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Home Explore MTG Editorial Board - ScoreMore 21 Sample Papers For CBSE Board Exam 2021-22 - Class 12 Chemistry-MTG Learning Media Pvt Ltd (2020)

MTG Editorial Board - ScoreMore 21 Sample Papers For CBSE Board Exam 2021-22 - Class 12 Chemistry-MTG Learning Media Pvt Ltd (2020)

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Description: MTG Editorial Board - ScoreMore 21 Sample Papers For CBSE Board Exam 2021-22 - Class 12 Chemistry-MTG Learning Media Pvt Ltd (2020)

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CHEMISTRY MTG Learning Media (P) Ltd. New Delhi | Gurugram

Price : ` 200 Revised Edition : 2021 Published by : MTG Learning Media (P) Ltd., New Delhi Corporate Office : Plot 99, Sector 44 Institutional Area, Gurugram, Haryana-122003 Phone : 0124 - 6601200. Web : mtg.in Email : [email protected] Registered Office : 406, Taj Apt., Ring Road, Near Safdarjung Hospital, New Delhi-110029 Information contained in this book has been obtained by mtg, from sources believed to be reliable. Every effort has been made to avoid errors or omissions in this book. In spite of this, some errors might have crept in. Any mistakes, error or discrepancy noted may be brought to our notice which shall be taken care of in the next edition. It is notified that neither the publishers nor the author or seller will be responsible for any damage or loss of action to anyone, of any kind, in any manner, therefrom. © MTG Learning Media (P) Ltd. Copyright reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the Publisher. All disputes subject to Delhi jurisdiction only. Visit www.mtg.in for buying books online.

PREFACE With each passing year CBSE Board incorporates some changes in the pattern of Board question papers to prepare the students for the competitive world ahead. These changes aim to discourage students from rote learning, develop their analytical skills and reasoning abilities to produce better results and to improve the academic quality of institutions. This year’s exam pattern includes Passage based (MCQs and Assertion & Reason), MCQs, Assertion & Reason, SA-I, SA-II and LA type questions. To keep you aligned with the latest pattern MTG brings you the completely revised edition of Score More 21 Sample Papers. These sample papers are prepared keeping in mind the various typology of questions like remembering, understanding, applying, analysing, evaluating and creating type. MTG Score More 21 Sample Papers are prepared, keeping all these aspects in mind. Solving more and more sample papers will increase your problem-solving speed and accuracy. By solving these sample papers, you can check your preparation level, and your strong and weak areas too. You can score more in the actual paper by getting to know your weak areas and working even harder on them. As you start attempting SQPs one by one your percentage scoring will improve. Your aim should be to reach the top position. After attempting all SQPs in exam-like environment the average score of 21 SQPs will give you the score which you are most likely to score in your actual exam. That is what MTG Guarantee. Practising these SQPs will definitely equip you to face your CBSE Board Class 12 exams with more confidence. Solutions of all SQPs are given as per the CBSE marking scheme. In MCQs section, students need to write only answer keys. Explanations for these MCQs are given for better understanding. The Self Evaluation Sheet provided after each Sample Question Paper (SQP) will help you to assess your performance. The performance analysis table will help you to check where you actually stand. Practice done with a proper planning promotes a person for a better performance. It is very necessary to practice in the right direction under a well guided guidance to reach to the goal. Practice means repeating an activity in the right direction which can sharpen the talent. Visit our website www.mtg.in to download the additional content associated with this book. Wishing our readers all the very best! MTG Editorial Board

MTG’s VALUABLE TIPS for BOARD EXAMS Exams are the part and parcel of our learning process. But most students suffer from exam phobia. They may be excellent at their work but when confronted by real exams, often they under-perform showing exam related stress. So, it’s time to strategize your learning and come out of the fear cropping up in your mind by combining self-confidence with disciplined study and by simply following these tips : GET SET STARTED Now with the ending session you are close to completing your syllabus; it is time for effective revision of the relevant content for better results. Preparing a well-planned timetable is the need of the hour. Timetable should be judiciously planned to give enough time to different subjects. It should have scope of learning, practising as well as recapitulation. DECK OUT WITH THE RIGHT TITLE Currently the market is flooded with many titles in different subjects. It calls for your wise decision to choose the right title. MTG gaining the trust of over 10 million readers across the country, now presents Score More 21 SAMPLE PAPERS covering all objective and subjective type of questions. The book in itself is a complete package and provides an opportunity to all the readers to assess their progress as it includes questions based on the latest changed pattern of CBSE exams. STRIVE FOR EXCELLENCE Start solving the sample papers provided in the book in an environment same as the real exam. Once you are done, check your answers to evaluate your learning. This will help you become aware of both your strong and weak areas. Work harder on your weak areas and repeat the process. You will master all concepts soon. IT’S SHOW TIME Remember, there is no shortcut to success. However, these tips would surely help you a lot, if followed truly for at least a month prior to the exams. Organising your studies like this will help you deliver your best during exams and there is no doubt that you’ll reap richer benefits. So, gird up your lions and give your best shot!!!

1SQP S Q P BLUE PRINT Time Allowed : 3 hours Maximum Marks : 70 S. No. Chapter Passage based/ SA-I SA-II LA Total MCQs/A & R (1 mark) (2 marks) (3 marks) (5 marks) 1. The Solid State 1(1) 1(2) – – 2. Solutions 3. Electrochemistry 1(4) 1(2) – – 4. Chemical Kinetics 5. Surface Chemistry 1(1) 1(2) – 1(5) 11(23) 6. The p-Block Elements 7. The d- and f-Block Elements 1(1) 1(2) – – 8. Coordination Compounds 9. Haloalkanes and Haloarenes 1(1) 1(2) – – 10. Alcohols, Phenols and Ethers 11. Aldehydes, Ketones and Carboxylic Acids 1(1) 1(2) – 1(5) 12. Amines 13. Biomolecules 1(1) – 2(6) – 8(19) Total 1(1) – 1(3) – 2(2) 1(2) 1(3) – 1(1) 1(2) – – 2(5) 1(2) – – 14(28) 1(1) – – 1(5) 2(2) – 1(3) – 16(22) 9(18) 5(15) 3(15) 33(70)

Subject Code : 043 SQP-1 CHEMISTRY Time allowed : 3 hours Maximum marks : 70 General Instructions : Read the following instructions carefully. (a) There are 33 questions in this question paper. All questions are compulsory. (b) Section A : Q. No. 1 to 16 are objective type questions. Q. No. 1 and 2 are passage based questions carrying 4 marks each while Q. No. 3 to 16 carry 1 mark each. (c) Section B : Q. No. 17 to 25 are short answer questions and carry 2 marks each. (d) Section C : Q. No. 26 to 30 are short answer questions and carry 3 marks each. (e) Section D : Q. No. 31 to 33 are long answer questions carrying 5 marks each. (f) There is no overall choice. However, internal choices have been provided. (g) Use of calculators and log tables is not permitted. SECTION - A (OBJECTIVE TYPE) 1. Read the passage given below and answer the following questions : Aromatic and aliphatic aldehydes are selectively reduced to primary alcohols under mild conditions with potassium formate in the presence of Pd/C catalyst. The high selectivity obtained is attributed to the presence of the basic potassium bicarbonate generated in situ from the formate salt via the hydrogen transfer process. Molecular hydrogen shows similar activity in the presence of potassium carbonate salts. Ketones do not react under these conditions, therefore this method is used to separate aldehydes selectively from a mixture of aldehydes and ketones. The following questions are multiple choice questions. Choose the most appropriate answer. (i) The hydrogenation process mentioned in the study above cannot be used to convert (a) benzaldehyde to benzyl alcohol (b) acetaldehyde to ethanol (c) propanone to propanol (d) propanal to propanol. (ii) This method can be used to separate a mixture of (I) CH3CHO and CH3CH2CHO (II) CH3COCH3 and CH3CHO (III) CHO COCH3 and (a) Only I and II (b) Only II (c) Only III (d) Only II and III 2 Class 12

(iii) AfonrmorsgcaonmicpcooumnpdoBun(Cd 3AH(8CO3)H. 6BOr)eoacntstrweaittmh eHnBt rwtiothfopromtastshieumcofmorpmouantedinCt.hCe presence of Pd/C catalyst with Mg forms Grignard reagent D which reacts with A to form a product which on hydrolysis gives hexan-3-ol. Hence A and B are respectively (a) propanone, 2-propanol (b) propanal, propanol (c) propanone, propanol (d) propanal, propanone. OR An organic compound X forms 2,4-DNP derivative and reduces Tollen’s reagent. X undergoes Cannizzaro reaction to give Y and Z. X also gives Y on treatment with potassium formate in the presence of Pd/C catalyst. Hence X is (a) acetone (b) benzaldehyde (c) propanone (d) ethanol. (iv) Compound Y is prepared by the treatment of potassium formate in the presence of Pd/C catalyst with compound X. Compound Y on oxidation iwnitthhealpkraelsiennecKe MofnHO24SiOs 4c,oint vperrotdeducteos compound Z. When compound Z is heated with compound Y a compound having fruity smell. To which families the compounds X, Y and Z belong to? (a) aldehyde, carboxylic acid and alcohol respectively (b) ketone, alcohol and carboxylic acid respectively (c) ketone, carboxylic acid and alcohol respectively (d) aldehyde, alcohol and carboxylic acid respectively. 2. Read the passage given below and answer the following questions : Raoult first proposed linear relationship between mole fraction and vapour pressure in 1887 and Raoult’s law has become a paradigm in chemistry. For dilute solutions, this relation holds good but deviations grow as the solution becomes more concentrated. Dilute solutions which obey Raoult’s law over the entire range of concentration are called ideal solutions. In 1908, Calendar explained some of the deviations from Raoult’s model in aqueous solutions, solute molecules are hydrated such that some of the water is bound to the solute, so it does not contribute to the vapour pressure. Such kind of deviations from Raoult’s model can be positive deviation and negative deviation. In positive deviation, the total vapour pressure of solution is greater than corresponding vapour pressure expected in case of ideal solution. In such solutions, Dmixing H and Dmixing V are positive. In negative deviation, total vapour pressure of solution is smaller than the corresponding vapour pressure expected in case of ideal solution. In these solutions, Dmixing H and Dmixing V are negative. In these questions (Q. No. i-iv), a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. (i) Assertion : An ideal solution obeys Raoult’s law. Reason : In an ideal solution, vapour pressure of the solute and solvent and their corresponding activities in solution are proportional to their mole fraction. (ii) Assertion : Highly concentrated solutions occur in a wide range of natural processes exhibit non-ideal behaviour. Reason : Non-ideal solutions do not obey Raoult’s law. (iii) Assertion : Dilute solutions are taken as ideal solutions. Reason : The solutions which follow the Raoult’s law at any temperature and concentration are known as ideal solutions. Chemistry 3

(iv) Assertion : Solutions having positive deviation from Raoult’s law shows Dmixing H = +ve. Reason : Energy is required to break solute-solute or solvent-solvent attractive forces. OR Assertion : Solutions having negative deviation from Raoult’s law shows Dmixing V = +ve. Reason : Solute-solvent interactions are stronger than solute-solute and solvent-solvent interactions. Following questions (Q. No. 3-11) are multiple choice questions carrying 1 mark each : 3. The reaction of chloroform with alcoholic KOH and p-toluidine forms (a) H3C CN (b) H3C N2Cl (c) H3C NHCHCl2 (d) H3C NC OR O O NH2 HN CH3 The reagents needed to convert are (a) KOH, Br2; LiAlH4 (b) KOH, Br2; CH3COCl (c) HONO, Cu2Cl2; (CH3CO)2O (d) KOH, Br2; Ni, H2; CH3COCl 4. The cell emf is independent of concentration of the species of the cell in (a) Fe | FeO(s) | KOH(aq) || NiO(s) | Ni2O3 | Ni (b) (Pt)H2 | HCl | Cl2(Pt) (c) Zn | Zn2+ || Cu2+ | Cu (d) Hg | Hg2Cl2 | KCl || AgNO3 | Ag 5. The following reaction, Anhydrous + CO + HCl AlCl3/ CuCl CHO is known as (b) Cannizzaro reaction (a) Perkin reaction (d) Gattermann-Koch reaction. (c) Kolbe reaction 6. Which of the processes is being shown in the figure? (a) Electrodialysis (b) Dialysis (c) Electroosmosis (d) Electrophoresis OR Identify the statement which is correct w.r.t. surface phenomenon. (a) Osmotic pressure of rubber sol will be same as that of sucrose solution having same mass mixed in same mass of H2O. (b) A gas may show physisorption at low temperature and chemisorption at higher temperature. (c) Soap sol of sodium palmitate will be coagulated near cathode on electrophoresis. (d) Gold sol on mixing with starch sol causes stabilisation of starch sol. 4 Class 12

7. The d-block elements form coloured ions because these elements (a) cannot absorb the radiation in the visible region (b) involve d-d transitions which fall in the visible region (c) allow d-s transition (d) absorb other colours except those required for d-d transition. ‑ ‑ OR Lanthanoid contraction is due to (b) efficient shielding effect of f electrons (a) poor shielding effect by 4f electrons (d) both (a) and (c). (c) diffused shape of f-orbitals 8. In the reaction, A + 3B 2C, the rate of formation of C is (a) the same as rate of consumption of A (b) the same as the rate of consumption of B (c) twice the rate of consumption of A (d) 3/2 times the rate of consumption of B. 9. The end product (Q) in the following sequence of reactions is (a) (b) (c) (d) 10. Glucose molecule reacts with ‘X’ number of phenylhydrazine to yield osazone. The value of ‘X’ is (a) four (b) one (c) two (d) three. OR Glucose when treated with CH3OH in presence of dry HCl gas gives a- and b-methyl glucosides because it contains (a) a –CHO group (b) a –CH2OH group (c) a ring structure (d) five –OH groups. 11. The reaction conditions leading to the best yields of C2H5Cl are (a) C2H6 (Excess) + Cl2 UV light (c) C2H6 + Cl2 (Excess) UV light (b) C2H6 + Cl2 dark room temp. (d) C2H6 + Cl2 UV light In the following questions (Q. No. 12-16), a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. 12. Assertion : Reaction of SO2 and H2S in the presence of Fe2O3 catalyst gives elemental sulphur. Reason : In this reaction H2S is acting as a reducing agent. 13. Assertion : Glycine exists as zwitter ion but o- and p-aminobenzoic acid do not. Reason : Due to the presence of –NH2 and –COOH groups within the same molecule, they neutralise each other and hence a-amino acids exist as dipolar ions or zwitter ions. 14. Assertion : K2O has fluorite structure. Reason : O2– are present at the corners and at the centre of all six faces. K+ ions occupy all the tetrahedral voids. Chemistry 5

15. Assertion : In complex [Cr(NH3)4BrCl]Cl, the ‘spin only’ magnetic moment is close to 2.83 B.M. Reason: Mononuclear complexes of chromium(III) in strong field ligand have three unpaired electrons. 16. Assertion : Phenol is more reactive than benzene. Reason: In case of phenol, the intermediate carbocation is more resonance stabilised. OR Assertion : tert-Butylmethylether is not prepared by the reaction of tert-butylbromide with sodium methoxide. Reason : Sodium methoxide is a weak nucleophile.  SECTION - B The following questions, Q. No. 17-25 are short answer type and carry 2 marks each. 17. Calculate the cell emf at 25°C for the following cell : Mg(s) | Mg2+ (0.01 M) || Sn2+ (0.10 M) | Sn(s) [Given : E°(Mg2+ / Mg) = – 2.34 V, E°(Sn2+/Sn) = – 0.136 V, 1 F = 96500 C mol–1]. 18. Calculate the order of the reaction for the decomposition of N2O5 at 30°C from the following rate data. S. No. Rate of reaction (mol L–1 hr–1) Concentration of N2O5 (mol L–1) 1. 0.10 0.34 2. 0.20 0.68 3. 0.40 1.36 19. The freezing point of pure nitrobenzene is 278.8 K. When 2.5 g of unknown substance is dissolved in 100 g of nitrobenzene, the freezing point of solution is found to be 276.8 K. If the freezing point depression constant for nitrobenzene is 8 K kg mol–1, what is the molar mass of unknown substance? OR A solution of glucose (C6H12O6) in water is labelled as 10% by weight. What would be the molality of the solution? (Molar mass of glucose = 180 g mol–1) 20. (a) Draw the molecular structure of the noble gas species which is isostructural with ICl–4. (b) Why is dioxygen a gas but sulphur a solid? 21. Identify the reaction and write the IUPAC name of the product formed : COONa (b) CH3COOH (a) + NaOH → OR How will you convert the following : (i) Propanone to propan-2-ol (ii) Ethanal to 2-hydroxypropanoic acid? 22. Explain why the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride? 6 Class 12

23. An element has a face centred cubic (fcc) structure with a cell edge of 0.2 nm. Calculate its density in g cm–3 if 400 g of this element contains 4.8 × 1024 atoms. [NA = 6 × 1023 mol–1]        24. A colloidal solution of ferric oxide is prepared by two different methods as shown below : FeCl3 FeCl3 Hot NaOH water (A) (B) (a) What is the charge on colloidal particles in two test tubes (A) and (B)? (b) Give reasons for the origin of charge. 25. Write the mechanism of dehydration of alcohol. OR (i) Write the IUPAC name of the following compound. CH3 CH CH CH CH2OH OH OH Br (ii) Write the distinction test for ethyl alcohol and 2-propanol. SECTION - C Q. No. 26-30 are short answer type II carrying 3 marks each. 26. Write the IUPAC nomenclature of the following complex along with its hybridisation and structure. K2[Cr(NO)(NH3)(CN)4], m = 1.73 BM 27. (a) A solution of KOH hydrolyses CH3CHClCH2CH3 and CH3CH2CH2CH2Cl. Which one of these is more easily hydrolysed? (b) Write the major product(s) in the following : (i) 2CH3 CH CH3 Na Dry ether Cl (ii) CH3 CH2 Br AgCN (c) RCl is hydrolysed slowly to ROH but the reaction is rapid if a catalytic amount of KI is added to the reaction mixture. Explain. OR A compound is formed by the substitution of two chlorine atoms for two hydrogen atoms in butane. Write the structures of the possible isomers. Give the IUPAC name of the isomer which can exhibit enantiomerism. 28. Give reasons for the following observations : (i) Cu+ ion is not stable in aqueous solution. (ii) Mn(II) ion shows maximum paramagnetic character amongst the bivalent ions of first transition series. (iii) Iron has higher enthalpy of atomisation than that of copper. Chemistry 7

29. (a) Why is Cr2+ reducing and Mn3+ is oxidising in nature when both have d4 configuration?            (b) Explain how [Ti(H2O)6]3+ becomes purple coloured. 30. (a) On which carbon atoms glycosidic linkage is present in sucrose? (b) Name the bases that are common in both DNA and RNA. (c) Despite having an aldehyde group glucose does not give 2,4-DNP test. Why? OR What is essential difference between a-glucose and b-glucose? What is meant by pyranose structure of glucose? SECTION - D Q. No. 31-33 are long answer type carrying 5 marks each. 31. (a) (i) Express the relation between the conductivity (k) and the molar conductivity (Lm) of a solution. (ii) Electrolytic conductivity of 0.30 M solution of KCl at 295 K is 3.72 × 10–2 S cm–1. Calculate the molar conductivity. (b) (i) Solutions of two electrolytes ‘A’ and ‘B’ are diluted. It is found that Lm value of ‘B’ increases 2 times while that of ‘A’ increases 20 times. Which of the two is a strong electrolyte? (ii) A galvanic cell has E°cell = 1.1 V. If an opposing potential of 1.1 V is applied to this cell, what will happen to the cell reaction and current flowing through the cell? (iii) How will the pH of brine solution be affected on electrolysis? OR (a) Calculate the cell emf and DG° for the cell reaction at 25°C for the cell : Zn(s) | Zn2+ (0.0004 M) || Cd2+ (0.2 M) | Cd(s) E° values at 25°C : Zn2+/ Zn = – 0.763 V; Cd2+/Cd = – 0.403 V; F = 96500 C mol–1; R = 8.314 J K–1 mol–1. (b) If E° for copper electrode is 0.34 V, how will you calculate its emf value when the solution in contact with it is 0.1 M in copper ions? How does emf for copper electrode change when concentration of Cu2+ ions in the solution is decreased? 32. (a) Explain the following : (i) Sulphur exhibits tendency for catenation but oxygen does not. (ii) SF6 is not easily hydrolysed though thermodynamically it should be. Why? (iii) Why HF acid is stored in wax coated glass bottles? (b) What happens (i) When SO2 is passed through acidified solution of K2Cr2O7 ? (ii) Chlorine gas is passed into a solution of NaI in water? OR (a) Explain why NH3 is basic while BiH3 is only feebly basic? (b) Arrange the following in the order of property indicated against each set giving reason : (i) H2O, H2S, H2Se, H2Te – increasing acidic character (ii) HF, HCl, HBr, HI – increasing acid strength. 8 Class 12

(c) Complete the following reactions :   (i) NCuH+3 + NaOCl (ii) HNO3(dilute) 33. (a) An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br2 and KOH forms a compound ‘C’ of molecular formula C6H7N. Write the structures and IUPAC names of compounds A, B and C. (b) Write distinguish tests for the following pairs of compounds : (i) Aniline and ethylamine (ii) Ethylamine and dimethylamine OR (a) Account for the following : (i) Aniline does not undergo Friedel–Crafts reaction. (ii) It is difficult to prepare pure amines by ammonolysis of alkyl halides. (b) Account for the following : (i) Like ammonia, amines are good nucleophiles. (ii) Aryl amines are weaker bases than alkyl amines. (c) Arrange the following in the decreasing order of their basic strength in aqueous solutions : CH3NH2, (CH3)2NH, (CH3)3N and NH3 Chemistry 9

1. (i) (c) : Ketones are not reduced under these (iii) (b) : A solution that obeys Raoult's law is known conditions. as ideal solution. Raoult's law only work for ideal solutions. α   (ii) (d) : This method can be used to separate aldehydes selectively from a mixture of aldehydes (iv) (a) and ketones since ketones do not react under these conditions. OR (d) : Solutions having negative deviation from Raoult’s lfaowrcsehsoowfsinDtmeirxaincgtiVon=in–vtehebescoaluutsieonof, the (iii) (b) : A is an aldehyde since ketones do not increase the reduced to alcohols on treatment with potassium formate in the presence of Pd/C. solution will be held more tightly. Therefore, there will be decrease in volume on mixing. CH(C3C3HH62OCH) O HCOOK CH3(CCH3H2C8OH)2OH Pd/C 3. (d) : H3C NH2 + CHCl3 + 3KOH H3C NC + 3KCl + 3H2O (A) (B) HBr –H2O CH3CH2CH2Br OR NH2 OH (C) Mg O (b) : CH3COCl CH3—CH2—C—C3H7 CH3CH2CHO CH3CH2CH2MgBr NH2 NHCOCH3 H Hydrolysis (D) Br2/KOH Hexan-3-ol 4. (a) : Cell reaction is OR Fe(s) + 2OH− → FeO(s) + H2O + 2e− (b) Since the compound forms 2,4-DNP derivative Ni2O3(s) + H2O + 2e− → 2NiO(s) + 2OH− and reduces Tollen’s reagent, it must be an aldehyde. Since X undergoes Cannizzaro reaction, hence it has Fe(s) + Ni2O3(s) → FeO(s) + 2NiO(s) no -hydrogen. Hence, emf is independent of OH– ions. CHO CH2OH COO– N+a 5. (d) Conc. NaOH + Cannizzaro reaction 6. (a) : When an electric field is applied to purify an impure colloidal solution, the process is known (X) (Y) (Z) as electrodialysis. The ions present in the colloidal solution migrate out to the oppositely charged Pd/C HCOOK electrode. CH2OH OR (b) (Y) 7. (b) : This is due to d-d transition. When visible (white) light falls on transition metal compounds, they (iv) (d) : RCHO HCOOK RCH2OH alk. KMnO4 RCOOH absorb certain radiation of visible light for transition Pd/C of electrons from lower d level to higher d level and (X) (Y) (Z) transmit the remaining ones. The colour observed corresponds to complementary colour of the light RCOOH + RCH2OH H2SO4 RCOOCH2R absorbed. Ester OR (Fruity smell) 2. (i) (a) (d) (ii) (b) : In non-ideal solution, solute-solute as well as solvent-solvent interactions are not similar 8. (c) : Rate = − d[A] = − 1 d[B] = 1 d[C] to solute-solvent interaction. dt 3 dt 2 dt 10 Class 12

19. Given, w1 = 2.5 g         9. (c) : DTf = 278.8 – 276.8 = 2 K w2 = 100 g, Kf = 8 K kg mol–1 10. (d) : One glucose molecule reacts with 3 molecules of phenylhydrazine to give osazone. ⇒ ∆T f = Kf × w1 × 1000 M1 w2 OR 2.5 1000 (c) : Because of the ring structure, C1 in glucose ⇒ 2 = 8 × M1 × 100 becomes chiral and hence glucose exists in two stereoisomeric forms, i.e., a– and b– corresponding ⇒ M1 = 4 × 2.5 × 10 ⇒ M1 = 10 × 10 to each stereoisomeric form, glucose forms two \\ M1 = 100 g mol–1 methyl glucosides, i.e., a– and b– methyl glucosides. 11. (a) OR 12. (b) : Mass of solution = 100 g Mass of solute = 10 g Mass of solvent = 100 – 10 = 90 g 90 = 1000 kg = 0.09 kg Number of moles of solute, n = 10 = 0.055 mol 180 13. (b) 0.055 mol 14. (d) : K2O has antifluorite structure. m = 0.09 kg = 0.61 m 15. (d) : Cr3+ having 3d3 configuration always have 3 unpaired electrons with strong field as well as weak 20. (a) Structure of ICl4– : I in ICl4– field ligands with three unpaired electrons thus, the has four bond pairs and two lone magnetic moment is 3.83 B.M. pairs. Therefore, according to VSEPR theory, it should be square planar as 16. (a) shown. Here, ICl4– has (7 + 4 × 7 + 1) = 36 OR valence electrons. A noble gas species having 36 (c) : Sodium methoxide is a strong nucleophile. valence electrons is XeF4 (8 + 4 × 7 = 36). Therefore, like ICl4–, XeF4 is also square planar. 1Sn72.+(Maq)g+(s)2e– MgS2n+((as)q)(C+a2teh–o(dAicnhoadlifcrheaalcftiroenac)tion) (b) O2 molecules are held together by weak van Mg(s) + Sn2+(aq) Mg2+(aq) + Sn(s) der Waals’ forces because of the small size and high E°cell = E°Sn2+/Sn – E°Mg2+/Mg electronegativity of oxygen. = – 0.136 + 2.34 = 2.204 V Sulphur shows catenation and the molecule is made Ecell = Ec°ell − 0.0591 log 0.01 M up of eight atoms, (S8) with strong intermolecular 2 0.10 M attractive forces. Hence ,sulphur exists as solid at = 2.24 – (0.0295 × –1) = 2.2335 V room temperature. 18. From the given data, 21. (a) COONa + NaOH → = 0.10 mol L–1 hr–1 = × (0.34 mol L–1)n r1 = 0.20 mol L–1 hr–1 = k × (0.68 mol L–1)n Sodium benzoate + Na2CO3 r2 = 0.40 mol L–1 hr–1 = k × (1.36 mol L–1)n r3 k Benzene r1 = 0.10 mol L−1 hr−1 = k(0.34 mol L−1)n (b) r2 0.20 mol L−1 hr−1 k(0.68 mol L−1)n or 1 = 1 n ⇒ n=1 2  2  OR Chemistry 11

(i) CH3 C O + H2 Ni or Pt CH3 CHOH Step 2 : Formation of carbocation CH3 CH3 H H H3C C O+ H Slow H3C C+            PropanonePropan-2-ol (ii) HH H Step 3 : Formation of ethene H2O + H2C+ CH3 H2C CH2 + H3O+ OR (i) 2-Bromopentane-1,3,4-triol. (ii) Treat the compound with Lucas reagent 22. There are two reasons : (conc. HCl + anhy. ZnCl2) : 2-propanol gives turbidity (a) In case of chlorobenzene, carbon to which in 5 min. whereas ethanol gives no turbidity at room chlorine is attached is sp2-hybridised and is more electronegative than the corresponding carbon in temperature. Conc. HCl + cyclohexyl chloride which is sp3- hybridised. So the Anhy. ZnCl2 net dipole moment is lower in chlorobenzene. CH3CH2OH + HCl No reaction (b) In chlorobenzene C Cl bond has some double bond character so its bond length is smaller. Hence, CH3CHCH3 + HCl Conc. HCl + dipole moment is lower than cyclohexyl chloride Anhy. ZnCl2 which has a longer C Cl single bond. OH CH3—CH—CH3 + H2O Cl Turbidity appears in 5 minute 23. Given, edge length a = 0.2 nm = 2 × 10–8 cm 26. µ = n (n + 2) = 1.73 which gives n = 1 This means tithiastpcrhesreonmt iausmCri+oonr has one unpaired Mass of substance = 400 g electron, i.e., Cr (I). This implies Number of atoms = 48 × 1023 atoms 1023 that NO is present as nitrosonium ion. Hence, the 48 4 atoms name will be potassium amminetetracyanonitrosonium \\ Number of unit cells = × = 12 × 1023 chromate (I). \\ Volume of one unit cell = a3 = (2 × 10–8 cm)3 Cr+ : 3d 4s = 8 × 10–24 cm3 \\ Total volume = Number of unit cells In the complex, as there is only one unpaired electron × Volume of one unit cell and coordination number is 6, it will become = 12 × 1023 × 8 × 10–24 = 9.6 cm3 Cr+ : \\ Density = Mass = 400 g = 41.7 g cm −3 d2sp3 hybridisation Volume 9.6 cm3 i.e., it will undergo d2sp3 hybridisation to give 24. (i) (a) Colloidal particles of test tube (A) are octahedral geometry. positively charged whereas colloidal particles of test tube (B) are negatively charged. NO 2 (b) In test tube (A), Fe3+ ions are adsorbed on the NC CN pInptt.eFste2tOub3e·x(HB2)O, O[oHr–Fieo2nOs3a⋅xreHa2Ods/oFreb3e+disofnortmheedp]p.t. Fe2O3⋅xH2O [or Fe2O3⋅xH2O/OH–is formed]. Cr CN NC NH3 27. (a) CH3CH2CHCH3 hydrolyses easily with KOH 25. Mechanism of dehydration of alcohol : Cl Step 1 : Formation of protonated alcohol because it is secondary alkyl halide. (b) (i) Fast (ii) CH3CH2NC (Protonated alcohol) Class 12 12

(c) Iodide is a powerful nucleophile and therefore, 29. (a) Cr2+ is reducing as its configuration changes it reacts rapidly with RCl to form RI from d4 to d3, a more stable half filled t2g configuration. KI → K+ + I– Mn3+ is oxidising because its configuration also changes from d4 to d5 (which is half-filled) and has I– + R — Cl → R — I + Cl– extra-stability. Also, I– ion is a better leaving group than Cl– ion (b) In [Ti(H2O)6]3+, titanium has 3d1 4s0 and therefore, RI is more readily hydrolysed to form configuration. In Ti3+ ion, the d-orbitals split up into ROH. two parts t2g and eg. The t2g part has three out of five d-orbitals (dxy, dyz, dzx) with energy lower than the HO– + R — I → ROH + I– atomic orbitals whereas eg have two orbitals (dx2 – y2 and dz2) having energy more than the atomic orbitals. OR Due to less energy difference between t2g and eg orbitals, the electrons transit between these two 1. CH3—CH2—CH2—CHCl2 orbital groups (called d-d transition). The energy Cl required for this purpose is sufficiently available in the visible region. The d-d transition absorbs yellow 2. CH3—CH2—C—CH3 region light and therefore colour observed is purple Cl (complementary colour). Cl 30. (a) In sucrose carbon no. 1 of a-D-glucose is 3. CH3—C1,H2-2d—ichCl*orHob—utaCneH2Cl joined together with carbon no. 2 of b-D-fructose. Cl (b) Adenine, guanine and cytosine are the bases common in DNA and RNA both. 4. CH3—C*1H, 3—-dicChlHor2o—butCanHe 2Cl (c) Although, glucose in its open chain structure contains free aldehydic group yet it does not give 5. ClCH2—CH2—CH2—CH2Cl 2, 4-DNP test a characteristic reaction of aldehyde, Cl Cl i.e., –CHO group. This is because glucose actually exists in the cyclic hemiacetal form with only a small 6. CH3—2, 3C*-dHic—hloCr*oHbu—tanCe H3 amount of the open chain form in equilibrium. Since the concentration of open chain form is very low Due to presence of four different groups around C* and its reaction with 2, 4-DNP is reversible thus, its in (3), (4) and (6), these are chiral compounds and 2, 4-DNP adduct is not observed. hence exhibit enantiomerism. OR 28. (i) In aqueous solutions, Cu+ undergoes In a-D-Glucose, the –OH group at C1 is towards disproportionation to form a more stable Cu2+ ion. right whereas in b-D-glucose, the –OH group at C1 2CCuu2++(aiqn) → Cu2+(aq) + Cu(s) is towards left. Such a pair of stereoisomers which aqueous solutions is more stable than Cu+ differ in the configuration only at C1 are called ion because hydration enthalpy of Cu2+ is higher than anomers. that of Cu+. It compensates the second ionisation enthalpy of Cu involved in the formation of Cu2+ 13 ions. (ii) This is due to the presence of maximum number of unpaired electrons in Mn2+ (3d5). (iii) Greater the number of unpaired electrons, stronger is the metallic bond and therefore, higher is the enthalpy of atomisation. Since, iron has greater number of unpaired electrons than copper hence iron has higher enthalpy of atomisation. Chemistry

Ecell = Ecell − 0.0591 log [Zn2+ ] 2 [Cd2+ ] = 0.36 − 0.0591 log 0.0004 2 0.2 = 0.36 − 0.0591 (−2.69) = 0.36 + 0.08 = 0.44 V 2 \\ DG = – nFEcell = – 2 × 96500 × 0.44 = – 84920 J/mol (b) Cu2+(aq) + 2e– → Cu(s) The six membered cyclic  structure of glucose is called ECu2+ /Cu = EC° u2+ /Cu − 0.059 log [Cu] ] pyranose structure (a- or b-), 2 [Cu2+ in analogy with heterocyclic compound pyran. = 0.34 − 0.059 log 1 = 0.34 − 0.059 log 10 2 0.1 2 31. (a) (i) κ × 103 = 0.34 − 0.059 × (1) = 0.34 – 0.0295 = 0.3105 V Λm = M 2 When the concentration of Cu2+ ions is decreased, where M is the concentration of solution in molarity. the potential for copper electrode decreases. (ii) Electrolytic conductivity, k = 3.72 × 10–2 S cm–1 Molar conductivity, κ(S cm−1) × 1000(cm3 L−1) 32. (a) (i) Single O—O bond is weaker than S—S bond because of high interelectronic repulsions Λm = between the lone pair and bond pair of O—O bond, M (mol L−1) as a result catenation property is weaker in oxygen. 3.72 × 10−2 ×1000 (ii) In SF6 molecule, sulphur is surrounded by six = 0.30 = 124 S cm2 mol–1 fluorine atoms which protect it from attack by reagents vw(bai)ltuhe(doi)ifleulteFicootnrro.sltSyriotnenc‘gBe e’ tolhencetrdreioluliysttienosno, .LwTmhiduines,ce‘rBfef’aeiscsetasossntlorowLnlmgy to such an extent that even thermodynamically most favourable reactions like hydrolysis do not occur. So, H2O cannot attack SF6 easily. electrolyte. (iii) HF does not attack on wax but attacks sodium (ii) If an external opposite potential is applied in the galvanic cell and increased slowly, reaction continues silicate which is the main constituent of glass. As a result, the glass bottles are slowly corroded or eaten to take place till the opposing voltage reaches the up by HF. value 1.1 V. After that reaction stops and no further chemical reaction takes place. Hence, no current Na2SiO3 + 6HF Na2SiF6 + 3H2O (b) (i) Acidified K2Cr2O7 turns green due to flows through the cell. formation of chromium sulphate. (iii) The overall reaction for electrolysis of brine solution can be written as K2Cr2O7 + 3SO2 + H2SO4 K2SO4 + 1 1 (Orange) Cr2(SO4)3 + H2O 2 2 NaCl(aq) + H2O(l) → NaOH(aq) + H2(g ) + Cl 2( g ) (ii) Cl will replace I. (Green) Hence, the pH of brine solution which is neutral will 2NaI(aq) + Cl2(g) → 2NaCl(aq) + I2(s) increase due to formation of NaOH. OR OR (a) The basic character decreases from NH3 to BiH3. The basic nature is due to the presence of lone pair of (a) E°cell = E°cathode – E°anode = – 0.403 – (– 0.763) electrons on the central atom. NH3 is the strongest = 0.36 V electron pair donor due to its small size as the electron density of the electron pair is concentrated The net cell reaction is over a small region. As the size increases the electron Zn(s) + Cd2(a+q) → Zn2(a+q) + Cd(s) Here, value of n = 2 14 Class 12

density gets diffused over a large region and hence (a) c(ait)alysItnwFhriiecdheils—a LCerwafitssarceiadc.tIiot nfo,rAmlCs la3siaslatdwdiethd the ability to donate the electron pair (basic nature) as a aniline due to which the nitrogen of aniline acquires decreases. positive charge. This positively charged nitrogen acts (b) (i) H2O < H2S < H2Se < H2Te As the atomic size increases down the group, the    as a strong deactivating group, hence aniline does not undergo Friedel — Crafts reaction. bond length increases and hence, the bond strength decreases. Consequently, the cleavage of E—H NH2 bond (E = O, S, Se, Te, etc.) becomes easier. As a result, the tendency to release hydrogen as proton + AlCl3 N+ H2AlCl–3 increases i.e., acidic strength increases down the (ii) The ammonolysis of alkyl halides with ammonia group. is a nucleophilic substitution reaction in which (ii) The acidic strength of the hydrohalic acids in ammonia acts as a nucleophile by donating the the order : electron pair on nitrogen atom to form primary HF < HCl < HBr < HI amine as the initial product. Now, the primary amine This order is a result of bond dissociation enthalpies can act as a nucleophile and combine with alkyl of H — X bond decreases from H — F to H — I as halide (if available) to give secondary amine and the the size of halogen atom increases. reaction continues in the same way to form tertiary (c) (i) 2NH3 + NaOCl N2H4 + NaCl + H2O amine and finally quaternary ammonium salt. Thus, (ii) 3Cu + 8HNO3 (dilute) 3Cu(NO3)2 + a mixture of products is formed and it is not possible 2NO + 4H2O to separate individual amines from the mixture. 33. (a) Formula of the compound ‘C ’ indicates it is RX R NH2 RX R2NH RX an amine. Since it is obtained by the rceoamctpioonunofdB‘Br2’  HX HX and KOH with the compound ‘B’ so can be an amide. As ‘B’ is obtained from compound R3N R X R4N+X– ‘A’ by reaction with ammonia followed by heating so, :(b) (i) Amines like ammonia are go o d nucleophiles. This is because alkyl group in an amine compound ‘A’ could be an aromatic acid. Formula shows electron releasing effect. This increases the of compound ‘C ’ shows it to be aniline, then ‘B’ is electron density on ‘N’ of amino group and hence, benzamide and compound ‘A’ is benzoic acid. The makes the amines very good nucleophiles. sequence of reactions can be written as follows : (ii) Due to resonance in aniline, the lone pair of electrons on nitrogen gets delocalised over COOH CONH2 NH2 the benzene ring and becomes less available for NH3 Br2/KOH protonation. D NH2 N+ H2 N+ H2 AB C (b) (i) Aniline gives white or brown precipitate – with bromine water. – NH2 N+ H2 – Ethylamine does not react with bromine water. In alkyl amine, alkyl group releases electrons and (ii) When heated with an alcoholic solution of KOH increases electron density on nitrogen, making it aisnodcyCanHidCel.3,Deitmheytlhaymlainmeingeivdeosefsonuolt smelling ethyl stronger base. give this test. (c) (CH3)2NH > CH3NH2 > (CH3)3N > NH3 OR  Chemistry 15

Self Evaluation Sheet Once you complete SQP-1, check your answers with the given solutions and fill your marks in the marks obtained column according to the marking scheme. Performance Analysis Table given at the bottom will help you to check your readiness. Q. No. Chapter Marks Per Question Marks Obtained 1 Aldehydes, Ketones and Carboxylic Acids 4×1 2 Solutions 4×1 ............... 3 Amines 1 ..............% 4 Electrochemistry 1 5 Aldehydes, Ketones and Carboxylic Acids 1 6 Surface Chemistry 1 7 The d- and f-Block Elements 1 8 Chemical Kinetics 1 9 Haloalkanes and Haloarenes 1 10 Biomolecules 1 11 Haloalkanes and Haloarenes 1 12 The p-Block Elements 1 13 Biomolecules 1 14 Solid State 1 15 Coordination Compounds 1 16 Alcohols, Phenols and Ethers 1 17 Electrochemistry 2 18 Chemical Kinetics 2 19 Solutions 2 20 The p-Block Elements 2 21 Aldehydes, Ketones and Carboxylic Acids 2 22 Haloalkanes and Haloarenes 2 23 Solid State 2 24 Surface Chemistry 2 25 Alcohols, Phenols and Ethers 2 26 Coordination Compounds 3 27 Haloalkanes and Haloarenes 3 28 The d- and f-Block Elements 3 29 The d- and f-Block Elements 3 30 Biomolecules 3 31 Electrochemistry 5 32 The p-Block Elements 5 33 Amines 5 70 Total Percentage Performance Analysis Table You are done! Keep on revising to maintain the position. You have to take only one more step to reach the top of the ladder. Practise more. If your marks is A little bit of more effort is required to reach the ‘Excellent’ bench mark. > 90% TREMENDOUS! Revise thoroughly and strengthen your concepts. 81-90% EXCELLENT! Need to work hard to get through this stage. 71-80% VERY GOOD! Try hard to boost your average score. 61-70% GOOD! 51-60% FAIR PERFORMANCE! Class 12 40-50% AVERAGE! 16

2SQP S Q P BLUE PRINT Time Allowed : 3 hours  Maximum Marks : 70 S. No. Chapter Passage based/ SA-I SA-II LA Total MCQs/A & R (1 mark) (2 marks) (3 marks) (5 marks) 1. The Solid State 1(1) 1(2) 1(3) – 2. Solutions 3. Electrochemistry 1(1) – – 1(5) 4. Chemical Kinetics 5. Surface Chemistry 1(1) 1(2) – – 10(23) 6. The p-Block Elements 7. The d- and f-Block Elements 1(4) – – – 8. Coordination Compounds 9. Haloalkanes and Haloarenes 1(1) – 1(3) – 10. Alcohols, Phenols and Ethers 11. Aldehydes, Ketones and Carboxylic Acids 2(2) 2(4) – – 12. Amines 13. Biomolecules 2(2) – – 1(5) 10(19) Total 1(1) 1(2) 1(3) – 1(1) – 1(3) – 2(2) 1(2) – – 1(1) 1(2) – 1(5) 13(28) 1(4) 1(2) 1(3) – 1(1) 1(2) – – 16(22) 9(18) 5(15) 3(15) 33(70)

Subject Code : 043 SQP-2 CHEMISTRY Maximum marks : 70 Time allowed : 3 hours General Instructions : Read the following instructions carefully. (a) There are 33 questions in this question paper. All questions are compulsory. (b) Section A : Q. No. 1 to 16 are objective type questions. Q. No. 1 and 2 are passage based questions carrying 4 marks each while Q. No. 3 to 16 carry 1 mark each. (c) Section B : Q. No. 17 to 25 are short answer questions and carry 2 marks each. (d) Section C : Q. No. 26 to 30 are short answer questions and carry 3 marks each. (e) Section D : Q. No. 31 to 33 are long answer questions carrying 5 marks each. (f) There is no overall choice. However, internal choices have been provided. (g) Use of calculators and log tables is not permitted. SECTION - A (OBJECTIVE TYPE) 1. Read the passage given below and answer the following questions : The synthesis of primary amines is of central importance because these compounds serve as a key precursors and central intermediates to produce value added fine and bulk chemicals as well as pharmaceuticals, agrochemicals and materials. Recently, a nickel-based homogeneous catalyst for both reductive amination of carbonyl compounds with ammonia and hydrogenation of nitroarenes to prepare all kinds of primary amines has been discovered. NH2 NH2 R1 R2(H) Ph R 4100-500-1b2a0r°HC2 Ph P Ph 401b20ar°CH2 O Ph P+ P Ph NO2 Ni(BF4)2.6H2O R1 R2(H) + NH3 R Remarkably, this Ni–complex enabled the synthesis of functionalized and structurally diverse benzylic, heterocyclic and aliphatic linear and branched primary amines as well as aromatic primary amines starting from inexpensive carbonyl compounds (aldehydes and ketones) and nitroarenes using ammonia and molecular hydrogen. The following questions are multiple choice questions. Choose the most appropriate answer : (i) The nickel-based homogeneous complex mention in the study above can be used to convert (a) nitrobenzene to aniline (b) nitrobenzene to benzylamine (c) benzamine to acetanilide (d) nitrite salt to nitrosoamine. 18 Class 12

(ii) The primary amine formed during reductive amination of acetaldehyde with ammonia can also be obtained by reduction of (a) acetamide (b) N–methylacetamide (c) methyl isocyanide (d) acetone. Which of the following is a primary amine? OR (a) Phenylaminomethane (b) N-methylaminobenzene (c) N-methyl-2-aminopropane (d) Triethylamine (iii) Nitrobenzene on treatment with this Ni-based homogeneous catalyst gives a compound, X which on reaction with warmed chloroform and alcoholic solution of KOH, forms a compound, Y. (Cao)mpanouilninde,YNo-nmtertehaytmlaneniltinwei,tPhhLeinAyllHis4ogciyvaens icdoempound, Z. Compounds X, Y and Z respectively are (b) aniline, phenyl isocyanide N-methylaniline (c) benzylamine, phenyl isocyanide N-methylaniline (d) N-methylaniline, benzylamine, phenyl isocyanide. (iv) An organic compound 'X' wcoitmhpmouonledcu'Yla'.rCfoomrmpuoluan, dC'2YH' o4On with this Ni-based homogeneous catalyst in presence of ammonia gives reaction with nitrous acid gives compound 'Z' with eCCvHHol32uB=trioCnHoBfrN2. What is the product of reaction of 'Z' with phosphorus bromide? (a) (b) CBrHC3HCH=2CBHr Br (c) (d) 2. Read the passage given below and answer the following questions: Chemical reactions require varying lengths of time for completion, depending upon the characteristics of the reactants and products and the conditions under which the reaction is taking place. Chemical kinetics is the study of reaction rates, how reaction rates change under varying conditions and by which mechanism the reaction proceeds. There are five general properties that can affect the rate of a reaction (i) Concentration of the reactant : The more concentrated the faster the rate; (ii) Temperature : usually reaction speed up with increasing temperature; (iii) Physical state of reactants: powder reacts faster than blocks as they have greater surface area and since the reaction occurs at the surface, we get a faster rate; (iv) Catalyst : A catalyst speeds up a reaction; (v) Light : Light of a particular wavelength may also speed up a reaction. Mathematically, rate of a reaction is defined as the change in concentration of reactant or product over time. In these questions (Q. No. i- iv), a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements, and reason is the correct explanation of the assertion. (b) Assertion and reason both are correct statements, but reason is not the correct explanation of the assertion. (c) Assertion is correct, but reason is wrong statement. (d) Assertion is wrong, but reason is correct statement. (i) Assertion : A catalyst increases the rate of a reaction. Reason : A catalyst forms an activated complex of lower potential energy. (ii) Assertion : Rate of reaction increases with increase in temperature. Reason : Number of collisions increases with increase in temperature. (iii) Assertion : Rate of reaction always decreases with time. Reason : The amount of reactant remaining always decreases with time in chemical reaction. (iv) Assertion : In rate law, unlike in the expression for equilibrium constants, the exponents for concentrations do not necessarily match the stoichiometric coefficients. Reason : It is the mechanism and not the balanced chemical equation for the overall change that governs the reaction rate. Chemistry 19

OR Assertion : A piece of wood burns slowly in air but rapidly burns in pure oxygen. Reason : Concentration of oxygen is less in air.               Following questions (Q. No. 3-11) are multiple choice questions carrying 1 mark each : 3. In orthorhombic crystal, the values of a, b and c are respectively 4.2 Å, 8.6 Å and 8.3 Å. The molecular mass of the solute is 155 g mol–1 and density is 3.3 g/cc. The number of formula units per unit cell is (a) 2 (b) 3 (c) 4 (d) 6 OR A binary solid (A+B–) has a zinc blende structure with B– ions constituting the lattice and A+ ions occupying 25% tetrahedral holes. The formula of solids is (a) AB (b) A2B (c) AB2 (d) AB4 4. X(ae)F6 on hydrolysis gives XeO (c) XeO2 (d) Xe XeO3 (b) 5. One kilogram of water contains 4 g of NaOH. The concentration of the solution is best expressed as (a) 0.1 molal (b) 0.1 molar (c) decinormal (d) about 0.1 mole. 6. Within the list shown below, the correct pair of structures of alanine in pH ranges 2-4 and 9-11 is respectively. I. HHHHI, 2323INNNNI ++ CCCCHHHH((CC((CCHHHH33))33CC))CCOO(bOO22–)H22–HI, III II. (c) II, III (d) III, IV III. IV. (a) OR The primary structure of a protein refers to (a) whether the protein is fibrous or globular (b) the amino acid sequence in the polypeptide (c) the orientation of the amino acid side chains in space (d) the presence or absence of an a-helix. 7. fAonrmetshceor m(Ap)oCun5Hd1(2BO) when heated with excess HI produced two alkyl iodide which on alkaline hydrolysis and (C). Oxidation of (B) gives acid and oxidation of (C) gives ketone. What is compound (A)? (a) CH3OCH2CH2CH2CH3 (b) C2H5OCH2CH2CH3 (c) C2H5OCH(CH3)2 (d) All of these 8. TChoencsiodoerrdtihneatfioolnlonwuinmgbceor,mopxliedxa,tiNona[nCurm(NbHer3,)n2(uomx)b2]e.r3oHf2dO-electrons, number of unpaired electrons on the metal and magnetic moment are respectively (a) 6, +2, 4, 0, 0 (b) 4, +3, 3, 3, 3.87 (c) 6, +3, 3, 3, 3.87 (d) 4, +2, 4, 0, 0 OR Which of the following statements are correct? (i) In octahedral complexes, dz2 , dx2 − y2 orbitals have higher energy than dxy, dyz and dzx orbitals. (ii) In tetrahedral complexes dxy, dyz, dzx orbitals have higher energy than dz2 and dx2 − y2 orbitals. (iii) The colours of complexes are due to electronic transitions from one set of d-orbitals to another set of orbitals. 9 (iv) Dtetrahedral = 4 × Doctahedral 20 Class 12

(a) Only (i), (ii) and (iii) (b) Only (i) and (iv) (c) Only (iii) and (iv) (d) Only (ii), (iii) and (iv) 9. On adding ofebwtadinreodp.sTohfedpilhuetenHomCelnoornFeisCkl3ntoowfnreashs ly precipitated ferric hydroxide a red coloured colloidal solution is (a) peptisation (b) dialysis (c) protective action (d) dissolution. 10. The lanthanoids contraction is responsible for the fact that (a) Zr and Y have about the same radius (b) Zr and Nb have similar oxidation state (c) Zr and Hf have about the same radius (d) Zr and Zn have same oxidation state. 11. The e.m.f. of the following galvanic cells : (i) Zn|Zn2+ (1 M) || Cu2+ (1 M) | Cu (ii) Zn|Zn2+ (0.1 M) || Cu2+ (1 M) | Cu (iii) Zn|Zn2+ (1 M) || Cu2+ (0.1 M) | Cu (iv) Zn|Zn2+ (0.1 M) || Cu2+ (0.1 M) | Cu are represented by E1, E2, E3 and E4 respectively. Which of the following orders is correct? (a) > > > (b) > > > (c) E1 > E2 = E3 > E4 (d) E3 > E2 = E1 > E4 E3 E1 E4 E2 E2 E1 E4 E3 OR If the conductivity and conductance of a solution is same then its cell constant is equal to (a) 1 (b) 0 (c) 10 (d) 1000 In the following questions (Q. No. 12 - 16) a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. 12. Assertion : Fluorine has a less negative electron affinity than chlorine. Reason : There is relatively greater effectiveness of 2p-electrons in the small F atom to repel the additional electron entering the atom than to 3p-electrons in the larger Cl atom. 13. Assertion : Addition reaction of water to but-1-ene in acidic medium yields butan-1-ol. Reason : Addition of water in acidic medium proceeds through the formation of secondary carbocation. 14. Assertion : Mn2+ is more stable than Mn3+. Reason : Mn2+ has half-filled configuration. 15. Assertion : Formaldehyde cannot be prepared by Rosenmund’s reduction. Reason : Acid chlorides can be reduced into aldehydes with hydrogen in boiling xylene using palladium or platinum as a catalyst supported on barium sulphate. This is known as Rosenmund’s reduction. OR Assertion : Benzaldehyde is more reactive than ethanal towards nucleophilic attack. Reason : The overall effect of –I and +R effect of phenyl group decreases the electron density on the carbon atom of > C = O group in benzaldehyde. 16. Assertion : The presence of nitro group facilitates nucleophilic substitution reactions in aryl halides. Reason : The intermediate carbanion is stabilised due to the presence of nitro group. Chemistry 21

SECTION - B The following questions, Q. No. 17-25 are short answer type and carry 2 marks each. 17. What happens when (i) Nitroethane is treated with LiAlH4? (ii) Ethanamine reacts with AgCl? 18. Write the final product(s) in each of the following reactions : (a) CH3CH2 CH CH3 Cu/573 K OH (b) C6H5 OH (i) CHH+Cl3 + aq. NaOH (ii) 19. Distinguish between : (i) Essential and non-essential amino acids. (ii) a-Glucose and b-glucose. OR What is DNA finger printing? What are the areas of its application? 20. Using crystal field theory, draw energy level diagram, write electronic configuration of the central metal atom/ion and determine the magnetic moment value for the following : [FeF6]3–, [Fe(H2O)6]2+. OR What will be the correct order of absorption of wavelength of light in the visible region for the complexes [Co(NH3)6]3+, [Co(CN)6]3– and [Co(H2O)6]3+ ? 21. (i) Tendency to show – 2 oxidation state diminishes from sulphur to polonium in group 16. Why? (ii) Only higher members of group-18 of the periodic table are expected to form compounds. Why? 22. An element with density 2.8 g cm–3 forms a fcc unit cell with edge length 4 × 10–8 cm. Calculate the molar mass of the element. (Given : NA = 6.022 × 1023 mol–1) 23. The E° values corresponding to the following two reduction electrode processes are : Cu+/Cu = + 0.52 V, Cu2+/Cu+ = + 0.16 V Formulate the galvanic cell for their combination. What will be the standard cell potential for it? Calculate DrG° for the cell reaction. (F = 96500 C mol–1) OR Determine the value of equilibrium constant V(Kc) and DG° for the following reaction : (N1i(Fs) + 926A5g0+0(aCq) m→oNl–1i2)+(aq) + 2Ag(s), E° = 1.05 = 24. How will you bring about the following conversions? (i) Ethanal to but-2-enal (ii) Propanone to propene 25. (a) Arrange halogens in increasing order of electron affinity. (b) Why does fluorine show abnormal behaviour? 22 Class 12

SECTION - C Q. No. 26-30 are short answer type II carrying 3 marks each. 26. (i) A compound forms hexagonal close-packed structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids? (ii) Why is Frenkel defect not found in pure alkali metal halides?         27. Write the chemical equations for the following conversions : (i) Aniline to N-phenylethanamide. (ii) Aniline to p-nitroaniline. OR Identify A, B and C in the following reactions : (a) (b) 28. Explain the following terms with a suitable example in each case : (i) Zeta potential (ii) Tanning (iii) Multimolecular colloids 29. (i) Allyl chloride is more reactive than n-propyl chloride towards nucleophilic substitution reaction. Explain why. (ii) What are the major products of the following reactions? Cl CH3 HBr (a) NaOCH3 (b) D Peroxide NO2 OR (i) Which of the two compounds, CH3CH CHCH2Br or CH3CH(Br)CH CH2 is achiral and which is chiral? (ii) Arrange each set of compounds in order of increasing boiling points : (a) Bromomethane, bromoform, chloromethane, dibromomethane (b) 1-Chloropropane, iso-propyl chloride, 1-chlorobutane. 30. Write the name, the structure and the magnetic behaviour of each one of the following complexes : (i) [Pt(NH3)2Cl(NO2)] (ii) [Co(NH3)4Cl2]Cl (iii) Ni(CO)4 (At. nos. Co = 27, Ni = 28, Pt = 78) SECTION - D Q No. 31-33 are long answer type carrying 5 marks each. 31. (i) Give one chemical test to distinguish between the following pairs of compounds : (a) Ethanal and propanal (b) Benzoic acid and ethyl benzoate (ii) How will you bring about the following conversions in not more than two steps? (a) Benzene to m-nitroacetophenone (b) 2-Methylpropanol to 2-methylpropene (iii) There are two —NH2 groups in semicarbazide. However, only one is involved in the formation of semicarbazone. Explain. Chemistry 23

OR An organic compound contains 69.77% carbon, 11.63% hydrogen and rest, oxygen. The molecular mass of the compound is 86. It does not reduce Tollens’ reagent but forms an addition compound with sodium hydrogen sulphite and give positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound and give reactions involved also. 32. (i) At 300 K, 36 g of glucose present per litre in its solution has an osmotic pressure of 4.98 bar. If osmotic pressure of solution is 1.52 bar at the same temperature, what would be its concentration? (ii) Osmotic pressure is more useful to calculate molecular mass than other colligative properties. Why?         (iii) If N2w92a3tgeKars?iiAss7sbs6uu.4bm8belketbdhaatrht. rNo2uegxherwtsataerpaartt2ia9l3pKre,shsuowre mofa0n.9y8m7ibllaimr. GolievsenoftNha2tgHaesnwroyu’slldawdicssoonlsvteanint 1 litre of for N2 at OR The vapour pressure of two pure liquids, A and B that form an ideal solution are 300 and 800 torr respectively, at temperature T. A mixture of the vapours of A and B for which the mole fraction of A is 0.25 is slowly compressed at temperature T. Calculate (a) the composition of the first drop of the condensate, (b) the total pressure when this drop is formed, (c) the composition of the solution whose normal boiling point is T, (d) the pressure when only the last bubble of vapour remains, and (e) the composition of the last bubble. 33. (a) Assign reason for the following : (i) The enthalpies of atomisation of transition elements are high. (ii) Scandium (Z = 21) does not exhibit variable oxidation states and yet it is regarded as transition element. (b) What may be the possible oxidation states of the transition metals with the following d electronic configurations in the ground state of their atoms : 3d34s2, 3d54s2 and 3d64s2. Indicate relative stability of oxidation states in each case. OR (a) How would you account for the following : (i) The atomic radii of the metals of the third (5d) series of transition elements are virtually the same as those of the corresponding members of the second (4d) series. (ii) The E° value for the Mn3+/Mn2+ couple is much more positive than that for Cr3+/Cr2+ couple or Fe3+/Fe2+ couple. (iii) The highest oxidation state of a Mn metal is exhibited in its oxide or fluoride. (b) Which of following cations are coloured in aqueous solutions and why? Sc3+, V3+, Ti4+, Mn2+ (At. Nos. Sc = 21, V = 23, Ti = 22, Mn = 25) 24 Class 12

1. (i)(a) 3. (c) : Z =V × N0 ×d     O M ×10−24 × × 1023 (ii) (a) : CH3 — C — H + NH3 Ni–based CHEt3hCylHam2iNneH2 = 4.2 × 8.6 × 8.3 155 6.023 × 3.3 = 3.84 ≈ 4 catalyst Acetaldehyde O LiAlH4 OR CH3 — C — NH2 +4[H] Ether CH3CH2NH2 (c) : No. of B– ( fcc) ions in unit cell 1 1 Acetamide 8 2 Ethylamine =8× +6× =4 OR 8 × 25 100 (a) : CH2NH2 Now A+ ion occupies 25% of tetrahedral holes = = 2 Thus, ratio of B– to A+ is 2 : 1 and formula is AB2. Phenylaminomethane 4. (a) : XeF6 + 3H2O → XeO3 + 6HF 4 (iii) (b) : NO2 NH2 5. (a) : Mole of NaOH = 40 = 0.1 mol Ni-based Mass of water = 1 kg Catalyst 0.1 Nitrobenzene An(Xili)ne CHCl3, KOH (alc). Molality = 1 = 0.1m NH – CH3 NC 6. (a) : H2N CH COOH LiAlH4 + KCl + H2O CH3 Alanine N-Methylaniline Phenyl isocyanide Acidic medium : (Z) (Y) (iv)(b) : H Basic medium : CH3 — C O + NH3 → CH3CH2NH2 HONO (X) (Y) H3PO3 + CH3CH2Br PBr3 CH3CH2OH + N2 OR (P + Br2) Bromoethane (Z) (b) 2. (i)(a) 7. (c) : C2H5OCH(CH3)2 + 2HI Heat (ii) (b) : An increase in temperature will raise the kinetic (A) energy of the reactant molecules. Therefore, a greater aq. C2H5I + (CH3)2CHI + H2O C2H5OH + KI proportion of reactant molecules will have minimum C2H5I + KOH energy necessary for an effective collision. Therefore C2H5OH oxidation (B) by increase in number of collisions, the rate of reaction (B) CH3COOH gets increased. (CH3)2CHI + KOH aq. (CH3)2CHOH + KI (iii) (d) : Rate of reaction not always decreases with time. (C) In zero order reaction, rate of reaction is independent of (CH3)2CHOH oxidation CH3COCH3 concentration. (C) (iv) (a) 8. (c) T: NhuHs,3ciosomrdoinnoatdioenntnatuemwbheirleofoxmiestaal bidentate ligand. is OR =2×1+2×2=6 (a) Rate of reaction is directly proportional to concentration Oxidation number of metal = +3 of oxygen. Chemistry 25

Cr = [Ar] 3d5, 4s1 ; Cr3+ = 3d3 (ii) a-Glucose and b-glucose differ in configuration of All the three electrons are unpaired. –OH group on the anomeric carbon w(Ch1i)l.eIinn a-glucose, the –OH ggrroouuppaattCC11isistotowwaradrds srilgehftt. b-glucose, Magnetic moment = n(n + 2) = 3 × 5 = 15= 3.87 B.M. the –OH OR OR (a) : Dtetrahedral = 4 × Doctahedral A sequence of bases on DNA is unique for a person and 9 information regarding this is called DNA finger printing. 9. (a) 10. (c) : It is a fact developed due to lanthanoid contraction, Applications : otherwise size of Hf should have been greater than Zr. (i) In forensic lab for identifying criminals. [Zn2+ (ii) To determine paternity of an individual. 11. (d) : E = E°− 0.0591 log [Cu2+ ] (iii)To identify the dead bodies in any accident by 2 ] comparing the DNA’s of parents or children. As Q ↓, E ↑ 20. [FeF6]3– : Fe3+ = 3d5 Q1 = 1, Q2 = 0.1, Q3 = 10, Q4 = 1 E2 > E1 = E4 > E3 OR (a) k = conductance × cell constant since k = conductance \\ cell constant = 1 Number of unpaired electrons = 5 12. (a) Magnetic moment = 5(5 + 2) = 5.92 B.M. [Fe(H2O)6]2+ : Fe2+ = 3d6 13. (d) : Addition of water to but-1-ene in acidic medium yields butan-2-ol because this reaction proceeds through the formation of secondary carbocation. 14. (a) : A half-filled or fully-filled orbital is more stable than incompletely filled orbital. 15. (b) Number of unpaired electrons = 4 Magnetic moment = 4(4 + 2) = 4.9 B.M. OR (a) OR 16. (a) As strength of ligand increases crystal field splitting 17. (i) Ethylamine is formed. energy (CFSE) increases. LiAlH4 C2H5NO2 dry ether C2H5 NH2 + 2H2O Order of strength of ligands : I– < Br– < Cl– <NO3– <o-Fp–h<enO<HN– O< 2o–x<<CHN2O– <<CpOy . Nitroethane Ethylamine = NH3 < en < dipy < (ii) Silver chloride dissolves in aqueous ethylamine due to the formation of the complex C2H5NH2 AgCl [Ag(C2H5NH2)2]+ Cl– Now, ∆E = hc . λ So, as CFSE increases, l decreases. 18. (a) CH3CH2 CH CH3 Cu/573 K O Thus, the correct order of absorption of wavelength of OH CH3CH2 C CH3 light in the visible region is : [Co(H2O)6]3+ > [Co(NH3)6]3+ > [Co(CN)6]3–. Butan-2-one 21. (i) From sulphur to polonium electronegativity OH (b) C6H5 OH (i) CHCl3 + aq. NaOH CHO decreases hence tendency to show negative oxidation (ii) H+ state decreases. Salicylaldehyde (ii) Higher members have smaller ionisation enthalpy 19. (i) The amino acids which can be synthesised in our than lower members and hence they are expected to body are called non-essential amino acids. form compounds. While the amino acids which cannot be synthesised 22. Given, density of solid, d = 2.8 g cm–3 in our body and must be taken through diet are called For fcc unit cell, Z = 4 essential amino acids. 26 Class 12

Edge length, a = 4 × 10–8 cm, 26. (i) No. of atoms in the close packing = 0.5 mol Molar mass, M = ?      = 0.5 × 6.022 × 1023 = 3.011 × 1023 NA = 6.022 × 1023 mol–1 or M = d × NA × a3 No. of octahedral voids = 1 × No. of atoms in the Using formula, d = Z×M Z packing = 3.011 × 1023 NA × a3 Substituting these values, we get No. of tetrahedral voids = 2 × No. of atoms in the 2.8 g cm−3 × 6.022 × 1023 mol−1 × (4 × 10−8 cm)3 packing = 2 × 3.011 × 1023 = 6.022 × 1023 M= 4 Total no. of voids = 3.011 × 1023 + 6.022 × 1023 or M = 2.8 × 6.022 × 6.4 = 26.98 g mol−1 = 9.033 × 1023 4 (ii) Frenkel defect is not found in pure alkali metal halides 23. At cathode : Cu+ + e– → Cu ; because alkali metal ions cannot fit into the interstitial sites. At anode : Cu+ → Cu2+ + e– ; E° = + 0.52 V 27. (i) E° = + 0.16 V (ii) Cell reaction : 2Cu+ → Cu + Cu2+ Cell representation is Cu+ | Cu2+ || Cu+ | Cu E°cell = E°cathode − E°anode = 0.52 – 0.16 = 0.36 V DrG° = – nE°F = – 1 × 0.36 × 96500 = –34740 J mol–1 OR HNie(rs)e+, n2=Ag2+(aq) → Ni2(+aq) + 2Ag(s), E° = 1.05 V Using formula, log Kc = nE°cell 0.059 2 ×1.05 OR or log Kc = 0.059 = 35.5932 (a) CH3Br KCN CH3CH2NH2 CH3CN LiAlH4 DAKGcg=a°in=an,–Dt2iGlo×°g9=365–5.n50F903E×2°c1eol.l0r5K=c = 3.92 × 1035 (B) – 202650 J (A) DG° = –202.65 kJ 273 K HNO2 CH3CH2OH (C) O OH– (b) CH3COOH NH3 CH3C(AO) NH2 Br2 + KOH Aldol ∆ 24. (i) 2CH3 C H Ethanal condensation OH CH3NC CHCl3 + NaOH CH3NH2 CH3 CH CH2 CHO (C) (B) D H+ 28. (i) The potential difference between the fixed layer and the diffused layer of opposite charge of colloidal CH3 CH CH CHO system is called Zeta potential. (ii) Animal hides are colloidal in nature. When a hide (ii) Propanone to propene But-2-enal which has positively charged particles is soaked in tannin, which contains negatively charged colloidal particles, O OH mutual coagulation takes place. This results in the NaBH4, CH3OH hardening of leather. This process is termed as tanning. CH3 C CH3 Reduction CH3 CH CH3 (iii)A large number of atoms or smaller molecules of a substance on dissolution aggregate together to form Propanone 2-Propanol species having size (diameter < 1 nm) in the colloidal range (1–1000 nm). Such species are known as multimolecular CH3 CH CH2 conc. H2SO4, 443 K colloids. For example, a sulphur sol consists of particles Dehydration containing a thousand or more of S8 sulphur molecules. Propene 25. (a) I2 < Br2 < F2 < Cl2 - electron affinity. (b) (i) Due to small size. (ii) Due to absence of vacant d-orbitals. (iii)Due to high electronegativity. Chemistry 27

29. (i) Allyl chloride is more reactive than n-propyl (ii) (a) chloride towards nucleophilic substitution reaction due to the greater stabilisation of allylic carbocation    intermediate formed by resonance. H2C CH H2C+ H2C+ CH CH2 : (ii) (a) CH3 CH (b) CH3 CH2OH SOCl2 2-Methylpropanol (b) CH3 HBr/Peroxide CH3 CH3 alc. KOH CH3 CH3 CH2 Cl CH3 C CH2 CH 1-Methylcyclohex(Aenneti-Madadriktioownn) ik1o-Bffr’somo-2-methyBlcryclohexane 2-Methylpropene OR (iii) (icsio)cnhtCairiHanlc3bCheiHcraaulcsCaerHibtoCcnoHnat2toBamirnswischhaeicrrehaalirscaCalrHbbo3eCnc*aHaut(soBemri).tCdHoesCnHo2t (ii) (a) Boiling point increases with increase in O– molecular mass. It also depends on the size and H2N C N+ H NH2 number of halogen atoms. Therefore, the arrangement of the given compounds in order of increasing boiling rAessoelneacntrcoenitddenoesistynoont aocntea—s aNnHuc2lgeroopuhpildeewcrheialesetshdeuloentoe points is as follows : pair of electrons on the ortehsoernNanHce2 group (i.e., attached to CH3Cl < CH3Br < CH2Br2 < CHBr3 NH) is not involved in and hence is available (b) For the same halogen, boiling point increases with for nucleophilic attack on the C O group of aldehydes increase in size of the alkyl group. Further the boiling and ketones. point decreases as the branching increases. Therefore, the arrangement of the given compounds in the order OR of increasing boiling points is as follows : (I) The given compound does not reduce Tollens’ (CH3)2CHCl < ClCH2CH2CH3 < ClCH2CH2CH2CH3 reagent which implies that it is not an aldehyde. 30. (i) [Pt(NH3)2Cl(NO2)] : Diamminechloridonitrito-N-platinum(II) O It is square planar and diamagnetic. (ii) [Co(NH3)4Cl2]Cl : (II) Positive iodoform test proves that it has a CH3 C Tetraamminedichloridocobalt (III) chloride group. It is octahedral and diamagnetic. (III) The oxidation products are indicative of the (iii) Ni(CO)4 : Tetracarbonylnickel(0) presence of 5 carbon atoms. It is tetrahedral and diamagnetic. To find the exact molecular formula %C = 69.77%, %H = 11.63% %O = 100 – (69.77 + 11.63) = 18.6 % C : H : O = 69.77 : 11.63 : 18.6 12 1 16 The structure based on inferences (I), (II) and (III) is O 31. (i) (a) Ethanal and propanal can be distinguished H3C C CH2CH2CH3 by iodoform test. Yellow precipitate of iodoform will Reactions involved are : be formed from ethanal on heating with iodine and sodium hydroxide solution whereas propanal does not give iodoform test. (b) Benzoic acid and ethyl benzoate can be distinguished by their reactions with sodium bicarbonate solution. Benzoic acid will give effervescence with NaHCO3 whereas ethyl benzoate does not react.  28 Class 12

(d) Now when the entire vapours are condensed and only last bubble of vapour remains, then we can assume that the mole fraction of A, which was in the vapour phase 32. (i) p = CRT originally, will now be in the liquid phase and the mole R and T are same in both cases hence fraction of Aprreesmsuarienwinhgeninlatshtebvuabpboleurofpvhaapsoeuxr′AreamndainPs′T. π1 C1 π 2C1 be the total π2 = C2 or, C2 = π1 = P75°A+y6A0+0 P=°B67y5B = (300 × 0.25) + (800 × 0.75) P′T = Torr. of partial Molarity of first solution P(e′)T According to Raoult’s law and Dalton’s law 36 C1 = 180 = 0.2 mol L−1 pressures, C2 = π 2C1 = 1.52 × 0.2 = 0.0610 M P′T x′A = P°A yA300 × 0.25 π1 4.98 = 675 (ii) Magnitude of osmotic pressure is large even for ∴ xA′ = P°A yA = 0.11 and xB′ = 0.89 PT′ very dilute solutions and it can be measured at room 33. (a) (i) As transition metals have a large number of temperature hence it is more useful for the calculation of molecular mass. unpaired electrons in the d-orbitals of their atoms they have strong interatomic attractions or metallic bonds. (iii)According to Henry’s law, pN2 = KH × xN2 Hence they have high enthalpy of atomization. 0.987 bar (ii) Scandium(Z=21)hasincompletelyfilled3d-orbitals xN2 = pN2 = 76480 bar = 1.29 ×10−5 in the ground state (3d1). Hence it is considered as a KH transition element. If moles of N2 are present in 1 L (i.e., 55.55 moles), (b) The possible oxidation states for 3d34s2 = +5, +4, x n = n+ n n +3, +2. 55.55 The possible oxidation states for 3d54s2 = +7, +6, +5, N2 55.55 +4, +3, +2 \\ n = 1.29 × 10 −5 55.55 or n = 1.29 × 10–5 × 55.55 moles The possible oxidation states for 3d64s2 = +6, +4, +3, +2. = 71.659 × 10–5 moles = 0.716 millimoles In a transition series the oxidation states which lead OR to exactly half filled or completely filled d-orbitals are more stable. v(aa)pLoeutrypAhaansde,yrBesbpeetchtievmelyo.le fraction of A and B in the OR PA P °A xA P °A xA (a) (i) This is due to lanthanoid contraction. PTotal P°A xA + P°B xB + (P °A − P °B ) xA (ii) Much larger third ionisation energy of Mn(where yA = = = P °B change is d5 to d4) is mainly responsible for this. This twhheecroen, xdAenissatthee. mole fraction of A in the first drop of also explains that +3 state of Mn is of little importance and from the relation, DG° = –nFE°. More p+oes–itiMvenis2+th; e valuFeeo3f+E°+, ree–actFioe2n+will be feasible. ∴ xA = P °A yAP °B Mn3+ − (P °A − P °B ) yA Substituting the given values, 3d4 3d5 3d5 3d6 more stable more stable 0.25 × 800 xA = [(300 − 800) = 0.47 (half filled) (half filled) 300 − × 0.25] Hence, Eth°vaalune for Mn3+/Mn+2 couple is much more positive that for Fe3+/Fe2+. (b) The total pressure when this drop is formed is given by (iii) Manganese can form pp - dp bond with oxygen by PPTToottaa=ll ==1P(43A10++0 P×B0=.4P7)°A+xA(8+00P°×Bx0B.53) utilising 2p-orbital of oxygen and 3d-orbital of manganese \\ 424 = 565 Torr due to which it can show highest oxidation state of +7. (c) Normal boiling point implies that the solution While with fluorine it cannot form such pp - dp bond boils at a temperature when vapour pressure of the thus, it can show a maximum of +4 oxidation state. (b) Only those ions will be coloured which have solution becomes equal to the atmospheric pressure partially filled d-orbitals facilitating d-d transition. (1 atm or 760 Torr). Ions with d0 and d10 will be colourless. Thus, PTotal = 760 = P°AxA + P°B(1 – xA) From electronic configuration of the ions, V3+(3d2) and 760 − P°B 760 800 Mn2+(3d5) are coloured while Ti4+(3d0) and Sc3+(3d0) xA = P°A − P°B = 300 − 800 = 0.08 and xB = 0.92 are colourless. −  Chemistry 29

Self Evaluation Sheet Once you complete SQP-2, check your answers with the given solutions and fill your marks in the marks obtained column according to the marking scheme. Performance Analysis Table given at the bottom will help you to check your readiness. Q. No. Chapter Marks Per Question Marks Obtained 1 Amines 4×1 2 Chemical Kinetics 4×1 ............... 3 The Solid State 1 ..............% 4 The p-Block Elements 1 5 Solutions 1 6 Biomolecules 1 7 Alcohols, Phenols and Ethers 1 8 Coordination Compounds 1 9 Surface Chemistry 1 10 The d- and f-Block Elements 1 11 Electrochemistry 1 12 The p-Block Elements 1 13 Alcohols, Phenols and Ethers 1 14 The d- and f-Block Elements 1 15 Aldehydes, Ketones and Carboxylic Acids 1 16 Haloalkanes and Haloarenes 1 17 Amines 2 18 Alcohols, Phenols and Ethers 2 19 Biomolecules 2 20 Coordination Compounds 2 21 The p-Block Elements 2 22 The Solid State 2 23 Electrochemistry 2 24 Aldehydes, Ketones and Carboxylic Acids 2 25 The p-Block Elements 2 26 The Solid State 3 27 Amines 3 28 Surface Chemistry 3 29 Haloalkanes and Haloarenes 3 30 Coordination Compounds 3 31 Aldehydes, Ketones and Carboxylic Acids 5 32 Solutions 5 33 The d- and f-Block Elements 5 70 Total Percentage Performance Analysis Table You are done! Keep on revising to maintain the position. You have to take only one more step to reach the top of the ladder. Practise more. If your marks is A little bit of more effort is required to reach the ‘Excellent’ bench mark. > 90% TREMENDOUS! Revise thoroughly and strengthen your concepts. 81-90% EXCELLENT! Need to work hard to get through this stage. 71-80% VERY GOOD! Try hard to boost your average score. 61-70% GOOD! 51-60% FAIR PERFORMANCE! 40-50% AVERAGE!

3SQP S Q P BLUE PRINT Time Allowed : 3 hours  Maximum Marks : 70 S. No. Chapter Passage based/ SA-I SA-II LA Total MCQs/A & R (1 mark) (2 marks) (3 marks) (5 marks) 1. The Solid State 1(1) – – 1(5) 2. Solutions 3. Electrochemistry 2(2) – 1(3) – 4. Chemical Kinetics 5. Surface Chemistry 1(4) – – – 11(23) 6. The p-Block Elements 7. The d- and f-Block Elements 2(2) – 1(3) – 8. Coordination Compounds 9. Haloalkanes and Haloarenes 1(1) 1(2) – – 10. Alcohols, Phenols and Ethers 11. Aldehydes, Ketones and Carboxylic Acids 2(2) – 1(3) – 12. Amines 13. Biomolecules 1(4) 1(2) – – 8(19) Total 1(1) 1(2) – 1(5) 1(1) 2(4) – 1(5) 1(1) 1(2) – – 1(1) 1(2) 1(3) – 14(28) 1(1) 1(2) 1(3) – 1(1) 1(2) – – 16(22) 9(18) 5(15) 3(15) 33(70)

Subject Code : 043 SQP-3 CHEMISTRY Time allowed : 3 hours Maximum marks : 70    General Instructions : Read the following instructions carefully. (a) There are 33 questions in this question paper. All questions are compulsory. (b) Section A : Q. No. 1 to 16 are objective type questions. Q. No. 1 and 2 are passage based questions carrying 4 marks each while Q. No. 3 to 16 carry 1 mark each. (c) Section B : Q. No. 17 to 25 are short answer questions and carry 2 marks each. (d) Section C : Q. No. 26 to 30 are short answer questions and carry 3 marks each. (e) Section D : Q. No. 31 to 33 are long answer questions carrying 5 marks each. (f) There is no overall choice. However, internal choices have been provided. (g) Use of calculators and log tables is not permitted. SECTION - A (OBJECTIVE TYPE) 1. Read the passage given below and answer the following questions : The study of the conductivity of electrolyte solutions is important for the development of electrochemical devices, for the characterisation of the dissociation equilibrium of weak electrolytes and for the fundamental understanding of charge transport by ions. The conductivity of electrolyte is measured for electrolyte solution with concentrations in the range of 10–3 to 10–1 mol L–1, as solution in this range of concentrations can be easily prepared. The molar conductivity (Λm) of strong electrolyte solutions can be nicely fit by Kohlrausch equation. Λm = Λ°m – K C '…(i) Where, Λ°m is the molar conductivity at infinite dilution and C is the concentration of the solution. K is an empirical proportionality constant to be obtained from the experiment. The molar conductivity of weak electrolytes, on the other hand, is dependent on the degree of dissociation of the electrolyte. At the limit of very dilute solution, the Ostwald dilution law is expected to be followed, 1 = 1 + Λm CA …(ii) Λm Λm (Λm )2 Kd Where, CA is the analytical concentration of the electrolyte and Kd is dissociation constant. The molar conductivity at infinite dilution can be decomposed into the contributions of each ion. Λ°m = n+l+ + n–l– …(iii) Where, l+ and l– are the ionic conductivities of positive and negative ions, respectively and v+ and v– are their stoichiometric coefficients in the salt molecular formula. The following questions are multiple choice questions. Choose the most appropriate answer : (i) Which statement about the term infinite dilution is correct? (a) Infinite dilution refers to hypothetical situation when the ions are infinitely far apart. 32 Class 12

(b) The molar conductivity at infinite dilution of NaCl can be measured directly in solution. (c) Infinite dilution is applicable only to strong electrolytes. (d) Infinite dilution refers to a real situation when the ions are infinitely far apart.   (ii) Which of the following is a strong electrolyte in aqueous solution? (a) HNO2 (b) HCN (c) NH3 (d) HCl (d) H2SO3 OR Which of the following is a weak electrolyte in aqueous solution? (a) K2SO4 (b) Na3PO4 (c) NaOH (iii) If the molar conductivities at infinite dilution for NaI, CH3COONa and (CH3COO)2Mg are 12.69, 9.10 and 18.78 S cm2 mol–1 respectively at 25°C, then the mola(br )con3d9u0c.5tiSvictymo2fmMogl–I21 at infinite dilution is (a) 25.96 S cm2, mol–1 (d) 3.89 × 10–2 S cm2 mol–1 (c) 189.0 S cm2 mol–1 (iv) Which of the following is the correct order of molar ionic conductivities of the following ions in aqueous solutions? (a) Li+ < Na+ < K+ < Rb+ (b) Li+ > Na+ > K+ > Rb+ (c) Rb+ < Na+ < Li+ < K+ (d) Li+ < Rb+< Na+ < K+ 2. Read the passage given below and answer the following questions : The lanthanide series is a unique class of 15 elements with relatively similar chemical properties. They have atomic number ranging from 57 to 71, which corresponds to the filling of the 4f orbitals with 14 electrons. This configuration leads to phenomenon known as lanthanide contraction. The lanthanides are sometimes referred to as the ‘rare earth elements’, leading to misconception that they are rare. In fact many of the rare earth elements are more common than gold, silver and in some cases, lead. The lanthanides are commonly found in nature as a mixture in a number of monazite (LnPO4) and bastnaesite (LnCO3F) in the +3 oxidation state. The chemical and physical properties of lanthanides provide the unique features that set them apart from other elements. Lanthanides are most stable in the +3 oxidation state. Yb and Sm though stable in the +3 state, also have accessible +2 oxidation states. The ease of accessibility of both oxidation states is quite important in chemical synthesis and these elements act as Lewis acid in the +3 oxidation state and single electron reductant in the +2 oxidation state. In these questions (Q. No. i-iv) , a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. (i) Assertion : The elements scandium and yttrium are called \"rare earths\". Reason : Scandium and yttrium are rare on earth's crust. (ii) Assertion : Separation of lanthanide elements is difficult. Reason : They have similar chemical properties. OR Assertion : There is continuous increase in size among lanthanides. Reason : Lanthanides show lanthanide contraction. (iii) Assertion : Yb2+ is more stable than Yb3+. Reason : Electronic configuration of Yb2+ is [Xe]4f 7. Chemistry 33

(iv) Assertion : All lanthanides have similar chemical properties. Reason : Because the lanthanoids differ only in the number of 4f - electrons. Following questions (Q. No. 3-11) are multiple choice questions carrying 1 mark each: 3. For a zero order reaction, the plot of conc. of reactants vs time is a straight line with (a) +ve slope and zero intercept (b) –ve slope and zero intercept (c) +ve slope and non-zero intercept (d) –ve slope and non-zero intercept. 4. The correct order of reactivity of aldehydes and ketones towards hydrogen cyanide is (a) CH3COCH3 > CH3CHO > HCHO (b) CH3COCH3 > HCHO > CH3CHO (c) CH3CHO > CH3COCH3 > HCHO (d) HCHO > CH3CHO > CH3COCH3 OR In a set of reactions m-bromobenzoic acid gave a product D. Identify the product D. COOH SOCl2 B NH3 C NaOH D Br Br2 A SO2NH2 COOH NH2 CONH2 (a) (b) (c) (d) Br NH2 Br Br 5. Which of the following reactions is most suitable for the preparation of n-propylbenzene? (a) Friedel-Crafts alkylation (b) Wurtz reaction (c) Wurtz-Fittig reaction (d) Grignard reaction 6. What are the products obtained when ammonia is reacted with excess of chlorine? (a) N2 and NCl3 (b) N2 and HCl (c) N2 and NH4Cl (d) NCl3 and HCl 7. Identify 'X'. (b) N, N-Dimethylbenzene C6H5COCl + (CH3)2NH Pyridine 'X' (d) N, N-Diphenylmethanamine (a) N, N-Dimethylbenzamide (c) N-Methyl-N-phenylamine OR Which of the following is the correct increasing order of basicity of amines in gaseous phase? (a) (CH3)2NH > CH3NH2 > (CH3)3N > NH3 (b) (CH3)3N > (CH3)2NH > CH3NH2 > NH3 (c) (CH3)2NH > (CH3)3N > CH3NH2 > NH3 (d) (CH3)3N > CH3NH2 > (CH3)2NH > NH3 8. Which of the following defect, if present, lowers the density of the crystal? (a) Frenkel (b) Schottky (c) Edge dislocation (d) Constitution of F- centres 9. Which plot is the adsorption isobar for chemisorption? (a) (b) (c) (d) 34 Class 12

OR If we add dilute aqueous solution of KI dropwise to AgNO3 aqueous solution like KI is in slight excess the AgI colloid formed will have (a) negative charge (b) positive charge ­ (c) neutral(d) nothing can be predicted. 10. Henry’s law constant for oxygen dissolved in water is 4.34 × 104 atm at 25° C. If the partial pressure of oxygen in air is 0.4 atm at 25° C, the oncentration (in moles per litre) of the dissolved oxygen in water in equilibrium with air at 25°C is (a) 1.3 × 10–3 M (b) 0.13 × 10–3 M (c) 0.28 × 10–4 M (d) 5.11 × 10–4 M 11. The given ether O CH2 CH2–OH I when reacts with cold HI gives (3) (4) OH CH2–I (1) (2) (a) mixture of 3 and 4 (b) mixture of 1 and 2 (c) mixture of 2 and 3 (d) mixture of 1 and 4. OR Phenol reacts with bromine in carbon disulphide at low temperature to give (a) m-bromophenol (b) p-bromophenol (c) o-and p-bromophenol (d) 2, 4, 6-tribromophenol. In the following questions (Q. No. 12-16) a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. 12. Assertion : Relative lowering of vapour pressure is directly proportional to the number of ions present in the solution. Reason : The relative lowering of vapour pressure of 0.1 M sugar solution is less than that of 0.1 M urea. 13. Assertion : Glucose when treated with CH3OH in presence of dry HCl gas gives a-and b-methyl glucosides. Reason : Glucose reacts with phenylhydrazine to form crystalline osazone. OR Assertion : A solution of sucrose in water is dextrorotatory but on hydrolysis in presence of little hydrochloric acid, it becomes laevorotatory. Reason : Sucrose on hydrolysis gives unequal amounts of glucose and fructose as a result of which change in sign of rotation is observed. 14. Assertion : Many reactions occuring on solid surface are zero order reactions. Pt Reason : 2NH3(g) 1130 K N2(g) + 3H2(g) rate = k 15. Assertion : OF2 is named as oxygen difluoride. Reason : In OF2, oxygen is less electronegative than fluorine. 16. Assertion : The [Ni(en)3]Cl2 has higher stability than [Ni(NH3)6]Cl2 Reason : In [Ni(en)3]Cl2, the geometry around Ni is octahedral. Chemistry 35

SECTION - B The following questions, Q. No. 17-25 are short answer type and carry 2 marks each : 17. (a) Why do the d-block elements exhibit a large number of oxidation states than the lanthanides? (b) How do the transition elements form interstitial compounds?   18. Write the IUPAC names of the products (A) and (B) in the following reactions : O (a) CH3COOH NH3 A (b) SeO2 B D OR How will you convert ethanal into the following compounds? (i) Butan-1,3-diol (ii) But-2-enal 19. Explain the following terms : (b) Electro-osmosis (a) Dialysis 20. [Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2– is diamagnetic. Explain why? 21. Identify A, B, C and D in the following reactions : CH3COOH NH3 A Br2 + KOH B CHCl3 + NaOH C LiAlH4 D D D OR (a) How are amines prepared from amides? (b) Acylation of amines with acid chloride is carried out in presence of stronger base. Why? 22. Identify the compound that on hydrogenation produces an optically inactive compound from the following compounds. CH2 Cl H CH3 (I) (II) OR Arrange the following in the increasing reactivity for SN1 reaction. Give reason. C6H5CH2Br, C6H5CH(C6H5)Br, C6H5CH(CH3)Br, C6H5C(CH3)(C6H5)Br 23. Account for the following : (i) Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. (ii) The boiling points of ethers are lower than isomeric alcohols. 24. Amino acids may be acidic, alkaline or neutral, how does this happen? What are essential and non-essential amino acids? Name one of each type. 25. Explain giving reason : (i) SN1 reaction of optically active compounds undergo racemisation whereas SN2 reaction led to inversion. (ii) No effect on reactivity of haloarene is observed by the presence of electron withdrawing group at meta position. 36 Class 12

SECTION - C Q. No. 26-30 are short answer type II carrying 3 marks each. 26. Hydrogen peroxide, H= 21O.026(a×q) decomposes to H2O(l) and O2(g) in a reaction that is first order in H2O2 and has a rate constant k 10–3 min–1.     (i) How long will it take for 15% of a sample of H2O2 to decompose? (ii) How long will it take for 85% of the sample to decompose? OR (i) For a reaction, A + B → Product, the rate law is given by, Rate = k[A]1[B]2. What is the order of the reaction? (ii) Write the unit of rate constant ‘k’ for the first order reaction. (iii) For the reaction A → B, the rate of reaction becomes twenty seven times when the concentration of A is increased three times. What is the order of reaction? 27. Describe the following about halogens (Group 17 elements): (i) Relative oxidising power of halogens. (ii) Relative acidic strength of the hydrogen halides. (iii) Perchloric acid is a stronger acid than chloric acid. 28. (a) Write a test to differentiate between pentan-2-one and pentan-3-one. (b) Compound ‘A’ was prepared by oxidation of compound ‘B’ with alkaline KMnO4. Compound ‘A’ on reduction with lithium aluminium hydride gets converted back to compound ‘B’. When compound ‘A’ is heated with compound ‘B’ in the presence of H2SO4 it produces fruity smell of compound ‘C’. To which families the compounds ‘A’, ‘B’ and ‘C’ belong to? Complete the following reactions : OR (i) CHO NaCN/HCl COOH (i) NaBH4 (ii) CH3COCH2COOC2H5 (ii) H+ (iii) OH CrO3 29. Explain why (a) Liquid ammonia bottle first cooled in ice before opening it. (b) aquatic species feel more comfortable in winter than in summer. (c) a solution of chloroform and acetone shows negative deviation from Raoult’s law. 30. (i) Account for the following : (a) Aqueous solution of methyl amine reacts with iron(III) chloride to precipitate iron(III) hydroxide. (b) The boiling points of amines are lower than those of corresponding alcohols. (ii) How is aminoethane obtained from ethanal? SECTION - D Q. No. 31-33 are long answer type carrying 5 marks each. 31. (a) Give the IUPAC name of [PtCl(NH2CH3)(NH3)2]Cl. [Fe(CN)6]4– and [FeF6]3–. [Atomic number (b) Compare the magnetic behaviour of the complex entities of Fe = 26]. (c) Tetrahedral complexes are always of high spin. Explain. Chemistry 37

OR (a) What is spectrochemical series? Explain the difference between a weak field ligand and a strong field         ligand. tChue2t+esiot no.f EFxe2p+laiionnwbhuyt?CuSO4 (b) FeSO4 solution mixed with (NH4)2SO4 solution in 1:1 molar ratio gives solution mixed with aqueous ammonia in 1 : 4 molar ratio does not give the test of 32. (a) An element crystallises in fcc structure. Its density is 7.2 g cm–3. 208 g of this element has 4.283 × 1024 atoms. Calculate the edge length of the unit cell. Ni2+ (b) Analysis shows that nickel oxide has the formula Ni0.98O1.00. What fractions of nickel exist as and Ni3+ ions? OR (a) Schottky defects generate an equal number of cation and anion vacancies while doping produces only cation vacancies and not anion vacancies. Why? (b) A metallic element crystallizes 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? (c) Why urea has a sharp melting point but glass does not? 33. (a) Compound A with molecular formula C4H9Br is treated with aq. KOH solution. The rate of this reaction depends upon the concentration of the compound A only. When another optically active isomer B of this compound was treated with aq. KOH solution, the rate of reaction was found to be dependent on concentration of compound and KOH both. (i) Write down the structural formula of both compounds A and B. (ii) Out of these two compounds, which one will be converted to the product with inverted configuration. (b) Why is it necessary to avoid even traces of moisture during the use of a Grignard reagent? OR (a) An alkyl halide, X, of formula C6H13Cl on treatment with potassium tertiary butoxide gives two isomeric alkenes Y and Z (C6H12). Both alkenes on hydrogenation give 2,3-dimethylbutane. Predict the structures of X, Y and Z. (b) Vinyl halides are less reactive than alkyl halides, but allyl halides are more reactive than alkyl halides. Explain. 38 Class 12

SOLUTIONS 1. (i)(a) (ii) (d) OR 5. (c): Friedel–Crafts alkylation of benzene with n-propyl (d) : Weak electrolytes do not dissociate in aqueousbromide will give isopropylbenzene as the major product. solution. Therefore, Wurtz-Fittig reaction is the right choice. (iii) (a) : According to Kohlrausch's law Λ°(MgI2) = 2Λ(°1[2(C.6H93)C–OO2()29M.1g0] )+=2Λ25°(.N96aI)S–c2mΛ2°(mCHo3l–C1OONa) Br + 2Na + Br — CH2CH2CH3 = 18.78 + Dry ether CH2CH2CH3 + 2NaBr (iv) (a) D 2. (i)(c) : The elements scandium and yttrium are 6. (d) : NH3 + 3Cl2 (excess) NCl3 + 3HCl called \"rare earths because they were originally discovered 7. (a) : together with lanthanides in rare minerals and isolated as oxides or \"earths\". Collectively, these metals are also called rare earth elements. (ii) (a) OR (d) : In lanthanide series, with increasing atomic number, there is a progressive decrease in the atomic as well as on radii of trivalent ions form La3+ to Lu3+. (iii) (c) : Yb2+ is more stable than Yb3+ because it will acquire stable configuration of completely filled 4f OR subshell after loosing 2 electrons. Electronic configuration (b) : Alkyl amines (1°, 2° and 3°) are stronger bases than of Yb2+ is [Xe]4f14. ammonia. This can be explained in terms of electron releasing inductive effect of alkyl group. As a result, the (iv) (a) electron density on the nitrogen atom increases and thus, they can donate the lone pair of electrons more 3. (d) : easily than ammonia. The electron releasing effect is maximum in tertiary 4. (d) : Alkyl group attached to carbonyl carbon amines and minimum in primary amines (in gas increases the electron density on the carbonyl carbon phase). 3° amine > 2° amine > 1° amine > NH3. and lowers its reactivity towards nucleophilic addition 8. (b) : In Schottky defect, equal number of cations reactions. and anions are missing from the lattice. Thus, mass of Also, as the number and size of alkyl group increases, the lattice becomes less and density gets lowered. the attack of nucleophile on the carbonyl group becomes more and more difficult due to steric hindrance. 9. (c) : The chemisorption isobar shows an initial Hence, the reactivity order will be : increase with temperature and then the expected decrease. HCHO > CH3CHO > CH3COCH3 The initial increase is because of the fact that the heat supplied acts as activation energy required in OR chemisorption. (c) : NH3 OR (a) AgNO3(aq) + KI(aq) → AgI + KNO3 Colloid AgI will absorb I– from KI electrolyte to form [AgI]I–. Therefore, it is negatively charged. 10. (d) : GivenHenry’slawconstant,KH=4.34×104atm. = 0.4 atm. ppO=2 KH . x pxOO22==KpKHOH.2xO=2 \\ 0.4 = 9.2 ×10−6 or 4.34 ×104 Chemistry 39

Moles of water (nH2O) = 1000 = 55.5 mol 16. (b) : The chelate complexes are more stable than 18 similar complexes containing unidentate ligands. Ni2+ nO2 with coordination number six forms octahedral complexes. Mole fraction of oxygen (xO2) = nO2 + nH2O 17. (a) All transition elements except the first and the last member in each series show a large number Since nO2 is very small in comparison to nH2O, of variable oxidation states. This is because difference nO2 of energy in the (n – 1) and ns-orbitals is very little. \\ xH2O = nH2O Hence, electrons from both the energy levels can be used for bond formation. or xO2 × nH2O = nO2 (b) Transition metals form a number of interstitial oprres9ne.On2t2×=in1501.–10610×0×5m51.05L–=4omfnOsoo2llusitniocne,5t.h11er×efo1r0e–,4mmoolal rairtye compounds in which small atoms such as H,C, B and = 5.11 × 10–4 M. N occupy the empty spaces in their lattices. 18. (a) 11. (b) : O SeO2 O (b) O OR (c) Phenol reacts with bromine in CS2 (or CHCl3) at (B) low temperature to form a mixture of ortho and para Cyclohexane-1, 2-dione bromophenols. OR 12. (c) : Relative lowering of vapour pressure is directly proportional to number of ions present in solution. (i) 2CH3CHO dil. NaOH Sugar and urea both being non electrolytes with same Aldol condensation molar concentration, will have same relative lowering Ethanal of vapour pressure. 13. (b) : Because of the ring structure C1 in glucose NaBH4 43 2 1 becomes chiral and hence glucose exists in two (Reduction) stereoisomeric forms, i.e., a- and b-corresponding to CH3CH(OH) CH2 CHO each stereoisomeric form, glucose forms two methyl glucosides, i.e., a- and b-methyl glucosides. 3-Hydroxybutanal OR 43 21 (c) : On hydrolysis sucrose gives equimolar mixture CH3 CH(OH) CH2CH2OH of D(+)-glucose and D(–)-fructose. Since the laureation on of fructose (–92°4°) is much more Butane-1, 3-diol than westernization of glucose +52.5°, therefore the resulting mixture becomes levorotatory. (ii) 2CH3CHO (i) dil. NaOH CH3CBuHt-2-enCaHl CHO 14. (b) (ii) H+, ∆ 15. (a) : The compounds of oxygen and fluorine are Ethanal not called oxides of fluorine but oxygen fluorides as fluorine is more electronegative than oxygen. 19. (a) Dialysis : It is a process to separate a crystalloid from a colloid by diffusion through a semipermeable 40 membrane. (b) Electro-osmosis : It is the movement of the molecules of the dispersion medium under the influence of an electric field whereas colloidal particles are not allowed to move. 2C0r.(2[4C)r:(N[AHr]3)36d]35+4:s1, Cr3+ : [Ar] 3d34s0 3d 4s 4p [Cr(NH3)6]3+ d2sp3 hybridisation It is paramagnetic due to presence of unpaired electrons. N[Ni(i(2C8)N:)[4A]2r–]3: d84s2, Ni2+ : [Ar]3d84s0 3d 4s 4p [Ni(CN)4]2– dsp2 hybridisation It is diamagnetic due to absence of unpaired electrons. Class 12

21. CH3COOH NH3 CH3C(AO) NH2 Br2 + KOH 23. (i) Acid dehydration of 2° and 3° alcohols give ∆ alkenes rather than ethers. Due to steric hindrance of nucleophilic attack by the alcohol molecule on the LiAlH4 CH3NC CHCl3 + NaOH CH3NH2 protonated alcohol molecule, does not occur. The (C) (B) protonated 2° and 3° alcohols lose water molecules to CH3 NH CH3 form stable 2° and 3° carbocations. (ii) The boiling points of ethers are much lower (D) OR than those of alcohol of comparable molar masses because like alcohols they cannot form intermolecular (a) Amides on reduction with LiAlH4 or by reaction with hydrogen bonds. Br2 and alkali give amines. O R C NH2 (i) LiAlH4 R CH2 NH2 ROHO OO O (ii) H2O HR R RR R R C NH2 Br2 + OH R NH2 Intermolecular Dipole-dipole interaction hydrogen bond (b) During acylation reaction HCl is formed. In order 24. Amino acids are classified as acidic, basic or to remove HCl and shift the equilibrium to the right neutral depending upon the relative number of amino hand side a stronger base like pyridine is added. and carboxyl groups in their molecules. R — NH2 + RCOCl Pyridine R — NHCOR + HCl (a) Equal number of amino and carboxyl groups makes it neutral (b) more number of amino groups 22. Compound(I) on hydrogenation produces an than carboxyl groups make it basic and (c) more optically inactive compound. Products obtained number of carboxyl groups as compared to amino by hydrogenation of compounds (I) and (II) are groups make it acidic. respectively Amino acids which cannot be synthesised in the body H Cl Cl H Chiral carbon and must be obtained through diet are known as CH3 CH3 CH3 * CH3 essential amino acids, e.g., valine and leucine. There are ten essential amino acids. Amino acids which can (Optically inactive) (Optically active) be synthesised in the body are known as non-essential amino acids, e.g., alanine and glutamic acids. OR For SN1, carbocation formed is resonance stabilised which is in the order : 25. (i) In SN1 reaction carbocation intermediate is C+ H2 < C+ H < formed which is a planar molecule so, an incoming nucleophile can attack from either side and a 1° benzyl 2° benzyl CH3 equilmolar mixture of two components are formed I II and resulting mixture is optically inactive. While in SN2 reaction an incoming nucleophile can C+ H < C+ CH3 attack from opposite side and inversion takes place. Cl 2° benzyl 3° benzyl (ii) N+ –O + –OH Slow step III IV II is more stable than I since CH3 group is electron donating group. O III (2° benzyl) is more stable than II due to delocalisation of +ve charge on two aromatic rings. Cl OH Cl OH Cl OH – N+ O– – N+ O– IV (3° benzyl) is more stable than III. – O– Thus, the reactivity order for SN1 is N+ << O OOH O < Cl OH N+ O– + Cl– N+ O– Fast step OO Chemistry 41

In case of meta-nitrobenzene, none of the resonating 28. (a) Pentan-2-one and pentan-3-one can be structures bear the negative charge on carbon atom differentiated by iodoform test. Pentan-2-one will give bearing the –NO2 group. Therefore, the presence of yellow precipitate of iodoform while pentan-3-one nitro group at meta- position does not stabilise the will not. negative charge and no effect on reactivity is observed by the presence of –NO2 group at meta-positions. Y   26. (i) Given, k = 1.06 × 10–3 min–1, [A]0 = 100 [A] 85 t = 2.303 log [A]0 (b) (A) is a carboxylic acid, (B) is an alcohol and (C) is k [A] an ester. 2.303 100 t = 1.06 ×10−3 min−1 log 85 t = 213.0063[2 log10 − log 85]min RCOOH LiAlH4 RCH2OH 213.0063[2 ×1−1.9294] 2303 × 0.0706 t= = 1.06 (A) (B) t = 153.39 min 153.4 min. RCOOH + RCH2OH H2SO4 RCOOCH2R (C) (ii) Given k = 1.06 × 10–3 min–1, [A]0 100 (A) (B) t= [A] 15 Fruity smell = OR 2.303 100 1.06 ×10−3 min−1 log 15 OH = 213.0063[2 log10 − log15] min (i) CHO NaCN/HCl CH CN 213.0063[2 ×1−1.1761]= COOH COOH = 2303× 0.8239 min = 1790 min. OO 1.06 (ii) H3C CH2 C OC2H5 (i) NaBH4 OR (ii) H + (i) Order of reaction is sum of powers of concentration OH terms. H3C CH CH2COOC2H5 \\ Order of reaction = 1 + 2 = 3 (ii) Unit of rate constant for first order reaction is s–1. (iii) OH CrO3 O (iii)Let r = k[A]n ...(i) Then, 27r = k[3A]n ...(ii) 29. (a) At room temperature, the vapour pressure of liquid ammonia is very high. On cooling, vapour If eqn. (ii) is divided by eqn. (i), we get pressure decreases. Hence, the liquid ammonia will 27r k[3A]n r = k[A]n or 33 = 3n not splash out. n=3 (b) In winter season at low temperature solubility of oxygen in water is higher than that in summer at \\ Thus, order = 3 high temperature. Hence, aquatic species feel more 27. (i) From top to bottom in group-17 oxidising comfortable in winter than in summer. power of halogens decreases. (c) A mixture of chloroform and acetone shows F2 > Cl2 > Br2 > I2. negative deviation from Raoult’s law because (ii) The acidic strength of the hydrohalic acids increases chloroform molecule forms H-bonding with acetone in the order : HF < HCl < HBr < HI molecule. As a result of this A B interaction becomes This order is a result of bond dissociation enthalpies of stronger than A A and B B interactions. This leads H — X bond, which decreases from H — F to H — I as to the decrease in vapour pressure and resulting in the size of halogen atom increases. negative deviation. (iii) The oxidation state of chlorine in perchloric acid H3C Cl (HOClO3) is +7 and in chloric acid (HOClO2) is +5. H3C C O H C Cl So, perchloric acid is a stronger acid than chloric acid. Cl 42 Class 12

30. (i) (a) Methyl amine reacts with water to form methyl ion. Some ligands are able to produce strong fields ∆ ∆     ∆ ∆ ammonium hydroxide whichismorebasicthanNH4OHin which, the splitting will be large whereas others and ionizes to give OH– ions. produce weak fields and consequently result in small CH3NH2 + H2O → CH3N+ H3 + OH− splitting of d-orbitals. In general, ligands can be FeCl3 + 3OH− → Fe(OH)3 ↓ + 3Cl− arranged in a series in the order of increasing field strength as given below : Red brown ppt . I– < Br– < SCN– < Cl– < S2– < F– < OH– < C2O42– < H2O < NCS– < EDTA4– < NH3 < en < CN– < CO. (b) As oxygen is more electronegative than nitrogen therefore, hydrogen bonding among alcohol molecules is stronger than among amine molecules. So, alcohols Such a series is termed as spectrochemical series. Ligands for which o < P are known as weak field have higher boiling points than amines. ligands and form high spin complexes. In this case o, (ii) the fourth electron 32egnete1gr. sLoignaenodfsthfoeregwohribcihtalsDgoiv>inPg the configuration t are known as strong field ligands and form low spin complexes. In this case it becomes more energetically favourable for the fourth electron to occupy a 31. (a) Diamminechlorido(methylamine) orbital with configuration t2g4 eg0. t2g platinum(II) chloride (b) When FeSO4 and (NH4)2SO4 solutions are mixed (b) (i) [Fe(CN)6]4– ion in 1 : 1 molar ratio, Mohr’s salt (a double salt) is formed. FeSO4(aq) + (NH4)2SO4(aq) → FeSO4·(NH4)2SO4·6H2O : FeSO4·(NH4)2SO4·6H2O Fe2+ ion is hybridised under the influence of strong Fe(2a+q) + 2NH4+(aq) + 2SO24(−aq) + 6H2O field ligand. Because Fe2+ ions are formed on dissolution of Mohr’s [Fe(CN)6]4– ion formation : salt, its aqueous solution gives the test of Fe2+ ions. When CuSO4 is mixed with ammonia, following reaction 4s 4p occurs : hybridisation CuSO4(aq) + 4NH3(aq) → [Cu(NH3)4]SO4 This complex does not produce Cu2+ ion, so the solution Since the complex ion does not contain any unpaired of CuSO4 and NH3 does not give the test of Cu2+ ion. electron, so it is diamagnetic. 32. (a) Given, structure = fcc, d = 7.2 g cm–3, m = 208 g (ii) [FeF6]3– ion N = 4.283 × 1024 atoms, a = ? 4.283 ×1024 : Number of unit cells = 4 =1.07 ×1024 Fe3+ ion is hybridised under the influence of weak Volume of unit cell = a3 \\ Total volume = Number of unit cells field ligand. × Volume of one unit cell [FeF6]3– ion formation : 4p 4d = 1.07 × 1024 × a3 Total mass = Total volume × density 4s 208 g = 1.07 ×1024 a3 × 7.2 g cm–3 As the complex ion contains five unpaired electrons, it or a3 = 1.07 208 g g cm−3 = 26.99 ×10−24 cm3 is highly paramagnetic in nature. ×1024 × 7.2 µs = 5(5 + 2) = 35 = 5.9 B.M. or a = 3 × 10–8 cm = 300 pm (c) For tetrahedral complexes, crystal field splitting energy (b) 98 Ni-atoms are associated with 100 O-atoms. Out of 98 Ni-atoms, suppose Ni present as Ni2+ = x t is always less than pairing energy. So, tetrahedral Then Ni present as Ni3+ = 98 – x complexes are always high spin. Total charge on x Ni2+ and (98 – x) Ni3+ should be equal to charge on 100 O2– ions. OR Hence, x × 2 + (98 – x) × 3 = 100 × 2 (a) The crystal field splitting, o, depends upon the or 2x + 294 – 3x = 200 or x = 94 field produced by the ligand and charge on the metal 43 Chemistry

Fraction of Ni present as N=i29+48=×991480×0 100 = 96% CH3 HOd– CH3 Bd–r Fraction of Ni present as Ni3+ =4 % O– H + C Br C     ∴ OR H (CB)H2CH3 H CH2CH3 (a) Schottky defects exist in pairs to maintain electrical Transition state CH3 +Br– neutrality. So it generates equal number of cation and HO C H anion vacancies. CH2CH3 Ionic solids are doped with metal ions of higher Inverted product valency. Therefore, some cations of lower valency are (b) Grignard reagents are highly reactive and react with displaced to maintain electrical neutrality. Hence, only cation vacancies are produced not anion vacancies. even traces of water to give corresponding hydrocarbons. RMgX + H2O RH + Mg(OH)X (b) In case of hcp unit cell, there are 6 atoms per unit cell. If the radius of metal atom is r1 then, OR Volume occupied by the metal atom = 6× 4 × πr 3 (a) 3 22 X 7 = 6 ×1.33 × × r3 = 25.08 r3 It has been shown geometrically that the base area of D hcp unit cell = 6 × 3 × 4 × r2 The two isomeric alkenes which on hydrogenation 4 yield 2,3-dimethylbutane are and the height = 4r × 2/3 \\ Volume of unit cell = Area × Height One of these is Y and the other is Z. = 6× 3 × 4r2 × 4r × 2 / 3 = 33.94 r3 These structures explain the given facts as follows : 4 Volume of the empty space of one unit cell = 33.94 r3 – 25.08 r3 = 8.86 r3 Hence, percentage void = 8.86 r3 × 100 or 26.10% 33.94 r3 (c) Urea is crystalline solid whereas glass is amorphous. Crystalline solids have sharp melting points where is amorphous solids do not possess sharp melting points. 33. (a) (i) As the rate of reaction depends upon the concentration of compound A (C4H9Br) only therefore, the reaction is proceeded by SN1 mechanism and the given DD compound will be tertiary alkyl halide, i.e., 2-bromo-2- methylpropane and the structure of (A) is (CH3)3CBr. (b) Vinyl halide itself shows resonance and is thus Optically active isomer oCfHA3CisH22-Cb*Hro(mBro)bCuHta3n. e and its stabilized. structural formula (B) is XX (ii) The rate of reaction of compound B depends both On the other hand, allyl carbonium ion formed after upon the concentration of compound B and KOH. Hence, removal of X – is stabilized by resonance and thus allyl the reaction follows SN2 mechanism. In SN2 reaction, halide is more reactive than alkyl halide. nucleophile attacks from the backside, therefore the product of hydrolysis will have opposite configuration.  44 Class 12

Self Evaluation Sheet Once you complete SQP-3, check your answers with the given solutions and fill your marks in the marks obtained column according to the marking scheme. Performance Analysis Table given at the bottom will help you to check your readiness. Q. No. Chapter Marks Per Question Marks Obtained 1 Electrochemistry 4×1 2 The d- and f-Block Elements 4×1 ............... 3 Chemical Kinetics 1 ..............% 4 Aldehydes, Ketones and Carboxylic Acids 1 5 Haloalkanes and Haloarenes 1 6 The p-Block Elements 1 7 Amines 1 8 The Solid State 1 9 Surface Chemistry 1 10 Solutions 1 11 Alcohols, Phenols and Ethers 1 12 Solutions 1 13 Biomolecules 1 14 Chemical Kinetics 1 15 The p-Block Elements 1 16 Coordination Compounds 1 17 The d- and f-Block Elements 2 18 Aldehydes, Ketones and Carboxylic Acids 2 19 Surface Chemistry 2 20 Coordination Compounds 2 21 Amines 2 22 Haloalkanes and Haloarenes 2 23 Alcohols, Phenols and Ethers 2 24 Biomolecules 2 25 Haloalkanes and Haloarenes 2 26 Chemical Kinetics 3 27 The p-Block Elements 3 28 Aldehydes, Ketones and Carboxylic Acids 3 29 Solutions 3 30 Amines 3 31 Coordination Compounds 5 32 The Solid State 5 33 Haloalkanes and Haloarenes 5 70 Total Percentage Performance Analysis Table You are done! Keep on revising to maintain the position. You have to take only one more step to reach the top of the ladder. Practise more. If your marks is A little bit of more effort is required to reach the ‘Excellent’ bench mark. > 90% TREMENDOUS! Revise thoroughly and strengthen your concepts. 81-90% EXCELLENT! Need to work hard to get through this stage. 71-80% VERY GOOD! Try hard to boost your average score. 61-70% GOOD! 51-60% FAIR PERFORMANCE! 45 40-50% AVERAGE! Chemistry


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