitrobenzene strongly deactivates the benzene ring due to – I and – M effects. This decreases the reactivity of benzene ring towards Friedel - Crafts alkylation. (iv) o-Nitrophenol shows intramolecular H-bonding and exits as a single molecule, while p-nitrophenol shows intermolecular H- bonding and thus its several molecules associate with each other. The ortho isomer goes with the steam due to its low boiling point. 22.

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23.

(–COR is a meta directing group)

(–R activates the ring at o- and p-positions for electrophilic aromatic substitution)

(–NO2 decrease the e– density at o- and p-positions for substitution)

aromatic

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24.

(i)

NOTE : NO2 group deactivates the ring and thus, decreases the electron density at o- and p-positions. This makes it for nucleophilic attack at these positions.

(ii)

(iii)

25.

(i)

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(ii)

+ HNO2 ——→

(iii)

26.

(i)

(ii)

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+

On fractional crystallization, para isomer crystallizes first. (iii)

1.

(b)

– OCH3 group increases electron density of ring at o- and p- position making (I) most basic. (III) is least basic due to –I effect of –CN group at meta position. Since, –I effect of –CN > –I effect of –OH group. Hence, correct basic strength will follow the order I > II > IV > III

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Basic strength Order of Kb value is, I < II < IV < III. 2.

(a) Gabriel phthalimide synthesis gives 1° amine in good yield.

3.

(c)

NaBH4 does not reduce R–CN.

4.

(b) (A)

(C)

(B)

(D)

Thus, increasing order of basicity is (B) < (A) < (D) < (C).

5.

(a)

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6.

(b)

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7.

(b)

8.

(c)

9.

(d) Basic strength of amines depends upon availability of lone pair of electrons. Aliphatic amines are more basic than aromatic amines. 10. (d) Conjugate acid of guanadine(B) is resonance stabilised and have 2 resonance structure.

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Similarly conjugate acid of (A) is also resonance stabilised and have one resonance structure. (C) does not exhibit resonance structure. therefore the basic order is, kb : (B) > (A) > (C) pkb : (B) < (A) < (C)

11. (a)

12. (c) Hinsberg’s reagent is benzenesulphonyl chloride (C6H5SO2Cl). It is used for detection of primary, secondary and tertiary amines. 13. (a) Primary amines are prepared by Gabriel phthalimide synthesis

14. (b) Higher the electronegativity (sp2 > sp3) of an atom, more will be its capacity to take proton. Thus N labelled as (b, (c) and (d) are better protonated than the N at (a) and (e). 15.

(b) Compound, III

is most basic as the lone pair of nitrogen is easily

available for the donation. In case of compound (I)

lone pair is not involved in resonance but

nitrogen atom is sp2 hybridsed, whereas in compound II the lone pair of nitrogen is involved in aromaticity which makes it least basic. 16. (b) pKb (A) EtNH2 3.29

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(B) (Et)2NH 3.00 (C) Me3N 4.22 (D) Ph – NH – Me 4.7 So, order of basic strength is:

(B) > (A) > (C) > (D) 17. (d) Aromatic diazonium salts are more stable than aliphatic diazonium salts. The higher stability of aryl diazonium salts is due to resonance. Electron donating substituents increase electron density on benzene ring. Hereby they increase the stability of diazonium salts. Electron withdrawing substituents decrease electron density on benzene ring. Hereby they decrease the stability of diazonium salts. –COCH3 group is electron withdrawing and hence, diazonium salts from (D) is less stable than that from (B). Although – O– COCH3 is electron donating substituent, but it is present in meta position. Hence, it will not have significant effect on stability. Therefore the increasing order of diazotisation is (A) < (D) < (C) < (B). 18.

(c) (a)

the lone pair of electron is less easily available for protonation. (b) the lone pair of electron is not available for proton. (c) +I more preferred than H-bonding, is due to steric hindrance to H-bonding in 1° amine.

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(d)

Unstable Thus, the correct order is (I) < (II) < (IV) < (III) 19. (c) Higher the stability of the conjugate acid, more is the basic character of the parent amine.

(IV)

The conjugate acid is stabilized by resonance with two different –NH2 groups.

(I)

The conjugate acid is stabilized by resonance with one –NH2 group. Hence, as compared to IV it is less basic.

(II)

Lone pair is not involvd in aromaticity. Hence, more available than III.

(III)

Lone pair is involved in aromatic sextet. Hence, not available. Hence, the correct order of basic strength is IV > I > II > III 20. (b) RCONH2 + Br2 + 4NaOH R–NH2 + K2CO3 + 2NaBr + 2H2O 21. (d) Primary, secondary and tertiary amines are distinguished by Hinsberg reagent test. The Hinsberg reagent is an alternative name for benzene sulphonyl chloride (C6H5SO2Cl).

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22.

(a)

23.

(c) Aliphatic amines are more basic than aromatic amines thus methylamine is most basic. Electron donating groups increase the basicity whereas electron withdrawing groups decrease the basicity of the aromatic amines. Thus p-methoxyaniline is more basic than aniline which is further more basic than p-nitroaniline.

24.

(d) R – CH2 – NH2 + CHCl3 + 3KOH (alc) R – CH2 – NC + 3KCl + 3H2O Alkyl isocyanide

25.

(b)

26.

(b)

Now since the molecular mass increases by 42 unit as a result of the reaction of one mole of CH3COCl with one-NH2 group and the given increase in mass

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is (390 – 180) 210. Hence the number of – NH2 groups = 210/42 = 5. 27. (c) The correct order of relative basicity of amines in the gas phase is 3° > 2° > 1° > NH3 The alkyl group releases electron and thus, tends to disperse the positive charge of the alkyl ammonium ion and therefore stabilises it. Since, NH+4 (from NH3) has no such alkyl group, it is not stabilised to such an extent as alkylammonium ion. 28. (c) Only primary aromatic amines undergo diazotisation followed by coupling.

29.

(a)

–Br is present on ring, hence less reactive. 30. (b) (i) Position (X) is most acidic due to – COOH group. (ii) –NH3+ group at position Y is more acidic than at Z because of presence of electron withdrawing – COOH group in close proximity. Hence – NH3+ group at position Z is least acidic. 31. (b) TIPS/Formulae :–NH is more activating than –CH3 group, whereas C = O group is a deactivating group. Hence, electrophilic substitution at paraposition will be governed by the ring having

group.

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32.

(a)

33.

(a) Reaction of (B) indicates that it is an aldehyde which thus should be C2H5CHO or CH3CH2CHO, hence (C) should be CH3CH2CH = NNHCONH2

34.

(b)

The protonated form of II would be III which is more stable because here the contributing structures (III) and (IV) are equivalent. In the availability of electron pair increases due to the +I effect of two CH3 groups while in CH3CH2NH2, +I effect of only one ethyl group is operative. In CH3 – CO – NH2, the electron availability on nitrogen decreases due to resonance as shown below

35.

Therefore, the order of basic strength would be 1 > 3 > 2 > 4. (d) Aliphatic amines are more basic than aromatic amines because in aliphatic amines electron pair on nitrogen is not involved in resonance.

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36.

(b)

37.

(a) In benzylamine, electron pair on nitrogen is not delocalised due to lack of conjugation; while in all other compounds it is delocalised and hence, lesser available for protonation.

38. (b) NOTE : Only primary aliphatic and aromatic amines give this test. CHX3 + RNH2 + 3KOH → RNC + 3 KX + 3H2O 39.

(b)

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This is Hoffmann bromomide degradation reaction. 40. (c) NOTE : Secondary amines (aliphatic as well as aromatic) react with nitrous acid to form N-nitrosoamines. (C2H5)2NH + HONO → (C2H5)2N – N = O + H2O N-Nitrosodiethylamine

41. (4.00) Scheme – I

Scheme – II

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Scheme – III

Number of Br atoms in (T) = 4 42. (18.60)

Molecular weight of (P) aniline = M.wt. of C6NH7 = 72 + 7 + 14 = 93

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Density of P = 1 g mL–1 ∴ 9.3 mL of P = 9.3 g P =

= 0.1 mole P

The mole ratio of PhNH2 : PhN2+ : =1:1:1 So the mole of Q formed will be 0.1 mole and extent of reaction is 100% but if it is 75% yield. Then amount of Q =

= 0.075 mol

The molecular formula of Q = C16H12ON2 So M.wt. of Q = 16 × 12 + 12 × 1 + 16 + 2 × 14 = 192 + 12 + 16 + 28 = 248 g So amount of Q = 248 × 0.075 = 18.6 g 43. (495)

Molecular mass of (D) = (6 × 12) + (4 × 1) + (1 × 14) + (3 × 80) = 330 g/mol Moles of (D) formed = 10 × 0.6 × 0.5 × 0.5 × 1 = 1.5 Mass of (D) formed = 1.5 × 330 = 495 g

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44.

Dipolar ion. (Salt like)

45. aniline. 46. (c, d) pKb difference between I and II is 0.53 and that of III and IV is 4.6. So option (b) is incorrect Correct statement are (c) and (d) The most basic compound in the given option is (II) and least basic compound is (III), so option (a) is also incorrect. In 2,4,6-trinitroaniline (III) due to strong –R effect of the three –NO2 groups, the .p. of electrons on is more involved with benzene ring hence it has least basic strength. Whereas (IV) N,N-dimethyl – 2,4,6-trinitroaniline, due to steric inhibition to resonance (SIR) effect; the lone pair of nitrogen is not in the plane of benzene, hence makes it ( .p.) more free to protonate

47.

(a)

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NOTE : Attack of NH3 will occur at aliphatic aldehyde group because the aromatic –CHO is in conjugation with benzene ring. 48. (a) group is acetylated by acetic anhydride in methylene chloride (solvent). Note that —CONH2 group does not undergo acetylation because here lone pair of electrons is delocalised.

49.

(a, b, d) In ice, water molecules are excessively H-bonded giving a cagelike structure which is lighter than water. Primary amines are more basic than tertiary amines, because the protonated 1° amines are extensively H-bonded and hence more stable than the corresponding protonated 3° amines.

Acetic acid undergoes dimerisation in benzene.

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50.

(a,b,c) Lower amines like NH3, CH3NH2 and (CH3)2NH break diborane molecule unsymmetrically, while larger amines like (CH3)3N, (C2H5)3N break diborane in symmetrical manner.

51.

(b, d) NOTE : Only primary amines give carbylamine test. Hence 2,4-dimethylanilinc and p-methyl–benzylamine both give this test.

52.

(a, d)

53.

(c) TIPS/Formulae : Anilinium hydrochloride has ionisable chlorine, whereas p-chloro-aniline has non ionizable chlorine. Thus, anilinium hydrochloride gives white precipitate of AgCl with AgNO3.

In chloroaniline, –Cl is directly attached to benzene ring, hence it is nonreactive. 54. (c) In Ist structure N has complete octet, whereas in IInd structure N has 10es in its valence shell, which is not possible. 55. (a, c) This is an example of Hoffmann degradation of amides.

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56.

(c) Scheme

(III)

:

Scheme (IV) :

Scheme (II) :

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Scheme (I) :

57. 58. 59. 60.

61.

(a) (b) (d) The reagent used in Hoffmann bromamide reaction is alkaline halogen (NaOH or KOH + X2). (d) Conversion of (iii) to (iv) involving rearrangement is the slowest step. Species (iii) is electron deficient (N has only 6 electrons), hence it has a tendency to get its octet completed by migration of alkyl group or aryl (Rearrangement). (b) Since the reaction is intramolecular, no cross product will be formed.

For 4, 5- The given reaction can be summarised as below.

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62.

(d) In strongly acidic conditions, aniline becomes protonated with the result lone pair of electrons is not available to produce +E and +M effects. Thus, here aniline becomes less reactive towards electrophilic substitution. Thus statement is false. Statement-2 is true.

63.

(i)

(ii) [Y], a 3° alcohol is optically inactive. (iii) Formation of [Y] from [X].

2° Carbocation

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64.

Test of phenolic group : Test of 1° amino group : Test of – COOH group : 65. The reaction proceeds via benzyne formation

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66.

Summary of the given facts.

From the above set, following conclusions can be drawn. (i) Since the oily compound F (C6H7N) reacts with acetyl chloride, it must have - NH2 or > NH group. Thus, (F) can be written as C6H5NH2 and hence, D is C6H5NHCOCH3.

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C6H5NH2 (F)

C6H5NHCOCH3 (D)

(ii) Compound (E) on treatment with alkali followed by acidification gives a white solid compound (G), C7H6O2. Thus, (G) seems to be an acid, hence it is C6H5COOH. (iii) Since (D) and (E) are isomers of the formula C8H9NO, and give C6H5NH2 and C6H5COOH respectively, both should be amides having different alkyl or aryl group. Thus, (D) should be C6H5NHCOCH3, and (E) must be CH3NHCOC6H5. (iv) Since compounds (D) and (E) are formed by the rearrangement of compounds (B) and (C) respectively. Compounds (B) and (C) should be oximes > C = NOH (recall that oximes rearrange to amides - Beckmann rearrangement). Further oximes having different alkyl (aryl) groups show geometrical isomerism (syn and anti), compounds (B) and (C) must have following structures.

Recall that Beckmann rearrangement involves migration of anti -alkyl or aryl group, i.e.,

Since (D) is formed from (B), and (E) from (C), (B) and (C) should have following structures.

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(B)

(C)

(v) Lastly, oximes (B) and (C) are formed from (A), the latter should be a ketone of the formula C6H5COCH3,

(A)

67.

(B) & (C) syn- and anti-isomors

(i)

(ii)

(iii)

— NH — CO —

68.

(i) TIPS/Formulae : Two factors operate in deciding the basicity of alkyl amines. (a) Inductive effect (b) Solvation effect (a) Inductive effect. The alkyl group being electron releasing, increases the charge density on nitrogen. This in turn increases the basicity of amines. The expected order of basicity is R3N > R2NH > RNH2 > NH3

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(b) Solvation effect. Because of the positive charge carried by the conjugate acid of an amine, it is stabilised by the hydrogen bonding with the solvent water. The larger the number of hydrogens attrached to the nitrogen in the conjugate acid, the larger is its stability and thus larger is the basicity of the corresponding base. The expected order of basicity of the alkylamines will be NH3 > RNH2 > R2NH > R3N The inductive and solvent effects predict the opposite trend in the basicity of alkyl amines. In going from R2NH to R3N the solvation effect plays a more dominating role as compared to the inductive effect making R2NH more basic than R3N. A simple explanation is that the steric factor in R3N makes the availability of a lone pair of electrons on nitrogen poor than in the dialkylamine, predicting R2NH a stronger base than R3N. (ii) Aniline is weak base than cyclohexylamine because of resonance while there is no resonance in cyclohexylamine.

69.

syn-

70.

anti-

C6H5NH2 + CHCl3 + 3KOH → Phenyl isonitrile (foul smelling)

71.

C6H5N ≡ C + 3KCl + 3H2O

A (C3H9N) is a 2º amine CH3 – NH – C2H5 (ethylmethylamine) because the product sulphonamide is solid and insoluble in alkali.

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Benzene sulphonyl chloride

72.

Let us summarise the given facts.

Reaction of the original compound with alcoholic potash and chloroform to give foul smelling gas indicates that it contains a primary –NH2 group. R – NH2 + CHCl3 + KOH ——→ R – NC ↑ (Basic compound)

Carbylamine (foul smelling)

Determination of mol. weight of the amine : 112 mL. of gas is evolved at S.T.P. by 0.295 g of amine 22400 mL. of gas is evolved by =

× 22400 = 59 g

Hence, the mol. wt. of the amine = 59 ∴ Mol. wt. of the alkyl group = 59 – 16 = 43 Nature of alkyl gp. of mol. wt. 43 = C3H7– Thus, the amine may be either CH3CH2CH2NH2 or The reaction of amine with NaNO2 at 0ºC and all other reactions may thus be written as below.

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CH3CH2CH2OH

+

N2↑ CH3CH2CH2OH

No yellow ppt. (CH3)2CHOH + N2 ↑ (CH3)2CHOH NOTE : Since the given reactions correspond to isopropylamine, the original compound is isopropylamine, (CH3)2CHNH2. 73. TIPS/Formulae : (i) Carbylamine reaction involves reaction of a primary amine with alc. KOH and CHCl3 which results a product containing isocyanide group with unpleasant smell. (ii) Riemer-Tiemann reaction of phenol produces o and p isomers. Solution of compound A in chloroform when treated with alcoholic KOH yields compound C (C7H5N) having an unpleasant odour which may be due to isocyanide. Hence, the above reaction may be an example of carbylamine reaction. Therefore, compound A must be aniline (C6H5NH2) and C must be phenylisocyanide (C6H5NC). C6H5NH2 + CHCl3 + 3KOH → C6H5NC + 3KCl + 3H2O Aniline, A

Phenyl isocyanide, C

Alkaline aqueous layer of B when heated with chloroform gives D and E which are isomeric with each other and have molecular formula C7H6O2. This indicates the possibility that Reimer-Tiemann reaction would have taken place. Hence, compound B must be phenol and compounds D and E are oand p-hydroxybenzaldehydes. This can be represented in the following way.

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+

74.

NOTE : Carbylamine reaction is a distinction test for primary amines. All primary amines (alphatic or aromatic) on heating with alcoholic KOH and CHCl3 give unpleasant or foul smell of isocyanide which is easily detected.

C2H5NH2 + CHCl3 + 3KOH (alc.) —→

+ 3KCl + 3H2O

Diethylamine , a 2° amine, does not respond this test. 75. Determination of empirical formula : C H N O % age

49.32

9.59

19.18

21.91

mole = 4.11 = 9.59 = 1.37 = 1.37 (Dividing by 1.37) Simplest ratio 3 7 1 1 ∴ Empirical formula of A will be C3H7NO and empirical formula wt = 73 Calculation of molecular wt. of (B) We know, meq. of Ag = meq. of Ag salt

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or or

= =

or Esalt = 181

Esalt = EAg + EAnion ⇒ Eanion = 181 – 108 = 73 For monobasic acid (B), adding one H, Mol. wt. = Eq. wt. + 1 = 73 + 1 = 74 B being monobasic can be represented as Cn H2n + 1 COOH ⇒ 74 = 12n + (2n + 1).1 + 12 + 16 + 16 + 1 ⇒ n = 2 ∴ B = C2H5COOH Nature of A : Since B is obtained by the action of A with NaOH followed by hydrolysis, so A is an amide, CH3CH2CONH2. Reaction : CH3CH2COONa

76.

(i) TIPS/Formulae : Presence of +I / +M group increases the basicity whereas presence of –I / –M group decreases the basicity.

(ii) The ease with which the lone pair of electron (unshared) on the N-atom coordinates with a proton determines the relative basic strength of amines. 77. Hoffmann degradation reaction.

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+ Br2 + 4KOH → + K2CO3 + 2KBr + 2H2O 78.

(i)

Consult (iv)

(ii)

(iii) C6H6

C6H5NO2

C6H5NH2

(iv)

(v) CH3CH2.Cl

CH3.CH2.CN CH3.CH2.CH2.NH2

1.

(c) In acidic medium, benzene diazonium chloride reacts with aniline (a basic compound) forming a30 dye (c).

(c) 2.

(c)

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3.

(c)

4.

(a) On reduction with Zn and HCl, C6H5N2Cl forms aniline as the main product.

5.

(c)

6.

(a)

7.

(d)

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8.

(b)

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9.

(a)

Coupling reaction will occur at ortho position of activated ring. 10.

(c) is a good leaving group but the formed carbocation will not be stable

due to –M effect of –COOH group. Thus, intramolecular attack of oxygen atom from rear side will take place at the carbocation.

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The formed 3 membered ring is not a stable compound. Now, the reaction will proceed with the attack of water molecule.

11.

(b) C6H5N2Cl + N=N–

OH

p-Hydroxyazobenzene

12.

(A) – (r), (s) ; (B) – (t) ; (C) – (p), (q) ; (D) – (r)

13.

(i)

Going backward, we can easily solve the problem.

(ii)

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(iii)

(iv)

(v)

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(vi) C6H5CHO

C6H5COOH C6H5CONH2

C6H5NH2

C6H5N2Cl 14.

15.

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16.

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1.

(a) Lactose contains and C4 of glucose.

-glycosidic linkage between C1 of galactose

-D-Galactose

2.

(c)

One acetal and one hemi acetal group is present in maltose. 3. (b) (i)

(ii)

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(iii) Thus x = 4, y = 6 and z = 5 [Resorcinol + Conc. HCl] 4. (c) Seliwanoff reagent It is used to distinguish aldoses and ketoses. Ketoses show red colour whereas aldoses show light pink colour with Seliwanoff Reagent. 5. (b) Glucose exists in cyclic form in which aldehyde group is not free, therefore it does not give Schiff’s test. 6. (d) Maltose on hydrolysis gives two moles of -D-glucose.

7.

(c) Gluconic acid is obtained by partial oxidation of glucose by mild oxidising agent e.g. Tollen’s reagent, Fehling solution, Br2 water.

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Gluconic acid can not form hemiacetal or acetal. 8. (a) Structure of glycogen is similar to amylopectin. It is found in yeast and fungi and stored in animal body. It contains aglycosidic linkages. 9. (a) Starch is a polymer of a-D-glucose. It has two components. (i) Amylose, which has only a-1, 4-glycosidic linkage and is a linear polymer. (ii) Amylopectin, which has a-1, 6-glycosidic linkage in addition to a-1, 4glycosidic linkage and is a cross-linked polymer. 10. (d) Linear structure of glucose,

Cyclic structure of glucose,

Here, * represents stereocenters. 11. (b) Sucrose contains glycosidic link between C1 of -D glucose and C2 of -D-fructose. Glucose + Fructose C12H22O11 + H2O

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12.

(a)

13. 14.

(d) In amylose 1, 4-α-glycosidic linkage is present. (a)

15.

(b)

16.

(d)

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During vigorous exercise sufficient oxygen is not available to meet the energy demand, so energy is derived through conversion of pyruvic acid to lactic acid. 17. (a) Starch is a mixture of amylose & amylopectin polysaccharides and monomer is glucose. Thus on complete hydrolysis it gives only glucose. 18. (d) Sucrose does not contain a free aldehydic or ketonic group, hence it does not show mutarotation. 19. (a) 6CO2 + 12NADPH + 18ATP → C6H12O6 + 12NADP + 18ADP 20. (c) Glycogen is called animal starch and is found in all animal cells. It constitutes the reserve food material. 21. (c) Fischer gave the prefix “D” to compounds whose bottom chiral has its OH to the right. So natural glucose is called D-glucose or dextrose. Structure of D-Glucose :

22.

(b)

At Cl, –OH group is above the plane, thus, it is β-pyranose. If hydrogen is present at the anomeric carbon (i.e. Cl), then it is an aldose.

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Hence, the given compound is an aldohexose. 23. (a) Ring (A) is pyranose (analogous to pyran) and ring (B) is furanose (analogous to furan). Before forming the disaccharide the monosaccharide (A) has –OH group at Cl, below the plane. The monosaccharide (B) has –OH group at Cl, above the plane.

Hence, ring (A) forms α-glycosidic linkage and ring (B) forms β-glycosidic linkage. 24. (a) Cellulose is a polysaccharide composed of only D-glucose units. Every adjacent glucose units are joined by β-glycosidic linkage between C1 of one glucose and C4 of the next. NOTE : Thus in every glucose units only three –OH groups are free to form triacetate.

Cellulose triacetate

25. 26.

(b) The two isomeric forms (α – and β –) of D-glucopyranose differ in configuration only at C–1; hence these are called anomers. (d) Glucose being an aldose responds to Tollen’s test while fructose (an α-hydroxy ketone), although a ketose, undergoes rearrangement in

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27.

presence of basic medium (provided by Tollen’s reagent) to form glucose, which then responds to Tollen’s test. (8)

28.

Thus, total number of stereoisomers in pyranose form of configuration = 23 = 8 (9)

D-

No. of chiral centres = 9. 29. (b, c, d) (a) Glucose

Gluconic acid.

Bromine water oxidises only aldehyde group to carboxylic group. It neither oxidises –OH group nor –CO group. (b) Sucrose is dextrorotatory (+ 66.6°) in nature but on hydrolysis it gives dextrorotatory glucose (+52.5) and laevorotatory fructose (–92.4°). Overall solution is laevorotatory since laevorotation is more than dextrorotation.

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(c)

and

are anomers

Because they differ in configuration only around C1 position. (d) Monosaccharides do not undergo further hydrolysis due to absence of glycosidic linkages. 30. (d)

Note : (i) In pyranose form, the groups which are in right side of Fischer projection will be below the plane. (ii) For β-form, –OH at Cl and –CH2OH at C5 will be on the same side. (iii) At C2, C3 and C3, C4; –H and –OH are trans to each other. Hence, the structure of β-L-glucopyranose will be :

Trick : Only option (c) and (d) are β-forms. However, in option (c) at C2 and C3, –OH groups are at the same side. Hence, only option (d)

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can be the answer. 31. (b, c) Invert sugar is an equimolar mixture of D-(+) glucose andD-(–) fructose.

Since the specific rotation of the hydrolysis product of (+) sucrose is inverted, i.e., (–), it is known as inverted sugar. •

Specific rotation of invert sugar =

•

D-glucose on oxidation with Br2-water produces gluconic acid and not saccharic acid.

32.

(a)

33.

(b, c)

Both the units of disaccharide X are bonded by C1 – C2 linkage. The reducing groups (aldehydic and ketonic respectively) are bonded together. Thus, X is non-reducing sugar. Also, X has α-glycosidic linkage.

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While Y has a hemiacetal group, hence it is reducing sugar. Also Y has βlinkage. 34. (i) – (c), (ii) – (d), (iii) – (a), (iv) – (b) 35. (c) Statement-1 is correct, but statement-2 is incorrect because glucose on reaction with Fehling solution gives Cu2O and not CuO. 36. In the two disaccharides, structure A will be reducing sugar since, both monosaccharides units are not linked through their reducing centers i.e., it has hemiacetal linkage, while in structure B, both the monosaccharide units are linked through their reducing centers, i.e., it has acetal linkage, hence it will be non-reducing. 37. L-Glucose is an enantiomer of D-glucose, hence

(i)

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(ii)

1. 2.

(a) Tyrosine is a non-essential amino acid. (c) Compounds having 1° amine give carbylamine reaction with CHCl3 and alc. KOH.

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3.

(a) Ceric ammonium nitrate test is used for detecting alcohols, while carbylamine test is for primary amines. Among the given peptides, only serine (Ser) has alcoholic group.

4.

(b) Lysine

5.

(c)

Note that lp of electrons on N labelled as N(a) are involved in delocalisation, hence not available for protonation. 6. (c) Structure of the given α-amino acids are:

Here, aspartic acid is acidic, glycine is neutral, while lysine and arginine are basic amino acids. Also, arginine is more basic due to stronger basic functional groups. The order of pKa value is directly proportional to the basic strength of amino acids, thus Arg > Lys > Gly > Asp. 7. (a)

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8. 9. 10.

(b) The N atom of amide is not basic so it will not exist in zwitter ionic form at pH 7.0. (b) Those amino acids that cannot be synthesized in our body and must be supplied in diet is called essential amino acid for ex. valine, histidine, isoluecine etc. (c) Ninhydrin is often used to detect ∝ – amino acids and also free amino and carboxylic acid groups on proteins and peptides. When about 0.5 mL of a 0.1% solution of ninhydrin is boiled for one or two minutes with a few mL of dilute amino acid or protein solution, a blue color develops. Ninhydrin degrades amino acids into aldehydes, ammonia, and CO2 through a series of reactions

to produce an intensely blue or purple pigment, sometimes called Ruhemann's purple. 11. (b) Glycosidic linkage is actually an ether bond as the linkage forming the rings in an oligosaccharide or polysaccharide is not just one bond, but the two bonds sharing an oxygen atom e.g. sucrose

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12.

(4)

* Represents chiral centre. 13. (2)

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No. of chiral centres = 2 14. (5) Asp – Glu – Lys tripeptide is :

No. of CO group = 5 15. (5) |z1| + |z2| + |z3| = 5

At pH = 2

and

of tyrosine and lysine are positively charged (+1

each) + 2 |z1| = 2 At pH = 6, NH2 of Lysine (+1), COOH (–1) of glutamic acid, so because of dipolar ion, |z2| = 0 At pH = 11, COOH of glutamic acid (–1) COOH of lysine (–1) OH of phenol ( –1) ∴ | z3 | = 3 16.

(1) The given structure has peptide bonds which on hydrolysis gives different amino acids, namely A, B, C and D. Out of these only A is

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naturally occurring which is glycine.

17.

(4) According to question, C – terminal must be alanine and N – terminal do have chiral carbon means it should not be glycine. So possible sequence is : Val Phe Gly Ala Val Gly Phe Ala Phe Val Gly Ala Phe Gly Val Ala NOTE: The –COOH group of alanine does not participate in forming peptide bond as given in the question. 18. (4) Peptides with isoelectric point (pI) more than 7 would exist as cation in neutral solution (pH = 7) which means the given polypeptide is of basic nature, so it must contain two or more amino groups. Hence IV, VI, VIII and IX are correct options. 19. (6) Molecular weight of decapeptide = 796 g/mol Total bonds to be hydrolysed = (10 – 1) = 9 per molecule

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Total weight of H2O added = 9 × 18 = 162 g/mol Total weight of hydrolysis products = 796 + 162 = 958 g Total weight % of glycine (given) = 47% Total weight of glycine in product Molecular weight of glycine = 75 g/mol Number of glycine molecules = NOTE: A dipeptide has one peptide bond. Thus, a decapeptide has 9 peptde bonds. 20.

(2) The basic groups in the given form of lysine is NH2 (not and

)

.

21. (37.84) Molecular formula of histamine is C5H9N3 Molecular mass of histamine = 5 × 12 + 9 × 1 + 3 × 14 = 111 Mass percentage of nitrogen in histamine = 22.

× 100 = 37.84%

The structure of the two possible dipeptides are

NOTE: A dipeptide structure has one

linkage.

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23.

Aspartame (i) Four functional groups present in it are (a) –NH2 (Amine) (b) –COOH (Carboxylic acid) (c) (d) (ii) Zwitterion structure is given as follows :

(iii)

Hence, on hydrolysis two amino acids (a) and (b) are obtained. (iv) Of the

above two amino acids,

is more

hydrophobic due to presence of non-polar and bulky benzyl group.

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24.

TIPS/Formulae : pH = 2 indicates acidic character whereas pH = 10 indicates basic character.

1.

(b) RNA has a single helix structure, whereas, DNA has a double helix structure. (b) RNA contains adenine (A), guanine (g), cytosine (c) and uracil (u). is cytosine, found in RNA and DNA, both. is uracil, found in RNA only. is thymine, found in DNA only. is not a pyrimidine, i.e., nat a constituents of RNA and DNA. (a) (a) Among 20 naturally occuring amino acids "cysteine" has '– SH' or thiol functional group. ⇒ General formula of amino acid

2. (1) (2) (3) (4) 3. 4.

⇒ Value of R = –CH2–SH in cysteine. 5. (c) Water soluble vitamins dissolve in water and are not stored by the body. The water soluble vitamins include the vitamin B-complex group and vitamin C. 6. (a) DNA contains ATGC bases Quinoline is not present in DNA or RNA. 7. (d) The two polynucleotide chains of DNA molecules are twisted around a common axis but run in opposite directions to form a right handed helix. The two chains are joined together by specific hydrogen bonds. Adenine-Thymine (two hydrogen bonds) and cytosine-guanine (three hydrogen bonds)

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8. 9. 10.

(c) Deficiency of vitamin D causes rickets. (d) The correct structure of thymine is option (d). (a) Maltose is obtained by partial hydrolysis of starch by the enzyme diastase present in malt. 2(C6H10O5)n + nH2O nC6H12O6 (maltose)

11.

(c) Vitamins Vitamin B1 (thiamine)

Deficiency Diseases Beriberi

Vitamin B2 (riboflavin) Cheilosis Vitamin B6 (pyridoxine) Convulsions Vitamin C (ascorbic acid) Scurvy

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1.

2.

(b) Buna-N is obtained by addition polymerisation.

Nylon-6, bakelite, nylon-6. 6 are obtained by condensation polymerisation. (a) Bakelite is an example of thermosetting polymer. PVC and Nylon 6 are thermoplastic and Buna-N is an elastomer.

3.

(d)

4.

(d) Nylon –6, 6 is obtained by condensation polymerisation of hexamethylenediamine and adipic acid.

So, nylon-6, 6 is a condensation polymer. Other given polymers are addition polymers. 5. (c) Nylon-6 is prepared from caprolactam.

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6.

(d) Teflon or polytetrafluoroethene is an addition polymer and its monomer is alkene derivative (CF2 = CF2) whereas terylene, melamine and nylon-6, 6 are condensation polymers.

7.

(a) Except dacron (terylene) all are additive polymers. Terephthalic acid condenses with ethylene glycol to give dacron.

HO – CH2 – CH2 – OH Ethylene glycol

8.

(d) Polymers which change irreversibly into hard and rigid material on heating are known as thermosetting polymers e.g bakelite.

9.

(b) Nylon is a polyamide fibre. It is prepared by the condensation polymerisation of adipic acid (HOOC.(CH2)4COOH) and hexamethylene diamine (H2N.(CH2)6.NH2).

10.

(c) Nylon is an aliphatic polyamide.

11. (i)-(f)-6; (ii)-(c)-2; (iii)-(d)-4; (iv)-(b)-1; (v)-(a)-5; (vi)-(e)-3. (i) Asbestos was used as an insulator I solid propellant rocket motors. (ii) Fluorocarbons were used as refrigerant which are carcinogens. (iii) Lithium metal an easily donate electron and therefore, can be used as a reducing agent. (iv) Nitric oxide is paramagnetic and an air pollutant.

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(v)

Zeolites have high porosity due to the presence of cavities of molecular dimensions. They are used as ion exchange for softening of water.

(vi) Zinc oxide is used as a semi-conductor. They have property of fluorescence.

1.

(d)

and

Both monomers of PHBV have chiral centre. 2. (c) Formation of Bakelite follows electrophilic substitution reaction of phenol with formaldehyde followed by dehydration.

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3.

4.

(d)

(b) The polymer

is urea-formaldehyde resin.

Its monomers are urea and formaldehyde. 5. (d) As the name of the polymer indicates, it is a copolymer of following two monomers.

6. 7.

(a) Adipic acid (having 6C) and hexamethylene diamine (having 6C). (c)

8.

(d)

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9.

(d) Mechanism for the formation of major product is as follows:

10.

(b) The statement (b) is not true. Chain growth polymerisation (or addition polymerisation) involves homopolymerisation only. Examples of such polymers include polythene, orlon and teflon.

11.

(a) Formation of nylon-6 involves hydrolysis of caprolactam, (its monomer) in initial state.

12.

(a)

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13.

(b) High density polythene is used in the manufacture of housewares like buckets, dustbins, bottles, pipes etc. Low density polythene is used for insulating electric wires and in the manufacture of flexible pipes, toys, coats, bottles etc.

14.

(a)

15. 16.

(d) Glyptal is used in the manufacture of paints and lacquers. (c) Chloroprene is a monomer of neoprene.

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Neoprene

17.

(b)

18.

(c) Phenol and formaldehyde undergo condensation polymerisation under two different conditions to give a cross linked polymer called bakelite.

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19.

(d) Polymethyl methacrylate is hard, fairly rigid. It is used for optical lenses. 20. (a) Degree of polymerization of Ist polymer = 100 Degree of polymerization of IInd polymer = 200 N1 = 2; N2 = 3 Molar mass of monomer (ethylene) of polythene = 28g/mol Now, Degree of polymerization is given as:

Thus, Molar mass of Ist polymer (M1) = 100 × 28 = 2800 g/mol Molar mass of IInd polymer (M2) = 200 × 28 = 5600 g/mol Now, the weight average molecular weight is given as:

21.

= 4900 g/mol (d) Nylon and cellulose, both have intermolecular hydrogen bonding, polyvinyl chloride has dipole-dipole interaction, while

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natural rubber has van der Waal forces which are weakest. 22. (6) Structure of melamine is as follows :

Total no. of lone pairs of electron is ‘6’. 23. cross-linking; 24. (a, c) (a) Nylon-6. It is obtained by heating caprolactam with water at high temperature and have amide linkage.

(b) Cellulose has only β-D-glucose units that are joined by glycosidic linkages between C–1 of one glucose unit and C–4 of the next glucose unit. (c) Teflon is prepared by heating tetrafluoroethene in presence of a persulphate catalyst at high pressure.

(d) Natural rubber is a linear polymer of isoprene (2-methyl-1, 3butadiene) containing cis alkene units. It is also called cis-1, 4polyisoprene. 25. (a,b,c,d) Condensation polymers are formed by condensation of a diol or diamine with a dicarboxylic acid.

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Hence, X may be

or

or

(b)

(c) H2N — (CH2)4 — NH2 Polyamide (d) N ≡ C – (CH2)4 – C ≡ N H2NCH2 – (CH2)4 – CH2NH2

26.

(c) (A) Natural rubber - Polymer of isoprene

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(B) Neoprene - Polymer of chloroprene (C) Buna N - Polymer of 1,3-butadiene and acrylonitrile (D) Buna S - Polymer of 1,3-butadiene and styrene 27.

(d) a. HDPE

- Ziegler-Natta catalyst

b. Polyacrylonitrile - Peroxide catalyst c. Novolac d. Nylon-6

- Catalysed by acid or base - Condensation at high T and P

28.

(d) Polymer Polystyrene Glyptal P.V.C. Bakelite 29. (A) : (p) and (s)

Use Manufacture of toys Paints and lacquers. Rain Coats Computer discs Cellulose is a natural polymer and has a C1 – C4 βglycosidic linkage. (B) : (q) and (r) Nylon-6, 6 is a synthetic polymer of hexamethylenediamine and adipic acid and has amide linkages. (C) : (p) and (r) Proteins are natural polymers of α amino acids joined by amide linkages (peptide bonds). (D) : (s) Sucrose is a disaccharide of α-D glucose and β-Dfructose and has an α, β-glycosidic linkage. 30. (c) High density polythene is formed when addition polymerisation of ethene takes place in the presence of a catalyst (Ziegler Natta catalyst) at a high temp. (333K to 349 K) and under low pressure (6-7 atm). High density polyethene is hard and chemically inert due to close packing thats why used to make buckets and dustbins. 31. (a)

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32.

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1. 2. 3.

4.

(b) If the level of noradrenaline is low, then the person suffers from depression. In such situations, anti-depressant drugs are required. (d) Synthetic drugs, brompheniramine (dimetapp) acts as antihistamines. (d) Seldane is an anti-histamine drug that has inhibitory action on histamine receptor.

(d)

Novestrol has phenolic, alcoholic and terminal alkyne groups, so it can react with Br2 water, ZnCl2l HCl as well as FeCl3. 5. (b) Noradrenaline, also called Norepinephrine, is a neurotransmitter. It plays a major role in mood change. 6. (d) Structure of histamine

Blood is slightly basic in nature (7.35 pH). At this pH, terminal NH2 will get protonated due to more basic nature. ∴ Predominant structure of histamine is

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7.

10.

(c) Enzyme inhibition can be either reversible or irreversible. In case of irreversible inhibition the inhibitor dissociates very slowly from its target enzyme as it has tightly bound to the enzyme either covalently or non-covalently. Some irreversible inhibitors are important drugs like Penicilline and Aspirin. Thus "drug induced poisoning" may bound irreversibly to the active site of the enzyme. (a) Bactericidal are the drugs that kills bacteria. Ofloxacin works by stopping the growth of bacteria. This antibiotic treats only bacterial infections. (d) Alitame is an artificial sweetner that is 2,000 times as sweet as sugar. (a) Phenelzine is an antidepressant, while others are antacids.

11.

(a)

12. 13. 14. 15.

(c) Diphenhydramine is used as antihistamine. (a) The aminoglycosides are among the oldest antibiotics. (d) Aspirin is an non-narcotics analgesic. (b) The term “antihistamine” refers only to H1 antagonists, which is also known as H1-receptor antagonists and H1-antihistamine. (5)

8.

9.

16.

is sucralose.

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(* = Chiral carbon)

17.

(3.00)

There are three chiral centres in penicillin. 18. (2.00)

19.

(9.00)

Structure of aspartame is shown above. It is a methyl ester of dipeptide formed from aspartic acid and phenylalanine. sp2 hybridised carbon atoms are shown by the star mark in the structure. 20. (b)

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21.

(a) A – Aspartame

B – Saccharine

C – Sucralose

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D – Alitame

(I)

A and D give positive test with ninhydrin because both have free carboxylic and amine groups. (II) C form precipitate with AgNO3 in the lassaigne extract of the sugar because it has chlorine atoms. (III) B and D give positive test with sodium nitroprusside because both have sulphur atoms. 22. (b) (a) Norethindrone is a progesterone derivative, so it is an antifertility drug (Q). (b) Ofloxacin is an antibiotic (P). (c) Equanil is a mild tranquilizer and hence used in hypertension (R). 23. (a) Allosteric Molecule bind to a site other than the active site of enzyme Competitive Molecule bind to the active site inhibitor of enzyme Receptor Molecule crucial for communication in the body. Poison Molecule binding to the enzyme covalently

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24.

(b) → Phenelzine contains hydrazine → Chloroxylenol contains phenol → Uracil is the pyrimidine base → Ranitidine contains furan ring

1.

(d) Sodium lauryl sulphate (C11H23CH2OSO3– Na+) is an anionic detergent. Glyceryl oleate is a glyceryl ester of oleic acid. Sodium stearate (C17H35COO–Na+) is a soap. Cetyltrimethyl ammonium bromide is a cationic detergent.

2.

(ii)

(i)

Animal charcoal/C, Ca3(PO4)2 /sugar refining; Invar/Fe, Ni/watch spring; Nichrome/Co, Ni/heating element; –1 Rydberg/cm /109677; Stainless steel/Fe, Cr, Ni, C/cutlery; Boltzmann/kJ deg–1/1.3805 × 10–26 Friedel-Craft/Lewis acid/anhydrous AlCl3; Fermentation/yeast/ethanol; Dehydrohalogenation/alcoholic alkali/alkene; Sandmeyer/cuprous chloride/chlorobenzene; Saponification/oil/soap.

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1.

5.

(d) In this acid base titration bunsen burner and measuring cylinder are of no use while other laboratory equipments will be required i.e., phenol phthalein, burette and pipette. (c) In toilet cleaning liquid, the main constituent is HCl, which can cause skin burn, so it should be treated with NaHCO3, which can easily neutralise the acid. (b) Oxalic acid is a primary standard solution whereas H2SO4 is a secondary standard solution. So it does not matter whether oxalic acid is taken in a burette or in conical flask. Therefore accurate measurement of concentration by titration depends on the nature of the solution. (d) Sodium metal on dissolution in liquid ammonia gives a deep blue solution due to the ammoniated electrons. (c)

6.

(c) pH range for methyl orange is

2.

3.

4.

7. 8.

Generally, weak bases have pH greater than 7. When methyl orange is added to a weak basic solution, solution becomes yellow. This solution is then titrated against a strong acid, at the end point pH will be less than 3.1. ∴ Solution becomes pinkish red. (c) Oxalic acid is used as a primary standard for NaOH standardizing. (d) In presence of HCl

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The above reaction path is confirming the presence of Zn2+ ion. 9. (d) Region 2 (blue flame) will be the hottest region of Bunsen flame as shown in given figure. 10. (c) Co2+ ion is precipitated by H2S in presence of NH4OH which is a group reagent of group IV in cationic analysis. 11. (a) If Na2CO3 is used in place of (NH4)2CO3. It will precipitate group V radicals as well as magnesium radicals. The reason for this is the high ionization of Na2CO3 in water into Na+ and CO2–3. Now the higher concentration of CO23 – is available which exceeds the solubility product of group V radicals as well as that of magnesium radicals. 12. (d) The group reagent of fourth group is ammoniacal H2S by which Zn2+ ion will be precipitated as ZnS, whereas Fe3+ ion and Al3+ ions will be precipitated as hydroxides. 13. (a) In presence of acid, ionisation of H2S is supressed, so less number of S2– are furnished. Hence, only those sulphides are precipitated which has low solubility product (Ksp); thus only CuS and HgS are precipitated. 14.

(b)

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15. (b) Precipitate of Zn(OH)2 formed at initial stage dissolves in excess of NH4OH due to the formation of tetrammine Zn (II) complex.

Note : The precipitate Al(OH)3 is not soluble in NH+4 solution because it does not form a complex compound. 16.

(b)

The metal ion is Bi3+. 17. (c) TIPS/Formulae : For precipitation, Ionic product > solubility product HgS having the lowest Ksp among the given compounds, will precipitate first. 18. (a) SO2 and H2S, both being reducing agents, can turn acidified dichromate solution green. SO2 can be obtained by the action of acid upon sulphite, while H2S is evolved by the action of acid upon sulphide. However, SO2 has a burning sulphur smell which is irritating. H2S has rotten egg like smell.

19.

(c) Since the saturated aqueous solution of (X) give white ppt with AgNO3, so (X) may be Cl2. Hence

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20.

(d)

21.

(d) Sodium salts are highly soluble. Cu2+ belongs to the IInd group in salt analysis and is precipitated as CuS, whereas Zn2+ belongs to the IV group and is precipitated as ZnS after CuS because of higher Ksp of ZnS. 22. (c) Only PbCl2 and Hg2Cl2 will precipitate as Pb2+ and Hg22+ as first group basic radicals. This is because their solubility product is less than that of other radicals. Note : dil. HCl is the first group reagent. 23. (a) Note : The ions of group II of salt analysis are precipitated by HCl and H2S whereas members of group IV are precipitated by H2S in alkaline medium. Bi3+ and Sn4+ both belong to group II They will be precipitated by HCl in presence of H2S. Both Bi3+ and Sn4+ belong to group II of qualitative inorganic analysis and will get precipitated by H2S. 24. (d) Sn2+ can be precipitated by H2S but not by HCl. 25. (7) All except MnS (buff coloured) and SnS2 (yellow) are black in colour. 26. (0.10)No. of eq. of oxalic acid = No. of eq. of NaOH Using average of the NaOH volume used in five experiments (VNaOH)avg. = 10 mL

MNaOH = 0.10 mol/L 27. (126) The balanced equations are 2MnCl2 + 5K2S2O8 + 8H2O →

2KMnO4 + 4K2SO4 + 6H2SO4 + 4HCl 2KMnO4 + 5H2C2O4 + 3H2SO4 →

...(1)

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K2SO4 + 2MnSO4 + 8H2O + 10CO2 Mass of oxalic acid added = 225 mg Milimoles of oxalic acid added =

...(2)

= 2.5

From equation (2) Milimoles of KMnO4 used to react with oxalic acid = 1 (5 m mole H2C2O4 2 m mole of KMnO4) and milimoles of MnCl2 required initially = 1 ∴ Mass of MnCl2 required initially = 1 × 126 = 126 mg (Molar mass of MnCl2 = 126) 28. Chromyl chloride (CrO2Cl2). 29. Fe3+; Without oxidation with HNO3, the Fe2+ ions present would not be converted into Fe3+. So, Fe(OH)2 will not be precipitated as its solubility product is higher than that of Fe(OH)3 and as NH4Cl suppresses the ionisation of NH4OH, this solubility product is not reached. 30. (True) : Ksp of CuS is less than Ksp of ZnS. On passing H2S in acidic medium, the dissociation of H2S is suppressed due to common ion effect and it provides [S2–] which is just sufficient to cross over Ksp of CuS and not Ksp of ZnS. Thus, only CuS gets precipitated. 31. (True) : Function of ammonium chloride is to suppress the ionisation of NH4OH and thus check the precipitation of Mg(OH)2 because the solubility product of Mg(OH)2 is high. This is used in salt analysis when 3rd group radicals are precipitated. The group reagent are NH4OH in presence of NH4Cl. 32. (b, c, d) Aqua regia is a mixture of conc. HCl and conc. HNO3 in 3:1 ratio. When gold dissolves in aqua regia, the species formed in which gold is in +3 oxidation state. In the absence of air the reaction between gold and aquaregia NO2 will not be produced. Yellow colour is due its decomposition into NOCl and Cl2.

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33.

(b, d) (a) Cu2+ shows characteristic green colour whereas Mn2+ shows yellowish green colour in flame test. Thus, they cannot be distinguished by flame test. (b) Only Cu2+ can give black precipitate of CuS in acidic medium on passing H2S. Mn2+ forms ppt. in ammonical medium. (c) Both Cu2+ and Mn2+ show the formation of precipitate by passing H2S in faintly basic medium. as per electrochemical

(d) 34.

series. (a)

35. (c, d)Only group II cations precipitate as sulphide with H2S in acidic medium that is (Cu2+, Pb2+) and (Hg2+, Bi3+) 36. (b, c) The blue precipitate of Fe2+ ions with potassium ferricyanide is due to formation of Turnbull’s blue KFell [Felll (CN)6] K. Fell [Felll (CN)6] + 2K+ Fe2+ + K3 [Fe(CN)6] Potassium ferro ferricyanide

The red colouration of Fe3+ ions with potassium thiocyanate is due to the formation of [Fe(CNS)3] + 3K+

Fe2+ + 3KCNS

37. (a, b, d) The reactions are 4NaCl + K2Cr2O7 + 6H2SO4 2CrO2Cl2 + 4NaHSO4 + 2KHSO4 + 3H2O (Red vapours)

CrO2Cl2 + 4NaOH chromyl

Na2CrO4 + 2NaCl + 2H2O yellow

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chloride

solution

38.

3+

39.

(a, b) Al (third group radical) and Ca2+ (fifth group radical) precipitate out as their hydroxide with NH4Cl and aq. NH3 (NH4OH) which are the group reagents. (d) [ZnCl4]2– and [Zn(CN4)]2– are tetrahedral complexes.

40.

(b)

41. (d) The filtrate on treatment with ammoniacal H2S gives a precipitate which dissolves in aqueous NaOH containing H2O2 giving a coloured solution. It contains Cr3+ ion.

42. (a) Lead salts give white precipitate of PbCl2 with dil. HCl which is soluble in hot water. Pb++ + 2Cl– ———→ PbCl2 (White ppt) soluble in hot water (P) Sol. (43-45) : Reaction of (Y) indicates that it is Fe3+ salt.

NOTE : Since the product formed (methylene blue) has sulphur in its structure, it should be supplied by the compound (X) which is thus Na2S.

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43. 44. 45. 46.

47.

48.

(d) (c) (b) (b) Sulphate is estimated as BaSO4 because of its insolubility in water. BaSO4 forms a white ppt. Therefore, reason is correct but do not explain the assertion. (b) Cd2+ is a 2nd group radical and Ni2+ is a 4th group radical. So, solubility product of NiS has to be more than CdS. Further, Cd2+ gives yellow colour of CdS with H2S, but Ni2+ gives black colour of NiS with H2S. Hence, assertion is correct and statement is wrong. ; ;

;

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49.

Let us summarise the given facts.

Note : Reaction of compound X with NaOH solution and subsequent treatments indicate that X has NH+4 radical. On the other hand, reaction of X with K2Cr2O7 solution, conc. H2SO4 and subsequent treatments indicate that A has Cl– radical. Thus, compound X is NH4Cl which explains all the above reactions.

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+ 7 KI + 2 H2O

50.

The above set

leads to following conclusions. (i) Since, the gas (B) is colourless and turns acidified K2Cr2O7 solution green, it should be H2S. (ii) Since, H2S gas is obtained by the reaction of dil. H2SO4 on (A), the latter must be sulphide. (iii) The white colour of the sulphide (A) points out towards ZnS. Thus, the various reactions can be written as given below.

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51.

Salt of Co

52.

CoO + gas

Summary of the given facts

The reaction corresponds to copper sulphate.

(Yellow solution) 53. Sodium iodide on reaction with HgI2 gives colourless complex salt, Na2[HgI4]. Na2[HgI4] HgI2 + 2NaI Colour is due to presence of residual HgI2. But on addition of excess NaI, it becomes colourless due to change of residual HgI2 into Na2[HgI4]. Na2[HgI4] (colourless) HgI2 + 2NaI (excess) The orange colour of HgI2 reappears due to conversion of Na2[HgI4] into HgI2 by means of NaOCl

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3Na2HgI4 + 2NaOCI + 2H2O

3HgI2

+ 2NaCl + 4NaOH + 2NaI3

Orange colour

54.

(i) The substance is Na2O2. When dissolved in water, the solution becomes alkaline with the liberation of H2O2 2NaOH + H2O2 Na2O2 + 2H2O Due to the alkaline solution, the red litmus paper will turn into blue, which subsequently changes into white due to oxidation caused by H2O2. (ii) The substance Na2O merely produces alkaline solution and thus the red litmus paper will turn into blue. 2NaOH Na2O + H2O Note : Upon dissolving in water, peroxides can produce hydrogen peroxide which is a bleaching agent. 55. The solubility products of CuS and ZnS are Ksp(CuS) 10-38 and Ksp (ZnS) 10-22 Since Ksp (CuS)