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ISC Class XII Analysis Of Pupil Performance 2018 : Chemistry

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Analysis of Pupil Performance CHEMISTRY Research Development and Consultancy Division Council for the Indian School Certificate Examinations New Delhi Year 2018 __________________________________________________________________________________ Published by: Research Development and Consultancy Division (RDCD) Council for the Indian School Certificate Examinations Pragati House, 3rd Floor 47-48, Nehru Place New Delhi-110019 Tel: (011) 26413820/26411706 E-mail: council@cisce.org Copyright, Council for the Indian School Certificate Examinations All rights reserved. The copyright to this publication and any part thereof solely vests in the Council for the Indian School Certificate Examinations. This publication and no part thereof may be reproduced, transmitted, distributed or stored in any manner whatsoever, without the prior written approval of the Council for the Indian School Certificate Examinations. FOREWORD This document of the Analysis of Pupils Performance at the ISC Year 12 and ICSE Year 10 Examination is one of its kind. It has grown and evolved over the years to provide feedback to schools in terms of the strengths and weaknesses of the candidates in handling the examinations. We commend the work of Mrs. Shilpi Gupta (Deputy Head) of the Research Development and Consultancy Division (RDCD) of the Council and her team, who have painstakingly prepared this analysis. We are grateful to the examiners who have contributed through their comments on the performance of the candidates under examination as well as for their suggestions to teachers and students for the effective transaction of the syllabus. We hope the schools will find this document useful. We invite comments from schools on its utility and quality. Gerry Arathoon Chief Executive & Secretary October 2018 i PREFACE The Council has been involved in the preparation of the ICSE and ISC Analysis of Pupil Performance documents since the year 1994. Over these years, these documents have facilitated the teaching-learning process by providing subject/ paper wise feedback to teachers regarding performance of students at the ICSE and ISC Examinations. With the aim of ensuring wider accessibility to all stakeholders, from the year 2014, the ICSE and the ISC documents have been made available on the Council s website www.cisce.org. The documents include a detailed qualitative analysis of the performance of students in different subjects which comprises of examiners comments on common errors made by candidates, topics found difficult or confusing, marking scheme for each answer and suggestions for teachers/ candidates. In addition to a detailed qualitative analysis, the Analysis of Pupil Performance documents for the Examination Year 2018 have a component of a detailed quantitative analysis. For each subject dealt with in the document, both at the ICSE and the ISC levels, a detailed statistical analysis has been done, which has been presented in a simple user-friendly manner. It is hoped that this document will not only enable teachers to understand how their students have performed with respect to other students who appeared for the ICSE/ISC Year 2018 Examinations, but also provide information on how they have performed within the Region or State, their performance as compared to other Regions or States, etc. It will also help develop a better understanding of the assessment/ evaluation process. This will help teachers in guiding their students more effectively and comprehensively so that students prepare for the ICSE/ ISC Examinations, with a better understanding of what is required from them. The Analysis of Pupil Performance document for ICSE for the Examination Year 2018 covers the following subjects: English (English Language, Literature in English), Hindi, History, Civics and Geography (History and Civics, Geography), Mathematics, Science (Physics, Chemistry, Biology), Commercial Studies, Economics, Computer Applications, Economic Applications, Commercial Applications. Subjects covered in the ISC Analysis of Pupil Performance document for the Year 2018 include English (English Language and Literature in English), Hindi, Elective English, Physics (Theory), Chemistry (Theory), Biology (Theory), Mathematics, Computer Science, History, Political Science, Geography, Sociology, Psychology, Economics, Commerce, Accounts and Business Studies. I would like to acknowledge the contribution of all the ICSE and the ISC examiners who have been an integral part of this exercise, whose valuable inputs have helped put this document together. I would also like to thank the RDCD team of, Dr. M.K. Gandhi, Dr. Manika Sharma, Mrs. Roshni George and Mrs. Mansi Guleria who have done a commendable job in preparing this document. Shilpi Gupta Deputy Head - RDCD October 2018 ii CONTENTS Page No. FOREWORD i PREFACE ii INTRODUCTION 1 QUANTITATIVE ANALYSIS 3 QUALITATIVE ANALYSIS 10 INTRODUCTION This document aims to provide a comprehensive picture of the performance of candidates in the subject. It comprises of two sections, which provide Quantitative and Qualitative analysis results in terms of performance of candidates in the subject for the ISC Year 2018 Examination. The details of the Quantitative and the Qualitative analysis are given below. Quantitative Analysis This section provides a detailed statistical analysis of the following: Overall Performance of candidates in the subject (Statistics at a Glance) State wise Performance of Candidates Gender wise comparison of Overall Performance Region wise comparison of Performance Comparison of Region wise performance on the basis of Gender Comparison of performance in different Mark Ranges and comparison on the basis of Gender for the top and bottom ranges Comparison of performance in different Grade categories and comparison on the basis of Gender for the top and bottom grades The data has been presented in the form of means, frequencies and bar graphs. Understanding the tables Each of the comparison tables shows N (Number of candidates), Mean Marks obtained, Standard Errors and t-values with the level of significance. For t-test, mean values compared with their standard errors indicate whether an observed difference is likely to be a true difference or whether it has occurred by chance. The t-test has been applied using a confidence level of 95%, which means that if a difference is marked as statistically significant (with * mark, refer to t-value column of the table), the probability of the difference occurring by chance is less than 5%. In other words, we are 95% confident that the difference between the two values is true. t-test has been used to observe significant differences in the performance of boys and girls, gender wise differences within regions (North, East, South and West), gender wise differences within marks ranges (Top and bottom ranges) and gender wise differences within grades awarded (Grade 1 and Grade 9) at the ISC Year 2018 Examination. The analysed data has been depicted in a simple and user-friendly manner. 1 Given below is an example showing the comparison tables used in this section and the manner in which they should be interpreted. The table shows comparison between the performances of boys and girls in a particular subject. The t-value of 11.91 is significant at Comparison on the basis of Gender Gender Girls Boys *Significant at 0.05 level N 2,538 1,051 Mean 66.1 60.1 SE 0.29 0.42 t-value 11.91* 0.05 level (mentioned below the table) with a mean of girls as 66.1 and that of boys as 60.1. It means that there is significant difference between the performance of boys and girls in the subject. The probability of this difference occurring by chance is less than 5%. The mean value of girls is higher than that of boys. It can be interpreted that girls are performing significantly better than boys. The results have also been depicted pictographically. In this case, the girls performed significantly better than the boys. This is depicted by the girl with a medal. Qualitative Analysis The purpose of the qualitative analysis is to provide insights into how candidates have performed in individual questions set in the question paper. This section is based on inputs provided by examiners from examination centres across the country. It comprises of question wise feedback on the performance of candidates in the form of Comments of Examiners on the common errors made by candidates along with Suggestions for Teachers to rectify/ reduce these errors. The Marking Scheme for each question has also been provided to help teachers understand the criteria used for marking. Topics in the question paper that were generally found to be difficult or confusing by candidates, have also been listed down, along with general suggestions for candidates on how to prepare for the examination/ perform better in the examination. 2 STATISTICS AT A GLANCE Total Number of Candidates: 41,895 Mean Marks: Highest Marks: 100 64.4 Lowest Marks: 02 3 PERFORMANCE (STATE-WISE & FOREIGN) West Bengal 65.9 Uttarakhand 63.7 Uttar Pradesh 63.1 Tripura 55.1 Telangana 61.3 Tamil Nadu 71.2 Sikkim 56.1 Rajasthan 58.8 Punjab 62.6 Puducherry 62.0 Odisha 59.7 New Delhi 71.1 Meghalaya 64.4 Manipur 51.2 Maharashtra 70.3 Madhya Pradesh 62.0 Kerala 69.4 Karnataka 71.9 Jharkhand 59.8 Himachal Pradesh 67.8 Haryana 71.8 Gujarat 67.2 Goa 61.6 Chhattisgarh 52.9 Chandigarh 70.5 Bihar 66.1 Assam 84.7 Andhra Pradesh 53.6 Foreign 77.3 The States of Assam, Karnataka and Haryana secured highest mean marks. Mean marks secured by candidates studying in schools abroad were 77.3. 4 GENDER-WISE COMPARISON BOYS GIRLS Mean Marks: 65.4 Mean Marks: 63.7 Candidates: 17,194 Candidates: 24,701 Number of Number of Comparison on the basis of Gender Gender Girls Boys *Significant at 0.05 level N Mean SE t-value 17,194 24,701 65.4 63.7 0.12 0.11 10.49* Girls performed significantly better than boys. 5 REGION-WISE COMPARISON East North Mean Marks: 64.1 Mean Marks: 63.2 Number of Candidates: 12,684 Number of Candidates: 22,509 Highest Marks: 100 Lowest Marks: 09 Highest Marks: 100 Lowest Marks: 02 REGION Mean Marks: 68.5 Mean Marks: 68.4 Number of Candidates: 4,510 Number of Candidates: 1,995 Highest Marks: 100 Lowest Marks: 19 South Mean Marks: 77.3 Number of Candidates: 197 Highest Marks: 100 Lowest Marks: 37 Foreign 6 Highest Marks: 100 Lowest Marks: 15 West Mean Marks obtained by Boys and Girls-Region wise 80.4 64.0 62.7 North 65.2 69.3 63.4 East 69.6 67.7 South 67.6 West 75.0 Foreign Comparison on the basis of Gender within Region Region North (N) East (E) South (S) West (W) Foreign (F) Gender Girls Boys Girls Boys Girls Boys Girls Boys Girls Boys N Mean SE 9,009 13,500 5,142 7,542 2,195 2,315 762 1,233 86 111 64.0 62.7 65.2 63.4 69.3 67.7 69.6 67.6 80.4 75.0 0.17 0.14 0.22 0.20 0.34 0.34 0.61 0.50 1.41 1.50 *Significant at 0.05 level The performance of girls was significantly better than that of boys in all the regions. 7 t-value 6.13* 5.96* 3.44* 2.57* 2.65* MARK RANGES : COMPARISON GENDER-WISE Comparison on the basis of gender in top and bottom mark ranges Marks Range Top Range (81-100) Bottom Range (0-20) Gender Girls Boys Girls Boys N Mean SE 3,554 4,811 2 10 88.2 88.3 16.0 13.0 0.08 0.07 1.00 2.04 Boys Girls t-value -0.21 1.32 All Candidates 88.3 81 - 100 88.2 88.2 69.9 61 - 80 70.1 70.0 51.0 41 - 60 51.3 51.2 36.3 21 - 40 36.5 36.3 13.0 0 - 20 16.0 13.5 8 GRADES AWARDED : COMPARISON GENDER-WISE Comparison on the basis of gender in Grade 1 and Grade 9 Grades Gender Girls Boys Girls Boys Grade 1 Grade 9 *Significant at 0.05 level N Mean SE 1,356 1,837 179 417 93.7 93.8 30.6 29.9 0.08 0.07 0.22 0.19 Boys Grade 1 Girls t-value -1.55 2.12* All Candidates 93.8 93.7 93.8 1 84.3 84.3 84.3 2 74.2 74.2 74.2 3 64.4 64.5 64.4 4 Grade 9 57.0 57.1 57.1 5 52.0 52.0 52.0 6 Grade 9 47.0 47.1 47.1 7 40.4 40.7 40.5 8 9 9 29.9 30.6 30.1 THEORY (PAPER 1) Question 1 (a) Fill in the blanks by choosing the appropriate word/words from those given in the [4 1] brackets: (square pyramidal, electrical, 74, 26, sp3d2, sp3d, chemical, 68, 32, tetrahedral, yellow, white, iodoform, Lucas) (i) A Galvanic cell converts _______ energy into ______ energy. (ii) (iii) (iv) The percentage of unoccupied spaces in bcc and fcc arrangements are _______ and ________ respectively. Propan-2-ol on reaction with iodine and sodium hydroxide gives ______ precipitate and the reaction is called ________test. The geometry of XeOF 4 molecule is ________ and the hybridisation of xenon atom in the molecule is __________. (b) Complete the following statements by selecting the correct alternative from the [4 1] choices given: (i) During the course of an S N 1 reaction, the intermediate species formed is: (1) (2) (3) (4) (ii) (iii) a carbocation a free radical a carbanion an intermediate complex Purification of aluminium by electrolytic refining is called: (1) Serpeck s process (2) Hoope s process (3) Hall s process (4) Baeyer s process An aqueous solution of urea freezes at 0 186oC, K f for water = 1 86 K kg mol-1, K b for water = 0 512 K kg mol-1. The boiling point of urea solution will be: (1) (2) 373 065 K 373 186 K 10 (3) (4) (iv) (c) 373 512 K 373 0512 K In the dehydration of alcohols to alkenes by heating with concentrated sulphuric acid, the initiation step is: (1) formation of carbocation (2) formation of an ester (3) protonation of alcohol molecule (4) elimination of water Match the following: [4 1] (i) Rate constant (a) Dialysis (ii) Biodegradable polymer (b) Glycine (iii) Zwitter ion (c) Arrhenius equation (iv) Purification of colloids (d) PHBV (d) Answer the following questions: (i) (1) Why does the density of transition elements increase from Titanium to [4 2] Copper? (at. no. Ti = 22, Cu = 29) (2) Why is zinc not regarded as a transition element? (at. no. Zn = 30) (ii) Identify the compounds A, B, C and D. 3 2 / + 3 2 / (iii) Calculate the osmotic pressure of a solution prepared by dissolving 0 025g of K 2 SO 4 in 2 0 litres of water at 25oC assuming that K 2 SO 4 is completely dissociated. (mol. wt. of K 2 SO 4 = 174 g mol-1) (iv) What type of isomerism is shown by the following coordination compounds: [Pt Cl 2 (NH 3 ) 4 )] Br 2 and [Pt Br 2 (NH 3 ) 4 ] Cl 2 . Write their IUPAC names. 11 Comments of Examiners (a) (i) Some candidates filled electrical and chemical in the blank instead of chemical and electrical. (ii) Many candidates wrote 68% and 74% instead of 32% and 26%. (iii) A few candidates wrote white precipitate instead of yellow precipitate and Lucas test instead of iodoform test. (iv) Some candidates wrote tetrahedral instead of square pyramidal in the first blank while a few candidates wrote d2sp3 instead of sp3d2 in the second blank. (b) (i) Some of the candidates wrote carbonation instead of a carbocation. (ii) Instead of Hoope s process, some candidates gave incorrect options such as Baeyer s process or Hall's process. (iii) Instead of 373.0512 K a few candidates wrote 373.512 K or 373.065 K which was not correct. (iv) Many candidates wrote the formation of carbocation or elimination of water instead of protonation of alcohol molecule. (c) Most of the candidates gave correct answers. In a few cases, purification of colloids was matched with glycine and zwitterion with dialysis. Suggestions for teachers Explain electrochemical cell and its working thoroughly. Discuss different types of unit cells and their packing in detail. Give adequate practice for different tests based on organic reactions. Explain the geometry and hybridization of compounds with examples. Explain the mechanism of organic reactions in a stepwise manner. Ask the students to learn colligative properties in detail. Interpret topics such as biodegradable, polymers, biomolecules, surface chemistry and chemical kinetics in detail. Ask students to learn periodic properties of transition elements. Give more practice in conversion of organic compounds. Every step of the conversion must be shown with proper conditions. Point out the importance of Van t Hoff factor while teaching abnormal molecular weight. Clarify isomers and isomerism of coordination compounds with examples. Explain rules of nomenclature with emphasis on correct spelling. (d) (i) (1) Most of the candidate did not write the correct explanation for increase of the density of transition elements. Their explanation was in terms of increase in atomic number instead of increase in nuclear change. (2) The reason for zinc not being regarded as a transition element was not written properly by majority of the candidates. (ii) Most of the candidates identified compounds A, B, C and D correctly but some candidates failed to identify B as CH 3 COONH 4. (iii) The value of osmotic pressure was not calculated correctly by most of the candidates. Van t Hoff factor was not considered, hence the answer obtained was one-third of the correct answer. (iv) The type of isomerism was correctly identified by most candidates, but many could not write the correct nomenclature of the compounds. 12 MARKING SCHEME Question 1 (a) (i) Chemical, electrical (ii) 32, 26 (iii) Yellow, iodoform (iv) Square pyramidal, sp3d2 (i) 1 or a carbocation (ii) 2 or Hoope s process (iii) 4 or 373 0512 K (iv) 3 or Protonation of alcohol molecule (i) Rate constant (c) Arrhenius equation (ii) Biodegradable polymer (d) PHBV (iii) Zwitter ion (b) Glycine (iv) Purification of colloids (a) Dialysis (b) (c) (d) (i) (ii) (1) On moving from Ti to Cu, the atomic radii decrease due to increase in nuclear charge. Therefore, atomic volume decreases with increase in atomic mass. Hence, density increases. (2) Zn has completely filled d-orbitals in its atomic as well as in its common oxidation state. (Zn+2) OR d10 configuration. 3 3 + 3 3 3 4 [B] [A] or [A] 3 / Acetic Acid [B] 3 4 / Ammonium Acetate [C] 3 2 / Acetamide / Ethanamide (iii) [D] 3 2 / Methyl Amine = = 3 0 025 174 2 0 0821 298 = 5 2728 10-3atm 13 3 2 2 + [C] 3 2 [D] (iv) Isomerism - Ionisation Isomerism IUPAC names - tetraamminedichloridoplatinum(IV) bromide and tetraamminedibromidoplatinum(IV) chloride [2] Question 2 (a) (i) Write the rate law expression for the reaction A + B + C D + E, if the order of reaction is first, second and zero with respect to A, B and C, respectively. (ii) How many times the rate of reaction will increase if the concentration of A, B and C are doubled in the equation given in (i) above? OR (b) The rate of reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation (E a ) of the reaction assuming that it does not change with temperature. (R = 8 314 J K-1 mol-1) Comments of Examiners (a) (i) Rate Law expression for the reaction A+ B+ C D+E was answered correctly by many candidates. However, some candidates gave the expression as: rate = [A]1[B]2[C] or rate K[A]1[B]2[C] , which was incorrect. (ii) The increase in the rate of reaction was calculated correctly by most of the candidates. (b) Some candidates used incorrect formula while a few others wrote wrong value of R . Instead of writing R= 8.314 J K-1 mol-1 they wrote R=0.0821. Some candidates could not write the correct unit along with the answer. Suggestions for teachers Explain the Rate Law expression with examples. Also give adequate practise for different order reactions. Make the relationship between change in concentration and rate of reaction for different order reactions clear to the students. Teach how Arrhenius equation is used to calculate the energy of activation (Ea) by using rate constant method. MARKING SCHEME Question 2 (a) (i) Rate law expression Rate = k [A]1 [B]2 [C]0 (ii) OR Rate = k [A]1 [B]2 Rate of reaction will increase 8 times if the concentration of A, B and C are doubled. OR (b) T 1 = 293 K T 2 = 313 K K2 / K1 = 4 R = 8 314 J K-1 mole-1 14 Log 2 = 1 Log 4 = 2 303 2 1 1 2 313 293 2 303 8 314 293 313 20 0 6021 = 19 147 91709 = 52862 94 J / mol or 52 863 kJ / mol [2] Question 3 (a) How do antiseptics differ from disinfectants? (b) State the role of the following chemicals in the food industry: (i) Sodium benzoate (ii) Aspartame Comments of Examiners (a) The difference between antiseptics and disinfectants was not brought out clearly by most of the candidates. Also, the examples given by many candidates were not correct. (b) (i) A few candidates could not write the role of Sodium benzoate in the food industry. (ii) For aspartame, some candidates wrote that it is used as a flavouring agent instead of an artificial sweetening agent. Suggestions for teachers Encourage students to read the Unit: Chemistry in Everyday life. The use of chemicals in medicine should be illustrated to the students with suitable examples. Discuss the use of various chemicals in the food industry with suitable examples. MARKING SCHEME Question 3 (a) Antiseptics Disinfectants (i) These chemicals prevent the growth of micro-organism or may even kill them without affecting the living tissues. (i) These chemicals destroy the microorganism but are harmful for living tissues. (ii) They are applied to living tissues such (ii) as ulcers, wounds, diseased stem surface. They are generally used to kill microorganisms and are applied to inanimate objects like floor, toilets. (Any one difference for each. An example will not be accepted as a difference.) (b) (i) Sodium benzoate: Food preservative (ii) Aspartame: artificial sweetening agent [2] 15 Question 4 An aromatic organic compound [A] on heating with NH 3 and Cu 2 O at high pressure gives [B]. The compound [B] on treatment with ice cold solution of NaNO 2 and HCl gives [C], which on heating with Cu/HCl gives compound [A] again. Identify the compounds [A], [B] and [C]. Write the name of the reaction for the conversion of [B] to [C]. Comments of Examiners Compounds (A) (B) and (C) were identified correctly by most of the candidates. A few candidates wrote (A) as benzene instead of chlorobenzene and (C) as chlorobenzene instead of benzenediazonium chloride. Some candidates wrote the name of the reaction as Sandmeyer s reaction instead of Diazotisation reaction. Suggestions for teachers Give ample practice for conversion of Organic compounds. Teach named organic reactions with proper conditions. MARKING SCHEME Question 4 Cl (A) O Or C6H5Cl or Chlorobenzene NH2 (B) O Or C6H5NH2 or Aniline Or C6H5N2Cl or Benzenediazonium chloride Diazotisation reaction N2Cl (C) O [2] Question 5 Write the names of the monomers for each of the following polymers: (a) Bakelite (b) Nylon 2 nylon 6 16 Comments of Examiners (a) The monomers of bakelite were named correctly by most candidates. A few candidates wrote benzene instead of phenol and acetaldehyde instead of formaldehyde. (b) Monomers of Nylon 2 nylon - 6 were not written correctly by most of candidates. Many candidates wrote monomers of Nylon 6 (i.e. caprolactum) or monomers of nylon 6, 6 (i.e. ethylene diamine and adipic acid) instead of monomers of Nylon - 2 nylon 6 (i.e. glycine and 5-aminocaproic acid). Suggestions for teachers Teach the correct pair of monomers to students. Give practice in writing names and structures of monomers and units of polymers. Elucidate biodegradable and nonbiodegradable polymers in class in detail. MARKING SCHEME Question 5 (a) Bakelite: phenol and formaldehyde (b) Nylon 2 nylon 6: glycine and 5 aminocaproic acid [2] Question 6 Name the purine bases and pyrimidine bases present in RNA and DNA. Comments of Examiners Many candidates did not specify the purine and the pyrimidine bases of DNA and RNA. They combined and wrote all the bases. Suggestions for teachers Teach structure of RNA and DNA with proper purines and pyrimidine bases along with diagrams. MARKING SCHEME Question 6 Purine bases Pyrimidine bases DNA Adenine and guanine Thymine and cytosine RNA Adenine and guanine Uracil and cytosine 17 Question 7 (a) [2] How will you obtain the following? (Give balanced equation.) (i) Picric acid from phenol. (ii) Ethyl chloride from diethyl ether. OR (b) How will you obtain the following? (Give balanced equation.) (i) Anisole from phenol (ii) Ethyl acetate from ethanol. Comments of Examiners (a)(i)Most of the candidates wrote unbalanced equations. A few of them did not write Suggestions for teachers concentrated HNO 3 or concentrated H 2 SO 4 . Explain organic reactions and Some candidates did not write the by-product i.e. conversions with proper reactants, H 2 O for the conversion of phenol to picric acid. catalysts and conditions. (ii) For the conversion of C 2 H 5 -O-C 2 H 5 to C 2 H 5 Cl, More practice should be given for the some candidates used Cl 2 instead of PCl 5 or conversion of organic compounds with balanced equations. SOCl 2 . A few candidates did not write the by Explain to the students to write the by product. product in all organic reactions. (b) (i) For the conversion of phenol to anisole, many candidates converted phenol directly by reacting with CH 3 Br. They did not convert phenol to phenoxide. (ii) For the conversion of ethanol to ethyl acetate, most of the candidates wrote incorrect structure of the product. Some candidates did not balance the equation. MARKING SCHEME Question 7 (a) (i) OH O + 3HNO3 conc. (ii) . 2 4 OH O2N O NO2 + 3H2O NO2 C 2 H 5 OC 2 H 5 + PCl 5 2C 2 H 5 Cl + POCl 3 or C 2 H 5 OC 2 H 5 + SOCl 2 2C 2 H 5 Cl + SO 2 18 OR OCH3 ONa (b) OH (i) O (ii) 2 O C 2 H 5 OH + CH 3 COOH 3 2 4 + NaBr O Anisole CH 3 COOC 2 H 5 + H 2 O Or C 2 H 5 OH + CH 3 COCl CH 3 COOC 2 H 5 + HCl Or C 2 H 5 OH + (CH 3 CO) 2 O CH 3 COOC 2 H 5 + CH 3 COOH [2] Question 8 40% of a first order reaction is completed in 50 minutes. How much time will it take for the completion of 80% of this reaction? Comments of Examiners Most of the candidates calculated the answer by using the unitary method instead of using the appropriate formula. Some candidates substituted the value of (a-x) as 40 instead of 60. Some candidates reported the time in seconds which was not required. Suggestions for teacher Give more practice in numerical problems based on order reaction. Emphasise on step by step calculations to calculate the value of k and t. Train students to solve problems in a step by step manner: formula substitution calculation answer with unit. MARKING SCHEME Question 8 k= = 2 303 2 303 50 2 303 log10 log 100 60 k = 50 0 2218 k = 0 0102 min-1 2 303 100 t = log 20 2 303 t = 0 0102 0 6989 t = 157 8 min 19 [3] Question 9 (a) The freezing point of a solution containing 5 85g of NaCl in 100 g of water is 3 348 oC. Calculate van t Hoff factor i for this solution.What will be the experimental molecular weight of NaCl? (K f for water = 1 86 K kg mol-1, at. wt. Na = 23, Cl = 35 5) OR (b) An aqueous solution containing 12 48g of barium chloride (BaCl 2 ) in 1000 g of water, boils at 100 0832oC. Calculate the degree of dissociation of barium chloride. (K b for water = 0 52 K kg mol-1, at. wt. Ba = 137, Cl = 35 5) Comments of Examiners (a) Most candidates calculated the value of van t Hoff factor (i) correctly but some failed to calculate the value of experimental molecular weight of NaCl. (b) Most of the candidates calculated the correct value of van t Hoff factor (i) but they were not able to calculate the number of particles formed after ionisation of BaCl 2 i.e. BaCl 2 Ba2+ + 2Cl-, n=3 They were not able to substitute the value of n correctly in the formula, hence, degree of dissociation of BaCl2 was not calculated correctly. Suggestions for teachers Teach abnormal molecular weight, using van t Hoff factor (i) along with degree of dissociation/ association comprehensively to the students. Give more practice in numerical problems based on van t Hoff factor. MARKING SCHEME Question 9 = (a) Or = 1000 5 85 1000 3 348 = 1 86 = 3 348 1 86 =1 8 58 5 100 = = or (b) 58 5 = 1 8 = 32 5 g mol-1 OR w2 = 12 48 g, Ts = 100 0832oC w1 = 1000 g, Kb for water = 0 52 K mol-1 M2 (BaCl2) = 208 20 Tb = Ts To = 100 0832 100 = 0 0832oC 1000 2 M2= = 1 1000 52 12 48 0 0832 1000 =78g mol-1 i= = 208 78 = 2 666 or 2 67 BaCl2 Ba+2 + 2Cl (3 ions) 1 2 67 1 1 67 = 1 = 3 1 = 2 = 0 835 or 83 5% [3] Question 10 Examine the defective crystal given below and answer the question that follows: A+ B- B- A+ B- A+ B- A+ B- A+ B- A+ B- A+ B- A+ A+ B- State if the above defect is stoichiometric or non-stoichiometric. How does this defect affect the density of the crystal? Also, write the term used for this type of defect. Comments of Examiners Most of the candidates were able to answer this part correctly. Some candidates wrote non- stoichiometric instead of stoichiometric defect. The density of crystal should decrease but some candidates wrote that there would be no change in density. The term used for this type of defect was Schottky defect, but some candidates wrote it as Frenkel defect. 21 Suggestions for teachers Explain various types of imperfections in solids to students. Also discuss how these imperfections affect the properties of the crystal. MARKING SCHEME Question 10 The defect is stoichiometric (because equal number of cations and anions are missing from lattice sites.) It lowers the density of the crystal. Schottky Defect. Question 11 [3] Give reason for each of the following: (a) For ferric hydroxide sol the coagulating power of phosphate ion is more than chloride ion. (b) Medicines are more effective in their colloidal form. (c) Gelatin is added to ice creams. Comments of Examiners (a) Most of the candidates did not write the correct reason. Many did not use Hardy Schulze law to explain the answer. (b) Many candidates gave incorrect reason regarding effectiveness of medicines in their colloidal form. (c) This part was also answered incorrectly by many candidates. Quite a few wrote that it acts as a flavouring or a sweetening agent. Suggestions for teachers Familiarise students with different laws/ principles/ key concepts and their applications thoroughly, especially those which are useful in our daily lives. Give more practice to students in answering reasoning type questions. MARKING SCHEME Question 11 (a) According to Hardy-Schulze law, phosphate ion has more negative charge as compared to chloride ion. (b) Assimilation is easy due to their colloidal size. (c) Gelatin when added to ice creams acts as an emulsifier and helps to stabilise the emulsion. (Protective colloid). 22 [3] Question 12 For the complex ion [Fe(CN)6]3- , state: (a) (i) the type of hybridisation. (ii) the magnetic behaviour. (iii) the oxidation number of the central metal atom. Write the IUPAC name of [Co(en)2Cl2]+ ion and draw the structures of its geometrical isomers. (b) Comments of Examiners (a) For the complex ion [Fe(CN)6]3-: Suggestions for teachers 2 3 the type of hybridisation was d sp but some candidates wrote sp3d2 hybridization. (ii) the magnetic behaviour of [Fe(CN)6]3- was paramagnetic but a few candidates wrote diamagnetic behavior. (iii) many candidates wrote oxidation state of central metal as +6 or -3 instead of +3. (b) Most of the candidates were unable to write the correct IUPAC name of the complex compound. They did not write the correct alphabetic order in case of name of Ligands. Oxidation number of central metal atom was not calculated correctly. Most of the candidates were unable to draw the structures of geometrical isomers. (i) Explain the method to determine the type of hybridization, magnetic behaviour and Oxidation state, by using valence bond theory. Discuss strong field, weak field, ligands, low spin complexes and high spin complexes with magnetic behaviour with students. Teach the rules of nomenclature of coordination compounds in detail and give adequate practice. Explain Geometrical isomers with examples. MARKING SCHEME Question 12 (a) (b) (i) d2sp3 (ii) Paramagnetic (iii) +3 dichloridobis(ethylenediamine) cobalt (III) ion en Cl Cl + en + Cl or Co Cl + en Co en Co Cl en Cl d-form l-form Cis 23 transform en [3] Question 13 (a) Explain why: (i) Mn2+ is more stable than Fe2+ towards oxidation to +3 state. (At. no. of Mn = 25, Fe = 26) (ii) Transition elements usually form coloured ions. (iii) Zr and Hf exhibit similar properties. (At. no. of Zr = 40, Hf = 72) OR (b) Complete and balance the following chemical equations: (i) KMnO4 + KI + H2SO4 _____ + ______ + ______ + _______ (ii) K2Cr2O7 + H2SO4 + H2S _____ + ______ + ______ + _______ (iii) KMnO4 + H2SO4 + FeSO4 _____ + ______ + ______ + _______ Comments of Examiners (a) (i) Many candidates wrote only the electronic configuration of Mn2+ and Fe2+ but they could not explain the reason for the stability of Mn2+ towards oxidation to + 3 state. (ii) Some candidates explained that formation of coloured ions of transition elements was due to vacant d orbital or partially filled d orbital which was incorrect. (iii)Most candidates were not able to give the reason for Zr and Hf exhibiting similar properties. (b) Unbalanced/ partially balanced/ incorrect equations were given by many candidates. Some candidates wrote incorrect formulae of products such as Mn(SO4)2 instead of MnSO4, Cr(SO4)3 instead of Cr2(SO4)3 and Fe2SO4 instead of Fe(SO4)3. Suggestions for teachers Explain properties of d Block elements in detail with the help of Orbital diagrams and suitable examples. Keywords such as d-d transition, half-filled sub shells, stability, etc. must be explained. Also explain Lanthanide contraction and its consequences. Give more practice in writing complete and correctly balanced chemical equations. Student should be explained the oxidising and reducing properties of K2 Cr2 CO7 and KMn O4 MARKING SCHEME Question 13 (a) (i) Electronic configuration of Mn+2 is 3d5, which is half filled and hence stable. Hence, it cannot lose third electron easily. In case of Fe+2 electronic configuration is 3d6. Hence,it can lose one electron easily to give stable configuration 3d5. (ii) It is due to d d transition by absorbing part of visible light. 24 (iii) (b) Due to Lanthanoid contraction they have same atomic and ionic size. or They have same general electronic configuration. (i) OR 2KMnO4 + 10KI + 8H2SO4 6K2SO4 + 2MnSO4 + 5I2 + 8H2O (ii) K2Cr2O7 +4H2SO4 + 3H2S K2SO4 + Cr2(SO4)3 + 3S + 7H2O (iii) 2KMnO4 +8H2SO4 + 10FeSO4 K2SO4+2MnSO4+5Fe2(SO4)3+8H2O [3] Question 14 (a) Arrange the following in the increasing order of their basic strength: C2H5NH2 , C6H5NH2 , (C2H5)2NH (b) Give a balanced chemical equation to convert methyl cyanide to ethyl alcohol. (c) What happens when benzene diazonium chloride reacts with phenol in weak alkaline medium? (Give balanced equation). Comments of Examiners (a) Instead of writing in increasing order of basic strength, a few candidates wrote the answer in decreasing order. Some candidates wrote incorrect order such as C2H5NH2 < C6H5NH2 < (C2H5)2NH instead of C6H5NH2 < C2H5NH2 < (C2H5)2NH. (b)Many candidates converted methyl cyanide to acetic acid or methyl cyanide to ethyl amine instead of methyl cyanide to ethyl alcohol. Many candidates did not write the by-product. (c) The reaction of benzene diazonium chloride with phenol in weak alkaline medium was not written by many candidates. Some candidates were neither able to write the correct product nor able to write the colour of the dye. Only a few could write the correct balanced equation. 25 Suggestions for teachers Explain clearly giving reasons how the basic strength of amines increases or decreases. Insist upon learning conversion of one organic compound to other and give ample practice in writing balanced chemical equations with proper conditions/reagents. Instruct students to read the observations of different organic reactions with colour or precipitate. Diazotisation, formation of azo dye should be explained properly to students. MARKING SCHEME Question 14 (a) The increasing order of basic strength given organic compounds: C6H5NH2 <C2H5NH2 < (C2H5)2NH (b) 3 Or 4[ ] 4 or 3 2 2 2 5 3 + 2 2 (c) O N2Cl + H O OH 2 3 2 + 2 + 2 3 2 2 O 2 3 2 + 2 + 2 O OH + HCl N=N P-hydroxyazobenzene (Orange dye) [3] Question 15 Name the sulphide ore of Copper. Describe how pure copper is extracted from this ore. Comments of Examiners Name of copper ore was mentioned correctly by most of the candidates. However, extraction of copper from the sulphide ore was not given stepwise by many candidates. Quite a few wrote the name of process such as Roasting, Smelting, Bessemerisation without giving the details and the chemical equation/s involved. Some forgot to write electro refining . Suggestions for teachers Advise students to learn metallurgy in detail. Explain principles and process of isolation of metals with the help of flowcharts. Interpret the extraction of metal with proper steps like concentration, roasting, smelting, etc. with complete balanced equations involved. The process of refining should also be illustrated completely. 26 MARKING SCHEME Question 15 The sulphide ore of copper is: Chalcopyrite / copper pyrite / CuFeS2 Extraction of pure copper from its ore: Froth flotation process Roasting concentrated ore is heated with excess of air or oxygen. 2CuFeS2 + O2 Cu2S + 2FeS + SO2 Smelting Roasted ore is mixed with coke and sand and fed into the blast furnace. 2Cu2S + 3O2 2Cu2O + 2SO2 2FeS + 3O2 2FeO + 2SO2 FeO + SiO2 FeSiO3 Auto reduction takes place in Bessemer Converter or Bessemerisation 2Cu2O + Cu2S 6Cu + SO2 Blister Copper Electrorefining gives pure Cu (99 9%) (either equation or statement) [5] Question 16 (a) (i) Calculate the emf and for the cell reaction at 25oC: 2+ 2+ Zn(s) ( ) ( ) ( ) ( 0 1M) (ii) (0 01M) Given 2+ = 0 763 2+ = 0 403 Define the following terms: (1) Equivalent conductivity (2) Corrosion of metals OR (b) (i) The specific conductivity of a solution containing 5 g of anhydrous BaCl2 (mol. wt. = 208) in 1000 cm3 of a solution is found to be 0 0058 ohm-1cm-1. Calculate the molar and equivalent conductivity of the solution. (ii) What is an electrochemical series? How is it useful in predicting whether a metal can liberate hydrogen from acid or not? 27 Comments of Examiners (a) (i) A few candidates calculated the value of E cell with negative sign i.e. - 0.36 V instead of +0. 36 V. Many candidates wrote incorrect formula for Nernst equation. Instead of calculating G , some candidates calculated G which was not asked. (ii) (1) Many candidates wrote incorrect definition of equivalent conductivity or wrote an incorrect formula and its relationship with specific conductance. (2) Some candidates wrote the definition of rusting instead of corrosion of metals. (b) (i) The values of molar conductivity and equivalent conductivity of the solution were calculated correctly but the answer with correct unit was not given by most of the candidates. (ii)Most candidates wrote the definition of metal activity series instead of electrochemical series. Some candidates wrote that metals above hydrogen can liberate hydrogen gas from acid. They did not write the answer in terms of reduction potential value. Suggestions for teachers Give adequate practise in solving numerical problems based on Nernst equations Explain the relationship between Gibbs free energy, G , E cell and differences between G and G clearly. Teach definitions of equivalent conductivity and corrosion of metal to the students. Also explain the related key concepts like factors affecting corrosion and prevention of corrosion. Give practice in numerical problems based on specific conductance, molar conductance and equivalent conductance. Instruct students to define electrochemical series in term of standard reduction potential values of elements. MARKING SCHEME Question 16 (a) (i) 0 0 0 = = 0 403 ( 0 763) = 0 36 V = 0 [ +2 ] 0 0591 log [ +2 ] Or =0 36 0 0591 2 0 1 log 0 01 28 =0 36 0 0295 log 10 = 0 36 0 0295 1 = 0 3305 V 0 = 0 = 2 96,500 0 36 = 69480 J /mol (ii) (1) (2) = 69 48 kJ/mol Equivalent Conductivity of Electrolyte is the conducting power of all the ions produced by dissolving one-gram equivalent in V cc. of solution. Corrosion of metals The slow and spontaneous process of the conversion of a metal into an undesirable compound (usually oxide) on exposure to atmospheric conditions is called corrosion of metals. OR 2 Molarity of BaCl2 = = 5 208 = 0 024 Molar conductivity (^ ) = Molar conductivity (^ ) = 0 0058 1000 1000 or 0 0058 1000 Molar conductivity (^ ) = 0 024 = 241 67 ohm-1 cm2 mol-1 208 Equivalent weight of BaCl2 = 2 = 104 5 Normality of BaCl2 = 104 = 048 Equivalent conductivity( ) = 1000 Or 0 0058 1000 = 0 048 = 120 83 ohm-1 cm2 eq-1 When various electrode systems are arranged in the order of their Eo values, the series obtained is called electrochemical series. Elements having negative value of Eo will liberate H2 from acids. (or any correct definition) [5] Question 17 (a) (i) Explain why: (1) Nitrogen does not form pentahalides. (2) Helium is used for filling weather balloons. 29 (3) (ii) ICl is more reactive than I2. Draw the structures of the following: (1) HClO4 (2) H3PO3 OR (b) (i) (ii) Explain why: (1) Mercury loses its meniscus in contact with ozone. (2) Halogens are coloured and the colour deepens on moving down in the group from fluorine to iodine. (3) Hydride of sulphur is a gas while hydride of oxygen is a liquid. Complete and balance the following reactions: (1) NaCl + MnO2 + H2SO4 _______ + _______ + ________ + _____ (2) KMnO4 + SO2 + H2O _______ + _______ + __________ Comments of Examiners (a)(i) (1) Most of the candidates did not mention the absence of d orbital in nitrogen. (2) Some candidates wrote that Helium is a noble gas but did not write that it is light, inert and non-inflammable. (3) Many candidates wrote that chlorine is more electronegative, but they did not mention that I-Cl bond is weaker than I-I Bond. (ii) (1) Many candidates drew correct structure of HClO4 but a few showed the bond between H and Cl instead of H and O(oxygen), which was not correct. (2) Many candidates drew incorrect structure they showed three bonds between -OH and P (phosphorus). (b)(i) (1) Mercury losses it meniscus due to tailing of Mercury or formation of mercurous oxide this was not specified by most of the candidates. (2) The absorption of light in visible region depends upon size of halogen atoms - this fact was not reported by many candidates. 30 Suggestions for teachers Explain the electronic configuration and properties of p-block elements with reasons for variable valency. Explain property of noble gases in detail. Guide students that inter halogen compounds are more reactive due to differences in electronegativity of atoms. They have low bond dissociation energy hence, more reactive. Show structure of oxyacids of halogens, sulphur and phosphorus in class. Teach characteristic properties of p- block elements in detail. Instruct students to write complete and balanced equations giving names of the reactant/s and product/s. (3) Some candidates did not explain the answer in terms of intermolecular hydrogen bonding. (ii) Many candidates wrote incomplete/incorrect or unbalanced equations. MARKING SCHEME Question 17 (a) (i) (ii) (1) Due to the absence of d-orbitals in its valence shell, nitrogen does not form pentahalides. (2) Helium being inert, non-inflammable and light gas is used in filling weather balloons. (3) I Cl bond is weaker than I I bond. I Cl bond breaks easily to form halogen atoms (I and Cl) (1) HClO4 (2) H3PO3 O H O Cl O O or O H O P H or O H O || H O Cl = O || O O H O P H O H OR (b) (i) (1) (2) (3) (ii) Mercury in presence of ozone is oxidised to sub oxide (Mercurous oxide) which dissolves in mercury. It starts sticking to glass and loses meniscus (Tailing of Hg). Or 2Hg + O3 Hg2O + O2 Halogens are coloured due to absorption of light in visible region. Fluorine being small absorbs violet colour and shows the complementary colour yellow whereas iodine absorbs yellow colour and shows complementary colour violet. Due to small size of oxygen atom H2O forms intermolecular H-bond and gets associated hence occurs as liquid. While H2S is simple covalent compound hence occurs as a gas. 2NaCl + MnO2 + 3H2SO4 2NaHSO4+ MnSO4 + Cl2 + 2H2O (1) OR (2) 2NaCl + MnO2 + 2H2SO4 2Na2SO4+ MnSO4 + Cl2 + 2H2O 2KMnO4 + 5SO2+2H2O K2SO4 +2MnSO4 + 2H2SO4 31 [5] Question 18 (a) (i) (ii) Give balanced equations for the following reactions: (1) Benzaldehyde reacts with hydrazine. (2) Acetic acid reacts with phosphorous pentachloride. (3) Acetone reacts with sodium bisulphite. Give one chemical test each to distinguish between the following pairs of compounds: (1) Ethanol and acetic acid (2) Acetaldehyde and benzaldehyde OR (b) (i) (ii) Write chemical equations to illustrate the following name reactions: (1) Clemmensen s reduction (2) Rosenmund s reduction (3) HVZ reaction Explain why: (1) Acetaldehyde undergoes aldol condensation, but formaldehyde does not. (2) Acetic acid is a weaker acid as compared to formic acid. 32 Comments of Examiners (a)(i)(1) Many candidates wrote incorrect formulae of reactants and product. (2) For the reaction of acetic acid with phosphorus pentachloride, most of the candidates gave correct equation. (3) Some candidates did not write the correct formula of product for the reaction of acetone with sodium bisulphite. (ii) (1) Iodoform test was mentioned for distinguishing ethanol from acetic acid but the observation was not written by many candidates. Some candidates mentioned esterification test to distinguish between ethanol and acetic acid, but this test is given by both the compounds. (2) Some candidates wrote Tollen s reagent test to distinguish between acetaldehyde and benzaldehyde - this is given by both the compounds. (b)(i) (1) A few candidates wrote unbalanced equations. Some wrote H2 instead of [H]. Several candidates did not write Zn/Hg and concentrated HCl. Many candidates did not write the by-product. (2) Many candidates wrote nascent hydrogen [H] instead of H2 or did not write the catalyst. (3) Some candidates did not write the by product. 33 Suggestions for teachers Give sufficient practice in writing organic equations, with name of reactants, products and correct structure. Tell students that the given chemical test should be positive for one compound and negative for the other compound. Instruct students to learn the named organic reactions with the name of specific catalyst used in the reaction and other necessary conditions in the presence of which the required product/s is/are formed. Teach mechanism of Aldol condensation to clear the concept of the presence of - H in CH3CHO. Explain the strength of acid and base with the help of inductive effect. MARKING SCHEME Question 18 (a) (i) (1) (2) (3) H H C6H5 C = N.NH2+H2O C6H5 C = O + NH2.NH2 Benzaldehyde hydrazone O CH3 C Cl+ POCl3+ HCl O CH3 C OH+ PCl5 CH3 C = O + NaHSO3 CH3 (ii) (1) CH3 C CH3 Ethanol + I2 + Alkali OH SO3Na Yellow precipitate of Iodoform, acetic acid does not respond to this test. or Acetic acid + neutral FeCl3 Blood red coloration, Ethanol does not respond to this test. (2) Acetaldehyde + I2 + Alkali (Or any other relevant test) Yellow precipitate of Iodoform Or Acetaldehyde + Fehling solution Red precipitate of Cu2O Benzaldehyde does not respond to the above test. (Or any other relevant test) OR (b) (i) (1) Clemmensen s Reduction: 3 + 4[ ] (2) / 3 3 + 4[ ] 3 3 + 2 or / 3 2 3 + 2 Rosenmund s Reduction: H O 4 R C = O + HCl R C Cl+ H2 34 (3) HVZ reaction: CH3COOH Cl2 /RedP Cl2 /RedP CH2ClCOOH CHCl2COOH HCl HCl HCl Cl2/RedP Cl3C. COOH (ii) (1) Aldol condensation is given by aldehydes having hydrogen atom. Acetaldehyde has H but HCHO does not have hydrogen atom. Hence, does not give aldol condensation. (2) Acetic acid has methyl (electron releasing) group which causes + I effect and decreases its acidic strength of acetic acid. Note: For questions having more than one correct answer/solution, alternate correct answers/solutions, apart from those given in the marking scheme, have also been accepted. 35 Topics found difficult by candidates The packing efficiency of different type of unit cells. Imperfection in solids. Numerical problem based on elevation of boiling point, depression in freezing point, van t Hoff factor and degree of dissociation. Numerical problem related to Arrhenius equation. Rate Law expression order of reaction and half-life period. Nernst equation, electrochemical series, equivalent and molar conductance corrosion of metals. Surface chemistry, application of colloids, purification of colloids. Coordination compounds nomenclature, isomerism hybridisation of coordination compounds. Chemical equations of inorganic compounds, extraction of copper, chemical equations of K2 Cr2 O7 and KMnO4. Named organic reactions, conversion of organic compounds reasoning type of questions of organic chemistry. Polymers, Biodegradable Polymers and their examples. Use of chemicals in Food Industry, antiseptics and disinfectants. Concepts in which candidates got confused Isomerism of coordination compounds (linkage and geometrical) and nomenclature of coordination compounds. Relationship between E cell and standard free energy change ( G ). Purines and pyrimidines present in RNA and DNA. Colloidal solution and their applications. Reactions of inorganic compounds with balanced equation. Tests to distinguish organic compounds. 36 Suggestions for candidates Be regular and systematic in your studies. Avoid selective study - questions are asked from every chapter. Name organic compounds preferably using the IUPAC Nomenclature. Always solve the numerical problems stepwise i.e., (a) formula (b) substitution (c) calculation (d) answer with correct unit. Practise adequate number of numerical problems. Always write balanced chemical equations with the essential conditions. Learn both positive and negative chemical tests of organic compounds. In each topic, learn the definitions with keywords with proper understanding. Write the keyword in your answer. Read the question paper carefully and understand what is required before attempting the question. Read the full chapter according to the topic given in the scope of syllabus. Do not waste time in attempting extra questions given as internal choice. Stress upon clarifying the concepts of each topic to answer the reasoning type of questions. 37

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Additional Info : ISC Class XII Analysis Of Pupil Performance 2018 : Chemistry
Tags : ISC Board, Class 11th, Class 12th, NDA/NA Entrance Examination  

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